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

by swle
Tags: 1 or r2, gravity, obey, photons
PF Gold
P: 5,027
 Quote by Barry_G Has such experiment been performed?
No, and it's not likely feasible, in practice. However, there is no doubt about what GR predicts for this. So called 'box of light' examples are used in classic GR exercises and papers.

I think there are astrophysical observations that provide evidence that radiating EM radiation cause as body to lose gravitational mass. That's probably the closest you can come to observational support.
Physics
PF Gold
P: 6,035
 Quote by Barry_G Thank you. I'd like to talk about that in more detail so I'll start a new thread in Classical Physics forum.
Yes, that's the right place to post questions about general properties of EM fields.
Mentor
P: 16,947
 Quote by Barry_G I don't see any numbers there, not even mention of photon, except for "Kinnersley–Walker photon rocket". If you know someone has calculated or measured this photon gravity filed, then please just tell me the number.
The reason I provided you the information about null dust and pp-waves is that those are the names you will find in the GR literature for the appropriate spacetimes that you are interested in. They won't use the word "photon" because the spacetimes are general for all massless radiation, not just EM radiation. Also, the authors in the literature generally know better than to mix quantum terminology with classical theories.

As far as the numbers go, unfortunately your question is too vague to answer with a concrete number. What photon energy are you considering, what is the geometry you are interested in, what measurement technique to you plan on using to generate the number of interest, etc.? Until you completely specify the problem then all that can be done is to provide you a link to the relevant concepts and solutions.
P: 68
 Quote by PeterDonis No, it will be twice as "influenced". The angular deflection of a light beam passing close to a massive object is given by: $$\delta \phi = \frac{4 G M}{c^2 b}$$ where b is the distance of closest approach. Since this is a function of 1/b, not 1/b^2, the bending is only doubled if b is halved. See here: http://en.wikipedia.org/wiki/Kepler_...ght_by_gravity Note that this formula is only valid for b very small compared to GM / c^2.
That's great. I think that answers the question, for me at least.

Thank you everyone, I'll leave it here and move to classical physics forum to discuss photon electromagnetic field components and Maxwell's Equations.
Emeritus
P: 7,596
 Quote by Barry_G Has such experiment been performed?
Not directly. Carlip argues in Kinetic energy and the equivalance principle that we do have good experimental reason to believe that electromagnetic binding energy contributes to gravity, and that this implies that electromangetic fields must gravitate. Classically (and GR is a classical theory), light is made up out of electromagnetic fields, so we know it's made up of something that's been observed to contribute to gravity (albeit indirectly).

Carlip also briefly discusses the "box of light". Carlip shows that in weak field gravity, the total system (box + light) must gravitate according to the total energy. There is a similar result for strong fields , but it requires that the metric be stationary (i.e. not a function of time). The argument is different in detail from Carlip's. While I'm not aware of any paper that specifically does the strong field calculation for a box of light, the calculations aren't hard to perform.

The non-technical summary of the strong field argument is that in some sense the interior of the box, the light, does "weigh" twice as much, but that the stress in the box walls compensates for this giving a negative contribution to the weight, due to the tension in the container walls.

As an aside, recall that tension and pressure are part of the stress-energy tensor - so here we see an example of stresses contributing to gravity.

The more technically accurate way removes the words "in some sense" by saying that it is the Komar mass of the interior of the box that doubles for the "box of light". This makes the argument more precise, at the cost of introducing a new term that seems to scare people away from understanding the point to be made. On the other hand, some "scariness" is perhaps warranted, at least if the fear induces some caution, for reasons which will be explained below.

As previously mentioned, even though the contents of the box weight twice as much, the stresses in the walls subtract from this "extra" mass, and you recover the value E/c^2 for the mass of contents + walls.

It's worth mentioning at this point, at the risk of confusion, that there are several definitions of "mass" in general relativity, and NONE of them is completely general (including the Komar mass). ALL of them require certain preconditions to be applied. Understanding the conditions where they are applicable may take some work, this is where the "scariness" factor comes in.

