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I know a photon has zero rest mass, but it does have plenty of energy. Since energy and mass are equivalent does this mean that a photon (or more practically, a light beam) exerts a gravitational pull on other objects? If so, does it depend on the frequency of the photon?

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Related: physics.stackexchange.com/q/18900/2451, physics.stackexchange.com/q/10612/2451 and links therein. –  Qmechanic Mar 27 '12 at 13:56
Not exactly related, but it would benefit by a merge. –  Manishearth Mar 27 '12 at 14:09
Related: physics.stackexchange.com/q/65361 –  Dimensio1n0 May 21 '13 at 16:08

2 Answers 2

up vote 19 down vote accepted

Yes, in fact one of the comments made to a question mentions this.

If you stick to Newtonian gravity it's not obvious how a photon acts as a source of gravity, but then photons are inherently relativistic so it's not surprising a non-relativistic approximation doesn't describe them well. If you use General Relativity instead you'll find that photons make a contribution to the stress energy tensor, and therefore to the curvature of space.

See the Wikipedia article on EM Stress Energy Tensor for info on the photon contribution to the stress energy tensor, though I don't think that's a terribly well written article.

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Yes I saw that question! That's what made me ask this question lol. After looking at the links what I got out of it too was that the higher the frequency of the photon the more relativistic mass it has and the stronger the gravitational pull it exerts. Does this mean that very technically, light doesn't only orbit objects but that they both orbit a center of mass like the earth and the moon? –  John Mar 27 '12 at 18:25
Except in special circumstances (a black hole) light doesn't orbit objects at all, though it is deflected by gravitational fields. However light will be contributing to the spacetime curvature of the bodies it passes, though it's contribution is likely to be insignificant. –  John Rennie Mar 28 '12 at 6:17
Yeah, I'm aware that light can't really orbit because it's velocity is higher than the escape velocity of anything but a singularity, but my point was that it contributes to determining the position of the center of gravity, right? (Even if it is practially negligable) –  John Mar 28 '12 at 13:43
Yes, sort of. Remember you can't describe photons without using GR, and the conventional centre of mass isn't a useful concept in GR. However the photon does deflect the planet, so I suppose it does shift the centre of mass. –  John Rennie Mar 28 '12 at 14:01
@Luaan: I'm reluctant to hedge every answer on GR with the qualification but quantum mechanics might change this. Also note that the stable photon orbit is at $1.5r_s$ not at the event horizon. –  John Rennie Oct 10 '14 at 14:58


You can show via conservation of energy arguments that photons confined within a volume (for the sake of argument, the inside of a sealed box with totally reflective surfaces) must produce the same gravitational effect as an amount of matter in the same volume which would have a mass equivalent to the energy of the photons.

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Well, almost right except that the photons also carry a nonzero pressure, the spatial components $T_{ii}$, which also influence the shape of the metric tensor i.e. the gravitational field. In general relativity, not only the total mass or mass density but also the rest of the stress-energy tensor (density of momentum, flux of momentum etc.) affects the resulting gravitational field. Whenever a component of the stress-energy tensor changes, the gravitational field changes as well. That's what Einstein's equations clearly say. At infinity away from the photons, only the total mass matters. –  Luboš Motl Mar 27 '12 at 14:56
Wow, I never realized that, thanks Lubos. I always heard of the stress energy tensor, but I didn't realize it involved more than just the mass. That helps me visualize how you can have things like frame dragging. –  John May 7 '12 at 16:07

protected by Qmechanic Feb 2 '13 at 17:11

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