If two bodies interact they interchange force carriers and as 3rd Newton law states as the one body influences the other in the same way the other body would be doing the same to the first body. So am I wrong in thinking that the black hole is losing some kind of energy forcing the photons to bend their trajectories as they posses some kind of kinetic mass/energy eqivalent?
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asked May 29, 2022 at 16:09
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$\begingroup$ The same conservation laws apply to black holes and billiard balls. If you know the final trajectories, you can easily calculate energies and momenta. $\endgroup$– safesphereCommented May 29, 2022 at 22:51
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A black hole does not lose energy because of gravity (at least, not unless we count extreme scenarios like the merging of two black holes). Instead, infalling photons and matter exchange potential energy for kinetic energy. As matter gets closer to the black hole and forms an accretion disc, some of this kinetic energy is converted into heat, light and other forms of radiation.
Newton's 3rd Law does tell us that gravity acts in both directions - so as the black hole attracts photons and matter, they in turn exert an equal and opposite force on the black hole. However, the effect on the black hole is insignificant because it is so much more massive than the infalling particles (once again, extreme scenarios like merging black holes are an exception).
answered May 29, 2022 at 16:23
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$\begingroup$ This answer is incorrect on many levels. There is no “potential energy” in GR. Potential energy relates to a force field, but gravity is not a force in GR. So a black hole doesn’t attract photons, but curves spacetime thus changing photon’s trajectories towards itself. The opposite is not true. Photons don’t curve spacetime and in this sense don’t affect the black hole the same way. Newton laws are not applicable to massless photons or in general to anything in GR. $\endgroup$ Commented May 29, 2022 at 21:21
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1$\begingroup$ @safesphere anything with a nonzero stress-energy tensor curves spacetime, electromagnetic fields have a nonzero stress-energy tensor, and thus curve spacetime. So yes photons do exert gravitational effects on massive bodies. $\endgroup$ Commented May 29, 2022 at 22:02
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$\begingroup$ @Triatticus “anything with a nonzero stress-energy tensor curves spacetime” - This is a widespread misconception based on a lack of understanding of what the stress energy tensor is and how it affects spacetime. A relativistic bullet (an object with a negligible mass, but a huge kinetic energy) moving very close to the speed of light does not curve spacetime based on the principle of relativity. The kinetic energy contribution to the tensor is exactly canceled out by the contribution of momentum (gravitomagnetism). All energy of photons is kinetic, so photons don’t curve spacetime. $\endgroup$ Commented May 29, 2022 at 22:23
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$\begingroup$ @safesphere so when I have two relativistic bullets that just barely miss each other, the spacetime is flat the whole time, but when I change an angle a little and they collide with kinetic energy getting transformed to heat, they become heavy and spacetime curves? Sounds very counterintuitive to me. Kinetic energy of any particular bullet depends on the observer, but when you have more objects there is an "irreducible" portion of kinetic energy corresponding to their kinetic energy in COM system (usually called inner energy). I would say this energy is what goes into stress-energy tensor. $\endgroup$– UmaxoCommented May 30, 2022 at 9:39
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$\begingroup$ @Umaxo All energy goes into the stress-energy-momentum tensor, including the kinetic energy. However momentum also goes to the same tensor with the opposite sign. It is easy to see from the relativity principle that their contributions cancel out exactly. When two relativistic bullets hit each other, their momenta cancel out, no longer contribute to the tensor, and no longer offset the contribution of the energy. The energy however conserves and changes its type from kinetic to something else. The energy contribution to the tensor remains and is no longer offset by the contribution of momenta. $\endgroup$ Commented May 30, 2022 at 22:45