Do Photons Have a Gravitational Effect?

[Tantalos]

According to GR energy creates gravity. Photons have no mass but have energy, so do they create gravity?

 

[Dale]

According to GR energy creates gravity.

No. According to GR the source of gravity is the stress-energy tensor. There are 10 independent components in the stress-energy tensor. Energy is only one of those 10 components.

http://en.wikipedia.org/wiki/Stress-energy_tensor

Photons have no mass but have energy, so do they create gravity?

We do not have a working theory of quantum gravity at the time so I cannot answer your question wrt photons, however I can answer it wrt classical pulses of light. Pulses of light have energy, they also have momentum, so several of the components of stress energy tensor will be non-zero. So light can be a source of gravity.

 

[Mentz114]

The electromagnetic field has a stress energy tensor so one would expect it gravitate. There are solutions to the EFE where electric fields contribute to the Einstein tensor, and presumably to the Riemann tensor. I don't know how physically justifiable these solutions are.

http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor

 

[bcrowell]

DaleSpam is correct. For a description of a laboratory test confirming that static electromagnetic fields produce gravitational effects, see http://www.lightandmatter.com/html_books/genrel/ch08/ch08.html , subsection 8.1.2, under "Experimental tests." When it comes to electromagnetic radiation rather than static fields, I don't know of any really direct empirical tests. However, cosmological models are sensitive to this effect, because the early universe was radiation-dominated. I believe that observations of the CMB are good enough these days that if EM waves didn't gravitate, discrepancies would have shown up by now. This is actually kind of an interesting question, so maybe I'll post in the cosmology forum and see if anyone knows.

 

[Mentz114]

Ben, I hope you don't think I was denying that EM gravitates. My last sentence only refers to a certain class of solutions and probably goes beyond what the OP asked.

 

[bcrowell]

Ben, I hope you don't think I was denying that EM gravitates. My last sentence only refers to a certain class of solutions and probably goes beyond what the OP asked.

No no, when I said "DaleSpam is correct," I didn't mean to imply that you were incorrect!

 

[Mentz114]

As it happens, my reply is irrelevant because I was talking about charge and static fields, not EM radiation. Whoops.

Solutions of the EFE for source free fields are called 'Electrovacuum' solutions. There's an excelent article here

http://en.wikipedia.org/wiki/Electrovacuum_solution

 

[threadmark]

No. According to GR the source of gravity is the stress-energy tensor. There are 10 independent components in the stress-energy tensor. Energy is only one of those 10 components.

http://en.wikipedia.org/wiki/Stress-energy_tensor

We do not have a working theory of quantum gravity at the time so I cannot answer your question wrt photons, however I can answer it wrt classical pulses of light. Pulses of light have energy, they also have momentum, so several of the components of stress energy tensor will be non-zero. So light can be a source of gravity.

Photons cannot be a source of gravity. Einstein’s field equations are dependant on mass and energy. Photons do not have mass so they cannot have gravity. TM

 

[bcrowell]

Photons cannot be a source of gravity. Einstein’s field equations are dependant on mass and energy. Photons do not have mass so they cannot have gravity. TM

No, this is incorrect. What DaleSpam wrote was correct.

 

[Rlam90]

I've got a question to further clarify this concept. Would the gravitational force produced by these photons be proportional to their energy divided by the speed of light, as in e=mc^2? If not, what would the algebraic solution to their gravitational field be?

 

[bcrowell]

I've got a question to further clarify this concept. Would the gravitational force produced by these photons be proportional to their energy divided by the speed of light, as in e=mc^2? If not, what would the algebraic solution to their gravitational field be?

If you could confine a very intense swarm of light rays of total energy E inside a spherical chamber, then GR predicts that you would get the same external field as you would get from a material sphere of mass E/c². (This follows from Birkhoff's theorem, which says that all spherically symmetric gravitational fields have the same form.) But in the general case without spherical symmetry, there is no simple algebraic solution; you have to solve the Einstein field equations. In particular, you can't find the gravitational field of a pencil beam of light just by simple application of E=mc².

 

[Rlam90]

If you could confine a very intense swarm of light rays of total energy E inside a spherical chamber, then GR predicts that you would get the same external field as you would get from a material sphere of mass E/c². (This follows from Birkhoff's theorem, which says that all spherically symmetric gravitational fields have the same form.) But in the general case without spherical symmetry, there is no simple algebraic solution; you have to solve the Einstein field equations. In particular, you can't find the gravitational field of a pencil beam of light just by simple application of E=mc².

I don't quite understand what you mean here, as I don't know any of the official theories themselves. Why can one not take that spherical volume withing the beam of protons, gather the amount of energy within the sphere, and calculate the equation using this? Why can one not take a spherical volume and calculate the amount of energy contained within and this energy's gravitational effects?

