How do photons affect each other gravitationally?

Question

Photons are energy. According to general relativity they should bend space. Assuming two photons pass one another in a large void of empty space how would they gravitationaly affect each other exactly? Would there be a change in their path, a change in color, both, neither or something entirely different?

Photons are not energy. Like all particles, photons have energy. — Xerxes, Sep 16 '20 at 15:34
If they are not energy and merely have it, when exactly are they? — Derek Seabrooke, Sep 17 '20 at 00:06
The photon field varies over the four coordinates of spacetime, so I guess they're whenever the t-coordinate says they are. — Xerxes, Sep 17 '20 at 13:02

G. Smith · Accepted Answer · 2020-09-14T19:59:18.023

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One can quantize linearized spacetime perturbations in General Relativity and compute the effect of photons scattering elastically by exchanging virtual gravitons. This theory isn’t consistent at Planck-scale photon energies but is believed to be fine at the energies of photons we observe... even very high-energy gamma rays.

All the energy coming in has to come out. In the center-of-momentum frame the two photons each enter with energy $E$ and exit with energy $E$. Thus in this frame there is no change in their frequency (“color”).

Their direction does change (but the effect is tiny). There is a probability of scattering through different angles, and this is described as usual by a differential cross-section $d\sigma/d\Omega$ which depends on the scattering angle $\theta$.

The details of the calculation are in this 1967 paper: Gravitational Scattering of Light by Light.

The differential cross section for unpolarized photons found in this paper — and then corrected in an erratum — is

$$\frac{d\sigma}{d\Omega}=\frac{32G^2E^2}{c^8\sin^4{\theta}}\left(1+\cos^{16}{\frac{\theta}{2}}+\sin^{16}{\frac{\theta}{2}}\right).$$

As you can guess, $G$ is Newton’s gravitational constant and $c$ is the speed of light.

Try computing the area $G^2E^2/c^8$ for a visible photon (or a gamma-ray photon) to see how tiny and unmeasurable this scattering effect is!

edited Sep 14 '20 at 19:59

answered Sep 14 '20 at 06:03

G. Smith

51,534

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I feel like if I got a $\cos^{16} (\theta/2)$ expression in my cross section calculation I would put down the pencil and back away slowly. – Michael Seifert Sep 14 '20 at 14:38
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Isn't it possible that while the overall energy stays constant it is differently distributed between the two photons when they leave? – Peter - Reinstate Monica Sep 14 '20 at 14:49
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@Peter-ReinstateMonica Not in the center-of-momentum frame. They exit like they entered, with equal and opposite momenta, and thus equal energy. All that happens is that the momenta change direction. – G. Smith Sep 14 '20 at 15:21
@G.Smith Ah, unless I miss something that is also immediately impossible otherwise from first principles, if I think about it... interesting. – Peter - Reinstate Monica Sep 14 '20 at 15:54
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That implies thatt in any frame other than the CM frame, the energies of the photons would change when they scatter. It's not unlike how a gravitational slingshot works: the particles just change their momentum vectors in the CM frame, but in another frame that manifests as a shift in energy. – Michael Seifert Sep 14 '20 at 17:58
The photons should experience an ever-so-slight gravitational redshift as they move apart, shouldn't they? Their kinetic energy should not be conserved as they separate, because it turns into gravitational potential energy.
(Of course, I am thinking about this in an unholy mixture of classical and relativistic framing, and I never took GR.)
– Glenn Willen Sep 14 '20 at 19:56
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@GlennWillen When you use quantum field theory to calculate scattering, you are only concerned with the initial and final states. If you want to think classically and argue that there should be a redshift as they separate, then you also have to argue that there should be a blueshift as they approach. They would cancel out. – G. Smith Sep 14 '20 at 20:36
@MichaelSeifert I came to this site from the HNQ. What's $\cos^{16}$? I've tried googling it, but it's all Klingon to me. – Nzall Sep 15 '20 at 21:03
@Nzall That is standard notation for the cosine raised to the 16th power: $(\cos{\frac{\theta}{2}})^{16}$ – G. Smith Sep 15 '20 at 21:07
@G.Smith The cosine is always between 1 and 0, right? And theta is the angle you're taking the cosine of right? Am I interpreting this right in that this angle is the number of degrees that the photon is deviating from the course due to the gravitational interaction of the other photon? – Nzall Sep 15 '20 at 21:28
@Nzall In general, the cosine is between -1 and 1 but in this case it is between 0 and 1. (This is related to the photons being indistinguishable.) You take the cosine of half the deflection angle. – G. Smith Sep 15 '20 at 21:34
Is there no mechanism for graviton mediated change of photon number? – Eric Towers Sep 15 '20 at 22:47
@EricTowers I think there is, but that would be an even-tinier effect. – G. Smith Sep 15 '20 at 23:08
@MichaelSeifert The best" comment I have read this year! Deserves to crack 100. – chux - Reinstate Monica Sep 16 '20 at 00:56
@G.Smith how could the photons blueshift as they approach? Gravity is limited to the speed of light so it cannot go any faster than the photons themselves. It seems it would be impossible for the photons to have any affect on each other prior to their collision so the whole of the affect would have to be after the fact! – Derek Seabrooke Jan 23 '21 at 03:59
@DerekSeabrooke In quantum field theory it is impossible to say what happens “before the collision” and “after the collision”. There isn't really a “collision”. There is an initial state of the fields and a final state of the fields, and no classical evolution of the former into the latter. – G. Smith Jan 23 '21 at 04:04
@G.Smith let's define collision as the crossing of each other's light cones. How could photons affect each other before their light cones overlap? – Derek Seabrooke Jan 23 '21 at 04:07
@DerekSeabrooke “Virtual particles” mediating interactions in QFT (in this case, gravitons) don’t have the same causal constraints that real particles do. I suggest that you try to have a mental picture of interacting quantized EM and gravitational fields filling space, not colliding point-like photons. Photons can seem to be point-like when we measure them (like in the photoelectric effect), but that doesn’t mean that they interact gravitationally in a point-like way. – G. Smith Jan 23 '21 at 04:25