The "big three" sorts of mass in GR are Komar mass, ADM mass, and Bondi mass - you'll see a brief discussion of them in the wiki at http://en.wikipedia.org/w/index.php?...ldid=514908524
 P: 68 Perhaps I should be more specific. I think what PeterDonis said answers the question as "yes, photons obey 1/r^2 gravity law". Because, I think what that equation describes is analogous to gravity potential which is 1/r, so that if we somehow worked out the force or acceleration we would get 1/r^2 relation. In other words, if instead of two beams of light there were two beams of electrons or two streams of dust, where one is passing at double the distance from the planet than the other, then further away beam of electrons or stream of dust would too be "influenced" two times more less than the closer one, just like with two beams of light, but the force or acceleration between the planet and each electron or dust particle would be function of 1/r^2.
Physics
PF Gold
P: 6,035
 Quote by Barry_G I think what that equation describes is analogous to gravity potential which is 1/r
The equation I gave describes the angular deflection of a light beam passing close to a massive object. I don't see any obvious analogy between that and gravitational potential.

 Quote by Barry_G if we somehow worked out the force or acceleration we would get 1/r^2 relation.
The problem with that approach is that viewing gravity as a "force" that causes "acceleration" is an approximation that only works when all the objects involved are moving very slowly compared to the speed of light. Obviously that's not the case for a light beam passing close to a massive object.

A better way to phrase the question asked in the OP would be: does light respond to gravity? Or, is the path of a light ray affected by gravity? The answer to that is clearly "yes". But trying to salvage an interpretation of the bending of light as responding to a 1/r^2 force law may not work, because that force law is a non-relativistic approximation only, and light is relativistic.

 Quote by Barry_G In other words, if instead of two beams of light it were two beams of electrons or two streams of dust, where one is passing at double the distance from the planet than the other, then further away beam of electrons or stream of dust would too be "influenced" two times more than the closer one, just like with two beams of light
I'm not sure this is true; the formula I gave is not exact, it's an approximation for the ultrarelativistic case where the object passing by is moving at or very close to the speed of light. The motion of a slower-moving object is more complicated.

 Quote by Barry_G but the force or acceleration between the planet and each electron or dust particle would be function of 1/r^2.
If the electrons or streams of dust are moving slowly enough, a "force" interpretation would work.
 Emeritus Sci Advisor P: 7,596 Carlip's paper http://arxiv.org/abs/gr-qc/9909014 has an expression for the defliction of a particle moving at velocity v, and a caution about the "force" interpretation. The defleciton angle is: $$\theta = \frac{2GM}{bv^2}\left(1+\frac{v^2}{c^2}\right)$$ where b is the impact parameter (it can be thought of as the distance of closests approach IIRC). and G,M, and v are the usual. The caution is "Not to ignore the curvature of space" when calculating light deflections. This spatial curvature produces effects in GR that can not really be well described as a force - though thinking of it as a force proportional to velocity^2 comes at least very close to working. (I don't think the resulting "force" transforms properly even so, you wind up with coordinate dependencies this way.) Hopefully it's obvious why dependence on coordinates is bad in this context, and if it's not obvious, I'm afraid I don't have the heart at the moment for another long discussion of why it is bad.
P: 68
 Quote by PeterDonis The equation I gave describes the angular deflection of a light beam passing close to a massive object. I don't see any obvious analogy between that and gravitational potential.
I find analogy between the two in the word "influenced", which I interpret as 'feeling gravity potential' (at some distance from the planet). If there is some object in space that emits electrons that would travel near the speed of light then we could measure deflection of that electron beam similarly how we do it for light, I guess, and then we could compare it with that of light and I think we would get similar result, not in regards to the amount of deflection, but in regards to that the electron beam twice as far from the planet than the other electron beam would be two times less "influenced", just as it the case with two beams of light.

Which leads me to another question. If we were to measure "influence" or deflection of two beams of particles passing near some planet at the same distance away from it and with the same speed, but one beam is made of electrons and the other of particles with greater mass, say neutrons, would we be able to measure any difference?

