Reflective photon rocket A space-craft, also of rest-mass $M_0$ and initially at rest, is propelled by
reflecting a plane-parallel beam of photons, generated at a rate $αM_0$ from a stationary
source mounted on the launch pad, off a perfectly reflecting mirror mounted on the rear
of the space-craft. Derive an expression for the acceleration of this second space-craft
as observed by astronauts on the space-craft as a function of its velocity relative to the
launch pad.

I have come up with several, different answers to this.  For all of them I use that the accelleration is proportional to the rate of incoming momentum ie $a\propto (\text{momentum per photon})\times (\text{rate at which they hit})$.
1) Poynting vector and momentum flux are proportional by a factor of $c$ and hence the momentum flux scales with energy flux.  Energy flux is proportional to the square of the frequency.  This implies $$a\propto \frac{1-\beta}{1+\beta}$$
2)assume photons are emitted every $\delta t_0$ in the frame of the launch pad.  This time, in the rocket frame becomes $\delta t= \gamma \delta t_0$.  In this time the launch pad moves away so the photon has further to travel, hence the time between photons hitting the rocket is $v\delta t$.  The momentum of each photon gets doppler shifted.  hence $$a \propto \sqrt{\frac{1-\beta}{1+\beta}}\frac{1}{\beta \gamma}$$
Which of these, if either, has the correct $\beta$ dependance?
 A: Answer 1 is right; answer 2 has the wrong rate of arrival of photons. Suppose the source emits some photons in a pulse which is $N$ wavelengths long. The time taken to emit them must be the time taken for light to travel the distance $N \lambda$, which is $N \lambda/c$. Equally, the time for those same photons to be received is $N \lambda'/c$ for any observer who finds the wavelength to be $\lambda'$. The wavelength transforms like the inverse frequency, so we get 
$$a \propto \sqrt{ \frac{1-\beta}{1+\beta} } \, \sqrt{ \frac{1-\beta}{1+\beta} } 
= \frac{1-\beta}{1+\beta}$$
If you don't want to go via the wavelength, then just consider the pulse itself as it catches up with the moving rocket, all observed in the launch pad frame. From this you will get the time the launch pad observer's clock registers for the photons to hit the rocket. The closing speed of pulse and rocket is $c-v$ so it takes $L/(c-v)$ for a pulse of length $L$ to be absorbed by the rocket (and the emitter took $\delta t_0 = L/c$ to emit it). The rocket observer finds a shorter time than this by $\gamma$, 
$$ \delta t = \frac{L}{c-v} \frac{1}{\gamma} = \frac{L}{c} \frac{\sqrt{1-\beta^2}}{1-\beta} = \delta t_0 \sqrt{ \frac{1+\beta}{1-\beta}}$$
After allowing also for the transformed energy-momentum of the radiation in the pulse
you thus end up with the same answer again. 
(A wrong answer to this question was for a long time shared on Oxford University's past exam paper 'suggested solutions' for internal consumption; I don't know if it is still there. Perhaps someone should have removed it; or perhaps it served as a prompt to students to think harder about their own work ...)
A: In the rest frame of the photon source the force is 2E/c.
