# Do photons exert gravity?

1. Jun 26, 2011

### AntonL

Do photons contribute to the gravity?

Are there any experimental results like two parallel laser beams attracting each other?

2. Jun 26, 2011

### Nabeshin

Within GR, yes.

None that I know of, as the effect would be extremely weak.

3. Jun 26, 2011

### Polyrhythmic

Anything that contributes to the stress-energy tensor, i.e. carries energy, contributes to the gravitational field. Photons carry energy, and therefore the answer is yes.

4. Jun 26, 2011

### cragar

Yes otherwise they could escape a black hole.

5. Jun 26, 2011

### bcrowell

Staff Emeritus
FAQ: Does light produce gravitational fields?

The short answer is yes. General relativity predicts this, and experiments confirm it, albeit in a somewhat more indirect manner than one could have hoped for.

Theory first. GR says that gravitational fields are described by curvature of spacetime, and that this curvature is caused by the stress-energy tensor. The stress-energy tensor is a 4x4 matrix whose 16 entries measure the density of mass-energy, the pressure, the flux of mass-energy, and the shear stress. In any frame of reference, an electromagnetic field has a nonvanishing mass-energy density and pressure, so it is predicted to act as a source of gravitational fields.

There are some common sources of confusion. (1) Light has a vanishing rest mass, so it might seem that it would not create gravitational fields. But the stress-energy tensor has a component that measures mass-energy density, not mass density. (2) One can come up with all kinds of goofy results by taking E=mc^2 and saying that a light wave with energy E should make the same gravitational field as a lump of mass E/c^2. Although this kind of approach sometimes suffices to produce order-of-magnitude estimates, it will not give correct results in general, because the source of gravitational fields in GR is not a scalar mass-energy density, it's the whole stress-energy tensor. However, there is one case of interest where this does happen to work. If a photon gas of total mass E is contained inside a spherical mirror, then the external spacetime is exactly the Schwarzschild solution for a mass E/c^2. The external field has a contribution from the photons that is double this amount, but half of that is canceled by the pressure at the mirror.

Experimentally, there are a couple of different ways that I know of in which light has been tested as a gravitational source. An order of magnitude estimate based on E=mc^2 tells us that the gravitational field made by an electromagnetic field is going to be extremely weak unless the EM field is extremely intense.

One place to look for extremely intense EM fields is inside atomic nuclei. Nuclei get a small but nonnegligible fraction of their rest mass from the static electric fields of the protons. According to GR, the pressure and energy density of these E fields should act as a source of gravitational fields. If it didn't, then nuclei with different atomic numbers and atomic masses would not all create gravitational fields in proportion to their rest masses, and this would cause violations of Newton's third law by gravitational forces. Experiments involving Cavendish balances[Kreuzer 1968] and lunar laser ranging[Bartlett 1986] find no such violations, establishing that static electric fields do act as sources of gravitational fields, and that the strength of these fields is as predicted by GR, to extremely high precision. The interpretation of these experiments as a test of GR is discussed in [Will 1976] and in section 3.7.3 of [Will 2006]; in terms of the PPN formalism, if E fields did not act as gravitational sources as predicted by GR, we would have nonzero values of the PPN zeta parameters, which measure nonconservation of momentum.

Another place to look for extremely intense EM fields is in the early universe. Simple scaling arguments show that as the universe expands, nonrelativistic matter becomes a more and more important source of gravitational fields compared to highly relativistic sources such as the cosmic microwave background. Early enough in time, light should therefore have been the dominant source of gravity. Calculations of nuclear reactions in the early, radiation-dominated universe predict certain abundances of hydrogen, helium, and deuterium. In particular, the relative abundance of helium and deuterium is a sensitive test of the relationships among a, a', and a'', where a is the scale-factor of the universe. The observed abundances confirm these relationships to a precision of about 5 percent.[Steigman 2007]

Kreuzer, Phys. Rev. 169 (1968) 1007

Bartlett and van Buren, Phys. Rev. Lett. 57 (1986) 21

Will, "Active mass in relativistic gravity - Theoretical interpretation of the Kreuzer experiment," Ap. J. 204 (1976) 234, available online at http://articles.adsabs.harvard.edu//full/1976ApJ...204..224W/0000224.000.html

Will, "The Confrontation between General Relativity and Experiment," http://relativity.livingreviews.org/Articles/lrr-2006-3/ [Broken], 2006

Steigman, Ann. Rev. Nucl. Part. Sci. 57 (2007) 463

Last edited by a moderator: May 5, 2017
6. Jun 26, 2011

### bcrowell

Staff Emeritus
I believe that in this special case, the deflection happens to be zero, whereas if the beams are antiparallel the deflection is nonzero. (This is all theory. The actual experiment is impractical.)

