SR fundamental contradiction

mlut...@wanadoo.fr

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Sep 27, 2006, 8:44:08 PM9/27/06

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SR fundamental contradiction
------------------------------------------

Luttgens:

Let x = ct.
Then x' = g(x - vt), where gamma = 1/sqrt(1-v^2/c^2), becomes
x' = g(c-v)t
What represents the length (c-v)t?
Is that length "dilated" by g?

Van de Moortel:

Consider the event E on the light signal with x = c t for some
chosen value of t.
Then c t - v t is the distance between the origin of S' (the 'moving
observer') and the light signal, as seen at time t in the S-frame
(the 'stationary frame'), and, by the way, so c - v is by definition
the closing velocity between the two.

For this event E, as seen in S', the light signal has covered the
distance
x' = c t' = g (c - v) t
This is a distance of the event E in the S' frame.

Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.
What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame.

Luttgens:

Any object (stick) measures shorter in terms of a frame relative to
which it is moving with velocity v that it does as measured in a frame
relative to which it is at rest, the ratio of shortening being
sqrt(1-v^2/c^2).
This is a relation between measurements referred to different frames.

If a stick of length x' = g(c-v)t is at rest in the S' frame,
it is moving at v relative to the frame S. So, measured in S, its
length is contracted by 1/g and becomes x = g(c-v)t * 1/g = (c-v)t.
This corresponds to Van de Moortel's reasoning, which is circular.
Indeed, x' = g(c-v)t has been obtained *by applying the LT* to
x = (c-v)t, when S' was considered as moving relative to S, and thus
relative to the stick. No wonder that one gets back x = (c-v)t
when S is afterwards considered as moving wrt S'.

One has instead to consider a stick of length x = (c-v)t at rest
in the frame S. Relative to the frame S', such stick is moving
at v, hence its length, measured in S', is shortened by 1/g wrt its
length measured in S. Thus, x' = (c-v)t / g.

But *according to the LT*, x' = (c-v)t * g !

Such contradiction demonstrates the falseness of the Lorentz
transformation, falseness whose origin lies in the postulate
that when x = ct, x' = ct'.

Without such postulate, the LT become

x' = (c-v)t / g
t' = t / g

and the stick of length (c-v)t at rest in S is indeed shortened by
sqrt(1-v^2/c^2) in S'.

Derivation of the correct transformation:
----------------------------------------

Let's consider two frames of reference, S and S', each in uniform
translatory motion relative to the other, the velocity of S' relative
to S being v.

Basis relations:

x' = ax + bt
t' = ex + kt

At the origin of S', x' = 0 and x = vt.
Hence, 0 = (av+b)t, whence b = -av

The basis relations are now

(1) x' = a(x - vt)
(2) t' = ex + kt

Now, let's suppose that a light signal, starting from the coincident

origins of frames S and S' at t = t' = 0, travels toward positive x.
After a time t, it will be at
x = ct + vt, and also at
x' = ct'

Substituting these values of x and x' in relations
(1) and (2), and eliminating t and t', one gets

(3) a - ec - ev - k = 0

If the signal travels toward negative x,
x = -ct + vt and x' = -ct'

Substituting these values of x and x' in relations
(1) and (2), and eliminating t and t', one gets

(4) a + ec - ev - k = 0

From (3) and (4), one gets e = 0

With e = 0, relations (3) or (4) reduce to a = k

Hence, relations (1) and (2) become

(1) x' = k (x - vt)
(2) t' = k t

Now, a light signal follow the y' axis. Relatively to S,
it travels obliquely, for, while the signal goes
a distance ct, the y'-axis advances a distance x = vt.
Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
But, also, y' = ct' = c * k * t.

Equating y' to y, one gets k = sqrt(1 - v^2/c^2)

The transforms obtained without the *bold* Einstein's postulate that
the speed of light is the same in all frames are thus

(5) x' = sqrt(1 - v^2/c^2) * (x - vt) = (x - vt) / g
(6) t' = sqrt(1 - v^2/c^2) * t = t / g,

(g corresponds to Einstein's gamma).

Transform (5) straightforwardly tells us that any body measures shorter
in terms of a frame relative to which it is moving with speed v than
it does as measured in a frame relative to which it is at rest.

Transform (6) implies that when two physical systems are in uniform
relative translation at speed v, the effects produced by system A
on system B are modified just as if all natural processes on A were
slowed down in the ratio sqrt(1 - v^2/c^2). (i.e., the so-called time
"dilation").

Those transforms, contrary to Einstein's LT, don't allow to claim that
simultaneity is relative (i.e., that events that are considered to
be simultaneous in one reference frame are not simultaneous in another
reference frame moving with respect to the first, cf. Wikipedia).

Marcel Luttgens

actioni...@yahoo.com

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Sep 27, 2006, 8:56:09 PM9/27/06

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mluttg...@wanadoo.fr wrote:
> SR fundamental contradiction
> ------------------------------------------
>
> Luttgens:
>
> Let x = ct.
> Then x' = g(x - vt), where gamma = 1/sqrt(1-v^2/c^2), becomes
> x' = g(c-v)t
> What represents the length (c-v)t?
> Is that length "dilated" by g?
>

> Marcel Luttgens

(c-v)t is the length I perceive between the moving guy and the photon.
That length is not dilated by g in my frame.

actioni...@yahoo.com

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Sep 27, 2006, 8:56:44 PM9/27/06

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mluttg...@wanadoo.fr wrote:
> SR fundamental contradiction
> ------------------------------------------
>
> Luttgens:
>
> Let x = ct.
> Then x' = g(x - vt), where gamma = 1/sqrt(1-v^2/c^2), becomes
> x' = g(c-v)t
> What represents the length (c-v)t?
> Is that length "dilated" by g?
>

Dirk Van de moortel

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Sep 27, 2006, 9:30:00 PM9/27/06

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<mlut...@wanadoo.fr> wrote in message news:1159389848.2...@i3g2000cwc.googlegroups.com...

which equals 1/g just like I explained.

> This is a relation between measurements referred to different frames.
>
> If a stick of length x' = g(c-v)t is at rest in the S' frame,
> it is moving at v relative to the frame S. So, measured in S, its
> length is contracted by 1/g and becomes x = g(c-v)t * 1/g = (c-v)t.

yes

> This corresponds to Van de Moortel's reasoning, which is circular.

Marcel calls the bleeding obvious "circular".

> Indeed, x' = g(c-v)t has been obtained *by applying the LT* to
> x = (c-v)t, when S' was considered as moving relative to S, and thus
> relative to the stick. No wonder that one gets back x = (c-v)t
> when S is afterwards considered as moving wrt S'.

Indeed no wonder.
The transformation is consistent with length contraction.
Do we have a breaktrough here?
Has Marcel's Precious Penny dropped?

>
> One has instead to consider a stick of length x = (c-v)t at rest
> in the frame S. Relative to the frame S', such stick is moving
> at v, hence its length, measured in S', is shortened by 1/g wrt its
> length measured in S. Thus, x' = (c-v)t / g.
>
> But *according to the LT*, x' = (c-v)t * g !
>
> Such contradiction demonstrates the falseness of the Lorentz
> transformation, falseness whose origin lies in the postulate
> that when x = ct, x' = ct'.

You had a light signal going at c and an object going at v
in some frame S and you wondered

"What represents the length (c-v)t?"

I gave you a possible physical object, namely one with length
g (c-v) t at rest in S', that can have this value (c-v) t as its length
in the S-frame. So your question was answered.

Now you imagine another possible physical object, namely one
with lenght (c-v) t at rest in S, which has of course lenght
1/g (c-v) t in frame S', according to the same rules.

Since we are talking about two different objects, there is no
contradiction. Quite on the contrary ;-)

You are suffering from your original syndrome again:
http://users.telenet.be/vdmoortel/dirk/Stuff/MarcelAtSchool.gif
See a doctor about it. Trust me.

Dirk Vdm

rambu...@yahoo.com

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Sep 28, 2006, 12:42:47 AM9/28/06

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Way to go, Dirk

Marcel needs a good kick in the pants now and then.

Tom Roberts

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Sep 28, 2006, 1:43:02 AM9/28/06

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mlut...@wanadoo.fr wrote:
> Let x = ct.
> Then x' = g(x - vt), where gamma = 1/sqrt(1-v^2/c^2), becomes
> x' = g(c-v)t
> What represents the length (c-v)t?

It is merely a mathematical artifact, and is not really the "length" of
anything. <shrug>

You are manipulating symbols without understanding what they represent
or mean. And you obtain nonsense. <shrug>

If you write down what you are trying to do more precisely, defining
each and every one of your symbols, you will see that the above is
nonsense because it intermixes symbols of different types.

For instance, when you said "Let x = ct", I imagine that you might have
meant:
Let us construct inertial coordinates {x,y,z,t} and ignore y and z.
In those coordinates let us consider a light pulse emitted from
the point (x=0,t=0) moving in the +x direction, so with x=f(t)
being the trajectory of this pulse parameterized by the time
coordinate t of this frame, the pulse has position x=f(t) = ct
for t>=0.

So your x refers to a specific light pulse and your t refers to a path
parameter of its trajectory.

On the other hand, when you wrote "x' = g(x - vt)" you really meant a
COORDINATE TRANSFORM between two inertial frames. That is, those symbols
DO NOT REFER TO THE LIGHT PULSE ABOVE, and t is NOT a path parameter, it
is a coordinate of an arbitrary point in the manifold.

If you spend the effort to fix up your overly loose terminology and
symbols, you will be able to answer your own question, and you will find
there is no "contradiction". If you don't bother to do that, you will
remain mystified. <shrug>

Yes, physicists are notoriously loose; but if you claim
to display a "contradiction", _YOU_ must be precise.

Tom Roberts

Ajay Sharma

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Sep 28, 2006, 4:29:41 AM9/28/06

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Sorcerer

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Sep 28, 2006, 9:17:52 AM9/28/06

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"Tom Roberts" <tjrobe...@sbcglobal.net> wrote in message
news:GMFSg.17769$Ij....@newssvr14.news.prodigy.com...

Don't do as I do, do as I tell you! Fuck you, ignorant Roberts!
It's your bowels that are notoriously loose.
Here's the precise contradiction, arsehole:
http://www.androcles01.pwp.blueyonder.co.uk/DominoEffect.GIF

Androcles

harry

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Sep 28, 2006, 9:38:46 AM9/28/06

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Just in addition to other comments:

<mlut...@wanadoo.fr> wrote in message
news:1159389848.2...@i3g2000cwc.googlegroups.com...

> SR fundamental contradiction
> ------------------------------------------
>
> Luttgens:
>
> Let x = ct.
> Then x' = g(x - vt), where gamma = 1/sqrt(1-v^2/c^2), becomes
> x' = g(c-v)t
> What represents the length (c-v)t?
> Is that length "dilated" by g?

SNIP

> Luttgens:
>
> Any object (stick) measures shorter in terms of a frame relative to
> which it is moving with velocity v that it does as measured in a frame
> relative to which it is at rest, the ratio of shortening being
> sqrt(1-v^2/c^2).
> This is a relation between measurements referred to different frames.
>
> If a stick of length x' = g(c-v)t is at rest in the S' frame,

Aargh!

Usually sticks are supposed to have a constant length. But in your equation,
presumably c=lightspeed, v may be constant thus g=constant while t changes.
Thus at constant speed, at t=2 your stick is twice as long as at t=1 while
it even has zero length at t=0. Your "stick" is perhaps made of rubber, with
someone pulling on it?!

x' is normally used for position coordinates, *not* for lengths. x' is used
in transformation equations (=between position coordinates) as well as in
trajectory equations (= position coordinate of something as function of
local time).

Tom Roberts explained that rather well except for one important point: you
can of course combine the two sets of equations in order to obtain the
position coordinate of the wave front ^^^ in the moving frame.

You might help yourself as well as this kind of discussions a lot by first
trying "Galilean" relativity: Starting with the equation of motion x' = w t
of a bowling ball that is thrown along the full length of a train wagon
relative to the train in motion, and the transformation equation between
train coordinates and embankment coordinates x' = x - v t, describe the
trajectory x(t) of the ball relative to the embankment. Now do the same for
the trajectory x(t) of the train wagon's rear end. Is there a contradiction?
Why not?

Harald

mlut...@wanadoo.fr

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Sep 28, 2006, 12:36:38 PM9/28/06

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Tom Roberts wrote:
> mlut...@wanadoo.fr wrote:
> > Let x = ct.
> > Then x' = g(x - vt), where gamma = 1/sqrt(1-v^2/c^2), becomes
> > x' = g(c-v)t
> > What represents the length (c-v)t?
>
> It is merely a mathematical artifact, and is not really the "length" of
> anything. <shrug>
>
> You are manipulating symbols without understanding what they represent
> or mean. And you obtain nonsense. <shrug>

Will you claim that Vdm didn't know what he wrote:

"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.
What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame. "

I used his reasoning to show that the correct transform is x' = (c-v)t
/g:

Now imagine a stick with this particular length

x = (c-v) t

at rest in the S frame.

What is the length of such a stick in the S'-frame?

If you apply length contraction, you find that this length
would be

x' = (c - v) t /g
in the S' frame, a length that is immediately given by the correct
transform.

The LT x' = g(c-v)t implies that such stick is *dilated* by g in the
S'-frame, which
is of course false. Notice that Vdm applied length contraction to that
*dilated* length
in order to get back the length (c-v)t in the S frame. By doing this,
he implicitely
recognized that the LT leads to length *dilation* in the S'-frame.

The rest of your post is mere quibbling.

Marcel Luttgens

mlut...@wanadoo.fr

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Sep 28, 2006, 1:25:20 PM9/28/06

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Tom Roberts wrote:
> mlut...@wanadoo.fr wrote:
> > Let x = ct.
> > Then x' = g(x - vt), where gamma = 1/sqrt(1-v^2/c^2), becomes
> > x' = g(c-v)t
> > What represents the length (c-v)t?
>
> It is merely a mathematical artifact, and is not really the "length" of
> anything. <shrug>
>
> You are manipulating symbols without understanding what they represent
> or mean. And you obtain nonsense. <shrug>

Vdm wrote:

"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.
What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame."

Notice that Vdm implicitely recognized that the length (c-v)t
is *dilated* in the S'-frame, according to the LT x' = g(c-v)t.
Such length *dilation* is of course false, a *contraction* being
expected.

I used exactly the same scenario:

Imagine a stick with the particular length
x = (c-v) t

at rest in the S frame.

What is the length of such a stick in the S'-frame?

If you apply length contraction, you find that this length
would be

x' = (c - v) t / g
in the S frame.

Such length corresponds to the solution given by the correct
transform x' = (c-v)t / g.

The rest of your post is mere quibbling.

Marcel Luttgens

>

mlut...@wanadoo.fr

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Sep 28, 2006, 1:34:01 PM9/28/06

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Tom Roberts wrote:
> mlut...@wanadoo.fr wrote:
> > Let x = ct.
> > Then x' = g(x - vt), where gamma = 1/sqrt(1-v^2/c^2), becomes
> > x' = g(c-v)t
> > What represents the length (c-v)t?
>
> It is merely a mathematical artifact, and is not really the "length" of
> anything. <shrug>
>
> You are manipulating symbols without understanding what they represent
> or mean. And you obtain nonsense. <shrug>

Vdm wrote:

"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.
What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame."

Notice that Vdm implicitely recognized that the length (c-v)t

is *dilated* in the S'-frame, according to the LT x' = g(c-v)t.
Such length *dilation* is of course false, a *contraction* being
expected.

I used exactly the same scenario:

Imagine a stick with the particular length

x = (c-v) t

at rest in the S frame.

What is the length of such a stick in the S'-frame?

If you apply length contraction, you find that this length
would be

x' = (c - v) t / g
in the S frame.

Such length corresponds to the solution given by the correct
transform x' = (c-v)t / g.

The rest of your post is mere quibbling.

Marcel Luttgens

>

mlut...@wanadoo.fr

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Sep 28, 2006, 2:32:56 PM9/28/06

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Tom Roberts wrote:
> mlut...@wanadoo.fr wrote:
> > Let x = ct.
> > Then x' = g(x - vt), where gamma = 1/sqrt(1-v^2/c^2), becomes
> > x' = g(c-v)t
> > What represents the length (c-v)t?
>
> It is merely a mathematical artifact, and is not really the "length" of
> anything. <shrug>

Vdm wrote:

"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.
What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame."

Notice that Vdm implicitely recognized that the length (c-v)t

is *dilated* in the S'-frame, according to the LT x' = g(c-v)t.
Such length *dilation* is of course false, a *contraction* being
expected.

I used exactly the same scenario:

Imagine a stick with the particular length

x = (c-v) t

at rest in the S frame.

What is the length of such a stick in the S'-frame?

If you apply length contraction, you find that this length
would be

x' = (c - v) t / g
in the S frame.

Such length corresponds to the solution given by the correct
transform x' = (c-v)t / g.

The rest of your post is mere quibbling. <shrug>

Marcel Luttgens

mlut...@wanadoo.fr

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Sep 28, 2006, 2:37:03 PM9/28/06

to

Tom Roberts wrote:
> mlut...@wanadoo.fr wrote:
> > Let x = ct.
> > Then x' = g(x - vt), where gamma = 1/sqrt(1-v^2/c^2), becomes
> > x' = g(c-v)t
> > What represents the length (c-v)t?
>
> It is merely a mathematical artifact, and is not really the "length" of
> anything. <shrug>
>
> You are manipulating symbols without understanding what they represent
> or mean. And you obtain nonsense. <shrug>

Vdm wrote:

"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.
What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame."

Notice that Vdm implicitely recognized that the length (c-v)t

is *dilated* in the S'-frame, according to the LT x' = g(c-v)t.
Such length *dilation* is of course false, a *contraction* being
expected.

I used exactly the same scenario:

Imagine a stick with the particular length

x = (c-v) t

at rest in the S frame.

What is the length of such a stick in the S'-frame?

If you apply length contraction, you find that this length
would be

x' = (c - v) t / g
in the S frame.

Such length corresponds to the solution given by the correct
transform x' = (c-v)t / g.

The rest of your post is mere quibbling.

Marcel Luttgens

Dirk Van de moortel

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Sep 28, 2006, 4:50:02 PM9/28/06

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"harry" <harald.vanlin...@epfl.ch> wrote in message news:115943...@sicinfo3.epfl.ch...

> Just in addition to other comments:
>
> <mlut...@wanadoo.fr> wrote in message news:1159389848.2...@i3g2000cwc.googlegroups.com...
>> SR fundamental contradiction
>> ------------------------------------------
>>
>> Luttgens:
>>
>> Let x = ct.
>> Then x' = g(x - vt), where gamma = 1/sqrt(1-v^2/c^2), becomes
>> x' = g(c-v)t
>> What represents the length (c-v)t?
>> Is that length "dilated" by g?
> SNIP
>
>> Luttgens:
>>
>> Any object (stick) measures shorter in terms of a frame relative to
>> which it is moving with velocity v that it does as measured in a frame
>> relative to which it is at rest, the ratio of shortening being
>> sqrt(1-v^2/c^2).
>> This is a relation between measurements referred to different frames.
>>
>> If a stick of length x' = g(c-v)t is at rest in the S' frame,
>
> Aargh!
>
> Usually sticks are supposed to have a constant length.

Doesn't matter.
Pick some fixed t, say 0.0001 and work from there.
My words were:

"Consider the event E on the light signal with x = c t for some
chosen value of t."

Dirk Vdm

Dirk Van de moortel

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Sep 28, 2006, 4:50:36 PM9/28/06

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<mlut...@wanadoo.fr> wrote in message news:1159454222.9...@i3g2000cwc.googlegroups.com...

>
> Tom Roberts wrote:
>> mlut...@wanadoo.fr wrote:
>> > Let x = ct.
>> > Then x' = g(x - vt), where gamma = 1/sqrt(1-v^2/c^2), becomes
>> > x' = g(c-v)t
>> > What represents the length (c-v)t?
>>
>> It is merely a mathematical artifact, and is not really the "length" of
>> anything. <shrug>
>>
>> You are manipulating symbols without understanding what they represent
>> or mean. And you obtain nonsense. <shrug>
>
> Vdm wrote:
>
> "Now imagine a stick with this particular length
> x' = g (c-v) t
> at rest in the S' frame.
> What is the length of such a stick in the S-frame?
> If you apply length contraction, you find that this length
> would be
> x' / g = (c - v) t
> in the S frame."
>
> Notice that Vdm implicitely recognized that the length (c-v)t
> is *dilated* in the S'-frame, according to the LT x' = g(c-v)t.

Imbecile, I did no such thing.
You have a maliciously deliberate reading comprehension
problem.

> Such length *dilation* is of course false, a *contraction* being
> expected.

Are you too much of a coward to reply directly to my reply?

Dirk Vdm

mlut...@wanadoo.fr

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Sep 28, 2006, 5:22:03 PM9/28/06

to

Tell that to Van de Moortel, who rightly wrote:
""Now imagine a stick with this particular length
x' = g (c-v) t

at rest in the S' frame. "

Marcel Luttgens

Dirk Van de moortel

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Sep 28, 2006, 6:00:37 PM9/28/06

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<mlut...@wanadoo.fr> wrote in message news:1159464123....@k70g2000cwa.googlegroups.com...

With his SNIP Harry missed the first line of my reply.
Just like you missed every single letter of it.

Anyway, here you go with yet another fumblamental contradiction
of yours:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Fumblamental.html
Congratulations with your persistent imbecility.

Dirk Vdm

Sorcerer

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Sep 28, 2006, 7:02:18 PM9/28/06

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"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:95USg.101253$lB5.1...@phobos.telenet-ops.be...

Persistent imbecility:
http://www.androcles01.pwp.blueyonder.co.uk/Fumble.htm

harry

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Sep 29, 2006, 9:36:18 AM9/29/06

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"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote

in message news:_2TSg.101154$dz7.1...@phobos.telenet-ops.be...

Right. From the way Lutgens formulated it, I have the impression that he
didn't copy that... It's helpful to write for example t1 for a value of t.

Sorcerer

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Sep 29, 2006, 12:41:38 PM9/29/06

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"harry" <harald.vanlin...@epfl.ch> wrote in message

news:115952...@sicinfo3.epfl.ch...

Read the fucking paper, you stoooopid piece of shit. Clueless Dork Van de
merde the local village dog tord can't help and it is NOT right.

"If we place x'=x-vt, it is clear that a point at rest in the system k must
have a system of values x', y, z, independent of time. We first define tau
as a function of x', y, z, and t.
http://www.fourmilab.ch/etexts/einstein/specrel/www/
Androcles

mlut...@wanadoo.fr

unread,

Sep 29, 2006, 1:12:50 PM9/29/06

to

Dirk Van de moortel wrote:

Of course, we are talking about two different objects, but the
one you used demonstrates the falseness of SR, there is no
doubt about that. Quite on the contrary ;-)

You wrote:

"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.
What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame."

Where does the length of the stick come from, if not from
the LT x' = g(c-v)t ?

Hence, you implicitely recognize that the length x = (c-v)t
is *dilated* in the S'-frame. Indeed, applying length contraction
by 1/g, you find back x = (c-v)t in the S-frame.
You are a parroting guru, who doesn't even understand the meaning
of the equations with which he is trying to defend SR. Otherwise,
you would have realized the falseness of SR, which, via the
LT x' = g (c-v) t, predicts a length *dilation* instead of
an expected length *contraction*.

You should put this demonstration of your stupidity into your
immortal fumbles.

Marcel Luttgens

mlut...@wanadoo.fr

unread,

Sep 29, 2006, 1:24:12 PM9/29/06

to

Dirk Van de moortel wrote:

Done. I repeat my response here:

Of course, we are talking about two different objects, but the
one you used demonstrates the falseness of SR, there is no
doubt about that. Quite on the contrary ;-)

You wrote:

"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.
What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame."

Where does the length of the stick come from, if not from

the LT x' = g(c-v)t ?

Hence, you implicitely recognize that the length x = (c-v)t
is *dilated* in the S'-frame. Indeed, applying length contraction
by 1/g, you find back x = (c-v)t in the S-frame.
You are a parroting guru, who doesn't even understand the meaning
of the equations with which he is trying to defend SR. Otherwise,
you would have realized the falseness of SR, which, via the
LT x' = g (c-v) t, predicts a length *dilation* instead of
an expected length *contraction*.

You should put this demonstration of your stupidity into your
immortal fumbles.

Marcel Luttgens

>
> Dirk Vdm

mlut...@wanadoo.fr

unread,

Sep 29, 2006, 1:25:11 PM9/29/06

to

Dirk Van de moortel wrote:

Done, here is my response again:

Of course, we are talking about two different objects, but the
one you used demonstrates the falseness of SR, there is no
doubt about that. Quite on the contrary ;-)

You wrote:

"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.
What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame."

Where does the length of the stick come from, if not from

mlut...@wanadoo.fr

unread,

Sep 29, 2006, 1:35:37 PM9/29/06

to

Dirk Van de moortel wrote:

You must be suffering from Alzheimer ! In that fumble, you demonstrated
your own imbecility !

Marcel Luttgens

>
> Dirk Vdm

Dirk Van de moortel

unread,

Sep 29, 2006, 4:24:38 PM9/29/06

to

<mlut...@wanadoo.fr> wrote in message news:1159535570.2...@m73g2000cwd.googlegroups.com...

Hey, retard, when I tell you to imagine a stick of length 5, do
you ask where 5 comes from?
Yes, that figures. Okay, I'll tell you a secret: it comes out of thin air.
How is that?

>
> Hence, you implicitely recognize that the length x = (c-v)t
> is *dilated* in the S'-frame.

No, I don't.
I take a value g (c-v) t out of your thin air.
That is what a sentence like

"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame."

means.

Marcel, you are TOOOOOO stupid for this. Really.
You should be embarrassed, but I guess that is one of those
feelings that autistic imbeciles can't have. Bad luck.

> Indeed, applying length contraction
> by 1/g, you find back x = (c-v)t in the S-frame.
> You are a parroting guru, who doesn't even understand the meaning
> of the equations with which he is trying to defend SR. Otherwise,
> you would have realized the falseness of SR, which, via the
> LT x' = g (c-v) t, predicts a length *dilation* instead of
> an expected length *contraction*.
>
> You should put this demonstration of your stupidity into your
> immortal fumbles.

The demonstration of yours is right here:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Fumblamental.html

Dirk Vdm

mlut...@wanadoo.fr

unread,

Sep 30, 2006, 10:37:17 AM9/30/06

to

You said:

"For this event E, as seen in S', the light signal has covered the
distance
x' = c t' = g (c - v) t
This is a distance of the event E in the S' frame.

Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame."

And now, you claim that its length comes out of thin air!

You are a stupid liar!

Marcel Luttgens

Dirk Van de moortel

unread,

Sep 30, 2006, 10:58:30 AM9/30/06

to

<mlut...@wanadoo.fr> wrote in message news:1159612637.0...@i42g2000cwa.googlegroups.com...

>
> Dirk Van de moortel wrote:
>> <mlut...@wanadoo.fr> wrote in message news:1159535570.2...@m73g2000cwd.googlegroups.com...

[snip repetitive demonstrations of your imbecility]

>> > Where does the length of the stick come from, if not from
>> > the LT x' = g(c-v)t ?
>>
>> Hey, retard, when I tell you to imagine a stick of length 5, do
>> you ask where 5 comes from?
>> Yes, that figures. Okay, I'll tell you a secret: it comes out of thin air.
>> How is that?
>
> You said:
>
> "For this event E, as seen in S', the light signal has covered the
> distance
> x' = c t' = g (c - v) t
> This is a distance of the event E in the S' frame.
>
> Now imagine a stick with this particular length
> x' = g (c-v) t
> at rest in the S' frame."
>
> And now, you claim that its length comes out of thin air!

Again, you forgot my opening line:

| Consider the event E on the light signal with x = c t for some
| chosen value of t.

"For some chosen value of t"... that's your thin air.

>
> You are a stupid liar!

I'm sorry, but you are too stupid to be qualified to know whether
someone is lying to you or not.
That is quite Amusing :-)

Dirk Vdm

Sue...

unread,

Sep 30, 2006, 5:19:33 PM9/30/06

to

Dirk Van de moortel wrote:

<< That is quite Amusing :-) >>

http://www.cs.cmu.edu/~rgs/alice-VII.html

Pay your debts...
...then blabber with a clean conscience.

Abstract
Einstein addressed the twin paradox in special relativity
in a relatively unknown, unusual and rarely cited paper
written in 1918, in the form of a dialogue between a
critic and a relativist. Contrary to most textbook versions
of the resolution, Einstein admitted that the special
relativistic time dilation was symmetric for the twins,
and he had to invoke, asymmetrically, the general relativistic
gravitational time dilation during the brief periods
of acceleration to justify the asymmetrical aging.
Notably, Einstein did not use any argument related to
simultaneity or Doppler shift in his analysis. I discuss
Einstein's resolution and several conceptual issues
that arise. It is concluded that Einstein's resolution using
gravitational time dilation suffers from logical and
physical flaws, and gives incorrect answers in a general
setting. The counter examples imply the need to reconsider
many issues related to the comparison of transported
clocks. The failure of the accepted views and
resolutions is traced to the fact that the special relativity
principle formulated originally for physics in empty
space is not valid in the matter-filled universe.

C. S. Unnikrishnan
Gravitation Group,
Tata Institute of Fundamental Research,
Homi Bhabha Road, Mumbai 400 005, India
http://www.iisc.ernet.in/currsci/dec252005/2009.pdf
-----

Sue...

Pay Dennis McCarthy c/o USNO

>
> Dirk Vdm

Dirk Van de moortel

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Sep 30, 2006, 7:02:27 PM9/30/06

to

"Sue..." <suzyse...@yahoo.com.au> wrote in message news:1159636773.3...@m73g2000cwd.googlegroups.com...

>
> Dirk Van de moortel wrote:
>
> << That is quite Amusing :-) >>

Isn't it, Dennis?

Dirk Vdm

G. L. Bradford

unread,

Oct 1, 2006, 2:59:20 AM10/1/06

to

"Sue..." <suzyse...@yahoo.com.au> wrote in message
news:1159636773.3...@m73g2000cwd.googlegroups.com...
>

But it is valid, though with a twist. 1) Wavelength. 2) Frequency. 3)
Speed of light constant : 1) Imaginary space (wavelength). 2) Imaginary time
(frequency). 3) Speed of light constant.

This is all that Special Relativity's observer is ever really dealing in,
whether space is empty space or matter-filled universe. Imaginary space,
imaginary time (wavelengths, frequencies, and the speed of light constant).

>
> Sue...

>>
>> Dirk Vdm
>

GLB

mlut...@wanadoo.fr

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Oct 1, 2006, 8:46:43 AM10/1/06

to

Of course, one can choose any value for t. What counts is the
formula x' = g (c-v) t, which means that the distance x = (c-v)t
measured in the S-frame is *dilated* by g in the S'-frame, whereas
it should be *contracted* by 1/g.
All your sophistry cannot hide the inherent contradiction of SR.

Marcel Luttgens

>
> Dirk Vdm

Dirk Van de moortel

unread,

Oct 1, 2006, 8:53:05 AM10/1/06

to

<mlut...@wanadoo.fr> wrote in message news:1159692403.8...@b28g2000cwb.googlegroups.com...

No, Marcel, it does not.
the formula x' = g (c-v) t is not what counts.
What counts is the meanings of the variables.
What counts is that you never understood them and you never
will. You invested too heavily in failing to understand, remember?
http://perso.orange.fr/mluttgens/

Dirk Vdm

mlut...@wanadoo.fr

unread,

Oct 1, 2006, 9:26:06 PM10/1/06

to

The meaning of the variables is clear to everybody, and should be
clear, even to you.

The reader has only to refer to what you wrote:

"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.

What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame."

Where does the length x' = g (c-v) t of the stick come from, if not
from the LT x' = g(c-v)t, that transformed the length x = (c-v)t of the
same stick measured in the S-frame ?

Hence, you implicitely recognize that the length x = (c-v)t
measured in the S-frame is *dilated* in the S'-frame. Indeed,
applying length contraction by 1/g to the length x' = g (c-v) t
at rest in the S'-frame, you find back x = (c-v)t in the S-frame.

Your quibbling and sophistry will not mask the falseness of the
Einsteinian LT, which predicts a length *dilation* instead of
an expected length *contraction*. The correct transform is
x' = (c-v)t/g, not x' = (c-v)t*g.

Marcel Luttgens

Dirk Van de moortel

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Oct 1, 2006, 9:38:54 PM10/1/06

to

<mlut...@wanadoo.fr> wrote in message news:1159737966.4...@m7g2000cwm.googlegroups.com...

Dirk Vdm

mlut...@wanadoo.fr

unread,

Oct 2, 2006, 12:15:47 PM10/2/06

to

Van de Moortel wrote in
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Fumblamental.html

"Consider the event E on the light signal with x = c t for some
chosen value of t.

Then c t - v t is the distance between the origin of S' (the 'moving
observer') and the light signal, as seen at time t in the S-frame
(the 'stationary frame'), and, by the way, so c - v is by definition
the closing velocity between the two.

For this event E, as seen in S', the light signal has covered the

distance
x' = c t' = g (c - v) t
This is a distance of the event E in the S' frame."

Let's imagine a stick with length
x = (c-v) t = ct - vt

at rest in the S frame.

Logically, the length of the stick corresponds to the distance
between two points fixed in S, which are occupied by the ends
of the stick simultaneously, i.e. at the same time t.

The coordinates of those two points in the S-frame are:

x2 = ct (the light signal, as seen at time t in the S-frame) and
x1 = vt (the origin of S', as seen at time t).

In S', the corresponding coordinates are, according to the LT:

x2' = ct' = g (c - v) t and
x1' = 0.

Hence the length of the stick in S' is given by
x2' - x1' = g (c - v) t.

Instead of being contracted by 1/g in the 'moving frame', the
stick is dilated by g!
Leading to a false result, the LT is necessarily false.

Marcel Luttgens

Dirk Van de moortel

unread,

Oct 2, 2006, 5:03:51 PM10/2/06

to

<mlut...@wanadoo.fr> wrote in message news:1159791347.8...@k70g2000cwa.googlegroups.com...

> x2 = c t (the light signal, as seen at time t in the S-frame) and
> x1 = v t (the origin of S', as seen at time t).

>
> In S', the corresponding coordinates are, according to the LT:
>

> x2' = c t' = g (c - v) t and

> x1' = 0.
>
> Hence the length of the stick in S' is given by
> x2' - x1' = g (c - v) t.

No. Length of a moving stick must be measured by taking the
distances to the end points simultaneously.
The events (t,x1) and (t,x2) are not simultanous in frame S':

{ x1' = g ( x1 - v t ) = g ( v t - v t ) = 0
{ t1' = g ( t - v x1 /c^2) = g ( t - v v t / c^2 ) = t / g

{ x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
{ t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2) = t sqrt(1-v/c) / sqrt(1+v/c)

Noone (in his right mind) would call x2' - x1' the length of the stick,
since x1' and x2' are distances at *different* times in the S'-frame,
as you can see.

I really don't know how many times this has been explained
to you.

>
> Instead of being contracted by 1/g in the 'moving frame', the
> stick is dilated by g!
> Leading to a false result, the LT is necessarily false.

Your understanding of it certainly is quite false ;-)

Dirk Vdm

mlut...@wanadoo.fr

unread,

Oct 3, 2006, 12:37:31 PM10/3/06

to

In your right mind, what is the length of the stick in the S'-frame, if
not
g (c - v) t ? You should of course demonstrate your solution.
Notice that if you find any value different from (c-v)t/g, the
Lt is false. And don't try to escape by telling me that SR has no
solution.

Marcel Luttgens

Dirk Van de moortel

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Oct 3, 2006, 4:25:26 PM10/3/06

to

<mlut...@wanadoo.fr> wrote in message news:1159879051....@e3g2000cwe.googlegroups.com...

(c-v) t / g

> You should of course demonstrate your solution.
> Notice that if you find any value different from (c-v)t/g, the
> Lt is false. And don't try to escape by telling me that SR has no
> solution.

Sigh.
So your stick has length in the S-frame = dx = (c-v) t, with some
chosen value for t. You want t = 5? You get t = 5.

Since the stick is at rest in S, the end-points can be measured
at any time, so dt for the measuring events doesn't matter.
Since the stick is moving in S', the end-points must be taken
simultaneously in S, so the measuring events must have dt' = 0.
Transformation:
{ dx' = g ( dx - v dt ) [1]
{ dt' = g ( dt - v dx / c^2 ) [2]
or
{ dx = g ( dx' + v dt' ) [3]
{ dt = g ( dt' + v dx' / c^2 ) [4]

You want a connection between dx' and dx, where dt' is known
to be 0, so the simplest way to go about is with equation [3], giving
dx = g dx'
and thus
dx' = dx / g

So the length in S' is (c-v) t / g.
So I don't find a value different from (c-v) t / g.

A stick has a length L in its rest frame.
When measured from a moving frame, that stick has length L / g.
What can be so difficult about that?

The fact that you have to get this spelled out in such trivial
detail, shows that - after at least 10 years - you *still*
haven't understood the meaning of the variables.
Aren't you *embarrassed* by that? You should be.

Dirk Vdm

mlut...@wanadoo.fr

unread,

Oct 3, 2006, 9:55:27 PM10/3/06

to

Your demonstration leads to the correct result, i.e.
the length in S' is (c-v) t / g, but from the *ad hoc* postulate that
"as dt for the measuring events doesn't matter, the measuring

events must have dt' = 0."

Remember that you wrote yesterday: "x1' and x2' are distances
at *different* times in the S'-frame":

"No. Length of a moving stick must be measured by taking the
distances to the end points simultaneously.
The events (t,x1) and (t,x2) are not simultanous in frame S':

{ x1' = g ( x1 - v t ) = g ( v t - v t ) = 0
{ t1' = g ( t - v x1 /c^2) = g ( t - v v t / c^2 ) = t / g

{ x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
{ t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2)
{ = t sqrt(1-v/c) / sqrt(1+v/c)

Noone (in his right mind) would call x2' - x1' the length of the stick,

since x1' and x2' are distances at *different* times in the S'-frame,
as you can see."

You should know (after how many years?) that the value of x' is
always zero. It is wholly independent of any value of t or t'.

On the other side, x2' = g(c-v)t at t2' = gt(1 - v/c).
But to t2' = gt(1 - v/c) corresponds t = t2' / g(1 - v/c), and
x1' is also independent of that specific value of t.

Iow, the value of x1' remains 0 at t2', hence the fact that
the events (t,x1) and (t,x2) are not simultaneous in frame S'
is irrelevant, and x2' - x1' = g (c - v) t is the length of
the stick in the S'-frame.

Thus, according to the Einsteinian LT, the stick is *dilated*,
instead of *contracted*, in the S'-frame.

But if you use the correct LT
x' = (x-vt) / g
t' = t /g
you get the expected length contraction.

Marcel Luttgens

Dirk Van de moortel

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Oct 3, 2006, 10:53:07 PM10/3/06

to

<mlut...@wanadoo.fr> wrote in message news:1159912526.9...@c28g2000cwb.googlegroups.com...

Yes, because YOU TOOK THEM at the same time t in S:

>> >> > x2 = c t (the light signal, as seen at time t in the S-frame) and
>> >> > x1 = v t (the origin of S', as seen at time t).

If two events are simultaneous in S then they are not so in S'.
If two events are simultaneous in S' then they are not so in S.
Big deal.

If you want to measure the length of a moving stick, you must
measure the distances to the end-points at the same time.
If you want to measure the length of a non-moving stick, you
can measure the distances to the end-points at any time, since
the end-points of a non-moving stick aren't going anywhere.
Big deal.
Too bad that you don't understand this :;-)

>
> "No. Length of a moving stick must be measured by taking the
> distances to the end points simultaneously.
> The events (t,x1) and (t,x2) are not simultanous in frame S':
>
> { x1' = g ( x1 - v t ) = g ( v t - v t ) = 0
> { t1' = g ( t - v x1 /c^2) = g ( t - v v t / c^2 ) = t / g
>
> { x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
> { t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2)
> { = t sqrt(1-v/c) / sqrt(1+v/c)
>
> Noone (in his right mind) would call x2' - x1' the length of the stick,
>
> since x1' and x2' are distances at *different* times in the S'-frame,
> as you can see."
>
> You should know (after how many years?) that the value of x' is
> always zero. It is wholly independent of any value of t or t'.
>

yes, that was what I wrote.

This on the other hand, I didn't write:

> On the other side, x2' = g(c-v)t at t2' = gt(1 - v/c).
> But to t2' = gt(1 - v/c) corresponds t = t2' / g(1 - v/c), and
> x1' is also independent of that specific value of t.

You are babbling.

I gave it on a platter:
For your chosen value of t and two events simultaneous in S
with resp. x2 = c t and x1 = v t you get

{ x1' = g ( x1 - v t ) = g ( v t - v t ) = 0
{ t1' = g ( t - v x1 /c^2) = g ( t - v v t / c^2 ) = t / g

and

{ x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
{ t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2) = t sqrt(1-v/c) / sqrt(1+v/c)

So x2' is *not* independent on the chosen value of t, so x1'
cannot be like you say "*also* independent" of t. Stop babbling.
You haven't got a clue, Marcel.

>
> Iow, the value of x1' remains 0 at t2', hence the fact that
> the events (t,x1) and (t,x2) are not simultaneous in frame S'
> is irrelevant, and x2' - x1' = g (c - v) t is the length of
> the stick in the S'-frame.

You have NO IDEA about events.
You have NO IDEA about the meaning of the variables.
You just don't know what you are babbling about.

>
> Thus, according to the Einsteinian LT, the stick is *dilated*,
> instead of *contracted*, in the S'-frame.

No. I proved to you that the length of a moving stick is
its proper lenght divided by gamma, i.e. *contracted*.
This has been shown to school kids since a *century*,
but I guess you are too stupid to understand that.

If you decide to measure the front and the rear of a
moving train at different times and then call the difference
of those distances the *dilated length* of the train, by all
means be my guest and entertain us some more.

>
> But if you use the correct LT
> x' = (x-vt) / g
> t' = t /g
> you get the expected length contraction.

Marcel, you must be the Ultimate Imbecile ;-)

Dirk Vdm

mlut...@wanadoo.fr

unread,

Oct 4, 2006, 10:54:20 AM10/4/06

to

You are so brainwashed by SR that you cannot think logically
anymore. In fact, you have become a crackpot.

Yes, by applying the LT's, one get

x1' = g ( x1 - v t ) = g ( v t - v t ) = 0
t1' = g ( t - v x1 /c^2) = g ( t - v v t / c^2 ) = t / g

which can be written

x1' = gt(v-v) = gt * 0
t1' = t/g

or

x1' = g^2 t1' * 0

Only a crackpot would deny that the value of x1' (=0) is independent
of t, or t1', or any other time at which one of the end of the stick
is measured in S'.

Otoh,

x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2) = t sqrt(1-v/c)

Notice that you made a stupid error, t2' is not t sqrt(1-v/c),
but t sqrt [(1-v/c)/(1+v/c)] !

One can also write

t2' = gt (c-v) / c, thus
t = c t2' / g (c-v)

Hence, x2' = g (c-v) t = c t2', the second end of the stick being
measured at a time t2'.

But at t2', x1', the first end of the stick, still measure 0.

If you deny this, you are a crackpot squared.

So, the length of the stick in the S'-frame is given by
x2' - x1' = g (c-v) t - 0 = g (c-v) t, or
= c t2' - 0 = c t2'

The length is thus *dilated* according to the Einsteinian LT.

You wrote:

"If you decide to measure the front and the rear of a
moving train at different times and then call the difference
of those distances the *dilated length* of the train, by all
means be my guest and entertain us some more."

This shows that you don't even understand the problem.
You are merely babbling.
The correct analogy is that of a car keeping the same
position after any time interval and another car moving away from
the first. After some time interval, the distance between the
two cars is of course increased, not reduced.

