Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Not exactly related, but it would benefit by a merge. – Manishearth Mar 27 '12 at 14:09
Related: – centralcharge May 21 '13 at 16:08
up vote 23 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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

As often, hasn't something been overlooked. Energy & mass cannot be interchangeable with regards having the ability to generate gravitational force, because otherwise 'binding energy' would also contribute, obviously it does not. Do we have an impasse, maybe not, if photons actually possess mass after all. I have calculated algebraically that when matter is decomposed directly into photons (which has now been observed) the resulting 'Relativistic Mass' is always a value that is twice that of the original 'Rest Mass' of the matter from which it formed. If !!!!!!! indeed photons do have a negligible barely detectable mass, and !!!! that mass generates 'Gravitational Energy' then it may well be that the mass is in it's purest form (no binding energy) & what's more it isn't just one mass, but many subdivisions of tiny particles all in the lowest possible state of being bound. Maybe even thousands if not millions of almost absolute zero mass particles, all in zero 'State of Entropy'. The next stop for them on the scale of subdivision can only be that of 'Phase Waves'. Photons of course are said to follow the 'Natural Geodesic' of curved space time, but what is overlooked is the possibility that vast numbers of photons generated in sufficient density can in fact exert a gravitational pull. This is still inconclusive, but if all the 'ifs' turn out to be positive in this scenario, we have a revolution in the way we think about matter & energy, especially photons. It should be noted that it is now know photons actually oscillate their speeds around 'slightly faster' & 'slightly slower' than the average speed of light. This is very important in the working of 'Photon Mass, indeed 'Photon Mass' may not work without this discovery!!! NB The 'Photon Mass' very likely exists inside an almost infinitely small volume, so therefore it's gravitational pull will only be detectable at exceedingly short range, well beyond detection of any available technology for considerable time to come. However, no matter how short this range might be even 'Probability Theory' predicts a finite chance that a gravitational force may be exerted at extremely long range, i.e. enough photons could exert a measurable 'Gravitational force' at very long ranges, especially if they're are a high density of them, as in the case of very dense bright stars. NOTE ALSO It is still not know for sure whether or not photons can exert pressure against a physical surface, something that may actually be almost impossible to detect with present technology.

share|cite|improve this answer
I have little idea what you are actually trying to say here, but it seems like you are trying to introduce your own personal theories. The gravitational effect of and on free photons is well understood and described within GR. It is also experimentally verified. There is no problem that indicates something was overlooked. – Jim Jun 15 '15 at 17:20
Could we see your work? Also – Asher Jun 15 '15 at 21:28
Binding energy does contribute to both inertial and gravitational mass. It is responsible for 99% of the mass of the proton. – John Rennie Jun 16 '15 at 7:38

protected by Qmechanic Feb 2 '13 at 17:11

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.