 

[jtbell]

Because gravitation depends not only on energy, but also on momentum, via the stress-energy tensor that was mentioned previously in this thread. A beam of light has non-zero net momentum in the direction the beam is traveling, and that comes into play in the Einstein field equations of GR.

 

[haushofer]

You can check Zee's "QFT in a nutshell" about this.

 

[pervect]

If you could confine a very intense swarm of light rays of total energy E inside a spherical chamber, then GR predicts that you would get the same external field as you would get from a material sphere of mass E/c². (This follows from Birkhoff's theorem, which says that all spherically symmetric gravitational fields have the same form.) But in the general case without spherical symmetry, there is no simple algebraic solution; you have to solve the Einstein field equations. In particular, you can't find the gravitational field of a pencil beam of light just by simple application of E=mc².

It's a little more complex than that. Let me outline an experiment in more detail:

If you imagine you had a hollow sphere, symmetry would mean that a probe just inside the surface of the sphere wouldn't be affected by the sphere.

If you fill the hollow sphere up with light, you would find the light was twice as good as creating a gravitational field than cold matter, due to the pressure terms, as measured by such a probe just inside the surface of the sphere.

You can think of the interior as a "photon gas", so the pressure is any direction is 1/3 the energy density. The Komar mass formula boils down to integrating rho + 3P for a small sphere, and since 3P = rho for a photon gas, you'd have twice the Komar mass and twice the gravity.

This translates into a measured acceleration of G (2E/c^2) / r^2.

But if you measure the gravity outside the sphere, the tension in the shell will essentially lower the Komar mass of the shell, and you'll only get an increment of E/c^2 from the original gravitational field you had outside the sphere.

For simplicity I'm assuming the shell doesn't expand when you fill it up with the photon gas. This is unrealistic, but it saves you from having to account for the work done by expanding the shell.

 

[Dale]

Would the gravitational force produced by these photons be proportional to their energy divided by the speed of light, as in e=mc^2? If not, what would the algebraic solution to their gravitational field be?

Again, we are talking not about photons but about classical EM radiation. The spacetimes produced by the presence of EM radiation are called pp-wave spacetimes, and there are several different solutions corresponding to different configurations of radiation:
http://en.wikipedia.org/wiki/Pp-wave_spacetime

 

[Dale]

But if you measure the gravity outside the sphere, ... you'll only get an increment of E/c^2 from the original gravitational field you had outside the sphere.

bcrowell specifically mentioned Birkhoff's theorem which is for the exterior solution, so I am sure that is what he meant.

 

[bcrowell]

Thanks, pervect, for your #15. That's helped me to understand this better. I think we're agreeing on the exterior results, but the issue of matching the boundary conditions properly had been bothering me, and I think your explanation helps clear that up.

While I was in the car this morning, I worked out the following somewhat more detailed and rigorous, but still nonmathematical, argument about the exterior field of the spherical box of photons.

Since the box has spherical symmetry, all you can measure about it, by any external observations, is some kind of scalar mass. In principle this could actually be three different scalars: the inertial mass, active gravitational mass, and passive gravitational mass. The claim that its gravitational fields can be parametrized by a scalar mass is consistent with Birkhoff's theorem. So basically I want to prove that E=m_i=m_p=m_a (in units where c=1).

E=m_i follows from geodesic motion and conservation of energy-momentum, when you put the box in a flat spacetime. If they were unequal, then you could have mechanisms inside the box that would convert the energy back and forth between, say, light and mechanical energy. If this changed its inertial mass, then it would either have to accelerate (violating the principle that small test bodies move along geodesics) or not accelerate (violating local conservation of energy-momentum). The geodesic motion principle does rely on the assumption of an energy condition, but I think the box satisfies every reasonable energy condition.

m_i=m_p follows from geodesic motion in a curved spacetime.

m_p=m_a follows from local conservation of energy-momentum, because otherwise Newton's third law is violated.

So this establishes that the box's active gravitational mass is equal to the energy of the photons, and the only necessary assumptions seem to have been that the box was small and spherically symmetric and that it obeyed an energy condition. In fact, I didn't even have to use the assumption that it contained light waves. It could contain Higg particles or pixie dust, and the argument would be exactly the same.

But what was bugging me about all this was precisely the kind of boundary-condition matching that pervect discussed. However, I'd still be interested in understanding this in a little more detail. I hadn't thought before about the possibility that matter under tension would count as negative pressure in the stress energy tensor. I assume that this doesn't violate any energy conditions...or maybe the energy conditions imply restrictions on how much tension matter can sustain?

 

[pervect]

The strong energy condition states T_00 + (1/2) T >=0 (by applying the more general expression $T_{ab} u^{a} u^{b} + (1/2)T > 0$) and essentially puts a limit on how much tension matter can sustain.