I guess you are too stupid to understand that.

Marcel Luttgens

Zoe

unread,

Oct 5, 2006, 6:47:06 PM10/5/06

to

<mlut...@wanadoo.fr> wrote in message news:1159959259.9...@b28g2000cwb.googlegroups.com...

>
> Dirk Van de moortel wrote:

Didn't notice this post yesterday - too big for my regular news server.

[snip]

>> Marcel, you must be the Ultimate Imbecile ;-)
>>
>> Dirk Vdm
>
> You are so brainwashed by SR that you cannot think logically
> anymore. In fact, you have become a crackpot.
>
> Yes, by applying the LT's, one get
>
> x1' = g ( x1 - v t ) = g ( v t - v t ) = 0
> t1' = g ( t - v x1 /c^2) = g ( t - v v t / c^2 ) = t / g
>
> which can be written
>
> x1' = gt(v-v) = gt * 0
> t1' = t/g
>
> or
>
> x1' = g^2 t1' * 0
>
> Only a crackpot would deny that the value of x1' (=0) is independent
> of t, or t1', or any other time at which one of the end of the stick
> is measured in S'.

:-)

>
> Otoh,
>
> x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
> t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2) = t sqrt(1-v/c)
>
> Notice that you made a stupid error, t2' is not t sqrt(1-v/c),
> but t sqrt [(1-v/c)/(1+v/c)] !

I wrote

{ x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
{ t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2) = t sqrt(1-v/c) / sqrt(1+v/c)

You have to look beyond the place where you break the lines apart.
What a malicious little twerp you are.
I have nothing to add to the fact that you are no doubt one of the
most disgusting and stupid imbeciles on the planet, so by all means,
continue to entertain us and try not to die too soon.

Dirk Vdm

Dirk Van de moortel

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Oct 5, 2006, 6:51:13 PM10/5/06

to

"Zoe" <zo...@symphu.com> wrote in message news:eg3jvo$v13$1...@emma.aioe.org...

sorry Zoe, for having used your system ;-P

Dirk Vdm

Zoe

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Oct 5, 2006, 6:55:34 PM10/5/06

to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message

news:BucVg.113625$Kc3.1...@phobos.telenet-ops.be...

>
> "Zoe" <zo...@symphu.com> wrote in message news:eg3jvo$v13$1...@emma.aioe.org...
>
> sorry Zoe, for having used your system ;-P

>
> Dirk Vdm

Geen probleem. Volgende keer naam aanpassen.

mlut...@wanadoo.fr

unread,

Oct 6, 2006, 1:00:27 PM10/6/06

to

Sorry, I overlooked the break.
But you are nevertheless a crackpot, as you are unable to realize that
x' is always zero. I repeat my last post, as it was difficult to find:

You are so brainwashed by SR that you cannot think logically
anymore. In fact, you have become a crackpot.

Yes, by applying the LT's, one get

x1' = g ( x1 - v t ) = g ( v t - v t ) = 0
t1' = g ( t - v x1 /c^2) = g ( t - v v t / c^2 ) = t / g

which can be written

x1' = gt(v-v) = gt * 0
t1' = t/g

or

x1' = g^2 t1' * 0

Only a crackpot would deny that the value of x1' (=0) is independent
of t, or t1', or any other time at which one of the end of the stick
is measured in S'.

Otoh,

x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2) = t sqrt(1-v/c)

One can also write

t2' = gt (c-v) / c, thus
t = c t2' / g (c-v)

Hence, x2' = g (c-v) t = c t2', the second end of the stick being
measured at a time t2'.

But at t2', x1', the first end of the stick, still measure 0.

If you deny this, you are a crackpot squared.

So, the length of the stick in the S'-frame is given by
x2' - x1' = g (c-v) t - 0 = g (c-v) t, or
= c t2' - 0 = c t2'

The length is thus *dilated* according to the Einsteinian LT.

Marcel Luttgens

Dirk Van de moortel

unread,

Oct 6, 2006, 3:43:59 PM10/6/06

to

<mlut...@wanadoo.fr> wrote in message news:1159959259.9...@b28g2000cwb.googlegroups.com...

[with more time and on my own system and ISP now]

>
> Dirk Van de moortel wrote:
>

[snip]

>> Marcel, you must be the Ultimate Imbecile ;-)
>>
>> Dirk Vdm
>
> You are so brainwashed by SR that you cannot think logically
> anymore. In fact, you have become a crackpot.
>
> Yes, by applying the LT's, one get
>
> x1' = g ( x1 - v t ) = g ( v t - v t ) = 0
> t1' = g ( t - v x1 /c^2) = g ( t - v v t / c^2 ) = t / g
>
> which can be written
>
> x1' = gt(v-v) = gt * 0
> t1' = t/g
>
> or
>
> x1' = g^2 t1' * 0
>
> Only a crackpot would deny that the value of x1' (=0) is independent
> of t, or t1', or any other time at which one of the end of the stick
> is measured in S'.

I said:
| "So x2' is *not* independent on the chosen value of t, so x1'
| cannot be like you say "*also* independent" of t."

I did not object to the independence. I objected to your usage
of the word *also* in your sentence

| > On the other side, x2' = g(c-v)t at t2' = gt(1 - v/c).
| > But to t2' = gt(1 - v/c) corresponds t = t2' / g(1 - v/c), and
| > x1' is also independent of that specific value of t.

You have to learn to read what you write and to properly
express yourself.

>
> Otoh,
>
> x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
> t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2) = t sqrt(1-v/c)
>
> Notice that you made a stupid error, t2' is not t sqrt(1-v/c),
> but t sqrt [(1-v/c)/(1+v/c)] !

As I already said, I wrote:
| x2' = g ( x2 - v t ) = g ( v t - c t ) = g (c-v) t
| t2' = g ( t - v x2 /c^2) = g (t - v c t / c^2) = t sqrt(1-v/c) / sqrt(1+v/c)

You have to learn to read what I write before you comment.

>
> One can also write
>
> t2' = gt (c-v) / c, thus
> t = c t2' / g (c-v)
>
> Hence, x2' = g (c-v) t = c t2', the second end of the stick being
> measured at a time t2'.
>
> But at t2', x1', the first end of the stick, still measure 0.

Imbecile.
x1' is a measurement made at time t1', so the first end of
the stick is at x1' at time t1' - not at t2'.
You have to learn to read what I write.
You have to learn to understand the meanings of the variables,
specially when people take the trouble to explain them.

>
> If you deny this, you are a crackpot squared.
>
> So, the length of the stick in the S'-frame is given by
> x2' - x1' = g (c-v) t - 0 = g (c-v) t, or
> = c t2' - 0 = c t2'

No, imbecile.
x1' is the distance at time t1'
x2' is the distance at time t2'
and t1' # t2' - just open your pig-eyes and use them to look
at the expressions.
All of this is by *your* design:

x2 = c t (the light signal, as seen at time t in the S-frame) and
x1 = v t (the origin of S', as seen at time t).

You have to learn to understand what you write before
you spout nonsense about my comments.

>
> The length is thus *dilated* according to the Einsteinian LT.

Only if you decide to measure the front and the rear of a

moving train at different times and then call the difference

of those distances the *dilated length* of the train.

You have to learn to try to understand what people try
to tell you.

>
> You wrote:
>
> "If you decide to measure the front and the rear of a
> moving train at different times and then call the difference
> of those distances the *dilated length* of the train, by all
> means be my guest and entertain us some more."
>
> This shows that you don't even understand the problem.
> You are merely babbling.
> The correct analogy is that of a car keeping the same
> position after any time interval and another car moving away from
> the first. After some time interval, the distance between the
> two cars is of course increased, not reduced.

Imbecile, no one is talking about two cars.
You originally asked a very stupid question and I gave
a trivial answer.
You have to learn to read with people write.
We ended up talking about one stick, the length of which
is measured in its own rest frame giving (t,x1) and (t,x2),
and in a moving frame giving (t1',x1') and (t2',x2').
x2' - x1' is not the length in the moving frame because
the distances to the end points are measured at different
times t1' and t2'.
You have to learn to understand what people are trying to
tell you.
If you want to measure its length in the moving frame, make
sure that you measure the distances simultaneously in that
frame. The they will not be simultaneous in the rest frame of
the stick, but that does not matter since the distances remain
constant in that frame.

But I guess you are too stupid to understand that.

> I guess you are too stupid to understand that.

:-)

Dirk Vdm

mlut...@wanadoo.fr

unread,

Oct 6, 2006, 8:32:07 PM10/6/06

to

The stupid Vdm wrote:

> x1' is a measurement made at time t1', so the first end of
> the stick is at x1' at time t1' - not at t2'.

The stupid Vdm ignores that the first end of the stick coincide
at *any* time with the origin of S'.
At the origin of S', one has always x1' = 0

Marcel Luttgens

Dirk Van de moortel

unread,

Oct 6, 2006, 10:30:23 PM10/6/06

to

<mlut...@wanadoo.fr> wrote in message news:1160166727....@m73g2000cwd.googlegroups.com...

The stick is at rest in the S-frame.
Here are *your* data

| "Let's imagine a stick with length
| x = (c-v) t = ct - vt
| at rest in the S frame."

and

| x2 = c t (the light signal, as seen at time t in the S-frame) and
| x1 = v t (the origin of S', as seen at time t).

for a fixed value of t from my first reponse to your silly question.

So the first end of the stick does *not* "coincide at *any* time

with the origin of S'."

My my.... you are a dishonest little creep, aren't you?
Or shall we keep it at Plain Stone Stupid? ;-)

Dirk Vdm

mlut...@wanadoo.fr

unread,

Oct 7, 2006, 3:39:31 PM10/7/06

to

Here is a classical explanation why a moving stick appears contracted
to an observer at rest:

Consider a stick which, when at rest in S, has a length Lo in the
direction of the x-axis.

Let the stick be set moving relative to S at such velocity that it is
at rest in S'. Its length as measured in S' will still be Lo,
because it must have a certain fixed value in any frame in which
the stick is at rest.

Let us see how the length now measure in S, relative to which the
stick is moving with a velocity v.
It seems reasonable to define the length as the distance between

two points fixed in S, which are occupied by the ends of the stick

simultaneously, i.e., at the same time t.
If the coordinates of these points are x1 and x2, the length is
then L = x2 - x1.

Since the stick is at rest in S', its ends have fixed coordinates
x1', x2' such as Lo = x2' - x1'.

If one substitutes in this last equation values of x2' and x1'
calculated from the LT x' = g(x - vt), one obtains, for a given
value of t,

Lo = g(x2 - x1) = gL, or L = Lo/g,

meaning that to an observer at rest, the length of a moving
stick appears shortened by 1/g.

For instance, x1' = 0 and x2' = g(c - v)t, thus
Lo = x2' - x1' = g(c - v)t.
L = Lo/g = (c - v)t = ct - vt.
Hence, x2 = ct and x1 = vt, meaning that according to S, one
end of the stick coincide with the origin of S', and the other
end corresponds to the distance travelled by a light signal
after a time t. Notice that in S', the origin of S' is always 0.
Notice also that a stick of length L = (c - v)t at
rest in S is *dilated* by g in S', according to the LT.
Indeed,
x1' = g(x1 - vt) = g(vt - vt) = 0
x2' = g(x2 - vt) = g(ct - vt) = g(c - v)t
x2' - x1' = g(c - v)t

Marcel Luttgens

Dirk Van de moortel

unread,

Oct 7, 2006, 3:47:45 PM10/7/06

to

<mlut...@wanadoo.fr> wrote in message news:1160235571....@c28g2000cwb.googlegroups.com...

>
> Dirk Van de moortel wrote:

[snip]

>>

>> My my.... you are a dishonest little creep, aren't you?
>> Or shall we keep it at Plain Stone Stupid? ;-)
>>
>> Dirk Vdm
>
> Here is a classical explanation why a moving stick appears contracted
> to an observer at rest:

Marcel, you are constipated. Try an enema.
I hear they can help you in Illinois.

Dirk Vdm

Brian Kennelly

unread,

Oct 7, 2006, 4:56:15 PM10/7/06

to

mlut...@wanadoo.fr wrote:
>
> Here is a classical explanation why a moving stick appears contracted
> to an observer at rest:
>
> Consider a stick which, when at rest in S, has a length Lo in the
> direction of the x-axis.
>
> Let the stick be set moving relative to S at such velocity that it is
> at rest in S'. Its length as measured in S' will still be Lo,
> because it must have a certain fixed value in any frame in which
> the stick is at rest.
>
> Let us see how the length now measure in S, relative to which the
> stick is moving with a velocity v.
> It seems reasonable to define the length as the distance between
> two points fixed in S, which are occupied by the ends of the stick
> simultaneously, i.e., at the same time t.
> If the coordinates of these points are x1 and x2, the length is
> then L = x2 - x1.
>
> Since the stick is at rest in S', its ends have fixed coordinates
> x1', x2' such as Lo = x2' - x1'.
>
> If one substitutes in this last equation values of x2' and x1'
> calculated from the LT x' = g(x - vt), one obtains, for a given
> value of t,
>
> Lo = g(x2 - x1) = gL, or L = Lo/g,
>
> meaning that to an observer at rest, the length of a moving
> stick appears shortened by 1/g.

Up to here, you are doing fine.

>
> For instance, x1' = 0 and x2' = g(c - v)t, thus

Now, you substitute a stick that is growing with time. That
will make it very hard to compare lengths between systems.

> Lo = x2' - x1' = g(c - v)t.
> L = Lo/g = (c - v)t = ct - vt.
> Hence, x2 = ct and x1 = vt, meaning that according to S, one
> end of the stick coincide with the origin of S', and the other
> end corresponds to the distance travelled by a light signal
> after a time t. Notice that in S', the origin of S' is always 0.
> Notice also that a stick of length L = (c - v)t at
> rest in S

At most, only one point of the stick can be at rest, because it
is expanding rapidly.

> is *dilated* by g in S', according to the LT.
> Indeed,
> x1' = g(x1 - vt) = g(vt - vt) = 0
> x2' = g(x2 - vt) = g(ct - vt) = g(c - v)t
> x2' - x1' = g(c - v)t

You are attempting to describe the length in S' by comparing the
end point locations at the same time in S. You must use the
same time in S' for a meaningful result. In this case it is
easy, because one end is at the origin, and the other end moves
at the speed of light

x1' = 0
x2' = ct'

So the length is always ct'. On the other hand, this tells us
nothing about the comparison of sticks with a fixed length.

Dirk Van de moortel

unread,

Oct 7, 2006, 5:52:44 PM10/7/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:Q_QVg.3648$gM1.1017@fed1read12...

Don't say I didn't warn you ;-)

Dirk Vdm

Sorcerer

unread,

Oct 7, 2006, 6:54:39 PM10/7/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message
news:Q_QVg.3648$gM1.1017@fed1read12...

How about up to here?

http://www.androcles01.pwp.blueyonder.co.uk/Smart/Smart.htm

How fine am I doing?

|
| >
| > For instance, x1' = 0 and x2' = g(c - v)t, thus
| Now, you substitute a stick that is growing with time.

That's called a tree.

That
| will make it very hard to compare lengths between systems.

Nah, you make wooden rulers out of trees, they stop growing with time
and makes it easy to compare lengths between systems.

|
| > Lo = x2' - x1' = g(c - v)t.
| > L = Lo/g = (c - v)t = ct - vt.
| > Hence, x2 = ct and x1 = vt, meaning that according to S, one
| > end of the stick coincide with the origin of S', and the other
| > end corresponds to the distance travelled by a light signal
| > after a time t. Notice that in S', the origin of S' is always 0.
| > Notice also that a stick of length L = (c - v)t at
| > rest in S

| At most, only one point of the stick can be at rest, because it
| is expanding rapidly.

Yes, it's the part just below the ground between the roots and the trunk
that's at rest.

|
| > is *dilated* by g in S', according to the LT.
| > Indeed,
| > x1' = g(x1 - vt) = g(vt - vt) = 0
| > x2' = g(x2 - vt) = g(ct - vt) = g(c - v)t
| > x2' - x1' = g(c - v)t
|
| You are attempting to describe the length in S' by comparing the
| end point locations at the same time in S. You must use the
| same time in S' for a meaningful result. In this case it is
| easy, because one end is at the origin, and the other end moves
| at the speed of light

Nah nah, sticks don't move as fast as light squirrels, or even heavy ones.

|
| x1' = 0
| x2' = ct'
|
| So the length is always ct'. On the other hand, this tells us
| nothing about the comparison of sticks with a fixed length.

Get a plastic ruler from Woolworths, they are lighter than
wooden sticks.

Brian Kennelly

unread,

Oct 7, 2006, 8:35:27 PM10/7/06

to

Sorcerer wrote:
> How about up to here?
>
> http://www.androcles01.pwp.blueyonder.co.uk/Smart/Smart.htm
>
> How fine am I doing?

I have not idea what that page is trying to say, but you still
appear to be misunderstanding Einstein's simple math.

Brian Kennelly

unread,

Oct 7, 2006, 8:59:45 PM10/7/06

to

Sorcerer wrote:
> How about up to here?
>
> http://www.androcles01.pwp.blueyonder.co.uk/Smart/Smart.htm
>
> How fine am I doing?
>

OK. I think I can condense your argument down to this:

You don't believe that it is possible to define t' (tau) in a
way that satisfies:

1/2(t'(0,20)) = t'(32,16)

Yours is an assertion easily disproved by example.

If we define
t'(x,t) = -3x/16 + t

then
t'(0,20) = 20
t'(32,16) = 10

and the equation is satisfied.

(Note, I didn't check your numbers, nor did I try to give the SR
equation. I simply provided an example to show that your
objection is without merit.)

Dirk Van de moortel

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Oct 7, 2006, 9:12:35 PM10/7/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:6zUVg.3669$gM1.3580@fed1read12...

mlut...@wanadoo.fr

unread,

Oct 7, 2006, 10:16:19 PM10/7/06

to

Thank you, even I was merely "parroting" SRists. Notice that
"parroting" is the only thing that gurus like Vdm can do. They simply
repeat
what they learned from other gurus. They are unable to think
by themselves.

>
> >
> > For instance, x1' = 0 and x2' = g(c - v)t, thus
> Now, you substitute a stick that is growing with time. That
> will make it very hard to compare lengths between systems.
>

No, like above, one has to restrict the solution "to a given value of
t".

> > Lo = x2' - x1' = g(c - v)t.
> > L = Lo/g = (c - v)t = ct - vt.
> > Hence, x2 = ct and x1 = vt, meaning that according to S, one
> > end of the stick coincide with the origin of S', and the other
> > end corresponds to the distance travelled by a light signal
> > after a time t. Notice that in S', the origin of S' is always 0.
> > Notice also that a stick of length L = (c - v)t at
> > rest in S

> At most, only one point of the stick can be at rest, because it
> is expanding rapidly.

I appreciate that, contrary to gurus like Vdm, you admit that the
origin of S' is always zero in S'.
Otoh, for a given value of t, the length of the stick is not expanding

>
> > is *dilated* by g in S', according to the LT.
> > Indeed,
> > x1' = g(x1 - vt) = g(vt - vt) = 0
> > x2' = g(x2 - vt) = g(ct - vt) = g(c - v)t
> > x2' - x1' = g(c - v)t
>
> You are attempting to describe the length in S' by comparing the
> end point locations at the same time in S. You must use the
> same time in S' for a meaningful result. In this case it is
> easy, because one end is at the origin, and the other end moves
> at the speed of light

Yes, but one has to take into account that the other end
of the stick has a well defined value for a given value of t.

>
> x1' = 0
> x2' = ct'
>
> So the length is always ct'. On the other hand, this tells us
> nothing about the comparison of sticks with a fixed length.

Don't forget that t' = g(t - vx/c^2), hence x2' = g(c - v)t.

The LT tells us that a stick of length (c - v)t in S has a length
g(c - v)t in S', for a given value of t.
Iow, following the LT, the observer in S will conclude that,
according to a S' observer, his stick is *dilated* by g.

I am looking forward to a physical interpretation of x' = g(x -vt), not
from parroting gurus, but from people who can think by themselves
and are genuinely interested in a thorough understanding of the LT's.
I think that the Einsteinian LT's are false, because they lead to
contradictions.
If one reject the the Einsteinian postulate that when x = ct, x' = ct',

the LT become
x' = (c-v)t / g
t' = t / g,
and then, there are no contradictory results any more.

Marcel Luttgens

Sorcerer

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Oct 7, 2006, 10:22:01 PM10/7/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:jcUVg.3663$gM1.328@fed1read12...

Oh... Well, it shows how the cuckoo malformations that every
shitheaded relativist blames Lorentz for are derived, is it too hard
for you?
Half of twenty is sixteen, the other half is four. That's Einstein's
simple math, I understand it very well.
What is it that you imagine I'm misunderstanding, fuckwit?
Androcles.

mlut...@wanadoo.fr

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Oct 7, 2006, 10:35:48 PM10/7/06

to

Dirk Van de moortel wrote:

Sorry, de gevraagde pagina kan niet gevonden worden.
Page not found - HTTP 404

How much do you get from Jiba?
If Jiba applies the Einsteinian LT's, you could better stay in your
kot.

Marcel Luttgens

Sorcerer

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Oct 7, 2006, 10:42:01 PM10/7/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:6zUVg.3669$gM1.3580@fed1read12...

And satisfy 1/2(t'(0,4)) = t'(32,16) as well? That's right, I don't
believe it is possible.

| Yours is an assertion easily disproved by example.

I didn't make any assertion, Einstein did.

That's not the answer, it should be 16.

tau = (t-vx/c^2)/ sqrt(1-v^2/c^2)
= (20 - 0.6*0) / sqrt( 1 -0.36)
= 20/0.8 = 16
Is Einstein's simple math too difficult for you?

|
| (Note, I didn't check your numbers, nor did I try to give the SR
| equation. I simply provided an example to show that your
| objection is without merit.)

Noted. You didn't bother to check and got the wrong answer.
Another arrogant shithead....
Fuck off, moron.
Androcles

Brian Kennelly

unread,

Oct 8, 2006, 3:29:59 AM10/8/06

to

Why do you want to satisfy that equation? You are objecting to
Einstein's derivation of the LT, and using your numbers, only
the equation I quoted corresponds to his derivation.

>
>
> | Yours is an assertion easily disproved by example.
>
> I didn't make any assertion, Einstein did.

No, you assert that Einstein's equation cannot be satisfied, and
that, consequently, it is nonsense. I disproved your assertion,
using your numbers.

>
>
> |
> | If we define
> | t'(x,t) = -3x/16 + t
> |
> | then
> | t'(0,20) = 20
> | t'(32,16) = 10
> |
> | and the equation is satisfied.
>
> That's not the answer, it should be 16.

No, the equation was:

> 1/2(t'(0,20)) = t'(32,16)

My proposed function for t' satisfies the equation:
1/2(20) = 10

16 is one of the function's arguments, not necessarily its
value. Einstein's functional equation only determines the
relationship up to a multiplicative constant, the value of which
is determined later in the paper.

If it will make you happier, use:
t'(x,t) = -3x/20 + 4t/5

Now t'(0,20) = 16
and t'(32,16) = 8
and, once again the equation is satisfied.
1/2/(16) = 8

> |
> | (Note, I didn't check your numbers, nor did I try to give the SR
> | equation. I simply provided an example to show that your
> | objection is without merit.)
>

I didn't check your numbers, because you didn't clearly state
where they came from.

Looking at your equations, I infer that:
c=5 (Light speed)
v=3 (Train speed)
x'=32 (Length of the moving train in the latin system)

Those numbers are consistent, and lead to the stated equations.

Your
> 1/2(t'(0,20) = t'(32,4)
doesn't correspond to anything in Einstein's paper, or in his
argument.

Brian Kennelly

unread,

Oct 8, 2006, 3:56:29 AM10/8/06

to

mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:
>
>>> For instance, x1' = 0 and x2' = g(c - v)t, thus
>> Now, you substitute a stick that is growing with time. That
>> will make it very hard to compare lengths between systems.
>>
>
> No, like above, one has to restrict the solution "to a given value of
> t"

What is t"?

You state that the length is proportional to the time. The
coefficient is positive, so it is increasing. At time t=0, the
length is zero, and is positive for all positive time values.

>
>>> Lo = x2' - x1' = g(c - v)t.
>>> L = Lo/g = (c - v)t = ct - vt.
>>> Hence, x2 = ct and x1 = vt, meaning that according to S, one
>>> end of the stick coincide with the origin of S', and the other
>>> end corresponds to the distance travelled by a light signal
>>> after a time t. Notice that in S', the origin of S' is always 0.
>>> Notice also that a stick of length L = (c - v)t at
>>> rest in S
>
>> At most, only one point of the stick can be at rest, because it
>> is expanding rapidly.
>
> I appreciate that, contrary to gurus like Vdm, you admit that the
> origin of S' is always zero in S'.
> Otoh, for a given value of t, the length of the stick is not expanding

I disagree. For a given value of t, the length of the stick has
a definite value, but it is increasing. (Shades of Zeno. Does
a moving arrow have a velocity at a given time?)

>
>>> is *dilated* by g in S', according to the LT.
>>> Indeed,
>>> x1' = g(x1 - vt) = g(vt - vt) = 0
>>> x2' = g(x2 - vt) = g(ct - vt) = g(c - v)t
>>> x2' - x1' = g(c - v)t
>> You are attempting to describe the length in S' by comparing the
>> end point locations at the same time in S. You must use the
>> same time in S' for a meaningful result. In this case it is
>> easy, because one end is at the origin, and the other end moves
>> at the speed of light
>
> Yes, but one has to take into account that the other end
> of the stick has a well defined value for a given value of t.
>
>> x1' = 0
>> x2' = ct'
>>
>> So the length is always ct'. On the other hand, this tells us
>> nothing about the comparison of sticks with a fixed length.
>
> Don't forget that t' = g(t - vx/c^2), hence x2' = g(c - v)t.

That is misleading you. To determine the length in S', you must
calculate both end points at the same value of t'.

>
> The LT tells us that a stick of length (c - v)t in S has a length
> g(c - v)t in S', for a given value of t.

First, the <length in S', for a given value of t> is
meaningless. You want the <length in S', for a given value of t'>.

Second, because the length is increasing, any perceived dilation
is accounted for by the growing stick, not the LT.

> Iow, following the LT, the observer in S will conclude that,
> according to a S' observer, his stick is *dilated* by g.

No, if we use a stick with a fixed length, at rest in S, with
end points at x1 and x2, then we can find the length in S' from
the LT:
x1 = g(x1'+vt1')
x2 = g(x2'+vt2')
Now, the length is calculated when t1'=t2', so we get:
x2-x1 = g(x2'-x1')
and the length is x2'-x1'=1/g(x2-x1)
It is contracted, not dilated.

>
> If one reject the the Einsteinian postulate that when x = ct, x' = ct',
>
> the LT become
> x' = (c-v)t / g

You are missing any dependence on x; the whole line collapses to
a point. Can I assume that you meant <x' = (x-vt)/g>?

> t' = t / g,
> and then, there are no contradictory results any more.

It does not form a group, even if we include the dependence on x.

Sorcerer

unread,

Oct 8, 2006, 4:56:40 AM10/8/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:Xg_Vg.3694$gM1.2041@fed1read12...

| Sorcerer wrote:
| > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > news:6zUVg.3669$gM1.3580@fed1read12...
| > | Sorcerer wrote:
| > | > How about up to here?
| > | >
| > | > http://www.androcles01.pwp.blueyonder.co.uk/Smart/Smart.htm
| > | >
| > | > How fine am I doing?
| > | >
| > |
| > | OK. I think I can condense your argument down to this:
| > |
| > | You don't believe that it is possible to define t' (tau) in a
| > | way that satisfies:
| > |
| > | 1/2(t'(0,20)) = t'(32,16)
| >
| >
| > And satisfy 1/2(t'(0,4)) = t'(32,16) as well? That's right, I don't
| > believe it is possible.
|
| Why do you want to satisfy that equation?

Because I want to send the light back again, moron.

"But the ray moves relatively to the initial point of k, when measured in
the stationary system, with the velocity c-v, so that

http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img31.gif

| You are objecting to
| Einstein's derivation of the LT, and using your numbers, only
| the equation I quoted corresponds to his derivation.

The guy was half-arsed, like you.
http://www.androcles01.pwp.blueyonder.co.uk/Rocket/eq22.A.GIF

|
| >
| >
| > | Yours is an assertion easily disproved by example.
| >
| > I didn't make any assertion, Einstein did.
| No, you assert that Einstein's equation cannot be satisfied,

I PROVE, I do NOT assert.

and
| that, consequently, it is nonsense. I disproved your assertion,
| using your numbers.

No you didn't, and you got the wrong answer anyway.

|
| >
| >
| > |
| > | If we define
| > | t'(x,t) = -3x/16 + t
| > |
| > | then
| > | t'(0,20) = 20
| > | t'(32,16) = 10
| > |
| > | and the equation is satisfied.
| >
| > That's not the answer, it should be 16.
| No, the equation was:
| > 1/2(t'(0,20)) = t'(32,16)
| My proposed function for t' satisfies the equation:
| 1/2(20) = 10

Wrong answer, shithead, it should be 16. Moving clocks run slow.

That's for the light time of x'/(c-v)
Coming back its t = x'/(c+v),
1/2(4) = 8.

|
| > |
| > | (Note, I didn't check your numbers, nor did I try to give the SR
| > | equation. I simply provided an example to show that your
| > | objection is without merit.)
| >
|
| I didn't check your numbers, because you didn't clearly state
| where they came from.

speed of light outgoing, track frame: 80/16 = 5
speed of light outgoing, train frame: 40/8 = 5
speed of light returning, track frame: 20/4 = 5
speed of light returning, train frame: 40/8 = 5

speed of light is the same ion all inertial frames of reference,
trains move by peristalsis.
"In the first place it is clear that the equations must be linear on account
of the properties of homogeneity which we attribute to space and time."
(linear in italics)

-- Shithead Einstein.

ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/

tau(4) = 8

tau(16) = 8.

Nice linear function, that.

That's because he left it out, fuckwit.

Androcles.

Brian Kennelly

unread,

Oct 8, 2006, 6:34:12 AM10/8/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message
> news:Xg_Vg.3694$gM1.2041@fed1read12...
> | Sorcerer wrote:
> | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > news:6zUVg.3669$gM1.3580@fed1read12...
> | > |

> | > | You don't believe that it is possible to define t' (tau) in a
> | > | way that satisfies:
> | > |
> | > | 1/2(t'(0,20)) = t'(32,16)
> | >
> | >
> | > And satisfy 1/2(t'(0,4)) = t'(32,16) as well? That's right, I don't
> | > believe it is possible.
> |
> | Why do you want to satisfy that equation?
>
> Because I want to send the light back again

The '20' in the left hand term includes the full round trip in
the latin frame. 16+4=20

Your new equation does not correspond to anything in the argument.

>
> "But the ray moves relatively to the initial point of k, when measured in
> the stationary system, with the velocity c-v, so that
>
> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img31.gif

The 'plus' sign on the right hand side does not belong. The
time in the right hand side includes only the out bound ray. It
represents the time of the reflection.

Einstein's equation simply says that, in the system moving with
the train, the reflection occurs at the mid point of the round
trip. It represents that in terms of the times and locations in
the other system.

What is your objection?

> |
> | >
> | >
> | > | Yours is an assertion easily disproved by example.
> | >
> | > I didn't make any assertion, Einstein did.
> | No, you assert that Einstein's equation cannot be satisfied,
>
> I PROVE, I do NOT assert.

Your proof must be flawed, because, as I showed you, a
counter-example exists.

State your proof.

>
> and
> | that, consequently, it is nonsense. I disproved your assertion,
> | using your numbers.
>
> No you didn't, and you got the wrong answer anyway.

Where is the flaw in my demonstration? I provided a function
for tau that satisfied the equation. I followed up by providing
a function that satisfied the equation and yielded the value you
wanted to see (below).

Coming back is already included in the equation (20=16+4). Your
new equation is meaningless.

>
> tau(4) = 8
>
> tau(16) = 8.
>
> Nice linear function, that.

If you include the 'x' values, you can satisfy both of these
equations with a linear function.

Using T for tau, and Einstein's x' (for consistency in our
discussion):
T(x',t) = 4/5t-x'*3/20

So,
t'(32,16) = 8
t'(-32,4) = 8

Nice linear function, indeed!

He left it out, because it does not correspond to anything in
the argument. If you want to introduce it, you will have to
explain its meaning and place in the argument.

Dirk Van de moortel

unread,

Oct 8, 2006, 9:04:29 AM10/8/06

to

<mlut...@wanadoo.fr> wrote in message news:1160260547....@e3g2000cwe.googlegroups.com...

[snip]

>> ... and check his diapers
>> http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Orgasm.html
>
> Sorry, de gevraagde pagina kan niet gevonden worden.
> Page not found - HTTP 404

Sorry.
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Androrgasm.html
Thanks for letting me know.

Dirk Vdm

Dirk Van de moortel

unread,

Oct 8, 2006, 9:09:43 AM10/8/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:NF_Vg.3700$gM1.3681@fed1read12...

> mlut...@wanadoo.fr wrote:
>> Brian Kennelly wrote:
>>> mlut...@wanadoo.fr wrote:
>>
>>>> For instance, x1' = 0 and x2' = g(c - v)t, thus
>>> Now, you substitute a stick that is growing with time. That
>>> will make it very hard to compare lengths between systems.
>>>
>>
>> No, like above, one has to restrict the solution "to a given value of
>> t"
> What is t"?

It is t follwed by the closing phrase quotation mark.

>
> You state that the length is proportional to the time. The coefficient is positive, so it is increasing. At time t=0, the length
> is zero, and is positive for all positive time values.

Not id t has a fixed initial value like we established higher
up the thread.

>
>>
>>>> Lo = x2' - x1' = g(c - v)t.
>>>> L = Lo/g = (c - v)t = ct - vt.
>>>> Hence, x2 = ct and x1 = vt, meaning that according to S, one
>>>> end of the stick coincide with the origin of S', and the other
>>>> end corresponds to the distance travelled by a light signal
>>>> after a time t. Notice that in S', the origin of S' is always 0.
>>>> Notice also that a stick of length L = (c - v)t at
>>>> rest in S
>>
>>> At most, only one point of the stick can be at rest, because it
>>> is expanding rapidly.
>>
>> I appreciate that, contrary to gurus like Vdm, you admit that the
>> origin of S' is always zero in S'.
>> Otoh, for a given value of t, the length of the stick is not expanding
>
> I disagree. For a given value of t, the length of the stick has a definite value, but it is increasing.

c = 1, v = 0.5, t = 2, L = (c-v) t.
Is L increasing?

> (Shades of Zeno. Does a moving arrow have a velocity at a given time?)

Yes.

Dirk Vdm

Sorcerer

unread,

Oct 8, 2006, 12:27:40 PM10/8/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:FZ0Wg.3714$gM1.2909@fed1read12...

| Sorcerer wrote:
| > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > news:Xg_Vg.3694$gM1.2041@fed1read12...
| > | Sorcerer wrote:
| > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > | > news:6zUVg.3669$gM1.3580@fed1read12...
| > | > |
| > | > | You don't believe that it is possible to define t' (tau) in a
| > | > | way that satisfies:
| > | > |
| > | > | 1/2(t'(0,20)) = t'(32,16)
| > | >
| > | >
| > | > And satisfy 1/2(t'(0,4)) = t'(32,16) as well? That's right, I don't
| > | > believe it is possible.
| > |
| > | Why do you want to satisfy that equation?
| >
| > Because I want to send the light back again
| The '20' in the left hand term includes the full round trip in
| the latin frame. 16+4=20

Yes, 16 is half of 20 and 4 is the other half.

| Your new equation does not correspond to anything in the argument.

tau(4) = 8, tau(16) = 8, hence 4 = 16.

| > "But the ray moves relatively to the initial point of k, when measured
in
| > the stationary system, with the velocity c-v, so that
| >
| > http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img31.gif
|
| The 'plus' sign on the right hand side does not belong.

Ok, what the fuck is it doing in
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif

See the thread title? "SR fundamental contradiction".
That wasn't me, that was someone else who realized something
was wrong, and a stooopid fuck like you with your head up your
arse can't see what it is.

| The
| time in the right hand side includes only the out bound ray. It
| represents the time of the reflection.

The time in the right hand side includes only half the total time.

It represents the time of the reflection.

|
| Einstein's equation simply says that, in the system moving with
| the train, the reflection occurs at the mid point of the round
| trip.

1/2 of 100 = 80, the other half is 20.

xi(20) = 40,
xi(80) = 40,
xi(32) = 40.
The distance in the right hand side includes only half the total distance.
It represents the distance of the reflection.

Nothing wrong with that, is there?

| It represents that in terms of the times and locations in
| the other system.
|
| What is your objection?

I have no objection, you are the shithead that

" I have not idea what that page is trying to say, but you still
appear to be misunderstanding Einstein's simple math."

but you still appear to be objecting Einstein's simple-minded math.

You got it right when you said "I have not idea", you have not brain,
you have not clue. Just believe what you are told to believe, shithead.

| > |
| > | >
| > | >
| > | > | Yours is an assertion easily disproved by example.
| > | >
| > | > I didn't make any assertion, Einstein did.
| > | No, you assert that Einstein's equation cannot be satisfied,
| >
| > I PROVE, I do NOT assert.
| Your proof must be flawed, because, as I showed you, a
| counter-example exists.
|
| State your proof.

I did. Trains move by peristalsis. A counter example exists,
some of them have wheels.
Wheels are not inertial frames of reference so maybe they don't count.
SR is not flawed, it works fine for earthworms.
1/2 (20) = 4, 1/2(20) = 16 and 16+4 = 20. Right, worm?

|
| >
| > and
| > | that, consequently, it is nonsense. I disproved your assertion,
| > | using your numbers.
| >
| > No you didn't, and you got the wrong answer anyway.
| Where is the flaw in my demonstration?

Nothing, it is fine. half of 10 is 8.

| I provided a function
| for tau that satisfied the equation. I followed up by providing
| a function that satisfied the equation and yielded the value you
| wanted to see (below).

Of course you did.
tau(4) = 8, tau(16) = 8.

See the thread title? "SR fundamental contradiction".
If a worm thinks tau(4) = tau(16) is a linear function
why should I object?
I have a tower in Paris, tall and made of iron. It's
a great tourist spot. Would you like to buy it? I can
arrange credit. All I need is a small deposit and a gullible worm.
There are some nice fish out in the river, do you like being
on my hook instead of Einstein's hook? He was such a
nice angler, too. Pity he can't reel you in, he's dead.
Of course being a REAL shitbag he'd be polite to you, but to me
you are just another dumbfuck I can waste, you are not going
anywhere with your life.

| > |
| > | If it will make you happier, use:
| > | t'(x,t) = -3x/20 + 4t/5
| > |
| > | Now t'(0,20) = 16
| > | and t'(32,16) = 8
| > | and, once again the equation is satisfied.
| > | 1/2/(16) = 8
| >
| > That's for the light time of x'/(c-v)
| > Coming back its t = x'/(c+v),
| > 1/2(4) = 8.
| Coming back is already included in the equation (20=16+4). Your
| new equation is meaningless.

Ok, worm. Stay on the hook.

|
| >
| > tau(4) = 8
| >
| > tau(16) = 8.
| >
| > Nice linear function, that.
| If you include the 'x' values, you can satisfy both of these
| equations with a linear function.
|
| Using T for tau, and Einstein's x' (for consistency in our
| discussion):
| T(x',t) = 4/5t-x'*3/20
|
| So,
| t'(32,16) = 8
| t'(-32,4) = 8
|
| Nice linear function, indeed!

Oh, I see, the distance is -32. Why not say t'(-32,-16) = 8 instead?
Time runs backwards, doesn't it?

| > |
| > | Looking at your equations, I infer that:
| > | c=5 (Light speed)
| > | v=3 (Train speed)
| > | x'=32 (Length of the moving train in the latin system)
| > |
| > | Those numbers are consistent, and lead to the stated equations.
| > |
| > | Your
| > | > 1/2(t'(0,20) = t'(32,4)
| > | doesn't correspond to anything in Einstein's paper, or in his
| > | argument.
| >
| > That's because he left it out,
|
| He left it out, because it does not correspond to anything in
| the argument. If you want to introduce it, you will have to
| explain its meaning and place in the argument.

How about this one instead:
http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm

No worm in that one, sick sock puppet.

How fine am I doing?

Androcles

Brian Kennelly

unread,

Oct 8, 2006, 4:15:13 PM10/8/06

to

Dirk Van de moortel wrote:

> "Brian Kennelly" <bwken...@cox.net> wrote in message news:NF_Vg.3700$gM1.3681@fed1read12...
>> mlut...@wanadoo.fr wrote:
>>> No, like above, one has to restrict the solution "to a given value of
>>> t"
>> What is t"?
>
> It is t follwed by the closing phrase quotation mark.

Sorry, you are right. I read and answered too quickly.

>
>> You state that the length is proportional to the time. The coefficient is positive, so it is increasing. At time t=0, the length
>> is zero, and is positive for all positive time values.
>
> Not id t has a fixed initial value like we established higher
> up the thread.

If t is fixed, then drop it from the argument, or use a
different symbol. It only serves to create confusion to use the
same symbol for a constant and a variable.

>
>>> I appreciate that, contrary to gurus like Vdm, you admit that the
>>> origin of S' is always zero in S'.
>>> Otoh, for a given value of t, the length of the stick is not expanding
>> I disagree. For a given value of t, the length of the stick has a definite value, but it is increasing.
>
> c = 1, v = 0.5, t = 2, L = (c-v) t.
> Is L increasing?

Yes, just as a moving object has a velocity at each time, with
which you agree below, an expanding object has an increasing
length at each time.

If you are using the equation, L=(c-v)t, to define the time of
the measurement in S, rather than to define the length of the
stick, it would be more clear to express it as T=L/(c-v). In
that case, however, it is not meaningful for expressing the
length in S'.

This answer does, however, highlight the problem with the
argument. A definite time in S, such as t=2, does not represent
the same definite time at both ends of the stick in S'.

Dirk Van de moortel

unread,

Oct 8, 2006, 4:19:24 PM10/8/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:lu9Wg.3721$gM1.426@fed1read12...

> Dirk Van de moortel wrote:
>> "Brian Kennelly" <bwken...@cox.net> wrote in message news:NF_Vg.3700$gM1.3681@fed1read12...
>>> mlut...@wanadoo.fr wrote:
>>>> No, like above, one has to restrict the solution "to a given value of
>>>> t"
>>> What is t"?
>>
>> It is t follwed by the closing phrase quotation mark.
> Sorry, you are right. I read and answered too quickly.
>
>>
>>> You state that the length is proportional to the time. The coefficient is positive, so it is increasing. At time t=0, the
>>> length is zero, and is positive for all positive time values.
>>
>> Not id t has a fixed initial value like we established higher
>> up the thread.
> If t is fixed, then drop it from the argument, or use a different symbol. It only serves to create confusion to use the same
> symbol for a constant and a variable.

Try explaining that to an imbecile like Marcel Luttgens ;-)
http://users.telenet.be/vdmoortel/dirk/Stuff/MarcelAtSchool.gif

>>
>>>> I appreciate that, contrary to gurus like Vdm, you admit that the
>>>> origin of S' is always zero in S'.
>>>> Otoh, for a given value of t, the length of the stick is not expanding
>>> I disagree. For a given value of t, the length of the stick has a definite value, but it is increasing.
>>
>> c = 1, v = 0.5, t = 2, L = (c-v) t.
>> Is L increasing?
> Yes, just as a moving object has a velocity at each time, with which you agree below, an expanding object has an increasing length
> at each time.

But t was fixed in the beginning of the thread:
| Consider the event E on the light signal with x = c t for some
| chosen value of t.
Remember, this is Marcel Luttgens you are dealing with.

