# Do photons obey the 1/r^2 gravity law?

1. Dec 8, 2012

### swle

Do photons obey the 1/r^2 gravity law?

This was a question that came up in our recent PhySoc meeting and no-one present seemed to know the answer (lecturers included!)

Does anyone have a unequivocal answer (and preferably citing a source)?

I have done a quick search through old posts, but I couldn't see anything.

Thanks!

2. Dec 8, 2012

### K^2

Do you mean gravity produced by photons or gravity experienced by photons?

3. Dec 8, 2012

### Barry_G

Gravity produced by photons? Aren't they supposed to be without mass?

Besides word "obey" I think strongly suggests it's about gravity experienced by photons.

4. Dec 8, 2012

### Barry_G

Since photons obeying gravity is about them following geodesics and curvature of space-time rather than obeying Newton's law and force of gravity, if you are looking for some explicit statement about it I think you should better ask whether curvatures of space-time curve according to inverse square law, and if they do then it would follow from that photon trajectories are influenced accordingly.

5. Dec 8, 2012

### HallsofIvy

Staff Emeritus
In order to talk about properties of photons, we are going to have to use general relativity. And in general relativity there is no "1/r^2" law.

6. Dec 8, 2012

### Dickfore

The "1/r2 gravity law" is called Newton's Law of Universal Gravitation. Strictly speaking, it is invalid within GR. Instead, it is supplanted by Einstein's field equations:
$$R_{\mu \nu} - \frac{1}{2} R \, g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}$$
(neglecting the cosmological constant term), where $R_{\mu \nu}$ is the Ricci curvature tensor, $R = R^{\mu}_{\mu} = g^{\mu \nu} R_{\mu \nu}$ is the scalar curvature, $T_{\mu \nu}$ is the stress-energy tensor, c is the speed of light in vacuum, and G is the Universal Gravitational Constant that also enters in Newton's Law.
In the limit of "weak gravitational fields" (when the metric tensor may be approximated as $g_{00} = 1 + 2 \phi/c^2, g_{0 i} = 0, g_{i k} = -\delta_{i k}$, where $\phi$ is the scalar gravitational potential, and this sets the condition what is meant by a weak gravitational field), the 00 component of Einstein's equations reduces to the Poisson equation for the gravitational potential:
$$\nabla^2 \phi = 4\pi G \, \rho$$
where $\rho$ is the mass density.

Because electromagnetic radiation ("photons") contribute to the stress-energy tensor, I would say they contribute to the curvature of spacetime. However, if you take the trace of Einestein's field equations:
$$g^{\mu \nu} R_{\mu \nu} - \frac{1}{2} R g^{\mu \nu} g_{\mu \nu} = \frac{8 \pi G}{c^4} g^{\mu \nu} T_{\mu \nu}$$
$$R = -\frac{8 \pi G}{c^4} \, T^{\mu}_{\mu}$$
and you remember that the electromagnetic stress-energy tensor is traceless, you see that electromagnetic fields do not contribute to the scalar curvature of spacetime!

Additionally, as claimed by previous posters, light follows geodesics (according to the rules of geometric optics) in curved spacetime. Or, if you write down Maxwell's equations, you need to use the covariant derivative:
$$D_{\nu} F^{\mu \nu} = -\mu_0 \, J^{\mu}$$
$$F_{\mu \nu} = D_{\mu} A_{\nu} - D_{\nu} A_{\mu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$
that is defined through the Christoffel symbols, which know about the spacetime curvature.

In any case, propagation of light is affected by the curvature of spacetime. This was experimentally proven by observing the deflection of starlight during a total eclipse of the Sun.

Last edited: Dec 8, 2012
7. Dec 8, 2012

### bobc2

Yes. For example a photon in the gravitational force field of the earth would result in a force of:

f = G(m1)(m2)/r^2

where G is Newton's gravitational constant, m1 is the mass of the earth, and m2 is the mass of the photon

f = G(m1)(0)/r^2 = 0

8. Dec 8, 2012

### pervect

Staff Emeritus
There's a famous result that shows that light deflects twice as much due to gravity than "Newtonian" predictions. This was one of the first tests of General Relativity.

