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

by swle
Tags: 1 or r2, gravity, obey, photons
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HomogenousCow
#55
Dec10-12, 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.
PeterDonis
#56
Dec10-12, 11:56 AM
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Quote Quote by Barry_G View Post
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?
Barry_G
#57
Dec10-12, 12:05 PM
P: 68
Quote Quote by PeterDonis View Post
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?
HomogenousCow
#58
Dec10-12, 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.
PAllen
#59
Dec10-12, 12:19 PM
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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.
Barry_G
#60
Dec10-12, 12:26 PM
P: 68
Quote Quote by HomogenousCow View Post
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.
Barry_G
#61
Dec10-12, 12:33 PM
P: 68
Quote Quote by HomogenousCow View Post
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.
HomogenousCow
#62
Dec10-12, 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 four-wave vector.
PAllen
#63
Dec10-12, 12:36 PM
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Quote Quote by Barry_G View Post
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.
Barry_G
#64
Dec10-12, 12:49 PM
P: 68
Quote Quote by PAllen View Post
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.
dextercioby
#65
Dec10-12, 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 e-m field and the linearized (aka Pauli-Fierz) gravity field is shortly discussed.
PAllen
#66
Dec10-12, 12:58 PM
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Quote Quote by Barry_G View Post
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.
Barry_G
#67
Dec10-12, 01:06 PM
P: 68
Quote Quote by dextercioby View Post
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?
dextercioby
#68
Dec10-12, 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 (?).
Barry_G
#69
Dec10-12, 01:15 PM
P: 68
Quote Quote by PAllen View Post
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.
Has such experiment been performed?
PeterDonis
#70
Dec10-12, 01:23 PM
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Quote Quote by Barry_G View Post
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?
I'm talking about an electromagnetic field, as described here:

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 3-vectors, the electric field and the magnetic field. Each 3-vector has 3 independent components, for 6 components total.

Electrons are not electromagnetic fields, so no, this doesn't apply to them.

Quote Quote by Barry_G View Post
Can you name those two components please?
The EM field associated with a photon (more precisely, associated with a classical electromagnetic wave, which is the best classical approximation to a photon) is a particular type of EM field called a "null electromagnetic field", as described, for example, here:

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.

Quote Quote by Barry_G View Post
How many electric fields and how many magnetic fields a single photon has, exactly?
Um, one of each?

Quote Quote by Barry_G View Post
What is the the strength of those fields?
It depends on the energy of the photon.

Quote Quote by Barry_G View Post
So, if net electric charge of a photon is zero, does that mean it contains both positive and negative electric fields, or what?
It means the electric and magnetic fields satisfy the source-free Maxwell's Equations:

http://en.wikipedia.org/wiki/Maxwell's_equations

See the section on Vacuum equations, electromagnetic waves, and the speed of light. "Source-free" means there is no charge or current present.
PeterDonis
#71
Dec10-12, 01:36 PM
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Quote Quote by Barry_G View Post
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?
No, it will be twice as "influenced". The angular deflection of a light beam passing close to a massive object is given by:

[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.
Barry_G
#72
Dec10-12, 01:39 PM
P: 68
Quote Quote by PeterDonis View Post
I'm talking about an electromagnetic field...
Thank you. I'd like to talk about that in more detail so I'll start a new thread in Classical Physics forum.


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