7. Jun 26, 2011

### PAllen

I believe there is a simple SR based plausibility argument for this result (not proof, just plausibility argument).

Consider two particles moving parallel at the same relativistic speed. In this frame, they each have large kinetic energy. Does it contribute to any coordinate independent curvature? No, because you can compute all such quantities in the center of momentum frame in which the particles have no kinetic energy.

Consider two particles moving anti-parallel. Now there is net kinetic energy in the center of momentum frame, and it does contribute to coordinate independent curvature.

While there is no center of momentum frame for photons, limiting arguments support the idea that parallel photons do not act gravitationally (none of their kinetic energy contributes to invariant mass, photons have no mass energy except their kinetic energy; if unconstrained, there is no pressure etc.), while other photon configurations do act gravitationally.

8. Jun 26, 2011

### Polyrhythmic

Could you please elaborate? Why would this be the case?

9. Jun 26, 2011

### Polyrhythmic

So that means that parallel moving photons actually create a gravitational field, they just don't feel it themselves?

10. Jun 26, 2011

### ardie

i must say this is a source of great debate. but we all know that Einstein famously proved by a thought experiment, that the source of gravity is Inertial mass and not just any mass which can be equivalent to energy. So if you were to say that photons have mass, then in the energy exansion describing your photon, you cannot ignore the second and higher order terms since the speed is infinately big, so you will find that the photon has infinite mass and excerts infinite gravity. This is why Einstein emphasised that it is Inertial mass that counts towards total gravitational field, which photon doesn't have any of.

11. Jun 26, 2011

### PAllen

No. If you consider the photon as the limit as mass goes to zero while holding momentum constant, parallel, unconstrained, photons would not be expected to be a source of gravity (metric curvature). They would respond to gravity (they follow null geodesics of the background geometry). Note that this argument is a heuristic justification of exact results (for parallel and anti-parallel light beams), and not to be taken too literally. For example, in GR, there is generally no notion of exactly parallel motion.

12. Jun 26, 2011

### bcrowell

Staff Emeritus
It is not a source of great debate, and your interpretation is incorrect.

13. Jun 26, 2011

### bcrowell

Staff Emeritus
We know that the energy of the electric fields inside the nucleus contributes to the nucleus's passive gravitational mass, which measures how much gravitational force it *feels*. If its active gravitational mass, which measures the gravitational forces it *exerts*, were different, then the forces it exerted would be unequal to the forces it felt.

14. Jun 26, 2011

### Polyrhythmic

Alright, I see what you mean. Thanks to the both of you!

15. Jun 26, 2011

### Nabeshin

This isn't true. The thing about photons which makes them unable to escape black holes is that they follow null geodesics. Even if photons did not contribute to the SET, they must still follow SOME geodesics, and the best they can do is null, i.e. they will still be stuck inside the event horizon.

16. Jun 26, 2011

### pervect

Staff Emeritus
If you consider first a pair of moving charges, you'll find that it's convenient to break down the force between them into an electric part and a magnetic part.

If you do the same for the photons, you can conclude that for photons moving in the same direction, parallel to one another, the part of the gravitational force that's analogous to the "electric" part of the force between charges is cancelled out by the gravitational force copponent that's analogous to the "magnetic" part, and there's no net force.

WHen the photons move in opposite directions, the two components add together rather than cancel each other, and so you get twice the expected force.

17. Jun 26, 2011

### cosmik debris

Yes, this is my understanding as well. Essentially the terms in the stress/energy tensor cancel for parallel beams. How this relates to individual photons I'm not sure.

18. Jun 26, 2011

### ardie

i recommend you read "On the influence of gravittation on the propagation of Light by A. Einstein" Annalen der Physik, 35, 1911

19. Jun 28, 2011

### cragar

Interesting, Why do photons need to follow a geodesic.

20. Jun 28, 2011

### PAllen

A null geodesic defines 'traveling at light speed'. All timelike paths (geodesic or not) are slower. A spacelike path is not a possible path for either matter or light (it would represent being in two places at once for some observer).