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This question follows from a schooling I received in this thread.

I figured that photons do not interact with gravity, except when they've spontaneously converted into a particle-antiparticle pair. So the equivalence $E=mc^2$ only means that a photon could transform into a massive particle.

Apparently I'm wrong; @JohnRennie succinctly writes:

The source of the curvature is the stress-energy tensor and this makes no distinction between mass and energy. They are treated as related by the famous $E=mc^2$.

This is cool because $E=mc^2$ can act as some sort of uncertainty relation; if you have a population of photons with energy $E$, they are engendered with a mass $\frac{E}{c^2}$, no matter what my intuition says. (Is that right?)

So my real question is, in any model of particle interactions, can a photon directly generate a gravitational field? For example, something like a Feynman diagram with a photon and a massive particle linked with a graviton. I suspect the answer is negative.

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    $\begingroup$ You seem to be hung up on the idea that mass produces gravity---which is a reasonable place to start as that is exactly what Newton said---but this idea is generalized in general relativity. The stress-energy tensor is the fundamental source of gravitation in GR and light affects it. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Nov 29, 2012 at 16:22
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    $\begingroup$ No, I'm hung up on the idea that energy directly produces gravity. $\endgroup$
    – Ryan
    Commented Nov 29, 2012 at 16:53
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    $\begingroup$ Let me put it like this: in nuclear and particle physics (my business) we know that binding energy acts like mass. We can measure these effects. In particular most of the mass of protons and neutrons is binding energy not quark mass, and the mass of the Earth is mostly the mass of protons and neutrons and Newtonian gravitation works fine with it. Energy is the source of gravitation and always has been even though Newton didn't know it. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Nov 29, 2012 at 17:01

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This is cool because $E=mc^2$ can act as some sort of uncertainty relation; if you have a population of photons with energy $E$, they are engendered with a mass $\frac{E}{c^2}$, no matter what my intuition says. (Is that right?)

No, it's not quite right. In relativity, it turns out that energy and momentum are parts of a single four-dimensional vector, imaginatively called "four-momentum", and mass is simply the length of that vector. The general relationship is $(mc^2)^2 = E^2 - (pc)^2$, which simplifies to $E = mc^2$ just in the case of zero momentum.

Thus, if you have photons in one direction, then each of them have energy $E = pc$, and there is no mass. But if you have photons in opposite directions but with equal energy, then the total momentum is zero, and you have have total mass.

Intuitively, think of a box of negligible mass, and fill it with lots of photons with zero total momentum. If you try to accelerate the box in some direction, then the photons coming to one side will be redshifted, thus impinging less momentum on that side, and on the other side will be blueshifted instead. Thus, accelerating it will require a force, so we can measure the positive mass of the filled box (more than an empty one).

However, your main mistake is that you're thinking that mass is the gravitational charge. It isn't, despite Newtonian thinking--it's actually energy. It's just every mass has energy $mc^2$ and this usually dominates all other forms of energy under normal conditions. Because the gravitational field is spin-2, not just energy is important, however--see stress-energy tensor.

So my real question is, in any model of particle interactions, can a photon directly generate a gravitational field? For example, something like a Feynman diagram with a photon and a massive particle linked with a graviton.

The Einstein field equation is completely classical, so it doesn't even mention gravitons. But a photon can generate a gravitational field because it has energy, which is the gravitational charge.

This is similar to how Maxwell's equations do not care about where the electrical charge comes from, whether it's continuous or made of discrete particles, or what kind of exchange interactions are going on under the hood--it just says that if you have such-and-such charge distribution, then the electromagnetic field will be related to it in such-and-such way.

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  • $\begingroup$ "However, your main mistake is that you're thinking that mass is the gravitational charge. It isn't, despite Newtonian thinking--it's actually energy. It's just every mass has energy $mc^2$ and this usually dominates all other forms of energy under normal conditions." This is it. This is what I was missing -- in GR, energy is the gravitational chage. $\endgroup$
    – Ryan
    Commented Nov 29, 2012 at 16:58
  • $\begingroup$ "Thus, if you have photons in one direction, then each of them have energy E=pc, and there is no mass. But if you have photons in opposite directions but with equal energy, then the total momentum is zero, and you have have total mass." That's a great way of explaining it! $\endgroup$
    – twistor59
    Commented Nov 29, 2012 at 20:56
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    $\begingroup$ @Ryan In fact, momentum and some other things are also part of the gravitational charge. In Einstein's General Relativity, it is the energy-momentum tensor that couples to the curvature of spacetime(aka gravitation). See en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor $\endgroup$
    – resgh
    Commented Nov 30, 2012 at 10:50
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    $\begingroup$ It couples, yes, though one wouldn't normally say that momentum is a gravitational charge any more than call electrical current an electric charge. Four-current $J^\mu$ and stress-energy $T^{\mu\nu}$ are analogous and for particles built in essentially the same way. $\endgroup$
    – Stan Liou
    Commented Nov 30, 2012 at 19:13
  • $\begingroup$ "But if you have photons in opposite directions but with equal energy, then the total momentum is zero, and you have have total mass", why can we use the total energy and total momentum for 2 photons in the formula $m^2=E^2-p^2$? $\endgroup$
    – M. Zeng
    Commented Oct 20, 2017 at 1:39
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There is a classical treatment of the gravitational field of a photon here. I have to confess that the details are beyond me, but basically the gravitational field of a photon looks like a gravity wave. This makes sense, since a photon travels at $c$ and therefore so must it's gravitational potential. It's reasonable that the reult would look like a gravity wave propagating at $c$.

If you're asking about a quantum description then I don't think there is an answer since we have no theory of quantum gravity.

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  • $\begingroup$ If it was a classical treatment, it should be dealing with EM waves, not photons :). (I haven't bothered to read the article. Just wondering) $\endgroup$
    – resgh
    Commented Nov 30, 2012 at 10:45
  • $\begingroup$ Classically a photon can be modelled as a soliton i.e. a localised EM wave. $\endgroup$
    – John Rennie
    Commented Nov 30, 2012 at 10:52
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What do you mean by "uncertainty relation"? I don't seem to get it.

Anyways, photons can directly generate a gravitational field, and should be able to be linked to massive(or any other particle with energy) particles via a graviton in Feynman diagrams. Everything with energy interacts with gravitation. Such a Feynman diagram should be possible. By the way, you probably are aware that there does not currently exist any universally accepted quantum theory of gravitation.

Also, photons are NOT engendered with mass. They interact through gravitation via their ENERGY.

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  • $\begingroup$ "What do you mean by "uncertainty relation"? I don't seem to get it." I knew that was a mistake to write! I meant it very loosely; basically, this thing that everyone keeps repeating, that energy produces a gravity field, is what I'm missing, and what I have to absorb. $\endgroup$
    – Ryan
    Commented Nov 29, 2012 at 16:54
  • $\begingroup$ @Ryan What are you relating from energy to 'uncertainty relation'? $\endgroup$
    – resgh
    Commented Nov 30, 2012 at 0:29
  • $\begingroup$ I'm not -- I was misunderstanding how intimately mass/energy were. Even with that misunderstanding, I wasn't confusing it with Heisenberg's concept. Just using loose language. $\endgroup$
    – Ryan
    Commented Nov 30, 2012 at 4:12

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