Dirk Vdm

Brian Kennelly

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Oct 8, 2006, 5:08:22 PM10/8/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message
> news:FZ0Wg.3714$gM1.2909@fed1read12...
> | Sorcerer wrote:
> | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > news:Xg_Vg.3694$gM1.2041@fed1read12...
> | > | Sorcerer wrote:
> | > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > | > news:6zUVg.3669$gM1.3580@fed1read12...
> | > | > |
> | > | > | You don't believe that it is possible to define t' (tau) in a
> | > | > | way that satisfies:
> | > | > |
> | > | > | 1/2(t'(0,20)) = t'(32,16)
> | > | >
> | > | >
> | > | > And satisfy 1/2(t'(0,4)) = t'(32,16) as well? That's right, I don't
> | > | > believe it is possible.
> | > |
> | > | Why do you want to satisfy that equation?
> | >
> | > Because I want to send the light back again
> | The '20' in the left hand term includes the full round trip in
> | the latin frame. 16+4=20
>
> Yes, 16 is half of 20 and 4 is the other half

Nobody but you is claiming that. 16 and 4 are the divisions of
the signal in the t times, but they are neither is 'half'. Only
the tau times are equated between the out and back portions of
the signal, so that they each represent half of the trip.

>
>
> | Your new equation does not correspond to anything in the argument.
>
>
> tau(4) = 8, tau(16) = 8, hence 4 = 16.

Those equations are yours, not Einstein's. Einstein included
the spatial dependence. When you do, then your conclusion does
not follow, even for linear equations.

tau(-32,4)=tau(32,16)= 8
No contradiction.

The correct form of your argument is:
if tau(x1',4)=tau(x2',16), then x1' \= x2'

>
>
> | > "But the ray moves relatively to the initial point of k, when measured
> in
> | > the stationary system, with the velocity c-v, so that
> | >
> | > http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img31.gif
> |
> | The 'plus' sign on the right hand side does not belong.
>
> Ok, what the fuck is it doing in
> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif

The plus sign is not on the right hand side. It is on the left.
The left hand side represents the out-bound and return trip.
The right hand side represents only the out-bound.

That is the reason for the 1/2 on the left. If the time is the
same in both directions, then the outbound time is half of the
full trip.

(Before you object that 16\=4, note that the times being equated
are the tau times, not the t times. Nobody is claiming that the
two t times are equal. Tau is a linear function of t *and* x',
so there is no contradiction, if the x' values are different.)

>
> See the thread title? "SR fundamental contradiction".
> That wasn't me, that was someone else who realized something

> was wrong, and a xxxxxxxxxxx like you with your head up your
> xxxx can't see what it is.
Then enlighten me.

Your arguments will carry more weight if you can refrain from
offensive language. The strength of your language is inversely
proportional to the strength of your argument.

>
> | The
> | time in the right hand side includes only the out bound ray. It
> | represents the time of the reflection.
>
> The time in the right hand side includes only half the total time.
> It represents the time of the reflection.
>
>
> |
> | Einstein's equation simply says that, in the system moving with
> | the train, the reflection occurs at the mid point of the round
> | trip.
>
> 1/2 of 100 = 80, the other half is 20.

The length of the train is the same in both directions. From
this obvious fact, we conclude that, in the train system, the
time for the light signals in both directions are the same.

>
> The distance in the right hand side includes only half the total distance.
> It represents the distance of the reflection.
>
> Nothing wrong with that, is there?

It is correct that the right hand side contains the distance of
the reflection (in Einstein's x'[=x-vt]).
It also contains the time of the reflection (in t time). The
value of the function is the tau time of the reflection.

>
> | It represents that in terms of the times and locations in
> | the other system.
> |
> | What is your objection?
>
> I have no objection,

Then you assent to the correctness of Einstein's equations?

> | > | > | Yours is an assertion easily disproved by example.
> | > | >
> | > | > I didn't make any assertion, Einstein did.
> | > | No, you assert that Einstein's equation cannot be satisfied,
> | >
> | > I PROVE, I do NOT assert.
> | Your proof must be flawed, because, as I showed you, a
> | counter-example exists.
> |
> | State your proof.
>
> I did. Trains move by peristalsis. A counter example exists,
> some of them have wheels.
> Wheels are not inertial frames of reference so maybe they don't count.
> SR is not flawed, it works fine for earthworms.
> 1/2 (20) = 4, 1/2(20) = 16 and 16+4 = 20. Right, worm?

State your proof, not your erroneous equations.

> | > | that, consequently, it is nonsense. I disproved your assertion,
> | > | using your numbers.
> | >
> | > No you didn't, and you got the wrong answer anyway.
> | Where is the flaw in my demonstration?
>
> Nothing, it is fine.

Thank you.

>
> How about this one instead:
> http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm
>
> No worm in that one, sick sock puppet.
> How fine am I doing?

As a criticism of Einstein, it is meaningless, because you have
light travelling at two different speeds in the same reference
system.

You also seem to be claiming, erroneously, that because the
clocks are synchronized in t time, they are synchronized in
train time. That is a proposition that you may want to prove,
but you cannot assume.

mlut...@wanadoo.fr

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Oct 8, 2006, 5:15:27 PM10/8/06

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alt="invisible hit counter" border="0"> </noscript>

Here is the cookie I got by clicking on
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session_1141211
1160326018%260
statcounter.com/
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Does Statcounter give you such infos?

Marcel Luttgens

Brian Kennelly

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Oct 8, 2006, 5:16:22 PM10/8/06

to

Dirk Van de moortel wrote:
> But t was fixed in the beginning of the thread:
> | Consider the event E on the light signal with x = c t for some
> | chosen value of t.
> Remember, this is Marcel Luttgens you are dealing with.

I was attempting to demonstrate the source of the error by
taking his equations at face value. If we do so, the source of
the dilation is found in the variable length, not in the LT
equations. The endpoints in S', determined at the same time t',
are at two different S times, t1 and t2. By his definition of
the length, the stick is longer when S' measures it, because it
is longer when S measures it. [If t2>t1, then (c-v)t2 > (c-v(t1)].

Dirk Van de moortel

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Oct 8, 2006, 5:52:25 PM10/8/06

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<mlut...@wanadoo.fr> wrote in message news:1160327726.9...@k70g2000cwa.googlegroups.com...

http://www.statcounter.com/privacy.html

Dirk Vdm

Dirk Van de moortel

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Oct 8, 2006, 5:54:37 PM10/8/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:HnaWg.3725$gM1.431@fed1read12...

> Dirk Van de moortel wrote:
>> But t was fixed in the beginning of the thread:
>> | Consider the event E on the light signal with x = c t for some
>> | chosen value of t.
>> Remember, this is Marcel Luttgens you are dealing with.

> I was attempting to demonstrate the source of the error by taking his equations at face value.

Big mistake.
You have to take his equations like he takes them, namely
without attatching *any* physical meaning to the variables
what-so-ever.
Trust me, that is how his peanut works ;-)

Dirk Vdm

Sorcerer

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Oct 8, 2006, 8:07:48 PM10/8/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:agaWg.3723$gM1.1723@fed1read12...

| Sorcerer wrote:
| > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > news:FZ0Wg.3714$gM1.2909@fed1read12...
| > | Sorcerer wrote:
| > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > | > news:Xg_Vg.3694$gM1.2041@fed1read12...
| > | > | Sorcerer wrote:
| > | > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > | > | > news:6zUVg.3669$gM1.3580@fed1read12...
| > | > | > |
| > | > | > | You don't believe that it is possible to define t' (tau) in a
| > | > | > | way that satisfies:
| > | > | > |
| > | > | > | 1/2(t'(0,20)) = t'(32,16)
| > | > | >
| > | > | >
| > | > | > And satisfy 1/2(t'(0,4)) = t'(32,16) as well? That's right, I
don't
| > | > | > believe it is possible.
| > | > |
| > | > | Why do you want to satisfy that equation?
| > | >
| > | > Because I want to send the light back again
| > | The '20' in the left hand term includes the full round trip in
| > | the latin frame. 16+4=20
| >
| > Yes, 16 is half of 20 and 4 is the other half
| Nobody but you is claiming that. 16 and 4 are the divisions of
| the signal in the t times, but they are neither is 'half'. Only
| the tau times are equated between the out and back portions of
| the signal, so that they each represent half of the trip.

Proof?

| >
| >
| > | Your new equation does not correspond to anything in the argument.
| >
| >
| > tau(4) = 8, tau(16) = 8, hence 4 = 16.
| Those equations are yours, not Einstein's.

Proof?

| Einstein included
| the spatial dependence. When you do, then your conclusion does
| not follow, even for linear equations.

Proof?

|
| tau(-32,4)=tau(32,16)= 8

| No contradiction.

Proof?

| The correct form of your argument is:
| if tau(x1',4)=tau(x2',16), then x1' \= x2'

Well done, shithead.
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif
See any x1' or x2' in that equation, fuckwit?
How about x1'/(c-v) and x2'/(c+v), moron?
Your blind faith in a huckster is pathetically stupid and ridiculous,
which is why I'm ridiculing you, stupid fuck.
Relativity has more holes in it than the cheese Einstein ate in Switzerland.
No contradiction.

| >
| > | > "But the ray moves relatively to the initial point of k, when
measured
| > in
| > | > the stationary system, with the velocity c-v, so that
| > | >
| > | >
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img31.gif
| > |
| > | The 'plus' sign on the right hand side does not belong.
| >
| > Ok, what the fuck is it doing in
| > http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif
| The plus sign is not on the right hand side. It is on the left.

Yes. That's why 1/2 of 20 is 16. The other half is 4.

| The left hand side represents the out-bound and return trip.
| The right hand side represents only the out-bound.

That's right. The right hand side represents only the out-bound.
We ignore the inbound, it won't produce the cuckoo malformations
and make a huckster famous. You are a so FUCKIN' stooopid.

|
| That is the reason for the 1/2 on the left. If the time is the
| same in both directions, then the outbound time is half of the
| full trip.
| (Before you object that 16\=4, note that the times being equated
| are the tau times, not the t times.

Yes, tau(16) = 8 and tau(4) =8.

| Nobody is claiming that the
| two t times are equal.

"we establish by definition that the ``time'' required by light to travel
from A to B equals the ``time'' it requires to travel from B to A. " --
Albert Nobody.

You are right. Your tin god is a nobody.

Tau is a linear function of t *and* x',
| so there is no contradiction, if the x' values are different.)

tau(4) = 8, tau(16) = 8

|
|
|
| >

| > See the thread title? "SR fundamental contradiction".
| > That wasn't me, that was someone else who realized something
| > was wrong, and a xxxxxxxxxxx like you with your head up your
| > xxxx can't see what it is.

What's up? Don't like plain English?

| Then enlighten me.

Sure:
http://www.androcles01.pwp.blueyonder.co.uk/

| Your arguments will carry more weight if you can refrain from
| offensive language. The strength of your language is inversely
| proportional to the strength of your argument.

I gave my arguments on my page without profanity. You objected, so
fuck you, I say it the way it is, cunt. The objective is to embarrass
the shit out of you, prude. I'm being deliberately offensive to an
arrogant fuckwit who doesn't know any mathematics so you can
whine about me farting in your church. What's good enough
for Monty Python is good enough for me. I don't rely on pretty
language for my arguments to carry weight, my arguments are
straightforward logical and mathematical, yours are based on your
faith in the Holey Church of Relativity and its Pope, Einstein the
Righteous Philanderer. (Not that I care about his philandering,
or Bill Clinton's for that matter.) This is about PHYSICS, you lunatic.

| >
| > | The
| > | time in the right hand side includes only the out bound ray. It
| > | represents the time of the reflection.
| >
| > The time in the right hand side includes only half the total time.
| > It represents the time of the reflection.
| >
| >
| > |
| > | Einstein's equation simply says that, in the system moving with
| > | the train, the reflection occurs at the mid point of the round
| > | trip.
| >
| > 1/2 of 100 = 80, the other half is 20.
|
| The length of the train is the same in both directions.

The speed of light is different in both directions, c-v and c+v.

| From
| this obvious fact, we conclude that, in the train system, the
| time for the light signals in both directions are the same.

Then you have a contradiction, shithead.
Either the train stretches and shrinks or the speed of light
isn't c. Pope Einstein says it shrinks, so does Archbishop
Lorentz. Which is it?

Say three Hail Aethers for your stupidity:

Hail Aether,
Full of Light,
Einstein is with thee.
Blessed art thou among absolute frames of reference,
and blessed is the fruit of thy tomb, Lorentz Transform.
Holey Aether,
Daughter of Lunacy,
prey on us morons now
and at the dilated hour of death.

| It also contains the time of the reflection (in t time). The
| value of the function is the tau time of the reflection.

I shine the light from the locomotive to the mirror at the rear, just
to be awkward and create a COUNTER EXAMPLE.
That makes the RHS tau(x', 0,0,0,x'/(c+v)).
Tough titty, cretin. Live with it.
Go ahead, derive a new set of cuckoo malformations.
Begin:
If x' be taken infinitesimally small,
1/2 [ 1/(c-v) + 1/(c+v) ] @tau/@t = ??

C'mon cretin, strut your stuff..

|
| >
| > | It represents that in terms of the times and locations in
| > | the other system.
| > |
| > | What is your objection?
| >
| > I have no objection,
| Then you assent to the correctness of Einstein's equations?

"The length of the train is the same in both directions" is WRONG.
As long as trains move my peristalis I will assent.

| > | > | > | Yours is an assertion easily disproved by example.
| > | > | >
| > | > | > I didn't make any assertion, Einstein did.
| > | > | No, you assert that Einstein's equation cannot be satisfied,
| > | >
| > | > I PROVE, I do NOT assert.
| > | Your proof must be flawed, because, as I showed you, a
| > | counter-example exists.
| > |
| > | State your proof.
| >
| > I did. Trains move by peristalsis. A counter example exists,
| > some of them have wheels.
| > Wheels are not inertial frames of reference so maybe they don't count.
| > SR is not flawed, it works fine for earthworms.
| > 1/2 (20) = 4, 1/2(20) = 16 and 16+4 = 20. Right, worm?
| State your proof, not your erroneous equations.

Your turn. Shine the light from the locomotive and derive the
cuckoo malformations, you fucking imbecile.

|
| > | > | that, consequently, it is nonsense. I disproved your assertion,
| > | > | using your numbers.
| > | >
| > | > No you didn't, and you got the wrong answer anyway.
| > | Where is the flaw in my demonstration?
| >
| > Nothing, it is fine.
| Thank you.
|
| >
| > How about this one instead:
| > http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm
| >
| > No worm in that one, sick sock puppet.
| > How fine am I doing?
|
| As a criticism of Einstein, it is meaningless, because you have
| light travelling at two different speeds in the same reference
| system.

Yes, t = x'/(c-v) with the light going forward and t = x'/(c+v)
with the light going backward. Einstein's old pocket watch had
two second hands.

| You also seem to be claiming, erroneously, that because the
| clocks are synchronized in t time, they are synchronized in
| train time. That is a proposition that you may want to prove,
| but you cannot assume.

Prove it then. Your claim, your burden of proof.

Androcles

Brian Kennelly

unread,

Oct 8, 2006, 9:15:29 PM10/8/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message
> news:agaWg.3723$gM1.1723@fed1read12...
> | Sorcerer wrote:
> | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > news:FZ0Wg.3714$gM1.2909@fed1read12...
> | > | Sorcerer wrote:
> | > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > | > news:Xg_Vg.3694$gM1.2041@fed1read12...
> | > | > | Sorcerer wrote:
> | > | > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > | > | > news:6zUVg.3669$gM1.3580@fed1read12...

> | > Yes, 16 is half of 20 and 4 is the other half
> | Nobody but you is claiming that. 16 and 4 are the divisions of
> | the signal in the t times, but they are neither is 'half'. Only
> | the tau times are equated between the out and back portions of
> | the signal, so that they each represent half of the trip.
>
>
> Proof?

That is the meaning of Einstein's equation.
The outbound tau time is half of the round trip tau time.

> |
> | tau(-32,4)=tau(32,16)= 8
>
> | No contradiction.
>
> Proof?

tau(x',t) = 4t/5-3x'/20
tau(32,16)= 4*16/5-3*32/20
= 64/5-96/20
= 64/5-24/5
= 40/5
= 8
tau(-32,4)= 4*4/5+3*32/20
= 16/5+96/20
= 16/5+24/5
= 40/5
= 8

Simple algebra.

>
>
> | The correct form of your argument is:
> | if tau(x1',4)=tau(x2',16), then x1' \= x2'
>

> Well done, xxxxxxxx.
> http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif
> See any x1' or x2' in that equation, xxxxxxxx?
I am not quoting that equation, but the equations that
correspond to your tau(4) and tau(16) when the space dependence
is included.

> How about x1'/(c-v) and x2'/(c+v), moron?

Irrelevant.

> Your blind faith in a huckster is pathetically stupid and ridiculous,
> which is why I'm ridiculing you,

I am afraid you the only on in this discussion with blind faith.
It shows in your unwillingness to deal with the actual arguments.

I don't feel ridiculed at all. You have attempted to insult me,
but you have done nothing to make /me/ look ridiculous.

>
>
> | The left hand side represents the out-bound and return trip.
> | The right hand side represents only the out-bound.
>
> That's right. The right hand side represents only the out-bound.
> We ignore the inbound, it won't produce the cuckoo malformations
> and make a huckster famous.

We don't need in the return ray on the right, because it
represents the time of reflection, which occurs before the
return trip.

>
> |
> | That is the reason for the 1/2 on the left. If the time is the
> | same in both directions, then the outbound time is half of the
> | full trip.
> | (Before you object that 16\=4, note that the times being equated
> | are the tau times, not the t times.
>
>
> Yes, tau(16) = 8 and tau(4) =8.
>
> | Nobody is claiming that the
> | two t times are equal.
>
> "we establish by definition that the ``time'' required by light to travel
> from A to B equals the ``time'' it requires to travel from B to A. " --
> Albert Nobody.

In this case, the definition applies to the tau times, which
express time in the moving system.

To apply them to the t times requires a different experimental
set up.

>
> | Your arguments will carry more weight if you can refrain from
> | offensive language. The strength of your language is inversely
> | proportional to the strength of your argument.
>

You demonstrate my point. You have no logical argument so you
resort to expletives.

>
>
>
> | >
> | > | The
> | > | time in the right hand side includes only the out bound ray. It
> | > | represents the time of the reflection.
> | >
> | > The time in the right hand side includes only half the total time.
> | > It represents the time of the reflection.
> | >
> | >
> | > |
> | > | Einstein's equation simply says that, in the system moving with
> | > | the train, the reflection occurs at the mid point of the round
> | > | trip.
> | >
> | > 1/2 of 100 = 80, the other half is 20.
> |
> | The length of the train is the same in both directions.
>
> The speed of light is different in both directions, c-v and c+v.

All right, it appears that your are using Einstein's
intermediate coordinates x'[=x-vt], t. Fair enough.

Your page then demonstrates that Einstein was right. Clocks
synchronized in the rest system are not synchronized in the
train system, if light speed is invariant.

The green ray is emitted at the same time as the red ray /in the
rest system/, and reaches the rear of the train before the red
ray reaches the front. Because, by postulate, the signal times
are the same in the train system, the green ray was emitted
first in the train system.

>
>
> | From
> | this obvious fact, we conclude that, in the train system, the
> | time for the light signals in both directions are the same.
>

> Then you have a contradiction, xxxxxxxx.
What is the contradiction? In the train system, the speed of
light is the same in both directions, so the signal times are

the same in both directions.

> | > The distance in the right hand side includes only half the total

> distance.
> | > It represents the distance of the reflection.
> | >
> | > Nothing wrong with that, is there?
> | It is correct that the right hand side contains the distance of
> | the reflection (in Einstein's x'[=x-vt]).
>
> | It also contains the time of the reflection (in t time). The
> | value of the function is the tau time of the reflection.
>
> I shine the light from the locomotive to the mirror at the rear, just
> to be awkward and create a COUNTER EXAMPLE.
> That makes the RHS tau(x', 0,0,0,x'/(c+v)).

That changes nothing, as it is equivalent to changing v to -v.
(The outbound ray is moving opposite to the direction of the train.)

The tau equation, using your numbers, becomes
tau(x',t)= 4t/5+3x'/20

Again, everything works.

> | >
> | > I have no objection,
> | Then you assent to the correctness of Einstein's equations?
>
>
> "The length of the train is the same in both directions" is WRONG.

Are you saying that distance depends on the direction of
measurement? Is the front of the train farther from the back
than the back is from the front?

No, the length of the train is the same in both directions.

>
> | You also seem to be claiming, erroneously, that because the
> | clocks are synchronized in t time, they are synchronized in
> | train time. That is a proposition that you may want to prove,
> | but you cannot assume.
>
> Prove it then. Your claim, your burden of proof.

No, it is your claim:
> "Now because all clocks are set to the same time, t, we can safely
> assume, in agreement with experience, they are synchronized, "

You seem to be applying that assumption to the train system
without proof. We have no reason to assume that.
Without that assumption, your green and red rays are not emitted
at the same time in the train system.

mlut...@wanadoo.fr

unread,

Oct 8, 2006, 9:17:08 PM10/8/06

to

The origin of the *dilation* lies in the LT x' = g(x-vt).
When, for instance, x = ct, x' = g(c-v)t.
For a given value of t, (c-v)t is of course the distance, as seen by S,
between
the point reached by a light signal and the origin of S'.
In the S'-frame, such distance becomes g(c-v)t, meaning that it is
dilated.

Marcel Luttgens

mlut...@wanadoo.fr

unread,

Oct 8, 2006, 9:21:41 PM10/8/06

to

You are monitoring all the visitors to your website. I don't think that

it is ethically allowed.

Marcel Luttgens

>
> Dirk Vdm

Brian Kennelly

unread,

Oct 8, 2006, 9:27:13 PM10/8/06

to

Of course the distance between any point and an out-going light
signal is increasing with time. That increase leads to the
dilation.

> In the S'-frame, such distance becomes g(c-v)t, meaning that it is
> dilated.

If you wish to calculate the S' length of a stick at rest in S,
with endpoints at vT and cT [so that L=(c-v)T], we can use the
LT as follows:

vT=g(x1'+vt')
cT=g(x2'+vt')

Subtracting we get:
(c-v)T=g(x2'-x1')

The length is:
x2'-x1'=(c-v)T/g
= L/g
It is contracted.

Dirk Van de moortel

unread,

Oct 8, 2006, 9:58:42 PM10/8/06

to

<mlut...@wanadoo.fr> wrote in message news:1160342501....@i42g2000cwa.googlegroups.com...

:-))
"A Crackpot's Sence of Ethics":
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/CrackpotEthics.html
Way to go, Marcel.

Dirk Vdm

Dirk Van de moortel

unread,

Oct 8, 2006, 10:14:11 PM10/8/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:R2eWg.3747$gM1.2434@fed1read12...

Remember, there is *no* way to make him understand the meanings
of the variables. We have been trying during a decade. It does not
work.
It also does not work with Sorcerer (John Parker, aka Androcles).
Actually, it is clear that they wouldn't even *want* it to work.
They would instantly lose the attention they are getting now.
Enjoy :-)

Dirk Vdm

Sorcerer

unread,

Oct 8, 2006, 11:41:55 PM10/8/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:RTdWg.3744$gM1.1846@fed1read12...

| Sorcerer wrote:
| > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > news:agaWg.3723$gM1.1723@fed1read12...
| > | Sorcerer wrote:
| > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > | > news:FZ0Wg.3714$gM1.2909@fed1read12...
| > | > | Sorcerer wrote:
| > | > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > | > | > news:Xg_Vg.3694$gM1.2041@fed1read12...
| > | > | > | Sorcerer wrote:
| > | > | > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > | > | > | > news:6zUVg.3669$gM1.3580@fed1read12...
| > | > Yes, 16 is half of 20 and 4 is the other half
| > | Nobody but you is claiming that. 16 and 4 are the divisions of
| > | the signal in the t times, but they are neither is 'half'. Only
| > | the tau times are equated between the out and back portions of
| > | the signal, so that they each represent half of the trip.
| >
| >
| > Proof?
| That is the meaning of Einstein's equation.
| The outbound tau time is half of the round trip tau time.

ROFL! I said "PROOF?"
All you gave me was assertion and I'm not a member
of the Holey Church of Relativity. I fart in it.

|
|
| > |
| > | tau(-32,4)=tau(32,16)= 8
| >
| > | No contradiction.
| >
| > Proof?
| tau(x',t) = 4t/5-3x'/20
| tau(32,16)= 4*16/5-3*32/20
| = 64/5-96/20
| = 64/5-24/5
| = 40/5
| = 8
| tau(-32,4)= 4*4/5+3*32/20
| = 16/5+96/20
| = 16/5+24/5
| = 40/5
| = 8

x' = -x', and that isn't a contradiction in your book?

|
| Simple algebra.

Very simple-minded, I agree. What happened to
the doppler shift?

|
| >
| >
| > | The correct form of your argument is:
| > | if tau(x1',4)=tau(x2',16), then x1' \= x2'
| >
| > Well done, xxxxxxxx.
| > http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img22.gif
| > See any x1' or x2' in that equation, xxxxxxxx?

| I am not quoting that equation, xxxxxxxxx,

I am.

| > How about x1'/(c-v) and x2'/(c+v), moron?
| Irrelevant.

|
| > Your blind faith in a huckster is pathetically stupid and ridiculous,
| > which is why I'm ridiculing you,
| I am afraid you the only on in this discussion with blind faith.
| It shows in your unwillingness to deal with the actual arguments.
|
| I don't feel ridiculed at all. You have attempted to insult me,
| but you have done nothing to make /me/ look ridiculous.

Irrelevant.

|
| >
| >
| > | The left hand side represents the out-bound and return trip.
| > | The right hand side represents only the out-bound.
| >
| > That's right. The right hand side represents only the out-bound.
| > We ignore the inbound, it won't produce the cuckoo malformations
| > and make a huckster famous.

| We don't need in the return ray on the right, xxxxxxx.

Irrelevant.

| >
| > |
| > | That is the reason for the 1/2 on the left. If the time is the
| > | same in both directions, then the outbound time is half of the
| > | full trip.
| > | (Before you object that 16\=4, note that the times being equated
| > | are the tau times, not the t times.
| >
| >
| > Yes, tau(16) = 8 and tau(4) =8.
| >
| > | Nobody is claiming that the
| > | two t times are equal.
| >
| > "we establish by definition that the ``time'' required by light to
travel
| > from A to B equals the ``time'' it requires to travel from B to A. " --
| > Albert Nobody.

| In this case, the definition applies xxxxxxx

It applies in all (inertial) frames of reference, by definition.

|
| To apply them to the t times requires a different experimental
| set up.

It applies in all frames of reference, by definition, shithead.

|
| >
| > | Your arguments will carry more weight if you can refrain from
| > | offensive language. The strength of your language is inversely
| > | proportional to the strength of your argument.
| >
| <rant deleted>
|

You lost the debate when you did that, you stooopid fuck!
Bye.
Androcles.

Brian Kennelly

unread,

Oct 9, 2006, 12:42:12 AM10/9/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message
> news:RTdWg.3744$gM1.1846@fed1read12...
> | Sorcerer wrote:
> | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > news:agaWg.3723$gM1.1723@fed1read12...
> | > | Sorcerer wrote:
> | >
> | >
> | > Proof?
> | That is the meaning of Einstein's equation.
> | The outbound tau time is half of the round trip tau time.
>
> ROFL! I said "PROOF?"
> All you gave me was assertion and I'm not a member
> of the Holey Church of Relativity. I fart in it.

What is to prove? The equation states that the one way time is
half the round trip, in the train system.

Once you see that, we can go on to prove that it is consistent
with the fact that the times in the other system are not equal.

> | > | tau(-32,4)=tau(32,16)= 8
> | >
> | > | No contradiction.
> | >
> | > Proof?
> | tau(x',t) = 4t/5-3x'/20
> | tau(32,16)= 4*16/5-3*32/20
> | = 64/5-96/20
> | = 64/5-24/5
> | = 40/5
> | = 8
> | tau(-32,4)= 4*4/5+3*32/20
> | = 16/5+96/20
> | = 16/5+24/5
> | = 40/5
> | = 8
>
> x' = -x', and that isn't a contradiction in your book?

I never claimed that x' = -x'. You offer two different t time
intervals, and I supplied the space components that you omitted.
The signals travel in opposite directions, so we expect the
intervals to be opposite (in these coordinates).

The out bound signal takes 16 units in the t coordinate, and
travels 32 units in the x' coordinate (it travels to the right).

The return signal takes 4 units in the t coordinate, and travels
-32 units in the x' coordinate (it travels to the left).

Both signals take 8 units in the tau coordinate.

No contradiction.

> | > Yes, tau(16) = 8 and tau(4) =8.
> | >
> | > | Nobody is claiming that the
> | > | two t times are equal.
> | >
> | > "we establish by definition that the ``time'' required by light to
> travel
> | > from A to B equals the ``time'' it requires to travel from B to A. " --
> | > Albert Nobody.
> | In this case, the definition applies xxxxxxx
>
> It applies in all (inertial) frames of reference, by definition.

Yes, but in the described set up, the signal does not return to
the starting point in the rest system, so we cannot invoke this
definition to equate the t times.

In the train system, the signal does return to the starting
point, so we can use the definition to equate the tau times.

>
>
> |
> | To apply them to the t times requires a different experimental
> | set up.
>
> It applies in all frames of reference, by definition,

Yes, but it does not apply to all experimental set ups; it only
applies when the signal returns to a fixed point. This
requirement is met in the train system, but not in the "rest"
system.

Sorcerer

unread,

Oct 9, 2006, 5:50:04 AM10/9/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:EVgWg.3754$gM1.118@fed1read12...

Fuck off, bigot, you lost. Game over.

http://www.androcles01.pwp.blueyonder.co.uk/GPS.htm

How am I doing?

Androcles

mlut...@wanadoo.fr

unread,

Oct 9, 2006, 10:30:01 AM10/9/06

to

Your site is a bunch of malicious crap, that illustrates your poor
sense of ethic. No wonder that you are now sending tracking cookies.

Marcel Luttgens

Brian Kennelly

unread,

Oct 9, 2006, 3:24:20 PM10/9/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message
> news:EVgWg.3754$gM1.118@fed1read12...
>
> <delete rant>
>
> Fuck off, bigot, you lost. Game over.
>

So, when you cannot answer an argument, you declare yourself the
winner and walk away?

I have tried to give you the benefit of the doubt, by assuming
that you simply did not understand Einstein's argument, rather
than that you refused to try. Perhaps I was wrong.

Sorcerer

unread,

Oct 9, 2006, 3:54:57 PM10/9/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:DQtWg.3776$gM1.1763@fed1read12...

That's exactly what you did!
You deleted because you had no answer to my argument, you lost.
You've even deleted the start of a new game.
Score :
Androcles --- 2 Fuckwit Kennelly --- 0.

Do you play chess by throwing the pieces in the air?

| x xxxx .... xxxx
| Perhaps I was wrong.

Yes, you were wrong. If you want a new game, start one.
I can kick your arse with logic any time I choose, you are out
of your league, bigotted punk.
Androcles.

Brian Kennelly

unread,

Oct 9, 2006, 4:28:04 PM10/9/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message
> news:DQtWg.3776$gM1.1763@fed1read12... | Sorcerer wrote: | >
> "Brian Kennelly" <bwken...@cox.net> wrote in message | >
> news:EVgWg.3754$gM1.118@fed1read12... | > | > <delete rant> |
> > | > Fuck off, bigot, you lost. Game over. | > | So, when
> you cannot answer an argument, you declare yourself the |
> winner and walk away?
>
> That's exactly what you did! You deleted because you had no
> answer to my argument, you lost.

No, I deleted the rant about your right to use blue language,
not because I could not answer it, but because it was off-topic
for this forum. (I acknowledge that it was a response to my
request to use civil language, but I chose not to argue about
your right to use offensive language. I simply accepted that
you chose not to comply with the request.)

On the other hand, this forum is set up for discussions of
relativity.

> You've even deleted the start of a new game.

That was unrelated to our discussion. If you want to discuss
it, start a new thread. If anyone is interested they will respond.

Restoring context:

> | I have tried to give you the benefit of the doubt, by
> | assuming that you simply did not understand Einstein's
> | argument, rather than that you refused to try.

> | Perhaps I was wrong.
>
> Yes, you were wrong.

If I was wrong about your attitude, then let us continue to work
through Einstein's argument, so we can find where you (or
Einstein) got it wrong.

I believe that your confusion results from ignoring the
possibility that tau (the train time) depends on both the time
and space coordinates in the other system. Are you assuming
that clocks synchronized in one inertial system are synchronized
in all other inertial systems?

Or are you simply confused by Einstein's x',t coordinates? They
do /not/ constitute an inertial system satisfying the POR, which
could lead to confusion, if you treat them as inertial.
Einstein only introduced those coordinates as an intermediate
step to obtaining the correct coordinates.

mlut...@wanadoo.fr

unread,

Oct 9, 2006, 4:34:46 PM10/9/06

to

Brian Kennelly wrote:
> mlut...@wanadoo.fr wrote:
> > Brian Kennelly wrote:
> >> Dirk Van de moortel wrote:
> >>> But t was fixed in the beginning of the thread:
> >>> | Consider the event E on the light signal with x = c t for some
> >>> | chosen value of t.
> >>> Remember, this is Marcel Luttgens you are dealing with.
> >> I was attempting to demonstrate the source of the error by
> >> taking his equations at face value. If we do so, the source of
> >> the dilation is found in the variable length, not in the LT
> >> equations. The endpoints in S', determined at the same time t',
> >> are at two different S times, t1 and t2. By his definition of
> >> the length, the stick is longer when S' measures it, because it
> >> is longer when S measures it. [If t2>t1, then (c-v)t2 > (c-v(t1)].
> >
> > The origin of the *dilation* lies in the LT x' = g(x-vt).
> > When, for instance, x = ct, x' = g(c-v)t.
> > For a given value of t, (c-v)t is of course the distance, as seen by S,
> > between
> > the point reached by a light signal and the origin of S'.

> Of course the distance between any point and an out-going light
> signal is increasing with time. That increase leads to the
> dilation.

I was not referring to that obvious expansion, but to the fact that
the ratio between a distance measured in S' and the corresponding
distance measured in S is always g, according to the LT (see below).

>
> > In the S'-frame, such distance becomes g(c-v)t, meaning that it is
> > dilated.
>
> If you wish to calculate the S' length of a stick at rest in S,
> with endpoints at vT and cT [so that L=(c-v)T], we can use the
> LT as follows:
>
> vT=g(x1'+vt')
> cT=g(x2'+vt')
>
> Subtracting we get:
> (c-v)T=g(x2'-x1')
>
> The length is:
> x2'-x1'=(c-v)T/g
> = L/g
> It is contracted.

Here is an exemple given by the tracking guru, which
indirectly proves the dilation:

"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.
What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame."

The length (c - v)t represents the distance in S between the
endpoints ct and vt (for a given value of t).

The particular length x' = g (c-v)t corresponds of course
to the dilated length obtained by applying the LT x' = g(x - vt)
to that case where x = ct in S.

Marcel Luttgens

Sorcerer

unread,

Oct 9, 2006, 4:40:24 PM10/9/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:mMuWg.3779$gM1.3014@fed1read12...

You are a fucking liar, too.

Androcles -- 3, Fuckwit Kennelly -- 0

Androcles

Dirk Van de moortel

unread,

Oct 9, 2006, 4:40:50 PM10/9/06

to

<mlut...@wanadoo.fr> wrote in message news:1160389801.2...@e3g2000cwe.googlegroups.com...

I don't send cookies. Statcounter does.
Block them if you like. I'm not at all interested in Unique Visitors.
I'm interested in liars like Andrcles and Golden Boar ;-)

Dirk Vdm

Dirk Van de moortel

unread,

Oct 9, 2006, 4:44:13 PM10/9/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:DQtWg.3776$gM1.1763@fed1read12...

Don't say I didn't warn you :-)
Nice guy, isn't he?

Dirk Vdm

Golden Boar

unread,

Oct 9, 2006, 5:06:25 PM10/9/06

to

I never lied Dirk, but you did, and was caught in the act.

Brian Kennelly

unread,

Oct 9, 2006, 5:26:46 PM10/9/06

to

Note this ^^^^^^^^^^^^^^^^. You switched the rest frame.

> What is the length of such a stick in the S-frame?
> If you apply length contraction, you find that this length
> would be
> x' / g = (c - v) t
> in the S frame."

So, when the stick is at rest in the S' frame, it is contracted
in the S frame. When the stick is at rest in the S frame, it is
contracted in the S' frame, as I showed above.

>
> The length (c - v)t represents the distance in S between the
> endpoints ct and vt (for a given value of t).
>
> The particular length x' = g (c-v)t corresponds of course
> to the dilated length obtained by applying the LT x' = g(x - vt)
> to that case where x = ct in S.

The case <x=ct> is a light signal, and increases with time. If
you do not want it expanding, choose a fixed time T, or simply
call the length L.

Sorcerer

unread,

Oct 9, 2006, 7:30:20 PM10/9/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:nDvWg.3780$gM1.71@fed1read12...
<delete rant>
Marcel has you beaten, fuckwit.

mlut...@wanadoo.fr

unread,

Oct 10, 2006, 12:47:52 PM10/10/06

to

If you consider a length g (c-v)T at rest in the S'-frame, you
get a length (c-v)T in the S-frame, by applying length contraction
(this is what the tracking guru did).

If you consider a stick of length (c-v)T at rest in the S-frame,
you get a length g(c-v)T in the S'-frame according to the LT.
It is dilated in the S'-frame. Indeed, if the ends of the stick
are x1 = vT and x2 = cT in the S-frame, x1' = 0 and x2' = g(c-v)T,
hence x2' - x1' = g(c-v)T. Notice that x1' is always zero, its
value is independent of time.

This is obvious: if L = L'/g (as in the guru exemple), L' is
necessarily gL. The one who got L = L'/g AND L' = L/g made a logical
mistake somewhere.

Marcel Luttgens

Sorcerer

unread,

Oct 10, 2006, 12:57:41 PM10/10/06

to

<mlut...@wanadoo.fr> wrote in message
news:1160484472.5...@e3g2000cwe.googlegroups.com...

Brian Kennelly wrote:
[anip]

You are wasting your time with that moron. He has the usual blind faith
in the Holey Church of Relativity that all relativists have, logic has no
effect on him.

Androcles

Brian Kennelly

unread,

Oct 10, 2006, 2:35:37 PM10/10/06

to

mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:
> If you consider a length g (c-v)T at rest in the S'-frame, you
> get a length (c-v)T in the S-frame, by applying length contraction
> (this is what the tracking guru did).
>
> If you consider a stick of length (c-v)T at rest in the S-frame,
> you get a length g(c-v)T in the S'-frame according to the LT.
> It is dilated in the S'-frame. Indeed, if the ends of the stick
> are x1 = vT and x2 = cT in the S-frame,

Okay, then from the LT, we have:
vT=g(x1'+vt1')
cT=g(x2'+vt2')
This allows us to find x2' and x1' at the same t'

> x1' = 0 and x2' = g(c-v)T,

There is no value of t' that makes both of these equations true.
When x1'=0, then (from the first equation):
vT=g(vt1')
So t1' = T/g

When x2'=g(c-v)T, then
cT=g(g(c-v)T+vt2')
So t2'=cT/vg - g(c-v)T/v
\=t1'

To find the length in S', we must measure the distance between
the endpoints at the same time in S' (the same t')

Setting t1'=t2' in the LT equations above and subtracting we get
cT-vT=g(x2'-x1')
x2'-x1'=(c-v)T/g

The length is contracted.

> hence x2' - x1' = g(c-v)T. Notice that x1' is always zero, its
> value is independent of time.

The only way that x1' can be always zero is if it is at rest in
S'. Because you stated that the stick is at rest in S, it, and
therefore its endpoints, will be moving in S':
vT=g(x1'+vt')
x1'=vT/g-vt'

In particular, when x2'=g(c-v)T, then t'=cT/vg - g(c-v)T/v and
x1' is:
x1' = vT/g-v(cT/vg - g(c-v)T/v)
= vT/g- cT/g +g(c-v)T
=g(c-v)T - (c-v)T/g

Subtracting this from x2' again gives the result derived above:
x2'-x1'=(c-v)T/g

>
> This is obvious: if L = L'/g (as in the guru exemple), L' is
> necessarily gL. The one who got L = L'/g AND L' = L/g made a logical
> mistake somewhere.

No, L=L'/g and L'=L/g apply to two different scenarios. In the
first, the stick is at rest in S', in the second, it is at rest
in S.

Sorcerer

unread,

Oct 10, 2006, 2:40:21 PM10/10/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:ZcOWg.3868$gM1.2629@fed1read12...

You are a stupid liar, Kennelly.

Androcles

Brian Kennelly

unread,

Oct 10, 2006, 2:48:47 PM10/10/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message
> news:ZcOWg.3868$gM1.2629@fed1read12...
>
> <delete rant>
>
> You are a stupid liar, Kennelly.
>
> Androcles
>
>
Apparently, I got under your skin.

Sorcerer

unread,

Oct 10, 2006, 3:30:39 PM10/10/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:jpOWg.3869$gM1.1817@fed1read12...

HAHAHA!
I got under yours and it was really easy, you were unable to answer me,
kook!
You reacted emotionally as all you fuckwits ever do. Where's your buddy
Phuckwit Duck these days, sock puppet?

http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm

How am I doing?

Hahaha!

Androcles

Brian Kennelly

unread,

Oct 10, 2006, 3:50:37 PM10/10/06

to

Sorcerer wrote:
>
> http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm
>
>
> How am I doing?

Poorly.

E.g., you state:
> the time for light to go from the camel's right ear (B) to a
> point in space off further to our left (A) is equal to the
> time for light to go from A to B in the camel's "stationary"
> (and inertial) frame of reference.

This is incorrect. The time for light to go from A to B is the
same as the time to go from B to A /in the train's frame/, not
in the camel's frame.

Einstein explicitly uses that fact in the derivation. If you
wish to understand his derivation, you must not introduce
contradictory assumptions that are not present in the original.

Sorcerer

unread,

Oct 10, 2006, 3:56:12 PM10/10/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:hjPWg.3876$gM1.2251@fed1read12...

| Sorcerer wrote:
| >
| > http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm
| >
| >
| > How am I doing?
| Poorly.

<delete rant>
Obviously you don't understand the first thing about relativity or debate.
Androcles -- 4, Fuckwit Kennelly -- 0

Dirk Van de moortel

unread,

Oct 10, 2006, 4:15:03 PM10/10/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:hjPWg.3876$gM1.2251@fed1read12...

He does not wish to understand his derivation.
He has invested a quarter of a century in miserably failing to
understand it, and another quarter in trying to rationalize his
miserable failure.

Loadofcrapalizing Special Relativity:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/LoadCrap.html

Dirk Vdm

Brian Kennelly

unread,

Oct 10, 2006, 4:34:10 PM10/10/06

to

You asked for criticism. Are you unwilling to defend your page
by answering the argument?

Brian Kennelly

unread,

Oct 10, 2006, 4:37:22 PM10/10/06

to

Dirk Van de moortel wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message news:hjPWg.3876$gM1.2251@fed1read12...
>> Sorcerer wrote:
>>> http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm
>>>
>>>
>>> How am I doing?
>> Poorly.
>>
>> E.g., you state:
>>> the time for light to go from the camel's right ear (B) to a
>>> point in space off further to our left (A) is equal to the
>>> time for light to go from A to B in the camel's "stationary"
>>> (and inertial) frame of reference.
>> This is incorrect. The time for light to go from A to B is the same as the time to go from B to A /in the train's frame/, not in
>> the camel's frame.
>>
>> Einstein explicitly uses that fact in the derivation. If you wish to understand his derivation, you must not introduce
>> contradictory assumptions that are not present in the original.
>
> He does not wish to understand his derivation.
> He has invested a quarter of a century in miserably failing to
> understand it, and another quarter in trying to rationalize his
> miserable failure.