score 13 · Answer 2 · answered Sep 14 '20 at 05:54

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Yes. It is possible to treat linearized gravity like a QFT, write the graviton propagator and study the scattering of two photon thought a graviton exchange. The amplitude, given the momentum of the two photons $k_{1}$ and $k_{2}$, in the limit where the graviton propagator momentum $q \rightarrow 0$ behaves proportionally to $$\sim \frac{(k_{1} \cdot p_{1})(k_{2} \cdot p_{2})}{q^{2}}$$

This means that two photons that are moving in the same direction do not interact gravitationally.

Further references can be found in the QFT book by Zee.

answered Sep 14 '20 at 05:54

Pipe

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It makes sense to me that two photons travelling in parallel would not interact gravitationally. After all gravity itself travels at the speed of the light which means the photon could only affect things it already passed. – Derek Seabrooke Sep 14 '20 at 06:57
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Wow, that parallel propagating photons do not interact gravitationally (and hence not at all -- @DerekSeabrooke is right: How could they!) is an interesting corollary. – Peter - Reinstate Monica Sep 14 '20 at 14:52
But one of the parallel photons could be in the light cone of the other if it were further behind, right? Is there still no effect in that case? – RBarryYoung Sep 14 '20 at 15:47
@RBarryYoung I'm sure that is correct. I was assuming two photons "next to each other". – Peter - Reinstate Monica Sep 14 '20 at 15:55
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@RBarryYoung if the leading photon affected the trailing photon but not vice versa, there would be Newton's third law violation (specifically the trailing photon would be gaining energy indefinitely without any source), and if they both affected each other, that would violate the light speed limit. – John Dvorak Sep 15 '20 at 05:11
@JohnDvorak OK, so ... what happens then? – RBarryYoung Sep 15 '20 at 05:14
@RBarryYoung The conclusion is inevitable: photons traveling in the same direction cannot affect each other gravitationally. – John Dvorak Sep 15 '20 at 05:18
Would two photons travelling anti-parallel also have no interaction? – Derek Seabrooke Jan 31 '21 at 00:05

joseph h · Answer 3 · 2020-09-14T06:13:20.490

Photons have energy and momentum, which means they will have terms in the stress-energy-momentum tensor (as per the Einstein Field equations of general relativity). This means they will cause spacetime curvature. So as they travel and approach each other, the local surrounding space-time will be warped and each photon will follow the path of least curvature, or a geodesic. That is probably all that will happen. Their paths will be deflected or curved. Of course this effect will be extremely small. You will never observe photons deflecting each other's paths - as far as I am aware.

Also, please see the article linked above by G.Smith which studies the "photon-photon interaction through the creation and annihilation of a virtual graviton" which uses the hypothetical graviton to investigate this question further. Very interesting!

score 2 · Answer 4 · answered Sep 16 '20 at 16:21

Photons do have stress-energy, and according to GR, they do create effects of gravity (like bending spacetime).

Given that photons have energy and momentum, it would surprise me if they do not induce curvature.

Do photons bend spacetime or not?

Just like for the static EM field, that we mathematically describe using virtual photons, we can analogously describe the static gravitational field around the photons using virtual gravitons.

To see that this can't happen with just the two photons, note that it's possible to switch frames of reference to a frame called the center of mass frame, in which the total momentum is zero. As they separate, their gravitational attraction may cause them to be red-shifted, but this redshift will approach some finite limit. The photons will not stop and come back together. Therefore, back in F1, the photons will be deflected, but not enough to reunite.

Gravitational interaction of two photons, initially separating

Based on what we know today, the photons' static gravitation fields will cause them to be deflected as they interact, and will cause them to be redshifted, though the effects are infinitesimal.

Describe in the answer from G. Smith, you see the effect of the photons on each other are extremely small at the energy levels we discuss here usually.

Just a not, the point is that your question does have merit, because we know that the effect is there, and if you go to theoretically much higher energy levels, you can see effects like geons.

https://en.wikipedia.org/wiki/Geon_(physics)

These nonsingular EM waves are held together in a confined region of space by their own gravitational attraction, though the energy levels to reach this would be extreme.

How do photons affect each other gravitationally?

4 Answers4