Graduate Photon speed for an observer at the photon sphere

[alexriemann]

TL;DR

Photon speed for an observer at the photon sphere

I am asked to compute the orbital period of a photon, in the Scwarzschild spacetime, at the photon sphere for an observer at the same radius, $r^\star=3M$. I have computed the result, $\Delta T=6\pi M$ where $c=G=1$ ,comparing with the proper time of an observer at infinity. However, as the result gives directly $\Delta T=2\pi r^\star$, I wonder if I can skip making the calculation by inferring that, for the observer sitting at the photon sphere, the speed of the photon is exactly $c=1$. If that is the case, how can I argue that this is true?

 

[Orodruin]

Please show your computation.

Note that you cannot have a time coordinate difference of $\Delta T = 6\pi M$ at the same time as you are saying that a local hovering observer at $r^\star$ observes light to pass by at $c = 1$ as the hovering observer is gravitationally time dilated relative to the observer at infinity.

 

[alexriemann]

$\Delta T$ is not the coordinate time difference, it is the orbital period of the photon for the observer sitting at the photon sphere. Actually, for the observer at infinity, the orbital period of the photon would be $6\sqrt3\pi M$. Maybe ny notation wasn't clear.

My guess is that I can argue that locally the speed of light is always c, and for the observer at the same radius as of the photon, this remains true because the photon radius coordinate remains the same throughout its motion.

 

[Orodruin]

ΔT is not the coordinate time difference

I’m not sure how else to read this:

comparing with the proper time of an observer at infinity

But anyway …

Yes, the definition of the $r$ coordinate relates to the area of the sphere*. As such, a great circle on such a sphere will have circumference $2\pi r$ and it will take $2\pi r/c$ for a light singnal to go around for the local observer.

* Consequently note that $r$ does not correspond to a radial distance of any sort.