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

 Quote by WannabeNewton This is obviously easier said than done to be shown in full in a thread. Why don't you try it yourself?
I wouldn't know how, touché! On the other hand my position is that it can not be done because photon electric and magnetic charge is zero, so it would be awkward if I even tried as people could think I'm being crazy arguing against myself.

 $L_{EM} = -\sqrt{-g}g^{ac}g^{bd}\triangledown_{[a}A_{b]}\triangledown_{[c}A_{d]}$ so take the total lagrangian density that includes this matter field lagrangian density and the einstein lagrangian density, vary the respective action to obtain the field equations and solve it if you want (=D). I don't see why you have a problem with the idea that the maxwell field can contribute to curvature. It isn't just mass density that contributes to the curvature. Look up the lens thirring effect as a correction to newtonian mechanics where the OTHER parts of the energy momentum tensor contribute to the non vanishing of the gravito - magnetic field.
I just don't see what electromagnetic stress–energy tensor has anything to do with photons since their electric and magnetic charge is zero.

I don't have any problem with photons having gravity field, but if you are going to claim it then I think you should also be able to point some actual number, or at least some estimation of upper limit like it was given for photon mass.

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 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.
No, mainstream theory would tell you that a general electromagnetic field has six independent components, three electric and three magnetic. A "photon", at least in the classical approximation that's appropriate here, is a special case of an EM field where there are only two independent components. But the EM field of a photon is not zero; that would require zero independent components. See below.

 Quote by Barry_G 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?
No, not ok. EM radiation has zero charge, but nonzero electric and magnetic fields. It has to, since it propagates electromagnetic disturbances from one place to another. How do you think a radio works?

 Quote by DaleSpam Yes, this is an important part of GR. The two types of solutions you will want to look at are null dust solutions and pp-wave spacetimes: http://en.wikipedia.org/wiki/Null_dust_solution http://en.wikipedia.org/wiki/Pp-wave_spacetime
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.
 Barry_G I have no idea what you are saying, in relativity we only talk bout the electromagnetic four tensor, not the electric field and magnetic field since it is not a frame independent idea. it seems to me that you might need to go learn physics first before discussing it qualitatively.

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 Quote by Barry_G If you know someone has calculated or measured this photon gravity filed, then please just tell me the number.
The calculations are just the stress-energy tensor for an EM field, applied to the EM field of a photon. If you don't understand the EM field of a photon, how do you expect to understand its stress-energy tensor?

I've already said that nobody has measured a photon's gravity field because it is far too weak. Nobody has measured an atom's gravity field either; do you think that means it doesn't have one?

 Quote by PeterDonis No, mainstream theory would tell you that a general electromagnetic field has six independent components, three electric and three magnetic.
Are you talking about photons or does that apply to electrons as well? What theory is that, can you point some reference where I can see what are those six components?

 A "photon", at least in the classical approximation that's appropriate here, is a special case of an EM field where there are only two independent components.
Can you name those two components please?

 EM radiation has zero charge, but nonzero electric and magnetic fields.
How many electric fields and how many magnetic fields a single photon has, exactly? What is the the strength of those fields?

So, if net electric charge of a photon is zero, does that mean it contains both positive and negative electric fields, or what?
 Barry_G please stop before you say something even more humiliating, you don't sound like you have any idea how GR and EM work.
 Blog Entries: 1 Recognitions: Gold Member Science Advisor As has been pointed out numerous times, there can't be a clear answer to 'gavity of one photon'. In QFT, there is no such thing as a isolated photon. In classical GR, there is no such thing as photons at all. However one classical object of interest would be the geon: a self gravitating, propagating, neutral, EM field clump with no rest rest mass. It has both self gravitation and external gravitation. It is presumed to decay over time. http://en.wikipedia.org/wiki/Geon_%28physics%29 So far as I know, there is no experiment that can be treated as evidence for geons except insofar as they follow from the GR field equation which is validated over a large domain.

 Quote by HomogenousCow Barry_G I have no idea what you are saying, in relativity we only talk bout the electromagnetic four tensor, not the electric field and magnetic field since it is not a frame independent idea. it seems to me that you might need to go learn physics first before discussing it qualitatively.
Really? But you somehow understand everyone else? Pay closer attention and realize it is K^2 who said: "You do realize that a photon IS electromagnetic field, right?". Now tell him to go and learn physics, will ya? Then go learn some physics yourself and realize electric charge of a photon is zero. After that you may try to bend a beam of light with magnetic fields and you will realize photons magnetic charge is zero as well.

