Androcles Misses the Main Point

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Daryl McCullough

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Feb 12, 2005, 2:21:18 PM2/12/05
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Androcles says...

>> As I said, Einstein assumes that 1 and 2 are both c:
>
>No he doesn't. NOWHERE can you find that in his paper

That is the most bizarre claim I've ever heard, Androcles.
http://www.fourmilab.ch/etexts/einstein/specrel/www/

We will raise this conjecture (the purport of which will
hereafter be called the ``Principle of Relativity'') to
the status of a postulate, and also introduce another
postulate, which is only apparently irreconcilable with
the former, namely, that light is always propagated in
empty space with a definite velocity c which is independent
of the state of motion of the emitting body.

He is clearly saying that light has constant speed c.

Later he writes:

We now have to prove that any ray of light, measured in the
moving system, is propagated with the velocity c, if, as we
have assumed, this is the case in the stationary system

Somehow you missed the fact that Einstein was assuming that
light propagates at speed c in both frames? After years of
arguing about relativity, you missed this, most important
fact about Einstein's theory?

Androcles, go take a course. Find a junior college that teaches
an introductory course in relativity. I'm sure it's not too expensive.

--
Daryl McCullough
Ithaca, NY

Androcles

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Feb 12, 2005, 9:04:42 PM2/12/05
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"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:culkv...@drn.newsguy.com...

> Androcles says...
>
>>> As I said, Einstein assumes that 1 and 2 are both c:
>>
>>No he doesn't. NOWHERE can you find that in his paper
>
> That is the most bizarre claim I've ever heard, Androcles.
> http://www.fourmilab.ch/etexts/einstein/specrel/www/
>
> We will raise this conjecture (the purport of which will
> hereafter be called the ``Principle of Relativity'') to
> the status of a postulate, and also introduce another
> postulate, which is only apparently irreconcilable with
> the former, namely, that light is always propagated in
> empty space with a definite velocity c which is independent
> of the state of motion of the emitting body.
>
> He is clearly saying that light has constant speed c.

Yep, the speed of mosquitoes, c, is 5 fps in all frames of reference.
Proof:
(c+v)/(1+ v/c) = c,
so
(5+1) / (1+ 1/5) = 5
Clearly.


>
> Later he writes:
>
> We now have to prove that any ray of light, measured in the
> moving system, is propagated with the velocity c, if, as we
> have assumed, this is the case in the stationary system

Ah... well, the word to look at is "prove".

>
> Somehow you missed the fact that Einstein was assuming that
> light propagates at speed c in both frames?

Funny, he also said "the velocity of light in our theory plays the
part, physically, of an infinitely great velocity"
AND
"But the ray moves relatively to the initial point of k, when measured
in the stationary system, with the velocity c-v",
so I fail to see any proof, just you selecting what you want to back up
your personal theory of relativity.

> After years of
> arguing about relativity, you missed this, most important
> fact about Einstein's theory?

The most important FACT is
"We now have to PROVE that any ray of light, measured in the
moving system, is propagated with the velocity c", MCullough,
and I gave PROOF that mosquitoes always fly in all brains of inertial
reference ( which is yours) with a definite velocity 5 fps which is
independent of the state of motion of Sam.

>
> Androcles, go take a course.

I already did. S100, M100, M201, M202, TM251, TS282, SM351.....


> Find a junior college that teaches
> an introductory course in relativity.

Nah... I prefer math, not imbecile.


> I'm sure it's not too expensive.

What waste good money I could spend on beer instead?
You are not being rational, psychotic McCullough.


Androcles.

Daryl McCullough

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Feb 12, 2005, 10:00:21 PM2/12/05
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Androcles says...

>"Daryl McCullough" <stevend...@yahoo.com> wrote

>> Androcles says...
>>
>>>> As I said, Einstein assumes that 1 and 2 are both c:
>>>
>>>No he doesn't. NOWHERE can you find that in his paper
>>
>> That is the most bizarre claim I've ever heard, Androcles.
>> http://www.fourmilab.ch/etexts/einstein/specrel/www/
>>
>> We will raise this conjecture (the purport of which will
>> hereafter be called the ``Principle of Relativity'') to
>> the status of a postulate, and also introduce another
>> postulate, which is only apparently irreconcilable with
>> the former, namely, that light is always propagated in
>> empty space with a definite velocity c which is independent
>> of the state of motion of the emitting body.
>>
>> He is clearly saying that light has constant speed c.
>
>Yep

So you admit now that you were wrong?

Dojyd

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Feb 12, 2005, 10:36:53 PM2/12/05
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.......... .morons .........


Daryl McCullough

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Feb 12, 2005, 10:31:06 PM2/12/05
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There are three quotes from Einstein:

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

1. ...light is always propagated in empty space with a


definite velocity c which is independent of the state
of motion of the emitting body.

2. ...the velocity of light in our theory plays the
part, physically, of an infinitely great velocity.

3. But the ray moves relatively to the initial point of k,


when measured in the stationary system, with the velocity c-v

Androcles has trouble understanding these three statements, so
I will explain them in more detail.

1. To say that light propagates at definite velocity c seems clear
enough. It means that in any inertial reference frame that uses
rectangular inertial coordinates, if a light signal is sent from
a point A with coordinates x_A, y_A, z_A at time t_A and propagates
to a point B with coordinates x_B, y_B, z_B at time t_B, then it
must be that

(x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2 = c^2 (t_B - t_A)^2

This is explained in Einstein's original paper, "On the Electrodynamics
of Moving Bodies" which can be found on the web at
http://www.fourmilab.ch/etexts/einstein/specrel/www/.

2. To say that light speed plays the role of an infinite velocity is to
say that it is a maximum velocity: nothing can travel faster than
light. Mathematically, the equations of special relativity approach
those of Galilean relativity in the limit as c-->infinity.

3. "the ray moves relatively to the initial point of k,


when measured in the stationary system, with the velocity c-v"

means that if light ray is travelling at speed c, and an object
(the "inital point of k") is travelling at speed v, then the
closing velocity between the light ray and the object is c-v.
That's clear enough.

The only difficulty, which is an insurmountable difficulty for poor
Androcles, is getting past the idea that

the closing velocity between the light ray and the object

and

the velocity of the light ray, as measured in the frame of the object

are *not* the same thing in Special Relativity.

RP

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Feb 12, 2005, 11:20:21 PM2/12/05
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Daryl McCullough wrote:

I think you've pinned down one of his errors in judgment here. Now
work on these misconceptions of his:

2) c can be arbitrarily chosen, that is, the speed of a mosquito can
be considered to be invariant, even though the guys carrying the
ladder are moving faster wrt the ground than the mosquito is moving
wrt them.

3) Light is a thing in motion, that is, c is a velocity, and not a
constant that can be set to 1.

4) Time is not a vector

*(He should consider the Minkowski relation, when c=1,

dx^2 + idt^2 = dx'2 + idt'^2

wherein it becomes evident that time is indeed a vector, in that the
expression on the right represents a simple rotation of the coordinate
frame in the same way that the expression below on the right

dx^2 + dy^2 = dx'^2 + dy^2

represents a rotation of the xy plane, in Galilean relativity.)

5) Einstein and Minkowski actually made all of the claims that are
posted here on their behalf :)


Richard Perry

Androcles

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Feb 13, 2005, 1:34:43 AM2/13/05
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"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:cumfs...@drn.newsguy.com...


You really are so fucking pathetically desperate to win a point, you'll
snip the tail off a sentence, won't you?

I have never once denied that the speed of light, RELATIVE TO THE SOURCE
at the TIME OF EMISSION [snip rest of sentence]

The speed of mosquitoes, c, is 5 fps in all frames of reference.


Proof:
(c+v)/(1+ v/c) = c,
so
(5+1) / (1+ 1/5) = 5
Clearly.

Now fuck off, you stupid moron, I've no more time for you.
Androcles.

Androcles

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Feb 13, 2005, 1:39:27 AM2/13/05
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I asked you why Einstein thinks 16 = 4, McCullough.
You have snipped and ignore the question.
There is no debate possible with a bigot like you, McCullough.

"Daryl McCullough" <stevend...@yahoo.com> wrote in message

news:cumhl...@drn.newsguy.com...

Androcles

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Feb 13, 2005, 1:43:47 AM2/13/05
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I asked McCullough why Einstein thinks 16 = 4, Perry.
He has repeatedly snipped and ignore the question.
There is no debate possible with a bigot like McCullough.
Now I'll ask you the same question to investigate how much
of a bigot you are.

Why does Einstein define time such that
x'/(c+v) = x'/(c-v), Perry?

Androcles.


"RP" <no_mail...@yahoo.com> wrote in message
news:3782l8F...@individual.net...

Tim Golden

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Feb 13, 2005, 1:52:08 AM2/13/05
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Geez. Such a fight. Are you sure you are not cro-magnon?
Or Electro-Magnon?
Is the simple answer doppler effect?

Androcles

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Feb 13, 2005, 4:17:35 AM2/13/05
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You no comprend anencephalitic?


Dirk Van de moortel

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Feb 13, 2005, 5:29:01 AM2/13/05
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"Daryl McCullough" <stevend...@yahoo.com> wrote in message news:cumhl...@drn.newsguy.com...

This has been explained to this idiot during more than *5* years.
It doesn't work. Just give it up, relax and enjoy :-)

Dirk Vdm


RP

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Feb 13, 2005, 7:35:22 AM2/13/05
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Androcles wrote:
> I asked McCullough why Einstein thinks 16 = 4, Perry.
> He has repeatedly snipped and ignore the question.
> There is no debate possible with a bigot like McCullough.
> Now I'll ask you the same question to investigate how much
> of a bigot you are.
>
> Why does Einstein define time such that
> x'/(c+v) = x'/(c-v), Perry?

He defines time as what a clock reads.
The intervals of time above are not equal as you've scripted them,
unless v=0. OTOH, unless I miss my guess, you are referring to two
expressions of time-like intervals that correspond to a detector
moving toward and then away from a source of light respectively. The
expressions (c+v) and (c-v) being the closing speeds of that detector
wrt a beam of light as measured by some inertial observer.

Wrt a given inertial observer light propagates at c wrt *him* alone,
and otherwise all relative velocities add as usual. If you want to
post as an objection the relativistic velocity composition formula
then first take note that it involves a transformation between frames
of reference, and thus a change in units of measure between the terms.
IOW, closing velocities add normally in any given frame just as they
are in the Galilean view, which is why you see expression such as
(c+v) and (c-v). Einstein even states on several occasions that the
lorentz transformation is a transformation between Galilean coordinate
systems.

Richard Perry

Androcles

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Feb 13, 2005, 7:57:31 AM2/13/05
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"RP" <no_mail...@yahoo.com> wrote in message
news:378vliF...@individual.net...

>
>
> Androcles wrote:
>> I asked McCullough why Einstein thinks 16 = 4, Perry.
>> He has repeatedly snipped and ignore the question.
>> There is no debate possible with a bigot like McCullough.
>> Now I'll ask you the same question to investigate how much
>> of a bigot you are.
>>
>> Why does Einstein define time such that
>> x'/(c+v) = x'/(c-v), Perry?
>
> He defines time as what a clock reads.

Hmm...You must be referring to some other Einstein. I'm talking about
the one that wrote

"If at the point A of space there is a clock, an observer at A can
determine the time values of events in the immediate proximity of A by
finding the positions of the hands which are simultaneous with these
events. If there is at the point B of space another clock in all
respects resembling the one at A, it is possible for an observer at B to
determine the time values of events in the immediate neighbourhood of B.
But it is not possible without further assumption to compare, in respect
of time, an event at A with an event at B. We have so far defined only
an ``A time'' and a ``B time.'' We have not defined a common ``time''
for A and B, for the latter cannot be defined at all unless 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"
in his paper
"ON THE ELECTRODYNAMICS OF MOVING BODIES".

Since you have the wrong Einstein, you probably can't answer my question


"Why does Einstein define time such that x'/(c+v) = x'/(c-v), Perry?"

Have a nice day.

Androcles.


RP

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Feb 13, 2005, 8:21:39 AM2/13/05
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Androcles wrote:

IOW, time is what a clock reads.

Have a nice day Andro.

Richard Perry


Androcles

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Feb 13, 2005, 8:49:18 AM2/13/05
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"RP" <no_mail...@yahoo.com> wrote in message
news:3792c4F...@individual.net...

Ah... seems we have more than one definition of time, then. There is
time as read by a clock, so if I say "What time is it?" and you say
"13:30 pm" we are talking about an instant, but if you say "How long
will it take to drive to London" and I reply "Oh, about 35 minutes" we
are talking about an interval of time.