If we have a Minkowski space-time with the sign convention that g_00 = -1 we can write:

T_00 = rho, T = -rho + Px + Py + Pz

where rho is the energy density, and Px, Py, and Pz are the pressures (which would be negative numbers for tension). The strong energy condition then becomes

(1/2) (rho + Px + Py + Pz) > 0

So it's the strong energy condition that forbids us from having a shell that's so strong that its tension would make the gravity weaker outside the shell than inside the shell, if the matter satisfies the strong energy condition it's contribution (rho + Px + Py + Pz) to the Komar mass integral must always be greater or equal than zero.

The Komar mass integral as usually stated (Wald, pg 289, 11.2.10 is)

$$M = 2 \int \left( T_{ab} - \frac{1}{2} T g_{ab} \right) n^{a} \xi^{b} dV$$

$n^a$ being a "unit future" of the volume, $\xi^b$ being the time-like killing vector. When the two are aligned, and you have a Minkowski space-time with g_00 = -1 the mass integral basically reduces to one of (rho + Px + Py + Pz)

 

[bcrowell]

Thanks, pervect, for #19 -- very informative!

 

[Rlam90]

It's a little more complex than that. Let me outline an experiment in more detail:

If you imagine you had a hollow sphere, symmetry would mean that a probe just inside the surface of the sphere wouldn't be affected by the sphere.

If you fill the hollow sphere up with light, you would find the light was twice as good as creating a gravitational field than cold matter, due to the pressure terms, as measured by such a probe just inside the surface of the sphere.

You can think of the interior as a "photon gas", so the pressure is any direction is 1/3 the energy density. The Komar mass formula boils down to integrating rho + 3P for a small sphere, and since 3P = rho for a photon gas, you'd have twice the Komar mass and twice the gravity.

This translates into a measured acceleration of G (2E/c^2) / r^2.

But if you measure the gravity outside the sphere, the tension in the shell will essentially lower the Komar mass of the shell, and you'll only get an increment of E/c^2 from the original gravitational field you had outside the sphere.

For simplicity I'm assuming the shell doesn't expand when you fill it up with the photon gas. This is unrealistic, but it saves you from having to account for the work done by expanding the shell.

Hah, I never thought about light forming pressure. I think I might learn a little bit more by interacting with this community than I ever could have alone.

I have a question on this, though, to help me better understand something. This "pressure" is caused by the random trajectories of these confined protons causing them to be more inclined to be located near the center of the sphere, correct? Do any forces come into play here? There must be some electromagnetic interaction between these photons? How about gravitational interaction between them? The gravitational interaction must be minute, but it's technically still there, correct?

 

[threadmark]

I thought this forum was to discus actual real science, not hypothetical nonsense. There is no evidence to suggest that photons create gravity. Not to mention that photons are considered particle and wave. This is comic book fantasy nonsense because it is nothing but ink on paper. Photons have no mass no charge no gravity. If your nonsense theories were somewhat correct what is stopping light from traveling in a circle. If they create gravity they are gravity affected. Although gravity does bend light it is not the photon itself affected it is the fabric of space time that perturbs the path that photons travel through.

 

[bcrowell]

I thought this forum was to discus actual real science, not hypothetical nonsense. There is no evidence to suggest that photons create gravity.

Step 1: Read this review article: http://relativity.livingreviews.org/Articles/lrr-2006-3/
Step 2: Lose the attitude.

 

[atyy]

Additional technical details here:
http://arxiv.org/abs/gr-qc/0510041
http://arxiv.org/abs/gr-qc/0505040

 

[Passionflower]

Step 1: Read this review article: http://relativity.livingreviews.org/Articles/lrr-2006-3/

Perhaps you could be a little more specific. Where in this article is there a reference to experimental evidence that photons create gravity (leaving for the moment in the middle what 'creating gravity' is actually supposed to mean)?

 

[bcrowell]

Perhaps you could be a little more specific. Where in this article is there a reference to experimental evidence that photons create gravity (leaving for the moment in the middle what 'creating gravity' is actually supposed to mean)?

Section 3.7.3. But you're going to need to understand the whole PPN discussion in the article before you'll understand why that's what 3.7.3 means.

 

[Passionflower]

Section 3.7.3. But you're going to need to understand the whole PPN discussion in the article before you'll understand why that's what 3.7.3 means.

Yes I know I am just a simple person, certainly not a smart as you.

Now I did a search and I did not even find the word photon in that chapter.

So perhaps we should end the conversation by concluding that the experimental evidence is clearly there but that it can only be understood by very intelligent people such as you?

 

[bcrowell]

Now I did a search and I did not even find the word photon in that chapter.

So perhaps we should end the conversation by concluding that the experimental evidence is clearly there but that it can only be understood by very intelligent people such as you?