What is the point of being condescending? I could be very well be older and more educated than you, so that's not only inappropriate but also very unnecessary. Just keep your personal comments to yourself and talk about the topic at hand, if you have anything to say about it.

 Quote by HomogenousCow Barry_G please stop before you say something even more humiliating, you don't sound like you have any idea how GR and EM work.
Pay attention and realize I'm not saying anything but ASKING FOR EXPLANATION about what is said by other people. Stop making fun of yourself and stop blaming me for your inability to understand.
 The electromagnetic field is indeed the analogous entity to the photon in classical theory, in quantum field theory we forsake the field concept and model the interactions through mediator particles, for the electromagnetic interaction it is the photon. Photons are not physical entities in classical theory, they are meaningless within the context, we use them informally because they are useful for discussion. For all sakes and purposes, the photon is just another term for an electromagnetic plane wave, which has the two propeties of momentum and energy, the four components of a four-wave vector.

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 Quote by Barry_G Really? But you somehow understand everyone else? Pay closer attention and realize it is K^2 who said: "You do realize that a photon IS electromagnetic field, right?". Now tell him to go and learn physics, will ya? Then go learn some physics yourself and realize electric charge of a photon is zero. After that you may try to bend a beam of light with magnetic fields and you will realize photons magnetic charge is zero as well. What is the point of being condescending? I could be very well be older and more educated than you, so that's not only inappropriate but also very unnecessary. Just keep your personal comments to yourself and talk about the topic at hand, if you have anything to say about it.
The two statements are consistent, not at odds. A pure E field in one frame has magnetic components in another, and vice versa. So just as 4-momentum combines Newtonian energy and momentum into one object, relativity combines the E+M field into the Faraday tensor. Charge is frame invariant, but what is E and what is M is frame dependent.

 Quote by PAllen As has been pointed out numerous times, there can't be a clear answer to 'gavity of one photon'.
No, that has not been pointed out, and if you are going to, then point it to K^2, who said in post #12: "Photons do generate gravity in GR".

 In classical GR, there is no such thing as photons at all.
Post #12, K^2 said: "Photons do generate gravity in GR". Take it with him or someone else who made claims about it in this thread, I'm mostly just asking questions in relation to those statements.
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Before the thread gets rightfully locked, I would add the following answer to the OP (and also thread's title) question: . As others have said, 1/r^2 law applies to classical electrostatics, magnetostatics and gravitostatics. Photons are not part of classical theories whatsoever. Nonetheless, the OP is invited to read through pages 427 and 428 of A. Zee's where a quantum setting for both the e-m field and the linearized (aka Pauli-Fierz) gravity field is shortly discussed.

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 Quote by Barry_G No, that has not been pointed out, and if you are going to, then point it to K^2, who said in post #12: "Photons do generate gravity in GR". Post #12, K^2 said: "Photons do generate gravity in GR". Take it with him or someone else who made claims about it in this thread, I'm mostly just asking questions in relation to those statements.
He was speaking loosely, and said so at the bottom of that post. In particular, use of the plural on photons is crucial. The EM field is the classical analog of a virtual photon field. There is no precise way in either QED or GR to discuss an 'isolated photon'.

Photons corresponding to light trapped in a mirrored box, for example, can be adequately represented both classically and via quantum theory. For this, you can state the box without the photons produces less gravity than with the photons; and, using a hypothetical 'exact' scale, the box with the photons weighs more than the empty box.

 Quote by dextercioby As others have said, 1/r^2 law applies to classical electrostatics, magnetostatics and gravitostatics. Photons are not part of classical theories whatsoever.
Two beams of light are passing next to some planet or a star, where one beam is at double the distance away than the other. Will trajectory of the closer beam not be four times as influenced compared to further away beam? And then regardless of what is your answer, if instead of beam of light there is a single photon, will it not follow the same trajectory?
 Blog Entries: 9 Recognitions: Homework Help Science Advisor For the photon, I don't know. For a classical monochromatic wave, in my mind there should be a difference since loosely speaking the spacetime is more curved as we approach a massive body (?).

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