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

by swle
Tags: 1 or r2, gravity, obey, photons
P: 2,470
 Quote by Barry_G No mention of any experiment or measurements there, they don't even mention photons. Plus, photons have zero electric charge, and they are magnetically neutral too, so I am sorry to inform you but that is nowhere near to even begin to support your claim how photons generate gravity. Never mind, just one more thing, what is supposed to be the strength of photon gravity field?
You do realize that a photon IS electromagnetic field, right? That isn't actually news to you, I hope. You see what happens when you ask for references without explanation? You don't understand it. Then I have to explain it. Then you don't understand that either. Where is an end to this? At what point do you accept that you need to step back and learn some fundamental theory?

 Quote by Barry_G It seems to me they say mass can be singled out and distinguished from the rest of the energy.
Yes. It's the energy that remains after you take away all of the kinetic energy, if you'd like to think of it that way. Yes, it's certainly special. But it's importance lies in particle propagation. Not in how it generates gravity. Gravity is generated by all of the energy.

Again, all of this is something you should know before you enter a discussion about gravity. SR is a prerequisite to GR, and we can't get past your confusion on that subject.

 Quote by Barry_G And so my point is that your statement "photons produce gravity" is just about as valid as if I say "photons have mass".
No. The difference is that if we say, "Photons have mass, but it's so absolutely tiny as not to be detectable by any experiment we have conducted, nor to show up as a side-effect in absolutely any theory," we can also say, "Who cares?" Our theory is more precise than experiment can refine with a massless photon. If photon mass has not manifested itself in all of this, the only sane assumption is to assume it is zero and keep going until it becomes a problem.

Photon producing no gravity would say that general relativity is absolutely wrong. It would fly in the face of all that we know about gravity. Possible? Technically. So are leprechauns. Technically.

 And while speaking of mass conservation in that sense, what about positron-electron annihilation?
Is e-p annihilation described by Lorentz transformation? "Mass conservation in that sense" has nothing to do with annihilation processes. Mass is not generally a conserved quantity. It's merely a frame-invariant one.
 P: 356 Oh, Now I see where the confusion has come in. Please re-read my post, a Photon is a quantum entity and thus cannot be described by a classical theory such as GR. Photons are the quantized "bits" of the electromagnetic field in quantum mechanics, in classical theory we still regard the EM field as a field. The Photon being changeless is irrelevant in general relativity.
Physics
PF Gold
P: 6,025
 Quote by Barry_G Article says: "Through all such conversions, however mass remains conserved..."
By "mass" here, the article means (at least, assuming it was actually trying to be correct) either total energy, or the total invariant mass of the *system*, as opposed to its components. Invariant mass is not additive: objects with zero invariant mass can form a system that has nonzero invariant mass. See below.

 Quote by Barry_G It seems to me they say mass can be singled out and distinguished from the rest of the energy.
No, they're not (at least, not if they're trying to be correct). See above.

 Quote by Barry_G And while speaking of mass conservation in that sense, what about positron-electron annihilation? They both have mass and yet they produce nothing else but photon which supposedly has no mass.
This is an example of how invariant mass is not additive. The total invariant mass of the system is conserved; it's the same before and after the annihilation. This is the relativistic version of saying that total energy is conserved; if we do our analysis in the center of mass frame of the total system, the invariant mass *is* the total energy (with a factor of c^2 in there if you want to measure mass and energy in different units). But stating it as conservation of the system's invariant mass lets us transform to other reference frames while the conservation law continues to hold.

Now look at the details in the center of mass frame. Before the annihilation, we have an electron and a positron, which are moving towards each other with equal and opposite velocity. The total energy of the system is:

* electron invariant mass + positron invariant mass + electron kinetic energy + positron kinetic energy

and the total momentum of the system is zero, because the momenta of the electron and positron are equal and opposite.

After the annihilation, we have two photons, which are moving in opposite directions. The total momentum is again zero (the photon momenta are equal and opposite and so cancel), and the total energy is

* photon 1 energy + photon 2 energy

By conservation of energy, we must have

* electron invariant mass + positron invariant mass + electron kinetic energy + positron kinetic energy = photon 1 energy + photon 2 energy = total energy of system = total invariant mass of system (because we're in the center of mass frame)

So the invariant mass of the system is conserved, even though the invariant masses of the components change.

 Quote by Barry_G Yeah, it's ambiguous enough not to be contradictory.
No, it's using terminology precisely enough to make it clear how it's not contradictory, because the term "mass" can mean different things. The apparent "contradiction" only arises if one is sloppy about terminology.

