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Do photons obey the 1/r^2 gravity law? 
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#37
Dec912, 01:06 PM

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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. 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. 


#38
Dec912, 01:13 PM

P: 356

Oh, Now I see where the confusion has come in.
Please reread 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. 


#39
Dec912, 01:20 PM

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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. 


#40
Dec912, 01:26 PM

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Why do all threads end this way...



#41
Dec912, 01:34 PM

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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 spacetime 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. 


#42
Dec912, 01:39 PM

P: 356

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? 


#43
Dec912, 03:13 PM

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And the reason this works is actually quite interesting. Are you familiar with pathintegral 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 worldline of a path is the distance that will affect final phase. And that means the shortest spacetime 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 spacetime. The whole thing becomes nonlinear, axioms of QFT fail, and we have no theory. There might be a way to formulate a selfconsistent nonlinear 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? 


#44
Dec912, 03:34 PM

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? 


#45
Dec912, 04:13 PM

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Wiki's treatment of the stressenergy 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. 


#46
Dec1012, 03:57 AM

P: 1,020

just realized barry is not the opener of thread



#47
Dec1012, 05:31 AM

P: 68

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? 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. http://en.wikipedia.org/wiki/Stressenergy_tensor : The stressenergy 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. 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 


#48
Dec1012, 06:23 AM

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.



#49
Dec1012, 07:20 AM

P: 68

 Chinese Proverb 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? 


#50
Dec1012, 10:45 AM

C. Spirit
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#51
Dec1012, 11:22 AM

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http://en.wikipedia.org/wiki/Null_dust_solution http://en.wikipedia.org/wiki/Ppwave_spacetime 


#52
Dec1012, 11:28 AM

P: 68

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
Dec1012, 11:29 AM

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#54
Dec1012, 11:44 AM

P: 68




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