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So the question states:

A's and B's systems are related by the standard Lorentz transform. B simultaneously fires off two photons from transmitters distance D apart along the x' axis, and in the direction of increasing x'. Show that A measures the distance between the photons as:

$$ D \sqrt{\frac{1 - \frac{v}{c}}{1 + \frac{v}{c}}} $$

So I put A's axis as $(t,x)$ and B's axis as $(t',x')$ and applied the Lorentz transform which is:

$\begin{bmatrix} t\\ x \\ \end{bmatrix}$ = $\gamma \begin{bmatrix} 1 & \frac v{{c}^2}\\ v & 1 \\ \end{bmatrix} \begin{bmatrix} t'\\ x' \\ \end{bmatrix}$

Where $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}.$$

So am I right in thinking that you would apply the Lorentz transformation with $x'=D, t'=t$ and then rearranging the resulting equations to get $x$ in the required form? I am asking because I have tried to rearrange the resultant equations but to no avail. I just want to make sure I am making the right substitutions before I attempt different manipulations of the resultant equation. Thanks.

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I tried my hand at editing the equations to display in a more reader-friendly way. You might double-check that I got the expressions you wanted. –  David H Oct 16 '13 at 17:17
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Shouldn't the off-diagonals be $\beta=v/c$ and not $v/c^2$ and $v$? –  Kyle Kanos Oct 16 '13 at 17:21
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1 Answer 1

up vote 3 down vote accepted

Let's start in B's frame. The spacetime points where the two photons are released are $P_1 = (0, 0)$ and $P_2 = (0, d)$. We'll choose A's frame so the origins coincide, in which case we just need to find the transform of the second point, $P_2$. Using the usual Lorentz transformations we find that:

$$ P_2' = (-\frac{\gamma v d }{c^2}, \gamma d) $$

So in A's frame $P_2'$ happened before $P_1'$. That means we need to wait a time $\gamma v d/c^2$ for the time coordinate of $P_2'$ to increase to zero, and because the photon is moving with a speed $c$ that means in this time the photon moves a distance $\gamma v d/c$, and the time zero position becomes:

$$ P_2'(t' = 0) = (0, \gamma d + \frac{\gamma v d }{c}) $$

The first photon is at the origin at $t' = 0$, so the separation $d'$ is simply:

$$ \begin{align} d' &= \gamma d + \frac{\gamma v d }{c} \\ &= \gamma d (1 + \frac{v}{c}) \\ &= d \frac{1 + v/c}{\sqrt{1 - v^2/c^2}} \\ &= d \frac{1 + v/c}{\sqrt{(1 - v/c)(1 + v/c)}} \\ &= d \frac{\sqrt{1 + v/c}}{\sqrt{1 - v/c}} \end{align}$$

Of course, only now do I realise I transformed from unprimed to primed rather than primed to unprimed, so swap the signs of $v$ to get the equation you're asked to prove.

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Thanks, for that I get it now. But I didn't want it solved, just wanted suggestions where I was going wrong. Thanks anyway, I know get where I was going wrong, it was thinking that the time in both frames was the same. –  user31135 Oct 16 '13 at 17:41
    
i was thinking that the time in both frames was the same. That is a common mistake. –  Kyle Kanos Oct 16 '13 at 17:45
    
@user31135: oops sorry :-( In my defence, I find Lorentz transform problems like this irrestible - they're like a physicist's Sudoku. Not too brain taxing, but there's a definite sense of satisfaction when you get the correct answer :-) –  John Rennie Oct 16 '13 at 17:51
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