I am an optimist. <g>

Sorcerer

unread,

Oct 10, 2006, 4:43:53 PM10/10/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:7YPWg.3878$gM1.682@fed1read12...

Yes. You poorly assumed the train was moving. Any fool can
see the camel moves and the train stays where it is. that's called
the "Principle of Relativity".
You are not just any fool, you are totally anencephalous and cannot
understand an equation or counter example where the light originates
at the locomotive. In short, you are a fucking moron who accuses
others of being "kooks".
Androcles

Dirk Van de moortel

unread,

Oct 10, 2006, 4:48:17 PM10/10/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:7%PWg.3879$gM1.3265@fed1read12...

I used to be one as well. Many of us used to be optimists.
It wears off with time. Those who become realist, stay and
have fun. Those who become pessimist, go away.
I hope you'll stay :-)

Dirk Vdm

Brian Kennelly

unread,

Oct 10, 2006, 5:19:04 PM10/10/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message
> news:7YPWg.3878$gM1.682@fed1read12...
> | Sorcerer wrote:
> | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > news:hjPWg.3876$gM1.2251@fed1read12...
> | > | Sorcerer wrote:
> | > | >
> | > | > http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm
> | > | >
> | > | >
> | > | > How am I doing?
> | > | Poorly.
> | > <delete rant>
> | > Obviously you don't understand the first thing about relativity or
> debate.
> |
> | You asked for criticism. Are you unwilling to defend your page
> | by answering the argument?
>
> Yes. You poorly assumed the train was moving. Any fool can
> see the camel moves and the train stays where it is. that's called
> the "Principle of Relativity".

In that case, you err by assigning different velocities to the
light rays. One of Einstein's postulates was the the velocity
of light is the same in any inertial system. In your
discussion, the velocity is not the same in two directions in
even the same system.

In Einstein's derivation, the different relative velocities
(c+v, c-v) appear because the train is moving in the reference
system that defines the time coordinate t.

> You are not just any fool, you are totally anencephalous and cannot
> understand an equation or counter example where the light originates
> at the locomotive.

Certainly, I can. It is actually quite easy, and introduces
nothing new.

> In short, you are a fucking moron who accuses
> others of being "kooks".

I don't recall that I have ever done so.

Sorcerer

unread,

Oct 10, 2006, 5:44:17 PM10/10/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:eCQWg.3881$gM1.3302@fed1read12...

| Sorcerer wrote:
| > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > news:7YPWg.3878$gM1.682@fed1read12...
| > | Sorcerer wrote:
| > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > | > news:hjPWg.3876$gM1.2251@fed1read12...
| > | > | Sorcerer wrote:
| > | > | >
| > | > | > http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm
| > | > | >
| > | > | >
| > | > | > How am I doing?
| > | > | Poorly.
| > | > <delete rant>
| > | > Obviously you don't understand the first thing about relativity or
| > debate.
| > |
| > | You asked for criticism. Are you unwilling to defend your page
| > | by answering the argument?
| >
| > Yes. You poorly assumed the train was moving. Any fool can
| > see the camel moves and the train stays where it is. that's called
| > the "Principle of Relativity".
| In that case, you err by assigning different velocities to the
| light rays.

I'd do it for a man walking the length of the train. What makes
light work by magic, shithead?
<delete rant>
Androcles

Brian Kennelly

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Oct 10, 2006, 6:08:42 PM10/10/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message
> news:eCQWg.3881$gM1.3302@fed1read12...
> | Sorcerer wrote:
> | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > news:7YPWg.3878$gM1.682@fed1read12...
> | > | Sorcerer wrote:
> | > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > | > news:hjPWg.3876$gM1.2251@fed1read12...
> | > | > | Sorcerer wrote:
> | > | > | >
> | > | > | > http://www.androcles01.pwp.blueyonder.co.uk/Rocket/Rocket.htm
> | > | > | >
> | > | > | >
> | > | > | > How am I doing?
> | > | > | Poorly.
> | > | > <delete rant>
> | > | > Obviously you don't understand the first thing about relativity or
> | > debate.
> | > |
> | > | You asked for criticism. Are you unwilling to defend your page
> | > | by answering the argument?
> | >
> | > Yes. You poorly assumed the train was moving. Any fool can
> | > see the camel moves and the train stays where it is. that's called
> | > the "Principle of Relativity".
> | In that case, you err by assigning different velocities to the
> | light rays.
>
> I'd do it for a man walking the length of the train.

What are your trying to say?

> What makes
> light work by magic,

If we are discussing Einstein's argument, we use Einstein's
postulates. You may disagree with the postulates, but if you
introduce a contrary postulate, your cannot conclude anything
about Einstein's argument. IOW, any inconsistency may have been
introduced by your postulate, so you don't know if it was
present in the original.

If you want to determine if Einstein's argument is flawed, you
cannot assume, a priori, that simultaneity is invariant, that
light speed is source dependent, or that light speed is fixed in
a single reference system. It may be interesting to explore the
consequences of each of those assumptions, but if do, you are
not discussing SR, because they each contradict the light speed
postulate.

(The same is true of any postulate equivalent to these, such as
length invariance.)

Sorcerer

unread,

Oct 10, 2006, 8:08:02 PM10/10/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:PkRWg.3885$gM1.3059@fed1read12...

That if I walk along a train my velocity w.r.t. the track is
the sum of my walking velocity and the velocity of the train.
What are you trying to splutter?

Ah... Blind faith in the Holey man, no connection with physics.
<delete rant>

Androcles -- 5, Shithead Kennelly -- 0.

Dirk Van de moortel

unread,

Oct 10, 2006, 8:31:12 PM10/10/06

to

"Sorcerer" <Headm...@hogwarts.physics_b> wrote in message news:C4TWg.123157$PD.6...@fe2.news.blueyonder.co.uk...

Yes, Brian should know that you have personally measured
your velocity w.r.t. the track when your walking speed w.r.t.
the train was 1 m/s and the train was going at 10 m/s. You
personally found the result to be 10.999999999999999 m/s
as opposed to the 11.000000000000000 m/s that keeps
popping up in your excuse for a brain, didn't you?

Dirk Vdm

Brian Kennelly

unread,

Oct 10, 2006, 8:53:18 PM10/10/06

to

In any inertial reference system, your speed relative to the
track is the sum of your speed relative to the train and the
train's speed relative to the track, *as long as both are
measured in the same reference system.*

We cannot, without further assumptions, compare your speed
relative to the train in two *different* reference systems.

Einstein's light postulate is one such assumption, that allows
us to compare light speed in different systems, and from that,
any other kinematic quantities, such as your velocity. One of
the results is that your speed relative to the train is
different in different reference systems.

> | > What makes
> | > light work by magic,
> | If we are discussing Einstein's argument, we use Einstein's
> | postulates.
>
> Ah... Blind faith in the Holey man, no connection with physics.

No. I did not base my argument on the truth of the postulate.
I only assert that you cannot claim that Einstein's arguments
are inconsistent if *you* introduce the inconsistency. If, as
you claim, the inconsistency is in SR, then try to demonstrate
it from the postulates of SR.

If you are not trying to claim that SR is internally
inconsistent, the forget Einstein's arguments, and move on to
the experimental data.

Sorcerer

unread,

Oct 10, 2006, 10:25:30 PM10/10/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:2LTWg.3893$gM1.1907@fed1read12...

I'll leave your rant intact.

The Lord Einstein said:
"But the ray moves relatively to the initial point of k, when measured in
the stationary system, with the velocity c-v"

Therefore you uttered heresy.
Say three Hail Aethers for penance.

Hail Aether,
Full of Light,
Einstein is with thee.
Blessed art thou among absolute frames of reference,
and blessed is the fruit of thy tomb, Lorentz Transform.
Holy Aether,
Daughter of Lunacy,
prey on us morons now,
and at the dilated hour of death.

| > | > What makes
| > | > light work by magic,
| > | If we are discussing Einstein's argument, we use Einstein's
| > | postulates.
| >
| > Ah... Blind faith in the Holey man, no connection with physics.
| No.

Yes, fuckin' liar. Blind faith in the Holey man, no connection with physics.

<rant deleted>
Father Androcles

Dirk Van de moortel

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Oct 10, 2006, 10:38:04 PM10/10/06

to

"Sorcerer" <Headm...@hogwarts.physics_b> wrote in message news:u5VWg.123557$PD.8...@fe2.news.blueyonder.co.uk...

Hey Brian, how long did you plan to remain optimistic?
Sorry, couldn't resist ;-)

Dirk Vdm

Brian Kennelly

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Oct 10, 2006, 10:40:46 PM10/10/06

to

Yes, if the point k is moving with speed v, the speed of light
relative to the moving point k is c-v. That also agrees with
what you stated above about adding velocities.

But, with the light postulate, the speed of light relative to k
is different in different systems. In particular, in the system
fixed to k, the speed of k is zero, and the relative speed is c.

In pre-relativity physics, the relative speed was assumed to be
invariant. SR replaces that assumption with the light postulate.

Brian Kennelly

unread,

Oct 10, 2006, 10:47:54 PM10/10/06

to

Dirk Van de moortel wrote:
>

> Hey Brian, how long did you plan to remain optimistic?
> Sorry, couldn't resist ;-)

I have a lot of patience.

I really want to believe that my interlocutor's confusion is the
result of unstated assumptions, rather than perversity or
bloody-mindedness.

If I am wrong, at least it is a good exercise to formulate
arguments in new ways.

Sorcerer

unread,

Oct 10, 2006, 11:54:41 PM10/10/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:OjVWg.3900$gM1.716@fed1read12...

Ok, so you lose again, checkmate.
<rant deleted>
Androcles 6, Fuckwit 0.
Wanna start a new game?
Androcles.

Brian Kennelly

unread,

Oct 11, 2006, 1:29:38 AM10/11/06

to

No, you and Einstein said the same thing and I agreed with you
both. If I erred, then so did you.

Brian Kennelly

unread,

Oct 11, 2006, 2:24:29 AM10/11/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message

> news:2LTWg.3893$gM1.1907@fed1read12...

> | Sorcerer wrote:
> | >
> | > That if I walk along a train my velocity w.r.t. the track is
> | > the sum of my walking velocity and the velocity of the train.

> | In any inertial reference system, your speed relative to the
> | track is the sum of your speed relative to the train and the
> | train's speed relative to the track, *as long as both are
> | measured in the same reference system.*
> |

> The Lord Einstein said:
> "But the ray moves relatively to the initial point of k, when measured in
> the stationary system, with the velocity c-v"
>

I went back to this post and trimmed it a little to show that
all three statements say the same thing. (Einstein was using
light rather then a walker, but that does not change the
essential meaning.)

If we designate your speed relative to the train, measured in
the track system, by /c-v/, and the speed of the train relative
to the track as /v/, then your speed relative to the track is
/c/. That agrees with our statement: c=(c-v)+v

It agrees with my answer, which simply restated yours, but
explicitly stated the restriction to a single reference system.

It agrees with Einstein's statement, with the speed of light
/c/, the speed of k /v/ and the speed of light relative to k
/c-v/. (His <(c-v)=c-v> is equivalent to <c=(c-v)+v>.)

Sorcerer

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Oct 11, 2006, 5:52:52 AM10/11/06

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"Brian Kennelly" <bwken...@cox.net> wrote in message

news:6OXWg.3913$gM1.2267@fed1read12...

Yes, fuckwit. You lost poorly because you are a poor loser, my
page accurately portrays the derivation of Einstein's cuckoo
malformations that he blames on Lorentz, arsehole.
<rant deleted>
Androcles 7, Shitheaded Piss-poor loser Kennelly 0.

Sorcerer

unread,

Oct 11, 2006, 6:09:07 AM10/11/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:yBYWg.3915$gM1.2336@fed1read12...

| Sorcerer wrote:
| > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > news:2LTWg.3893$gM1.1907@fed1read12...
| > | Sorcerer wrote:
| > | >
| > | > That if I walk along a train my velocity w.r.t. the track is
| > | > the sum of my walking velocity and the velocity of the train.
|
| > | In any inertial reference system, your speed relative to the
| > | track is the sum of your speed relative to the train and the
| > | train's speed relative to the track, *as long as both are
| > | measured in the same reference system.*

What's this, Animal Farm?
All animals are equal, but some are more equal than others.
Who the fuck do you think you are, moron?

<rant deleted>
I trimmed your trimming. You are doing poorly, you didn't answer my
question, you dumb cunt.
What makes light magic if the same laws of electrodynamics and optics will
be valid for all frames of reference for which the equations of mechanics
hold good?
Androcles

Brian Kennelly

unread,

Oct 11, 2006, 6:19:46 AM10/11/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message
> news:yBYWg.3915$gM1.2336@fed1read12...
> | Sorcerer wrote:
> | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > news:2LTWg.3893$gM1.1907@fed1read12...
> | > | Sorcerer wrote:
> | > | >
> | > | > That if I walk along a train my velocity w.r.t. the track is
> | > | > the sum of my walking velocity and the velocity of the train.
> |
> | > | In any inertial reference system, your speed relative to the
> | > | track is the sum of your speed relative to the train and the
> | > | train's speed relative to the track, *as long as both are
> | > | measured in the same reference system.*
> What's this, Animal Farm?
> All animals are equal, but some are more equal than others.
> Who the fuck do you think you are, moron?

Do you disagree with my statement agreeing with your statement?

>
>
> | > |
> | > The Lord Einstein said:
> | > "But the ray moves relatively to the initial point of k, when measured
> in
> | > the stationary system, with the velocity c-v"
> | >
> | I went back to this post and trimmed it a little
> <rant deleted>
> I trimmed your trimming. You are doing poorly, you didn't answer my
> question, you dumb cunt.
> What makes light magic if the same laws of electrodynamics and optics will
> be valid for all frames of reference for which the equations of mechanics
> hold good?

There is nothing magic. Do you disagree with Einstein's
statement agreeing with your statement?

Sorcerer

unread,

Oct 11, 2006, 6:49:41 AM10/11/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:620Xg.3920$gM1.540@fed1read12...

| Sorcerer wrote:
| > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > news:yBYWg.3915$gM1.2336@fed1read12...
| > | Sorcerer wrote:
| > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > | > news:2LTWg.3893$gM1.1907@fed1read12...
| > | > | Sorcerer wrote:
| > | > | >
| > | > | > That if I walk along a train my velocity w.r.t. the track is
| > | > | > the sum of my walking velocity and the velocity of the train.
| > |
| > | > | In any inertial reference system, your speed relative to the
| > | > | track is the sum of your speed relative to the train and the
| > | > | train's speed relative to the track, *as long as both are
| > | > | measured in the same reference system.*
| > What's this, Animal Farm?
| > All animals are equal, but some are more equal than others.
| > Who the fuck do you think you are, moron?
| Do you disagree with my statement agreeing with your statement?

Your rant is like Einstein's luminiferous aether, superfluous, and
you are doing poorly.

|
| >
| >
| > | > |
| > | > The Lord Einstein said:
| > | > "But the ray moves relatively to the initial point of k, when
measured
| > in
| > | > the stationary system, with the velocity c-v"
| > | >
| > | I went back to this post and trimmed it a little
| > <rant deleted>
| > I trimmed your trimming. You are doing poorly, you didn't answer my
| > question, you dumb cunt.
| > What makes light magic if the same laws of electrodynamics and optics
will
| > be valid for all frames of reference for which the equations of
mechanics
| > hold good?
| There is nothing magic.

Good.

| Do you disagree with Einstein's
| statement agreeing with your statement?

Einstein can agree with me anytime, it is the kook Kennelly I have trouble
with.
Where am I doing poorly? My pages accurately depict the derivation
of the cuckoo malformations the huckster Einstein blamed on Lorentz, kook.
I'm doing far better than you ever will.
Next game:
http://www.androcles01.pwp.blueyonder.co.uk/GPS.htm
How am I doing?

Androcles

Dirk Van de moortel

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Oct 11, 2006, 7:28:43 AM10/11/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:vqVWg.3901$gM1.3808@fed1read12...

> Dirk Van de moortel wrote:
>>
>> Hey Brian, how long did you plan to remain optimistic?
>> Sorry, couldn't resist ;-)
> I have a lot of patience.
>
> I really want to believe that my interlocutor's confusion is the result of unstated assumptions, rather than perversity or
> bloody-mindedness.

Trust me, it is the result of stupidity and perversity and
bloody-mindedness. No kidding.

>
> If I am wrong, at least it is a good exercise to formulate arguments in new ways.

Absolutely.
We've all been there and we've had a truly excellent
argument formulating school. That is the big bonus you get
on these guys :-)
Enjoy!

Dirk Vdm

mlut...@wanadoo.fr

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Oct 11, 2006, 10:57:34 AM10/11/06

to

Brian Kennelly wrote:
> mlut...@wanadoo.fr wrote:
> > Brian Kennelly wrote:
> >> mlut...@wanadoo.fr wrote:
> > If you consider a length g (c-v)T at rest in the S'-frame, you
> > get a length (c-v)T in the S-frame, by applying length contraction
> > (this is what the tracking guru did).
> >
> > If you consider a stick of length (c-v)T at rest in the S-frame,
> > you get a length g(c-v)T in the S'-frame according to the LT.
> > It is dilated in the S'-frame. Indeed, if the ends of the stick
> > are x1 = vT and x2 = cT in the S-frame,
> Okay, then from the LT, we have:
> vT=g(x1'+vt1')
> cT=g(x2'+vt2')
> This allows us to find x2' and x1' at the same t'
>
> > x1' = 0 and x2' = g(c-v)T,
> There is no value of t' that makes both of these equations true.
> When x1'=0, then (from the first equation):
> vT=g(vt1')
> So t1' = T/g
>
> When x2'=g(c-v)T, then
> cT=g(g(c-v)T+vt2')
> So t2'=cT/vg - g(c-v)T/v
> \=t1'
>
> To find the length in S', we must measure the distance between
> the endpoints at the same time in S' (the same t')
>
> Setting t1'=t2' in the LT equations above and subtracting we get
> cT-vT=g(x2'-x1')
> x2'-x1'=(c-v)T/g
>
> The length is contracted.
>
> > hence x2' - x1' = g(c-v)T. Notice that x1' is always zero, its
> > value is independent of time.
> The only way that x1' can be always zero is if it is at rest in
> S'. Because you stated that the stick is at rest in S, it, and
> therefore its endpoints, will be moving in S':
> vT=g(x1'+vt')
> x1'=vT/g-vt'
>
> In particular, when x2'=g(c-v)T, then t'=cT/vg - g(c-v)T/v and
> x1' is:
> x1' = vT/g-v(cT/vg - g(c-v)T/v)
> = vT/g- cT/g +g(c-v)T
> =g(c-v)T - (c-v)T/g
>
> Subtracting this from x2' again gives the result derived above:
> x2'-x1'=(c-v)T/g
>
> >
> > This is obvious: if L = L'/g (as in the guru exemple), L' is
> > necessarily gL. The one who got L = L'/g AND L' = L/g made a logical
> > mistake somewhere.
> No, L=L'/g and L'=L/g apply to two different scenarios. In the
> first, the stick is at rest in S', in the second, it is at rest
> in S.

Brian:

"If you wish to calculate the S' length of a stick at rest in S,
with endpoints at vT and cT [so that L=(c-v)T], we can use the
LT as follows:

vT=g(x1'+vt')
cT=g(x2'+vt')

Subtracting we get:
(c-v)T=g(x2'-x1')

The length is:
x2'-x1'=(c-v)T/g
= L/g
It is contracted."

By using the 'inverse' LT x = g(x'+vt), you imply that the stick
is at rest in S', not in S.
Indeed, you first calculate x1 and x2 from x1' and x2', which is
possible only if you know the endpoints of the stick in S':
x1 = vT = g(x1'+vt')
x2 = cT = g(x2'+vt')
Then, by substrating, you calculate the length of the stick in S:
x2 - x1 = g(x2'-x1'), or
L = gL'
>From this, you calculate back the length in S' as L'= L/g !
Don't you realize that you reasoning is circular?

Luttgens:

Here is an exemple given by the tracking guru, which
indirectly proves the dilation:

"Now imagine a stick with this particular length
x' = g (c-v) t
at rest in the S' frame.
What is the length of such a stick in the S-frame?
If you apply length contraction, you find that this length
would be
x' / g = (c - v) t
in the S frame."

The length (c - v)t represents the distance in S between the
endpoints ct and vt (for a given value of t).

The particular length x' = g (c-v)t corresponds of course
to the dilated length obtained by applying the LT x' = g(x - vt)
to that case where x = ct in S.

Brian:

"So, when the stick is at rest in the S' frame, it is contracted
in the S frame."

So, you are claiming that L = L'/g.

Brian:

"When the stick is at rest in the S frame, it is
contracted in the S' frame, as I showed above."

Now, you claim that L' = L/g, hence, according to you,
L = gL'. Or you also claimed that L = L'/g.
Don't you see the contradiction?

Brian:

"Okay, then from the LT, we have:
vT=g(x1'+vt1')
cT=g(x2'+vt2')
This allows us to find x2' and x1' at the same t'
There is no value of t' that makes both of these equations true.
To find the length in S', we must measure the distance between
the endpoints at the same time in S' (the same t')
Setting t1'=t2' in the LT equations above and subtracting we get
cT-vT=g(x2'-x1')
x2'-x1'=(c-v)T/g
The length is contracted."

Another contradiction of you:
- There is no value of t' that makes both of these equations true.
- Setting t1'=t2' in the LT equations above ..., the length is
contracted.

On the other side, you use again the 'inverse' transform, making
the same circular reasoning as above.
Hocus pocus !

Luttgens:

Notice that x1' is always zero, its value is independent of time.

Brian:

"The only way that x1' can be always zero is if it is at rest in
S'. Because you stated that the stick is at rest in S, it, and
therefore its endpoints, will be moving in S':
vT=g(x1'+vt')
x1'=vT/g-vt'

In particular, when x2'=g(c-v)T, then t'=cT/vg - g(c-v)T/v and
x1' is:
x1' = vT/g-v(cT/vg - g(c-v)T/v)
= vT/g- cT/g +g(c-v)T
=g(c-v)T - (c-v)T/g

Subtracting this from x2' again gives the result derived above:
x2'-x1'=(c-v)T/g "

The end vT of the stick in S always coincide with the S' origin,
so in S', such end is independent of T, thus its coordinate x1'
is always 0.
By applying the 'direct' LT, which is the correct procedure, one gets
L' = x2' - 0 = g(c-v)T = gL
The length is thus dilated in S', according to the Einsteinian LT.

By using the correct LT (obtained without the postulate that
when x = ct, x' = ct')
x' = (x-vt)/g
t' = t/g,
one gets from x1 = vT and x2 = cT (thus a stick of length (c-v)T in S,
x1' = 0
x2' = (c-v)T/g
L' = x2' - x1' = L/g, the correct result.

Marcel Luttgens

Brian Kennelly

unread,

Oct 11, 2006, 4:41:05 PM10/11/06

to

mlut...@wanadoo.fr wrote:
>
> Brian:
>
> "If you wish to calculate the S' length of a stick at rest in S,
> with endpoints at vT and cT [so that L=(c-v)T], we can use the
> LT as follows:
>
> vT=g(x1'+vt')
> cT=g(x2'+vt')
>
> Subtracting we get:
> (c-v)T=g(x2'-x1')
>
> The length is:
> x2'-x1'=(c-v)T/g
> = L/g
> It is contracted."
>
> By using the 'inverse' LT x = g(x'+vt), you imply that the stick
> is at rest in S', not in S.

No. It only implies that I am using the S' coordinates to use
calculate the length. Using the S' coordinates allows me to
work with x' and t' without the need to calculate any t values.

> Indeed, you first calculate x1 and x2 from x1' and x2', which is
> possible only if you know the endpoints of the stick in S':
> x1 = vT = g(x1'+vt')
> x2 = cT = g(x2'+vt')

No. <vT> and <cT> are constants, so those equations allow us to
calculate the endpoints in S' from the fixed values in S, for
any time t'.
If it is easier, you can express them as:
x1'=vT/g-vt'
x2'=cT/g-vt'

> Then, by substrating, you calculate the length of the stick in S:
> x2 - x1 = g(x2'-x1'), or
> L = gL'
>>From this, you calculate back the length in S' as L'= L/g !
> Don't you realize that you reasoning is circular?

Where is the circularity?
I used the fixed endpoints in S and the LT to calculate the
length in S'.

If you prefer, we can do it the long way:
x1'=g(vT-vt1)
x2'=g(cT-vt2)
t1'=g(t1-v^2T/c^2) (x1=vT)
t2'=g(t2-vT/c) (x2=cT)

Setting t1'=t2', we get
t2=t1-v^2T/c^2+vT/c

Choosing t1=T, so that x1'=0
t2=T(1-v^2/c^2+v/c)
So:
x2'=g(cT-vT(1-v^2/c^+v/c))
=g(cT-v^2T/c-vT(1-v^2/c^2))
= g(cT(1-v^2/c^2)-vT(1-v^2/c^2))
=g((c-v)T(1/g^2)
=(c-v)T/g

Again, we find that the length is contracted in S'.

>
> Luttgens:
>
> Here is an exemple given by the tracking guru, which
> indirectly proves the dilation:
>
> "Now imagine a stick with this particular length
> x' = g (c-v) t
> at rest in the S' frame.
> What is the length of such a stick in the S-frame?
> If you apply length contraction, you find that this length
> would be
> x' / g = (c - v) t
> in the S frame."
>
> The length (c - v)t represents the distance in S between the
> endpoints ct and vt (for a given value of t).
>
> The particular length x' = g (c-v)t corresponds of course
> to the dilated length obtained by applying the LT x' = g(x - vt)
> to that case where x = ct in S.
>
> Brian:
>
> "So, when the stick is at rest in the S' frame, it is contracted
> in the S frame."
>
> So, you are claiming that L = L'/g.

Yes, the stick is at rest in the S' frame.

>
> Brian:
>
> "When the stick is at rest in the S frame, it is
> contracted in the S' frame, as I showed above."
>
> Now, you claim that L' = L/g, hence, according to you,

Yes, when the stick is at rest in the S frame.

> L = gL'. Or you also claimed that L = L'/g.

> Don't you see the contradiction?

No, because the stick cannot be at rest in both frames, unless
v=0, in which case g=1, and L=L'

>
> Brian:
>
> "Okay, then from the LT, we have:
> vT=g(x1'+vt1')
> cT=g(x2'+vt2')
> This allows us to find x2' and x1' at the same t'
> There is no value of t' that makes both of these equations true.
> To find the length in S', we must measure the distance between
> the endpoints at the same time in S' (the same t')
> Setting t1'=t2' in the LT equations above and subtracting we get
> cT-vT=g(x2'-x1')
> x2'-x1'=(c-v)T/g
> The length is contracted."
>
> Another contradiction of you:
> - There is no value of t' that makes both of these equations true.
> - Setting t1'=t2' in the LT equations above ..., the length is
> contracted.

You took the statement out of context. I said that, given the
LT, there is no single value of t' that allows both <x1'=0> and
<x2'=g(c-v)T> to both be true.
IOW, this system of equations is inconsistent:

vT=g(x1'+vt1')
cT=g(x2'+vt2')

x1'=0
x2'=g(c-v)T
t1'=t2'

The first two equations are the LT, and the last is a
requirement for measuring length in S'. You must drop either
the third or fourth equation. In either case, you find that the
length is contracted.
E.g., if we drop the fourth equation, then we get
vT=g(vt1')
cT=g(x2'+vt1')
t1'=T/g
cT=gx2'+vT
x2'=(c-v)T/g

Earlier, I showed the calculation obtained by dropping the third
equation.

>
> On the other side, you use again the 'inverse' transform, making
> the same circular reasoning as above.

The inverse transformation has the same content as the direct
equations.

>
> Luttgens:
>
> Notice that x1' is always zero, its value is independent of time.
>
> Brian:
>
> "The only way that x1' can be always zero is if it is at rest in
> S'. Because you stated that the stick is at rest in S, it, and
> therefore its endpoints, will be moving in S':
> vT=g(x1'+vt')
> x1'=vT/g-vt'
>
> In particular, when x2'=g(c-v)T, then t'=cT/vg - g(c-v)T/v and
> x1' is:
> x1' = vT/g-v(cT/vg - g(c-v)T/v)
> = vT/g- cT/g +g(c-v)T
> =g(c-v)T - (c-v)T/g
>
> Subtracting this from x2' again gives the result derived above:
> x2'-x1'=(c-v)T/g "
>
> The end vT of the stick in S always coincide with the S' origin,
> so in S', such end is independent of T, thus its coordinate x1'
> is always 0.

So the stick is at rest in S', and moving in S

> By applying the 'direct' LT, which is the correct procedure, one gets
> L' = x2' - 0 = g(c-v)T = gL
> The length is thus dilated in S', according to the Einsteinian LT.

Because you stated above that the stick is at rest in S', we
conclude that it is contracted in S.

The LT never implies that a stick will have a greater length
than it does in its rest system.

Sorcerer

unread,

Oct 11, 2006, 5:19:25 PM10/11/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:A89Xg.3928$gM1.2408@fed1read12...
| mlut...@wanadoo.fr wrote:

| The LT never implies that a stick will have a greater length
| than it does in its rest system.

HAHAHA!
Nobody on the train realises it shrinks and stretches,
only we on the trackside see that. The rider of the train
sees the fence stretching and shrinking.
http://www.androcles01.pwp.blueyonder.co.uk/Smart/Smart.htm

Never mind, the men in white coats will be along soon,
they have a nice soft padded room for you in case you hurt
yourself with your bloody-minded perversions.
I thought nobody could be a stupid as Dork Van de merde,
but you come pretty close.
Let's see, you need a name... Kennel... ly...lie.. sleeping dogs..
You must be the pee-puppy's doghouse he's alway in.

Brian Kennelly

unread,

Oct 11, 2006, 5:34:00 PM10/11/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message
> news:620Xg.3920$gM1.540@fed1read12...
> | Sorcerer wrote:
> | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > news:yBYWg.3915$gM1.2336@fed1read12...
> | > | Sorcerer wrote:
> | > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > | > news:2LTWg.3893$gM1.1907@fed1read12...
> | > | > | Sorcerer wrote:
> | > | > | >
> | > | > | > That if I walk along a train my velocity w.r.t. the track is
> | > | > | > the sum of my walking velocity and the velocity of the train.
> | > |
> | > | > | In any inertial reference system, your speed relative to the
> | > | > | track is the sum of your speed relative to the train and the
> | > | > | train's speed relative to the track, *as long as both are
> | > | > | measured in the same reference system.*
> | > What's this, Animal Farm?
> | > All animals are equal, but some are more equal than others.
> | > Who the fuck do you think you are, moron?
> | Do you disagree with my statement agreeing with your statement?
>
> Your rant is like Einstein's luminiferous aether, superfluous, and
> you are doing poorly.

So, are you saying that my statement disagrees with yours, or
that they were both wrong?

> | > | > |
> | > | > The Lord Einstein said:
> | > | > "But the ray moves relatively to the initial point of k, when
> measured
> | > in
> | > | > the stationary system, with the velocity c-v"
> | > | >
> | > | I went back to this post and trimmed it a little
> | > <rant deleted>
> | > I trimmed your trimming. You are doing poorly, you didn't answer my
> | > question, you dumb cunt.
> | > What makes light magic if the same laws of electrodynamics and optics
> will
> | > be valid for all frames of reference for which the equations of
> mechanics
> | > hold good?
> | There is nothing magic.
> Good.
>
> | Do you disagree with Einstein's
> | statement agreeing with your statement?
>
> Einstein can agree with me anytime, it is the kook Kennelly I have trouble
> with.
> Where am I doing poorly? My pages accurately depict the derivation
> of the cuckoo malformations the huckster Einstein blamed on Lorentz, kook.

That is what we are are trying to determine. So far, I think
we have agreed that the speed of light relative to k, in the
stationary system, is c-v when it is moving to the right, and
c+v when moving to the left. (I.e., the speed of light in the
stationary system, relative to the train, is 2 and 8, going and
returning.)

Do you agree that the distance travelled in the stationary
system, relative to the left endpoint, is x'? (I.e., do you
agree that the length of the train in the stationary system is 32?)

> Next game:
> http://www.androcles01.pwp.blueyonder.co.uk/GPS.htm
We should finish the first game, unless you have decided to
forfeit again.

Dirk Van de moortel

unread,

Oct 11, 2006, 5:46:43 PM10/11/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:A89Xg.3928$gM1.2408@fed1read12...
> mlut...@wanadoo.fr wrote:

[snip]

> The LT never implies that a stick will have a greater length than it does in its rest system.

Keep in mind that you are explaining this to someone who has
no understanding of coordinates, transformations, elementary
analytic geometry or ditto linear algebra.
A full decade we have been trying to explain this stuff to him.
Just like Androcles, the man has decided that it can't be, so
it can't be.
I think it's a combination of lack of brain power and a
lifetime investment in getting used to that idea.

But you are doing just fine :-)

Dirk Vdm

Sorcerer

unread,

Oct 11, 2006, 6:35:46 PM10/11/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:dW9Xg.3933$gM1.2969@fed1read12...

No, relative to kappa. K and k are roman frames, kappa is the greek frame.
The train is the kappa-frame and the k-frame, K is the camel frame.
That's just more of Einstein's deliberate bloody-minded perversions
to confuse fuckwits like you. He gets into Hebrew in GR.

It's like this:
"If we place x'=x-vt, it is clear that a point at rest in the system k
must have a system of values x', y, z, independent of time."
That's the k-frame.
Now... the kappa-frame is the k-frame transformed by beta (now called
gamma just to confuse you further with bloody-minded perversions).
kappa is in fact:
xi = x'/sqrt(1-v^2/c^2) because x' = x-vt,
and
tau = t*sqrt(1-v^2/c^2)
which does not include the bloody-minded perversion of x.

Proof:
tau = (t-vx/c^2)/sqrt(1-v^2/c^2)
but x = vt, so
tau = (t-v^2 * t/c^2)/sqrt(1-v^2/c^2)
= t(1-v^2/c^2)/sqrt(1-v^2/c^2)
= t*sqrt(1-v^2/c^2)

This is because the bastard was a bloody-minded pervert, just like
your pal Dork Van de merde, the local village dog tord.
I'm pessimistic about you being able to stand up against a
knuckle-dragging Neanderthal like the pee-puppy. He warned you
about me, I like to live up to my reputation.

| Do you agree that the distance travelled in the stationary
| system, relative to the left endpoint, is x'?

Stationary systems do not travel, they are stationary.
x' is the length of the train, independent of time.
The distance travelled by the moving system in the stationary
system is x in time t at velocity v = dx/dt.
I do not agree with your bloody-minded perversity, only a
fuckwitted kook would claim a stationary system travelled.

(I.e., do you
| agree that the length of the train in the stationary system is 32?)

It is 32, independent of time. God said so (your god, not mine).

Forfeit is when you "<delete rant>" because you lose your temper
and can't answer, you stupid cunt. You've forfeited 7 times,
this is the eighth game I'm winning. If you want to continue
the first game go back and respond like a gentleman and
apologise for calling others "kooks". You are the kook.

Androcles

Sorcerer

unread,

Oct 11, 2006, 6:36:19 PM10/11/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:dW9Xg.3933$gM1.2969@fed1read12...

No, relative to kappa. K and k are roman frames, kappa is the greek frame.

| Do you agree that the distance travelled in the stationary
| system, relative to the left endpoint, is x'?

Stationary systems do not travel, they are stationary.

x' is the length of the train, independent of time.
The distance travelled by the moving system in the stationary
system is x in time t at velocity v = dx/dt.
I do not agree with your bloody-minded perversity, only a
fuckwitted kook would claim a stationary system travelled.

(I.e., do you
| agree that the length of the train in the stationary system is 32?)

It is 32, independent of time. God said so (your god, not mine).
|
|

| > Next game:
| > http://www.androcles01.pwp.blueyonder.co.uk/GPS.htm
| We should finish the first game, unless you have decided to
| forfeit again.

Forfeit is when you "<delete rant>" because you lose your temper

and can't answer, you stupid cunt. You've forfeited 7 times,
this is the eighth game I'm winning. If you want to continue
the first game go back and respond like a gentleman and
apologise for calling others "kooks". You are the kook.

I CAN piss higher than you.

Androcles

Brian Kennelly

unread,

Oct 11, 2006, 6:57:48 PM10/11/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message

> news:dW9Xg.3933$gM1.2969@fed1read12...

> | That is what we are are trying to determine. So far, I think
> | we have agreed that the speed of light relative to k, in the
> | stationary system, is c-v when it is moving to the right, and
> | c+v when moving to the left. (I.e., the speed of light in the
> | stationary system, relative to the train, is 2 and 8, going and
> | returning.)
>
> No, relative to kappa. K and k are roman frames, kappa is the greek frame.
> The train is the kappa-frame and the k-frame, K is the camel frame.
> That's just more of Einstein's deliberate bloody-minded perversions
> to confuse fuckwits like you. He gets into Hebrew in GR.

Don't anticipate. We are not working with kappa. We will
introduce it later.

>
> It's like this:
> "If we place x'=x-vt, it is clear that a point at rest in the system k
> must have a system of values x', y, z, independent of time."
> That's the k-frame.

Accepted.

If the speed of light is c in K, and the speed of k is v, in K,
then the speed of light relative to k is c-v, in K.

Agreed? If not, please explain.

> | Do you agree that the distance travelled in the stationary
> | system, relative to the left endpoint, is x'?
>
> Stationary systems do not travel, they are stationary.
> x' is the length of the train, independent of time.

Okay, let me rephrase it. x' is the length of the train,
measured in K.
Agreed?

>
>
> (I.e., do you
> | agree that the length of the train in the stationary system is 32?)
>
> It is 32, independent of time. God said so (your god, not mine).

Agreed.

Note: all the measurements so far are made in the K system.

Sorcerer

unread,

Oct 11, 2006, 7:20:50 PM10/11/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:M8bXg.3936$gM1.130@fed1read12...

You snipped.
Androcles 8, Fuckhead Doghouse 0.

Brian Kennelly

unread,

Oct 11, 2006, 7:50:50 PM10/11/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message
> news:M8bXg.3936$gM1.130@fed1read12...
> | Sorcerer wrote:
> | > "Brian Kennelly" <bwken...@cox.net> wrote in message
> | > news:dW9Xg.3933$gM1.2969@fed1read12...
> | > | That is what we are are trying to determine. So far, I think
> | > | we have agreed that the speed of light relative to k, in the
> | > | stationary system, is c-v when it is moving to the right, and
> | > | c+v when moving to the left. (I.e., the speed of light in the
> | > | stationary system, relative to the train, is 2 and 8, going and
> | > | returning.)
> | >
> | > No, relative to kappa. K and k are roman frames, kappa is the greek
> frame.
> | > The train is the kappa-frame and the k-frame, K is the camel frame.
> | > That's just more of Einstein's deliberate bloody-minded perversions
> | > to confuse fuckwits like you. He gets into Hebrew in GR.
>
> You snipped.

???
No answer?

If, in K, the speed of light is c, and the speed of k is v, then
the speed of light relative to k is c-v.

Agreed? If not, please explain.

x' is the length of the train, measured in K.

Agreed?

Using the numbers from your example, in the reference frame of
the fence, with c=5 and v=3.
The speed of light relative to the train is 2
The length of the train is 32

Agreed?

mlut...@wanadoo.fr

unread,

Oct 11, 2006, 8:08:02 PM10/11/06

to

Your reasoning is more and more interesting:

"I used the fixed endpoints in S and the LT to calculate the
length in S'."

>From x1 = vT and x2 = cT, values of the endpoints of the stick in S,
that are admittedly known, you now calculate x1' and x2' by using
the 'direct' LT

x1'=g(vT-vt')
x2'=g(cT-vt'), thus obtaining

x1'=vT/g-vt'
x2'=cT/g-vt'

Then, with the help of the 'inverse' transform, you *recalculate*
x1 and x2 !

x1 = vT = g(x1'+vt')
x2 = cT = g(x2'+vt')

Substracting x1 from x2, you then get

(c-v)T=g(x2'-x1')

and finally claim that L' is contracted, because

L' = x2'-x1 = (c-v)T/g
= L/g

forgetting that

(c-v)T = g((cT/g-vt') - (vT/g-vt'))
= (c-v)T

You should better not grasp an LT stick, short or long, it is
full of shit.

Marcel Luttgens

Dirk Van de moortel

unread,

Oct 11, 2006, 8:19:29 PM10/11/06

to

"Sorcerer" <Headm...@hogwarts.physics_b> wrote in message news:6QaXg.109532$aP3....@fe3.news.blueyonder.co.uk...

[snip]

>
> It's like this:
> "If we place x'=x-vt, it is clear that a point at rest in the system k
> must have a system of values x', y, z, independent of time."
> That's the k-frame.
> Now... the kappa-frame is the k-frame transformed by beta (now called
> gamma just to confuse you further with bloody-minded perversions).
> kappa is in fact:
> xi = x'/sqrt(1-v^2/c^2) because x' = x-vt,
> and
> tau = t*sqrt(1-v^2/c^2)
> which does not include the bloody-minded perversion of x.
>
> Proof:
> tau = (t-vx/c^2)/sqrt(1-v^2/c^2)
> but x = vt, so
> tau = (t-v^2 * t/c^2)/sqrt(1-v^2/c^2)
> = t(1-v^2/c^2)/sqrt(1-v^2/c^2)
> = t*sqrt(1-v^2/c^2)

Ha yes, for one specific point x = vt, so x must be vt for
all points, because if x has had some value, it must keep
that value for the rest of its life :-)
http://users.telenet.be/vdmoortel/dirk/Stuff/AndrolesAtSchool.gif
Unfortunately Androcles At School always forgets that if x
must remain vt, then
x' = x-vt = 0
and then of course
xi = x'/sqrt(1-v^2/c^2) = 0
but that is slightly inconvenient, so he never mentions it.
It's very simple and basic, but he wouldn't understand
it anyway, so why bother?
The reason is of course that Androcles At School does
not have the brain power to understand the concepts of
variables, constants, equations, coordinates, events,
transformations etc. All Androcles at School can do, is
juggle with equations without ever having the faintest
clue about the meanings of any of the variables.
That is one of the main reasons why we all love
Androcles At School :-)

>
> This is because the bastard was a bloody-minded pervert, just like
> your pal Dork Van de merde, the local village dog tord.
> I'm pessimistic about you being able to stand up against a
> knuckle-dragging Neanderthal like the pee-puppy. He warned you
> about me, I like to live up to my reputation.

Yes, that is another very important reason why we love
Androcles At School so much - it likes to live up to its
reputation, specially when it drinks itself inside out :-)
Are its poor grandchildren familiar with the concept of liver
cirrhosis?

>
> | Do you agree that the distance travelled in the stationary
> | system, relative to the left endpoint, is x'?
>
> Stationary systems do not travel, they are stationary.
> x' is the length of the train, independent of time.

Ha yes, and according to Androcles At School, because
x = vt, so x' must be 0.
Big Train!

> The distance travelled by the moving system in the stationary
> system is x in time t at velocity v = dx/dt.

Ouch.... another usage of the dreadful symbol x :-)

> I do not agree with your bloody-minded perversity, only a
> fuckwitted kook would claim a stationary system travelled.

Only a drunk would think that "distance travelled in the stationary
system" implies that the system travels :-)

Dig, Androfart, dig :-)

Dirk Vdm

Dirk Van de moortel

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Oct 11, 2006, 8:20:17 PM10/11/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:wWbXg.3939$gM1.1418@fed1read12...

Somehow I think it's time to become slightly realistic.
Sorry, I know I shouldn't do this, but I still can't resist.
I hope you forgive me my little weakness :-)

Dirk Vdm

Dirk Van de moortel

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Oct 11, 2006, 8:22:26 PM10/11/06

to

<mlut...@wanadoo.fr> wrote in message news:1160597281....@h48g2000cwc.googlegroups.com...

[snip]

Marcel reminds me of Androcles.
Only slightly less dense.
We love you too, Marcel!

