Question

Compute the orbital period of a photon, in the Scwarzschild spacetime, at the photon sphere for an observer at the same radius,r⋆=3Mr⋆=3M. You have computed the result, ΔT=6πMΔT=6πM where c=G=1c=G=1, comparing with the proper time of an observer at infinity. However, as the result gives directlyΔT=2πr⋆ΔT=2πr⋆, You wonder if you can skip making the calculation by inferring that, for the observer sitting at the photon sphere, the speed of the photon is exactly c=1 c=1. If that is the case, how can you argue that this is true?

Answer 1

In the Schwarzschild spacetime, the speed of light is constant and does not depend on the observer's position relative to the black hole. Therefore, for an observer sitting at the photon sphere, the speed of the photon can be argued to be exactly c=1.

Step-by-step explanation:

In the Schwarzschild spacetime, the orbital period of a photon at the photon sphere for an observer at the same radius, r⋆=3M, is given by ΔT = 6πM, where c=G=1.

However, we can argue that for the observer sitting at the photon sphere, the speed of the photon is exactly c=1.

This can be inferred because the speed of light, c, is constant in all frames of reference and does not depend on the observer's position relative to the black hole.

Therefore, the observer at the photon sphere would measure the speed of the photon to be c=1.