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So just a precursor, my main focus is math and I don't fully understand everything in physics so I apologize if this has an obvious answer or the wrong tags are used.

That being said, I do know that mass-less particles travel at the speed of light and they have energy, meaning they exert a gravitational force. I also know that gravity travels at the speed of light. By knowing these 2 things can we conclude that light makes a sort of sonic boom with gravity being the pressure wave and the mass-less particle being the supersonic object?

Qmechanic
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Jacob Claassen
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    Note that for light, one doesn't get Cherenkov radiation (roughly the equivalent of a Mach cone for sound) until the emitter's speed exceeds the local speed of light. So you'd have to either have a warp bubble or a matter distribution with a gravitational index of refraction. – rob May 04 '18 at 00:46
  • @rob, I'm saying that photons exert gravitational waves in the direction they're traveling and since the wave will travel at the speed of the photon it will be like a sonic boom but for gravity. – Jacob Claassen May 04 '18 at 00:59
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    For a massless photon traveling through vacuum, it can't create a "gravity boom" because the energy would have to come from somewhere, and we know from astronomy that photons don't lose energy traveling through vacuum. – Peter Shor May 04 '18 at 02:18
  • If you consider electromagnetism, no electromagnetic waves are produced for a moving charged object that is not accelerating. Using that as an analogy I would imagine a light pulse would carry some miniscule gravitational field but wouldn't lose energy due to radiation. Maybe someone has done the calculation for classical GR coupled to classical EM. – octonion May 04 '18 at 02:21
  • @PeterShor, would creating this gravity boom lower the energy of the photon? Because it interacts with gravity regardless. This would just be the gravitational effect clumping together. If that makes sense... – Jacob Claassen May 04 '18 at 02:29
  • @JacobClaassen both gravitons and photons are quantum mechancial entities. In quantum mechanics energy can only be exchanged through interactions of real particles. A photon running along with velocity c is not interacting , as surely as an electron running with velocity v in space is not interacting and nothing will happen because there is no energy input or output. – anna v May 04 '18 at 05:39
  • The gravitational field of a photon has been calculated. It is a gravitational wave that appears as if emanating from the point of the emission of the photon and thus associated with the event of the emission. This implies that the gravitational energy comes from the emission. @PeterShor is correct that a photon does not lose energy in flight. This is easy to understand, because time does not move for the photon at the speed of light, so the photon cannot decay emitting gravitons and losing energy. – safesphere May 04 '18 at 06:52

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To start with, in order to study photon graviton interactions one must have a quantized gravitation which will give the appropriate interaction vertices and diagrams. There does seem to be a "graviton compton scattering" according to the link. As graviton is spin two and photon is spin 1 it is not simple like electron photon. This is for a real interaction.

The gravitational field that a particle carries due to its energy momentum vector is described with virtual exchanges. The electric field of an electron moving with velocity v can be described in quantum field theory as built up by virtual photons, this means that no energy is lost until there is an interaction. I presume that the gravitational field of a particle moving with velocity v will be described by virtual gravitons.

The virtual gravitons and the real photon, whose gravitational field creates the virtual gravitons, are not exchanging any energy. A sonic boom type of mathematics cannot be built up because virtual means no energy is exchanged, and a sonic boom consumes energy.

So no, the analogy cannot hold.

anna v
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Massless particles are known to experience the same gravitational acceleration as other particles (which provides empirical evidence for the equivalence principle) because they do have relativistic mass, which is what acts as the gravity charge.

D.Sarkar
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