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

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    Gravity Law Photons
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SUMMARY

Photons do not obey the 1/r² gravity law as described by Newton's Law of Universal Gravitation. Instead, their behavior in a gravitational field is explained by general relativity, which describes gravity as the curvature of spacetime. Photons follow geodesics in this curved spacetime, and while they contribute to the stress-energy tensor, they do not have mass in the traditional sense. The deflection of light due to gravity is twice the amount predicted by Newtonian physics, confirming the principles of general relativity.

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  • Understanding of general relativity and Einstein's field equations
  • Familiarity with the concept of geodesics in curved spacetime
  • Knowledge of the stress-energy tensor and its role in gravity
  • Basic grasp of Newton's Law of Universal Gravitation
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  • Study Einstein's field equations in detail, focusing on their implications for gravity
  • Explore the concept of geodesics and how they apply to massless particles like photons
  • Investigate the stress-energy tensor and its significance in general relativity
  • Examine experimental evidence supporting light's deflection in gravitational fields
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Physicists, students of general relativity, and anyone interested in the interaction between light and gravity will benefit from this discussion.

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  • #32
K^2 said:

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?
 
  • #33
Barry_G said:
a nonzero photon mass could be so small that present-day experiments cannot probe it. The experimental results just serve to set an upper bound to the photon mass...

Yes, that's true. The current upper bound (at least as of 2003, this is the latest experiment I'm aware of) is 10^-51 grams:

http://www.aip.org/pnu/2003/split/625-2.html

But note that that is an upper bound: the results, as with all previous experiments done to test this, are consistent with the photon being massless.

Also, you appear to be confused about the usage of the term "mass". When we say a photon has zero mass, we mean invariant mass:

http://en.wikipedia.org/wiki/Invariant_mass

When you say that "mass" is what produces gravity, you really mean (whether you realize it or not) energy:

http://en.wikipedia.org/wiki/Mass–energy_equivalence

The stress-energy tensor is just the correct relativistic generalization of "energy produces gravity": it includes contributions from energy, momentum, pressure, and stress, because you have to include all those things for the source of gravity to transform correctly when you change coordinates.

So there is nothing contradictory about saying that photons have zero mass (meaning invariant mass) but still produce gravity (because they have energy, or more precisely a nonzero stress-energy tensor).

As far as experiments being done to directly test that photons produce gravity, their gravity is far too weak to directly detect. But that's also true of plenty of objects that have nonzero invariant mass, like atoms. I believe modern Cavendish-type experiments can detect the gravity from masses on the order of a kilogram, but I haven't been able to find a reference online.
 
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  • #34
Keep it civil people.
In relativity we have to write all equations in a coordinate independent manner, called general covariance. Some scalars are not coordinate independent, some are, for example the inner product between two four vectors is, mass density measured in an arbitrary frame is not.
Thus we require an entity that is coordinate independent to describe the effect of matter on space time, the Stress-Energy tensor is chosen for Einstein's Relativity. The Stress-Energy tensor is a second order tensor that describes the flux of four-momentum across a slice of constant four-space. For example the 00 component is the flux of the 0th component of the momentum four vector, energy across a slice of constant 0, time, hence it is just the energy density.
Also, be aware that in GR, when we speak of the photon we are being very informal about it, GR is not a quantum theory and only a quantum theory can quantize the EM field, thus putting a good definition on what we mean by a "Photon".
Now here's where your question gets answered, I will put this in two parts, the effect of "photons" on spacetime, and the effect of spacetime on the "photons"
In GR we usually speak of photons as particles, however GR is a classical theory, hence we still have to formally treat photons in their usual field sense. An electromagnetic field has a stress energy tensor known as the electromagnetic-stress-energy tensor. Simply feed this into the Einstein field equation, (actually, not so simple, I don't think we actually have any closed form solutions of this) and "in principle", you can get the metric for that system.
2.If you already have a prescribed metric for the system, you can use this to solve Maxwell's equations in curved space time. (Again, "In principal", whenever curved space time is involved things do indeed get very nonlinear, and even worse, coupled)
 
  • #35
PeterDonis said:

Article says: "Through all such conversions, however mass remains conserved..."

It seems to me they say mass can be singled out and distinguished from the rest of the energy. 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. That doesn't sound as if mass is conserved in the way that part of the article seems to suggests.


So there is nothing contradictory about saying that photons have zero mass (meaning invariant mass) but still produce gravity (because they have energy, or more precisely a nonzero stress-energy tensor).

Yeah, it's ambiguous enough not to be contradictory.


As far as experiments being done to directly test that photons produce gravity, their gravity is far too weak to directly detect.

And so my point is that your statement "photons produce gravity" is just about as valid as if I say "photons have mass".
 
  • #36
HomogenousCow said:
An electromagnetic field has a stress energy tensor known as the electromagnetic-stress-energy tensor. Simply feed this into the Einstein field equation...

If photons are electrically and magnetically neutral, what numbers do we feed in that equation?
 
  • #37
Barry_G said:
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?

Barry_G said:
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.

Barry_G said:
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.
 
  • #38
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.
 
  • #39
Barry_G said:
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.

Barry_G said:
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.

Barry_G said:
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 let's 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.

Barry_G said:
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.

Barry_G said:
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.
 
  • #40
Why do all threads end this way...
 
  • #41
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.
 
  • #42
K^2 said:
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?
 
  • #43
HomogenousCow said:
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? Let's 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?
 
  • #44
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
Barry_G said:
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?title=Stress–energy_tensor&oldid=517465899 is not bad.

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

http://upload.wikimedia.org/wikipedia/commons/3/37/StressEnergyTensor.svg
 
  • #46
just realized barry is not the opener of thread:-p
 
  • #47
K^2 said:
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 referred 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#Experimental_checks_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
 
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  • #48
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
PeterDonis said:
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?
 
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  • #50
Barry_G said:
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.
 
  • #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.
 

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