Lorentz transformation - need a little clarification 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.
 A: 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.
