Distance Between Two Photons Calculated in Different Inertial Frames

Question

I am a self-studier. This is a question from a text I am studying:

The distance between two photons traveling along the $x$-axis of an inertial frame, $S$, is always $l$. Show that in a second inertial frame, $S'$, moving at constant speed $V = \beta c$ along the $x$-axis, the separation between them is given by: $$\Delta x' = l\left(\frac{1 + \beta}{1 - \beta}\right)^{1/2}$$

I would appreciate help solving this. I was also wondering what $l$ is in the frame of the photons.

Also I was wondering how to deal with the length $l$ in the $S$ frame, as it seems to be moving at the speed of light.

Thanks for your help and patience with an enthusiastic beginner.

Answer 1

We don't normally answer homework question in full, but I'll answer because I'm very eager to make the point that the best, by far, method of answering questions like this is to choose relevant spacetime points in $S$, apply the Lorentz transformations and find where the points are in $S'$. This normally makes the answer obvious.

In this case we have a pair of photons travelling along in $S$ with a distance of $\ell$ between them. Let's choose our $x$ axis so that the photons are travelling along it, and we'll choose the origin so that in $S$ the rear photon is at the point $(t = 0, x = 0)$. This means that at time zero the front photon is at $(0, \ell)$:

All we have to do is find the two points in $S'$, and work out the spacing between them and we'll have our answer. The Lorentz transformations are:

$$ t' = \gamma (t - \frac{vx}{c^2}) $$

$$ x' = \gamma (x - vt) $$

If we feed in the first point $(t = 0, x = 0)$ we just get $(0, 0)$, so in $S'$ the rear photon is at the origin. If we feed in the position of the front photon $(0, \ell)$ we get:

$$ t' = \gamma (t - \frac{vx}{c^2}) = -\gamma\frac{v\ell}{c^2} $$

$$ x' = \gamma (x - vt) = \gamma\ell $$

So in $S'$ the front photon is at $(-\gamma\frac{v\ell}{c^2}, \gamma\ell)$. But we don't want to know the position of the front photon at time $t' = -\gamma\frac{v\ell}{c^2}$, we want to know it at time $t' = 0$. So we wait a time $-\gamma\frac{v\ell}{c^2}$ during which the photon (travelling at $c$) moves a distance $c\gamma\frac{v\ell}{c^2} = \gamma\frac{v\ell}{c}$. So at time zero in $S'$ the position of the front photon is:

$$ \left(0, \gamma\ell + \gamma\frac{v\ell}{c}\right) $$

and the spatial separation between the two points is just the $x'$ coordinate of the front photon:

$$ \ell' = \gamma\ell + \gamma\frac{v\ell}{c} \tag{1} $$

The rest is just algebra. The question defines $\beta = \frac{v}{c}$, and the Lorentz factor $\gamma$ is then given by:

$$ \gamma = \frac{1}{\sqrt{1 - \beta^2}} $$

Substitute this into equation (1) and we get:

$$\begin{align} \ell' &= \ell\left( \frac{1 + \beta}{\sqrt{1 - \beta^2}} \right) \\ &= \ell\left( \frac{1 + \beta}{\sqrt{(1 - \beta)(1 + \beta)}} \right) \\ &= \ell\left( \frac{\sqrt{1 + \beta}}{\sqrt{1 - \beta}} \right) \\ &= \ell\left( \frac{1 + \beta}{1 - \beta}\right)^{\tfrac{1}{2}} \end{align}$$

Which is what we were asked to prove.

Answer 2

First consider what you're transforming. The $\Delta x$ interval is $l$ while the $\Delta t$ interval is $l/c$ (if you can't see why draw a space-time diagram). The interval $l$ in the $S$ frame is simply the distance between two points. It's easy to picture if you imagine them as stationary particles, and easy to picture if they're moving slowly. It holds that the interval is the same even when the points are travelling at the speed of light. The speed only changes the time interval, which is made easier to see with a space-time diagram (try drawing up all three cases!).

Now you can use the Lorentz transformation: $$ \Delta x' = \gamma(\Delta x -v\Delta t) $$

And simplify to get the answer in the correct form.

Further, the idea of $l$ in the frame of the photons is not possible. This is because photons do not have a rest frame. Since one of the fundamental principles of special relativity state that light speed is constant in all inertial reference frames, there cannot exist a frame where the speed of a photon is zero; it must be $c$. Hope that clears some things up for you.

Answer 3

I have been thinking more about this problem. Of course, my understanding of the application of Lorentz transformations was substantially enhanced by the two answers kindly provided.

I was able to solve it using proper time.

Also, due to the nature of the problem, I can solve it using just the time equation of Lorentz transformations:

$t' = \gamma (t - \frac{vx}{c^2})$. In this case, $t'= \frac{\ell'}{c}$.

Then $\frac{\ell'}{c} = \gamma (\frac{\ell}{c} - \frac{v\ell}{c^2})$

The rest follows as in (1) in the first of the answers above.