Dirk Vdm

Brian Kennelly

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Oct 11, 2006, 8:35:55 PM10/11/06

to

Those equations are incorrect. I used 'inverse' equations.

x1 = vT = g(x1'+vt')
x2 = cT = g(x2'+vt')

To obtain:
>
> x1'=vT/g-vt'
> x2'=cT/g-vt'
E.g.
Start with vT=g(x1'+vt')
Divide by g: vT/g=x1'+vt'
Subtract vt' from both sides: vT/g-vt'=x1'
Swap sides to obtain: x1'=vT/g-vt'

Any questions?

(Note that these equations give <x2'-x1'=(c-v)T/g> at all times t'.)

>
> Then, with the help of the 'inverse' transform, you *recalculate*
> x1 and x2 !
>
> x1 = vT = g(x1'+vt')
> x2 = cT = g(x2'+vt')
>
> Substracting x1 from x2, you then get
>
> (c-v)T=g(x2'-x1')
>
> and finally claim that L' is contracted, because
>
> L' = x2'-x1 = (c-v)T/g
> = L/g

Yes, that is correct.

>
> forgetting that
>
> (c-v)T = g((cT/g-vt') - (vT/g-vt'))
> = (c-v)T

Excuse me? I forgot that (c-v)T=(c-v)T?
You will need to try harder than that.

Sorcerer

unread,

Oct 11, 2006, 10:26:42 PM10/11/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:wWbXg.3939$gM1.1418@fed1read12...

| Sorcerer wrote:
| > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > news:M8bXg.3936$gM1.130@fed1read12...
| > | Sorcerer wrote:
| > | > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > | > news:dW9Xg.3933$gM1.2969@fed1read12...
| > | > | That is what we are are trying to determine. So far, I think
| > | > | we have agreed that the speed of light relative to k, in the
| > | > | stationary system, is c-v when it is moving to the right, and
| > | > | c+v when moving to the left. (I.e., the speed of light in the
| > | > | stationary system, relative to the train, is 2 and 8, going and
| > | > | returning.)
| > | >
| > | > No, relative to kappa. K and k are roman frames, kappa is the greek
| > frame.
| > | > The train is the kappa-frame and the k-frame, K is the camel frame.
| > | > That's just more of Einstein's deliberate bloody-minded perversions
| > | > to confuse fuckwits like you. He gets into Hebrew in GR.
| >
| > You snipped.
| ???
| No answer?

Nothing to respond to, the game was over when you snipped and failed
to answer. You snipped again, too.
Androcles 9, Arrogant bloody-minded perverted Kennelly 0.

Sorcerer

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Oct 11, 2006, 10:27:40 PM10/11/06

to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:llcXg.124155$tF5.2...@phobos.telenet-ops.be...

|
| "Sorcerer" <Headm...@hogwarts.physics_b> wrote in message
news:6QaXg.109532$aP3....@fe3.news.blueyonder.co.uk...
|
| [snip]

Gladly.

Brian Kennelly

unread,

Oct 11, 2006, 10:38:38 PM10/11/06

to

Sorcerer wrote:
> "Brian Kennelly" <bwken...@cox.net> wrote in message

> news:wWbXg.3939$gM1.1418@fed1read12...

> | Sorcerer wrote:
> | > You snipped.
> | ???
> | No answer?
>
> Nothing to respond to, the game was over when you snipped and failed
> to answer. You snipped again, too.
> Androcles 9, Arrogant bloody-minded perverted Kennelly 0.
>

It is considered good netiquette to trim responses, and remove
old text that is not directly relevant to the reply. If I
removed something that you felt was relevant you can certainly
quote it, and state why it is important to the point being made.

I try to make a point of leaving more than enough quoted text to
establish the context of my replies.

Sorcerer

unread,

Oct 12, 2006, 1:13:01 AM10/12/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:OneXg.3947$gM1.525@fed1read12...

| Sorcerer wrote:
| > "Brian Kennelly" <bwken...@cox.net> wrote in message
| > news:wWbXg.3939$gM1.1418@fed1read12...
| > | Sorcerer wrote:
| > | > You snipped.
| > | ???
| > | No answer?
| >
| > Nothing to respond to, the game was over when you snipped and failed
| > to answer. You snipped again, too.
| > Androcles 9, Arrogant bloody-minded perverted Kennelly 0.
| >
| It is considered good netiquette to trim responses

It is etiquette to respond before snipping, arrogant bloody-minded
perverted kook, don't try to teach me manners, you cunt.

Androcles.

Brian Kennelly

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Oct 12, 2006, 1:57:12 AM10/12/06

to

Perhaps. I did not consider the possibility of mental illness.

> Sorry, I know I shouldn't do this, but I still can't resist.
> I hope you forgive me my little weakness :-)

Certainly.

Sorcerer

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Oct 12, 2006, 6:46:21 AM10/12/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:ZhhXg.3965$gM1.1912@fed1read12...
<delete rant>

| Dirk Van de moortel wrote:

[anip]

I "tink" you morons are shitheads.

mlut...@wanadoo.fr

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Oct 13, 2006, 1:37:36 PM10/13/06

to

No, only a couple of remarks:

>From the direct transform, one gets x1'=0, so, in
your relation x1'=vT/g-vt', vT/g = vt', thus t' = T/g.

And indeed, from the 'direct' time transform
t' = g(T-vx/c^2), one gets
t' = g(T-vT*v/c^2)
= gT(1-v^2/c^2)
= T/g

>From the 'direct' transform, one also gets
x2' = g(x2-vT) = g(cT-vT) = g(c-v)T
x2'-x1' = g(c-v)T
L' = gL

You got cT=g(x2'+vt')

According to the 'direct' time transform,
t' = g(T-cT*v/c^2),
= gT(1-v/c), hence, by substitution in your formula

cT = g(gT(c-v) + v*gT(1-v/c))
= g^2 * T(c-v + v-v^2/c)
= g^2 * Tc(1-v^2/c^2)
= cT

So, all what you demonstrated is that cT = cT !

Btw, do you reallly need the help of the tracking guru? Everybody knows
that he is a mere malicious parrot!

Marcel Luttgens

Brian Kennelly

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Oct 13, 2006, 3:51:53 PM10/13/06

to

mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:

No, that is not correct. You get
x2'=g(cT-vt)

>
> You got cT=g(x2'+vt')
>
> According to the 'direct' time transform,
> t' = g(T-cT*v/c^2),
> = gT(1-v/c), hence, by substitution in your formula

No, the direct transform (when x2=cT) yields
t'=g(t-cT*v/c^2)
=g(t-vT/c)

From this equation, and t'=T/g (with which you agreed above),
you get
T/g=g(t-vT/c)
T/g^2=t-vT/c
T(1-v^2/c^2)=t-vT/c
t=T(1-v^2/c^2)+vT/c

Substituting this value into x2'=g(cT-vt) gives
x2'=g(cT-vT(1-v^2/c^2)-v^2T/c)
=g(cT(1-v^2/c^2)-vT(1-v^2/c^2))
=g*(1/g^2)*(c-v)T
=(c-v)T/g

(You can see why I chose to work with the inverse transforms.
With them, we don't need to calculate the values of t).

Your fundamental error seems to be that you assume that "the
same time" in S' is the same as "the same time" in S. It is
important to use the values of the endpoints "at the same time"
in S', if you want to measure the length according to S'

If x1=vT, and x2=cT, for all times in S, then x1'=0 and
x2'=g(c-v)T occur at different times in S'.

--New approach--
We can turn this around. Let us use your coordinates in S' to
calculate the coordinates in S. We will use the 'inverse'
equations, so that we can put all the known quantities on the
right hand side.

Assume that x1'=0 and x2'=g(c-v)T, when t'=T/g. Then
x1=g(x1'+vt')
x1=vT (1)

So far, so good.

x2=g(x2'+vt')
=g^2(c-v)T+vT
=g^2((c-v)T+g^2vT) (2)
\=cT

Either you erred when you placed right endpoint at cT in S, or
you erred when you claimed that x2'=g(c-v)T when x1'=0.

From (1) and (2), we get x2-x1=g^2(c-v)T, and, again, L=gL' or
L'=L/g

Sorcerer

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Oct 13, 2006, 4:33:02 PM10/13/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:sCOXg.4048$gM1.2753@fed1read12...

HAHAHAHA!

c = v
c^2 = cv
c^2-v^2 = cv - v^2
(c+v)(c-v) = v(c-v)
c+v = v
But v = c, given, so
c+c = c
2c = c
2 =1.
(which you agreed with above)
Androcles

mlut...@wanadoo.fr

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Oct 14, 2006, 2:26:46 PM10/14/06

to

Those times are given by the direct transform as

t' = T/g (from x1 = vT)
t' = g(t-cT*v/c^2)
= g(t-vT/c) (from x2 = cT)

Let's call those times in S'

t1' = T/g
t2' = g(t-vT/c)

Boldly equating t1' to t2', you got

t = T(1-v^2/c^2)+vT/c
= T/g^2 + vT/c,
t2' = g(T/g^2 + vT/c - vT/c)
= T/g

You had to assume that t1' = t2' = T/g to get the correct solution
L'=L/g.

Notice that from the 'inverse' transform t = g(t' + vx'/c^2), one gets
- from x1' = 0,
t1 = gt'
- from x2' = (c-v)T/g,
t2 = gt' + gvx'/c^2
= gt' + v(c-v)T/c^2,
hence t1<>t2 for a same value of t'.
So, if x1'=0 and x2'=(c-v)T/g, for all times in S', then x1=vT
and x2=cT occur at different times in S

With the Einsteinian LT, there is no coherent solution.

The correct LT is

x' = (x-vt)/g
t' = t/g

Marcel Luttgens

Brian Kennelly

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Oct 14, 2006, 5:45:25 PM10/14/06

to

Yes! You finally got it.

>
> Notice that from the 'inverse' transform t = g(t' + vx'/c^2), one gets
> - from x1' = 0,
> t1 = gt'
> - from x2' = (c-v)T/g,
> t2 = gt' + gvx'/c^2
> = gt' + v(c-v)T/c^2,
> hence t1<>t2 for a same value of t'.

Correct.

> So, if x1'=0 and x2'=(c-v)T/g, for all times in S', then x1=vT
> and x2=cT occur at different times in S
>
> With the Einsteinian LT, there is no coherent solution.

Can you explain that statement? We have obtained a coherent
solution, and you followed the derivation.

>
> The correct LT is
> x' = (x-vt)/g
> t' = t/g

What is your derivation?
Have you noticed that if you apply your equations, first with v,
then with -v (which should get you back to the original system),
your time equation results in
t''=t/g^2
It should be
t''=t

More generally, your equations only represent a group when g=1.

Sorcerer

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Oct 14, 2006, 6:55:36 PM10/14/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:Um9Yg.4110$gM1.617@fed1read12...

Of course t" = t' = t.
What are you trying to say by posting this garbage? 2 = 1?

Androcles

mlut...@wanadoo.fr

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Oct 15, 2006, 8:42:34 AM10/15/06

to

Your solution (the LT solution) is correct, only if you
equate t1' with t2', but then t1 is different from t2. I call
this an incoherent solution.

>
> >
> > The correct LT is
> > x' = (x-vt)/g
> > t' = t/g
> What is your derivation?

I gave it in the first message of this thread. It follows closely
Einstein's derivation,
but doesn't assume that when x=ct, x'=ct'. Here is it again:

Derivation of the correct transformation:
-------------------------------------------------------

Let's consider two frames of reference, S and S', each in uniform
translatory motion relative to the other, the velocity of S' relative
to S being v.

Basis relations:

x' = ax + bt
t' = ex + kt

At the origin of S', x' = 0 and x = vt.
Hence, 0 = (av+b)t, whence b = -av

The basis relations are now

(1) x' = a(x - vt)
(2) t' = ex + kt

Now, let's suppose that a light signal, starting from the coincident

origins of frames S and S' at t = t' = 0, travels toward positive x.

After a time t, it will be at
x = ct + vt, and also at
x' = ct'

Substituting these values of x and x' in relations
(1) and (2), and eliminating t and t', one gets

(3) a - ec - ev - k = 0

If the signal travels toward negative x,
x = -ct + vt and x' = -ct'

Substituting these values of x and x' in relations
(1) and (2), and eliminating t and t', one gets

(4) a + ec - ev - k = 0

From (3) and (4), one gets e = 0

With e = 0, relations (3) or (4) reduce to a = k

Hence, relations (1) and (2) become

(1) x' = k (x - vt)
(2) t' = k t

Now, a light signal follow the y' axis. Relatively to S,
it travels obliquely, for, while the signal goes
a distance ct, the y'-axis advances a distance x = vt.
Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
But, also, y' = ct' = c * k * t.

Equating y' to y, one gets k = sqrt(1 - v^2/c^2)

The transforms obtained without the *bold* Einstein's postulate that

the speed of light is the same in all frames are thus

(5) x' = sqrt(1 - v^2/c^2) * (x - vt) = (x - vt) / g
(6) t' = sqrt(1 - v^2/c^2) * t = t / g,

(g corresponds to Einstein's gamma).

Transform (5) straightforwardly tells us that any body measures shorter

in terms of a frame relative to which it is moving with speed v than
it does as measured in a frame relative to which it is at rest.

Transform (6) implies that when two physical systems are in uniform
relative translation at speed v, the effects produced by system A
on system B are modified just as if all natural processes on A were
slowed down in the ratio sqrt(1 - v^2/c^2). (i.e., the so-called time
"dilation").

Those transforms, contrary to Einstein's LT, don't allow to claim that
simultaneity is relative (i.e., that events that are considered to
be simultaneous in one reference frame are not simultaneous in another
reference frame moving with respect to the first, cf. Wikipedia).

> Have you noticed that if you apply your equations, first with v,
> then with -v (which should get you back to the original system),
> your time equation results in
> t''=t/g^2
> It should be
> t''=t

No, the value of g is independent of the sign of v, as v is squared.

>
> More generally, your equations only represent a group when g=1.

I agree, but why should it represent a group?

Marcel Luttgens

Dirk Van de moortel

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Oct 15, 2006, 9:01:47 AM10/15/06

to

<mlut...@wanadoo.fr> wrote in message news:1160901753.9...@i42g2000cwa.googlegroups.com...

>
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:

[snip]

>> > The correct LT is
>> > x' = (x-vt)/g
>> > t' = t/g
>> What is your derivation?
>
> I gave it in the first message of this thread. It follows closely
> Einstein's derivation,
> but doesn't assume that when x=ct, x'=ct'. Here is it again:
>
> Derivation of the correct transformation:
> -------------------------------------------------------

Ha, so you found a new victim.
I'm sure Brian will love it :-)

Dirk Vdm

mlut...@wanadoo.fr

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Oct 15, 2006, 9:34:55 AM10/15/06

to

Dirk Van de moortel wrote:

Stupid spammer!

Marcel Luttgens

Dirk Van de moortel

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Oct 15, 2006, 9:43:52 AM10/15/06

to

<mlut...@wanadoo.fr> wrote in message news:1160904895.7...@f16g2000cwb.googlegroups.com...

Not a all.
I'm very curious as to whether Brian will find a new way
of confusing you even more :-)
Enjoy, Marcel - I'll be watching closely.

Dirk Vdm

Brian Kennelly

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Oct 15, 2006, 4:01:42 PM10/15/06

to

mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:
>>> Brian Kennelly wrote:
>
> Your solution (the LT solution) is correct, only if you
> equate t1' with t2', but then t1 is different from t2. I call
> this an incoherent solution.

The fact that simultaneity is not invariant is a feature of SR,
and probably the hardest to accept for many people. It follows
from the light postulate.

But, it is not incoherent.

>
>>> The correct LT is
>>> x' = (x-vt)/g
>>> t' = t/g
>> What is your derivation?
>
> I gave it in the first message of this thread. It follows closely
> Einstein's derivation,
> but doesn't assume that when x=ct, x'=ct'. Here is it again:
>
> Derivation of the correct transformation:
> -------------------------------------------------------
>
> Let's consider two frames of reference, S and S', each in uniform
> translatory motion relative to the other, the velocity of S' relative
> to S being v.
>
> Basis relations:
>
> x' = ax + bt
> t' = ex + kt
>
> At the origin of S', x' = 0 and x = vt.
> Hence, 0 = (av+b)t, whence b = -av
>
> The basis relations are now
>
> (1) x' = a(x - vt)
> (2) t' = ex + kt
>
> Now, let's suppose that a light signal, starting from the coincident
>
> origins of frames S and S' at t = t' = 0, travels toward positive x.
>
> After a time t, it will be at
> x = ct + vt, and also at

Where did you get this equation? Why does light travel with
speed c+v in S?

> x' = ct'
>
> Substituting these values of x and x' in relations
> (1) and (2), and eliminating t and t', one gets
>
> (3) a - ec - ev - k = 0
>
> If the signal travels toward negative x,
> x = -ct + vt and x' = -ct'

Again, where did you get the first equation? It appears that
you are assuming that light 'drifts' with speed *v* in S, so you
are using the vector sum of *c* and *v*.

>
> Substituting these values of x and x' in relations
> (1) and (2), and eliminating t and t', one gets
>
> (4) a + ec - ev - k = 0
>
> From (3) and (4), one gets e = 0
>
> With e = 0, relations (3) or (4) reduce to a = k
>
> Hence, relations (1) and (2) become
>
> (1) x' = k (x - vt)
> (2) t' = k t
>
> Now, a light signal follow the y' axis. Relatively to S,
> it travels obliquely, for, while the signal goes
> a distance ct, the y'-axis advances a distance x = vt.
> Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
> But, also, y' = ct' = c * k * t.

Now, you have light travelling at c in both S and S'. (You
derived the y equation by assuming that the signal goes a
distance ct). To be consistent with your two x equations,
assuming vector addition between *c* and *v*, you should have
used y=ct.

>
> Equating y' to y, one gets k = sqrt(1 - v^2/c^2)

With y=ct, you get k=1

>
> The transforms obtained without the *bold* Einstein's postulate that

What assumption are you using in place of that postulate?

>
> the speed of light is the same in all frames are thus
>
> (5) x' = sqrt(1 - v^2/c^2) * (x - vt) = (x - vt) / g
> (6) t' = sqrt(1 - v^2/c^2) * t = t / g,

No, using consistent vector addition you get the Galilean
transformations:
x'=x-vt
t'=t

>
> (g corresponds to Einstein's gamma).
>
> Transform (5) straightforwardly tells us that any body measures shorter
>
> in terms of a frame relative to which it is moving with speed v than
> it does as measured in a frame relative to which it is at rest.
>
> Transform (6) implies that when two physical systems are in uniform
> relative translation at speed v, the effects produced by system A
> on system B are modified just as if all natural processes on A were
> slowed down in the ratio sqrt(1 - v^2/c^2). (i.e., the so-called time
> "dilation").
>
> Those transforms, contrary to Einstein's LT, don't allow to claim that
> simultaneity is relative (i.e., that events that are considered to
> be simultaneous in one reference frame are not simultaneous in another
> reference frame moving with respect to the first, cf. Wikipedia).
>
>
>> Have you noticed that if you apply your equations, first with v,
>> then with -v (which should get you back to the original system),
>> your time equation results in
>> t''=t/g^2
>> It should be
>> t''=t
>
> No, the value of g is independent of the sign of v, as v is squared.

That is why applying your transform twice with v and -v inserts
a g^2 term. The time coordinate is multiplied by the same g in
both directions.

>
>> More generally, your equations only represent a group when g=1.
>
> I agree, but why should it represent a group?

Because the transformations should have the same form for any
value of v. In addition, the inverse of each transform should
have that form.

>
> Marcel Luttgens
>

Sorcerer

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Oct 15, 2006, 8:03:06 PM10/15/06

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"Brian Kennelly" <bwken...@cox.net> wrote in message

news:GXsYg.4152$gM1.2005@fed1read12...

Yes you are, you bloody minded pervert.

mlut...@wanadoo.fr

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Oct 17, 2006, 9:54:04 AM10/17/06

to

Brian Kennelly wrote:
> mlut...@wanadoo.fr wrote:
> > Brian Kennelly wrote:
> >> mlut...@wanadoo.fr wrote:
> >>> Brian Kennelly wrote:
> >
> > Your solution (the LT solution) is correct, only if you
> > equate t1' with t2', but then t1 is different from t2. I call
> > this an incoherent solution.
> The fact that simultaneity is not invariant is a feature of SR,
> and probably the hardest to accept for many people. It follows
> from the light postulate.
>
> But, it is not incoherent.

You probably remember your following derivation:

"If you prefer, we can do it the long way:

x1'=g(vT-vt1)
x2'=g(cT-vt2)
t1'=g(t1-v^2T/c^2) (x1=vT)
t2'=g(t2-vT/c) (x2=cT)

Setting t1'=t2', we get
t2=t1-v^2T/c^2+vT/c

Choosing t1=T, so that x1'=0
t2=T(1-v^2/c^2+v/c)
So:
x2'=g(cT-vT(1-v^2/c^+v/c))
=g(cT-v^2T/c-vT(1-v^2/c^2))
= g(cT(1-v^2/c^2)-vT(1-v^2/c^2))
=g((c-v)T(1/g^2)
=(c-v)T/g

Again, we find that the length is contracted in S'."

Choose t2 = T instead of t1 = T, and you will see that the LT
solution is incoherent.

Thank you for your pertinent analysis.
Contrary to those critics, who transformed this NG into a hornet's
nest,
you adopted a positive attitude, the only one that makes scientific
progress possible.
Anyhow, I probably had a big headache when I wrote my derivation.
Here is the revised version:

Derivation of the correct transformation:
----------------------------------------

Let's consider two frames of reference, S and S', each in uniform

translatory motion relative to the other, the velocity of S' relative
to S being v.

Basis relations:

x' = ax + bt
t' = ex + kt

At the origin of S', x' = 0 and x = vt.
Hence, 0 = (av+b)t, whence b = -av

The basis relations are now

(1) x' = a(x - vt)
(2) t' = ex + kt

Now, let's suppose that a light signal, starting from the coincident

origins of frames S and S' at t = t' = 0, travels toward positive x.
After a time t, it will be at

x = ct, and also at
x' = (c-v)t'
(Notice that Einstein postulated here that x' = ct')

Substituting these values of x and x' in relations

(1) and (2), one gets:

(c-v)t' = a(c-v)t, thus t' = at
t' = (ec+k)t, hence

(3) ec + k -a = 0

If the signal travels toward negative x,

x = -ct and x' = -(c+v)t'
(Here, Einstein postulated that x' = -ct')

Substituting these values of x and x' in relations

(1) and (2), one gets

-(c+v)t' = a(-ct-vt) = -a(c+v)t, thus t' = at
t' = (-ec+k)t, hence

(4) -ec + k - a = 0

From (3) and (4), one gets e = 0 and a = k

Hence, relations (1) and (2) become

(1) x' = k (x - vt)
(2) t' = k t

Now, a light signal follow the y' axis. Relatively to S,
it travels obliquely, for, while the signal goes
a distance ct, the y'-axis advances a distance x = vt.
Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
But, also, y' = ct' = c * k * t.

Equating y' to y, one gets k = sqrt(1 - v^2/c^2)

The transform obtained without the *bold* Einstein's postulate that
the speed of light is the same in all frames is thus

(5) x' = sqrt(1 - v^2/c^2) * (x - vt) = (x - vt) / g
(6) t' = sqrt(1 - v^2/c^2) * t = t / g,

(g corresponds to Einstein's gamma).

The inverse transform is

(7) x = g(x'+vt')
(8) t = gt'

You cannot apply the transform twice.
t' = t/g gives the time in a moving frame wrt the time observed
in a frame at rest.
If you know t', you must use the inverse transform t = gt' to
calculate t, but not apply t/g twice.

>
> >
> >> More generally, your equations only represent a group when g=1.
> >
> > I agree, but why should it represent a group?
> Because the transformations should have the same form for any
> value of v. In addition, the inverse of each transform should
> have that form.

You automatically get a group when you use the Einstein's postulate
to derive the LT. Iow, the existence of such group has no objective
physical justification.

Marcel Luttgens

>
> >
> > Marcel Luttgens
> >

Dirk Van de moortel

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Oct 17, 2006, 2:30:44 PM10/17/06

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<mlut...@wanadoo.fr> wrote in message news:1161078844.8...@h48g2000cwc.googlegroups.com...
>
> Brian Kennelly wrote:

[snip]

>> That is why applying your transform twice with v and -v inserts
>> a g^2 term. The time coordinate is multiplied by the same g in
>> both directions.
>
> You cannot apply the transform twice.
> t' = t/g gives the time in a moving frame wrt the time observed
> in a frame at rest.
> If you know t', you must use the inverse transform t = gt' to
> calculate t, but not apply t/g twice.

And it doesn't even hurt, does it? ;-)

Dirk Vdm

Nico

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Oct 17, 2006, 3:34:39 PM10/17/06

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Dirk Van de moortel ha scritto:
Excuese me sir, may you see my last post " problem: real or immaginary
part " at sci.physics ?

Dirk Van de moortel

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Oct 17, 2006, 4:00:52 PM10/17/06

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"Nico" <nico...@yahoo.it> wrote in message news:1161099279.6...@h48g2000cwc.googlegroups.com...

>
> Dirk Van de moortel ha scritto:
> Excuese me sir, may you see my last post " problem: real or immaginary
> part " at sci.physics ?

Sorry, I had lost sight of it. I replied.

Dirk Vdm

Brian Kennelly

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Oct 20, 2006, 9:49:53 AM10/20/06

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No, if we choose t2=T, then we get
T=t1-v^2T/c^2+vT/c
t1=T+v^2T/c^2-vT/c

x1'=g(vT-vt1)
=g(vT-vT-v*v^2T/c^2+v*vT/c)
=g((c-v)(v^2T/c^2))
x2'=g(cT-vt2')
=g(cT-vT)
=g(c-v)T
Subtracting, we get
x2'-x1'=g(c-v)T(1-v^2/c^2)
=(c-v)T/g

This is the same result we got when t1=T, so the solution is
still coherent.

It appears that, again, you are assuming a vector sum for the
velocity vector.

>
> Substituting these values of x and x' in relations
> (1) and (2), one gets:
>
> (c-v)t' = a(c-v)t, thus t' = at
> t' = (ec+k)t, hence
>
> (3) ec + k -a = 0
>
> If the signal travels toward negative x,
> x = -ct and x' = -(c+v)t'
> (Here, Einstein postulated that x' = -ct')
>
> Substituting these values of x and x' in relations
> (1) and (2), one gets
>
> -(c+v)t' = a(-ct-vt) = -a(c+v)t, thus t' = at
> t' = (-ec+k)t, hence
>
> (4) -ec + k - a = 0
>
> From (3) and (4), one gets e = 0 and a = k
>
> Hence, relations (1) and (2) become
>
> (1) x' = k (x - vt)
> (2) t' = k t
>
> Now, a light signal follow the y' axis. Relatively to S,
> it travels obliquely, for, while the signal goes
> a distance ct, the y'-axis advances a distance x = vt.
> Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
> But, also, y' = ct' = c * k * t.

I don't see where you obtained y'=ct'. If your are using a
vector sum for the velocity, you obtain y'=sqrt(c^2-v^2)t'

>
> Equating y' to y, one gets k = sqrt(1 - v^2/c^2)

Again, using the vector sum for velocity, I get k=1.

>
> The transform obtained without the *bold* Einstein's postulate that
> the speed of light is the same in all frames is thus
>
> (5) x' = sqrt(1 - v^2/c^2) * (x - vt) = (x - vt) / g
> (6) t' = sqrt(1 - v^2/c^2) * t = t / g,
>
> (g corresponds to Einstein's gamma).
>
> The inverse transform is
>
> (7) x = g(x'+vt')
> (8) t = gt'
>
>
>

> You cannot apply the transform twice.
> t' = t/g gives the time in a moving frame wrt the time observed
> in a frame at rest.

So, you are proposing that there is only one "rest frame"?

> If you know t', you must use the inverse transform t = gt' to
> calculate t, but not apply t/g twice.
>
>>>> More generally, your equations only represent a group when g=1.
>>> I agree, but why should it represent a group?
>> Because the transformations should have the same form for any
>> value of v. In addition, the inverse of each transform should
>> have that form.
>
> You automatically get a group when you use the Einstein's postulate
> to derive the LT. Iow, the existence of such group has no objective
> physical justification.
>

If there is not preferred frame, then the transformations
between any two frames, including the inverse, will have the
same form. This is the physical justification for the existence
of a group.

mlut...@wanadoo.fr

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Oct 20, 2006, 1:13:20 PM10/20/06

to

Mathematically coherent, in the sense that when t1'=t2', you
always get x2'-x1' = (c-v)T/g when x1=vT and x2=cT, as long as
you use the relation t2-t1 = T(v/c - v^2/c^2).
But notice that with t2=T, you get x1'=g((c-v)(v^2T/c^2))
instead of zero, which is physically false, because when x1=vT
and x2=cT, the origin of S' is at x1=vT and at x1'=0.
Only setting t1=T leads to a physical coherent result (perhaps
would it be better to say physically 'correct').

Not at all, the factor (c-v) doesn't mean that light speed
is not c anymore. According to S, light is at ct and S' at vt,
hence in S', after a time t, light has travelled a distance
(c-v)t. To express such distance in S', one has to use t', thus
x' = (c-v)t'.

>
> >
> > Substituting these values of x and x' in relations
> > (1) and (2), one gets:
> >
> > (c-v)t' = a(c-v)t, thus t' = at
> > t' = (ec+k)t, hence
> >
> > (3) ec + k -a = 0
> >
> > If the signal travels toward negative x,
> > x = -ct and x' = -(c+v)t'
> > (Here, Einstein postulated that x' = -ct')
> >
> > Substituting these values of x and x' in relations
> > (1) and (2), one gets
> >
> > -(c+v)t' = a(-ct-vt) = -a(c+v)t, thus t' = at
> > t' = (-ec+k)t, hence
> >
> > (4) -ec + k - a = 0
> >
> > From (3) and (4), one gets e = 0 and a = k
> >
> > Hence, relations (1) and (2) become
> >
> > (1) x' = k (x - vt)
> > (2) t' = k t
> >
> > Now, a light signal follow the y' axis. Relatively to S,
> > it travels obliquely, for, while the signal goes
> > a distance ct, the y'-axis advances a distance x = vt.
> > Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
> > But, also, y' = ct' = c * k * t.

> I don't see where you obtained y'=ct'. If your are using a
> vector sum for the velocity, you obtain y'=sqrt(c^2-v^2)t'

Fistly, I don't use a vector sum (see above), and secondly,
there is no physical reason why y' should be different from y,
at least according to Einstein, with whom I agree here.

>
> >
> > Equating y' to y, one gets k = sqrt(1 - v^2/c^2)
> Again, using the vector sum for velocity, I get k=1.

But I didn't!

>
> >
> > The transform obtained without the *bold* Einstein's postulate that
> > the speed of light is the same in all frames is thus
> >
> > (5) x' = sqrt(1 - v^2/c^2) * (x - vt) = (x - vt) / g
> > (6) t' = sqrt(1 - v^2/c^2) * t = t / g,
> >
> > (g corresponds to Einstein's gamma).
> >
> > The inverse transform is
> >
> > (7) x = g(x'+vt')
> > (8) t = gt'
> >
> >
> >
> > You cannot apply the transform twice.
> > t' = t/g gives the time in a moving frame wrt the time observed
> > in a frame at rest.
> So, you are proposing that there is only one "rest frame"?

I didn't say in 'the' frame at rest!

>
> > If you know t', you must use the inverse transform t = gt' to
> > calculate t, but not apply t/g twice.
> >
> >>>> More generally, your equations only represent a group when g=1.
> >>> I agree, but why should it represent a group?
> >> Because the transformations should have the same form for any
> >> value of v. In addition, the inverse of each transform should
> >> have that form.
> >
> > You automatically get a group when you use the Einstein's postulate
> > to derive the LT. Iow, the existence of such group has no objective
> > physical justification.
> >
> If there is not preferred frame, then the transformations
> between any two frames, including the inverse, will have the
> same form. This is the physical justification for the existence
> of a group.

Could you give a concrete exemple using figures?

Marcel Luttgens

Brian Kennelly

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Oct 20, 2006, 3:58:33 PM10/20/06

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mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> No, if we choose t2=T, then we get
>> T=t1-v^2T/c^2+vT/c
>> t1=T+v^2T/c^2-vT/c
>>
>> x1'=g(vT-vt1)
>> =g(vT-vT-v*v^2T/c^2+v*vT/c)
>> =g((c-v)(v^2T/c^2))
>> x2'=g(cT-vt2')
>> =g(cT-vT)
>> =g(c-v)T
>> Subtracting, we get
>> x2'-x1'=g(c-v)T(1-v^2/c^2)
>> =(c-v)T/g
>>
>> This is the same result we got when t1=T, so the solution is
>> still coherent.
>
> Mathematically coherent, in the sense that when t1'=t2', you
> always get x2'-x1' = (c-v)T/g when x1=vT and x2=cT, as long as
> you use the relation t2-t1 = T(v/c - v^2/c^2).
> But notice that with t2=T, you get x1'=g((c-v)(v^2T/c^2))
> instead of zero, which is physically false, because when x1=vT
> and x2=cT, the origin of S' is at x1=vT and at x1'=0.

The stick is at rest in S, so x1=vT and x2=cT at *all* times in
S. The origin of S' is at x1=vT only when t1=T. When t2=T, and
t2'=t1', then x1<>vT, because t1<>T

> Only setting t1=T leads to a physical coherent result (perhaps
> would it be better to say physically 'correct').

No, that would be incorrect, because you are asking S' to mark
ends of the stick at two different times. In between the stick
moves, so you cannot determine the length.

I don't follow.
You state that:
x'=(c-v)t
and
x'=(c-v)t'

From these two equations, we find that t'=t.

Then where do you get y'=ct'?

Einstein got that equation from his light postulate. Because of
that postulate, a light signal travelling along the y' access
must have speed 'c'.

You reject that postulate, so you cannot use it here.

Dirk Van de moortel

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Oct 20, 2006, 4:41:44 PM10/20/06

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<mlut...@wanadoo.fr> wrote in message news:1161350000.6...@e3g2000cwe.googlegroups.com...

Perhaps you should explain what you think an event is,
and what you think coordinates of an event are.
Perhaps then someone could help you understand why
after a decade on Usenet you still haven't got a clue what
you are talking about.
You see, before you start juggling with *transformation*
equations, you should at least have some kind of grasp of
what it is that is being transformed to begin with.
Embarrassment? Ever heard of it?

Dirk Vdm

mlut...@wanadoo.fr

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Oct 20, 2006, 5:41:50 PM10/20/06

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In the first scenario, the light arrived at ct and S' arrived at vt
after some
specific time t.
>From x' = g(x-vt), x'1 = g(x1-vt), hence x' =0 as x1=vt..
Aftrwards, you complicated the problem by introducing t1 and t2.

>
> > Only setting t1=T leads to a physical coherent result (perhaps
> > would it be better to say physically 'correct').

> No, that would be incorrect, because you are asking S' to mark
> ends of the stick at two different times. In between the stick
> moves, so you cannot determine the length.
>

Following the original scenario, the only correct result is obtained
with t1 = T. But I agree that from t2-t1 = T(v/c - v^2/c^2), one gets
an infinity of other solutions.

I thought that my sentence was clear enough: According to S, etc...
Thus, I stated that x = (c-v)t and x' = (c-v)t'.

I don't reject the hypothesis that the speed of light is independent
of the motion of its source.

But I keep claiming that the Einsteinian LT are false:

Here again is a classical explanation why a moving stick appears
contracted to an observer at rest.
Remember that you considered it as correct.

BEGIN

Consider a stick which, when at rest in S, has a length Lo in the
direction of the x-axis.

Let the stick be set moving relative to S at such velocity that it is
at rest in S'. Its length as measured in S' will still be Lo,
because it must have a certain fixed value in any frame in which
the stick is at rest.

Let us see how the length now measure in S, relative to which the
stick is moving with a velocity v.
It seems reasonable to define the length as the distance between
two points fixed in S, which are occupied by the ends of the stick
simultaneously, i.e., at the same time t.
If the coordinates of these points are x1 and x2, the length is
then L = x2 - x1.

Since the stick is at rest in S', its ends have fixed coordinates
x1', x2' such as Lo = x2' - x1'.

If one substitutes in this last equation values of x2' and x1'
calculated from the LT x' = g(x - vt), one obtains, for a given
value of t,

x2' = g(x2 - vt)
x1' = g(x1 - vt)

Lo = g(x2 - x1) = gL, or L = Lo/g

END

Such demonstration amounts to sophistry.

Indeed, it began by calling Lo the length of a stick at rest in S,
whose end coordinates are x1 and x2. Now, it calls deceitfully
that length L.
To avoid confusion, such length should be called Lo(S).

Moreover, it calls Lo the length x2'-x1', i.e. the length of the
stick as measured in S'. Again, to avoid confusion, let's call
such length Lo(S').

So, the end formula should be written

Lo(S') = gLo(S),

meaning that, according to the Einsteinian LT x' = g(x - vt), a moving
stick is *dilated* relative to a stick at rest.

As the MMX has shown that the moving stick is in fact *contracted*,
one has to conclude that the Einsteinian LT is false.

Marcel Luttgens

mlut...@wanadoo.fr

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Oct 20, 2006, 6:08:23 PM10/20/06

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I would like to make clearer why, after a time t, the light signal will
be at
x' = (c-v)t' in S'.
Indeed, according to S, the distance between the endpoint of the signal
and the origin of S' is ct - vt after a time t. In S', the
corresponding distance
is (c-v)t', hence the S'-coordinate of the endpoint is x' = (c-v)t'.
In S, the endpoint is of course at x = ct.
But I am convinced that you already grasped what I meant.

Marcel Luttgens

Brian Kennelly

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Oct 20, 2006, 6:09:47 PM10/20/06

to

Where does the light come in? The ends of the stick are at vT
and cT at *all* times in S.
The origin of S' is at vT only when t=T (and t'=T/g).

>>From x' = g(x-vt), x'1 = g(x1-vt), hence x' =0 as x1=vt..
> Aftrwards, you complicated the problem by introducing t1 and t2.

No, you complicated the problem by ignoring them, and expecting
S' to mark the ends at two different times.

S and S' measure time differently.

>
>>> Only setting t1=T leads to a physical coherent result (perhaps
>>> would it be better to say physically 'correct').
>
>> No, that would be incorrect, because you are asking S' to mark
>> ends of the stick at two different times. In between the stick
>> moves, so you cannot determine the length.
>>
>
> Following the original scenario, the only correct result is obtained
> with t1 = T. But I agree that from t2-t1 = T(v/c - v^2/c^2), one gets
> an infinity of other solutions.

To measure the stick in S', you must mark the ends at the same
time in S'. You get the correct result for any single value of
t'.

>
>>>>>
>>> Not at all, the factor (c-v) doesn't mean that light speed
>>> is not c anymore. According to S, light is at ct and S' at vt,
>>> hence in S', after a time t, light has travelled a distance
>>> (c-v)t. To express such distance in S', one has to use t', thus
>>> x' = (c-v)t'.
>
>> I don't follow.
>> You state that:
>> x'=(c-v)t
>> and
>> x'=(c-v)t'
>>
>> From these two equations, we find that t'=t.
>>
>
> I thought that my sentence was clear enough: According to S, etc...
> Thus, I stated that x = (c-v)t and x' = (c-v)t'.

If you look at the text quoted above, you stated > in S', after

a time t, light has travelled a distance
> (c-v)t.

Does that not say that x'=(c-v)t?

>
>>>>> Now, a light signal follow the y' axis. Relatively to S,
>>>>> it travels obliquely, for, while the signal goes
>>>>> a distance ct, the y'-axis advances a distance x = vt.
>>>>> Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
>>>>> But, also, y' = ct' = c * k * t.
>>>> I don't see where you obtained y'=ct'. If your are using a
>>>> vector sum for the velocity, you obtain y'=sqrt(c^2-v^2)t'
>>> Fistly, I don't use a vector sum (see above), and secondly,
>>> there is no physical reason why y' should be different from y,
>>> at least according to Einstein, with whom I agree here.
>
>> Then where do you get y'=ct'?
>>
>> Einstein got that equation from his light postulate. Because of
>> that postulate, a light signal travelling along the y' access
>> must have speed 'c'.
>>
>> You reject that postulate, so you cannot use it here.
>
> I don't reject the hypothesis that the speed of light is independent
> of the motion of its source.

That part of the postulate was not used to derive y'=ct'. The
primary assertion of the postulate is that the speed of light is
a universal constant; it has the same value for every observer,
in every direction.

You have not explained how you derived y'=ct'.

No, it states that the length of the stick, in the system in
which it is at rest, is Lo.

> To avoid confusion, such length should be called Lo(S).

Only if the stick is at rest in S.

>
> Moreover, it calls Lo the length x2'-x1', i.e. the length of the
> stick as measured in S'. Again, to avoid confusion, let's call
> such length Lo(S').
>
> So, the end formula should be written
>
> Lo(S') = gLo(S),

No, you stated that the stick was at rest in S', so Lo(S')
represents the rest length of the stick: Lo. You did not define
Lo(S) when the stick is in motion, but we will simply call it L.
Lo=gL
L=Lo/g

>
> meaning that, according to the Einsteinian LT x' = g(x - vt), a moving
> stick is *dilated* relative to a stick at rest.

No, the stick is at rest in S', so it is moving in S. The
moving stick is shorter that at rest.

Dirk Van de moortel

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Oct 20, 2006, 6:17:49 PM10/20/06

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<mlut...@wanadoo.fr> wrote in message news:1161367703....@e3g2000cwe.googlegroups.com...

[snip]

> I would like to make clearer why, after a time t, the light signal will
> be at
> x' = (c-v)t' in S'.

A typical Luttgens statement :-)
You either say "After a time t, the signal is at some x in S",
or you say "After a time t', the signal is at some x' in S'".

> Indeed, according to S, the distance between the endpoint of the signal
> and the origin of S' is ct - vt after a time t. In S', the
> corresponding distance
> is (c-v)t',

In S' the corresponding distance is c t'.

> hence the S'-coordinate of the endpoint is x' = (c-v)t'.
> In S, the endpoint is of course at x = ct.

and in S' the end point is at x' = c t'.

> But I am convinced that you already grasped what I meant.

We are convinced that you don't grasp the first letter of
what you say :-)

Dirk Vdm

Brian Kennelly

unread,

Oct 20, 2006, 6:19:09 PM10/20/06

to

mlut...@wanadoo.fr wrote:
> mlut...@wanadoo.fr wrote:
>> Brian Kennelly wrote:

I don't follow this step. The distance between the light signal
and the origin of S' is (c-v)t in S. How do you get from there
to the distance in S' having a distance (c-v)t'.

Are you assuming, a priori, that the relative velocity is invariant?

> hence the S'-coordinate of the endpoint is x' = (c-v)t'.
> In S, the endpoint is of course at x = ct.
> But I am convinced that you already grasped what I meant.

Yes, but I did not follow the leap to x'=(c-v)t'.

mlut...@wanadoo.fr

unread,

Oct 20, 2006, 9:18:34 PM10/20/06

to

Dirk Van de moortel wrote:

> <mlut...@wanadoo.fr> wrote in message news:1161367703....@e3g2000cwe.googlegroups.com...
>
> [snip]
>
> > I would like to make clearer why, after a time t, the light signal will
> > be at
> > x' = (c-v)t' in S'.
>
> A typical Luttgens statement :-)
> You either say "After a time t, the signal is at some x in S",
> or you say "After a time t', the signal is at some x' in S'".
>
> > Indeed, according to S, the distance between the endpoint of the signal
> > and the origin of S' is ct - vt after a time t. In S', the
> > corresponding distance
> > is (c-v)t',
>
> In S' the corresponding distance is c t'.
>
> > hence the S'-coordinate of the endpoint is x' = (c-v)t'.
> > In S, the endpoint is of course at x = ct.
>
> and in S' the end point is at x' = c t'.