The planet would have so much more mass compared to that of those particles that it would be kind of like "hammer and feather" thing, but then again, even a small difference in the change of angle when they pass next to the planet would grow larger with the distance, and so at the end we could actually measure even the smallest differences in mass of those particles in such beams. And if all this was true and possible, then I guess that would give us a way to measure photon gravity field (mass) too, or at least put it in some perspective compared to that of an electron.
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PF Gold
P: 6,035
 Quote by Barry_G the electron beam twice as far from the planet than the other electron beam would be two times less "influenced", just as it the case with two beams of light.
As you can see from the formula pervect posted, the result for an electron beam moving at v < c would be inversely proportional to b, yes. So I was too pessimistic when I said that result would be more complicated. I'm pretty sure pervect's formula still requires that b is much greater than GM / c^2.

 Quote by Barry_G Which leads me to another question. If we were to measure "influence" or deflection of two beams of particles passing near some planet at the same distance away from it and with the same speed, but one beam is made of electrons and the other of particles with greater mass, say neutrons, would we be able to measure any difference?
The mass of the particles in the beam doesn't appear anywhere in the formula, so it wouldn't make a difference. Only the velocity of the particles in the beam matters. This is a manifestation of the fact that, in Newtonian language, all objects fall with the same acceleration in a gravitational field, regardless of their mass.

 Quote by Barry_G The planet would have so much more mass compared to that of those particles that it would be kind of like "hammer and feather" thing
The "hammer and feather" thing doesn't depend on the mass of the planet being so much larger than the mass of the particles in the beam. It's always true for gravity, according to our best current theories, regardless of the mass of the gravitating object or the mass of the objects being deflected.

 Quote by Barry_G I guess that would give us a way to measure photon gravity field (mass) too, or at least put it in some perspective compared to that of an electron.
As I noted above, the mass of the particles in the beam doesn't appear in the formula, so you can't use bending of the beam by a massive object to measure anything about the mass of the particles in the beam.
P: 68
 Quote by PeterDonis The "hammer and feather" thing doesn't depend on the mass of the planet being so much larger than the mass of the particles in the beam. It's always true for gravity, according to our best current theories, regardless of the mass of the gravitating object or the mass of the objects being deflected.
Oops. You are right, force would be greater for particles with greater mass, but acceleration would be the same as it gets divided by proportionally greater mass. I got confused thinking about J. J. Thomson experiment and trying to make a parallel with measurement of electron mass in cathode ray tube. Which now makes me wonder how could that kind of thing measure any mass since the principle would be the same and so deflection of particles with different mass would be the same, having the same charge. Apparently I need to revisit that one.

On the other hand that perhaps makes for even more conclusive comparison regarding this topic. It seems it would mean that any beam of anything would be exactly deflected as much as any other beam of anything else, regardless of the strength of gravity field (mass) of the particles constituting any such beam. Which than means, I suppose, if everything else follows inverse square gravity law, and if beam of light bends exactly the same as a beam of anything else would, then photons too obey the same law. Does that follow?
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PF Gold
P: 6,035
 Quote by Barry_G Oops. You are right, force would be greater for particles with greater mass, but acceleration would be the same as it gets divided by proportionally greater mass.
Yes, but that's only true of gravity. It isn't true for other forces.

 Quote by Barry_G I got confused thinking about J. J. Thomson experiment and trying to make a parallel with measurement of electron mass in cathode ray tube. Which now makes me wonder how could that kind of thing measure any mass since the principle would be the same and so deflection of particles with different mass would be the same, having the same charge.
No, it wouldn't, because Thomson's experiment was using the electromagnetic force, not gravity, to move electrons. So he was really measuring the ratio of the electron's charge to its mass; but since there were already independent measurements of the electron's charge, measuring the charge/mass ratio allowed him to calculate the electron's mass. If he had done the same type of experiment with, say, a proton, he would have measured a different charge/mass ratio and therefore a different mass.

