I’m not quite convinced. If I understand you correctly, what you’re saying is the following:
Suppose we send out one photon ν of momentum p_ν. We hence gain momentum -p_ν. This photon then randomly hits a particle X of a particle-antiparticle pair and transfers its momentum/energy to the particle X, creating/altering it into a particle X'. When X' and ¬X collide again, this momentum/energy is lost and doesn’t hit the opposite wall of the cavity (where it would be transferred onto us again, causing us to lose the previously-gained momentum -p_ν).
But this would require that when X' and ¬X annihilate each other, the extra momentum of X' is also annihilated. Intuitively, I’d expect it to be sent out as a photon – with momentum p_ν.
> But this would require that when X' and ¬X annihilate each other, the extra momentum of X' is also annihilated. Intuitively, I’d expect it to be sent out as a photon – with momentum p_ν.
I.e. annihilation of the extra momentum would be required for this to work on the basis of background fluctuations (in my understanding), but I don’t think that this happens.
Suppose we send out one photon ν of momentum p_ν. We hence gain momentum -p_ν. This photon then randomly hits a particle X of a particle-antiparticle pair and transfers its momentum/energy to the particle X, creating/altering it into a particle X'. When X' and ¬X collide again, this momentum/energy is lost and doesn’t hit the opposite wall of the cavity (where it would be transferred onto us again, causing us to lose the previously-gained momentum -p_ν).
But this would require that when X' and ¬X annihilate each other, the extra momentum of X' is also annihilated. Intuitively, I’d expect it to be sent out as a photon – with momentum p_ν.