"We have not defined a common ``time''
for A and B, for the latter cannot be defined at all unless 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"

In other words it will take the same INTERVAL of time to drive home FROM
London as it will to drive TO London.
And of course we are unable to know what the clock in London is reading
until we get there so we cannot arrive there at precisely 14:05 pm. That
has to play hell with railway time tables, no wonder the trains never
run on time.

What I still don't understand, though, is why the time at the front of
the train
is different from the time at the back of the train and gets further
apart the faster the train goes.

>
> Have a nice day Andro.

Thanks, I'd like to, but I can't quite work out what a day is, you see,
and its bothering me. I've got a rough idea, but I'd like to know
precisely. I thought it as an interval of time, but now I'm told time is
what clock reads...
Have a nice 14:00 pm :-)
Androcles.


DarylMcCullough

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Feb 13, 2005, 8:35:16 AM2/13/05
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Androcles asks

>>> Why does Einstein define time such that
>>> x'/(c+v) = x'/(c-v), Perry?

>> He defines time as what a clock reads.

This is an important point that Androcles has trouble
understanding. First of all, "what a clock reads"
only defines time between two events that are colocated
in the rest frame of that clock. In order to compare
the times of *distant* events, it is necessary to use
some clock synchronization convention. Einstein's
synchronization convention is this: two clocks
at rest in a frame can be synchronized by adjusting
them so that the time required for light to travel
from A to B is equal to the time required to travel
from B to A. This synchronizes the clocks *for* that
frame. Other frames will judge them *not* synchronized.

If a light signal is sent from point A
to point B, and reflected back to A, then

1. In the frame in which A and B are at rest, the time
for light to travel from A to B is equal to the time
for light to travel from B back to A.

2. In the frame in which A and B are moving at speed v
(with B a constant distance L in front of A), the time
required for light to travel from A to B is L/(c+v). The
time required for light to travel from B back to A is
L/(c-v). These times are *not* equal, of course.

Androcles

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Feb 13, 2005, 10:03:51 AM2/13/05
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"DarylMcCullough" <stevend...@yahoo.com> wrote in message
news:cunl2...@drn.newsguy.com...

> Androcles asks
>
>>>> Why does Einstein define time such that
>>>> x'/(c+v) = x'/(c-v), Perry?
>
>>> He defines time as what a clock reads.
>
> This is an important point that Androcles has trouble
> understanding. First of all, "what a clock reads"
> only defines time between two events that are colocated
> in the rest frame of that clock.

Oh, I didn't know a clock could record two events.
I thought it only recorded the corrent event, "now".
How does it define the time between breakfast and lunch, then?

> In order to compare
> the times of *distant* events, it is necessary to use
> some clock synchronization convention.

Oh yeah, that's right. Supernova 1987A took place
in 1987, 18 years ago, and not 170,000 years ago as I stupidly supposed
because "the velocity of light in our theory plays the part, physically,

of an infinitely great velocity."

How could I possible mistake 18 years for 170,000 years? Silly me.


> Einstein's
> synchronization convention is this: two clocks
> at rest in a frame can be synchronized by adjusting
> them so that the time required for light to travel
> from A to B is equal to the time required to travel
> from B to A.

Yeah, that's it. And since it takes 16 seconds one way and 4 seconds
back a clock at 32 * 60,000 km away can be synchronized by adjusting it
to Big Ben's clock in St Stephens Tower, Westminster, right? Or should
we set Big Ben to the time of the distant clock, 16 seconds early or 4
seconds late?

This synchronizes the clocks *for* that
> frame.

Well yeah, of course. Can't do it if the other clock is moving, we'd
have dopper shift.

Other frames will judge them *not* synchronized.

Well sure, and since just about every planet and moon in the universe is
moving, we can spend eternity synchronizing them.
Seems as if we'll never synchronize any of them.


> If a light signal is sent from point A
> to point B, and reflected back to A, then
>
> 1. In the frame in which A and B are at rest, the time
> for light to travel from A to B is equal to the time
> for light to travel from B back to A.
>
> 2. In the frame in which A and B are moving at speed v
> (with B a constant distance L in front of A), the time
> required for light to travel from A to B is L/(c+v). The
> time required for light to travel from B back to A is
> L/(c-v). These times are *not* equal, of course.

Ah, right. So to make them equal, we create a function tau() using
6 variables that includes apples, oranges, bananas and plums, x and t,
so that
稼tau(0,apple, orange, plum, banana, t)+tau(0,apple, orange, plum,
banana t+x'/(c-v)+x'/(c+v))] = tau(x',apple, orange, plum, banana,
t+x'/(c-v))

and then we combine the x with the t to give
t' = ( t - vx/c^2) / sqrt( 1 - v^2/c^2)
apple' = apple
orange' = orange
plum' = plum
banana' = banana
x' = (x-vt) / sqrt( 1 - v^2/c^2)
because x' is a function of tau, right?

Now, in MY theory of relativity,

banana' = ( banana - v * apple/ c^2) / sqrt( 1 - v^2/c^2)
apple' = ( apple - v * banana ) / sqrt( 1 - v^2/c^2)

where v is defined as d(apple)/d(banana),
and the value of c is 300,000,000 apples per banana
but plays the part, physically, of an infinitely great number of apples
per banana.

Androcles.


Daryl McCullough

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Feb 13, 2005, 10:59:22 AM2/13/05
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Androcles says...

>"DarylMcCullough" <stevend...@yahoo.com> wrote

>> This is an important point that Androcles has trouble
>> understanding. First of all, "what a clock reads"
>> only defines time between two events that are colocated
>> in the rest frame of that clock.
>
>Oh, I didn't know a clock could record two events.
>I thought it only recorded the corrent event, "now".
>How does it define the time between breakfast and lunch, then?

What you do is this: When you have breakfast, you look at your
clock, and write down the time (in minutes) on a piece of paper.
Suppose that it says 8:30, which is 510 minutes past midnight.
So you write down 510. Later, when you have lunch, you look at your clock
again, and write down the time. Suppose that it says 11:45. That's
705 minutes past midnight. So you write down 705.

Now subtract the two times to get 705 - 510 = 195 minutes. That's
the time between breakfast and lunch.

Any time you have a question, Androcles, just ask. That's what
we're here for.

Androcles

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Feb 13, 2005, 12:17:18 PM2/13/05
to

"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:cuntg...@drn.newsguy.com...

> Androcles says...
>
>>"DarylMcCullough" <stevend...@yahoo.com> wrote
>
>>> This is an important point that Androcles has trouble
>>> understanding. First of all, "what a clock reads"
>>> only defines time between two events that are colocated
>>> in the rest frame of that clock.
>>
>>Oh, I didn't know a clock could record two events.
>>I thought it only recorded the corrent event, "now".
>>How does it define the time between breakfast and lunch, then?
>
> What you do is this: When you have breakfast, you look at your
> clock, and write down the time (in minutes) on a piece of paper.
> Suppose that it says 8:30, which is 510 minutes past midnight.
> So you write down 510. Later, when you have lunch, you look at your
> clock
> again, and write down the time. Suppose that it says 11:45. That's
> 705 minutes past midnight. So you write down 705.
>
Oh, so the mosquito writes down 16, then he writes down 4, and (16 +
4)/2 = 16. Gotcha.


> Now subtract the two times to get 705 - 510 = 195 minutes. That's
> the time between breakfast and lunch.

What if I have breakfast at home and lunch in London, though?
Is London time the same as home time? After all, I am whizzing through
empty space at 0.0001c, just like Sam and Joe are walking and it is 32
miles
to London, so perhaps it shrank.

What's even more worrying is Joe. His mosquito flies to Sam in 4 seconds
and takes 16 seconds to come back, so then we have (4 + 16)/2 = 4.

>
> Any time you have a question, Androcles, just ask. That's what
> we're here for.

I did have one, but I can't seem to get an answer.
Why does Einstein define time such that (16+4)/2 = 16 ?

Androcles


Dirk Van de moortel

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Feb 13, 2005, 12:19:56 PM2/13/05
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"Daryl McCullough" <stevend...@yahoo.com> wrote in message news:cuntg...@drn.newsguy.com...

The thing is, he doesn't trust you since you are what he
calls a "relativist". He is convinced that the answer you
just gave is deviously wrong, and that you are preparing
to nail him to the ground, hich is of course not necessary
since he has nailed himself to the ground some 5 decades
ago.

Enjoy :-)

Dirk Vdm


Daryl McCullough

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Feb 13, 2005, 12:33:20 PM2/13/05
to
Androcles says...

>"Daryl McCullough" <stevend...@yahoo.com> wrote

>>>> This is an important point that Androcles has trouble
>>>> understanding. First of all, "what a clock reads"
>>>> only defines time between two events that are colocated
>>>> in the rest frame of that clock.

>What if I have breakfast at home and lunch in London, though?

As I said, a clock can be used to compute the time between


events that are colocated in the rest frame of that clock.

That means that they are at the same location.

Jesse Mazer

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Feb 13, 2005, 1:30:54 PM2/13/05
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Androcles wrote:

>"Daryl McCullough" <stevend...@yahoo.com> wrote in message
>news:cuntg...@drn.newsguy.com...
>
>
>>Androcles says...
>>
>>
>>
>>>"DarylMcCullough" <stevend...@yahoo.com> wrote
>>>
>>>
>>>>This is an important point that Androcles has trouble
>>>>understanding. First of all, "what a clock reads"
>>>>only defines time between two events that are colocated
>>>>in the rest frame of that clock.
>>>>
>>>>
>>>Oh, I didn't know a clock could record two events.
>>>I thought it only recorded the corrent event, "now".
>>>How does it define the time between breakfast and lunch, then?
>>>
>>>
>>What you do is this: When you have breakfast, you look at your
>>clock, and write down the time (in minutes) on a piece of paper.
>>Suppose that it says 8:30, which is 510 minutes past midnight.
>>So you write down 510. Later, when you have lunch, you look at your
>>clock
>>again, and write down the time. Suppose that it says 11:45. That's
>>705 minutes past midnight. So you write down 705.
>>
>>
>>
>Oh, so the mosquito writes down 16, then he writes down 4, and (16 +
>4)/2 = 16. Gotcha.
>
>

Daryl, you should note that Androcles doesn't understand that the
Lorentz transformation only works for inertial frames and isn't meant to
be used to determine the reading of clocks carried by observers who
change velocities, like the mosquito in this example. If you really
think he is arguing in good faith and that it is worth your time to try
to point out his misunderstandings, try explaining this to him and see
what he says.

Jesse

Androcles

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Feb 13, 2005, 7:57:20 PM2/13/05
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Aww, you snipped my question...
Androcles


Androcles

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Feb 13, 2005, 10:35:26 PM2/13/05
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"Jesse Mazer" <vze2...@mail.verizon.net> wrote in message
news:420FC6D3...@mail.verizon.net...

>
>
> Androcles wrote:
>
>>"Daryl McCullough" <stevend...@yahoo.com> wrote in message
>>news:cuntg...@drn.newsguy.com...

What a great ID for a message from McCullough!

I wonder how many people actually read the [W]hole the message...
LOL!

Jesse, you should try understand that McCullough isn't arguing in good
faith, he's taking the piss because he's a cretin and doesn't know what
else
to do.

The Lorentz transformation came from somewhere, it didn't just magically
appear out of thin air. Here's how.

The Lorentz transformation tau = (t-vx/c^2)
was derived, by Einstein, from the equation

tau = a * ( t - (vx' / (c^2-v^2))).

He obtained tau = a * ( t - (vx' / (c^2-v^2)))

by integrating the equation

dtau/dx' + v/(c^2-v^2) * dtau/dt = 0,

and he got that from

½ * [1/(c-v) + 1/(c+v)] * dtau/dt = dtau/dx' + 1/(c-v) * dtau/dt.

To obtain that, he differentiated the equation

½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))


The sequence to derive the Lorentz transformation is therefore

1)
Define t = x'/(c-v) = x'/(c+v) because the time for light to go from A
to B equals the time it takes to travel from B to A

2)
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))

3)
½[1/(c-v)+1/(c+v)] * dtau/dt = dtau/dx' + 1/(c-v) * dtau/dt

4)
dtau/dx' + v/(c^2-v^2) * dtau/dt = 0

5)
tau = a * ( t - (vx' / (c^2-v^2)))

6)
tau = (t-vx/c^2) / sqr(1-v^2/c^2)

(which you can verify at
http://www.fourmilab.ch/etexts/einstein/specrel/www/
section 3 )
and that comes from the mosquito example, so we cannot put the
Lorentz equations into the mosquito example to derive the Lorentz
equations, which the idiot McCullough tried to do.

