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

  • #51
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  • #52
WannabeNewton said:
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.
 
  • #53
Barry_G said:
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.

Barry_G said:
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?
 
  • #54
DaleSpam said:
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.
 
  • #55
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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  • #56
Barry_G said:
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?
 
  • #57
PeterDonis said:
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?
 
  • #58
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
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_(physics)

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
HomogenousCow said:
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.
 
  • #61
HomogenousCow said:
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.
 
  • #62
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.
 
  • #63
Barry_G said:
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.
 
  • #64
PAllen said:
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.
 
  • #65
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.
 
  • #66
Barry_G said:
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.
 
  • #67
dextercioby said:
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?
 
  • #68
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
PAllen said:
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?
 
  • #70
Barry_G said:
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.

Barry_G said:
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/Classification_of_electromagnetic_fields

As shown in that article, the two invariants that are used to classify EM fields are

<br /> P = E^2 - B^2 \\<br /> Q = \vec{E} \cdot \vec{B}<br />

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.

Barry_G said:
How many electric fields and how many magnetic fields a single photon has, exactly?

Um, one of each?

Barry_G said:
What is the the strength of those fields?

It depends on the energy of the photon.

Barry_G said:
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.
 
  • #71
Barry_G said:
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:

\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_problem_in_general_relativity#Bending_of_light_by_gravity

Note that this formula is only valid for b very small compared to GM / c^2.
 
  • #72
PeterDonis said:
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.
 
  • #73
Barry_G said:
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.
 
  • #74
Barry_G said:
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.
 
  • #75
Barry_G said:
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.
 
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  • #76
PeterDonis said:
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_problem_in_general_relativity#Bending_of_light_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.
 
  • #77
Barry_G said:
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?title=Mass_in_general_relativity&oldid=514908524
 
  • #78
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.

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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 [strike]more[/strike] 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.
 
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  • #79
Barry_G said:
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.

Barry_G said:
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.

Barry_G said:
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.

Barry_G said:
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.
 
  • #80
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.
 
  • #81
PeterDonis said:
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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  • #82
Barry_G said:
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.

Barry_G said:
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.

Barry_G said:
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.

Barry_G said:
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.
 
  • #83
PeterDonis said:
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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  • #84
Barry_G said:
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.

Barry_G said:
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.

Barry_G said:
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.
 
  • #85
PeterDonis said:
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.
 
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  • #86
PeterDonis said:
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.
 
  • #87
Barry_G said:
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.
 
  • #88
K^2 said:
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.
 
  • #89
Barry_G said:
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.

I can completely understand your desire of wanting to understand these things better and being able to see where these things come from. You are several years of rigorous education short of being able to actually follow most of these derivations. When I give you links confirming results, you do not understand them. When I try to explain, you do not understand explanation. There is not a whole lot that I can do. I have been at your state many years ago. The only way you are going to understand these things is if you start learning the underlying theory. And you have to start with basics. You have no concepts of relativity or quantum, and your understanding of electrodynamics is rudimentary. I suspect your mechanics isn't much better. That's where you have to start. Not asking questions about advanced topics in General Relativity. I mean, it's fine if you are just curious and want to know the answer, and I'm happy to give you answers as I have been for several pages, but you need to accept that neither me nor anybody else can explain these things to you on the level you will understand without learning a whole lot more physics than you know now.
 
  • #90
Trajectories, that's all that matters, to me at least. What I wanted is something concrete I can relate to classical mechanics so I can compare it with upper limits given for estimation of photon mass. I was hoping someone would spill some numbers related to photon gravity field, but I'm just as happy with those equations given by PeterDonis and pervect related to light bending. All that so I can keep my understanding that photons actually do have intrinsic mass.
 
  • #91
We don't discuss metaphysics here, I have no idea where you got that part about "2 electric fields and six magnetic fields".
 
