Qed

[Peron]

Is the magnetic field of a magnet made of virtual particles?

[Bob_for_short]

I would answer no. It is a static filed having no relation to propagating particles - photons. It is similar to static Coulomb field.

 

It is the photon propagator which is called a virtual photon actually. The main part of this propagator is the usual Coulomb interaction between charges, so it is not a photon at all.

 

There is also some retardation effect involved in this propagator and this makes it look like a "radiated wave". In the first Born approximation, however, one does not calculate effects of really radiated photons although they accompany any scattering experiment. As soon as this part of retardation effect is not included, the meaning of the propagator retardation is vague and cannot be "absolutized". It is better to keep in mind the main, Coulomb part of charge interaction. It is just a potential energy of charge interaction. In QM there are also quantum mechanical effects connected with non-local (smeared) nature of "particles" as de Broglie waves. This makes the selective (propagator) retardation even more vague. Similarly for the electron magnetic field.

Edited November 30, 2009 by Bob_for_short

[Severian]

It is a rather hard question to answer because the technically correct answer ('yes') can be so misconstrued.

 

Technically, every particle you observe has to be be virtual. By that I mean it is very slightly off-mass-shell, so doesn't obey [math]E^2 = p^2 + m^2[/math]. If it were on-mass-shell, it would live forever, so you couldn't observe it. Of course, since most measurements are made on relatively large timescales, it is very very nearly on-mass-shell, and indeed any macroscopic magnetic field is composed of relatively long lived photons, so they are very very nearly on-mass-shell too. (I think it was in this sense that Bob said 'no'.)

 

The most on-shell photons we can observe are from the Cosmic Microwave Background.

[Bob_for_short]

It is a rather hard question to answer because the technically correct answer ('yes') can be so misconstrued...

 

I disargee. It is quite clear question. Real photons are transversal. They are solutions of free equations. Their role is, rughly speaking, to get into the charge equation motion as external fileds. Photons cannot compensate each other.

 

The electric and magnetic fileds of charges are not transversal. They are solutions of (often static) equations with sources. These fileds cannot propagate independently, theiy are "attached" to charges. In many cases they are compenstated: the negative and positive tensions cancel each other and we observe neutral non-magnetic bodies.

 

The term "virtual particles" implies some propagation. In fact, it is just charge interactions without radiation: the energy-momentum in scattering is conserved without taking the radiation losses into account.

 

Another thing - it is not physical approximation and it leads to IR divergences. So I think it is better to speak of real charges and their interactions without "virtual" particles.

 

Anyway, static fileds are not virtual particles. They are properties of charges and they have no independent meaning.

[Peron]

Anyway, static fileds are not virtual particles. They are properties of charges and they have no independent meaning.

 

So, what is a magnetic field made out of then?

[Bob_for_short]

So, what is a magnetic field made out of then?

 

As I said previously, the magnetic field is a way to say that bodies interact in this or that way. One needs two bodies in order to describe (measure) the magnetic field quantitatively. So it is a long-distance interaction of bodies (charges).

 

Often one speaks of magnetic filed itself, I admit, but anyway what is meant is interaction. For example, in magneto-statics one can express the current-current interaction without appealing to the notion of magnetic filed because the filed equations can be resolved and their solutions (fields) can be injected into the equations of motion of charges. So the charge interaction is described with their positions and velocities.

 

The notion of fields becomes sensible while considering radiation of charges - the kinetic energy loss due to emission of EM waves, i.e., when the fields are real, not virtual.

Edited December 1, 2009 by Bob_for_short

[Severian]

I disargee. It is quite clear question. Real photons are transversal. They are solutions of free equations. Their role is, rughly speaking, to get into the charge equation motion as external fileds. Photons cannot compensate each other.

 

So what definition of 'real' and 'virtual' are you using? The usual definition, in my experience, is whether they are on-mass-shell or not - ie. whether or not they satisfy the mass-shell condition. Do you disagree with this?

 

Anyway, static fileds are not virtual particles. They are properties of charges and they have no independent meaning.

 

What is a 'static field' exactly? If I have a charge, I can produce a static field, yes? But now I suddenly switch off the charge, what happens to the field? It doesn't suddenly disappear everywhere at once. The effect must propagate out from the charge.