So what I want to know is why Einstein thinks (16 + 4)/2 = 16
or even ½[tau(16+4) = tau (16)
or even ½[tau(16) + tau(4) ] = tau(16)
or even ½[tau(32,0,0, 16) + tau(0,0,0, 4) ] = tau(32,0,0,0, 16)
or even
½[tau(0,0,0,0)+tau(0,0,0,16+4)] = tau(32,0,0,16),

to derive the Lorentz equations, and of course McCullough is unable to
answer, so he refuses to answer and snips instead, which is standard
procedure for a relativist that continues to insist that we must start
at the Lorentz equations.

According to McCullough, (80,0,0,16) = (0,0,0,16)
by his synchronized magic markers with clocks attached stretching out
along the road at equal intervals from the big sign that says "Origin",
so he's unable to make up his mind what the time is even in one
stationary frame, let alone
in a moving frame, tau(80,0,0,16) = tau(0,0,0,16)

Now he's reduced to sarcasm and snippety snip snip to try to save face,
he imagines that I can't outdo him in a pissing contest after he's been
defeated in the logical contest. I must admit I did start the
pissing contest in a moment of boredom to goad him into it. He
fell for it.

Now, you may ask why I would continue to pummel McCullough into
the ground with logic instead of just plonking him as I normally would.
As it turns out, McCullough is quite a useful and imaginitive bloke,
even if he's hopeless at math.
It was McCullough that came up with Sam, Joe and a mosquito, also a big
sign saying "Origin" and a set of magic markers with clocks on them. I
added a ladder and turned "Origin" into "Origins", a fast food
restaurant, to add some colour to the the story.

To this end, then, McCullough is useful to me and I've realized (see, I
have learned something) that there may be other babies that I've thrown
out with the bathwater. So I deleted all the names on my kill-file
(you've been resurrected)
and am only putting back those that are truly useless, such as moortel.
You may come up with a tidbit I can make use of.


He that knows not, and knows not that he knows not, he is a fool.
Shun him.
He that knows not, and knows that he knows not, he is a simple man.
Teach him.
He that knows, and knows not that he knows, he is a tool.
Use him.
He that knows, and knows that he knows, he is a wise man.
Follow him.

Persian or Arabian Proverb.

Sometimes the third stanza is quoted as

He that knows, and knows not that he knows, he is a sleep. Wake him.

but I'm not sure what a "sleep" is when used as a noun. :-)

McCullough knows, and knows not that he knows.
Mazer I'm not sure about. Maybe you are a "sleep", because you seem
to be advising the ostler McCullough that he can teach the blacksmith
Androcles how to make a horseshoe. Wake up.

Androcles.


Dirk Van de moortel

unread,
Feb 14, 2005, 3:19:29 AM2/14/05
to

"Androcles" <Androcles@ MyPlace.org> wrote in message news:QHSPd.101495$K7.5...@fe2.news.blueyonder.co.uk...

>
> Aww, you snipped my question...
> Androcles

You see Daryl, whenever you think you can help this imbecile,
he gets suspicious and he shuts down communication.
The man is severely sick with paranoia :-)

Dirk Vdm


Dirk Van de moortel

unread,
Feb 14, 2005, 4:09:15 AM2/14/05
to

"Androcles" <Androcles@ MyPlace.org> wrote in message news:20VPd.102559$K7.9...@fe2.news.blueyonder.co.uk...

Jesse, you should understand that (1) some people really
honestly think that Androcles' condition *can* be cured...

>
> Jesse, you should try understand that McCullough isn't arguing in good
> faith, he's taking the piss because he's a cretin and doesn't know what
> else
> to do.
>

... and (2) that Androcles really honestly thinks that *everyone*
in the world is trying to put him in a straightjacket.

> The Lorentz transformation came from somewhere, it didn't just magically
> appear out of thin air. Here's how.

You see?
You explain how we use a clock and he panics ;-)
[snip panic escape]

Dirk Vdm


Daryl McCullough

unread,
Feb 14, 2005, 11:19:27 AM2/14/05
to
Androcles says...

>"Daryl McCullough" <stevend...@yahoo.com> wrote

>As for deriving the Lorentz tranformations, here's a derivation, going
>back to Sam and Joe. Let's introduce two new characters, Sally and Jane.
>
>Sam and Joe are at rest relative to frame B (for Boys).
>
>Sally and Jane are at rest relative to frame G (for Girls).
>
>Let the speed of frame B as measured in frame G be v.
>
>Androcles:
>_______________________________ --> v
>_|_|_|_|_|_|_|B|_|_|_|_|_|_|_|_
>
>
>_______________________________
>_|_|_|_|_|_|_|G|_|_|_|_|_|_|_|_
>
>
>McCullough:
>Also, by symmetry let the speed of frame G as measured in frame B also
>be v.
>
>Androcles:
>_______________________________
>_|_|_|_|_|_|_|B|_|_|_|_|_|_|_|_
>
>
>_______________________________ --> v
>_|_|_|_|_|_|_|G|_|_|_|_|_|_|_|_
>
>
>v = 0

Obviously, I should have been more explicit. In frame G,
Sam and Joe are travelling at speed v to the right. In
frame B, Sally and Jane are travelling at speed v to the
*left*. So

In frame G

_______________________________ --> v
_|_|_|_|_|_|_|B|_|_|_|_|_|_|_|_


_______________________________
_|_|_|_|_|_|_|G|_|_|_|_|_|_|_|_

In frame B

_______________________________
_|_|_|_|_|_|_|B|_|_|_|_|_|_|_|_


<-- v _______________________________
_|_|_|_|_|_|_|G|_|_|_|_|_|_|_|_

Let e(i,j) be the event at which rung number i of the G-ladder passes
rung number j of the B-ladder. Let x(i,j) be the location of this
event, in G-coordinates, and let t(i,j) be the time of this event,
in G-coordinates. Let x'(i,j) and t'(i,j) be the location and time
of this event in B-coordinates. Let's assume that we pick our origin
so that x(0,0) = t(0,0) = x'(0,0) = t'(0,0) = 0. Let's figure out
the coordinates of e(i,j) for arbitrary i and j.

Obviously, rung i of the G ladder is always at location x=iL, as
measured in frame G, so we have

1. x(i,j) = iL

To compute t(i,j), note that in the G-frame, at time t=0, rung i of
the G-ladder is at location x = i L and rung j of the B-ladder is
at location x = j l. Therefore, the distance between these rungs
is (iL - jl). Since the B-ladder is travelling at speed v, these
two rungs will pass each other at time

2. t(i,j) = (iL - jl)/v

The computation for frame B is similar, except that in frame B,
it is the B-rung number j that is stationary, so we have

3. x'(i,j) = jL

In frame B, rung i of the G-ladder at time t=0 is at location
x' = il, while the location of rung j of the B-ladder is always
x' = jL. So the distance between these rungs is (il - jL), and so
they will pass at time

4. t'(i,j) = (il - jL)/v

Since x(i,j) = iL, and x'(i,j) = jL, we can rewrite t
(for fixed i and j) as follows:

5. t = (x - x' l/L)/v (from equation 2)

which gives us x' in terms of x and t:

6. x' = (x - vt)L/l

Similarly, equation 4 can be rewritten (for fixed i and j) as

7. t' = (x l/L - x')/v

Substituting form x' from equation 6 gives us

8. t' = (x l/L - (x-vt) L/l)/v
= L/l t + x/v (l/L - L/l)

Let's let g be the ratio L/l. Then our transformation equations are:

9. x' = g (x-vt)
10. t' = g t + x/v (1/g - g)
= g (t + x/v (1/g^2 - 1))

Notice that so far, we haven't made any physical assumptions about
the speed of light or the length of moving objects, or the slowing
of moving clocks. We've only invoked the relativity principle, that
Sally and Jane have just as much right to consider themselves at rest
as Sam and Joe. So our equations so far work for Galilean relativity,
as well as Einstein's relativity. We haven't said anything
at all about the factor g = L/l. Galilean relativity is the special
case g = 1, which means that

x' = x-vt
t' = t

To get Einstein's relativity, we now impose another assumption:
the speed of light is c in all inertial reference frames, regardless
of the motion of the source.

So a light signal sent from Sam or Sally
at time t=0 will follow the path x = ct in frame G. In frame B,
we have

x' = g (x - vt)
= g (c - v) t
= cgt (1 - v/c)

t' = g (t + x/v (1/g^2 - 1))
= g t (1 + c/v(1/g^2 - 1))

So, in order for x' to equal c t', we must have

cgt (1-v/c) = cgt (1 + c/v (1/g^2 - 1))

which has the solution

1/g^2 = 1-v^2/c^2

or g = 1/square-root(1-(v/c)^2)

McCullough: Putting it altogether,

x' = 1/square-root(1-v^2/c^2) (x - vt)
t' = 1/square-root(1-v^2/c^2) (t + x/v (1-v^2/c^2 - 1))
= 1/square-root(1-v^2/c^2) (t - vx/c^2)

Androcles

unread,
Feb 14, 2005, 10:07:21 AM2/14/05
to

"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:cuq3r...@drn.newsguy.com...
> Androcles says...

>
>>Jesse, you should try understand that McCullough isn't arguing in good
>>faith
>
> I'm not arguing at all---I'm trying to explain.

Failed.

>
>>The Lorentz transformation came from somewhere, it didn't just
>>magically
>>appear out of thin air.
>

> I gave you a derivation, and you ignored it.


Actually I responded to it as a seperate post, but I'm not googling for
it again, so here's the copy. You ignored the response.

McCullough wrote:

As for deriving the Lorentz tranformations, here's a derivation, going
back to Sam and Joe. Let's introduce two new characters, Sally and Jane.

Sam and Joe are at rest relative to frame B (for Boys).

Sally and Jane are at rest relative to frame G (for Girls).

Let the speed of frame B as measured in frame G be v.

Androcles:
_______________________________ --> v
_|_|_|_|_|_|_|B|_|_|_|_|_|_|_|_


_______________________________
_|_|_|_|_|_|_|G|_|_|_|_|_|_|_|_


McCullough:
Also, by symmetry let the speed of frame G as measured in frame B also
be v.

Androcles:
_______________________________
_|_|_|_|_|_|_|B|_|_|_|_|_|_|_|_


_______________________________ --> v
_|_|_|_|_|_|_|G|_|_|_|_|_|_|_|_


v = 0

McCullough:
For simplicity, let's assume that the direction from Sam
to Joe is along the +x axis in both frames, and that the direction
from Sally to Jane is also along the +x axis in both frames.

Assume that Sam and Joe are holding a long ladder (the B-ladder)
between them. There is also a long ladder (the G-ladder) between
Sally and Jane. Let the distance between rungs of the B-ladder
be L, as measured in frame B, and let the distance be l, as measured
in frame G (we'll work out what l is in terms of L a bit later).
Symmetrically, let the distance between rungs of the G-ladder be L
as measured in frame G, and l as measured in frame B.

Let e(i,j) be the event at which rung number i of the G-ladder passes
rung number j of the B-ladder. Let x(i,j) be the location of this
event, in G-coordinates, and let t(i,j) be the time of this event,
in G-coordinates. Let x'(i,j) and t'(i,j) be the location and time
of this event in B-coordinates. Let's assume that we pick our origin
so that x(0,0) = t(0,0) = x'(0,0) = t'(0,0) = 0. Let's figure out
the coordinates of e(i,j) for arbitrary i and j.

Obviously, rung i of the G ladder is always at location x=iL, as
measured in frame G, so we have

1. x(i,j) = iL

To compute t(i,j), note that in the G-frame, at time t=0, rung i of
the G-ladder is at location x = i L and rung j of the B-ladder is
at location x = j l. Therefore, the distance between these rungs
is (iL - jl). Since the B-ladder is travelling at speed v, these
two rungs will pass each other at time

2. t(i,j) = (iL - jl)/v

Androcles:
Division by zero detected.
The motion of B in frame G was given as v
The motion of G in frame B was given as v.