Searching for a key word is not the same as reading something and making an effort to understand it. There is no reason for the word "photon" to appear there, because the article is about classical physics, not quantum mechanical physics.

 

[PeterDonis]

...general relativity has already done it for you. Space time is the reason photons are affected by gravity. Photons do not bend space time like mass.

General relativity says they do. Look up "radiation dominated universe". This is a model in general relativity where spacetime is bent by nothing but photons--nothing else is present.

 

[pervect]

See for instance the original paper by Tolman, Ehrenfest, Podolsky (written in 1931, it may be hard to get a hold of, but you can get a lot from the abstract)

http://prola.aps.org/abstract/PR/v37/i5/p602_1

or the more accessible

http://books.google.com/books?id=GE...=onepage&q=Tolman-Ehrenfest-Podolsky.&f=false

Einstein said that light is deflected by a massive object - but is light gravitationally deflected by light? Tolman, Ehrenfest and Podolsky discovered that in the weak field limit, two light beams moving in the same direction do not interact gravitationally, but two light beams moving in the opposite direction do.

TEP analyzed pencils and pulses of light, not "photons" as GR is not a quantum theory.

The above reference shows the same result from perturbative quantum gravity point of view, which however is outside the forum (and incidentally my personal knowledge). Followups up on the perturbative quantum gravity explanation would have the best chance of finding someone knowledgeable in the "beyond the standard model" forum.

 

[bcrowell]

Here is a new FAQ I've written up on this topic.

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 about this. (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.

Experimentally, there are a couple of different ways that I know of in which this has been tested. An order of magnitude estimate based on E=mc^2 tells us that the gravitational fields 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 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, "The Confrontation between General Relativity and Experiment," http://relativity.livingreviews.org/Articles/lrr-2006-3/ , 2006

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

 

[atyy]

@bcrowell, do you think it would be very cheating to use this as a "proof": if we assume G=T, with T being the electromagnetic stress tensor, then we get Maxwell's equations and the Lorentz force law automatically by the covariant conservation of T implied by G (along the lines of section 20.6 in MTW)?

 

[bcrowell]

@bcrowell, do you think it would be very cheating to use this as a "proof": if we assume G=T, with T being the electromagnetic stress tensor, then we get Maxwell's equations and the Lorentz force law automatically by the covariant conservation of T implied by G (along the lines of section 20.6 in MTW)?

Well, I might say it in a somewhat different way. If you believe in $G\propto T$, then you get local conservation of energy-momentum. Therefore by conservation of momentum, if light is acted on by gravitational fields, it must also create gravitational fields. The argument doesn't depend on the specific properties of electromagnetic waves at all. You can substitute any other field in place of the word "light," and the answer is the same.

IMO the theoretical side of all this is much more straightforward than the experimental side. The experimental side is complicated and requires a lot more effort for interpretation. E.g., Kreuzer didn't interpret his results as a test of GR; Will did that years later.

 

[threadmark]

A photon has zero mass, zero charge, and zero gravity. Let me explain why.
A photon being single she likes to walk on the beach alone. A Photon vector is not affected if two photons cross paths. Because a photon is a boson, it likes to be left alone. So much that it can occupy the same point in space if a physical possibility that two photons could be confine in a symmetric vector. Meaning if a photon could be seen in faze with another photon, it would not affect the two photons. Photons do not have a charge so the electric field in light must be caused by time varying magnetic fields which photons produce. But the electric properties of light are caused in my opinion by the W and Z bosons. However I cannot produce a formula you would understand, I think it’s correct but not relevant to prove a photon can not produce gravity. The magnetic field produced by a photon is a property of the wave function and of course we all know gravity is the affect on space (TIME) being time is a key factor hear. A Photon has a rest mass of zero if E=mc2. it can be seen as a particle in this state being at rest, so the idea that it can produce a gravitation affect in this state is humorous at best. The particle at rest does not produce a magnetic field, thus does not have time varying magnetic fields to produce electric fields. And if you check your GR because I am sure we didn’t read the same notes, I’m sure there is a notation that energy is required to produce gravity. Photons at rest do not have energy because they have no momentum. To further my disappointment in myself for actually trying to help this filtered opinion based discussion, photons as a wave is not a body that can be compressed and mass density is apart of many calculations working with the production of gravity.
Sorry if this doesn’t make sense to you. But its like I am trying to prove the moons not made of cheese.

TM

 

[bcrowell]

A photon being single she likes to walk on the beach alone.

This is not a scientific argument.

A Photon has a rest mass of zero if E=mc2. it can be seen as a particle in this state being at rest

There is no rest frame in which a light wave is a rest.

To further my disappointment in myself for actually trying to help this filtered opinion based discussion,

In #31, I provided four references to peer-reviewed scientific papers. Have you read any of them?