I knew that you were stupid, but not to that point!
Your x' = ct' is the Einstein's postulate!
Unless you are a masochist, but more probably a sadomasochist!

Marcel Luttgens

mlut...@wanadoo.fr

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Oct 20, 2006, 9:27:17 PM10/20/06

to

Are you assuming that it is not invariant?

Marcel Luttgens

mlut...@wanadoo.fr

unread,

Oct 20, 2006, 10:42:26 PM10/20/06

to

Yes, that was intended in the scenario, reread the first posts of the
sadomaso,
even *he* had no doubt about this. He was the first to specify "at some
specific
time", which means, of course, t = T.

>
> >>From x' = g(x-vt), x'1 = g(x1-vt), hence x' =0 as x1=vt..
> > Aftrwards, you complicated the problem by introducing t1 and t2.

> No, you complicated the problem by ignoring them, and expecting
> S' to mark the ends at two different times.
>
> S and S' measure time differently.
>

Of course, in the general case, but not in the scenario.

> >
> >>> Only setting t1=T leads to a physical coherent result (perhaps
> >>> would it be better to say physically 'correct').
> >
> >> No, that would be incorrect, because you are asking S' to mark
> >> ends of the stick at two different times. In between the stick
> >> moves, so you cannot determine the length.
> >>
> >
> > Following the original scenario, the only correct result is obtained
> > with t1 = T. But I agree that from t2-t1 = T(v/c - v^2/c^2), one gets
> > an infinity of other solutions.
>
> To measure the stick in S', you must mark the ends at the same
> time in S'. You get the correct result for any single value of
> t'.

And the correct result is straightforwardly given with my
transformation.

> >
> >>>>>
> >>> Not at all, the factor (c-v) doesn't mean that light speed
> >>> is not c anymore. According to S, light is at ct and S' at vt,
> >>> hence in S', after a time t, light has travelled a distance
> >>> (c-v)t. To express such distance in S', one has to use t', thus
> >>> x' = (c-v)t'.
> >
> >> I don't follow.
> >> You state that:
> >> x'=(c-v)t
> >> and
> >> x'=(c-v)t'
> >>
> >> From these two equations, we find that t'=t.
> >>
> >
> > I thought that my sentence was clear enough: According to S, etc...
> > Thus, I stated that x = (c-v)t and x' = (c-v)t'.
> If you look at the text quoted above, you stated > in S', after
> a time t, light has travelled a distance
> > (c-v)t.
> Does that not say that x'=(c-v)t?

The answer is given in mine (and your) almost simultaneous post.

> >>>>> Now, a light signal follow the y' axis. Relatively to S,
> >>>>> it travels obliquely, for, while the signal goes
> >>>>> a distance ct, the y'-axis advances a distance x = vt.
> >>>>> Thus c^2t^2 = v^2t^2 + y^2, whence y = sqrt(c^2 - v^2) * t.
> >>>>> But, also, y' = ct' = c * k * t.
> >>>> I don't see where you obtained y'=ct'. If your are using a
> >>>> vector sum for the velocity, you obtain y'=sqrt(c^2-v^2)t'
> >>> Fistly, I don't use a vector sum (see above), and secondly,
> >>> there is no physical reason why y' should be different from y,
> >>> at least according to Einstein, with whom I agree here.
> >
> >> Then where do you get y'=ct'?
> >>
> >> Einstein got that equation from his light postulate. Because of
> >> that postulate, a light signal travelling along the y' access
> >> must have speed 'c'.
> >>
> >> You reject that postulate, so you cannot use it here.
> >
> > I don't reject the hypothesis that the speed of light is independent
> > of the motion of its source.

> That part of the postulate was not used to derive y'=ct'. The
> primary assertion of the postulate is that the speed of light is
> a universal constant; it has the same value for every observer,
> in every direction.

I agree that you accurately quoted the fanciful Einstein's postulate.
I disagree with your first sentence.

>
> You have not explained how you derived y'=ct'.
>

y' = ct' is a direct consequence of the hypothesis that c is
independent
of the motion of S'.

Yes, it states that its length is Lo when it is at rest in S, and Lo
when it is at rest in S'.

>
> > To avoid confusion, such length should be called Lo(S).

> Only if the stick is at rest in S.

And Lo(S') when it is at rest in S'. Notice that Lo(S') = Lo(S),
according
to the premise.

>
> >
> > Moreover, it calls Lo the length x2'-x1', i.e. the length of the
> > stick as measured in S'. Again, to avoid confusion, let's call
> > such length Lo(S').
> >
> > So, the end formula should be written
> >
> > Lo(S') = gLo(S),

> No, you stated that the stick was at rest in S', so Lo(S')
> represents the rest length of the stick: Lo. You did not define
> Lo(S) when the stick is in motion, but we will simply call it L.
> Lo=gL
> L=Lo/g

You should closely follow the reasoning:

"Let us see how the length now measure in S, relative to which the
stick is moving with a velocity v.
It seems reasonable to define the length as the distance between
two points fixed in S, which are occupied by the ends of the stick
simultaneously, i.e., at the same time t.
If the coordinates of these points are x1 and x2, the length is
then L = x2 - x1."

We agreed to call the length of the stick at rest in S, Lo(S).
Now, it is called L. Iow, L = Lo(S).

"Since the stick is at rest in S', its ends have fixed coordinates
x1', x2' such as Lo = x2' - x1'."

We also agreed to call Lo(S') the length of the stick at rest in S'.
Hence, Lo = x2' - x1' sould be written Lo(S') = x2' - x1'.

"If one substitutes in this last equation values of x2' and x1'
calculated from the LT x' = g(x - vt), one obtains, for a given
value of t,

x2' = g(x2 - vt)
x1' = g(x1 - vt)

Lo = g(x2 - x1) = gL, or L = Lo/g"

Logically, as Lo represents in fact Lo(S') and (x2-x1) represents
Lo(S),
Lo = g(x2 - x1) = gL should be written Lo(S') = gLo(S), meaning, as I
said,
that the length of a moving stick is *dilated* relative to the length
of the
same stick at rest.

I realize that it is no easy to detect the logical 'hocus pocus' that
Einstein had to use in order to sell his postulate that "that the speed

of light is a universal constant; it has the same value for every
observer,
in every direction."

>
> >

> > meaning that, according to the Einsteinian LT x' = g(x - vt), a moving
> > stick is *dilated* relative to a stick at rest.

> No, the stick is at rest in S', so it is moving in S. The
> moving stick is shorter that at rest.

Are you an Einstein's integrist? I hope not.

Marcel Luttgens

Brian Kennelly

unread,

Oct 20, 2006, 11:13:08 PM10/20/06

to

I am not making any assumption about it. I am trying to follow
your derivation, and determine your assumptions.

With Einstein's light postulate, relative velocity is not
invariant. Because you do not accept that postulate, I expect
that you have another idea about how to compare the two systems.
Assuming that relative velocity is invariant is one such bridge.

Brian Kennelly

unread,

Oct 20, 2006, 11:25:34 PM10/20/06

to

mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:
>>> Brian Kennelly wrote:

>>>> mlut...@wanadoo.fr wrote:
>
>>>>> Not at all, the factor (c-v) doesn't mean that light speed
>>>>> is not c anymore. According to S, light is at ct and S' at vt,
>>>>> hence in S', after a time t, light has travelled a distance
>>>>> (c-v)t. To express such distance in S', one has to use t', thus
>>>>> x' = (c-v)t'.
>>>> I don't follow.
>>>> You state that:
>>>> x'=(c-v)t
>>>> and
>>>> x'=(c-v)t'
>>>>
>>>> From these two equations, we find that t'=t.
>>>>
>>> I thought that my sentence was clear enough: According to S, etc...
>>> Thus, I stated that x = (c-v)t and x' = (c-v)t'.
>> If you look at the text quoted above, you stated > in S', after
>> a time t, light has travelled a distance
>>> (c-v)t.
>> Does that not say that x'=(c-v)t?
>
> The answer is given in mine (and your) almost simultaneous post.

As far as I can see, you answered in the affirmative, that
x'=(c-v)t. From that and your postulated x'=(c-v)t', we derive
t'=t when x=ct.

>>>> Einstein got that equation from his light postulate. Because of
>>>> that postulate, a light signal travelling along the y' access
>>>> must have speed 'c'.
>>>>
>>>> You reject that postulate, so you cannot use it here.
>>> I don't reject the hypothesis that the speed of light is independent
>>> of the motion of its source.
>
>> That part of the postulate was not used to derive y'=ct'. The
>> primary assertion of the postulate is that the speed of light is
>> a universal constant; it has the same value for every observer,
>> in every direction.
>
> I agree that you accurately quoted the fanciful Einstein's postulate.
> I disagree with your first sentence.
>
>> You have not explained how you derived y'=ct'.
>>
>
> y' = ct' is a direct consequence of the hypothesis that c is
> independent
> of the motion of S'.

Then why do you reject x'=ct'? If c is independent of the
motion of S', it follows as directly as y'=ct'.

If you disagree, please explain how y'=ct' is a direct
consequence, but x'=ct' is not.

>
> You should closely follow the reasoning:
>
> "Let us see how the length now measure in S, relative to which the
> stick is moving with a velocity v.
> It seems reasonable to define the length as the distance between
> two points fixed in S, which are occupied by the ends of the stick
> simultaneously, i.e., at the same time t.
> If the coordinates of these points are x1 and x2, the length is
> then L = x2 - x1."
>
> We agreed to call the length of the stick at rest in S, Lo(S).
> Now, it is called L. Iow, L = Lo(S).

Don't lose track of the fact that you are looking for the length
of the stick in motion in S. This is not Lo(S), according to
your definition.

>
> "Since the stick is at rest in S', its ends have fixed coordinates
> x1', x2' such as Lo = x2' - x1'."
>
> We also agreed to call Lo(S') the length of the stick at rest in S'.
> Hence, Lo = x2' - x1' sould be written Lo(S') = x2' - x1'.
>
> "If one substitutes in this last equation values of x2' and x1'
> calculated from the LT x' = g(x - vt), one obtains, for a given
> value of t,
>
> x2' = g(x2 - vt)
> x1' = g(x1 - vt)
>
> Lo = g(x2 - x1) = gL, or L = Lo/g"
>
> Logically, as Lo represents in fact Lo(S') and (x2-x1) represents
> Lo(S),

No, according to your definition, Lo(S) is the length of the
stick, if it were at rest in S. You cannot assume that it is
the same as L, the length of the stick in motion.

> Lo = g(x2 - x1) = gL should be written Lo(S') = gLo(S), meaning, as I
> said,
> that the length of a moving stick is *dilated* relative to the length
> of the
> same stick at rest.

You cannot use that equation, because we did not agree that
L=Lo(S). In fact, you just demonstrated that it is not equal.

>
> Are you an Einstein's integrist? I hope not.

I am not familiar with that term.

Brian Kennelly

unread,

Oct 21, 2006, 1:37:49 AM10/21/06

to

mlut...@wanadoo.fr wrote:
> Yes, that was intended in the scenario, reread the first posts of the
> sadomaso,
> even *he* had no doubt about this. He was the first to specify "at some
> specific
> time", which means, of course, t = T.

Reading the early posts, I got the impression that you were
describing the distance between the light signal and another
point moving at a constant velocity.

Of course, in that case, the length is increasing with time, so
it is not straightforward to compare the lengths between systems.

The answer is that the distance at any given time in S is
(c-v)t, and the distance at any given time in S' is ct'. *But*,
the two distances do not correspond to the same fixed length,
because the times cannot be directly compared.

If we pick time t'=T', and calculate the locations of the origin
of S' and the light signal in the S system, we get:
x1=g(vT')
x2=g(cT'+vT')

If these points correspond to the ends of an interval at rest in
S, the length is x2-x1=gcT' or cT'=(x2-x1)/g.

So, the length in S', cT', is contracted when compared to a
fixed length in S whose endpoints correspond to the ends of the
interval at time T'.

If you measure the distance in S, at time T, and compare it to
an interval at rest in S', you get:
x1'=g(vT-vT)
=0
x2'=g(cT-vT)
=g(c-v)T

So
x2'-x1'=g(c-v)T
or (c-v)T=(x2'-x1')/g
So, the length in S is contracted when compared to a fixed
length in S' whose endpoints correspond to the ends of the
interval at time T.

The situation is entirely symmetrical, and in both cases, the
length is contracted when compared to a fixed length at rest in
the other system.

>
>>> >From x' = g(x-vt), x'1 = g(x1-vt), hence x' =0 as x1=vt..
>>> Aftrwards, you complicated the problem by introducing t1 and t2.
>
>> No, you complicated the problem by ignoring them, and expecting
>> S' to mark the ends at two different times.
>>
>> S and S' measure time differently.
>>
>
> Of course, in the general case, but not in the scenario.

Please explain.

mlut...@wanadoo.fr

unread,

Oct 22, 2006, 9:34:46 AM10/22/06

to

Yes, x'=(c-v)t *according* to S, but *in S'* -as S' is moving at v
relative to S-, its time should be t'<>t, hence x'=(c-v)t'.
Notice that x' represents the difference between the endpoint of
the light signal and the origin of S', the same endpoint having
the coordinate x=ct in S.
At this point of my derivation, the link between t' and t
is not yet known.

You found yourself such relations in your demonstration

"If you prefer, we can do it the long way:

Setting t1'=t2', choosing t1=T, so that x1'=0, etc.
x2 = cT
x2'=(c-v)T/g",
where T/g represents T'.

>
> >>>> Einstein got that equation from his light postulate. Because of
> >>>> that postulate, a light signal travelling along the y' access
> >>>> must have speed 'c'.
> >>>>
> >>>> You reject that postulate, so you cannot use it here.
> >>> I don't reject the hypothesis that the speed of light is independent
> >>> of the motion of its source.
> >
> >> That part of the postulate was not used to derive y'=ct'. The
> >> primary assertion of the postulate is that the speed of light is
> >> a universal constant; it has the same value for every observer,
> >> in every direction.
> >
> > I agree that you accurately quoted the fanciful Einstein's postulate.
> > I disagree with your first sentence.
> >
> >> You have not explained how you derived y'=ct'.
> >>
> >
> > y' = ct' is a direct consequence of the hypothesis that c is
> > independent
> > of the motion of S'.
> Then why do you reject x'=ct'? If c is independent of the
> motion of S', it follows as directly as y'=ct'.
>
> If you disagree, please explain how y'=ct' is a direct
> consequence, but x'=ct' is not.

As shown above, one has x'=(c-v)t', not x'=ct'.
The case of y' is wholly different, because the light signal
follows the y' axis, which is moving at v wrt the y axis.
Iow, the vector v is perpendicular to the vector c, not
parallel as in the case of x,x'.

> > You should closely follow the reasoning:
> >
> > "Let us see how the length now measure in S, relative to which the
> > stick is moving with a velocity v.
> > It seems reasonable to define the length as the distance between
> > two points fixed in S, which are occupied by the ends of the stick
> > simultaneously, i.e., at the same time t.
> > If the coordinates of these points are x1 and x2, the length is
> > then L = x2 - x1."
> >
> > We agreed to call the length of the stick at rest in S, Lo(S).
> > Now, it is called L. Iow, L = Lo(S).

> Don't lose track of the fact that you are looking for the length
> of the stick in motion in S. This is not Lo(S), according to
> your definition.

You are right, it should be called Lo(S)'.

> >
> > "Since the stick is at rest in S', its ends have fixed coordinates
> > x1', x2' such as Lo = x2' - x1'."
> >
> > We also agreed to call Lo(S') the length of the stick at rest in S'.
> > Hence, Lo = x2' - x1' sould be written Lo(S') = x2' - x1'.

This is correct.

> >
> > "If one substitutes in this last equation values of x2' and x1'
> > calculated from the LT x' = g(x - vt), one obtains, for a given
> > value of t,
> >
> > x2' = g(x2 - vt)
> > x1' = g(x1 - vt)
> >
> > Lo = g(x2 - x1) = gL, or L = Lo/g"
> >
> > Logically, as Lo represents in fact Lo(S') and (x2-x1) represents
> > Lo(S),

> No, according to your definition, Lo(S) is the length of the
> stick, if it were at rest in S. You cannot assume that it is
> the same as L, the length of the stick in motion.

So, we have Lo = Lo(S'), which is the length of the stick at rest in
S',
and L = Lo(S)', which is its length in S when S is moving wrt S'.
L = Lo/g could thus be written
Lo(S)' = Lo(S')/g.
As Lo(S) = Lo(S'), this relation means that the length of a moving
stick
is shortenend by 1/g relative to a stick at rest.

>
> > Lo = g(x2 - x1) = gL should be written Lo(S') = gLo(S), meaning, as I
> > said,
> > that the length of a moving stick is *dilated* relative to the length
> > of the
> > same stick at rest.
> You cannot use that equation, because we did not agree that
> L=Lo(S). In fact, you just demonstrated that it is not equal.
>

Yes.

Marcel Luttgens

Dirk Van de moortel

unread,

Oct 22, 2006, 9:56:14 AM10/22/06

to

<mlut...@wanadoo.fr> wrote in message news:1161509686.0...@e3g2000cwe.googlegroups.com...

An expression like "its time should be" shows that you
have no idea what events and coordinates are.

> Notice that x' represents the difference between the endpoint of
> the light signal and the origin of S', the same endpoint having
> the coordinate x=ct in S.
> At this point of my derivation, the link between t' and t
> is not yet known.

At no point in your derivation, the meanings of x, t, x' and t'
is understood by you :-)

Dirk Vdm

mlut...@wanadoo.fr

unread,

Oct 22, 2006, 11:02:46 AM10/22/06

to

Brian Kennelly wrote:
> mlut...@wanadoo.fr wrote:
> > Yes, that was intended in the scenario, reread the first posts of the
> > sadomaso,
> > even *he* had no doubt about this. He was the first to specify "at some
> > specific
> > time", which means, of course, t = T.
> Reading the early posts, I got the impression that you were
> describing the distance between the light signal and another
> point moving at a constant velocity.
>
> Of course, in that case, the length is increasing with time, so
> it is not straightforward to compare the lengths between systems.
>

The length is indeed increasing with time, but the scenario
stated that after a specific time t=T, x is at cT and x' at g(c-v)T.

>
> The answer is that the distance at any given time in S is
> (c-v)t, and the distance at any given time in S' is ct'.

This is according to Einstein. As for me, x'=(c-v)t', not ct'.
But I will accept ct' for the sake of this discussion.

> *But*,
> the two distances do not correspond to the same fixed length,
> because the times cannot be directly compared.
>
> If we pick time t'=T', and calculate the locations of the origin
> of S' and the light signal in the S system, we get:
> x1=g(vT')
> x2=g(cT'+vT')

With the LT, one is obliged to start from the moving frame
to get meaningful results. This amounts to consider that the
S frame is moving away at v relative to the S' frame at rest.
We did the same with the sticks in our other post ("Let us
see how the length now measure in S, etc.").
As the LT have been derived by assuming that S' is the moving
frame, this appears contradictory.

>
> If these points correspond to the ends of an interval at rest in
> S, the length is x2-x1=gcT' or cT'=(x2-x1)/g.

You have in fact calculated the length in S by assuming that
S is the moving frame.

>
> So, the length in S', cT', is contracted when compared to a
> fixed length in S whose endpoints correspond to the ends of the
> interval at time T'.

And this is the result of a sophism.

>
> If you measure the distance in S, at time T, and compare it to
> an interval at rest in S', you get:
> x1'=g(vT-vT)
> =0
> x2'=g(cT-vT)
> =g(c-v)T

The equations represent the coordinates in S' of the origin
of S' and of the light endpoint, when S' is moving away from S
at v.

>
> So
> x2'-x1'=g(c-v)T
> or (c-v)T=(x2'-x1')/g

g(c-v)T is a length measured in the moving frame S'.
The corresponding length measured in the rest frame S is (c-v)T.
The length comoving with S' is thus *dilated* relative to the length
'at rest'. Conversely, the length 'at rest' is contracted
relative to the 'moving' length.

> So, the length in S is contracted when compared to a fixed
> length in S' whose endpoints correspond to the ends of the
> interval at time T.

Here, you are implying that S is the moving frame.

>
> The situation is entirely symmetrical, and in both cases, the
> length is contracted when compared to a fixed length at rest in
> the other system.

The correct conclusion is the opposite: a length at rest is contracted
when compared to a moving length.
This is a direct consequence ot the LT.

>
> >
> >>> >From x' = g(x-vt), x'1 = g(x1-vt), hence x' =0 as x1=vt..
> >>> Aftrwards, you complicated the problem by introducing t1 and t2.
> >
> >> No, you complicated the problem by ignoring them, and expecting
> >> S' to mark the ends at two different times.
> >>
> >> S and S' measure time differently.
> >>
> >
> > Of course, in the general case, but not in the scenario.
> Please explain.

I think that the matter is amply settled above.

Marcel Luttgens

Sorcerer

unread,

Oct 22, 2006, 11:38:29 AM10/22/06

to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote
in message news:2fH_g.142487$m11.2...@phobos.telenet-ops.be...

| At no point in your derivation, the meanings of x, t, x' and t'
| is understood by you :-)

is - singular.
are - plural.
Didn't they teach you to write English in Belgium?
How old are you?

Brian Kennelly

unread,

Oct 22, 2006, 5:17:00 PM10/22/06

to

You have <x-vt=(c-v)t> when x=ct. You previously derived
<x'=a(x-vt)> From these, we find <x'=a(c-v)t> when x=ct.

You then assert that <x'=(c-v)t'> when x=ct. I assume that this
is based on keeping the relative velocity invariant for the
light signal.

This yields <t'=at> when x=ct.

*******************************************************
Note that if the origin of S is moving to the left at v,
according to S', we can derive:
(-vt')=a(-vt)
t'=at when x=0
Using your time equation, we get
at=kt
So, a=k

From that result, and the previous equation, we find
at=ect+at
e=0
*******************************************************

Generally, if the relative velocity is invariant for three
different velocities, the transformation equations become:
x'=a(x-vt)
t'=at

We can see that all relative velocities in the x direction are
invariant.

In your original derivation, you required this condition for v,
c and -c. I derived it from 0, v, and c. (There is nothing
special about c in this case.)

> Notice that x' represents the difference between the endpoint of
> the light signal and the origin of S', the same endpoint having
> the coordinate x=ct in S.
> At this point of my derivation, the link between t' and t
> is not yet known.

Actually, your equation requires that t'=at when x=ct, and, as I
showed, t'=at for all transformations.

>>>> You have not explained how you derived y'=ct'.
>>>>
>>> y' = ct' is a direct consequence of the hypothesis that c is
>>> independent
>>> of the motion of S'.
>> Then why do you reject x'=ct'? If c is independent of the
>> motion of S', it follows as directly as y'=ct'.
>>
>> If you disagree, please explain how y'=ct' is a direct
>> consequence, but x'=ct' is not.
>
> As shown above, one has x'=(c-v)t', not x'=ct'.
> The case of y' is wholly different, because the light signal
> follows the y' axis, which is moving at v wrt the y axis.
> Iow, the vector v is perpendicular to the vector c, not
> parallel as in the case of x,x'.

You had <y=sqrt(c^2-v^2)t> and <x=vt> for the light signal along
the y' axis.

(Assuming y'=y, from symmetry arguments, and using A for the
sqrt to shorten the equations)
y'=At
x'=0
t'=at

y'=At'/a

Now the relative velocity between the signal and the S' origin
is A. You assert that in S', it is c, so relative velocity is
invariant along the x axis, but not in other directions? Why is
the speed 'c' along the y' axis in S'? At what angle does the
relative velocity cease to be invariant, and the light signal
move with speed 'c'?

If we assume that relative velocity is invariant, we get
<y'=At'>, which results in a=1.

Sorcerer

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Oct 22, 2006, 5:58:15 PM10/22/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:3IN_g.6010$gM1.2754@fed1read12...

I knew you couldn't go the distance.

x'=At
y'=0
t'=at

x'=At'/a, which results in a=1.

Hahahaha! Fuckwit.
Androcles

Brian Kennelly

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Oct 22, 2006, 8:19:40 PM10/22/06

to

mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:
>>> Yes, that was intended in the scenario, reread the first posts of the
>>> sadomaso,
>>> even *he* had no doubt about this. He was the first to specify "at some
>>> specific
>>> time", which means, of course, t = T.
>> Reading the early posts, I got the impression that you were
>> describing the distance between the light signal and another
>> point moving at a constant velocity.
>>
>> Of course, in that case, the length is increasing with time, so
>> it is not straightforward to compare the lengths between systems.
>>
>
> The length is indeed increasing with time, but the scenario
> stated that after a specific time t=T, x is at cT and x' at g(c-v)T.
>

>> The answer is that the distance at any given time in S is
>> (c-v)t, and the distance at any given time in S' is ct'.
>
> This is according to Einstein. As for me, x'=(c-v)t', not ct'.
> But I will accept ct' for the sake of this discussion.

You are trying to find a contradiction in SR, so you must work
with the LT. From them, we get <x'=ct'>.

>
>> *But*,
>> the two distances do not correspond to the same fixed length,
>> because the times cannot be directly compared.
>>
>> If we pick time t'=T', and calculate the locations of the origin
>> of S' and the light signal in the S system, we get:
>> x1=g(vT')
>> x2=g(cT'+vT')
>
> With the LT, one is obliged to start from the moving frame
> to get meaningful results.

No, you can start from either frame. The forward and inverse LT
equations contain the same physical and mathematical content.
You use whichever equations are simplest. I started from the
inverse equations, because we started with quantities known in
S', and wanted to find the corresponding coordinates in S.

Each frame is moving with respect to the other, so there is no
simple meaning to the phrase "start from the moving frame".

> This amounts to consider that the
> S frame is moving away at v relative to the S' frame at rest.
> We did the same with the sticks in our other post ("Let us
> see how the length now measure in S, etc.").
> As the LT have been derived by assuming that S' is the moving
> frame, this appears contradictory.

There is no contradiction. Each frame is moving with respect to
the other.

>
>> If these points correspond to the ends of an interval at rest in
>> S, the length is x2-x1=gcT' or cT'=(x2-x1)/g.
>
> You have in fact calculated the length in S by assuming that
> S is the moving frame.

I calculated the S coordinates corresponding to the points where
the interval was marked in S'. The difference between the S
coordinates is the rest length in S.

>
>> So, the length in S', cT', is contracted when compared to a
>> fixed length in S whose endpoints correspond to the ends of the
>> interval at time T'.
>
> And this is the result of a sophism.

Please explain. I identified the ends of the interval in S',
calculated the corresponding points in S, and subtracted to get
the distance in S, assuming that the marked points were not
moving in S.

>
>> If you measure the distance in S, at time T, and compare it to
>> an interval at rest in S', you get:
>> x1'=g(vT-vT)
>> =0
>> x2'=g(cT-vT)
>> =g(c-v)T
>
> The equations represent the coordinates in S' of the origin
> of S' and of the light endpoint, when S' is moving away from S
> at v.

Yes, but they represent those points at two different times in
S'. If we assume the the points are at the ends of a rod at
rest in S', then the difference is the rest length of the
interval in S'.

>
>> So
>> x2'-x1'=g(c-v)T
>> or (c-v)T=(x2'-x1')/g
>
> g(c-v)T is a length measured in the moving frame S'.

It is the rest length of the interval defined in S at time T.

> The corresponding length measured in the rest frame S is (c-v)T.

Note that both endpoints are moving in S, so this is the length
in motion.

> The length comoving with S' is thus *dilated* relative to the length
> 'at rest'.

'Comoving with' is another way to say 'at rest in'. The rest
length in S is longer than the moving length in S.

> Conversely, the length 'at rest' is contracted
> relative to the 'moving' length.

No, you simply confused the rest and moving lengths. It is much
simpler if you use a fixed length rather than one that is
increasing with time.

>
>> So, the length in S is contracted when compared to a fixed
>> length in S' whose endpoints correspond to the ends of the
>> interval at time T.
>
> Here, you are implying that S is the moving frame.

Either frame can be seen as moving. The question is whether the
marked interval is moving in each frame.

>
>>>>> >From x' = g(x-vt), x'1 = g(x1-vt), hence x' =0 as x1=vt..
>>>>> Aftrwards, you complicated the problem by introducing t1 and t2.
>>>> No, you complicated the problem by ignoring them, and expecting
>>>> S' to mark the ends at two different times.
>>>>
>>>> S and S' measure time differently.
>>>>
>>> Of course, in the general case, but not in the scenario.
>> Please explain.
>
> I think that the matter is amply settled above.

No, it is not. What do you mean when you say that S and S'
measure time the same in this scenario?

mlut...@wanadoo.fr

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Oct 23, 2006, 1:25:04 PM10/23/06

to

I assume that y'=ct', simply because the vector v is then
perpendicular to the vector c. This justify the velocity c
along the y' axis in S'.

The resulting transform is

x' = (x-vt)/g
t' = t/g,

the inverse transform being

x = g(x'+vt')

t = gt'

Allow me to get back to your 'long way' derivation of length
contraction:

BEGIN

x1'=g(vT-vt1)
x2'=g(cT-vt2)
t1'=g(t1-v^2T/c^2) (x1=vT)
t2'=g(t2-vT/c) (x2=cT)

Setting t1'=t2', we get
t2=t1-v^2T/c^2+vT/c

Choosing t1=T, so that x1'=0
t2=T(1-v^2/c^2+v/c)
So:
x2'=g(cT-vT(1-v^2/c^+v/c))
=g(cT-v^2T/c-vT(1-v^2/c^2))
= g(cT(1-v^2/c^2)-vT(1-v^2/c^2))
=g((c-v)T(1/g^2)
=(c-v)T/g

END

You got x1'=0
To get x2'=(c-v)T/g, you had to assume that t1'=t2'.
Then x2'-x1'=(c-v)T/g
As (c-v)T represents x2-x1, which is the length L of the
stick in S, you demonstrated length contraction. Indeed,
L' = L/g

The 'short way' use the inverse LT

x1 = g(x1'+vt1')
x2 = g(x2'+vt2')

x2-x1 = g(x2'-x1') + gv(t2'-t1'), or
L = gL' + gv(t2'-t1')

To get the correct result L = gL', or L' = L/g, you
*have* to assume that t1' = t2'!

Otoh, length contraction is straightforwardly obtained with
my transformation x'=(x-vt)/g:

x1' = (vT-vT)/g = 0
x2' = (cT-vT)/g
x2'-x1' = (cT-vT)/g = (x2-x1)/g
L' = L/g

So, why should one prefer the Einsteinian LT, who *needs* an
assumption (t2'=t1') whose only justification is to obtain the correct
result, to my transformation ?

Marcel Luttgens

Brian Kennelly

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Oct 23, 2006, 3:28:16 PM10/23/06

to

It seems rather ad-hoc and inconsistent. What is the speed if,
in S', v and c form an angle of 89.9 degrees, or 45 degrees?

For the moment let us ignore the y direction. We agree that,
with the assumption of an invariant relative velocity we have
reached:
x'=a(v)(x-vt)
t'=a(v)t

'a' is a function of v, a(0)=1, and from symmetry a(v)=a(-v).

From these equations, with x=ut, we get
x'=a(ut-vt)
=(u-v)t'

So, for any velocity in the x direction, the relative velocity
between that point and the S' origin is invariant.

The transformation from S' to S" is
x"=a(u-v)(x'-(u-v)t')
t"=a(u-v)t'

Combining these, we get the transformation:
x"=a(u-v)a(v)(x-vt-(u-v)t)
=a(u-v)a(v)(x-ut)
t"=a(u-v)a(v)t

The direct equations give:
x"=a(u)(x-ut)
t"=a(u)t

Consistency requires that a(u-v)*a(v)=a(u).

Setting u=0, we find that a(v)*a(-v)=1. This gives a(v)^2=1, or
a(v)=1 (the negative root is excluded, because a(0)=1).

We do not need to examine the y direction to conclude that your
transformation must reduce to:
x'=x-vt
t'=t

Your equations also rely on the t1'=t2' assumption, because you
have absolute simultaneity. t1=t2 implies t1'=t2'.

The real problem with your equations is that they do not satisfy
the principle of relativity.

mlut...@wanadoo.fr

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Oct 24, 2006, 10:23:03 AM10/24/06

to

I will simply remind that the proof of the pudding is in the eating.
Iow, my equations, even if you quibbled about their derivation, lead
straightforwardly to two physically well attested results, i.e.
length contraction and time dilation. And luckily, they have absolute
simultaneity.

Otoh, the Einstein transformation (the LT), precisely because they
predict the so-called relativity of simultaneity, a phenomenon that
by the way has never been proved, need the ad-hoc assumption that
t1'=t2' to derive length contraction. If one choose t1=t2, t1'<>t2'
and one doesn't get length contraction. Their predicted 'relative
simultaneity' is thus their rehibitory defect. But SRists,
unscientifically and illogically, will play down such lack of coherence
and continue to defend the validity of the LT.

Marcel Luttgens

Brian Kennelly

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Oct 24, 2006, 3:44:12 PM10/24/06

to

Your equations put in length contraction and time dilation by
hand, while SR shows that they follow from the light postulate
and the principle of relativity.

Your transformations cannot be derived from any consistent
assumptions, unless g=1, in which case they reduce to the
Galilean.

Assuming g <>1 in your equations leads to transformation
equations that are different for each observer, and which
predict different effects for positive and negative velocities
for most observers. IOW, they have a built in preferred
reference system.

>
> Otoh, the Einstein transformation (the LT), precisely because they
> predict the so-called relativity of simultaneity, a phenomenon that
> by the way has never been proved, need the ad-hoc assumption that
> t1'=t2' to derive length contraction.

That is not an ad-hoc assumption. You must mark both ends of a
moving object at the same time to measure its length.

It is also possible to measure the length of a moving object, if
you know its speed, by measuring the time it takes to pass a
fixed point. Using this very natural method to measure length,
the LT lead to contraction of moving objects without the need to
invoke t1'=t2' or t1=t2.

> If one choose t1=t2, t1'<>t2'
> and one doesn't get length contraction.

You don't get anything directly meaningful. It is similar to
measuring the time of a trip from LA to NY by using the local
time of departure and arrival. There is a systematic error that
skews the results. Each observer must consistently use his own
time reference.

> Their predicted 'relative
> simultaneity' is thus their rehibitory defect.

I do not agree. It would be a defect if it was possible to
demonstrate absolute simultaneity, but this has not been done.

> But SRists,
> unscientifically and illogically, will play down such lack of coherence
> and continue to defend the validity of the LT.

The validity of SR must stand on its proven internal
consistency, and on its agreement with observation. On those
standards, it continues to be accepted.

If you have an alternative theory, you must enumerate its
assumptions, and demonstrate that it can explain the
observations as well as, or better than, SR.

It appears that your suggested transformations are inconsistent
with the principle of relativity. This alone does not make them
wrong, but you will need to provide some very strong arguments
for such a move.

Dirk Van de moortel

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Oct 24, 2006, 5:49:52 PM10/24/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:zuq%g.14$U%2...@newsfe16.phx...
> mlut...@wanadoo.fr wrote:

[snip]

>> Otoh, the Einstein transformation (the LT), precisely because they
>> predict the so-called relativity of simultaneity, a phenomenon that
>> by the way has never been proved, need the ad-hoc assumption that
>> t1'=t2' to derive length contraction.

:-))

> That is not an ad-hoc assumption. You must mark both ends of a moving object at the same time to measure its length.

No no no, you don't get it.
Marcel Luttgens measures his distances to the front and the
rear of the moving train with a 10 minutes time interval. That
is why his trains look so long in the stationary system. You
see, Marcel is more into length *dilation* ;-)

Dirk Vdm

mlut...@wanadoo.fr

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Oct 25, 2006, 1:20:48 PM10/25/06

to

Brian Kennelly wrote:
> mlut...@wanadoo.fr wrote:
> > Brian Kennelly wrote:

> > I will simply remind that the proof of the pudding is in the eating.
> > Iow, my equations, even if you quibbled about their derivation, lead
> > straightforwardly to two physically well attested results, i.e.
> > length contraction and time dilation. And luckily, they have absolute
> > simultaneity.

> Your equations put in length contraction and time dilation by
> hand, while SR shows that they follow from the light postulate
> and the principle of relativity.
>
> Your transformations cannot be derived from any consistent
> assumptions, unless g=1, in which case they reduce to the
> Galilean.
>
> Assuming g <>1 in your equations leads to transformation
> equations that are different for each observer, and which
> predict different effects for positive and negative velocities
> for most observers. IOW, they have a built in preferred
> reference system.

Why not?

>
> >
> > Otoh, the Einstein transformation (the LT), precisely because they
> > predict the so-called relativity of simultaneity, a phenomenon that
> > by the way has never been proved, need the ad-hoc assumption that
> > t1'=t2' to derive length contraction.

> That is not an ad-hoc assumption. You must mark both ends of a
> moving object at the same time to measure its length.

If one measures the ends at the same time in the moving frame,
one gets for the same ends different times in the rest frame.
Imo, this proves that Einstein's postulate of the constancy of
light speed leads to incoherent results.

>
> It is also possible to measure the length of a moving object, if
> you know its speed, by measuring the time it takes to pass a
> fixed point. Using this very natural method to measure length,
> the LT lead to contraction of moving objects without the need to
> invoke t1'=t2' or t1=t2.

Yes, but then, it is not so easy to show that the LT are inconsistent.

>
> > If one choose t1=t2, t1'<>t2'
> > and one doesn't get length contraction.

> You don't get anything directly meaningful. It is similar to
> measuring the time of a trip from LA to NY by using the local
> time of departure and arrival. There is a systematic error that
> skews the results. Each observer must consistently use his own
> time reference.
>
> > Their predicted 'relative
> > simultaneity' is thus their rehibitory defect.

> I do not agree. It would be a defect if it was possible to
> demonstrate absolute simultaneity, but this has not been done.

Neither has the funny relative simultaneity been demonstrated.

>
> > But SRists,
> > unscientifically and illogically, will play down such lack of coherence
> > and continue to defend the validity of the LT.

> The validity of SR must stand on its proven internal
> consistency, and on its agreement with observation. On those
> standards, it continues to be accepted.

Let us start with x1=vT, x2=cT, x1'=0, t1'=t2' and t1=T.

>From t1'= g(t1-v^2T/c^2) and t1=T, one gets t1'=T/g, or t1'=t1/g, or
t1=gt1', meaning that the time measured by a clock situated at the
first end a stick situated in the rest frame S is 'expanded'
by a factor g relative to a clock at rest in the moving frame S'.
As t2'=t1', the time measured in S by a clock situated at the second
end of the stick is expected to be 'expanded' by the same factor g
relative to the corresponding time t2' measured in S'.
In SR, this is not the case.

You have already calculated t2, for instance in your 'long way'
derivation: t2=T(1-v^2/c^2+v/c).

One can now calculate the time 'expansion' ratio in S for the second
end of the stick.
It is expected that such ratio, that could be called g(LT),
should be identical to g, or, at least, that g(LT)/g should
coherently depends on v.

This is not the case:

With v=0.01, one gets g=1.00 and g(LT)=1.01.
With v=0.999, one gets g=22.37 and g(LT)=22.39.
So, for small and very big velocities, g and gLT are similar.

But for intermediate velocities, g(LT)/g varies noticeably with v:

v g(LT) g g(LT)/g
_ ___ _ _______

0.01 1.01 1.00 1.01
0.1 1.10 1.01 1.09
0.3 1.27 1.05 1.21
0.5 1.44 1.15 1.25
0.7 1.69 1.40 1.21
0.9 2.50 2.29 1.09
0.999 22.39 22.37 1.00

This suffices to demonstrate the internal inconsistency of SR.

> If you have an alternative theory, you must enumerate its
> assumptions, and demonstrate that it can explain the
> observations as well as, or better than, SR.
>
> It appears that your suggested transformations are inconsistent
> with the principle of relativity. This alone does not make them
> wrong, but you will need to provide some very strong arguments
> for such a move.

An argument like the queer behaviour of clocks in the rest frame
is strong enough to 'falsify' SR.

An alternative theory is one that rejects Einstein's postulate,
and leads to time dilation and length contraction, which is the case
of my transformation.

Marcel Luttgens

Dirk Van de moortel

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Oct 25, 2006, 3:01:37 PM10/25/06

to

<mlut...@wanadoo.fr> wrote in message news:1161782448.4...@f16g2000cwb.googlegroups.com...

[snip]

> An argument like the queer behaviour of clocks in the rest frame
> is strong enough to 'falsify' SR.

I think it's time for Brian to become slightly less optimistic about
Marcell Luttgens ;-)

Dirk Vdm

Brian Kennelly

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Oct 25, 2006, 3:02:32 PM10/25/06

to

mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:
>>> Brian Kennelly wrote:
>
>> Your equations put in length contraction and time dilation by
>> hand, while SR shows that they follow from the light postulate
>> and the principle of relativity.
>>
>> Your transformations cannot be derived from any consistent
>> assumptions, unless g=1, in which case they reduce to the
>> Galilean.
>>
>> Assuming g <>1 in your equations leads to transformation
>> equations that are different for each observer, and which
>> predict different effects for positive and negative velocities
>> for most observers. IOW, they have a built in preferred
>> reference system.
>
> Why not?

I assume you are asking why your equations cannot be derived
from consistent assumptions. I demonstrated that, from your
assumption that there is a non-zero relative velocity that is
the same in both systems, the POR requires that a=1 for consistency.

Your equations can be used, but they single out one system as
preferred, and you lose the POR.

It is also a problem that your equations lead to the prediction
of *dilation* of moving objects.

Assume a stick of length L' at rest in S'. Let the left end be
at the S' origin. Then your equations give:
0=(x1-vt)/g
:: x1=vt
L'=(x2-vt)/g
:: x2=gL'+vt

Subtracting, we find the that length in S is L=gL'. The length
is expanded compared to its length at rest.

>
>>> Otoh, the Einstein transformation (the LT), precisely because they
>>> predict the so-called relativity of simultaneity, a phenomenon that
>>> by the way has never been proved, need the ad-hoc assumption that
>>> t1'=t2' to derive length contraction.
>
>> That is not an ad-hoc assumption. You must mark both ends of a
>> moving object at the same time to measure its length.
>
> If one measures the ends at the same time in the moving frame,
> one gets for the same ends different times in the rest frame.
> Imo, this proves that Einstein's postulate of the constancy of
> light speed leads to incoherent results.

It is unexpected, even counter-intuitive, but not incoherent.

>
>> It is also possible to measure the length of a moving object, if
>> you know its speed, by measuring the time it takes to pass a
>> fixed point. Using this very natural method to measure length,
>> the LT lead to contraction of moving objects without the need to
>> invoke t1'=t2' or t1=t2.
>
> Yes, but then, it is not so easy to show that the LT are inconsistent.

That is because they are not. The LT are internally consistent.

>
>>> If one choose t1=t2, t1'<>t2'
>>> and one doesn't get length contraction.
>
>> You don't get anything directly meaningful. It is similar to
>> measuring the time of a trip from LA to NY by using the local
>> time of departure and arrival. There is a systematic error that
>> skews the results. Each observer must consistently use his own
>> time reference.
>>
>>> Their predicted 'relative
>>> simultaneity' is thus their rehibitory defect.
>
>> I do not agree. It would be a defect if it was possible to
>> demonstrate absolute simultaneity, but this has not been done.
>
> Neither has the funny relative simultaneity been demonstrated.

It follows from the light postulate, which is consistent with
the observation.

It is also, as Lorentz showed in the 1890's, a possible
interpretation of Fresnel's equation.

Not until you explain how you calculated these numbers.

As far as I can see, g depends only on the velocity, and is the
same at every point.

To calculate the time dilation factor, you need to compare time
intervals between the systems.

Let us look at a stick of length L', at rest in S', and
calculate the S times when t' is 0 and T'.

