# Does a photon exert a gravitational pull?

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?

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.

• 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, 2012 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. Mar 28, 2012 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, 2012 at 13:43
• @John I'd be a bit more careful with the word "singularity" :D Black holes don't necessarily contain singularities (time dilation combined with hawking radiation might imply that before the singularity could ever form, the black hole "evaporates"), and they certainly aren't ones when taken as "everything behind the event horizon". And since the only "stable orbit" would be exactly at the event horizon, this does indeed mean the black hole as a whole, rather than just the singularity (whether it even exists or not). Oct 10, 2014 at 13:02
• @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. Oct 10, 2014 at 14:58

Yes.

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.

• 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. Mar 27, 2012 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, 2012 at 16:07

In general relativity, the source of gravity is the so-called four-momentum vector, which combines momentum and energy. Light has momentum and energy, so it is a source of gravity.

Rest mass is the length of the four-momentum vector. It's an oddity of the geometry of spacetime that a vector can have a length of zero even when its components are nonzero, and that turns out to be the case for a coherent beam of light. But it's the whole vector that is the source of gravity, not just the length.

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.

• 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, 2015 at 17:20
• Could we see your work? Also en.wikipedia.org/?title=Radiation_pressure Jun 15, 2015 at 21:28
• Binding energy does contribute to both inertial and gravitational mass. It is responsible for 99% of the mass of the proton. Jun 16, 2015 at 7:38