 Quote by Barry_G And so my point is that your statement "photons produce gravity" is just about as valid as if I say "photons have mass".
But the difference is that "photons produce gravity" is unambiguous, while "photons have mass" is ambiguous; it depends on what you mean by "mass". Photons have zero invariant mass, but nonzero energy. So "photons have mass" can be true or false depending on how you interpret it, while "photons produce gravity" is unambiguous and true. It's you who are insisting on ambiguous terminology, not me.
 P: 356 Why do all threads end this way...
 Sci Advisor P: 2,470 HomogenousCow, if we have a single photon, and all we do not consider how it interacts with anything else, it is fully described by Maxwell's Equations. Quantization, and indeed, linearity, are not necessary until we start considering a second particle in the same space. There should not be a problem with describing a photon in space-time it itself curves in a manner that is 100% consistent with both GR and QM. As soon as you throw in a second photon, or any other particle, yes, we start making assumptions.
P: 356
 Quote by K^2 HomogenousCow, if we have a single photon, and all we do not consider how it interacts with anything else, it is fully described by Maxwell's Equations. Quantization, and indeed, linearity, are not necessary until we start considering a second particle in the same space. There should not be a problem with describing a photon in space-time it itself curves in a manner that is 100% consistent with both GR and QM. As soon as you throw in a second photon, or any other particle, yes, we start making assumptions.
I apologize, my knowledge of photons is not all that thorough.
I am a bit confused here myself, in GR we can calculate classical trajectories for photons (null paths), however isn't this a direct violation of quantum mechanics?
P: 2,470
 Quote by HomogenousCow I am a bit confused here myself, in GR we can calculate classical trajectories for photons (null paths), however isn't this a direct violation of quantum mechanics?
While we can't do full-on quantum gravity, we can do field theory in curved space-time. So if you have a macroscopic massive body that produces curvature which we can assume to be otherwise irrelevant on quantum level, we can talk about trajectories of quantized particles in the resulting space-time.

And the reason this works is actually quite interesting. Are you familiar with path-integral formulation of field theory? Lets look at classical optics. Light is said to take the shortest path. But that's if you think of light as beams. Furthermore, how would it know which path is shortest? Enter wave optics. You can look at light propagation and note that it's equivalent to point sources throughout space emitting spherical waves. Each source excites the neighbors, and so it propagates.

But what path does light take in such description? All of them. From every point it goes in all possible directions, and from each following point it goes in all directions available there. However, what happens when we add all of these paths together? The phase of the EM wave will depend on distance traveled up to that point. With infinitely many ways to travel from one point to another, aren't all of the phases going to be random? Almost. The paths that are local minima will have zero derivative with respect to perturbation, and so will have infinitely many paths of infinitesimal variation with identical length. All of the random phases cancel out. The phase corresponding to shortest path persists. And we get optics.

Turns out, same thing works for particles in field theory. Things are a little more complicated because the relationship between k and ω is a bit more complex, but you still have some frequency with respect to proper time. That means that distance along the world-line of a path is the distance that will affect final phase. And that means the shortest space-time distance is the path the particle will take. Ergo, particles of quantum field theory follow geodesics of GR.

Problems happen when we decide that we want these particles to influence the curvature of space-time. The whole thing becomes non-linear, axioms of QFT fail, and we have no theory.

There might be a way to formulate a self-consistent non-linear field theory, but what are you going to do with it, when all of your QFT framework depends on quantization, which, in turn, depends on linearity?
 P: 356 I am familiar with the path integral formulation (Although I dread it), but as I understand we have to do a functional integral over all possible paths with the phase being the action of that path. I do understand your point, it is a good approximation to just assume the stationary path since it is there that the phases add constructively. On a side note, what is the current status of quantum gravity? How is string theory coming along?
Emeritus
P: 7,594
 Quote by Barry_G If photons are electrically and magnetically neutral, what numbers do we feed in that equation?
The components of the stress energy tensor most relevant to light are momentum density, energy density, and pressure.

Wiki's treatment of the stress-energy tensor http://en.wikipedia.org/w/index.php?...ldid=517465899 is not bad.

The following diagram is from the wiki and has all the components of the tensor.

 P: 1,020 just realized barry is not the opener of thread
P: 68
 Quote by K^2 You do realize that a photon IS electromagnetic field, right?
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?

Electron, for example, is not neutrally charged, it has both electric and magnetic fields measured to be greater than zero, so are you trying to say electron gravity field is due to not only its intrinsic mass but also due to its electric and magnetic fields?

 That isn't actually news to you, I hope. You see what happens when you ask for references without explanation? You don't understand it. Then I have to explain it. Then you don't understand that either. Where is an end to this? At what point do you accept that you need to step back and learn some fundamental theory?
I asked about experiments and actual measurements, never mind that. Now, let me explain what is happening here. You claimed photons generate gravity and then you refered to electromagnetic stress–energy tensor as the source for this photon gravity field. So what you need to address is how do you imagine photon electric and magnetic zero charge have anything to do with electromagnetic stress–energy tensor, not to mention any gravity even.

