1.1) Pphoton = E/c
The box, of mass M, will recoil slowly in the opposite direction to the photon with speed v. The momentum of the box is:
1.2) Pbox = Mv
The photon will take a short time, Δt, to reach the other side of the box. In this time, the box will have moved a small distance, Δx. The speed of the box is therefore given by
1.3) v = del x / del t
By the conservation of momentum, we have
1.4) M*(del x/del t) = E/c
If the box is of length L, then the time it takes for the photon to reach the other side of the box is given by:
1.5) del t = L/c
Substituting into the conservation of momentum equation (1.4) and rearranging:
1.6) M*del x = EL/c^2
Now suppose for the time being that the photon has some mass, which we denote by m. In this case the centre of mass of the whole system can be calculated. If the box has position x1 and the photon has position x2, then the centre of mass for the whole system is:
1.7) x-bar = (Mx1 +mx2)/(M+m)
We require that the centre of mass of the whole system does not change. Therefore, the centre of mass at the start of the experiment must be the same as the end of the experiment. Mathematically:
1.8) (Mx1+mx2)/(M+m) = (M(x1-del x) + mL)/(M+m)
The photon starts at the left of the box, i.e. x2 = 0. So, by rearranging and simplifying the above equation, we get:
1.9) m*L = M*del x
Substituting (1.4) into (1.9) gives:
1.10) m*L = EL/c^2
E/c^2 = m
E = mc^2
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