Special Relativity - Distance between photons

1. Jul 6, 2011

Pi-Bond

1. The problem statement, all variables and given/known data
(Not homework, just a textbook question I'm confused about)

Two photons travel along the x axis of S, with a constant distance L between them. Prove that in S' (moving at velocity v w.r.t S) the distance between these photons is $L\sqrt{\frac{c+v}{c-v}}$

2. Relevant equations
The Lorentz transforms

3. The attempt at a solution
I don't understand why the length does not simply contract by the gamma factor, i.e. why the distance in S' is not $\gamma L$. To obtain this I used the Lorentz transform for the coordinates of the positions of the photons. Setting the first at x=0 and the second at x=L yielded the coordinates in S' for the first as $x'=-\gamma vt$ and the second as $x'=\gamma(L-vt)$. Subtracting these two yielded the above result. What am I doing wrong?

2. Jul 6, 2011

George Jones

Staff Emeritus

3. Jul 6, 2011

Pi-Bond

But why would you need t', doesn't the x transformation suffice for calculating the distance in S'?

4. Jul 6, 2011

George Jones

Staff Emeritus
You need the simultaneous positions of the photons. In the primed frame, this means the positions at the same t'.

5. Jul 6, 2011

Pi-Bond

Ok, that makes sense. But where do I put the values of t'?

6. Jul 6, 2011

George Jones

Staff Emeritus
Let's move back to the unprimed frame.
This gives the positions of the photons at one instant in time in the unprimed frame, say t = 0. What are the equations of the worldlines of the photons, i.e., what are the positions of the photons at an arbitrary time t in the unprimed frame?

7. Jul 7, 2011

Pi-Bond

Assuming the first photon starts at x=0 at t=0, it's position would be given by x=ct. And similarly for the second photon, the position would be given by x=L+ct. So should I transform these two coordinates and substitute for t in terms of t'?

8. Jul 7, 2011

George Jones

Staff Emeritus
Yes, transform x and t in both of these equations.

9. Jul 7, 2011

Pi-Bond

Ok, I finally got it!

If the position of the first photon is given by x=ct in S, it's position in S' is given by:

$x'=t'\frac{c-v}{1-\frac{v}{c}}$

Similarly for the other photon, whose position in S is x=L+ct, it's position in S' is:

$x'=t'\frac{c-v}{1-\frac{v}{c}}+\gamma L(1+\frac{v}{c})$

The distance in S' is the difference in these positions, and thus the answer is obtained:

$\delta x'=L\sqrt{\frac{c+v}{c-v}}$

Just a follow up question - why can't I use the formula for length contraction to obtain this result?

10. Jul 7, 2011

lanedance

the length contraction formula gives the observed length contraction of the proper length of an object measured in the objects rest frame, when the proper length is measured from a frame of reference in relative motion to the object. the photons do not have a valid rest frame from which to measure the distance between them.

Last edited: Jul 7, 2011
11. Jul 7, 2011

George Jones

Staff Emeritus
Great!
lancedance gave you one answer. If I have the stamina, tomorrow I will give you a different and more detailed answer. In the meantime, can you write
$$\frac{c-v}{1-\frac{v}{c}}$$
much more simply? This won't change the answer you got, but it might make the physical situation clearer.

12. Jul 7, 2011

lanedance

hope i didn't confuse things, it has been a while since I've played with relativity...

13. Jul 8, 2011

Pi-Bond

I see - I can't use the length contraction formula because the photons themselves are in motion relative to the frame S.

@George Jones:
The first photon's position in S' can be written as:

$x'=ct'$

And the second one can be written as:

$x'=ct'+\gamma L(1+\frac{v}{c})$

I just copied my working here so I left the formulas as I wrote them. I would like to see what answer you can give!

14. Jul 10, 2011

George Jones

Staff Emeritus
As lanedance has said, in Lorentz contraction, there are two frames of reference; a ruler is at rest in one frame of reference and moves with respect to the other reference frame. In this situation, things (photons) move in two frames. This suggests considering a ruler that moves in two different frames.

In the unprimed frame, let a ruler have length $L$ and move (in the positive x direction) with speed $V$. Let the worldlines of ends of the ruler be given by $x = Vt$ and $x = Vt + L$.

Let the primed frame move with respect to the unprimed frame with speed $v$. Using Lorentz transformations to express these worldlines with respect to the primed coordinates gives (after rearrangement) $x' = V't'$ and $x' = V't' + L'$, where
$$V' = \frac{V - v}{1 - \frac{Vv}{c^2}}$$
(relative speed!), and
$$L' = \frac{L}{\gamma \left( 1 - \frac{Vv}{c^2} \right)} .$$
Here, $\gamma = \left(1 - v^2 / c^2 \right)^{-1/2}$.

In the special case $V = v$, then the ruler is at rest in the primed frame, and $V' = 0$ and $L' = \gamma L$ in the above equations. Lorentz contraction!

In the special case $V = c$, then there isn't a ruler, but $L$ and $L'$ as spatial distances between photons, and $V' = c$ and
$$L' = L \sqrt{\frac{1 + \frac{v}{c}}{1 - \frac{v}{c}}} .$$
This shows that Lorentz contraction and the situation given in the original post are quite different special cases of a more general situation.

15. Jul 10, 2011

Pi-Bond

How did you get the formula for L'?

$L'=\large \frac{L}{\gamma (1-\frac{Vv}{c^{2}})}$

16. Jul 11, 2011

Pi-Bond

After solving a similar question, I think I understand this situation. The formula for L' simply comes from the Lorentz transforms. In summary, this procedure is necessary because the length to be measured is itself in motion in the original frame. Thus to transform the length to another frame, the motion of the length must be accounted for - hence the Lorentz transforms are to be used.

Thanks a lot for guiding me through!