McCullough:


The computation for frame B is similar, except that in frame B,
it is the B-rung number j that is stationary, so we have

3. x'(i,j) = jL

In frame B, rung i of the G-ladder at time t=0 is at location
x' = il, while the location of rung j of the B-ladder is always
x' = jL. So the distance between these rungs is (il - jL), and so
they will pass at time

4. t'(i,j) = (il - jL)/v

Androcles:
Division by zero detected


McCullough:


Since x(i,j) = iL, and x'(i,j) = jL, we can rewrite t
(for fixed i and j) as follows:

5. t = (x - x' l/L)/v (from equation 2)

which gives us x' in terms of x and t:

6. x' = (x - vt)L/l

Similarly, equation 4 can be rewritten (for fixed i and j) as

7. t' = (x l/L - x')/v

Substituting form x' from equation 6 gives us

8. t' = (x l/L - (x-vt) L/l)/v
= L/l t + x/v (l/L - L/l)

Let's let g be the ratio L/l. Then our transformation equations are:

9. x' = g (x-vt)
10. t' = g t + x/v (1/g - g)
= g (t + x/v (1/g^2 - 1))

Notice that so far, we haven't made any physical assumptions about
the speed of light or the length of moving objects, or the slowing
of moving clocks. We've only invoked the relativity principle, that
Sally and Jane have just as much right to consider themselves at rest
as Sam and Joe. So our equations so far work for Galilean relativity,
as well as Einstein's relativity. We haven't said anything
at all about the factor g = L/l. Galilean relativity is the special
case g = 1, which means that

x' = x-vt
t' = t

Androcles :
v= 0, x' = x.

McCullough:


To get Einstein's relativity, we now impose another assumption:
the speed of light is c in all inertial reference frames, regardless
of the motion of the source.


Androcles:
THIS NOT EINSTEIN'S POSTULATE.
Einstein DERIVED (c+v)/(1+v/c) = c in section 5 of "Electrodynamics"
Reference :
http://www.fourmilab.ch/etexts/einstein/specrel/www/
and BASED that derivation on section 3, from which McCullough
is now attempting to derive the Lorentz transforms.
Quote:
"with the help of the equations of transformation developed in § 3 "
Unquote.

McCullough's argument is entirely and competely circular.


McCullough:


So a light signal sent from Sam or Sally
at time t=0 will follow the path x = ct in frame G. In frame B,
we have

x' = g (x - vt)
= g (c - v) t
= cgt (1 - v/c)

Androcles:
McCullough has not stated what 'g' is.


McCullough:

t' = g (t + x/v (1/g^2 - 1))
= g t (1 + c/v(1/g^2 - 1))

So, in order for x' to equal c t', we must have

cgt (1-v/c) = cgt (1 + c/v (1/g^2 - 1))

which has the solution

1/g^2 = 1-v^2/c^2

or g = 1/square-root(1-(v/c)^2)

Androcles:
But since v = 0, g = 1.


McCullough: Putting it altogether,

x' = 1/square-root(1-v^2/c^2) (x - vt)
t' = 1/square-root(1-v^2/c^2) (t + x/v (1-v^2/c^2 - 1))
= 1/square-root(1-v^2/c^2) (t - vx/c^2)

--
Daryl McCullough
Ithaca, NY


>


>>and that comes from the mosquito example, so we cannot put the
>>Lorentz equations into the mosquito example to derive the Lorentz
>>equations, which the idiot McCullough tried to do.
>

> I never tried to use the mosquito example to derive the Lorentz
> transformations. It was just used to derive the mathematical
> result
>
> If objects A and B are travelling at speed v in a straight line,
> and a signal travels from one to the other and back at speed c,
> then the round trip time will be 2cL/(c^2 - v^2).
>
> This is a mathematical result which is independent of whether one
> assumes Galilean relativity or Einstein's relativity.


>
>>So what I want to know is why Einstein thinks (16 + 4)/2 = 16
>

> Since Einstein doesn't think that, and that doesn't follow from
> anything said by Einstein, your question is loony.


>
> --
> Daryl McCullough
> Ithaca, NY

You "derivation" is looney, McCullough.


Androcles.


Daryl McCullough

unread,
Feb 14, 2005, 6:59:44 AM2/14/05
to
Androcles says...

>Jesse, you should try understand that McCullough isn't arguing in good

>faith

I'm not arguing at all---I'm trying to explain.

>The Lorentz transformation came from somewhere, it didn't just magically

>appear out of thin air.

I gave you a derivation, and you ignored it.

>and that comes from the mosquito example, so we cannot put the


>Lorentz equations into the mosquito example to derive the Lorentz
>equations, which the idiot McCullough tried to do.

I never tried to use the mosquito example to derive the Lorentz


transformations. It was just used to derive the mathematical
result

If objects A and B are travelling at speed v in a straight line,
and a signal travels from one to the other and back at speed c,
then the round trip time will be 2cL/(c^2 - v^2).

This is a mathematical result which is independent of whether one
assumes Galilean relativity or Einstein's relativity.

>So what I want to know is why Einstein thinks (16 + 4)/2 = 16

Since Einstein doesn't think that, and that doesn't follow from


anything said by Einstein, your question is loony.

--
Daryl McCullough
Ithaca, NY

Daryl McCullough

unread,
Feb 14, 2005, 7:12:38 AM2/14/05
to
Androcles says...

>I did have one, but I can't seem to get an answer.
>Why does Einstein define time such that (16+4)/2 = 16 ?

He didn't, as has been explained many times. In the example
at hand, 16 seconds is the time required for light to travel
80 units in going from Sam to Joe as measured in the
*stationary* frame, not in Sam's frame. 4 seconds is the time
required for light to travel 20 units in going from Joe to Sam
in the *stationary* frame.

What Einstein said was that in *Sam's* frame, the time required
for light to go from Sam to Joe is the same as the time required
to go from Joe to Sam.

Your nonsensical result, (16+4)/2 = 16, follows from your assuming
that time as measured in Sam's frame is the same as time as measured
in the stationary frame. Why do you assume that? Einstein didn't
assume that. So your nonsensical result follows from an extra
assumption on your part. Yes, your assumption leads to a contradiction,
which is why you shouldn't make it.

The real question is: Why does Androcles define time so that
(16+4)/2 = 16?

Dirk Van de moortel

unread,
Feb 14, 2005, 10:43:35 AM2/14/05
to

"Androcles" <Androcles@ MyPlace.org> wrote in message news:J83Qd.102980$K7.3...@fe2.news.blueyonder.co.uk...

Yes, we have seen it before:
http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ExplainDonkey.html
It means "Please go away now, don't you understand
that I don't even *want* you to explain something?"

Dirk Vdm


Androcles

unread,
Feb 14, 2005, 11:48:40 AM2/14/05
to

"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:cuq4j...@drn.newsguy.com...

> Androcles says...
>
>>I did have one, but I can't seem to get an answer.
>>Why does Einstein define time such that (16+4)/2 = 16 ?
>
> He didn't, as has been explained many times.

Yeah yeah. He didn't say


"If at the point A of space there is a clock, an observer at A can
determine the time values of events in the immediate proximity of A by
finding the positions of the hands which are simultaneous with these
events. If there is at the point B of space another clock in all
respects resembling the one at A, it is possible for an observer at B to
determine the time values of events in the immediate neighbourhood of B.
But it is not possible without further assumption to compare, in respect
of time, an event at A with an event at B. We have so far defined only
an ``A time'' and a ``B time.'' We have not defined a common ``time''
for A and B, for the latter cannot be defined at all unless 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. "
and that was entirely made up by me.

Note that A and B are points in space. Not points along a road.

In the example
> at hand, 16 seconds is the time required for light to travel
> 80 units in going from Sam to Joe as measured in the
> *stationary* frame, not in Sam's frame.

You really do have a problem with A and B being points in space,
right?


4 seconds is the time
> required for light to travel 20 units in going from Joe to Sam
> in the *stationary* frame.

Yeah, we know. Space is stationary <yawn>.

>
> What Einstein said was that in *Sam's* frame, the time required
> for light to go from Sam to Joe is the same as the time required
> to go from Joe to Sam.


That's right. Silly, but that IS what he said. That's why I want to know
why he thinks 4 seconds equals 16 seconds.

>
> Your nonsensical result, (16+4)/2 = 16, follows from your assuming
> that time as measured in Sam's frame is the same as time as measured
> in the stationary frame. Why do you assume that?

I didn't assume it. Why shoud I assume anything as nonsensical
as x'/(c-v) = x'/(c+v) ?


> Einstein didn't
> assume that.

No of course not, he DEFINED it instead.


> So your nonsensical result follows from an extra
> assumption on your part.

What assumption is that, then?
I do agree that the result is nonsensical, of course, but
I put that down to the nonsensical definition that
the time for a mosquito to fly from Sam to Joe equals
the time it takes to fly from Joe to Sam.
Show my assumption.


> Yes, your assumption leads to a contradiction,
> which is why you shouldn't make it.

Well, what is it then?

> The real question is: Why does Androcles define time so that
> (16+4)/2 = 16?

Ah, I see your difficulty.
Ok, I can explain where you make an incorrect assumption.
Note that I'm carefully using the function tau() with all the correct
coordinates and not the time in the stationary frame.


The time it takes to fly from Sam to Joe is
tau(32,0,0,16) - tau(0,0,0,0)
The time it takes to fly from Joe to Sam is
tau(0,0,0,20) - tau(32,0,0,16)

Now we must determine what the function tau() is.

> There is a big sign along the side of the road saying "Origin".
> There is a clock beside the sign, which gives the time to the
> nearest 1/100 second.
> (Let's assume that the clocks "flip over"
> back to 0.00 when they reach a time of 99.99 seconds.)
>
> There are markers along the side of the road, with a spacing
> of one unit between them. Each marker has written on it a number
> telling the distance of that marker from the Origin sign. So
> the first marker has a "1" on it, the second has a "2" on it,
> etc.
>
> Also, at each marker is a clock. The clocks are all identical
> in construction. The clocks are synchronized such that Joe
> passes the clock under the Origin sign, the time reads 89.33.
> When Joe passes the clock at marker 1, the time on that clock reads
> 89.67.
> When Joe passes the clock at marker 2, that clock reads 90.00.
> When Joe passes the clock at marker 3, that clock reads 90.33.
> ...
> When Joe passes the clock at marker 32, that clock reads 0.00.
> ...
> When Joe passes the clock at marker 80, that clock reads 16.00
> ...
> When Joe passes the clock at marker 92, that clock reads 20.00
>
> Sam is behind Joe. When Sam passes the clock under the Origin
> sign, that clock reads 0.00 (which is 10.66 seconds after Joe
> passed it, according to that clock).
> When Sam passes the clock under marker 1, that clock reads 0.33.
> When Sam passes the clock under marker 2, that clock reads 0.66.
> ...
> When Sam passes the clock under marker 48, that clock reads 16.00.
> ...
> When Sam passes the clock under marker 60, that clock reads 20.00.
>
Now let's suppose that just as Sam passed the Origin sign, he
sent a homing mosquito towards Joe.

The mosquito passes marker 1 when the clock there reads 0.20.
> It passes marker 2 when the clock there reads 0.40.
> It passes marker 3 when the clock there reads 0.60.
> ...
> Finally, it reaches marker 80 when the clock there reads 16.00,
> which is exactly when Joe reaches that marker. Joe is holding

a paint can , and the mosquito is reflected back towards Sam.
> It passes marker 79 when the clock there reads 16.20.
> It passes marker 78 when the clock there reads 16.40.
> ...
> Finally, it reaches marker 60 when the clock there reads 20.00,
> which is exactly when Sam reaches that marker.
>
> So, in the coordinate system determined by the clocks and markers,
>
> 1. Sam passes the Origin at the same time (0.00) when Joe passes
> 32.

So we now write (32,0,0,0)

> 2. Sam passes marker 48 at the same time (16.00) when Joe passes
> 80,
which is the same time that the mosquito reaches Joe.

So we now write ( 48,0,0,16)


> 3. Sam passes marker 60 at the same time (20.00) when Joe passes
> 92.

So we write (60,0,0,20)

>
> In each of these events, the difference between Joe's marker number
> and
> Sam's marker number is 32.

Now we determine what the function tau() is.

tau(0,0,0,0) = tau(32,0,0,0)

because Sam's time is equal to Joes time, 0, but they are separated by
32 ft.
Now, they could be separated by 1,000,000 miles, and we'd still have
tau(0,0,0,0) = tau(1000000,0,0,0)
or in general,
tau(0, 0, 0, 0) = tau(L, 0, 0, 0)

We could even include some extra parameters,
tau(0, orange, apple, 0) = tau(L, orange, apple, 0)
but no matter, we rearrange to give
tau(0, 3 oranges, 5 apples, 0) - tau(L, 1 orange, 2 apples, 0) = 0
seconds
because the clock at Origin is 0 in the moving frame.
We do have another function, xi(),
and
xi(L, 3 oranges, 5 apples, 0) - xi(L, 1 orange, 2 apples, 0) = L ft
Also we have a function orange() and a function apple()

so that
apple(0, 3 oranges, 5 apples, 0) - apple(L, 1 orange, 2 apples, 0) = 3
(apples).

and
orange(0, 3 oranges, 5 apples, 0) - orange(L, 1 orange, 2 apples, 0) = 2
(oranges).

