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

[Barry_G]

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?

 

[K^2]

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?

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.

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.

 

[HomogenousCow]

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.

 

[PeterDonis]

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.

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.

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.

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.

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.

 

[HomogenousCow]

Why do all threads end this way...

 

[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.

 

[HomogenousCow]

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?

 

[K^2]

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?

 

[HomogenousCow]

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?

 

[pervect]

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

 

[andrien]

just realized barry is not the opener of thread

 

[Barry_G]

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

 

[andrien]

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.

 

[Barry_G]

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?

 

[WannabeNewton]

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.

 

[Dale]

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

 

[Barry_G]

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.

 

[PeterDonis]

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.

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?

 

[Barry_G]

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.

 

[HomogenousCow]

Barry_G I have no idea what you are saying, in relativity we only talk bout the electromagnetic four tensor, not the electric field and magnetic field since it is not a frame independent idea.
it seems to me that you might need to go learn physics first before discussing it qualitatively.

 

[PeterDonis]

If you know someone has calculated or measured this photon gravity filed, then please just tell me the number.

The calculations are just the stress-energy tensor for an EM field, applied to the EM field of a photon. If you don't understand the EM field of a photon, how do you expect to understand its stress-energy tensor?

I've already said that nobody has measured a photon's gravity field because it is far too weak. Nobody has measured an atom's gravity field either; do you think that means it doesn't have one?

 

[Barry_G]

No, mainstream theory would tell you that a general electromagnetic field has six independent components, three electric and three magnetic.

Are you talking about photons or does that apply to electrons as well? What theory is that, can you point some reference where I can see what are those six components?

A "photon", at least in the classical approximation that's appropriate here, is a special case of an EM field where there are only two independent components.

Can you name those two components please?

EM radiation has zero charge, but nonzero electric and magnetic fields.

How many electric fields and how many magnetic fields a single photon has, exactly? What is the the strength of those fields?

So, if net electric charge of a photon is zero, does that mean it contains both positive and negative electric fields, or what?

 

[HomogenousCow]

Barry_G please stop before you say something even more humiliating, you don't sound like you have any idea how GR and EM work.

 

[PAllen]

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.

 

[Barry_G]

Barry_G I have no idea what you are saying, in relativity we only talk bout the electromagnetic four tensor, not the electric field and magnetic field since it is not a frame independent idea.
it seems to me that you might need to go learn physics first before discussing it qualitatively.

Really? But you somehow understand everyone else? Pay closer attention and realize it is K^2 who said: "You do realize that a photon IS electromagnetic field, right?". Now tell him to go and learn physics, will ya? Then go learn some physics yourself and realize electric charge of a photon is zero. After that you may try to bend a beam of light with magnetic fields and you will realize photons magnetic charge is zero as well.

What is the point of being condescending? I could be very well be older and more educated than you, so that's not only inappropriate but also very unnecessary. Just keep your personal comments to yourself and talk about the topic at hand, if you have anything to say about it.

 

[Barry_G]

Barry_G please stop before you say something even more humiliating, you don't sound like you have any idea how GR and EM work.

Pay attention and realize I'm not saying anything but ASKING FOR EXPLANATION about what is said by other people. Stop making fun of yourself and stop blaming me for your inability to understand.

 

[HomogenousCow]

The electromagnetic field is indeed the analogous entity to the photon in classical theory, in quantum field theory we forsake the field concept and model the interactions through mediator particles, for the electromagnetic interaction it is the photon.
Photons are not physical entities in classical theory, they are meaningless within the context, we use them informally because they are useful for discussion. For all sakes and purposes, the photon is just another term for an electromagnetic plane wave, which has the two propeties of momentum and energy, the four components of a four-wave vector.

 

[PAllen]

Really? But you somehow understand everyone else? Pay closer attention and realize it is K^2 who said: "You do realize that a photon IS electromagnetic field, right?". Now tell him to go and learn physics, will ya? Then go learn some physics yourself and realize electric charge of a photon is zero. After that you may try to bend a beam of light with magnetic fields and you will realize photons magnetic charge is zero as well.

What is the point of being condescending? I could be very well be older and more educated than you, so that's not only inappropriate but also very unnecessary. Just keep your personal comments to yourself and talk about the topic at hand, if you have anything to say about it.

The two statements are consistent, not at odds. A pure E field in one frame has magnetic components in another, and vice versa. So just as 4-momentum combines Newtonian energy and momentum into one object, relativity combines the E+M field into the Faraday tensor. Charge is frame invariant, but what is E and what is M is frame dependent.

 

[Barry_G]

As has been pointed out numerous times, there can't be a clear answer to 'gavity of one photon'.

No, that has not been pointed out, and if you are going to, then point it to K^2, who said in post #12: "Photons do generate gravity in GR".

In classical GR, there is no such thing as photons at all.

Post #12, K^2 said: "Photons do generate gravity in GR". Take it with him or someone else who made claims about it in this thread, I'm mostly just asking questions in relation to those statements.

 

[dextercioby]

Before the thread gets rightfully locked, I would add the following answer to the OP (and also thread's title) question:

<We don't know, since we don't have a fully working and rigorous (i.e. renormalizable) quantum theory of electromagnetism in the presence of gravity>. As others have said, 1/r^2 law applies to classical electrostatics, magnetostatics and gravitostatics. Photons are not part of classical theories whatsoever.

Nonetheless, the OP is invited to read through pages 427 and 428 of A. Zee's <Quantum Field Theory in a Nutshell> where a quantum setting for both the e-m field and the linearized (aka Pauli-Fierz) gravity field is shortly discussed.

 

[PAllen]

No, that has not been pointed out, and if you are going to, then point it to K^2, who said in post #12: "Photons do generate gravity in GR".

Post #12, K^2 said: "Photons do generate gravity in GR". Take it with him or someone else who made claims about it in this thread, I'm mostly just asking questions in relation to those statements.

He was speaking loosely, and said so at the bottom of that post. In particular, use of the plural on photons is crucial. The EM field is the classical analog of a virtual photon field. There is no precise way in either QED or GR to discuss an 'isolated photon'.

Photons corresponding to light trapped in a mirrored box, for example, can be adequately represented both classically and via quantum theory. For this, you can state the box without the photons produces less gravity than with the photons; and, using a hypothetical 'exact' scale, the box with the photons weighs more than the empty box.

 

[Barry_G]

As others have said, 1/r^2 law applies to classical electrostatics, magnetostatics and gravitostatics. Photons are not part of classical theories whatsoever.

Two beams of light are passing next to some planet or a star, where one beam is at double the distance away than the other. Will trajectory of the closer beam not be four times as influenced compared to further away beam? And then regardless of what is your answer, if instead of beam of light there is a single photon, will it not follow the same trajectory?

 

[dextercioby]

For the photon, I don't know. For a classical monochromatic wave, in my mind there should be a difference since loosely speaking the spacetime is more curved as we approach a massive body (?).

 

[Barry_G]

Photons corresponding to light trapped in a mirrored box, for example, can be adequately represented both classically and via quantum theory. For this, you can state the box without the photons produces less gravity than with the photons; and, using a hypothetical 'exact' scale, the box with the photons weighs more than the empty box.

Has such experiment been performed?

 

[PeterDonis]

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.

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

$$P = E^2 - B^2 \\ Q = \vec{E} \cdot \vec{B}$$

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.

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

Um, one of each?

What is the the strength of those fields?

It depends on the energy of the photon.

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.