When x'=0:

t1=0 (when t'=0)
t2=g(T') (when t'=T')
So T=gT' or T'=T/g

When x'=L':
t1=g(vL'/c^2) (when t'=0)
t2=g(T'+vL'/c^2) (when t'=T')

Subtracting, we find again that T=gT' or T'=T/g

It is the same factor at both ends.

>
>> If you have an alternative theory, you must enumerate its
>> assumptions, and demonstrate that it can explain the
>> observations as well as, or better than, SR.
>>
>> It appears that your suggested transformations are inconsistent
>> with the principle of relativity. This alone does not make them
>> wrong, but you will need to provide some very strong arguments
>> for such a move.
>
> An argument like the queer behaviour of clocks in the rest frame
> is strong enough to 'falsify' SR.
>
> An alternative theory is one that rejects Einstein's postulate,
> and leads to time dilation and length contraction, which is the case
> of my transformation.
>

No, you do not have an alternate theory. You have a set of
ad-hoc equations, and they predict length expansion for moving
objects.

Brian Kennelly

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Oct 25, 2006, 3:09:57 PM10/25/06

to

Dirk Van de moortel wrote:

No, I think Marcell is making an effort to defend his position,
and to understand the LT.

The biggest problem is that he wants to compare lengths and
times of expanding intervals, rather that sticking to fixed
lengths.

Dirk Van de moortel

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Oct 25, 2006, 3:19:10 PM10/25/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:e5L%g.206$yy2...@newsfe11.phx...

> Dirk Van de moortel wrote:
>> <mlut...@wanadoo.fr> wrote in message news:1161782448.4...@f16g2000cwb.googlegroups.com...
>>
>> [snip]
>>
>>> An argument like the queer behaviour of clocks in the rest frame
>>> is strong enough to 'falsify' SR.
>>
>> I think it's time for Brian to become slightly less optimistic about
>> Marcell Luttgens ;-)
>>
>
> No, I think Marcell is making an effort to defend his position, and to understand the LT.

ahem ;-)

>
> The biggest problem is that he wants to compare lengths and times of expanding intervals, rather that sticking to fixed lengths.

I think his biggest problem, is the time he has spent to get this
together:
http://perso.orange.fr/mluttgens/
Most notorious:
http://perso.orange.fr/mluttgens/LTfalse.htm
http://perso.orange.fr/mluttgens/mmx.htm
http://perso.orange.fr/mluttgens/SR%20FLAW.htm
http://perso.orange.fr/mluttgens/twinpdx1.htm
which have been debunked *to death* on this forum.

He is a troll and he is making an effort to test your optimism ;-)

Cheers,
Dirk Vdm

Brian Kennelly

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Oct 25, 2006, 4:00:59 PM10/25/06

to

Dirk Van de moortel wrote:
> I think his biggest problem, is the time he has spent to get this
> together:
> http://perso.orange.fr/mluttgens/
> Most notorious:
> http://perso.orange.fr/mluttgens/LTfalse.htm
> http://perso.orange.fr/mluttgens/mmx.htm
> http://perso.orange.fr/mluttgens/SR%20FLAW.htm
> http://perso.orange.fr/mluttgens/twinpdx1.htm
> which have been debunked *to death* on this forum.
>

Thanks for the links. I had not seen them yet. The do reveal
his confusion, which is mostly rooted in the silent assumption
of invariance for relative velocity (or, equivalently,
simultaneity).

Dirk Van de moortel

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Oct 25, 2006, 4:20:29 PM10/25/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:gQL%g.95$U%2....@newsfe16.phx...

Yes, perhaps strongly rooted.
But I really think that his confusion is mostly rooted in his
silent but persistent failing to understand the meaning of the
variables in the equations he uses.

Good luck :-|

Dirk Vdm

mlut...@wanadoo.fr

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Oct 25, 2006, 5:01:58 PM10/25/06

to

Brian Kennelly wrote:
> mlut...@wanadoo.fr wrote:
> > Brian Kennelly wrote:
> >> mlut...@wanadoo.fr wrote:
> >>> Brian Kennelly wrote:
> >
> >> Your equations put in length contraction and time dilation by
> >> hand, while SR shows that they follow from the light postulate
> >> and the principle of relativity.
> >>
> >> Your transformations cannot be derived from any consistent
> >> assumptions, unless g=1, in which case they reduce to the
> >> Galilean.
> >>
> >> Assuming g <>1 in your equations leads to transformation
> >> equations that are different for each observer, and which
> >> predict different effects for positive and negative velocities
> >> for most observers. IOW, they have a built in preferred
> >> reference system.
> >
> > Why not?

> I assume you are asking why your equations cannot be derived
> from consistent assumptions. I demonstrated that, from your
> assumption that there is a non-zero relative velocity that is
> the same in both systems, the POR requires that a=1 for consistency.

I reject the POR.

>
> Your equations can be used, but they single out one system as
> preferred, and you lose the POR.
>
> It is also a problem that your equations lead to the prediction
> of *dilation* of moving objects.
>
> Assume a stick of length L' at rest in S'. Let the left end be
> at the S' origin. Then your equations give:
> 0=(x1-vt)/g
> :: x1=vt
> L'=(x2-vt)/g
> :: x2=gL'+vt
>
> Subtracting, we find the that length in S is L=gL'. The length
> is expanded compared to its length at rest.
>

What my transformation predicts is that a moving stick is contracted
wrt a
stick at rest, i.e. L'=L/g.
Knowing the length L' of the moving stick, it is clear that the length
of the
stick at rest is L = gL', Iow, it is dilated in S.
Calling the moving stick a stick at rest, because it is at rest in the
moving frame S',
but forgetting that the rest frame remains S, is the type of logical
error systematically
made by SRists.

>
> >
> >>> Otoh, the Einstein transformation (the LT), precisely because they
> >>> predict the so-called relativity of simultaneity, a phenomenon that
> >>> by the way has never been proved, need the ad-hoc assumption that
> >>> t1'=t2' to derive length contraction.
> >
> >> That is not an ad-hoc assumption. You must mark both ends of a
> >> moving object at the same time to measure its length.
> >
> > If one measures the ends at the same time in the moving frame,
> > one gets for the same ends different times in the rest frame.
> > Imo, this proves that Einstein's postulate of the constancy of
> > light speed leads to incoherent results.
> It is unexpected, even counter-intuitive, but not incoherent.
>
> >
> >> It is also possible to measure the length of a moving object, if
> >> you know its speed, by measuring the time it takes to pass a
> >> fixed point. Using this very natural method to measure length,
> >> the LT lead to contraction of moving objects without the need to
> >> invoke t1'=t2' or t1=t2.
> >
> > Yes, but then, it is not so easy to show that the LT are inconsistent.

> That is because they are not. The LT are internally consistent.

Ha! Ha!

One has g=1/sqrt(1-v^2/c^2) and t1'=t1/g, thus g corresponds to t1/t1'.
Similarly, g(LT) corresponds to t2/t2', with t2'= T/g.
We can calculate g(LT) from t2 = T(1 + v/c - v^2/c^2)
g(LT) = t2/(T/g) = g(1 + v/c - v^2/c^2)
Now we can compare the ratio g(LT) for different values of v.

>
> As far as I can see, g depends only on the velocity, and is the
> same at every point.

Both g and g(LT) depend only on velocity.

>
> To calculate the time dilation factor, you need to compare time
> intervals between the systems.
>
> Let us look at a stick of length L', at rest in S', and
> calculate the S times when t' is 0 and T'.
>

You make again the same logical mistake, because S' remains the moving
frame.
Iow, the stick continue to move wrt the S-frame, even if one considers
it at rest
in the S'-frame.

> When x'=0:
>
> t1=0 (when t'=0)
> t2=g(T') (when t'=T')
> So T=gT' or T'=T/g
>
> When x'=L':
> t1=g(vL'/c^2) (when t'=0)
> t2=g(T'+vL'/c^2) (when t'=T')
>
> Subtracting, we find again that T=gT' or T'=T/g
>
> It is the same factor at both ends.
>
>
> >
> >> If you have an alternative theory, you must enumerate its
> >> assumptions, and demonstrate that it can explain the
> >> observations as well as, or better than, SR.
> >>
> >> It appears that your suggested transformations are inconsistent
> >> with the principle of relativity. This alone does not make them
> >> wrong, but you will need to provide some very strong arguments
> >> for such a move.
> >
> > An argument like the queer behaviour of clocks in the rest frame
> > is strong enough to 'falsify' SR.
> >
> > An alternative theory is one that rejects Einstein's postulate,
> > and leads to time dilation and length contraction, which is the case
> > of my transformation.

> No, you do not have an alternate theory. You have a set of
> ad-hoc equations, and they predict length expansion for moving
> objects.

They are not ad hoc and they predict time 'dilation' and length
contraction,
not length expansion.

Perhaps don't you realize that, when a moving clock shows a time
interval of
1 second when a clock at rest shows a time interval of 2 seconds, the
observer moving with the clock and reading 1 second will rightly claim
that
the observer at rest will read 2 seconds, even if the moving observer
considers
himself at rest.

Marcel Luttgens

Brian Kennelly

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Oct 25, 2006, 5:50:42 PM10/25/06

to

mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:
>>> Brian Kennelly wrote:
>

> I reject the POR.
On what basis? The principle of relativity has been a
fundamental part of physics since at least the time of Newton.
It is consistent with all observations.

>
>> Your equations can be used, but they single out one system as
>> preferred, and you lose the POR.
>>
>> It is also a problem that your equations lead to the prediction
>> of *dilation* of moving objects.
>>
>> Assume a stick of length L' at rest in S'. Let the left end be
>> at the S' origin. Then your equations give:
>> 0=(x1-vt)/g
>> :: x1=vt
>> L'=(x2-vt)/g
>> :: x2=gL'+vt
>>
>> Subtracting, we find the that length in S is L=gL'. The length
>> is expanded compared to its length at rest.
>>
>
> What my transformation predicts is that a moving stick is contracted
> wrt a
> stick at rest, i.e. L'=L/g.
> Knowing the length L' of the moving stick, it is clear that the length
> of the
> stick at rest is L = gL', Iow, it is dilated in S.
> Calling the moving stick a stick at rest, because it is at rest in the
> moving frame S',
> but forgetting that the rest frame remains S, is the type of logical
> error systematically
> made by SRists.

I am afraid you just lost a lot of ground. The rest length is
the length measured from a reference frame moving with the stick
(IOW, in which the stick is at rest).

But, let's do it your way. A stick of length L in S is moving
with speed v to the right. The left end is x1=vt, and the right
end is x2=L+vt
The length of the moving stick is x2-x1=L. There is no change
in length for a moving stick.

>
>
>> That is because they are not. The LT are internally consistent.
>
> Ha! Ha!

You have yet to demonstrate a single inconsistency.

>
>> Not until you explain how you calculated these numbers.
>
> One has g=1/sqrt(1-v^2/c^2) and t1'=t1/g, thus g corresponds to t1/t1'.
> Similarly, g(LT) corresponds to t2/t2', with t2'= T/g.
> We can calculate g(LT) from t2 = T(1 + v/c - v^2/c^2)
> g(LT) = t2/(T/g) = g(1 + v/c - v^2/c^2)
> Now we can compare the ratio g(LT) for different values of v.

Your g(LT) cannot be compared to g, because it was computed from
a different S time interval.

You computed g from the interval from 0 to T, but you computed
g(LT) from the S interval from 0 to T(1+v/c-v^2/c^2), then
divided it by T. Your numbers are meaningless.

>
>> As far as I can see, g depends only on the velocity, and is the
>> same at every point.
>
> Both g and g(LT) depend only on velocity.

No, you demonstrated a difference due to position. Your error
introduced a dependence on position.

>
>> To calculate the time dilation factor, you need to compare time
>> intervals between the systems.
>>
>> Let us look at a stick of length L', at rest in S', and
>> calculate the S times when t' is 0 and T'.
>>
>
> You make again the same logical mistake, because S' remains the moving
> frame.
> Iow, the stick continue to move wrt the S-frame, even if one considers
> it at rest
> in the S'-frame.

I fully agree that a stick at rest in S' is moving in S. If I
made a logical mistake, then we made it together.

>
>
>> No, you do not have an alternate theory. You have a set of
>> ad-hoc equations, and they predict length expansion for moving
>> objects.
>
> They are not ad hoc and they predict time 'dilation' and length
> contraction,
> not length expansion.

When you tried to derive your equations, you had to put in the g
factor by hand. You made an ad-hoc assumption about light speed
along y', without physical justification.

You obtained a result that is inconsistent with the results we
can obtain from considering the x direction alone (which
requires that a=1).

>
> Perhaps don't you realize that, when a moving clock shows a time
> interval of
> 1 second when a clock at rest shows a time interval of 2 seconds, the
> observer moving with the clock and reading 1 second will rightly claim
> that
> the observer at rest will read 2 seconds, even if the moving observer
> considers
> himself at rest.

You must be careful about your use of 'when', because they may
disagree about its meaning. I will make the disambiguating
assumption that the clocks are adjacent at the moment designated
by when for each of them.

Assuming the clocks are adjacent, they will agree about each
other's readings, but that does not allow them to compare
intervals. You must add to your example comparison of readings
at another time, if you want to compare intervals.

mlut...@wanadoo.fr

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Oct 26, 2006, 9:12:54 AM10/26/06

to

Brian Kennelly wrote:
> mlut...@wanadoo.fr wrote:
> > Brian Kennelly wrote:
> >> mlut...@wanadoo.fr wrote:
> >>> Brian Kennelly wrote:
> >
> > I reject the POR.
> On what basis? The principle of relativity has been a
> fundamental part of physics since at least the time of Newton.
> It is consistent with all observations.
>

I don't reject Galilean relativity, what I reject is Einsteinian
relativity, based on
the hypothesis of the constancy of light speed.

This is exactly what I said: a stick comoving with S' is at rest in S'.

> But, let's do it your way. A stick of length L in S is moving
> with speed v to the right. The left end is x1=vt, and the right
> end is x2=L+vt
> The length of the moving stick is x2-x1=L. There is no change
> in length for a moving stick.

Sure, in S, its length is L, in S', it is L'=L/g.

> >
> >
> >> That is because they are not. The LT are internally consistent.
> >
> > Ha! Ha!
> You have yet to demonstrate a single inconsistency.
>
>
> >
> >> Not until you explain how you calculated these numbers.
> >
> > One has g=1/sqrt(1-v^2/c^2) and t1'=t1/g, thus g corresponds to t1/t1'.
> > Similarly, g(LT) corresponds to t2/t2', with t2'= T/g.
> > We can calculate g(LT) from t2 = T(1 + v/c - v^2/c^2)
> > g(LT) = t2/(T/g) = g(1 + v/c - v^2/c^2)
> > Now we can compare the ratio g(LT) for different values of v.

> Your g(LT) cannot be compared to g, because it was computed from
> a different S time interval.

I compared g(LT) to g, simply because t2 = g(TL)*t2', by analogy with
t1 = gt1'.

>
> You computed g from the interval from 0 to T, but you computed
> g(LT) from the S interval from 0 to T(1+v/c-v^2/c^2), then
> divided it by T. Your numbers are meaningless.
>

They are not meaningless, as t2 = g(TL)*t2'.
g(TL) should at least have a relation with g, that whould have a
physical meaning.

> >
> >> As far as I can see, g depends only on the velocity, and is the
> >> same at every point.
> >
> > Both g and g(LT) depend only on velocity.

> No, you demonstrated a difference due to position. Your error
> introduced a dependence on position.

One can explain the variation with v of the ratio g(LT)/g by referring
to the LT,
which is evident, as g(LT) has been derived from them, but you can't
give
it a coherent physical meaning. This proves that the LT are physically
incoherent.

> >> To calculate the time dilation factor, you need to compare time
> >> intervals between the systems.
> >>
> >> Let us look at a stick of length L', at rest in S', and
> >> calculate the S times when t' is 0 and T'.
> >>
> >
> > You make again the same logical mistake, because S' remains the moving
> > frame.
> > Iow, the stick continue to move wrt the S-frame, even if one considers
> > it at rest
> > in the S'-frame.

> I fully agree that a stick at rest in S' is moving in S. If I
> made a logical mistake, then we made it together.

Then what is the purpose of stressing that the stick can be considered
at rest in the moving frame? Jumping from one frame to another, make
logical mistakes and draw wrong conclusions?

> >> No, you do not have an alternate theory. You have a set of
> >> ad-hoc equations, and they predict length expansion for moving
> >> objects.
> >
> > They are not ad hoc and they predict time 'dilation' and length
> > contraction,
> > not length expansion.
> When you tried to derive your equations, you had to put in the g
> factor by hand. You made an ad-hoc assumption about light speed
> along y', without physical justification.
>

It is now the nth time that I claim that the physical justification for
y'=ct' is that
the vector v is perpendicular to the vector c, hence, vectorially,
c+v=c.

> You obtained a result that is inconsistent with the results we
> can obtain from considering the x direction alone (which
> requires that a=1).

There must be some flaw in your assertion, as I obtain physically
meaningful
transformations.

>
> >
> > Perhaps don't you realize that, when a moving clock shows a time
> > interval of
> > 1 second when a clock at rest shows a time interval of 2 seconds, the
> > observer moving with the clock and reading 1 second will rightly claim
> > that
> > the observer at rest will read 2 seconds, even if the moving observer
> > considers
> > himself at rest.
>
> You must be careful about your use of 'when', because they may
> disagree about its meaning. I will make the disambiguating
> assumption that the clocks are adjacent at the moment designated
> by when for each of them.
>
> Assuming the clocks are adjacent, they will agree about each
> other's readings, but that does not allow them to compare
> intervals. You must add to your example comparison of readings
> at another time, if you want to compare intervals.

I didn't specify the details, as I was convinced that you would
perfectly
understand what I meant.
In order to get rid of the complications brought about by the fanciful
relativity of simultaneity, the direct consequence of the postulate of
light
speed constancy, let's take the well known exemple of the GPS clocks.
Neglecting the effects of the difference of gravitational potential
between
the satellite and the Earth, and assuming a non-rotating Earth, one is
left with a pure SR effect on the satellite clock, that I call clock1.
As clock1 is moving at v relative to an Earth clock0, its time is
slowed down by g, according to the relation t(clock1)=t(clock0)/g.
Notice that nobody ever used the 'time' LT t'=g(t-vx/c^2) to calculate
t(clock1) from t(clock0). In fact, the 'time' LT is useless.
Clock1 is of course at rest in the satellite. This doesn't change the
fact that
it is moving at v relative to clock0, hence that t(clock1) remains
equal
to t(clock0)/g, or that t(clock0) is still equal to g*t(clock(1).
Of course, SRists are tempted, illogically, to claim that, as clock1 is
at rest in the satellite, the satellite itself is at rest and the Earth
itself
is moving relative to the satellite, which allow them to calculate that
t(clock0)=t(clock1)/g, hence that t(clock1)=g*t(clock0), i:n plai:n
contradiction
with the former relation t(clock1)=t(clock0)/g, the only one which has
a physical meaning.

Marcel Luttgens

Brian Kennelly

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Oct 26, 2006, 2:47:15 PM10/26/06

to

mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:
>>> Brian Kennelly wrote:
>>>> mlut...@wanadoo.fr wrote:
>>>>> Brian Kennelly wrote:
>>> I reject the POR.
>> On what basis? The principle of relativity has been a
>> fundamental part of physics since at least the time of Newton.
>> It is consistent with all observations.
>>
>
> I don't reject Galilean relativity, what I reject is Einsteinian
> relativity, based on
> the hypothesis of the constancy of light speed.

Your proposed equations are inconsistent with Galilean relativity.

So you have the remarkable prediction that a moving stick has
no change of length when measured from the system in which it
has a non-zero velocity, but is contracted when measured from a
system in which it has a zero velocity.

Of course, the measuring instruments are also moving with S', so
they will be contracted as well, and S' will not measure any
change of length.

>>>> Not until you explain how you calculated these numbers.
>>> One has g=1/sqrt(1-v^2/c^2) and t1'=t1/g, thus g corresponds to t1/t1'.
>>> Similarly, g(LT) corresponds to t2/t2', with t2'= T/g.
>>> We can calculate g(LT) from t2 = T(1 + v/c - v^2/c^2)
>>> g(LT) = t2/(T/g) = g(1 + v/c - v^2/c^2)
>>> Now we can compare the ratio g(LT) for different values of v.
>
>> Your g(LT) cannot be compared to g, because it was computed from
>> a different S time interval.
>
> I compared g(LT) to g, simply because t2 = g(TL)*t2', by analogy with
> t1 = gt1'.

Interval comparisons require two times. t1=gt1' is just a
convenient shorthand for (t1-0)=g(t1'-0), when the clocks were
synchronized at zero. You need to identify the starting and
ending times in both systems for your g(LT).

>
>> You computed g from the interval from 0 to T, but you computed
>> g(LT) from the S interval from 0 to T(1+v/c-v^2/c^2), then
>> divided it by T. Your numbers are meaningless.
>>
>
> They are not meaningless, as t2 = g(TL)*t2'.
> g(TL) should at least have a relation with g, that whould have a
> physical meaning.

Using o2 and o2' to designate the starting times, the correct
equation is
(t2-o2)=g(t2'-o2')

Identify o2 and o2', and you will find that g(LT)=g

>
>>>> As far as I can see, g depends only on the velocity, and is the
>>>> same at every point.
>>> Both g and g(LT) depend only on velocity.
>
>> No, you demonstrated a difference due to position. Your error
>> introduced a dependence on position.
>
> One can explain the variation with v of the ratio g(LT)/g by referring
> to the LT,
> which is evident, as g(LT) has been derived from them, but you can't
> give
> it a coherent physical meaning. This proves that the LT are physically
> incoherent.

You are correct. Your g(LT) has no coherent physical meaning,
but, because it is a symptom of your confusion, and nothing to
do with the LT, you have proven nothing.

>
>>>> To calculate the time dilation factor, you need to compare time
>>>> intervals between the systems.
>>>>
>>>> Let us look at a stick of length L', at rest in S', and
>>>> calculate the S times when t' is 0 and T'.
>>>>
>>> You make again the same logical mistake, because S' remains the moving
>>> frame.
>>> Iow, the stick continue to move wrt the S-frame, even if one considers
>>> it at rest
>>> in the S'-frame.
>
>> I fully agree that a stick at rest in S' is moving in S. If I
>> made a logical mistake, then we made it together.
>
> Then what is the purpose of stressing that the stick can be considered
> at rest in the moving frame? Jumping from one frame to another, make
> logical mistakes and draw wrong conclusions?

Any frame can be considered as the rest frame; it is a feature
of Galilean relativity. The stick is at rest in the frame in
which it has zero velocity.

The purpose of the transformation equations is to allow us to
change our descriptions from one rest frame to another.

>
>> When you tried to derive your equations, you had to put in the g
>> factor by hand. You made an ad-hoc assumption about light speed
>> along y', without physical justification.
>>
>
> It is now the nth time that I claim that the physical justification for
> y'=ct' is that
> the vector v is perpendicular to the vector c, hence, vectorially,
> c+v=c.

From that equation, then the light speed in the original frame
is sqrt(c^2+v^2), but you stated that it was c. (Remember that
you affirmed that relative velocity was invariant).

>
>> You obtained a result that is inconsistent with the results we
>> can obtain from considering the x direction alone (which
>> requires that a=1).
>
> There must be some flaw in your assertion, as I obtain physically
> meaningful
> transformations.

Only when g=1. I demonstrated earlier that, from the x
direction alone, we can conclude a(u-v)*a(v)=a(u). That
requires that a=exp(bv), with b constant. From a(v)=a(-v), we
find that b=0, so a=1.

>
>>> Perhaps don't you realize that, when a moving clock shows a time
>>> interval of
>>> 1 second when a clock at rest shows a time interval of 2 seconds, the
>>> observer moving with the clock and reading 1 second will rightly claim
>>> that
>>> the observer at rest will read 2 seconds, even if the moving observer
>>> considers
>>> himself at rest.
>> You must be careful about your use of 'when', because they may
>> disagree about its meaning. I will make the disambiguating
>> assumption that the clocks are adjacent at the moment designated
>> by when for each of them.
>>
>> Assuming the clocks are adjacent, they will agree about each
>> other's readings, but that does not allow them to compare
>> intervals. You must add to your example comparison of readings
>> at another time, if you want to compare intervals.
>
> I didn't specify the details, as I was convinced that you would
> perfectly
> understand what I meant.

No, I did not. Please specify the details.

> In order to get rid of the complications brought about by the fanciful
> relativity of simultaneity, the direct consequence of the postulate of
> light
> speed constancy, let's take the well known exemple of the GPS clocks.

You are introducing complications from accelerated motion. You
have not yet demonstrated that you understand simple inertial
motion, so we will not go there.

Dirk Van de moortel

unread,

Oct 26, 2006, 3:03:50 PM10/26/06

to

<mlut...@wanadoo.fr> wrote in message news:1161853974....@m7g2000cwm.googlegroups.com...
> Brian Kennelly wrote:

[snip]

>> You must be careful about your use of 'when', because they may
>> disagree about its meaning. I will make the disambiguating
>> assumption that the clocks are adjacent at the moment designated
>> by when for each of them.
>>
>> Assuming the clocks are adjacent, they will agree about each
>> other's readings, but that does not allow them to compare
>> intervals. You must add to your example comparison of readings
>> at another time, if you want to compare intervals.
>
> I didn't specify the details, as I was convinced that you would
> perfectly
> understand what I meant.
> In order to get rid of the complications brought about by the fanciful
> relativity of simultaneity, the direct consequence of the postulate of
> light
> speed constancy, let's take the well known exemple of the GPS clocks.
> Neglecting the effects of the difference of gravitational potential
> between
> the satellite and the Earth, and assuming a non-rotating Earth, one is
> left with a pure SR effect on the satellite clock, that I call clock1.
> As clock1 is moving at v relative to an Earth clock0, its time is
> slowed down by g, according to the relation t(clock1)=t(clock0)/g.

No. Its time is *not* slowed down.
Hundreds of times people have told you that this is the
wrong way to think about this, and the ideal way to never
say or do anything right.
The time you (on Earth) measure between two of *its* ticks
is longer than the time it indicates between these ticks.

> Notice that nobody ever used the 'time' LT t'=g(t-vx/c^2) to calculate
> t(clock1) from t(clock0). In fact, the 'time' LT is useless.

This one is only useless because if want to get rid of the x in
the equation, the clock must be at rest on *Earth*. But you
are talking about a clock at rest on the *satellite*, with x' = 0,
so you must use the equation that links t, t, and x', namely the
equation t = g ( t' + v x'/c^2 ), which gives you your t' = t / g.

> Clock1 is of course at rest in the satellite. This doesn't change the
> fact that
> it is moving at v relative to clock0, hence that t(clock1) remains
> equal
> to t(clock0)/g, or that t(clock0) is still equal to g*t(clock(1).

Yes, since x' = 0 you have t = g t'

> Of course, SRists are tempted, illogically, to claim that, as clock1 is
> at rest in the satellite, the satellite itself is at rest and the Earth
> itself
> is moving relative to the satellite, which allow them to calculate that
> t(clock0)=t(clock1)/g

No. That is valid for a clock on *Earth*, with x = 0.
The time someone on the satellite measures between two of
*your* ticks on your *Earth* clock is longer than the time it
indicates between these ticks, with the equation
t' = g ( t - v x/c^2 ) giving t' = g t, or t = t' / g.

>, hence that t(clock1)=g*t(clock0), i:n plai:n
> contradiction
> with the former relation t(clock1)=t(clock0)/g, the only one which has
> a physical meaning.

Not in contradiction.
In perfect symmetry and accordance:

The time someone measures between two ticks on a
moving clock, is longer than the time the clock indicates
between these ticks.

Did I say 'hundreds' of times?
I'm sure it must be *thousands* of times now.
Is it really so difficult, or are you just a dishonest troll who
pretends to fail to understand it?

Dirk Vdm

mlut...@wanadoo.fr

unread,

Oct 26, 2006, 5:01:15 PM10/26/06

to

Brian Kennelly wrote:
> mlut...@wanadoo.fr wrote:
> > Brian Kennelly wrote:
> >> mlut...@wanadoo.fr wrote:
> >>> Brian Kennelly wrote:
> >>>> mlut...@wanadoo.fr wrote:
> >>>>> Brian Kennelly wrote:
> >>> I reject the POR.
> >> On what basis? The principle of relativity has been a
> >> fundamental part of physics since at least the time of Newton.
> >> It is consistent with all observations.
> >>
> >
> > I don't reject Galilean relativity, what I reject is Einsteinian
> > relativity, based on
> > the hypothesis of the constancy of light speed.
> Your proposed equations are inconsistent with Galilean relativity.

I didn't claim that they were.

You misunderstood, I didn't refer to the constancy of the length of
the stick at rest in S' and measured in S'. I clearly said " in S, its
length is L, in S', it is L'=L/g", meaning that the length of the stick
measured in S' is contracted wrt its length measured in S.

>
>
> >>>> Not until you explain how you calculated these numbers.
> >>> One has g=1/sqrt(1-v^2/c^2) and t1'=t1/g, thus g corresponds to t1/t1'.
> >>> Similarly, g(LT) corresponds to t2/t2', with t2'= T/g.
> >>> We can calculate g(LT) from t2 = T(1 + v/c - v^2/c^2)
> >>> g(LT) = t2/(T/g) = g(1 + v/c - v^2/c^2)
> >>> Now we can compare the ratio g(LT) for different values of v.
> >
> >> Your g(LT) cannot be compared to g, because it was computed from
> >> a different S time interval.
> >
> > I compared g(LT) to g, simply because t2 = g(TL)*t2', by analogy with
> > t1 = gt1'.
>
> Interval comparisons require two times. t1=gt1' is just a
> convenient shorthand for (t1-0)=g(t1'-0), when the clocks were
> synchronized at zero. You need to identify the starting and
> ending times in both systems for your g(LT).
>

It is implicit that the clocks have to be beforehand synchronized.

> >
> >> You computed g from the interval from 0 to T, but you computed
> >> g(LT) from the S interval from 0 to T(1+v/c-v^2/c^2), then
> >> divided it by T. Your numbers are meaningless.
> >>
> >
> > They are not meaningless, as t2 = g(TL)*t2'.
> > g(TL) should at least have a relation with g, that whould have a
> > physical meaning.
> Using o2 and o2' to designate the starting times, the correct
> equation is
> (t2-o2)=g(t2'-o2')
>
> Identify o2 and o2', and you will find that g(LT)=g

g(LT) is a ratio, which is a function of v:

g(LT) = t2/(T/g) = g(1 + v/c - v^2/c^2)

And g(LT)/g = 1 + v/c - v^2/c^2.
Could you show how you get g(LT)/g = 1, when v<>0 ?

> >
> >>>> As far as I can see, g depends only on the velocity, and is the
> >>>> same at every point.
> >>> Both g and g(LT) depend only on velocity.
> >
> >> No, you demonstrated a difference due to position. Your error
> >> introduced a dependence on position.
> >
> > One can explain the variation with v of the ratio g(LT)/g by referring
> > to the LT,
> > which is evident, as g(LT) has been derived from them, but you can't
> > give
> > it a coherent physical meaning. This proves that the LT are physically
> > incoherent.
> You are correct. Your g(LT) has no coherent physical meaning,
> but, because it is a symptom of your confusion, and nothing to
> do with the LT, you have proven nothing.
>

g(LT) has nothing to do with the LT? The confused one is you.

> >
> >>>> To calculate the time dilation factor, you need to compare time
> >>>> intervals between the systems.
> >>>>
> >>>> Let us look at a stick of length L', at rest in S', and
> >>>> calculate the S times when t' is 0 and T'.
> >>>>
> >>> You make again the same logical mistake, because S' remains the moving
> >>> frame.
> >>> Iow, the stick continue to move wrt the S-frame, even if one considers
> >>> it at rest
> >>> in the S'-frame.
> >
> >> I fully agree that a stick at rest in S' is moving in S. If I
> >> made a logical mistake, then we made it together.
> >
> > Then what is the purpose of stressing that the stick can be considered
> > at rest in the moving frame? Jumping from one frame to another, make
> > logical mistakes and draw wrong conclusions?

> Any frame can be considered as the rest frame; it is a feature
> of Galilean relativity. The stick is at rest in the frame in
> which it has zero velocity.

No, physically, not any frame can be considered as the rest frame.
For instance, no sane person would consider that a falling stone is
at rest, but the Earth, and in a sense the whole universe, is moving
relative to the stone. Only SRists believe that all frames are
physically
equivalent. For instance, they claim that the Earth can be considered
as moving wrt muons, or that the Earth is moving wrt a spaceship,
forgetting that their theory is only valid for inertial frames, frames
that exist nowhere in the universe. It is intellectually interesting
to discuss about some aspects of SR like its relativity of
simultaneity,
probably more than about the gender of angels.

>
> The purpose of the transformation equations is to allow us to
> change our descriptions from one rest frame to another.

Of course, what would be their purpose otherwise?

>
>
> >
> >> When you tried to derive your equations, you had to put in the g
> >> factor by hand. You made an ad-hoc assumption about light speed
> >> along y', without physical justification.
> >>
> >
> > It is now the nth time that I claim that the physical justification for
> > y'=ct' is that
> > the vector v is perpendicular to the vector c, hence, vectorially,
> > c+v=c.
> From that equation, then the light speed in the original frame
> is sqrt(c^2+v^2), but you stated that it was c. (Remember that
> you affirmed that relative velocity was invariant).

What is the original frame? I presume that you mean the rest frame.
Yes, in S, the velocity ot the light signal following the y'-axis is
sqrt(c^2+v^2), but it remains c in S'.

>
> >
> >> You obtained a result that is inconsistent with the results we
> >> can obtain from considering the x direction alone (which
> >> requires that a=1).
> >
> > There must be some flaw in your assertion, as I obtain physically
> > meaningful
> > transformations.
> Only when g=1. I demonstrated earlier that, from the x
> direction alone, we can conclude a(u-v)*a(v)=a(u). That
> requires that a=exp(bv), with b constant. From a(v)=a(-v), we
> find that b=0, so a=1.
>

Did you try your demonstration in the case where x'=ct' instead
of x'=(c-v)t' ?

> >
> >>> Perhaps don't you realize that, when a moving clock shows a time
> >>> interval of
> >>> 1 second when a clock at rest shows a time interval of 2 seconds, the
> >>> observer moving with the clock and reading 1 second will rightly claim
> >>> that
> >>> the observer at rest will read 2 seconds, even if the moving observer
> >>> considers
> >>> himself at rest.
> >> You must be careful about your use of 'when', because they may
> >> disagree about its meaning. I will make the disambiguating
> >> assumption that the clocks are adjacent at the moment designated
> >> by when for each of them.
> >>
> >> Assuming the clocks are adjacent, they will agree about each
> >> other's readings, but that does not allow them to compare
> >> intervals. You must add to your example comparison of readings
> >> at another time, if you want to compare intervals.
> >
> > I didn't specify the details, as I was convinced that you would
> > perfectly
> > understand what I meant.
> No, I did not. Please specify the details.

The main detail is that the clocks have to be synchronized beforehand.
This is so obvious, that I thought you didn't need such precision.

>
> > In order to get rid of the complications brought about by the fanciful
> > relativity of simultaneity, the direct consequence of the postulate of
> > light
> > speed constancy, let's take the well known exemple of the GPS clocks.
> You are introducing complications from accelerated motion. You
> have not yet demonstrated that you understand simple inertial
> motion, so we will not go there.

SRists seem to ignore the meaning of inertial frames when they consider
that the road is moving relative to a car, or that the Earth is moving
relative
to a spaceship, etc. I don't.

Marcel Luttgens

Dirk Van de moortel

unread,

Oct 26, 2006, 5:04:58 PM10/26/06

to

<mlut...@wanadoo.fr> wrote in message news:1161882075.1...@b28g2000cwb.googlegroups.com...

[snip]

>
> SRists seem to ignore the meaning of inertial frames when they consider
> that the road is moving relative to a car, or that the Earth is moving
> relative
> to a spaceship, etc. I don't.

What you don't, is "understanding what you are dealing with".
That, Marcel, is the only important thing that you don't :-)

Dirk Vdm

Brian Kennelly

unread,

Oct 26, 2006, 6:25:21 PM10/26/06

to

mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:
>>> Brian Kennelly wrote:
>>>> mlut...@wanadoo.fr wrote:
>>>>> Brian Kennelly wrote:
>>>>>> mlut...@wanadoo.fr wrote:
>>>>>>> Brian Kennelly wrote:
>>>>> I reject the POR.
>>>> On what basis? The principle of relativity has been a
>>>> fundamental part of physics since at least the time of Newton.
>>>> It is consistent with all observations.
>>>>
>>> I don't reject Galilean relativity, what I reject is Einsteinian
>>> relativity, based on
>>> the hypothesis of the constancy of light speed.
>> Your proposed equations are inconsistent with Galilean relativity.
>
> I didn't claim that they were.

You can't have it both ways. Either you reject Galilean
relativity or you don't. Above, you claimed that you do not
reject it, and now you acknowledge that your equations are
inconsistent with it. Of what use are your equations?

Using your equations, we conclude that the length of the stick
is independent of is speed for any observer. We do find that
its length depends on the speed of the observer, but that the
observer will not be able to measure the change. Of what use
are your equations?

You did not identify the starting times of the intervals.
Because your analysis is based on t1'=t2', you must also use
o1'=o2' (as you wrote above, "the clocks have to be beforehand
synchronized"). The length of the interval is L, and the left
end is located at the origin of S':
x1'=0
x2'=L

At o1'=0, we find:
o1=g(o1'+v0/c^2)
=0
o2=g(o2'+vL/c^2)
=g(vL/c^2)

At t1'=T', we find, as before:
t1=gT'
t2=g(T'+vL/c^2)

It is obvious that t2-o2=g(t2'-o2') and t1-01=g(t1'-01'), with
the same value of g at both ends.

(You can also do this directly from the equation t2=t1+vL/c^2.
When o1=0, o2=vL/c^2. If you insist on using L=(c-v)T, get
o2=v^2T/c^2-vT/c. Either way, the L drops out when comparing
time intervals.)

>
>>>>>> As far as I can see, g depends only on the velocity, and is the
>>>>>> same at every point.
>>>>> Both g and g(LT) depend only on velocity.
>>>> No, you demonstrated a difference due to position. Your error
>>>> introduced a dependence on position.
>>> One can explain the variation with v of the ratio g(LT)/g by referring
>>> to the LT,
>>> which is evident, as g(LT) has been derived from them, but you can't
>>> give
>>> it a coherent physical meaning. This proves that the LT are physically
>>> incoherent.
>> You are correct. Your g(LT) has no coherent physical meaning,
>> but, because it is a symptom of your confusion, and nothing to
>> do with the LT, you have proven nothing.
>>
>
> g(LT) has nothing to do with the LT? The confused one is you.

No, because you used clocks that were not synchronized at the
beginning of the interval, but are synchronized at the end, it
is obvious that they are running at different rates, and the
comparison is meaningless.

If you use synchronized clocks, you find that g has the same
value everywhere.

>
>>>>>> To calculate the time dilation factor, you need to compare time
>>>>>> intervals between the systems.
>>>>>>
>>>>>> Let us look at a stick of length L', at rest in S', and
>>>>>> calculate the S times when t' is 0 and T'.
>>>>>>
>>>>> You make again the same logical mistake, because S' remains the moving
>>>>> frame.
>>>>> Iow, the stick continue to move wrt the S-frame, even if one considers
>>>>> it at rest
>>>>> in the S'-frame.
>>>> I fully agree that a stick at rest in S' is moving in S. If I
>>>> made a logical mistake, then we made it together.
>>> Then what is the purpose of stressing that the stick can be considered
>>> at rest in the moving frame? Jumping from one frame to another, make
>>> logical mistakes and draw wrong conclusions?
>
>> Any frame can be considered as the rest frame; it is a feature
>> of Galilean relativity. The stick is at rest in the frame in
>> which it has zero velocity.
>
> No, physically, not any frame can be considered as the rest frame.

What, then, is the physical content of Galilean relativity? (As
Newton put it, "The motions of bodies included in a given space
are the same among themselves, whether that space is at rest, or
moves uniformly forwards in a right line without any circular
motion.")

How about if I more clearly state that the frames must be
inertial?

>>
>>>> When you tried to derive your equations, you had to put in the g
>>>> factor by hand. You made an ad-hoc assumption about light speed
>>>> along y', without physical justification.
>>>>
>>> It is now the nth time that I claim that the physical justification for
>>> y'=ct' is that
>>> the vector v is perpendicular to the vector c, hence, vectorially,
>>> c+v=c.
>> From that equation, then the light speed in the original frame
>> is sqrt(c^2+v^2), but you stated that it was c. (Remember that
>> you affirmed that relative velocity was invariant).
>
> What is the original frame? I presume that you mean the rest frame.
> Yes, in S, the velocity ot the light signal following the y'-axis is
> sqrt(c^2+v^2), but it remains c in S'.

Previously, you stated that the speed of the light signal in the
rest system was 'c'. Which is it?

>
>>>> You obtained a result that is inconsistent with the results we
>>>> can obtain from considering the x direction alone (which
>>>> requires that a=1).
>>> There must be some flaw in your assertion, as I obtain physically
>>> meaningful
>>> transformations.
>> Only when g=1. I demonstrated earlier that, from the x
>> direction alone, we can conclude a(u-v)*a(v)=a(u). That
>> requires that a=exp(bv), with b constant. From a(v)=a(-v), we
>> find that b=0, so a=1.
>>
>
> Did you try your demonstration in the case where x'=ct' instead
> of x'=(c-v)t' ?

I used x=ut and x'=(u-v)t' to derive the relation for any value
of u.

Brian Kennelly

unread,

Oct 26, 2006, 6:27:04 PM10/26/06

to

Dirk Van de moortel wrote:

I think you are right about him. two steps forward, and three
steps back....

Dirk Van de moortel

unread,

Oct 26, 2006, 7:33:27 PM10/26/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message news:9370h.145$U%2....@newsfe16.phx...

Indeed, it's been like that since a good part of a decade.
While some of us advanced quite a few miles by honing
our explanations to sheer objective irresistibility, our friend
gladly and open-mindedly took giant leaps and joyfully ran
the other way.

A retrograde Echternach procession ;-)

Dirk Vdm

mlut...@wanadoo.fr

unread,

Oct 27, 2006, 1:50:48 PM10/27/06

to

Brian Kennelly wrote:
> mlut...@wanadoo.fr wrote:
> > Brian Kennelly wrote:
> >> mlut...@wanadoo.fr wrote:
> >>> Brian Kennelly wrote:
> >>>> mlut...@wanadoo.fr wrote:
> >>>>> Brian Kennelly wrote:
> >>>>>> mlut...@wanadoo.fr wrote:
> >>>>>>> Brian Kennelly wrote:
> >>>>> I reject the POR.
> >>>> On what basis? The principle of relativity has been a
> >>>> fundamental part of physics since at least the time of Newton.
> >>>> It is consistent with all observations.
> >>>>
> >>> I don't reject Galilean relativity, what I reject is Einsteinian
> >>> relativity, based on
> >>> the hypothesis of the constancy of light speed.
> >> Your proposed equations are inconsistent with Galilean relativity.
> >
> > I didn't claim that they were.
> You can't have it both ways. Either you reject Galilean
> relativity or you don't. Above, you claimed that you do not
> reject it, and now you acknowledge that your equations are
> inconsistent with it. Of what use are your equations?
>

Starting from Galimean relativity, I got equations that can be
considered
as an extension of it. Like most SRists, you like to quibble.

Quibbling again, unless you didn't understand what I said.

And what 'g' do you get in between?

Yes, but those SRists who consider that the Earth can be considered
as orbiting around an Earth satellite forget that in SR, the frames
must be inertial.