So we say "It is clear that the function apple is independent of time. "
We also say
"It is clear that the function orange is independent of time. "
"It is clear that the function orange is independent of apple "
because we don't mix apples with oranges or oranges with time.
(unless we are cretins)
Now comes the really hard part for you to grasp.

"It is clear that the function tau is independent of distance."

Clear as mud, huh? Well, read through again about the magic markers
and synchronized clocks, it may come to you, but the bottom line
is tau(x,y,z,t) = a * t.

and since Joe(32,0,0,16) (moving frame)
is equal to Joe(80,0,0, a*t) (stationary frame)
and it is clear (to some people but not all) that the function tau is
independent of distance and a = 1.

Now, by Einstein's definition (not mine, that's where you are confused)
(16 + 4)/2 = 16. but if you are not clear about 'a' being 1,
(16a + 4a)/ 2 = 16a.

So why does EINSTEIN DEFINE 16 = 4 ?
do {
McCullough : No he doesn't.
Androcles: Yes he does.
} forever and ever, Amen.

Androcles.

Daryl McCullough

unread,
Feb 14, 2005, 12:43:27 PM2/14/05
to
Androcles says...

>> Your nonsensical result, (16+4)/2 = 16, follows from your assuming
>> that time as measured in Sam's frame is the same as time as measured
>> in the stationary frame. Why do you assume that?
>
>I didn't assume it. Why shoud I assume anything as nonsensical
>as x'/(c-v) = x'/(c+v)?

Einstein never said that. That's your addition.

>> So your nonsensical result follows from an extra
>> assumption on your part.
>
>What assumption is that, then?

You are assuming that

x'/(c-v) = time in Sam's frame for light to travel from Sam to Joe

You are assuming that

x'/(c+v) = time in Sam's frame for light to travel from Joe to Sam

Those two assumptions lead to your nonsensical result.

>The time it takes to fly from Sam to Joe is
>tau(32,0,0,16) - tau(0,0,0,0)
>The time it takes to fly from Joe to Sam is
>tau(0,0,0,20) - tau(32,0,0,16)

That's already not correct. First compute the coordinates in the
stationary frame:

Light signal leaves Sam
x=0
y=0
z=0
t=0

Light signal arrives at Joe and is reflected
x=80
y=0
z=0
t=16

Return signal arrives at Sam
x=60
y=0
z=0
t=20

The time it takes to fly from Sam to Joe is

tau(80,0,0,16) - tau(0,0,0,0)

The time it takes to fly from Joe to Sam is

tau(0,0,0,60) - tau(80,0,0,16)

Daryl McCullough

unread,
Feb 14, 2005, 12:33:39 PM2/14/05
to
Androcles says...

>> What Einstein said was that in *Sam's* frame, the time required
>> for light to go from Sam to Joe is the same as the time required
>> to go from Joe to Sam.
>
>That's right. Silly, but that IS what he said. That's why I want to know
>why he thinks 4 seconds equals 16 seconds.

He doesn't. In Sam's frame, the time required for light to go from
Sam to Joe is 8 seconds. The time required for light to go from Joe
back to Sam is 8 seconds.

16 and 4 are the times in the *stationary* frame. In the moving
frame, the times are 8 and 8.

Androcles

unread,
Feb 14, 2005, 3:22:12 PM2/14/05
to

"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:cuqj2...@drn.newsguy.com...

> Androcles says...
>
>>"Daryl McCullough" <stevend...@yahoo.com> wrote
>
>>As for deriving the Lorentz tranformations, here's a derivation, going
>>back to Sam and Joe. Let's introduce two new characters, Sally and
>>Jane.
>>
>>Sam and Joe are at rest relative to frame B (for Boys).
>>
>>Sally and Jane are at rest relative to frame G (for Girls).
>>
>>Let the speed of frame B as measured in frame G be v.
>>
>>Androcles:
>>_______________________________ --> v
>>_|_|_|_|_|_|_|B|_|_|_|_|_|_|_|_
>>
>>
>>_______________________________
>>_|_|_|_|_|_|_|G|_|_|_|_|_|_|_|_
>>
>>
>>McCullough:
>>Also, by symmetry let the speed of frame G as measured in frame B also
>>be v.
>>
>>Androcles:
>>_______________________________
>>_|_|_|_|_|_|_|B|_|_|_|_|_|_|_|_
>>
>>
>>_______________________________ --> v
>>_|_|_|_|_|_|_|G|_|_|_|_|_|_|_|_
>>
>>
>>v = 0
>
> Obviously, I should have been more explicit.

Yes, it does help to be obviously explicit instead of obviously
misleading.


In frame G,
> Sam and Joe are travelling at speed v to the right. In
> frame B, Sally and Jane are travelling at speed v to the
> *left*. So
>
> In frame G
>
> _______________________________ --> v
> _|_|_|_|_|_|_|B|_|_|_|_|_|_|_|_
>
>
> _______________________________
> _|_|_|_|_|_|_|G|_|_|_|_|_|_|_|_
>
> In frame B
>
> _______________________________
> _|_|_|_|_|_|_|B|_|_|_|_|_|_|_|_
>
>
> <-- v _______________________________
> _|_|_|_|_|_|_|G|_|_|_|_|_|_|_|_
>

So in the ground frame, then, the B frame is moving at v/2
and the G frame is moving at -v/2. That IS good news. Perhaps
we can now cease being concerned about A and B being points
in space in some god-forsaken rest frame.

Err... well, g = L/l, right? and l = L, right? so g = 1, right?

> We haven't said anything
> at all about the factor g = L/l.

Oh, I thought we did. Isn't g = 1, then?


Galilean relativity is the special
> case g = 1, which means that
>
> x' = x-vt
> t' = t

Yes, of course, but you said earlier "by symmetry"


>
> To get Einstein's relativity, we now impose another assumption:
> the speed of light is c in all inertial reference frames, regardless
> of the motion of the source.

Now I have to stop you dead in your tracks right there.
Why one Earth would you expect me (or any other sane person)
to make such an absurd assumption?
You do not have a shred of physical evidence to support it,
and Einstein certainly didn't in 1905, because in 1920 he was whining
(we have to depart from the original paper for a moment)

" w= c-v
The velocity of propagation of a ray of light relative to the
carriage thus comes out smaller than c.
But this result comes into conflict with the principle of
relativity set forth in Section V. For, like every other general law of
nature, the law of the transmission of light in vacuo must, according to
the principle of relativity, be the same for the railway carriage as
reference-body as when the rails are the body of reference. But, from
our above consideration, this would appear to be impossible. If every
ray of light is propagated relative to the embankment with the velocity
c, then for this reason it would appear that another law of propagation
of light must necessarily hold with respect to the carriage—a result
contradictory to the principle of relativity."

http://www.bartleby.com/173/7.html

And also
"Prominent theoretical physicists were therefore more inclined to
reject the principle of relativity, in spite of the fact that no
empirical data had been found which were contradictory to this
principle."


Not real physicists, mind you. No real physicist would give up the PoR.
He's only got the support of prominent THEORETICAL physicists, who he
doesn't name. And just because he hasn't found the empirical data
(because he was never an astronomer and didn't look) doesn't mean it
didn't exist.
Sitting in an armchair or playing as violin and dreaming up how Nature
works isn't physics. He stopped being a physicist when he became a
theoretical physicist.
The empirical data IS there.
Understanding it is another matter.
Here is the empirical data.
http://www.britastro.org/vss/gifc/00918-ck.gif


> So a light signal sent from Sam or Sally
> at time t=0 will follow the path x = ct in frame G.

Yes. It will, I fully agree.

In frame B,
> we have
>
> x' = g (x - vt)
> = g (c - v) t
> = cgt (1 - v/c)
>
> t' = g (t + x/v (1/g^2 - 1))
> = g t (1 + c/v(1/g^2 - 1))
>
> So, in order for x' to equal c t', we must have

But x' is NOT equal to ct'.
x' = (c-v)t' and x' = (c+v)t' because that Einstein's 1905 version,
given in the words, "But the ray moves relatively to the initial point
of k, when measured in the stationary system, with the velocity c-v, so
that x'/(c-v) = t."
from which (c-v) t = (c+v) t.

Now, I have also show elsewhere that tau is NOT a function of length
and time, but a function of time only, by your magic markers
(0,0,0,t) = (x', 0,0,0,t) for ANY value of x'.

So
(c-v) * tau1 = (c+v) * tau2 = x'
or
(5-3) * 16 = (5+3) * 4 = 32.

And what I want to know is why YOU think 16 = 4.

do
{
McCullough: No I don't.
Androcles: Have you changed your mind then?
McCullough: No.
Androcles: Have you got some other idea about synchronizing
clocks for your magic markers then?
McCullough: No
Androcles: Yes you do think 16 = 4.
} until (McCullough, theoretical physicist, changes his mind);

Androcles, real physicist.

Androcles

unread,
Feb 14, 2005, 5:08:25 PM2/14/05
to

"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:cuqnd...@drn.newsguy.com...

> Androcles says...
>
>>> What Einstein said was that in *Sam's* frame, the time required
>>> for light to go from Sam to Joe is the same as the time required
>>> to go from Joe to Sam.
>>
>>That's right. Silly, but that IS what he said. That's why I want to
>>know
>>why he thinks 4 seconds equals 16 seconds.
>
> He doesn't. In Sam's frame, the time required for light to go from
> Sam to Joe is 8 seconds. The time required for light to go from Joe
> back to Sam is 8 seconds.

Oh, ok.

>
> 16 and 4 are the times in the *stationary* frame. In the moving
> frame, the times are 8 and 8.

tau = (t-vx/c^2) /sqrt( 1-v^2/c^2)
= (16 - 3 * 80/5^2) /sqrt(1- 9/25)
= 8.
Have you got any more good ones I can tell my grandchildren?
BTW, I'd be happy to tell YOUR grandchildren jokes like that
(for a fee, of course, and I'm not expensive by any means, but I do rate
baby sitter's fees and have to eat.)

Androcles.


Randy Poe

unread,
Feb 14, 2005, 5:16:23 PM2/14/05
to

Androcles wrote:
> "Daryl McCullough" <stevend...@yahoo.com> wrote in message
> news:cuqj2...@drn.newsguy.com...

> > So, in order for x' to equal c t', we must have


>
> But x' is NOT equal to ct'.

t' is a time measured in the moving system.

> x' = (c-v)t' and x' = (c+v)t' because that Einstein's 1905 version,
> given in the words, "But the ray moves relatively to the initial
point
> of k, when measured in the stationary system,

Read those last five words. Do they say "measured in
the moving system"? Read the second to last word. Is
it "moving"?

> with the velocity c-v, so
> that x'/(c-v) = t."

What is that symbol on the right. Is that t'?

> from which (c-v) t = (c+v) t.

Where in that equation did (c+v) occur?

- Randy

Jesse Mazer

unread,
Feb 14, 2005, 5:20:38 PM2/14/05
to

Androcles wrote:

I've seen derivations of the Lorentz transformation before, but inspired
by your post, I decided to look over the original derivation at
http://www.fourmilab.ch/etexts/einstein/specrel/www/ which you are
referring to. There were some parts that I was confused by at first, but
by modifying Einstein's notation a little, without changing the
structure of his derivation, I was able to follow along. If you're
arguing in good faith, feel free to point out any parts of my analysis
you disagree with; if not, feel free to just insult or ignore my post,
it was an interesting exercise for me anyway.

First Einstein assumes you have to coordinate systems defined by
measurements on a system of rigid measuring rods and clocks, K and k,
with all the spatial axes parallel and with k moving at velocity v along
K's x-axis. K's coordinates are (x,y,z,t) while k's coordinates are (xi,
eta, zeta, tau). Then he defines a new coordinate x'=x-vt, and says "it
is clear that a point at rest in the system k must have a system of
values x', y, z, independent of time". Although he doesn't state it this
way, what he has effectively done is to introduce a *third* coordinate
system Kg, with y,z,t coordinates identical to K, but with x' coordinate
given by x-vt. To make this a little clearer, I'm going to modify his
notation and say that coordinate system Kg uses coordinates x',y',z',t',
with these coordinates related to K's coordinates x,y,z,t by a Galilei
transform:

x'=x-vt
y'=y
z'=z
t'=t

An important thing to notice is that, unlike k and K, Kg doesn't
necessarily correspond to the measurements of any observer using a
system of measuring-rods and clocks; it isn't really an inertial
reference frame at all. So, the postulate that all observers must
measure the speed of light to be c in their own rest frame doesn't apply
to Kg. In fact, since we know light must travel at c in both directions
in K, and Kg is related to K by a Galilei transform, it must be true
that light travels at (c-v) in the +x' direction of Kg, and (c-v) in the
-x' direction of Kg.