It's not too clear how to interpret the original question so that it's testable by experiment, but if you take it as an expression of "is light deflection calculated to have the samve value in GR as it is in Newtonian theory" the answer is no.

9. Dec 8, 2012

### Bill_K

The famous "Newtonian calculation" that gets half the correct value has no legitimate basis, it's simply wrong. In Newtonian theory there is no light deflection at all, because light obeys Maxwell's Equations and travels in straight lines.

10. Dec 8, 2012

### pervect

Staff Emeritus
The only good derivation I can think of for the "Newtonian" light deflection is in the context of the PPN approximation.

Here you can recover the "Newtonian" results by setting the PPN paramater gamma=0. I believe this corresponds to the predictions of Newton-Cartan theory, a geometrical reformulation of Newtonian gravity. I think you can still characterize Newton-Cartan theory with a 1/r^2 force law.

I'm not sure of the history of all this - but at least at the time, people seem to be convinced of the validity of the test.

I suppose I should mention http://en.wikipedia.org/w/index.php?title=Parameterized_post-Newtonian_formalism&oldid=523184874 for the other readers of this thread who might not all be familiar with the acronym PPN. Anyway, because gamma is the only PPN pamarmeter important to light deflection, I view this as convincing demonstration that "space curvature" is the explanation for the extra deflection of light in GR, as gamma can be interpreted as the "amount of space curvature per unit mass". Occasionally this line of argument of mine causes some argument, I think it's reasonable though.

Last edited: Dec 8, 2012
11. Dec 8, 2012

### bcrowell

Staff Emeritus
The 1/r^2 law is a Newtonian approximation to general relativity. As others have noted, there is no 1/r^2 law in GR.

A photon gas inside a spherical mirror produces the same external gravitational field as an equivalent amount of mass-energy in any other form: https://www.physicsforums.com/showpost.php?p=2956775&postcount=15 If the fields are weak, then the Newtonian approximation is good, and the field is approximately 1/r^2.

FAQ: If light is massless, why is it affected by gravity?

General relativity describes gravity as curvature of spacetime. Any sufficiently small particle (massive or massless) traveling through a curved spacetime moves along a geodesic, which means a "line" that is as straight as possible.

Another thing to realize is that "mass" has a specialized technical meaning in relativity; it means $m=\sqrt{E^2-p^2}$ (in units where c=1). When we say that a photon is massless, that's what we mean. But mass in GR doesn't have all the properties you might think. For example, mass isn't additive, and a box full of photons has a nonzero contribution to its mass coming from the photons, even though the photons individually have zero mass.

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 6, 2017
12. Dec 8, 2012

### K^2

Gravity isn't generated by mass. It's generated by stress-energy tensor, which is certainly non-zero for photons, which have both energy and momentum. Photons do generate gravity in GR.
Strictly speaking? Sure. But it does hold loosely within reasonable limits, and will be followed by light as well as matter. If you pretend that photons are particles traveling at speed c having a mass p/c, and you are looking at "acceleration" due to gravity in perpendicular direction, you'll only be a factor of 2 off. So 1/r² dependence can still be said to hold. It's an entirely valid question to ask.

Of course, the answer comes in with many caveats. But you are not going to be able to explain them all to a person who asks such a question.

13. Dec 8, 2012

### Barry_G

Two beams of light are passing next to some massive planet or a star, where one beam is at double the distance away than the other. Will trajectory of the closer beam be four times as influenced compared to further away beam?

What do you mean by "weak", anything less than black hole? But isn't that strength proportional to square of the distance? How do you measure the strength of gravity field?

Perhaps because photons actually do have mass?

