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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?

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    $\begingroup$ Photons are not energy. Like all particles, photons have energy. $\endgroup$ – Xerxes 2 days ago
  • $\begingroup$ If they are not energy and merely have it, when exactly are they? $\endgroup$ – Derek Seabrooke 2 days ago
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    $\begingroup$ The photon field varies over the four coordinates of spacetime, so I guess they're whenever the t-coordinate says they are. $\endgroup$ – Xerxes yesterday
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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!

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    $\begingroup$ 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. $\endgroup$ – Michael Seifert Sep 14 at 14:38
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    $\begingroup$ Isn't it possible that while the overall energy stays constant it is differently distributed between the two photons when they leave? $\endgroup$ – Peter - Reinstate Monica Sep 14 at 14:49
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    $\begingroup$ @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. $\endgroup$ – G. Smith Sep 14 at 15:21
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    $\begingroup$ 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. $\endgroup$ – Michael Seifert Sep 14 at 17:58
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    $\begingroup$ @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. $\endgroup$ – G. Smith Sep 14 at 20:36
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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. Scattering of two photons 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.

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    $\begingroup$ 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. $\endgroup$ – Derek Seabrooke Sep 14 at 6:57
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    $\begingroup$ Wow, that parallel propagating photons do not interact gravitationally (and hence not at all -- @DerekSeabrooke is right: How could they!) is an interesting corollary. $\endgroup$ – Peter - Reinstate Monica Sep 14 at 14:52
  • $\begingroup$ 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? $\endgroup$ – RBarryYoung Sep 14 at 15:47
  • $\begingroup$ @RBarryYoung I'm sure that is correct. I was assuming two photons "next to each other". $\endgroup$ – Peter - Reinstate Monica Sep 14 at 15:55
  • $\begingroup$ @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. $\endgroup$ – John Dvorak Sep 15 at 5:11
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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!

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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.

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