So now a light beam is sent from the origin of k at tau0, reflected by a
mirror at rest in k at tau1, and returned to the origin at tau2. As
Einstein said, any point which is at rest in k must also be at rest in
this new coordinate system which I am calling Kg, so neither the point
of origin nor the mirror are moving in Kg. So if the origin of Kg
coincides with the origin of k, and if we say the mirror is at position
x'=xm' in Kg, then since light travels at (c-v) in the +x' direction of
Kg, the light will take time xm'/(c-v) to reach the mirror in Kg, and
since light travels at (c+v) in the -x' direction of Kg, it will take an
additional time of xm'/(c+v) to return to the origin. Thus, if the light
is emitted at t'=t0' in Kg's frame, it is reflected at t'=t0' +
xm'/(c-v), and returns to the origin at t'=t0' + xm'/(c-v) + xm'/(c+v)

So, if k's coordinate tau is expressed as a function of Kg's coordinates
like tau(x',y',z',t') then we have:

tau0 = tau(0, 0, 0, t0')
tau1 = tau(xm', 0, 0, t0' + xm'/(c-v))
tau2 = tau(0, 0, 0, t0' + xm'/(c-v) + xm'/(c+v))

Now, since k represents the actual measurements of a non-accelerating
observer using his measuring-rods and clocks, we know that light must
travel at c in both directions in this coordinate system, and since the
origin and the mirror are at rest in k, the light must take the same
amount of time to reach the mirror as it takes to be reflected back to
the origin in k. So, this gives 1/2(tau0 + tau2) = tau1, which
substituting in the expressions above gives

1/2[tau(0, 0, 0, t0') + tau(0, 0, 0, t0' + xm'/(c-v) + xm'/(c+v))] =
tau(xm', 0, 0, t0' + xm'/(c-v))

Then he goes from this to the equation 1/2(1/(c-v) + 1/(c+v))*(dtau/dt')
= dtau/dx' + (1/c-v)*(dtau/dt'), which also confused me for a little
while because I didn't know what calculus rule he was using to go from
the last equation to this one. But then I realized that if you just
ignore the y' and z' coordinates and look at tau(x',t'), then since he
says "if x' is chosen infinitesimally small", you can just assume tau is
a slanted plane in the 3D space with x',t' as the horizontal axes and
tau as the vertical axes. The general equation for a slanted plane in
these coordinates which goes through some point xp', tp', and taup would be:

tau(x',t') = Sx'*(x' - xp') + St'*(t' - tp') + taup

Where Sx' is the slope of the plane along the x' axis and St' is the
slope of the plane along the t' axis. If we say this plane must go
through the three points tau0, tau1 and tau2 earlier, then we can use
tau0's coordinates for xp', tp' and taup, giving:

tau(x',t') = Sx'*x' + St'*t' + tau0

So, plugging in tau1 = tau(xm', t0' + xm'/(c-v)) gives

tau1 = Sx'*xm' + St'*(t0' + xm'/(c-v)) + tau0

And plugging in tau2 = tau(0, t0' + xm'/(c-v) + xm'/(c+v)) gives:

tau2 = St'*(t0' + xm'/(c-v) + xm'/(c+v)) + tau0

So plugging these into 1/2(tau0 + tau2) = tau1 gives:

1/2(tau0 + St'*(t0' + xm'/(c-v) + xm'/(c+v)) + tau0 ) = Sx'*xm' +
St'*(t0' + xm'/(c-v)) + tau0

With a little algebra, this reduces to:

(1/2)*St'*(1/(c-v) + 1/(c+v)) = Sx' + St'*(1/(c-v))

And since tau(x',t') is just a plane, of course St' = dtau/dt' and Sx' =
dtau/dx', so this gives the equation 1/2(1/(c-v) + 1/(c+v))*(dtau/dt') =
dtau/dx' + (1/c-v)*(dtau/dt') which Einstein got.

He then reduces this to dtau/dx' + dtau/dt'*(v/(c^2 - v^2)) = 0, which
is just algebra. He also says that light moves along the y'-axis and
z'-axis of Kg at velocity squareroot(c^2 - v^2), which isn't too hard to
see--a light beam moving vertically along the zeta-axis of k will also
be moving vertically along the z'-axis of Kg since these coordinate
systems aren't moving wrt one another, but in K the light beam must be
moving diagonally since k is moving at v in k, so if you look at a
triangle with ct as the hypotenuse and vt as the horizontal side, the
vertical side must have length t*squareroot(c^2 - v^2), and since z'=z
and t'=t the light beam must also travel that distance in time t' in
Kg's coordinate system. The same kind of argument shows the velocity is
also squareroot(c^2 - v^2) in the y'-direction.

Since tau(x',y',z',t') is a linear function (ie tau(x',y',z',t') = Ax' +
By' + Cz' + Dt'), then from this you can conclude that if dtau/dx' +
dtau/dt'*(v/(c^2 - v^2)) = 0 for a light ray moving along the x'-axis,
tau(x',t') must have the form tau = a(t' - vx'/(c^2 - v^2)) where a is
some function of v (so that dtau/dt' = a and dtau/dx' = -av/(c^2 - v^2),
which means dtau/dx' = (-v/(c^2 - v^2))*dtau/dt').

Next he says that in the k coordinate system the light ray's position as
a function of time would just be xi(tau)=c*tau, so plugging that
expression for tau in gives xi=ac(t' - vx'/(c^2 - v^2)). But in system
Kg, this light ray is moving at velocity (c-v), so its t' coordinate as
a function of x' is t'(x') = x'/(c-v). Plugging this in gives xi =
ac(x'/(c-v) - vx'/(c^2 - v^2)) = a*(c^2/(c^2 - v^2))*x'.

Similarly, if light is going in the eta-direction then eta(tau)=c*tau,
so plugging in tau = a(t' - vx'/(c^2 - v^2)) gives eta=ac(t' - vx'/(c^2
- v^2)). In Kg this ray is moving at squareroot(c^2 - v^2) in the
y'-direction, so t'(y')=y'/squareroot(c^2 - v^2), and plugging this in
gives eta= a(y'/squareroot(c^2 - v^2) - vx'/(c^2 - v^2))...since x'=0
for this ray, this reduces to eta=(ac/squareroot(c^2 - v^2))* y'. The
relation between zeta and z' is exactly the same, so we have:

tau = a(t' - vx'/(c^2 - v^2))
xi = a*(c^2/(c^2 - v^2))*x'
eta = (ac/squareroot(c^2 - v^2))* y'
zeta=(ac/squareroot(c^2 - v^2))* z'

Since the relation between Kg coordinates (x',y',z',t') and K
coordinates (x,y,z,t) is just x'=x-vt, y'=y, z'=z, t'=t, we can plug in
and simplify to get:

tau = Phi(v) * Beta * (t - vx/c^2)
xi = Phi(v) * Beta * (x - vt)
eta = Phi(v) * y
zeta = Phi(v) * z

Where where Beta = c/squareroot(c^2 - v^2) = 1/squareroot(1 - v^2/c^2),
and Phi(v) = ac/squareroot(c^2 - v^2) (since a was an undetermined
function of v in the first place he just writes Phi(v)).

To find Phi(v), he then imagines a coordinate system K' which is moving
at -v relative to k (and unlike Kg, this coordinate system is supposed
to correspond to the measurements made on a system of measuring-rods and
clocks, so it's a valid inertial reference frame). He uses
(x',y',z',t') for the K' coordinate system, but since I've already used
that for Kg, let's call the K'-coordinates (x",y",z",t"). Then the
transform from k-coordinates to K'-coordinates would have to be:

t" = Phi(-v) * Beta(-v) * (tau + v*xi/c^2) = Phi(v)*Phi(-v)*t
x" = Phi(-v) * Beta(-v) * (xi + v*tau) = Phi(v)*Phi(-v)*x
y" = Phi(-v) * eta = Phi(v)*Phi(-v)*y
z" = Phi(-v) * zeta = Phi(v)*Phi(-v)*z

If K' is moving at -v in the k-system, and k is moving at +v in the
K-system, then K and K' should really be the same system, so
Phi(v)*Phi(-v) should be 1. Then he argues that if you have a rod of
lenght l lying along the eta-axis of k and at rest in that system, then
if you transform the coordinates of its ends into the K-system, you find
that its length in the K-system is l/Phi(v), and by symmetry he argues
that the length of a vertical rod moving horizontally can only depend on
the magnitude of the velocity and not the direction, so l/Phi(v) =
l/Phi(-v), which means Phi(v) = Phi(-v)...combining with Phi(v)*Phi(-v)
= 1 which he obtained earlier, he concludes that Phi(v)=1, which gives
him the Lorentz transform.

>
>The Lorentz transformation tau = (t-vx/c^2)
>was derived, by Einstein, from the equation
>
>tau = a * ( t - (vx' / (c^2-v^2))).
>
>He obtained tau = a * ( t - (vx' / (c^2-v^2)))
>
>by integrating the equation
>
>dtau/dx' + v/(c^2-v^2) * dtau/dt = 0,
>
>and he got that from
>
> ½ * [1/(c-v) + 1/(c+v)] * dtau/dt = dtau/dx' + 1/(c-v) * dtau/dt.
>
>To obtain that, he differentiated the equation
>
>½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
>
>
>The sequence to derive the Lorentz transformation is therefore
>
>1)
>Define t = x'/(c-v) = x'/(c+v) because the time for light to go from A
>to B equals the time it takes to travel from B to A
>

Well, keep in mind that the x'-coordinate is equal to x-vt, so if you're
using x' you're in a coordinate system which is different from K or k,
the system I called Kg above. And since this system was just obtained by
a Galilean transform of K, it does not necessarily correspond to any
actual measurements made by an observer using measuring rods and clocks,
so the postulate that light must be measured to travel at c by all
observers doesn't apply to it. In fact, if K measures light to travel at
c in both directions, then in Kg-coordinates light must travel at (c-v)
in one direction and (c+v) in the other, so in Kg coordinates the time
for light to go from the source to the mirror is *not* the same as the
time for the light to return from the mirror to the source.

>
>2)
>½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
>
>3)
>½[1/(c-v)+1/(c+v)] * dtau/dt = dtau/dx' + 1/(c-v) * dtau/dt
>
>4)
>dtau/dx' + v/(c^2-v^2) * dtau/dt = 0
>
>5)
>tau = a * ( t - (vx' / (c^2-v^2)))
>
>6)
>tau = (t-vx/c^2) / sqr(1-v^2/c^2)
>
>(which you can verify at
> http://www.fourmilab.ch/etexts/einstein/specrel/www/
>section 3 )
>and that comes from the mosquito example, so we cannot put the
>Lorentz equations into the mosquito example to derive the Lorentz
>equations, which the idiot McCullough tried to do.
>
>So what I want to know is why Einstein thinks (16 + 4)/2 = 16
>or even ½[tau(16+4) = tau (16)
>or even ½[tau(16) + tau(4) ] = tau(16)
>or even ½[tau(32,0,0, 16) + tau(0,0,0, 4) ] = tau(32,0,0,0, 16)
>or even
>½[tau(0,0,0,0)+tau(0,0,0,16+4)] = tau(32,0,0,16),
>

You're using those equations wrong. Like I said, tau is a function not
of K-coordinates but of a new coordinate system I called Kg, which is
obtained by making the substitution x'=x-vt. In terms of your mosquito
example, if the K-frame is the one where the ladder is moving at v=3
fps, then the coordinate of different events in this frame are:

Mosquito leaves Joe: x=0, t=0

Mosquito reaches Sam: x=80, t=16

Mosquito returns to Joe: x=60, t=20

So in the Kg frame defined by x'=x-vt, t'=t the coordinates of these
same events are:

Mosquito leaves Joe: x'=0, t'=0

Mosquito reaches Sam: x'=32, t'=16

Mosquito returns to Joe: x'=0, t'=20

So plugging these into Einstein's equation for 1/2[tau0 + tau2] = tau1,
where tau0 is the coordinate of the first event in the k-frame, tau1 is
the coordinate of the second event and tau2 is the coordinate of the
third, we have:

1/2[tau(0,0,0,0) + tau(0,0,0,20)] = tau(80,0,0,16)

But of course, we would only assume *this* if the speed of the mosquito,
5 fps, were actually the speed of light. And the reason Einstein wanted
to assume the speed of light is the same when measured by all observers
is just that he wanted Maxwell's equations to work for all observers,
and light must travel at c in all directions in any frame where
Maxwell's equations hold, as I've explained to you in previous posts.

