Interpretation of red shift We see light as having travelled in a straight line from stars or galaxies light years away from us. However it's path is more likely of multiple curves as a result of gravity along the journey (gravitational lensing) . How can we be sure that the accelerations thus experienced have not produced braking radiation (bremsstrahlung) thereby reducing the energy and therefore frequency of the light spectrum leading to an apparent red shift?
 A: Bremsstrahlung (yes, that's the correct english term) is only emitted if a charged particle is deflected - as a photon has no charge it won't ever emit bremsstrahlung.
Another way to see that a photon cannot simply emit another photon is by conservation of energy-momentum. If the process $\gamma \rightarrow \gamma +\gamma $ were possible, one should be able to go to the center of momentum frame of the two final particles. However, conservation of momentum tells us that then, the momentum of the initial particle (and hence the energy, applying $E=pc$) must vanish, an absurd situation!
A: As has been pointed out by Benedikt, photons never emit Bremsstrahlung since they do not carry charge. However, more fundamentally important to this question is the fact that particles in free fall do not experience any proper acceleration at all - the covariant derivative of their 4-velocity with respect to some appropriately chosen affine parameter vanishes at all points, meaning these particles trace out geodesics in space-time. In the special case of light these will be null geodesics - so, even if photons did carry charge, there still would not be any bremsstrahlung present.
