SR fundamental contradiction

mlut...@wanadoo.fr

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Sep 27, 2006, 4:44:08 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, 4:56:09 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, 4:56:44 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, 5:30:00 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 27, 2006, 8:42:47 PM9/27/06

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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 27, 2006, 9:43:02 PM9/27/06

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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, 12:29:41 AM9/28/06

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Sorcerer

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Sep 28, 2006, 5:17:52 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, 5:38:46 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, 8:36:38 AM9/28/06

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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, 9:25:20 AM9/28/06

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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, 9:34:01 AM9/28/06

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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, 10:32:56 AM9/28/06

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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, 10:37:03 AM9/28/06

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, 12:50:02 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, 12:50:36 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, 1:22:03 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, 2:00:37 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, 3:02:18 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, 5:36:18 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, 8:41:38 AM9/29/06

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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, 9:12:50 AM9/29/06

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, 9:24:12 AM9/29/06

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, 9:25:11 AM9/29/06

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, 9:35:37 AM9/29/06

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, 12:24:38 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, 6:37:17 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, 6:58:30 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, 1:19:33 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, 3:02:27 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,

Sep 30, 2006, 10:59:20 PM9/30/06

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

unread,

Oct 1, 2006, 4:46:43 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, 4:53:05 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, 5:26:06 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, 5:38:54 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, 8:15:47 AM10/2/06

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, 1:03:51 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, 8:37:31 AM10/3/06

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

unread,

Oct 3, 2006, 12:25:26 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, 5:55:27 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, 6:53:07 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, 6:54:20 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, 2:47:06 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, 2:51:13 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, 2:55:34 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, 9:00:27 AM10/6/06

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, 11:43:59 AM10/6/06

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, 4:32:07 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, 6:30:23 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, 11:39:31 AM10/7/06

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, 11:47:45 AM10/7/06

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, 12:56:15 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, 1:52:44 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

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Oct 7, 2006, 2:54:39 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, 4:35:27 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, 4:59:45 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, 5:12:35 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, 6:16:19 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, 6:22:01 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, 6:35:48 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, 6:42:01 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 7, 2006, 11:29:59 PM10/7/06

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 7, 2006, 11:56:29 PM10/7/06

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, 12:56:40 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, 2:34:12 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, 5:04:29 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, 5:09:43 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, 8:27:40 AM10/8/06

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, 12:15:13 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, 12:19:24 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, 1:08:22 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, 1:15:27 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, 1:16:22 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, 1:52:25 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, 1:54:37 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, 4:07:48 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, 5:15:29 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, 5:17:08 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, 5:21:41 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, 5:27:13 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, 5:58:42 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, 6:14:11 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, 7:41:55 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 8, 2006, 8:42:12 PM10/8/06

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, 1:50:04 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, 6:30:01 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, 11:24:20 AM10/9/06

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, 11:54:57 AM10/9/06

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, 12:28:04 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, 12:34:46 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, 12:40:24 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, 12:40:50 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, 12:44:13 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, 1:06:25 PM10/9/06

to

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

Brian Kennelly

unread,

Oct 9, 2006, 1:26:46 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, 3:30:20 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, 8:47:52 AM10/10/06

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, 8:57:41 AM10/10/06

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, 10:35:37 AM10/10/06

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, 10:40:21 AM10/10/06

to

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

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

You are a stupid liar, Kennelly.

Androcles