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If yes, then how does this accord with relativity: the laws of physics are the same in all reference frames? We can move from a reference frame in which the photon has near zero energy density, to a reference frame where it has near infinite energy density.

If no, why are there no gravitational effects for a photon with an arbitrarily high energy density? For some reference frames, a photon can have more energy that the mass of the sun. How are these reference frames consistent with each other?

user1247
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The answer to this question is through logical positivism--- the principle that the questions be formulated in a way that they give an experimentally realizable situation and ask "what happens"?

In this case, it depends on how you scatter things off the photon. If you scatter two of these things head on, they will combine to make a mini-black hole, and you need a full quantum gravity theory to describe the decay. But if you chase the photon and study it in a frame where its energy is very sub-Planckian, you will discover it is a photon. This is also the answer for a very glancing collision of two of these photons, moving nearly parallel.

The quantity which determines the regime which is appropriate is the center of mass energy of the collision.

Further, since all objects will produce black holes which are difficult to tell apart from one another, any highly boosted trans-Planckian point-particle undergoing a collision whose center-of-mass Schwartschild radius is bigger than the particle radius will look indistinguishable statistically from any other particle. This principle is the universality of the gravitational interaction at high energies, and it is an important focus of string theory research. Susskind investigated highly boosted strings on the path to black hole complementarity, concluding that a highly boosted string falling into a black hole will lengthen and wind around the horizon, and is indistinguishable from a surface excitation of the black hole.

Ron Maimon
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  • This is also the answer for a very glancing collision of two of these photons, moving nearly parallel. --

    But if one photon has more energy than the mass of the sun (for example) then in this reference frame won't the parallel photon be attracted by a gravitational field, and hence distinguishable from a reference frame where it wasn't attracted?

     – user1247 Jan 15 '12 at 05:36
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    I gave you +1, Ron, despite your misspelling of Karl Schwarzschild's name. – Luboš Motl Jan 15 '12 at 07:53
  • @user1247: Two photons moving in parallel photons do not attract no matter how boosted--- this is a consequence of special relativity, but it is also confirmed in GR, and it is the basis of plane wave exact solutions which propagate in a fixed direction and do not collapse. – Ron Maimon Jan 15 '12 at 08:22
  • @Ron Maimon -- But the essence of my question (for which I'm itching for an answer) is why. I thought that in GR energy is a source of gravity. Why is the energy of a photon an exception? – user1247 Jan 15 '12 at 16:05
  • @Ron Maimom -- According to accepted answers to another stackexchange question you aren't correct about this. – user1247 Jan 15 '12 at 16:14
  • @user1247: Two nonparallel beams of light do attract, there is no contradiction. Only parallel light beams are nonattractive. The energy of the photon is attractive, but gravity is a tensor force, and a photon also has momentum along its direction, and the analog of magnetic forces in gravity make the photon beams stable. The same is true in electromagnetism--- two electrons highly boosted in the same direction have an electric repulsion exactly balanced by a magnetic attraction. – Ron Maimon Jan 15 '12 at 16:20
  • @Ron Maimon -- OK, just to make sure I understand: you are saying that if we have two photons moving anti-parallel (approaching each other from opposite directions), their deflection angle is invariant with respect to the photon energy. If the two photons are 500nm they will be deflected the same angle as if they are blue shifted to have the energy equivalent of a supermassive black hole. The invariance is because the deflection due to the gravitational field of the supermassive black hole is compensated by gravitomagnetic effect. Is this right? I'll accept your answer if you edit to add this. – user1247 Jan 15 '12 at 16:33
  • @user1247: This is not right, because angle is not relativistically invariant. Also, boosting photons doesn't change their center of mass energy, because one is blueshifted and the other redshifted. If you boost them perpendicular to their motion, they become nearly parallel, and time dilation slows the collision. The momentum transfer squared is the relativistically invariant analog of angle. The compensation is for parallel beams. I don't know what you want me to add, because what you ask me to add is wrong. – Ron Maimon Jan 15 '12 at 17:35
  • OK, I will re-state my question. If we are in the CM frame watching two photons come in from opposite directions, does the angle of deflection due to the gravitational interaction depend on the CM energy? – user1247 Jan 15 '12 at 18:52
  • @user1247: Yes, of course. The angle of deflection is equal to the center of mass energy divided impact parameter (the distance between the pencils), up to constant factors, for small angles. – Ron Maimon Jan 15 '12 at 19:04
  • @Ron Maimon -- If two photons are moving parallel and one has more energy than the black hole at the center of the milky way, then I understand from your comments that gravitomagnetic effects cancel the gravitational attraction. OK, but what about hawking radiation, and other effects one would expect in GR of an object with that much energy density? – user1247 Jan 15 '12 at 19:05
  • @user1247: Hawking radiation time dilated with boost, so it only depends on the invariant rest mass of the object, in this case, zero. There is no Hawking radiation from a transplanckian photon, although there will be as soon as you scatter it off a stationary object. – Ron Maimon Jan 15 '12 at 23:09
  • @Ron Maimon: So the rate of hawking radiation is exactly compensated for by the gamma factor? I've never seen a formula for hawking radiation consistent with that. Also, if there is Hawking radiation between a transplanckian photon and a stationary object, then any stationary observer should see Hawking radiation from a transplankian photon? Or are you somehow distinguishing between a "hard scatter" in which the hawking radiation is highly off-shell vs elastic scattering? – user1247 Jan 15 '12 at 23:59
  • Dude, stop trying to find inconsistencies in what I am saying, and learn this stuff. If you boost a black hole, the Hawking radiation rate goes down by time dilation, and this is so obvious it isn't written in the formulas. Your'e supposed to be competent enough to figure this out by yourself. The hawking radiation is not "between" the transplanckian photon and the stationary object, but the two will collide at a finite impact parameter and form a black hole, which then decays by normal Hawking radiation (in its rest frame). – Ron Maimon Jan 16 '12 at 04:25
  • I'm not trying to find inconsistencies, but your statement about hawking radiation seems to be obviously flawed. The time dilation factor varies continuously and doesn't have a "turn on". Hawking radiation behaves totally differently. I don't understand. I will continue in chat! – user1247 Jan 16 '12 at 05:52