Re: Two photons... relative distance question

In article <1115403945.075565.323200@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Curious <anthonyroseuk-curious@xxxxxxxxxxx> wrote:
>Note that the two frames are aligned on photon A so that photon A's
>position = Ax = Ax' = 0;
>and they are aligned with respect to time so that t = t' = 0.

They are "aligned with respect to time" (i.e. synchronized) so that t = t'
= 0 only at x = x' = 0. At other locations (with different x and x') t
and t' are not equal.

To simplify the arithmetic, let's use light-seconds as our unit of
distance so in frame 1, Ax = Ax' = 0 and At = At' = 0; and Bx = 1.5;
and Bt = 0; and we want to find Bx' and Bt'. In these units, c = 1
light-second per second (which simplifies the Lorentz transformation
equations), beta = 0.5, and gamma = 1/sqrt(1 - beta^2) = 1.1547.

Bx' = gamma * (Bx - beta*Bt)
= 1.1547 * (1.5 - 0.5*0)
= 1.73205

Bt' = gamma * (Bt - beta*Bx)
= 1.1547 * (0 - 0.5*1.5)
= -0.86603

This says that in your frame 2, at t' = -0.86603 seconds, photon B is
located at x' = 1.73205 light-seconds. Applying the conversion factor of
1 light-second = 299 792 458 m/sec, this agrees with your calculation
below (which is only half the story because you didn't calculate Bt').

>I think that the answer is as follows, where Bx' = photon B's position
>in Frame 1 and Bx = photon B's position in Frame 2:
>
>Bx = gamma Bx' + gamma v t'
> = gamma 449688687m + gamma 0.5c 0
> = gamma 449688687m
> = 1 / sqrt(1 - v squared/c squared) * 449688687m
> = 1 / sqrt(1 - 0.5c squared/c squard) * 449688687m
> = 1 / sqrt(1 - 0.25) * 449688687m
> = 1 / sqrt(0.75) * 449688687m
> = 1 / 0.86602540378443864676372317075294 * 449688687m
> = 1.1547005383792515290182975610039 * 449688687m
> = 519255768.9819587m

--
Jon Bell <jtbell@xxxxxxxxxx> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA
.