Hard problem in special relativity and two dim lorenz transformation

Two light sources are at rest and at distance D apart on the x-axis of some inertial frame. They emit photons simultaneously in that frame in the positive x direction. Show that in a frame in which the sources have velocity u along the x-axis, the photons are separated by a constant distance

$\displaystyle D\sqrt\frac{c-u}{c+u}$

In frame O' in which sources are at rest

Photon 1

$\displaystyle x_1'= ct'$

Photon 2

$\displaystyle x_2'=ct'+D$

then applying the two dim lorentz transformation into frame O which is going speed u in negative x direction

$\displaystyle x_1=\gamma(c+u)t'$

$\displaystyle x_2=\gamma(c+u)t'+\gamma D$

However

$\displaystyle x_2-x_1=\gamma D$ which is not right

I have tried to eliminate the $\displaystyle t'$ but this did not help

not sure where to go now

Re: Hard problem in special relativity and two dim lorenz transformation

Hi, FGT12.

I will use unprimed coordinates to denote the frame in which the light sources A and B are stationary, and primed coordinates to denotes the frame in which A and B have velocity u.

If we start out making the assumption that the source A is at the origin in the unprimed coordinates, then

$\displaystyle x_{A}=0$ and $\displaystyle x_{B}=D.$

Without losing generality, we will also assume that the pulses of light are emitted right as the original coordinate system's origin aligns with the origin of the primed system. With this assumption we have

The position of the photon emitted by A in primed coordinates = $\displaystyle ct'_{A}$

and similarly

The position of the photon emitted by B in primed coordinates = Distance from A to B in primed coordinates + $\displaystyle ct'_{B}.$

By the postulates relativity, c is constant for all inertial reference frames. However, we must remember that time will now be different at A and B according to the primed coordinate system.

So we must determine the following three quantities:

1) $\displaystyle t'_{A}$

2) $\displaystyle t'_{B}$

3) Distance from A to B as seen from the primed coordinates

The Lorentz Eqautions tell us

1) $\displaystyle t'_{A}=\frac{t}{\sqrt{1-\frac{u^{2}}{c^{2}}}}$ (remember, we have $\displaystyle x_{A}=0$)

2) $\displaystyle t'_{B}=\frac{t-\frac{uD}{c^{2}}}{\sqrt{1-\frac{u^{2}}{c^{2}}}}$ (using $\displaystyle x_{B}=D$ here)

3) If you use the Lorentz Equations to find $\displaystyle x'_{A}$ and $\displaystyle x'_{B}$, then compute $\displaystyle x'_{B}-x'_{A}$ (which is the distance between A and B in the primed coordinates) we should get

$\displaystyle \frac{D}{\sqrt{1-\frac{u^{2}}{c^{2}}}}.$

Then the distance between the photons relative to the primed system is

$\displaystyle \frac{D}{\sqrt{1-\frac{u^{2}}{c^{2}}}}+ct'_{B}-ct'_{A}.$

If we do a little algebra to the last line, we should get the desired result.

Does this help? Let me know if anything is unclear.

Good luck!