 Quote by Barry_G On the other hand that perhaps makes for even more conclusive comparison regarding this topic. It seems it would mean that any beam of anything would be exactly deflected as much as any other beam of anything else, regardless of the strength of gravity field (mass) of the particles constituting any such beam. Which than means, I suppose, if everything else follows inverse square gravity law, and if beam of light bends exactly the same as a beam of anything else would, then photons too obey the same law. Does that follow?
Photons do obey "the same law"; it's the formula that pervect wrote down. But although the law does not depend on the mass of the particles in the beam, it does depend on their velocity. So photons, moving at c, will be deflected differently than particles moving slower than c, like electrons.
P: 68
 Quote by PeterDonis Yes, but that's only true of gravity. It isn't true for other forces. No, it wouldn't, because Thomson's experiment was using the electromagnetic force, not gravity, to move electrons. So he was really measuring the ratio of the electron's charge to its mass; but since there were already independent measurements of the electron's charge, measuring the charge/mass ratio allowed him to calculate the electron's mass. If he had done the same type of experiment with, say, a proton, he would have measured a different charge/mass ratio and therefore a different mass.
Uh, yes, I was too haste to write that. It all makes sense now.

 Photons do obey "the same law"; it's the formula that pervect wrote down. But although the law does not depend on the mass of the particles in the beam, it does depend on their velocity. So photons, moving at c, will be deflected differently than particles moving slower than c, like electrons.
That settles it then. Thank you for your patience.
Emeritus
P: 7,596
 Quote by PeterDonis As you can see from the formula pervect posted, the result for an electron beam moving at v < c would be inversely proportional to b, yes. So I was too pessimistic when I said that result would be more complicated. I'm pretty sure pervect's formula still requires that b is much greater than GM / c^2.
Yes, the formula is for large b, not exact - the paper mentions that.
P: 2,470
 Quote by Barry_G Field? I'd say eight fields, two electric and six of them magnetic, but no, that's not what mainstream theory would tell you. A photon is quanta of electromagnetic radiation, and despite the name, despite there are, I mean could very well be, magnetic and electric fields constituting a photon, it is still electrically and magnetically neutral, which means its electric and magnetic field is measured to be zero. Ok?
Where do you get this nonsense? I am really curious. There seems to be the system to this stuff, but it has nothing to do with any kind of established science.

We are talking about electrodynamics. Quantum electrodynamics specifically, but that distinction is only necessary to establish second quantization since we are not looking at interactions. In electrodynamics, unlike electrostatics, you do not need a charge to have an electric field. Maxwell's laws allow electric field to arise as response to charge or as response to a changing magnetic field. Hence the electromagnetic wave, quantization of which is the photon.

The exact electric field of a photon is given by $E = E_0 e^{ik_{\mu}x^{\mu}}$. Magnetic field is given by similar expression with ||B||=||E||/c. E0 is normalized so that total energy is $\hbar \omega$. That normalization goes to zero for an exact value of k. If you take a photon in superposition of values of k, it behaves as a wave packet, having finite amplitude in a finite region of space.

Gravity being fundamentally non-linear, the exact effect this has on gravity depends on the exact wave packet. But for any distribution in k, you can write down the exact equation for E and B fields and that will give you an equation to solve for space-time curvature.
 Mhm, so what's the number, what is the strength of a single photon gravity field?
Number? Gravity isn't described by a number. It's not even described by a vector, like in classical theory. It is described by a metric tensor, which can be represented by a 4x4 matrix that is a function of space and time. To solve for that metric tensor, you have to solve the Einstein Field Equation, which in it's most general form is a system of sixteen non-linear differential equations. There are not many exact solutions known. A general photon solution isn't even close, as it has an almost infinite complexity.

But there are a few special cases. There is plane wave solution. There is the Vaidya Metric which ignores some specifics of EM radiation, but otherwise models a radiating star. The later belongs to a class of null dust solutions which assume a uniform flux of massless particles.

