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Do photons obey the 1/r^2 gravity law? 
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#55
Dec1012, 11:53 AM

P: 366

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. 


#56
Dec1012, 11:56 AM

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PF Gold
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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? 


#57
Dec1012, 12:05 PM

P: 68

So, if net electric charge of a photon is zero, does that mean it contains both positive and negative electric fields, or what? 


#58
Dec1012, 12:17 PM

P: 366

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.



#59
Dec1012, 12:19 PM

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PF Gold
P: 5,083

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. 


#60
Dec1012, 12:26 PM

P: 68

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. 


#61
Dec1012, 12:33 PM

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#62
Dec1012, 12:34 PM

P: 366

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 fourwave vector. 


#63
Dec1012, 12:36 PM

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PF Gold
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#64
Dec1012, 12:49 PM

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#65
Dec1012, 12:52 PM

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Before the thread gets rightfully locked, I would add the following answer to the OP (and also thread's title) question:
<We don't know, since we don't have a fully working and rigorous (i.e. renormalizable) quantum theory of electromagnetism in the presence of gravity>. 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 <Quantum Field Theory in a Nutshell> where a quantum setting for both the em field and the linearized (aka PauliFierz) gravity field is shortly discussed. 


#66
Dec1012, 12:58 PM

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PF Gold
P: 5,083

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. 


#67
Dec1012, 01:06 PM

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#68
Dec1012, 01:12 PM

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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 (?).



#69
Dec1012, 01:15 PM

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#70
Dec1012, 01:23 PM

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PF Gold
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http://en.wikipedia.org/wiki/Electromagnetic_field As you can see from the section on the mathematical description of the EM field, it can be described by two 3vectors, the electric field and the magnetic field. Each 3vector has 3 independent components, for 6 components total. Electrons are not electromagnetic fields, so no, this doesn't apply to them. http://en.wikipedia.org/wiki/Classif...agnetic_fields As shown in that article, the two invariants that are used to classify EM fields are [tex] P = E^2  B^2 \\ Q = \vec{E} \cdot \vec{B} [/tex] For a null electromagnetic field, P = Q = 0; this constrains the components of the electric and magnetic field vectors so that there are only two independent ones. You should be able to work that out from the equations above. http://en.wikipedia.org/wiki/Maxwell's_equations See the section on Vacuum equations, electromagnetic waves, and the speed of light. "Sourcefree" means there is no charge or current present. 


#71
Dec1012, 01:36 PM

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PF Gold
P: 6,250

[tex]\delta \phi = \frac{4 G M}{c^2 b}[/tex] 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. 


#72
Dec1012, 01:39 PM

P: 68




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