Mass is supposed to be additive, and you just said yourself photons actually do contribute to the mass of the box.

I'm not sure what you just said. Are atomic masses exact sum of their protons, neutrons and electrons masses or not?

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14. Dec 8, 2012

### Barry_G

I'd say gravity is a property proportional to what we call mass, rather than "generated" by mass. Anyway, your statement sounds as if GR invalidates all the rest of the physics. You surely meant to say that specifically and only in theory of General Relativity gravity is not 'generated' by mass?

How do you know this is not just semantics? How do you know what you call stress-energy tensor is not just another name for mass? How do you measure stress-energy tensor, and how does it relate to measurements of mass?

Isn't gravity just another name for mass? I mean when we measure mass are we not actually measuring the force of gravity? How do you separate one from another? Is there any reason to conclude mass and gravity are not one and the same, one property rather than two separate ones?

So this perpendicular acceleration of photons toward some mass they pass next to, is two times less or two times more compared to would be for, say, electrons?

Last edited: Dec 8, 2012
15. Dec 8, 2012

### K^2

Short answer is that mass doesn't transform correctly under change of frame of reference. The only quantity that does transform correctly, and happens to be the conserved quantity under the transformations, is the stress-energy tensor.

16. Dec 8, 2012

### Barry_G

But that doesn't mean stress-energy tensor does not refer to the same property as what we call mass in the rest of the physics. It seems to me they are the same, where GR equations just embed some "corrections" to the classical concept of mass, just like SR equations do, but they all refer to the same thing, the same property. Wouldn't you agree? In any case, how do we measure this stress-energy tensor? How do we even get to compare the two?

17. Dec 8, 2012

### WannabeNewton

How can they be the same? The SET is a 2 - tensor field and mass density is a scalar field. The point is that there are factors other than just mass that contribute to curvature in general. If you consider the newtonian limit then the 00 component will get the usual, familiar result involving the scalar potential and mass density.

18. Dec 8, 2012

### K^2

GR equations aren't corrections. They are a total overhaul. While classical approximation exists for many phenomena, such as planetary orbits and to an extent even frame dragging, via gravitomagnetic interactions, there are also situations where classical gravity fails absolutely. Gravitational field of a photon is one such example. Photon distorts space-time despite having zero mass.

19. Dec 8, 2012

### Barry_G

Their description is not the same, but both stress–energy tensor and mass refer to the same physical phenomena.

http://en.wikipedia.org/wiki/Stress-energy_tensor : The stress-energy tensor is the source of the gravitational field in the Einstein field equations of general relativity, just as mass density is the source of such a field in Newtonian gravity.

Like what for example? You can not say "other than mass" if you are referring to GR because there is no concept of mass in that theory to start with, and instead there is stress-energy tensor.

20. Dec 8, 2012

### Barry_G

You can not say "despite having zero mass" because in GR there is no concept of mass to start with. It's like saying in USA there are no lifts because there are elevators. And when you say photons have a property that you quantify with the description called "stress–energy tensor", translated to classical physics you just said photons have mass.

http://en.wikipedia.org/wiki/Stress-energy_tensor : The stress-energy tensor is the source of the gravitational field in the Einstein field equations of general relativity, just as mass density is the source of such a field in Newtonian gravity.

Can we measure this "stress-energy_tensor", how do we measure it?

Can you give me some practical example relating to some actual measurements?

Why pretend? A photon is an elementary particle, is it not? Is it any less real or any less of a particle than electron for example? We can emit individual single photons in double-slit experiments, and whenever we actually get to measure them we measure discrete energy quanta, just like we do when we measure electrons, right? Where by "measure" I mean when they impact a sensor at specific location with specific energy.

Now, as you said, when a photon passes next to some mass it will experience acceleration towards that mass, but this acceleration will be off by the factor of two. What experiment are you referring to, can you point me to some paper or internet article where I can see some actual numbers regarding this perpendicular acceleration of photons?