Let me stress that. Entire class of solutions that deal with gravity due to massless particles. You seem to be very confused on what mass is. It is not something that is required to generate gravity or people wouldn't be wasting time on such things.
P: 68
 Quote by K^2 Where do you get this nonsense? I am really curious. There seems to be the system to this stuff, but it has nothing to do with any kind of established science.
I suppose you are referring only to the first sentence. As I said it indeed is not not part of, although it has everything to do with, established science. Just my personal understanding and interpretation based on experimental measurements. On the other hand the rest of what I said there is well established. Photons are magnetically and electrically neutral, meaning that you can not deflect a beam of light by either electric or magnetic fields.

I made a new thread in Classical Physics forums to move this discussion away from here, I'll respond to the rest of what you said about it there. The title of the thread is: "Photon is electromagnetic field, right?"

 Gravity being fundamentally non-linear, the exact effect this has on gravity depends on the exact wave packet. But for any distribution in k, you can write down the exact equation for E and B fields and that will give you an equation to solve for space-time curvature.
Why distribution, does it not apply to a single photon? I wish you would provide some links along with you're saying, so I know what 'k' means and what equations you are talking about, and also to check you have not made any mistakes. Obviously I have not studied SR or GR and that you know a lot about it, but that still does not mean I should take for granted whatever you say. I like to check everything and make my own opinions.

 Number? Gravity isn't described by a number. It's not even described by a vector, like in classical theory. It is described by a metric tensor, which can be represented by a 4x4 matrix that is a function of space and time.
That's why the concept of mass is useful. But even in GR that metric tensor thing surely has to be able to assume some numerical values, which then should translate to other concepts and descriptions used in other theories, more or less. For example, numerical value(s) of metric tensor relating to Earth and Sun should translate to values describing their masses and the difference should stay proportional. Yes? After all both are descriptions relating to the same physical phenomena - gravity field.

 Let me stress that. Entire class of solutions that deal with gravity due to massless particles. You seem to be very confused on what mass is. It is not something that is required to generate gravity or people wouldn't be wasting time on such things.
You seem to interpret concepts and descriptions too literally, which can restricts your ability to see connections and the bigger picture.
P: 2,470
 Quote by Barry_G Why distribution, does it not apply to a single photon?
It is a single photon. We are talking about Quantum here. Single photon will be described by a probability distribution. That corresponds to a particular configuration of the EM field. You can have a photon that's highly localized, or you can have a photon that's highly delocalized. Photon with fixed k is infinitely delocalized, and so it will have a zero amplitude everywhere. That means, it will have no contribution to gravity. But that's not physical. Nothing is infinitely distributed. A photon with some probability distribution for k can give you a physical picture of photon being mostly at certain place, that place moving through space at the speed of light.

k is the wave vector. Specifically, look under the SR section that describes wave 4-vector.

 But even in GR that metric tensor thing surely has to be able to assume some numerical values
No. It's a tensor. A tensor cannot be a number. You can describe it with a matrix of four numbers, sure. Like I said, it's not going to be possible to solve for a general photon. There are some related solutions which I have provided.

 For example, numerical value(s) of metric tensor relating to Earth and Sun should translate to values describing their masses and the difference should stay proportional. Yes? After all both are descriptions relating to the same physical phenomena - gravity field.
The only thing that ends up being the same (roughly) are the trajectories for the bodies. There is no actual gravitational field in GR, so what else should stay the same?

 You seem to interpret concepts and descriptions too literally, which can restricts your ability to see connections and the bigger picture.
Physics is interpreted literally. That's what makes it different from the new age crap. It's a different question of whether the models are correct. And we can discuss that. But keep in mind that GR and QM are tested on the limits of our ability to perform the experiments with no indication of either one failing any time soon. While we do have some potential boundaries to these models, such as Plank's scales, there is absolutely nothing we can interact with beyond or even close to these boundaries.

The only serious known limitation of modern physics is incompatibility between QM and GR. There are people working on it. I have acknowledged these limitations in this thread, and have outlined the conditions under which they are irrelevant to this or that discussion.

When I was in middle school, I did have roughly your understanding of physics. I did want to verify all these things for myself. So I read books and I studied physics. Things I'm telling you are verifiable, and I've tested many of them myself. I make mistakes, certainly, but I can actually recognize them and look for problems, because I have certain amount of background on the subjects.