Don't explain, please, just take your claims and put them into practice. Use electromagnetic stress–energy tensor and calculate the strength of a single photon gravity field, if you can. Just show me, that's all.

 Yes. It's the energy that remains after you take away all of the kinetic energy, if you'd like to think of it that way. Yes, it's certainly special. But it's importance lies in particle propagation. Not in how it generates gravity. Gravity is generated by all of the energy.
Mhm, so what's the number, what is the strength of a single photon gravity field?

 Again, all of this is something you should know before you enter a discussion about gravity. SR is a prerequisite to GR, and we can't get past your confusion on that subject.
This is forum where people talk about stuff and ask questions, it's not a competition or some vanity contest, wake up! Besides, no one is forcing you to talk to me or back up your claims, suit yourself.

 No. The difference is that if we say, "Photons have mass, but it's so absolutely tiny as not to be detectable by any experiment we have conducted, nor to show up as a side-effect in absolutely any theory," we can also say, "Who cares?" Our theory is more precise than experiment can refine with a massless photon. If photon mass has not manifested itself in all of this, the only sane assumption is to assume it is zero and keep going until it becomes a problem.
I care, because I am nice, friendly and caring person. Plus, photon mass has manifested itself as much as photon gravity field, they are concepts describing one and the same physical phenomena.

http://en.wikipedia.org/wiki/Stress-energy_tensor : The stress-energy tensor is the source of the gravitational field in the Einstein field equations of general relativity, just as mass density is the source of such a field in Newtonian gravity.

 Photon producing no gravity would say that general relativity is absolutely wrong. It would fly in the face of all that we know about gravity. Possible? Technically. So are leprechauns. Technically.
I did not say photons do not have gravity field, they obviously do. What I said is how your statement "photons produce gravity" is just about as valid as if I say "photons have mass".

http://en.wikipedia.org/wiki/Photon#...on_photon_mass : The photon is currently understood to be strictly massless, but this is an experimental question. If the photon is not a strictly massless particle, it would not move at the exact speed of light... Relativity would be unaffected by this
 P: 1,020 photons are result of quantizing the electromagnetic field using say creation and annihilation operator.They don't have charge,so they are not affected by EM field.if you are using stress energy tensor then it need to expressed for a single photon.It is however possible to interpret but it is certainly useful to go with the EM field rather than a single photon.
P: 68
 Quote by PeterDonis No, it's using terminology precisely enough to make it clear how it's not contradictory, because the term "mass" can mean different things. The apparent "contradiction" only arises if one is sloppy about terminology.
The beginning of wisdom is to call things by their right names.
- Chinese Proverb

 But the difference is that "photons produce gravity" is unambiguous, while "photons have mass" is ambiguous; it depends on what you mean by "mass". Photons have zero invariant mass, but nonzero energy. So "photons have mass" can be true or false depending on how you interpret it, while "photons produce gravity" is unambiguous and true. It's you who are insisting on ambiguous terminology, not me.
By "mass" I of course mean intrinsic mass. The one that is given for elementary particles where, for example, mass of an electron is 9.109x10^-31 kg, and for photon is estimated to be less than 1x10^-18 eV/c^2, according to Wikipedia. Where 1x10^-18 eV/c^2 = 1.783x10^-54 kg, I think.

I don't agree I am being ambiguous, at least I am trying to be very specific, and what I am saying here is that energy is ambiguous concept, especially if you can not pinpoint what part of that energy belongs to frequency, velocity or intrinsic mass. Now, if you'll forgive me I'd prefer to ignore the rest of your post so we can concentrate on what we started to talk about. So let me ask you, if not measured, has anyone calculated what is supposed to be the strength of a single photon gravity field? What's the number?
C. Spirit
Thanks
P: 5,402
 Quote by Barry_G Don't explain, please, just take your claims and put them into practice. Use electromagnetic stress–energy tensor and calculate the strength of a single photon gravity field, if you can. Just show me, that's all.
This is obviously easier said than done to be shown in full in a thread. Why don't you try it yourself? $L_{EM} = -\sqrt{-g}g^{ac}g^{bd}\triangledown_{[a}A_{b]}\triangledown_{[c}A_{d]}$ 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.
Mentor
P: 16,942
 Quote by Barry_G So let me ask you, if not measured, has anyone calculated what is supposed to be the strength of a single photon gravity field?
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
P: 68
 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.
Physics
PF Gold
P: 6,025
 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?
P: 68
 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.

 Related Discussions Special & General Relativity 16 Special & General Relativity 32 General Physics 6 Special & General Relativity 12 Special & General Relativity 13