answer to YBM's bell problem - sci.physics.relativity

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Message from discussion answer to YBM's bell problem

YBM

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More options Sep 9 2008, 3:46 pm

Newsgroups: sci.physics.relativity

From: YBM <ybm...@nooos.fr>

Date: Tue, 09 Sep 2008 21:46:04 +0200

Local: Tues, Sep 9 2008 3:46 pm

Subject: Re: answer to YBM's bell problem

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rbwinn a écrit :

> YBM proposed the folowing problem.

Here it is :

>> two light rays are emitted from (-a,0,0) and (a,0,0) (a>0)
>> at the same time in A in the direction of the origine
>> of A (so velocities are respectively (c,0,0) and (-c,0,0).
>> When both arrive at O it makes a bell ring.

>> Try to seriously answer these questions :
>> In frame A and in frame B :
>> - What are the coordinates of both photons at a
>> given time ?
>> - What are the coordinates of the event "the bell ring"
>> in A and in B.

>> At every step explain why you choose t, t' or n' (and
>> which n') as time coordinate.
> ... Since he did not
> specify whether the bell was at the origin of A or the origin of B, we
> will work the problem with the bell at the origin of A.

At least you've guess one thing right in your life. The bell is
at rest in A frame. You seem to have guess right, as well, that
both light rays were emitted à t=t'=0.

> In A, the
> light is emitted at x=a and x=(-a).
> In B the light is emitted at
> x'=a and x'=(-a).

It is what you'd get if you use Galilean transformations.

> Light emitted at -a has a velocity of c relative
> to both frames of reference.

This is NOT what you'd get if you use Galilean transformations.

> Light emitted at a has a velocity of -c

This is NOT what you'd get if you use Galilean transformations.

> relative to both frames of reference. If the bell is at the origin of
> A, it will ring when t=a/c.

> The mathematics for this is, for the light from -a, x1=-a, x2=0
> x2-x1 = 0-(-a) = a
> x=wt=ct =a
> t=a/c

> For the light from a, x1=a, x2=0 w=(-c)
> x2-x1= 0 - a
> x=wt=(-c)t = -a
> t=a/c

You forget to say that this is in frame A. Then this is ok.

> Light from both directions reaches the bell at a time of a/c, and the
> bell rings.
> To determine what time a clock in B reads when the bell rings,

> n'= t(1-v/w)
> n'=a/c(1-v/c)

You have a very fancy way to use your own "thoery". There is
TWO light rays. One with velocity -c, the other one with
velocity c. Here you only use w=c. Is there something
special with the light ray comming from (-a,0,0) ? Moreover
if you compute the speed of the other light ray, using
n'=a/c(1-v/c), you won't get c.

Do you see why I asked you a problem with *two* light rays to consider ?
To point out that this force you to arbitrarely choice one of
the rays to setup your "slower clocks", but then the other light
ray won't propagate at c.

If you had followed your *own words* you'd have used w=c
for the ray emitted at (-a,0,0), giving for you "slower
clock" : n'=a/c(1-v/c), then when considering the
light ray emitted at (a,0,0), at velocity -w you'd have
used n'=a/c(1+v/c).

Then you'd have found that your "theory" does not
provite a unique time for a single event... In other
words, what is an event in frame B, is *two* event
in frame B : both light rays, according to you theory,
do not arrive on the bell at the time... so the bell
does not ring in B. But it rings in A : contradiction !

There is no such absurdities in SR : for SR, the bell
rings in both frames... It just happens that in frame
B they were emitted at coordinates
(-a/sqrt(1-v^2/c^2),0,0) at time va/(c^2*sqrt(1-v^2/c^2))
for the "left" light ray, and :
(a/sqrt(1-v^2/c^2),0,0) at time -va/(c^2*sqrt(1-v^2/c^2))

Next time you'll pretend I've proven something, please
don't lie and write that I've proven Robert Winn's modified
"Galilean Transformation" to be absurd.

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