>
> >>
> >>>> When you tried to derive your equations, you had to put in the g
> >>>> factor by hand. You made an ad-hoc assumption about light speed
> >>>> along y', without physical justification.
> >>>>
> >>> It is now the nth time that I claim that the physical justification for
> >>> y'=ct' is that
> >>> the vector v is perpendicular to the vector c, hence, vectorially,
> >>> c+v=c.
> >> From that equation, then the light speed in the original frame
> >> is sqrt(c^2+v^2), but you stated that it was c. (Remember that
> >> you affirmed that relative velocity was invariant).
> >
> > What is the original frame? I presume that you mean the rest frame.
> > Yes, in S, the velocity ot the light signal following the y'-axis is
> > sqrt(c^2+v^2), but it remains c in S'.
>
> Previously, you stated that the speed of the light signal in the
> rest system was 'c'. Which is it?

Then I was inattentive, sorry.

>
> >
> >>>> You obtained a result that is inconsistent with the results we
> >>>> can obtain from considering the x direction alone (which
> >>>> requires that a=1).
> >>> There must be some flaw in your assertion, as I obtain physically
> >>> meaningful
> >>> transformations.
> >> Only when g=1. I demonstrated earlier that, from the x
> >> direction alone, we can conclude a(u-v)*a(v)=a(u). That
> >> requires that a=exp(bv), with b constant. From a(v)=a(-v), we
> >> find that b=0, so a=1.
> >>
> >
> > Did you try your demonstration in the case where x'=ct' instead
> > of x'=(c-v)t' ?
> I used x=ut and x'=(u-v)t' to derive the relation for any value
> of u.

So, it doesn't apply to x'=ct'. Anyhow, it is wrong:

> For the moment let us ignore the y direction. We agree that,
> with the assumption of an invariant relative velocity we have
> reached:

> x'=a(v)(x-vt)
> t'=a(v)t

> 'a' is a function of v, a(0)=1, and from symmetry a(v)=a(-v).

When v = 0, a(0)=1, x'=x and t'=t. End of the story!

> From these equations, with x=ut, we get

> x'=a(ut-vt)
=(u-v)t'

Yes.

> So, for any velocity in the x direction, the relative velocity
> between that point and the S' origin is invariant.

> The transformation from S' to S" is
> x"=a(u-v)(x'-(u-v)t')
> t"=a(u-v)t'

What is S"?
What is the velocity S" relative to S'? It seems to be u-v.

> Combining these, we get the transformation:
> x"=a(u-v)a(v)(x-vt-(u-v)t)
> =a(u-v)a(v)(x-ut)
> t"=a(u-v)a(v)t

????

> The direct equations give:
> x"=a(u)(x-ut)
> t"=a(u)t

????????

> Consistency requires that a(u-v)*a(v)=a(u).

> Setting u=0, we find that a(v)*a(-v)=1. This gives a(v)^2=1, or
> a(v)=1 (the negative root is excluded, because a(0)=1).

Setting u=0, we have
x'= (u-v)t'= -vt'
t'= t/g

> We do not need to examine the y direction to conclude that your
> transformation must reduce to:
> x'=x-vt

> t'=t .

Marcel Luttgens

Brian Kennelly

unread,

Oct 27, 2006, 2:52:53 PM10/27/06

to

mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:
>>> Brian Kennelly wrote:
>>>> mlut...@wanadoo.fr wrote:
>>>>> Brian Kennelly wrote:
>>>> Of course, the measuring instruments are also moving with S', so
>>>> they will be contracted as well, and S' will not measure any
>>>> change of length.
>>> You misunderstood, I didn't refer to the constancy of the length of
>>> the stick at rest in S' and measured in S'. I clearly said " in S, its
>>> length is L, in S', it is L'=L/g", meaning that the length of the stick
>>> measured in S' is contracted wrt its length measured in S.
>> Using your equations, we conclude that the length of the stick
>> is independent of is speed for any observer. We do find that
>> its length depends on the speed of the observer, but that the
>> observer will not be able to measure the change. Of what use
>> are your equations?
>>
>
> Quibbling again, unless you didn't understand what I said.

Apparently, you don't understand your equations.

>
>>>> Identify o2 and o2', and you will find that g(LT)=g
>>> g(LT) is a ratio, which is a function of v:
>>> g(LT) = t2/(T/g) = g(1 + v/c - v^2/c^2)
>>> And g(LT)/g = 1 + v/c - v^2/c^2.
>>> Could you show how you get g(LT)/g = 1, when v<>0 ?
>> You did not identify the starting times of the intervals.
>> Because your analysis is based on t1'=t2', you must also use
>> o1'=o2' (as you wrote above, "the clocks have to be beforehand
>> synchronized"). The length of the interval is L, and the left
>> end is located at the origin of S':
>> x1'=0
>> x2'=L
>>
>> At o1'=0, we find:
>> o1=g(o1'+v0/c^2)
>> =0
>> o2=g(o2'+vL/c^2)
>> =g(vL/c^2)
>>
>> At t1'=T', we find, as before:
>> t1=gT'
>> t2=g(T'+vL/c^2)
>>
>> It is obvious that t2-o2=g(t2'-o2') and t1-01=g(t1'-01'), with
>> the same value of g at both ends.
>
> And what 'g' do you get in between?

You get the same value of g at any point you choose. The
position, represented by <L>, drops out of the result.

>
>> (You can also do this directly from the equation t2=t1+vL/c^2.
>> When o1=0, o2=vL/c^2. If you insist on using L=(c-v)T, get
>> o2=v^2T/c^2-vT/c. Either way, the L drops out when comparing
>> time intervals.)
>>
>>

>> If you use synchronized clocks, you find that g has the same
>> value everywhere.
>>

>>> What is the original frame? I presume that you mean the rest frame.
>>> Yes, in S, the velocity ot the light signal following the y'-axis is
>>> sqrt(c^2+v^2), but it remains c in S'.
>> Previously, you stated that the speed of the light signal in the
>> rest system was 'c'. Which is it?
>
> Then I was inattentive, sorry.

Are you now saying that, in S, you have <y=ct>?

>
>>>>>> You obtained a result that is inconsistent with the results we
>>>>>> can obtain from considering the x direction alone (which
>>>>>> requires that a=1).
>>>>> There must be some flaw in your assertion, as I obtain physically
>>>>> meaningful
>>>>> transformations.
>>>> Only when g=1. I demonstrated earlier that, from the x
>>>> direction alone, we can conclude a(u-v)*a(v)=a(u). That
>>>> requires that a=exp(bv), with b constant. From a(v)=a(-v), we
>>>> find that b=0, so a=1.
>>>>
>>> Did you try your demonstration in the case where x'=ct' instead
>>> of x'=(c-v)t' ?
>> I used x=ut and x'=(u-v)t' to derive the relation for any value
>> of u.
>
> So, it doesn't apply to x'=ct'. Anyhow, it is wrong:

By choosing u=c+v, you get the case of x'=ct'. It applies for
any velocity.

>
>> For the moment let us ignore the y direction. We agree that,
>> with the assumption of an invariant relative velocity we have
>> reached:
>
>> x'=a(v)(x-vt)
>> t'=a(v)t
>
>> 'a' is a function of v, a(0)=1, and from symmetry a(v)=a(-v).
>
> When v = 0, a(0)=1, x'=x and t'=t. End of the story!
>
>> From these equations, with x=ut, we get
>
>> x'=a(ut-vt)
> =(u-v)t'
>
> Yes.
>
>> So, for any velocity in the x direction, the relative velocity
>> between that point and the S' origin is invariant.
>
>> The transformation from S' to S" is
>> x"=a(u-v)(x'-(u-v)t')
>> t"=a(u-v)t'
>
> What is S"?

S" is the frame moving with the object with speed u.

> What is the velocity S" relative to S'? It seems to be u-v.

Yes, that follows from the equations we already have.

>
>> Combining these, we get the transformation:
>> x"=a(u-v)a(v)(x-vt-(u-v)t)
>> =a(u-v)a(v)(x-ut)
>> t"=a(u-v)a(v)t
>
> ????

OK. I will do it again slowly.
The transformation from S' to S", with speed <u-v> is
x"=a(u-v)(x'-(u-v))t
t"=a(u-v)t'

Substituting the equations we already have for the
transformation from S to S':

x"=a(u-v)(a(v)(x-vt)-(u-v)a(v)t)
=a(u-v)a(v)(x-vt-(u-v)t)
=a(u-v)a(v)(x-vt-ut+vt)
x"=a(u-v)a(v)(x-ut)

t"=a(u-v)a(v)t

>
>> The direct equations give:
>> x"=a(u)(x-ut)
>> t"=a(u)t
>
> ????????

These are simply your equations for the case v=u, giving the
transformation from S to S".

>
>> Consistency requires that a(u-v)*a(v)=a(u).
This is a functional equation, that must be satisfied by a, for
all values of u and v.

>
>> Setting u=0, we find that a(v)*a(-v)=1. This gives a(v)^2=1, or
>> a(v)=1 (the negative root is excluded, because a(0)=1).
>
> Setting u=0, we have
> x'= (u-v)t'= -vt'

I don't know if you were trying to advance your objection, or
just posting an equation. Was there a point?

> t'= t/g
You ignored the functional equation, and substituted a=1/g,
without any justification. (Of course, the functional equation
can be expressed in terms of g, and, after inverting, gives
g(u-v)g(v)=g(u), leading to g=1.)

Androcles

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Oct 27, 2006, 5:42:26 PM10/27/06

to

"Brian Kennelly" <bwken...@cox.net> wrote in message

news:70p0h.5154$%l3....@newsfe13.phx...
| mlut...@wanadoo.fr wrote:

| > Setting u=0, we have
| > x'= (u-v)t'= -vt'
| I don't know

Learn it then.

Setting u=0, we have
x'= (u-v)t'= -vt'

Is that so fuckin' hard you have to be so bloody-minded as to
say you don't know, ignoramus? Learn it, shithead.
Your fuckin' wires are crossed, you bloody-minded cunt.
Androcles

mlut...@wanadoo.fr

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Oct 27, 2006, 5:58:10 PM10/27/06

to

Brian Kennelly wrote:
> mlut...@wanadoo.fr wrote:
> > Brian Kennelly wrote:
> >> mlut...@wanadoo.fr wrote:
> >>> Brian Kennelly wrote:
> >>>> mlut...@wanadoo.fr wrote:
> >>>>> Brian Kennelly wrote:
> >>>> Of course, the measuring instruments are also moving with S', so
> >>>> they will be contracted as well, and S' will not measure any
> >>>> change of length.
> >>> You misunderstood, I didn't refer to the constancy of the length of
> >>> the stick at rest in S' and measured in S'. I clearly said " in S, its
> >>> length is L, in S', it is L'=L/g", meaning that the length of the stick
> >>> measured in S' is contracted wrt its length measured in S.
> >> Using your equations, we conclude that the length of the stick
> >> is independent of is speed for any observer. We do find that
> >> its length depends on the speed of the observer, but that the
> >> observer will not be able to measure the change. Of what use
> >> are your equations?
> >>
> >
> > Quibbling again, unless you didn't understand what I said.
> Apparently, you don't understand your equations.

It is clear that I meant that the length L' of a moving stick is
contracted by g
relative to a stick of length L at rest. In S', the stick is at rest,
so its length is L.

Can't you read? I just said that in S, the velocity ot the light signal

following the y'-axis is sqrt(c^2+v^2), but it remains c in S'.

Hence (cf. my derivation), y = sqrt(c^2 - v^2) * t and y'=ct'.

You don't seem to understand the meaning of the transformation
x'=(x-vt)/g
t' = t/g,
where x is the position of some object in frame S,
v is the velocity of a frame S' relative to S,
x-vt is the difference in S between x and the origin of frame S'
after a time t,
x'=(x-vt)/g is the coordinate of the object in S'.

In the former discussions, we used x=ct, but you can of course
choose x=ut. Then x'=(u-v)t/g. But notice that x' is the coordinate
of the object in S'.

Now you consider the frame S" of the object, and claim that
the transformation from S' to S", with speed <u-v> is

x"=a(u-v)(x'-(u-v))t"
t"=a(u-v)t'

This is plainly wrong, because the coordinate of the object
in S" is *zero*, thus x"=0.

Your demonstration must be a hoax!

Marcel Luttgens

Brian Kennelly

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Oct 27, 2006, 10:03:18 PM10/27/06

to

OK. In S', its length is L, so we have for the endpoints:
0=(x1-vt)/g
:: x1=vt

L=(x2-vt)/g
:: x2=gL+vt

x2-x1=gL.
Conclusion: The moving stick is measured to be longer in the S
system than in its rest system.

>
>>>>> What is the original frame? I presume that you mean the rest frame.
>>>>> Yes, in S, the velocity ot the light signal following the y'-axis is
>>>>> sqrt(c^2+v^2), but it remains c in S'.
>>>> Previously, you stated that the speed of the light signal in the
>>>> rest system was 'c'. Which is it?
>>> Then I was inattentive, sorry.
>> Are you now saying that, in S, you have <y=ct>?
>
> Can't you read? I just said that in S, the velocity ot the light signal
>
> following the y'-axis is sqrt(c^2+v^2), but it remains c in S'.
> Hence (cf. my derivation), y = sqrt(c^2 - v^2) * t and y'=ct'.

Your math fails. If y=sqrt(c^2-v^2)t and x=vt, then the speed
of the signal is c, not sqrt(c^2+v^2). OTOH, that is the
correct speed, under your assumptions, if y=ct and x=vt.

>> OK. I will do it again slowly.
>> The transformation from S' to S", with speed <u-v> is
>> x"=a(u-v)(x'-(u-v))t
>> t"=a(u-v)t'
>
> You don't seem to understand the meaning of the transformation
> x'=(x-vt)/g
> t' = t/g,
> where x is the position of some object in frame S,
> v is the velocity of a frame S' relative to S,
> x-vt is the difference in S between x and the origin of frame S'
> after a time t,
> x'=(x-vt)/g is the coordinate of the object in S'.

Correct, but I will continue to use the multiplier a(v), because
using 1/g implies something that you have not demonstrated.

>
> In the former discussions, we used x=ct, but you can of course
> choose x=ut. Then x'=(u-v)t/g. But notice that x' is the coordinate
> of the object in S'.

I demonstrated before that, if x=ct implies x'=(c-v)t', x=ut
implies x'=(u-v)t' for any u. And yes, it is understood that
x' is the coordinate in S', and x is the coordinate in S, and x"
is the coordinate in S".

>
> Now you consider the frame S" of the object, and claim that
> the transformation from S' to S", with speed <u-v> is
> x"=a(u-v)(x'-(u-v))t"
> t"=a(u-v)t'
>
> This is plainly wrong, because the coordinate of the object
> in S" is *zero*, thus x"=0.

What is wrong? I used your equations.
Why would the be wrong when x"=0, if your equations are correct
when x'=0?

>
> Your demonstration must be a hoax!

Where is the error?

Androcles

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Oct 28, 2006, 2:03:06 AM10/28/06

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"Brian Kennelly" <bwken...@cox.net> wrote in message

news:Mkv0h.18$Hc...@newsfe11.phx...

Dependent on which way the light is going, shithead.
http://www.androcles01.pwp.blueyonder.co.uk/Smart/Smart.htm
How am I doing?
Androcles

mlut...@wanadoo.fr

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Oct 28, 2006, 10:21:21 AM10/28/06

to

You really don't understand what you are speaking about.
Till now, one considered that S' -and the stick- are moving away
at v from the S-frame, hence L'=L/g. Knowing L', the length of the
stick
in S', one conclude that the corresponding length in S is L=gL'.
Now L' is called L by you ("In S', the stick is at rest, so its length
is L")
so L must of course be called L'. Hence, its length in S is L'=gL,
according to your new terminology.
I am becoming rather annoyed by your sophistry.

.
> >
> >>>>> What is the original frame? I presume that you mean the rest frame.
> >>>>> Yes, in S, the velocity ot the light signal following the y'-axis is
> >>>>> sqrt(c^2+v^2), but it remains c in S'.
> >>>> Previously, you stated that the speed of the light signal in the
> >>>> rest system was 'c'. Which is it?
> >>> Then I was inattentive, sorry.
> >> Are you now saying that, in S, you have <y=ct>?
> >
> > Can't you read? I just said that in S, the velocity ot the light signal
> >
> > following the y'-axis is sqrt(c^2+v^2), but it remains c in S'.
> > Hence (cf. my derivation), y = sqrt(c^2 - v^2) * t and y'=ct'.

> Your math fails. If y=sqrt(c^2-v^2)t and x=vt, then the speed
> of the signal is c, not sqrt(c^2+v^2). OTOH, that is the
> correct speed, under your assumptions, if y=ct and x=vt.

This has nothing to do with math, it's a matter of geometry.
According to S, the signal travels *obliquely*. Apply the Pythagorean
theorem, and you will get y=sqrt(c^2-v^2)t.
I am now more than annoyed by your unceasing quibbling.
Unless quibbling is not the right word, it could be incompetence
instead.

If you had understood my equations, you would have concluded
that the way you use them is irrelevant, as your S" is nothing more
than another name for the object. When you use sophistry, you should
at least show some intelligence.

Marcel Luttgens

Dirk Van de moortel

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Oct 28, 2006, 10:27:00 AM10/28/06

to

<mlut...@wanadoo.fr> wrote in message news:1162030881.2...@m7g2000cwm.googlegroups.com...

[snip]

> You really don't understand what you are speaking about.
> Till now, one considered that S' -and the stick- are moving away
> at v from the S-frame, hence L'=L/g.

Hence L = L' / g.
This is called "length contraction", Marcel.

Dirk Vdm

mlut...@wanadoo.fr

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Oct 28, 2006, 1:30:04 PM10/28/06

to

Dirk Van de moortel wrote:

No, that's what is called sophistry, the usual reasoning mode of
SRists.
But they are too dense to realize it. They are really brainwashed
by 100 years of Einsteinism.

Marcel Luttgens

Dirk Van de moortel

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Oct 28, 2006, 1:56:10 PM10/28/06

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<mlut...@wanadoo.fr> wrote in message news:1162042204.5...@h48g2000cwc.googlegroups.com...

>
> Dirk Van de moortel wrote:
>> <mlut...@wanadoo.fr> wrote in message news:1162030881.2...@m7g2000cwm.googlegroups.com...
>>
>> [snip]
>>
>> > You really don't understand what you are speaking about.
>> > Till now, one considered that S' -and the stick- are moving away
>> > at v from the S-frame, hence L'=L/g.
>>
>> Hence L = L' / g.
>> This is called "length contraction", Marcel.
>>
>> Dirk Vdm
>
> No, that's what is called sophistry, the usual reasoning mode of
> SRists.

Well, perhaps for some strange reason you choose to call L
the proper length as measured in S', and L' the measured
lenght in S. To confuse the Russians, right?

> But they are too dense to realize it. They are really brainwashed
> by 100 years of Einsteinism.

Ah, well, a brain wash wouldn't be of much use to you, would it?
Something as tiny as that doesn't get all that filthy, does it?

Dirk Vdm

Brian Kennelly

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Oct 28, 2006, 4:22:29 PM10/28/06

to

It was your condition. If you go back and read your statement
of the conditions, you will find, "In S', the stick is at rest,

so its length is L."

No matter what you call the length, your equations predict that
a moving stick will be measured to be longer than in its rest
system.

> .
>>>>>>> What is the original frame? I presume that you mean the rest frame.
>>>>>>> Yes, in S, the velocity ot the light signal following the y'-axis is
>>>>>>> sqrt(c^2+v^2), but it remains c in S'.
>>>>>> Previously, you stated that the speed of the light signal in the
>>>>>> rest system was 'c'. Which is it?
>>>>> Then I was inattentive, sorry.
>>>> Are you now saying that, in S, you have <y=ct>?
>>> Can't you read? I just said that in S, the velocity ot the light signal
>>>
>>> following the y'-axis is sqrt(c^2+v^2), but it remains c in S'.
>>> Hence (cf. my derivation), y = sqrt(c^2 - v^2) * t and y'=ct'.
>
>> Your math fails. If y=sqrt(c^2-v^2)t and x=vt, then the speed
>> of the signal is c, not sqrt(c^2+v^2). OTOH, that is the
>> correct speed, under your assumptions, if y=ct and x=vt.
>
> This has nothing to do with math, it's a matter of geometry.
> According to S, the signal travels *obliquely*. Apply the Pythagorean
> theorem, and you will get y=sqrt(c^2-v^2)t.
> I am now more than annoyed by your unceasing quibbling.
> Unless quibbling is not the right word, it could be incompetence
> instead.

I am trying to find a rational reason to accept your <y'=ct'>.
It makes perfect sense if <y=ct>, or if we accept Einstein's
light postulate, but I don't understand where you get it.
Sometimes you assert that the light speed in S is c, sometimes
you assert that it is sqrt(c^2+v^2).

Your argument that the direction is perpendicular to the motion
does not work, because, in S', the velocity is zero. All
directions are perpendicular to a zero vector. In S, where the
velocity is directed along the x axis, the light signal is *not*
perpendicular to the motion.

>
>>> This is plainly wrong, because the coordinate of the object
>>> in S" is *zero*, thus x"=0.
>> What is wrong? I used your equations.
>> Why would the be wrong when x"=0, if your equations are correct
>> when x'=0?
>
> If you had understood my equations, you would have concluded
> that the way you use them is irrelevant, as your S" is nothing more
> than another name for the object. When you use sophistry, you should
> at least show some intelligence.

S" is the frame moving with the speed u relative to S, just as
S' is the frame moving with speed v relative to S. It is also
true that, with your equations, S" is moving with speed u-v
relative to S'.

How did I misuse the equations?

Your equations give directly:
x'=a(v)(x-vt)
t'=a(v)t

and

x"=a(u)(x-ut)
t"=a(u)t

I wanted the transformation from S' to S". Assuming that the
transformations will have the same form, I got:
x"=a(w)(x'-wt')
t"=a(w)t'

Where is the error? What did I misunderstand?

mlut...@wanadoo.fr

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Oct 28, 2006, 5:02:15 PM10/28/06

to

If you can, use my equations correctly.
x' = (x-vt)/g
t' = t/g
For instance, x=cT, thus x'=(c-v)T/g.
Consider a stick of length (c-v)t in S, such as x1=vT and x2=cT
x1' = (vT-vT)/g = 0
x2' = (cT-vT)/g
x2'-x1' = (cT-vT)/g = (x2-x1)/g
L' = L/g, L' being the length of the stick in S'. It is obviously
contracted.
I will not repeat such elementary demonstration anymore.

I can't help you. How many time did I tell you and explained to you
that in S', y'=ct', and in S, y=sqrt(c^2-v^2)t ?

>
> >
> >>> This is plainly wrong, because the coordinate of the object
> >>> in S" is *zero*, thus x"=0.
> >> What is wrong? I used your equations.
> >> Why would the be wrong when x"=0, if your equations are correct
> >> when x'=0?
> >
> > If you had understood my equations, you would have concluded
> > that the way you use them is irrelevant, as your S" is nothing more
> > than another name for the object. When you use sophistry, you should
> > at least show some intelligence.
> S" is the frame moving with the speed u relative to S, just as
> S' is the frame moving with speed v relative to S. It is also
> true that, with your equations, S" is moving with speed u-v
> relative to S'.
>
> How did I misuse the equations?

You case is desperate. Apply my equations to the object, which has
been done more than once, then call the object S" if you want.
This change nothing to the result.

To conclude (for good) those discussions about the so-called relativity
of simultaneity, an Einsteinian hoax, I invite you to try to understand
the following point of view from

Ben Rudiak-Gould, Oct 26 2006, in the thread "question (length
contraction)"

"In Newtonian physics, there's no difference between you alone
accelerating
and the whole rest of the universe accelerating in the opposite
direction.
The only thing that matters is the relative distances and angles
between
different objects at each instant of absolute time, and those will be
the
same either way. It's easy to define a reference frame which gives all
distances and all angles at all times in terms of some rigid reference
body,
and it's just like an inertial frame except that there are extra
"fictitious
forces" whose sole purpose is to make sure that the rest of the
universe
accelerates in the proper way. Someone once suggested that Newton's
second
law should be changed to

F = (Mm/(M-m)) a

where m is the mass of the object that's apparently being acted upon,
and M
is the mass of the rest of the universe. This is indistinguishable from
F =
ma when m << M, and it has the nice property that the force needed to
accelerate an object is the same as the force needed to accelerate
everything else in the opposite direction.

In special relativity this doesn't work, because of the way that space
and
time are mixed together. If you naively try to define a global
reference
frame around an arbitrarily moving object, it has all kinds of
pathological
properties: in general there are surfaces where time stops, and regions

where time runs backwards, and branch cuts where the time coordinate of
a
point is discontinuous with the time coordinate of neighboring points,
and
regions of spacetime that don't have coordinates at all. The laws of
physics
become a mess, and you can't just patch up the situation by adding
fictitious forces.

So the answer to your question is that the contraction of the universe
when
you accelerate is an artifact of the pathological coordinate system
that
people have to use in order to claim that the universe contracts when
you
accelerate. The universe couldn't really accelerate like that. In fact
the
universe can't accelerate at all in special relativity, at least not
rigidly, because the rigid acceleration of an object of length L can
never
exceed c^2 / L."

Thank you nevertheless,

Marcel Luttgens

Dirk Van de moortel

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Oct 28, 2006, 5:11:01 PM10/28/06

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<mlut...@wanadoo.fr> wrote in message news:1162054935....@k70g2000cwa.googlegroups.com...

>
> If you can, use my equations correctly.
> x' = (x-vt)/g
> t' = t/g

Stick of lenght L' in S'.
Want to measure it in S?
Measure front and back at same time t, so
{ dt = 0
{ dx' = ( dx - v dt ) / g
gives
dx' = dx / g
so
dx = g dx'
so
L = g L'
so it becomes elongated in the S-frame.
Nothing you can do about it.
Except perhaps if you measure the distances to the
end points at different times and then subtract the distances.
Marcel has very long trains ;-)

> For instance, x=cT, thus x'=(c-v)T/g.
> Consider a stick of length (c-v)t in S, such as x1=vT and x2=cT
> x1' = (vT-vT)/g = 0
> x2' = (cT-vT)/g
> x2'-x1' = (cT-vT)/g = (x2-x1)/g
> L' = L/g, L' being the length of the stick in S'. It is obviously
> contracted.

If it is at rest in S', then it is measured to be longer in S.
So it is obviously elongated.

> I will not repeat such elementary demonstration anymore.

You better don't ;-)

Dirk Vdm

Brian Kennelly

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Oct 28, 2006, 5:31:05 PM10/28/06

to

You changed the conditions. From those equations, the stick is
at rest in S, not in S'.

Why do you insist on confusing yourself by using endpoint
coordinates like vT and cT? If you use 0 and L, you can avoid
confusing T and t.

> x1' = (vT-vT)/g = 0
> x2' = (cT-vT)/g
> x2'-x1' = (cT-vT)/g = (x2-x1)/g
> L' = L/g, L' being the length of the stick in S'. It is obviously
> contracted.
> I will not repeat such elementary demonstration anymore.

The truth is that your equations predict a result that is
independent of the speed of the stick.
Assume that its speed is u, then, if the endpoints are 0 and L
when t=0, then:

x1=ut
x2=L+ut

x1'=(ut-vt)/g
x2'=(L+ut-vt)/g

If x2'-x1'=L', then L'=L/g for any value of u. In other words,
it doesn't matter if the stick is moving or not, L' will be less
than L.

The real problem is that L' is the difference of coordinates,
but the coordinates in S' are using a different scale, and do
not represent length in the same units.

Consider a moving meter stick in S. S measures the same
difference of coordinates for any speed, and concludes that the
there is no change of length. (If the coordinate unit is the
meter, it covers one unit at any time.) S' will see the same
meter stick covering 1/g units of his coordinates. One unit of
his coordinates therefore measures g meters, not 1 meter. The
difference is simply a difference of scale, not of length.

If you assume that the meter stick covers one coordinate unit in
its rest system, then your equations predict that the moving
meter stick is longer than when it is at rest.

Apparently, you cannot identify a misuse or error, and I
conclude that you have no answer. It was fun while it lasted.

>
> To conclude (for good) those discussions about the so-called relativity
> of simultaneity, an Einsteinian hoax, I invite you to try to understand
> the following point of view from
>
> Ben Rudiak-Gould, Oct 26 2006, in the thread "question (length
> contraction)"
>
>

Brian Kennelly

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Oct 28, 2006, 9:59:26 PM10/28/06

to

mlut...@wanadoo.fr wrote:
> If you can, use my equations correctly.
> x' = (x-vt)/g
> t' = t/g
> For instance, x=cT, thus x'=(c-v)T/g.
> Consider a stick of length (c-v)t in S, such as x1=vT and x2=cT
> x1' = (vT-vT)/g = 0
> x2' = (cT-vT)/g
> x2'-x1' = (cT-vT)/g = (x2-x1)/g
> L' = L/g, L' being the length of the stick in S'. It is obviously
> contracted.
> I will not repeat such elementary demonstration anymore.
>

I decided to go back and review this at a more basic level.
What is the right hand coordinate value of a meter stick on the
x' axis, at rest in S', if its left end is at 0?

How many meter sticks does it take to cover the interval from 0
to L' along the x' axis?

How long is the interval from 0 to L'?

mlut...@wanadoo.fr

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Oct 29, 2006, 10:13:31 AM10/29/06

to

Dirk Van de moortel wrote:

> <mlut...@wanadoo.fr> wrote in message news:1162054935....@k70g2000cwa.googlegroups.com...
> >
> > If you can, use my equations correctly.
> > x' = (x-vt)/g
> > t' = t/g
>
> Stick of lenght L' in S'.
> Want to measure it in S?
> Measure front and back at same time t, so
> { dt = 0
> { dx' = ( dx - v dt ) / g
> gives
> dx' = dx / g
> so
> dx = g dx'
> so
> L = g L'
> so it becomes elongated in the S-frame.

The S' observer, knowing that his stick is contracted
relative to a stick at rest, will conclude that the stick of the
observer at rest is *dilated" wrt its own stick.
Conversely, the S observer, kowing the S' is moving wrt him,
concludes that a stick in S' is contracted.

Can't SRists think logically?

Marcel Luttgens

Dirk Van de moortel

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Oct 29, 2006, 10:25:45 AM10/29/06

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<mlut...@wanadoo.fr> wrote in message news:1162116811.0...@e64g2000cwd.googlegroups.com...

>
> Dirk Van de moortel wrote:
>> <mlut...@wanadoo.fr> wrote in message news:1162054935....@k70g2000cwa.googlegroups.com...
>> >
>> > If you can, use my equations correctly.
>> > x' = (x-vt)/g
>> > t' = t/g
>>
>> Stick of lenght L' in S'.
>> Want to measure it in S?
>> Measure front and back at same time t, so
>> { dt = 0
>> { dx' = ( dx - v dt ) / g
>> gives
>> dx' = dx / g
>> so
>> dx = g dx'
>> so
>> L = g L'
>> so it becomes elongated in the S-frame.
>
> The S' observer, knowing that his stick is contracted
> relative to a stick at rest, will conclude that the stick of the
> observer at rest is *dilated" wrt its own stick.
> Conversely, the S observer, kowing the S' is moving wrt him,
> concludes that a stick in S' is contracted.
>
> Can't SRists think logically?

Here's some razor sharp logical thinking:
If and only if you express yourself in terms of measurements,
events and coordinates, you will understand.
Conclusion: you will never understand.

>
> Marcel Luttgens
>
>> Nothing you can do about it.
>> Except perhaps if you measure the distances to the
>> end points at different times and then subtract the distances.
>> Marcel has very long trains ;-)

Marcel can have *extremely* long trains ;-)

Dirk Vdm

mlut...@wanadoo.fr

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Oct 29, 2006, 10:54:32 AM10/29/06

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I don't confuse T and t, as T corresponds to a specific time t.

>
>
> > x1' = (vT-vT)/g = 0
> > x2' = (cT-vT)/g
> > x2'-x1' = (cT-vT)/g = (x2-x1)/g
> > L' = L/g, L' being the length of the stick in S'. It is obviously
> > contracted.
> > I will not repeat such elementary demonstration anymore.
> The truth is that your equations predict a result that is
> independent of the speed of the stick.
> Assume that its speed is u, then, if the endpoints are 0 and L
> when t=0, then:
>
> x1=ut
> x2=L+ut
>
> x1'=(ut-vt)/g
> x2'=(L+ut-vt)/g
>
> If x2'-x1'=L', then L'=L/g for any value of u. In other words,
> it doesn't matter if the stick is moving or not, L' will be less
> than L.

For a specific time t =T, the stick ends occupy in S the definite
positions x1=uT and x2=L+uT. It cannot be considered as moving
in S.

But the answer is straightforward! Your demonstration uses two
different entities, the object and its frame, wheras they are
identical.
Physically, the frame *is* the object under another name.
So, your further development makes no sense. It can be called
a logical misuse.

Bye, Brian, and continue to enjoy your LT and their incoherences.

Marcel Luttgens

Dirk Van de moortel

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Oct 29, 2006, 11:33:34 AM10/29/06

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<mlut...@wanadoo.fr> wrote in message news:1162119272....@i42g2000cwa.googlegroups.com...
>
> Brian Kennelly wrote:

[snip]

>> Apparently, you cannot identify a misuse or error, and I
>> conclude that you have no answer. It was fun while it lasted.
>
> But the answer is straightforward! Your demonstration uses two
> different entities, the object and its frame, wheras they are
> identical.
> Physically, the frame *is* the object under another name.
> So, your further development makes no sense. It can be called
> a logical misuse.
>
> Bye, Brian, and continue to enjoy your LT and their incoherences.

>
> Marcel Luttgens

Ah... those infamous incoherences of analytic geometry and
linear algebra, discovered by Marcel Luttgens All By Himself.

Dirk Vdm

Brian Kennelly

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Oct 29, 2006, 2:46:51 PM10/29/06

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mlut...@wanadoo.fr wrote:
>
> For a specific time t =T, the stick ends occupy in S the definite
> positions x1=uT and x2=L+uT. It cannot be considered as moving
> in S.

So, the stick is at rest in S *and* in S'? Conclusion, S' is
not moving, v=0.

> But the answer is straightforward! Your demonstration uses two
> different entities, the object and its frame, wheras they are
> identical.
> Physically, the frame *is* the object under another name.
> So, your further development makes no sense. It can be called
> a logical misuse.

My demonstration uses /three/ reference frames, not two.

These equations are yours:
x'=a(v)(x-vt}
t'=a(v)t

x"=a(u)(x-ut}
t"=a(u)t

Obviously, I did not make any errors in them that cannot be
attributed to you.

This equation is a generalization of your equations, using S' as
the starting point, rather than S:
x"=a(u-v)(x-(u-v)t')
t"=a(u-v)t'

If that is a misuse, then what are the correct equations to
transform from S' to S"?

The Ghost In The Machine

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Oct 29, 2006, 5:08:05 PM10/29/06

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In sci.physics.relativity, mlut...@wanadoo.fr
<mlut...@wanadoo.fr>
wrote
on 29 Oct 2006 02:13:31 -0800
<1162116811.0...@e64g2000cwd.googlegroups.com>:

>
> Dirk Van de moortel wrote:
>> <mlut...@wanadoo.fr> wrote in message news:1162054935....@k70g2000cwa.googlegroups.com...
>> >
>> > If you can, use my equations correctly.
>> > x' = (x-vt)/g
>> > t' = t/g
>>
>> Stick of lenght L' in S'.
>> Want to measure it in S?
>> Measure front and back at same time t, so
>> { dt = 0
>> { dx' = ( dx - v dt ) / g
>> gives
>> dx' = dx / g
>> so
>> dx = g dx'
>> so
>> L = g L'
>> so it becomes elongated in the S-frame.
>
> The S' observer, knowing that his stick is contracted
> relative to a stick at rest, will conclude that the stick of the
> observer at rest is *dilated" wrt its own stick.
> Conversely, the S observer, kowing the S' is moving wrt him,
> concludes that a stick in S' is contracted.
>
> Can't SRists think logically?

How does one measure a moving stick?

There are several methods I can think of, all using light as an
intermediary. Obviously one cannot simply go outside and
try to match one's own stick with the moving stick, especially
if one wants to stay in one's own coordinate-space.

Some setup. (x,t)_O is the observer's space; (x', t')_M
the stick's space. (The subscript identifies the space,
to avoid ambiguity.) The observer has a stick of
known length, a mirror, and a precise clock. The moving
coordinate system has two lights, one at each end of
the stick.

And of course the Lorentz dictates that

x_M = g*(x_O-vt_O)
t_M = g*(t_O-vx_O/c^2)
or (g*(x_O-vt_O), g*(t_O-vx_O/c^2))_M

and the inverse Lorentz:

x_O = g*(x_M+vt_M)
t_O = g*(t_M+vx_M/c^2)
or (g*(x_M+vt_M), g*(t_M+vx_M/c^2))_O

(Note that the rod is overtaking the observer, if the observer is facing
in the positive x direction. This may affect some signs in the
calculations below.)

Reference: How to calculate v.

Affix a mirror to O's rod's endpoint, and observe one of the remote
lights passing by O's rod's endpoints, then calculate

v = L_O/(t_1 - t_0 - L_O/c)

where L_O is the observer's rod's length. This works because lightspeed
is invariant. There's no Lorentz here, either.

(Note: In the usual setups, t_0 = 0 since the origins of observer and
stick are coincident at time 0.)

Method #1: Pass the rod.

Observe both lights passing by. These are events (0,0)_M
and (L_M, t_M)_M for some L_M (currently unknown) and t_M.
Because the observer's at x_O = 0, we know that

(0,0)_M = (0,0)_O
(L_M,t_M)_M = (0,t_O)_O

In other words, x_O = g*(L_M+v*t_M) = 0, or t_M = -L_M/v.
The problem is t_M is in the stick's space, and therefore
not directly observable. We're stuck with

t_O = g*(t_M+v*L_M/c^2) = g*(t_M+v^2*t_M/c^2)
= g*t_M*(-1+v^2/c^2) = -t_M*sqrt(1-v^2/c^2)

and since we know v we can naively calculate L = -v*t_M,
or L = L_M*sqrt(1-v^2/c^2).

The rod is measured as being too short.

Method #2: Reflect.

Sometime after the rod passes us by, say at time t_F (in the
observer's space), send light pulses that will reflect off the rod's
endpoints, using a variant of radar. The difference between the
reflected pulses should give us an idea of the rod's length.

Or does it?

The radar pulse can be modeled as:

(c*(t_O-t_F),t_O)_O

where t_O >= t_F.

In M-space this transforms into

(g*(c*(t_O-t_F)-v(t_O)),g*((t_O)-vc*(t_O-t_F)/c^2) )_M
= ( g*(c*t_O-c*t_F-v*t_O), g*(t_O-v/c*t_O+v/c*t_F) )_M
= ( c*(g*t_O-gv/c*t_O-g*t_F), g*t_O-gv/c*t_O+gv/c*t_F) )_M
= ( c*(t_M - t_G), t_M)_M

where t_M = g*t_O-gv/c*t_O+gv/c*t_F and
t_G = g*(1+v/c)*t_F = t_F*sqrt(1+v/c)/sqrt(1-v/c)
which, as it turns out, mathematically proves lightspeed invariance,
though it's clear that the target thinks the radar pulse originated
at a different time. (Of course, it's in a different coordinate
system anyway.)

The pulse will hit [*] the near endpoint at (0,t_M1)_M
and the far endpoint at (L_M,t_M2)_M. These are easy
to calculate:

t_M1 = t_G
t_M2 = t_G+L_M/c

but again they're in the wrong space, so we have to transform back.

At this point it's probably simpler just to set t_F = 0 (which means the
radar pulse starts just as the origin of M passes the origin of O), and
therefore t_G = t_M1 = t_O1 = 0. Hence t_M2 = L_M/c and we can compute
t_O2 = g*(t_M2+vL_M/c^2) = g*(L_M/c+vL_M/c^2) = g*L_M/c*(1+v/c)
= L_M/c*sqrt(1+v/c)/sqrt(1-v/c)

and if we naively calculate
L = (t_O2-t_O1)*c = L_M*sqrt(1+v/c)/sqrt(1-v/c)
the rod is now measured as being too *long*.

If one instead sets t_F to be a large enough negative value
(i.e., we aim the radar behind us as the rod approaches),
the pulse travels backwards with the equation

(-c*(t_O-t_F),t_O)_O

to hit the rod; this results in the mismeasurement of
L=L_M*sqrt(1-v/c)/sqrt(1+v/c), or the rod is too short again. I
leave the mathematical verification of this to the interested reader.

Three measurements, three different results, none of them "correct".

>
> Marcel Luttgens
>
>> Nothing you can do about it.
>> Except perhaps if you measure the distances to the
>> end points at different times and then subtract the distances.
>> Marcel has very long trains ;-)
>>
>> > For instance, x=cT, thus x'=(c-v)T/g.
>> > Consider a stick of length (c-v)t in S, such as x1=vT and x2=cT
>> > x1' = (vT-vT)/g = 0
>> > x2' = (cT-vT)/g
>> > x2'-x1' = (cT-vT)/g = (x2-x1)/g
>> > L' = L/g, L' being the length of the stick in S'. It is obviously
>> > contracted.
>>
>> If it is at rest in S', then it is measured to be longer in S.
>> So it is obviously elongated.
>>
>> > I will not repeat such elementary demonstration anymore.
>>
>> You better don't ;-)
>>
>> Dirk Vdm
>

[*] reflectance is generally ignored here since the pulse
originates in observer-space; therefore total transit time
is simply double the time from origin to endpoint, if one
assumes isotropic space.

--
#191, ewi...@earthlink.net
fortune: not found

--
Posted via a free Usenet account from http://www.teranews.com

mlut...@wanadoo.fr

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Oct 30, 2006, 2:20:18 PM10/30/06

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Brian Kennelly wrote:
> mlut...@wanadoo.fr wrote:
> >
> > For a specific time t =T, the stick ends occupy in S the definite
> > positions x1=uT and x2=L+uT. It cannot be considered as moving
> > in S.
> So, the stick is at rest in S *and* in S'? Conclusion, S' is
> not moving, v=0.

Of course, the stick is moving according to x1=ut and x2=L+ut.
But its *instantaneous* position when t=T is x1=uT and x2=L+uT.

> > But the answer is straightforward! Your demonstration uses two
> > different entities, the object and its frame, wheras they are
> > identical.
> > Physically, the frame *is* the object under another name.
> > So, your further development makes no sense. It can be called
> > a logical misuse.
> My demonstration uses /three/ reference frames, not two.
>

The equations give the same results for the object or S".

> These equations are yours:
> x'=a(v)(x-vt}
> t'=a(v)t
>
> x"=a(u)(x-ut}
> t"=a(u)t
>
> Obviously, I did not make any errors in them that cannot be
> attributed to you.
>
> This equation is a generalization of your equations, using S' as
> the starting point, rather than S:
> x"=a(u-v)(x-(u-v)t')
> t"=a(u-v)t'
>
> If that is a misuse, then what are the correct equations to
> transform from S' to S"?

The position in S' of the object, or of its frame S", is given by
x' = (x-vt)/g
For instance, if x = ct,
x' = (c-v)t/g
Period.

Marcel Luttgens

Brian Kennelly

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Oct 30, 2006, 2:50:25 PM10/30/06

to

mlut...@wanadoo.fr wrote:
> Brian Kennelly wrote:
>> mlut...@wanadoo.fr wrote:
>>> For a specific time t =T, the stick ends occupy in S the definite
>>> positions x1=uT and x2=L+uT. It cannot be considered as moving
>>> in S.
>> So, the stick is at rest in S *and* in S'? Conclusion, S' is
>> not moving, v=0.
>
> Of course, the stick is moving according to x1=ut and x2=L+ut.
> But its *instantaneous* position when t=T is x1=uT and x2=L+uT.

I agree, but you went further by stating that it "cannot be
considered as moving". So, we agree; the stick is located
between uT and L+uT, and is moving to the right with speed u.

This predicts that there is no change of length for a moving stick.

You did not answer the question, so I will rephrase it for you.

How do I find x" and t", if I know x' and t', if I know that S"
is moving with speed w, relative to S'?