Jesse

Timo Nieminen

unread,
Feb 14, 2005, 6:06:59 PM2/14/05
to
On Mon, 14 Feb 2005, Jesse Mazer wrote:

> And the reason Einstein wanted
> to assume the speed of light is the same when measured by all observers
> is just that he wanted Maxwell's equations to work for all observers,
> and light must travel at c in all directions in any frame where
> Maxwell's equations hold,

At which point one notes that if one has a theory of optics and
electromagnetism in which light does not necessarily travel at c in free
space in all frames (such as, for example, pretty much all ballistic
emission theories), then that theory of optics and electromagnetism must
disagree with classical electrodynamics (= the Maxwell equations and
invariant permittivity and permeability of free space).

Given the general success of classical electrodynamics, the theory would
need to agree with classical electrodynamics on most matters. Depending on
the theory, that's no necessarily a problem at all, and it might usefully
direct one towards experiments to usefully test the validity of classical
EM vs the alternative theory offered.

Of course, it's possible that it isn't possible to formulate such an
alternate theory as a field theory without making bizarre assumptions,
which, in the absence of fields with energy, momentum, and angular
momentum, might be an even bigger departure from Newtonianism than SR.

That could all be quite fun and interesting to discuss in-depth. But,
alas, the anti-relativists don't appear to be willing to discuss!

PS: And, yes, falsification of classical electrodynamics and its
replacement by an alternative theory of optics and EM doesn't necessarily
do squat to SR.

--
Timo

Daryl McCullough

unread,
Feb 14, 2005, 6:18:46 PM2/14/05
to
Androcles says...

Not necessarily. Here were the definitions:

L is the distance between rungs of the G-ladder
as measured in the G-frame.

l is the distance between rungs
of the B-ladder as measured in the G-frame.

v is the speed of the B-ladder as measured in the G-frame.

L' is the distance between rungs of the G-ladder
as measured in the B-frame.

l' is the distance between rungs
of the B-ladder as measured in the B-frame.

u' is the speed of the G-ladder as measured in the B-frame.

I said that by symmetry,

L = l'
l = L'
v = u'

I did not assume that L = l.



>> We haven't said anything
>> at all about the factor g = L/l.
>
>Oh, I thought we did. Isn't g = 1, then?

No.

>> To get Einstein's relativity, we now impose another assumption:
>> the speed of light is c in all inertial reference frames, regardless
>> of the motion of the source.
>
>Now I have to stop you dead in your tracks right there.
>Why one Earth would you expect me (or any other sane person)
>to make such an absurd assumption?

Because Maxwell's equations predict that light has speed c. If
Maxwell's equations are valid in every rest frame, then it follows
that light has speed c in every rest frame.

The principle of relativity (there is no standard for rest) plus
the validity of Maxwell's equations in vacuum implies that light
has the same speed in every reference frame.

Androcles

unread,
Feb 15, 2005, 12:28:13 AM2/15/05
to

"Jesse Mazer" <vze2...@mail.verizon.net> wrote in message
news:42114E2A...@mail.verizon.net...

The x-axis, not "K's x-axis". A minor point, perhaps...

> K's coordinates are (x,y,z,t) while k's coordinates are (xi, eta,
> zeta, tau).

Yes.

Then he defines a new coordinate x'=x-vt, and says "it
> is clear that a point at rest in the system k must have a system of
> values x', y, z, independent of time".

Note (with hindsight) that
xi = x'/sqrt(1-v^2/c^2).


> Although he doesn't state it this way, what he has effectively done is
> to introduce a *third* coordinate system Kg, with y,z,t coordinates
> identical to K, but with x' coordinate given by x-vt.

Ah, the ghost of k.
Why not give it its real name, the Galilean moving frame?
Now your notation would look much better if it were kG instead
of Kg.

> To make this a little clearer,

Who to, yourself? I'm quite comfortable with the way Einstein has
written his paper, I know what he says. He swaps his notation around
all over the place.
In section 5 he writes
V = (c+w)/ (1 + w/c) instead of
V = (c+v)/ (1+v/c),
but it doesn't fool a mathematician. Now you are going to make
it "clearer" by changing notation. Who do you think you are fooling?
Still, if you insist on calling a spade a spoon....

You changing notation can only
muddy the waters, it will never make anything clearer.


> I'm going to modify his notation and say that coordinate system Kg
> uses coordinates x',y',z',t', with these coordinates related to K's
> coordinates x,y,z,t by a Galilei transform:
>
> x'=x-vt
> y'=y
> z'=z
> t'=t

To make this a little clearer, right?


> An important thing to notice is that, unlike k and K, Kg doesn't
> necessarily correspond to the measurements of any observer using a
> system of measuring-rods and clocks; it isn't really an inertial
> reference frame at all.

Not really, huh?

To make this a little clearer, try a search on Einstein's paper for the
word
"inertial" and give me the sentence in which you found it, because
damnly my frank, I really don't give a dear whether you think kG or Kg
is an inertial frame or not. It is, but that's entirely irrelevant. You
are not making things a little
clearer, you are muddying the waters with irrelevancies.


> So, the postulate that all observers must measure the speed of light
> to be c in their own rest frame doesn't apply to Kg.

Which postulate is that, then?
Show me Einstein's postulate that says "the speed of light to be c in
their own rest frame". You are not making things clearer, you are
muddying.

> In fact,

Oh goody, a FACT is coming up. I just love facts.


> since we know

Oh, a fact we KNOW... how exciting.


> light must travel at c in both directions in K,


Must it? I never knew that was a fact. I thought it was a working
hypothesis, but I never knew it was a fact. To tell you the truth,
I still don't think it is a fact. You can try to persuade me it's a
fact,
and that helps to muddy the waters, but it isn't a fact and "we" don't
know that it is a fact, in fact.

and Kg is related to K by a Galilei transform,

Yes.

> it must be true that light travels at (c-v) in the +x' direction of
> Kg, and (c-v) in the -x' direction of Kg.


Yes indeed. So to make this little muddier, there is no third frame
and the frame you are calling Kg is called k by Einstein, IN FACT.
In FACT, I KNOW that when a relativist says "In fact, we know"
he's about to lie through his fucking teeth.

>
> So now a light beam is sent from the origin of k at tau0, reflected by
> a mirror at rest in k at tau1,


Whoa! Who says the mirror is at rest in k?
Who says there is a mirror?

What we have is ONE light ray, seen from TWO points of view.
We do NOT have any third frame, nothing has to be inertial, and
as far as WE are concerned the reflector is irrelevant, but it could
be a falling raindrop which is not at rest in either K or k. In FACT.


> and returned to the origin at tau2.

> As Einstein said, any point which is at rest in k must also be at rest
> in this new coordinate system which I am calling Kg, so neither the
> point of origin nor the mirror are moving in Kg.

Einstein said you had a new coordinate system called Kg?
Nah, I don't think he did. I don't think Einstein is alive to say that.
I think you are making it up. In fact, WE know you are.


> So if the origin of Kg coincides with the origin of k, and if we say
> the mirror is at position x'=xm' in Kg, then since light travels at
> (c-v) in the +x' direction of Kg, the light will take time xm'/(c-v)
> to reach the mirror in Kg, and since light travels at (c+v) in the -x'
> direction of Kg, it will take an additional time of xm'/(c+v) to
> return to the origin.


Yes, it will, I fully agree.
I particularly like your clarification that light travels at c-v.
In fact, we know that light travels at c-v and c+v.

Now we need to clarify that Einstein DEFINED
' the "time" required by light to travel from A to B equals the "time"
it requires to travel from B to A. ' to which I wlll add, to make it a
little clearer,
"in all inertial frames of reference".
In fact, we know this because he makes use of

稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))

and x', c-v, c+v are all terms belonging to the frame that you call Kg
and Einstein calls k.

To make this a little clearer, in fact we know that according to
Einstein

x'/(c-v) = x'/(c+v),

from which an intelligent 12-year-old can figure out that v = 0
but a blind and faithful relativist cannot.

[snip remainder unread. Too many assumptions already]

Androcles.


Androcles

unread,
Feb 15, 2005, 12:40:11 AM2/15/05
to

"Timo Nieminen" <ti...@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0502150848290.2116-100000@localhost...

[snip response to Mazer]

> That could all be quite fun and interesting to discuss in-depth. But,
> alas, the anti-relativists don't appear to be willing to discuss!


Not true. Discuss this.


"It is known that Maxwell's electrodynamics--as usually understood at
the present time--when applied to moving bodies, leads to asymmetries
which do not appear to be inherent in the phenomena. Take, for example,
the reciprocal electrodynamic action of a magnet and a conductor. The
observable phenomenon here depends only on the relative motion of the
conductor and the magnet, whereas the customary view draws a sharp
distinction between the two cases in which either the one or the other
of these bodies is in motion. For if the magnet is in motion and the
conductor at rest, there arises in the neighbourhood of the magnet an
electric field with a certain definite energy, producing a current at
the places where parts of the conductor are situated. But if the magnet
is stationary and the conductor in motion, no electric field arises in
the neighbourhood of the magnet. In the conductor, however, we find an
electromotive force, to which in itself there is no corresponding
energy, but which gives rise--assuming equality of relative motion in
the two cases discussed--to electric currents of the same path and
intensity as those produced by the electric forces in the former case.
" - Einstein.

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

I'll kick off the discussion with

It is known that Einstein's electrodynamics--as usually understood at
the present time--when applied to moving bodies, leads to asymmetries
which do not appear to be inherent in the phenomena.

For if "the velocity W of the man relative to the embankment is here
replaced by the velocity of light relative to the embankment. w is the
required velocity of light with respect to the carriage, and we have w =
c-v. "

Reference http://www.bartleby.com/173/7.html

Androcles.


Androcles

unread,
Feb 15, 2005, 1:01:20 AM2/15/05
to

"Randy Poe" <poespa...@yahoo.com> wrote in message
news:1108419383.6...@c13g2000cwb.googlegroups.com...


Differentiate this for me, Poe :

稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/ something_not_there )]
= tau(x',0,0,t+x'/(c-v))

I get:

稼1/(c-v) + 1/something_not_there ] dtau/dt = dtau/dx' + 1/(c-v) *
dtau/dt,

but I'm having difficulty proceeding on.

Show the derivation of the Lorentz transforms, Poe.


Is it
tau = (t-vx/c^2) / something_not_there?

Androcles.


Timo Nieminen

unread,
Feb 15, 2005, 1:02:49 AM2/15/05
to
On Tue, 15 Feb 2005, Androcles wrote:

> "Timo Nieminen" <ti...@physics.uq.edu.au> wrote:
>
> [snip response to Mazer]
>
> > That could all be quite fun and interesting to discuss in-depth. But,
> > alas, the anti-relativists don't appear to be willing to discuss!
>
> Not true.

You're willing to discuss classical electrodynamics vs alternatives?
About a month ago, you refused to discuss Ritzian electrodynamics,
reflection from moving mirrors, reflection of light from moving sources
from stationary mirrors, etc. Are you willing to discuss such matters now?

> Discuss this.
>
> "It is known that Maxwell's electrodynamics--as usually understood at
> the present time--when applied to moving bodies, leads to asymmetries
> which do not appear to be inherent in the phenomena. Take, for example,
> the reciprocal electrodynamic action of a magnet and a conductor. The
> observable phenomenon here depends only on the relative motion of the
> conductor and the magnet, whereas the customary view draws a sharp
> distinction between the two cases in which either the one or the other
> of these bodies is in motion. For if the magnet is in motion and the
> conductor at rest, there arises in the neighbourhood of the magnet an
> electric field with a certain definite energy, producing a current at
> the places where parts of the conductor are situated. But if the magnet
> is stationary and the conductor in motion, no electric field arises in
> the neighbourhood of the magnet. In the conductor, however, we find an
> electromotive force, to which in itself there is no corresponding
> energy, but which gives rise--assuming equality of relative motion in
> the two cases discussed--to electric currents of the same path and
> intensity as those produced by the electric forces in the former case.
> " - Einstein.
>

> I'll kick off the discussion with
>
> It is known that Einstein's electrodynamics--as usually understood at
> the present time--when applied to moving bodies, leads to asymmetries
> which do not appear to be inherent in the phenomena.
>
> For if "the velocity W of the man relative to the embankment is here
> replaced by the velocity of light relative to the embankment. w is the
> required velocity of light with respect to the carriage, and we have w =
> c-v. "
>
> Reference http://www.bartleby.com/173/7.html

And your point is? Einstein is just demonstrating (in a popular work, at
that, not a technical work) that the invariant c hypothesis and the PoR
are incompatible if space-time coordinates in inertial frames are related
by Galilei transforms.

