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Does a photon lose energy or redshift as it circles?

Will it's wavelength be the rest wavelength with centripetal and gravitational forces exactly cancelling?

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Conceptually yes. There is an orbit called the photon sphere with a 1.5 gravitational radius. A photon can circle around a black hole on this orbit indefinitely without losing energy. In reality however this is not possible, because the orbit is unstable. Any deviation of the direction of light from this exact orbit (e.g. due to dispersion) would send the photon either to the black hole or away from it. So in reality, light may make a number of circles, but eventually will all disperse away from the photon sphere.

The wavelength is not important, but it cannot be longer than the circumference of the orbit (very long radio waves).

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  • $\begingroup$ I interpret your answer as "conceptually no". ;) $\endgroup$ – m4r35n357 Jan 14 at 18:24
  • $\begingroup$ I was curious since binaries emit gravitational radiation. But photons don't... $\endgroup$ – cumfy Jan 14 at 20:47
  • $\begingroup$ @m4r35n357 I hear you and for any practical purpose it makes sense. However, a photon is a quantum object described by a quantum wave of probability. As such, its full treatment in strong gravity is a matter of the theory of quantum gravity that has not yet been developped. Although intuitively we may expect that the quantum wave of a photon in the photon sphere would disperse over time. This means the probability of the photon to be found in the photon sphere would decrease with time, but never become exactly zero. Therefore "conceptually" the photon can circle indefinitely. $\endgroup$ – safesphere Jan 14 at 20:59
  • $\begingroup$ @cumfy Whether photons can emit gravitational waves is a question for quantum gravity that has not yet been developed. However, Special Relativity prohibits a decay of massless particles, such as photons. Thus, if gravitons exist, then a photon in flight cannot emit gravitons. Some preliminary results show that the gravitational field of a photon is expected to be a gravitational wave emanating from the points of photon's emission and absorption. In any case this is rather moot, because photons would disperse away from the photon sphere long before losing energy as gravitational waves. $\endgroup$ – safesphere Jan 14 at 21:14
  • $\begingroup$ "The wavelength is not important, but it can not be longer than the circumference of the orbit (very long radio waves)." Was this argument, then, a law that prevents the formation of black holes smaller than the Planck length? Because black holes need to have a photon sphere, don't need?. $\endgroup$ – João Bosco Jan 16 at 2:52
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A photon, being a quantum of EM radiation, is not only a particle but also a wave. Basic solutions of wave equation around the black hole are quasinormal modes, which are exponentially decaying in time. They correspond to a wave either being absorbed by a black hole horizon or escaping to infinity. So no, a photon could not be orbiting black hole indefinitely even in principle. For a given value of angular momentum number $\ell$ there would be a quasinormal mode with the lowest imaginary part of an eigenvalue (this is a generalization of a circular orbit on a photon sphere), but this imaginary part (which is a decay constant of that mode) would approach a constant value for a large $\ell$.

Additionally, if a photon has high enough energy, its back-reaction on the gravitational field would cause it to lose energy via emission of gravitons. Conceptually the process is similar to an electron in excited state in an atom losing energy via emission of a photon.

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  • $\begingroup$ It sounds like your answer and mine state the same using different words. The physical meaning of the photon, as a wave, getting away from the photon sphere is dispersion. So we are describing the same process, no? While the probability of the photon to be found on the photon sphere is decreasing in time, it is never zero, is it? Thus conceptually, (although not practically) the photon can stay indefinitely on the photon sphere. Is this not true? Dispersion applies to straight lines too, yet we can detect photons coming from galaxies 13 billion years away. Just very few such photons. $\endgroup$ – safesphere Jan 14 at 21:37
  • $\begingroup$ Furthermore, the analogy with an electron in an atom does not hold, because the photon is massless and per special relativity cannot decay, e.g. emit gravitons. For this to happen, a massive particle must be involved (even if virtual, although it is debatable) like in the photon-photon interaction. Is this why you are referring to this process happening only at a high enough energy? $\endgroup$ – safesphere Jan 14 at 21:47
  • $\begingroup$ Agreed to the first part, and an analysis of the wave equations in the WKB approximation could be formulated in terms of 'particles' near a photon sphere. Probability of a photon staying in orbit "indefinitely" (and not for a given long time) is exactly zero, although here is somewhat philosophical stance should be taken on how to treat events of zero probability. $\endgroup$ – A.V.S. Jan 14 at 21:48
  • $\begingroup$ Agreed. Thanks for your insight! +1 $\endgroup$ – safesphere Jan 14 at 21:50
  • $\begingroup$ photon is massless and per special relativity cannot decay, e.g. emit gravitons this is not a decay of an isolated photon. It is a decay of a bound (if only marginally) state of a black hole and a photon, as a result of this decay we have a photon occupying lower number state and an escaping graviton. Just like kinematic forbids isolated electron to emit a photon, bound electron could emit a photon by transition into lower energy state. $\endgroup$ – A.V.S. Jan 14 at 21:51

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