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?
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$\begingroup$ locally the speed of light is always c, and since the photon's orbital radius doesn't change it stays that way in the frame of the observer, since his gravitational time dilation is the same at the same radius. $\endgroup$– YukterezCommented Nov 10, 2023 at 20:35
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$\begingroup$ Thank you very much for your answer. $\endgroup$– Alexandre ZagaraCommented Nov 10, 2023 at 20:51
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$\begingroup$ In Schwarzschild coordinates this works since the angular components gθθ and gφφ are euclidean so that the circumference measured with stationary rulers is simply 2πr. $\endgroup$– YukterezCommented Nov 10, 2023 at 22:16
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$\begingroup$ That was exactly my missing point! thank you so much! $\endgroup$– Alexandre ZagaraCommented Nov 10, 2023 at 23:09
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$\begingroup$ @Yukterez According to the latest research of solving Maxwell equations in the photon sphere limit,- solution of wave equation isn't stable. Which means that photon soon either fall into black hole or escape from photon sphere. I don't remember exact paper where I've read about this instability. Though, I believe you can google it. $\endgroup$– Agnius VasiliauskasCommented Nov 10, 2023 at 23:35
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