Irrelevant to your 1st quote from Einstein 1905. How is it meant to kick
off a discussion related to the 1st quote?

--
Timo

Androcles

unread,
Feb 15, 2005, 1:08:09 AM2/15/05
to

"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:curbk...@drn.newsguy.com...
[something I've snipped]


You snipped

"So in the ground frame, then, the B frame is moving at v/2
and the G frame is moving at -v/2. That IS good news. Perhaps
we can now cease being concerned about A and B being points
in space in some god-forsaken rest frame."

so I'm returning the compliment, McCullough.

Androcles.


Androcles

unread,
Feb 15, 2005, 2:38:52 AM2/15/05
to

"Timo Nieminen" <ti...@physics.uq.edu.au> wrote in message
news:Pine.LNX.4.50.0502151546340.5359-100000@localhost...

Of course it isn't a technical work. There was nothing technical in
the 1905 paper, either.

' the "time" required by light to travel from A to B equals the "time"
it requires to travel from B to A. ' is an assertion, it isn't
technical.
What's your point?


that the invariant c hypothesis and the PoR
> are incompatible if space-time coordinates in inertial frames are
> related
> by Galilei transforms.

Well then, the invariant c hypothesis fails, and we are done.

Androcles.

Androcles

unread,
Feb 15, 2005, 3:13:15 AM2/15/05
to

"Daryl McCullough" <stevend...@yahoo.com> wrote in message
news:cuqnv...@drn.newsguy.com...

> Androcles says...
>
>>> Your nonsensical result, (16+4)/2 = 16, follows from your assuming
>>> that time as measured in Sam's frame is the same as time as measured
>>> in the stationary frame. Why do you assume that?
>>
>>I didn't assume it. Why shoud I assume anything as nonsensical
>>as x'/(c-v) = x'/(c+v)?
>
> Einstein never said that. That's your addition.

They are the time for the light to reach B from A and the time for light
tor reach A from B, and Einstein DID say ' the "time" required by light

to travel from A to B equals the "time" it requires to travel from B to

A. '
Your petty denials can't change that.


>
>>> So your nonsensical result follows from an extra
>>> assumption on your part.
>>
>>What assumption is that, then?
>
> You are assuming that
>
> x'/(c-v) = time in Sam's frame for light to travel from Sam to Joe

Bullshit.
Einstein said


"But the ray moves relatively to the initial point of k, when measured

in the stationary system, with the velocity c-v, so that x'/(c-v) = t."

Obviously he doesn't mean t = x/c = 80/5 = 16 in the stationary frame
or he would have said so. He means what he says, 32/(5-3) = 16.
I've made no assumption, you are a liar.

>
> You are assuming that
>
> x'/(c+v) = time in Sam's frame for light to travel from Joe to Sam

No I am not, x'/(c+v) appears in the equation and can only mean the
time for the light to travel from Joe to Sam, Einstein states

"and at the time tau1 be reflected thence to the origin of the
co-ordinates, arriving there at the time tau2; we then must have
(tau0+tau2)/2 = tau1."

You are a liar.


>
> Those two assumptions lead to your nonsensical result.

Ok, liar, let's see you derive the LTs without making assumptions.

>>The time it takes to fly from Sam to Joe is
>>tau(32,0,0,16) - tau(0,0,0,0)
>>The time it takes to fly from Joe to Sam is
>>tau(0,0,0,20) - tau(32,0,0,16)
>
> That's already not correct. First compute the coordinates in the
> stationary frame:

Who gives a flying fuck about the stationary frame?
You are doing that to obfuscate, liar.

> Light signal leaves Sam
> x=0
> y=0
> z=0
> t=0
>
> Light signal arrives at Joe and is reflected
> x=80
> y=0
> z=0
> t=16
>
> Return signal arrives at Sam
> x=60
> y=0
> z=0
> t=20
>
> The time it takes to fly from Sam to Joe is
> tau(80,0,0,16) - tau(0,0,0,0)
>
> The time it takes to fly from Joe to Sam is
> tau(0,0,0,60) - tau(80,0,0,16)

tau(0,0,0,60) indeed!

At least put the 60 in the x position, for Einstein's sake,
and what the fuck is the function tau supposed to be doing in
the stationary frame?

You've had too much wacky baccy, McCullough.

Androcles.

Timo Nieminen

unread,
Feb 15, 2005, 4:22:00 AM2/15/05
to
On Tue, 15 Feb 2005, Androcles wrote:

> "Timo Nieminen" <ti...@physics.uq.edu.au> wrote:
> > On Tue, 15 Feb 2005, Androcles wrote:
> >
> >> "Timo Nieminen" <ti...@physics.uq.edu.au> wrote:
> >>
> >> [snip response to Mazer]
> >>
> >> > That could all be quite fun and interesting to discuss in-depth.
> >> > But,
> >> > alas, the anti-relativists don't appear to be willing to discuss!
> >>
> >> Not true.
> >
> > You're willing to discuss classical electrodynamics vs alternatives?
> > About a month ago, you refused to discuss Ritzian electrodynamics,
> > reflection from moving mirrors, reflection of light from moving
> > sources
> > from stationary mirrors, etc. Are you willing to discuss such matters
> > now?

Absence of reply noted.

> >> Discuss this.
> >>
> >> "It is known that Maxwell's electrodynamics--as usually understood at
> >> the present time--when applied to moving bodies, leads to asymmetries
> >> which do not appear to be inherent in the phenomena. Take, for
> >> example,

[cut Einstein quote from introduction to 1905 paper, basically irrelevant
to discussion of classical EM vs alternatives]

> >> I'll kick off the discussion with
> >>
> >> It is known that Einstein's electrodynamics--as usually understood at
> >> the present time--when applied to moving bodies, leads to asymmetries
> >> which do not appear to be inherent in the phenomena.
> >>
> >> For if "the velocity W of the man relative to the embankment is here
> >> replaced by the velocity of light relative to the embankment. w is
> >> the
> >> required velocity of light with respect to the carriage, and we have
> >> w =
> >> c-v. "
> >>
> >> Reference http://www.bartleby.com/173/7.html
> >
> > And your point is? Einstein is just demonstrating (in a popular work,
> > at
> > that, not a technical work)
>
> Of course it isn't a technical work. There was nothing technical in
> the 1905 paper, either.
>
> ' the "time" required by light to travel from A to B equals the "time"
> it requires to travel from B to A. ' is an assertion, it isn't
> technical.
> What's your point?

My point was simply that your quotes and your comment on them are
irrelevant to the issue that you implied you were willing to discuss.

Your lack of clarification of any relevant point is noted. Did you have
one?

> that the invariant c hypothesis and the PoR
> > are incompatible if space-time coordinates in inertial frames are
> > related
> > by Galilei transforms.
>
> Well then, the invariant c hypothesis fails, and we are done.

Unwillingness to discuss classical EM vs alternative EM theories noted.

QED.

--
Timo

Androcles

unread,
Feb 15, 2005, 5:03:28 AM2/15/05
to

"Timo Nieminen" <uqtn...@mailbox.uq.edu.au> wrote in message
news:Pine.OSF.4.58.05...@dingo.cc.uq.edu.au...

"Of course, it's possible that it isn't possible to formulate such an


alternate theory as a field theory without making bizarre assumptions,
which, in the absence of fields with energy, momentum, and angular
momentum, might be an even bigger departure from Newtonianism than SR."


Change of subject noted.

>
> Your lack of clarification of any relevant point is noted. Did you
> have
> one?

Did you?


>
>> that the invariant c hypothesis and the PoR
>> > are incompatible if space-time coordinates in inertial frames are
>> > related
>> > by Galilei transforms.
>>
>> Well then, the invariant c hypothesis fails, and we are done.
>
> Unwillingness to discuss classical EM vs alternative EM theories
> noted.

Unwillingness to remain on topic, "departure from Newtonianism than SR."
noted.
Androcles,

> QED.
>
> --
> Timo


Randy Poe

unread,
Feb 15, 2005, 9:58:07 AM2/15/05
to

Androcles wrote:
> "Randy Poe" <poespa...@yahoo.com> wrote in message
> news:1108419383.6...@c13g2000cwb.googlegroups.com...

> > Where in that equation did (c+v) occur?


>
>
> Differentiate this for me, Poe :

With respect to what?

>
> ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/ something_not_there )]
> = tau(x',0,0,t+x'/(c-v))
>

Are the "something_not_there"'s supposed to refer
to x/(c+v)?

> I get:
>
> ½[1/(c-v) + 1/something_not_there ] dtau/dt = dtau/dx' + 1/(c-v) *


> dtau/dt,
>
> but I'm having difficulty proceeding on.

That's because it appears you haven't decided what
you're differentiating with respect to, because
what you've done is wrong regardless of what
variable you're differentiating wrt.

>
> Show the derivation of the Lorentz transforms, Poe.

I can't. Not to you. It follows from accepting a single
meaning of the symbols above (something you won't do),
and taking as an assumption that light travels at
c when measured by both moving and stationary
observers (something you won't do).

It is a consequence of those things. If you really
want to ask "show me how the Lorentz transform is
a consequence of these definitions" then you should
suspend disbelief and start from those definitions.
But if you say "show me how it follows from those
definitions" and then say "I'm not going to let you
make those definitions", then it's pretty obvious
you're just trolling.

There are still a couple of people on this newsgroup
trying to fight the first issue, for instance
trying to point out to you that "the light moves
relative to the mirror at c-v AS SEEN IN THE
STATIONARY FRAME" and "the light moves relative
to the mirror at c AS SEEN IN THE MOVING FRAME"
are not contradictory statements.

Do you need glasses? Is there some reason that
you keep quoting Einstein's statement about how
things are measured IN THE STATIONARY FRAME and
pretending they are measurements IN THE MOVING
FRAME?

Why is it you keep quoting that passage but can't
seem to see the words "when measured in the stationary
system"? Is it some odd brain dysfunction? Did you have
a stroke? Is this just confined to the written word or
do you have similar processing difficulties with the
spoken word?

Perhaps if you count backward from the end, you'll
see the s-word:

"But the ray moves relatively to the initial point

of k, when measured in the stationary system..."

OK, focus on the last word. It's "system". Now look
just before that word. There's another s-word there.
The phrase "when measured in the [Androcles blanks
out here]" is telling you that observers in the
K system will measure the ray as closing on the
mirror at a rate of c-v, but it's not telling you
anything about what observers in the k system see.

- Randy

Dirk Van de moortel

unread,
Feb 15, 2005, 10:04:05 AM2/15/05
to

"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message news:42120f00$1...@usenet01.boi.hp.com...
>
> "Randy Poe" <poespa...@yahoo.com> wrote in message news:1108479487.0...@o13g2000cwo.googlegroups.com...

>
> Androcles wrote:
> > "Randy Poe" <poespa...@yahoo.com> wrote in message
> > news:1108419383.6...@c13g2000cwb.googlegroups.com...
>
> > > Where in that equation did (c+v) occur?
> >
> >
> > Differentiate this for me, Poe :
>
> With respect to what?
>
> >
> > 稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/ something_not_there )]

> > = tau(x',0,0,t+x'/(c-v))
> >
>
> Are the "something_not_there"'s supposed to refer
> to x/(c+v)?
>
> > I get:
> >
> > 稼1/(c-v) + 1/something_not_there ] dtau/dt = dtau/dx' + 1/(c-v) *

Up to here was actually written by Randy

>
> Nice try, but you *know* it won't work ;-)

This line was written by me.

So now we have to reply and reply again to ourself
to get proper quoting when someone has used Google
Beta ;-)
Thanks guys, nice job!

Dirk Vdm


Dirk Van de moortel

unread,
Feb 15, 2005, 10:01:21 AM2/15/05
to

"Randy Poe" <poespa...@yahoo.com> wrote in message news:1108479487.0...@o13g2000cwo.googlegroups.com...

Androcles wrote:
> "Randy Poe" <poespa...@yahoo.com> wrote in message
> news:1108419383.6...@c13g2000cwb.googlegroups.com...

> > Where in that equation did (c+v) occur?
>
>
> Differentiate this for me, Poe :

With respect to what?

>
> 稼tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/ something_not_there )]
> = tau(x',0,0,t+x'/(c-v))
>

Are the "something_not_there"'s supposed to refer
to x/(c+v)?

> I get:
>
> 稼1/(c-v) + 1/something_not_there ] dtau/dt = dtau/dx' + 1/(c-v) *