  • #92
Barry_G said:
Trajectories, that's all that matters, to me at least. What I wanted is something concrete I can relate to classical mechanics so I can compare it with upper limits given for estimation of photon mass. I was hoping someone would spill some numbers related to photon gravity field, but I'm just as happy with those equations given by PeterDonis and pervect related to light bending. All that so I can keep my understanding that photons actually do have intrinsic mass.
My second post in this thread.
K^2 said:
If you pretend that photons are particles traveling at speed c having a mass p/c, and you are looking at "acceleration" due to gravity in perpendicular direction, you'll only be a factor of 2 off.
Which is consistent with factor of 2 you get from pervect's equation in the limit v->c.

But this is light following gravitational field of a spherical body. There is an exact solution for that in GR. That's a much simpler problem than photon generating gravity.
 
  • #93
If you pretend that photons are particles traveling at speed c having a mass p/c, and you are looking at "acceleration" due to gravity in perpendicular direction, you'll only be a factor of 2 off.
are you talking about the deflection of light which is twice as large as compared to Newtonian theory?
 
  • #94
K^2 said:
My second post in this thread.

Which is consistent with factor of 2 you get from pervect's equation in the limit v->c.

I know you said that, I especially like how you defined mass, and I like that factor of two because I think I might be able to explain it, but you gave no links nor did you respond to my question about it. Ok, so where do you see factor of two? More specifically, what value do you take for momentum since it varies according to wavelength, and what equations are you comparing?
 
  • #95
If you only look at the deflection angle, which is the best way to go about it, you can drop the mass. In classical theory it doesn't matter, and in GR, it means something completely different.

So method one. Take a particle of arbitrary mass m (you can take limit m->0 in the end, as it doesn't matter) and shoot it past, say, a star of mass M. You fire it originally distance b from radial, and infinitely far away the particle's initial velocity (hyperbolic excess velocity) is c. You compute classic trajectory, and observe the particle leave the star with the same distance b from radial, same velocity c, but heading in a slightly different direction, making angle θ with original.

\theta = sin^{-1}\left(2\sqrt{\frac{GM^2}{GM^2+b^2 v^4}}\right)

This is general formula that works for any v, so you can take v->c. It can be derived from formualae on this page by keeping in mind that b*v = v(r)*r at closest approach r = -a(1-e) due to conservation of angular momentum. Note, also, that what I'm calling θ is the deflection angle. The angle in the article is angle between asymptotes. Hence the inverse sine instead of inverse cosine and the factor of 2. In the limit bv² >> GM and v->c, the deflection angle is small, and the above simplifies to the following.

\theta = \frac{2GM}{bc^2}

Method two. You find a null-geodesic that corresponds to the parameter b above. It also yields you an angle, some θ'. I am not going to point to derivation of that, because it involves Christoffel Symbols of Schwarzschild Metric. But the result is given in this article on Gravitational Lensing. Note that what I call b they call r.

\theta = \frac{4GM}{bc^2}

Both of these describe a "photon". First method describes it in classical approximation, second under GR. The difference in the deflection angle is exactly factor of 2.

This lead to one of the original tests of GR described on this page. The deflection of the light from the stars due to gravity of the Sun can be measured, and does, in fact agree with GR rather than the classical result.

However, because the difference in deflection between GR and classical result is by a constant factor, it can be said that influence on the gravity on light does drop as 1/r as it does in classical theory. In fact, both formulae give deflection as 1/r from distance to the star.

And that was my original point. I just didn't expect to have to explain it in that much detail in the GR section of this forum.
 
  • #96
  • #97
andrien said:
here is a link in which both Newtonian result and prediction of einstein theory is given.Calculation is discussed also
http://www.mathpages.com/rr/s6-03/6-03.htm

Thank-you andrien, that is a really interesting link - who is Kevin Brown?
 
  • #98
swle said:
Thank-you andrien, that is a really interesting link - who is Kevin Brown?
A man,who just does not want to reveal himself.
http://www.numericana.com/fame/
 
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