 

So although a field may appear static, it is a continual transference of momenta from one object to another.

[Bob_for_short]

So what definition of 'real' and 'virtual' are you using?

I avoid using the term "virtual particle". For me it is not a particle at all. Strictly speaking, a “virtual particle” is the corresponding propagator, nothing else. Physically it is a synonym of quasi-static interaction of charges. Giving an independent sense to “virtual particles” is unforgivable sloppiness. It does not create any insight but a great confusion.

 

A real particle, a photon, for example, is indeed on mass-shell only approximately. It is, however, not because of its being “virtual” but because any photon has a limited wave-train (10000-100000 vibrations, for example). Thus it is a wave packet with slightly different frequencies. In many practical applications one does not care about it and uses E = hf, as if it were a one-frequency thing.

The usual definition, in my experience, is whether they are on-mass-shell or not - ie. whether or not they satisfy the mass-shell condition. Do you disagree with this?

I disagree with the sloppiness mentioned above. Besides, those variables of integration are dumb and independent. We denote them as “momentum” and “energy” and this is the cause of confusion. As soon as they are independent Fourier variables, there is no mass-shell relationship between them. It is natural. Finally, nothing is left after integration. So if we work in terms of final expressions, no notion of “virtual particles” appears.

What is a 'static field' exactly? If I have a charge, I can produce a static field, yes? But now I suddenly switch off the charge, what happens to the field? It doesn't suddenly disappear everywhere at once. The effect must propagate out from the charge.

It is not really correct to imagine a point-like charge and its field as a real entity. According to the strict QED results, a point-like charge is an inclusive, illusory, approximate picture rather than the primary one. It is another subject. If you like, I can explain it in another thread.

 

It is also better to imagine real processes - scattering, bound states, excitation and radiation of real QM systems. Then there no problems in description appear. For example, when one says that a charge is surrounded with a field 1/r, one always means that if you put another charge at r_2, their potential energy of interaction will be 1/|r - r_2| rather than 1/r + 1/r_2. What is counted is the interaction energy, not the "self-action" one. In your example you speak of the "self-field".

[ajb]

The definition of virtual that I am aware of is that the classical equations of motion (and/or constraints) are not obeyed. In this case, the mass-shell constraint is not obeyed.

 

This is the standard definition, irrespective of the "sloppiness" or associated misconceptions.

 

 

Technically, every particle you observe has to be be virtual.

 

 

This is correct. Imagine a photon from the Sun entering your eye. Is it virtual or real?

 

I think it is virtual. Loosely, it is not "free" and connects the Sun to your eye. It corresponds to a propagator.

[Bob_for_short]

The definition of virtual that I am aware of is that the classical equations of motion (and/or constraints) are not obeyed. In this case, the mass-shell constraint is not obeyed.

Wre can do without even mentioning the mass-shell constraints if we work in the coordinate space rather with the propagator Fourier images.

This is correct. Imagine a photon from the Sun entering your eye. Is it virtual or real? I think it is virtual. Loosely, it is not "free" and connects the Sun to your eye. It corresponds to a propagator.

Such a photon is, of course, free, as soon as it "leaves" the Sun. The Earth-Sun distance is much larger than the "photon size" - the length of its wave-tarin or wave-packet. It is never described as a photon propagator.

[ajb]

So, you do agree on the definition of virtual?

 

Such a photon is, of course, free, as soon as it "leaves" the Sun. The Earth-Sun distance is much larger than the "photon size" - the length of its wave-tarin or wave-packet. It is never described as a photon propagator.

 

My thinking is that it needs to interact with electrons in your eye to be detected. Thus virtual.

[Bob_for_short]

I understand you. Unfortunately your language is wrong. A real photon is absorbed by my eye molecule and the molecule excitation is then transformed (dissipates) into an electric signal. Why do you need to call the photon a virtual one?

 

Let us take a charged oscillator for simplicity. In an external plane EM wave it gets excited. If the wave is not limited in time and is a resonant to the oscillator, the latter accumulates the energy. Solve the corresponding classical equation with the following initial conditions: before t = 0 there is no filed, and after t = 0 the oscillator obtains an external periodic pumping filed: ma = -kx +eEsin(wt).

 

If the wave is limited in time, the energy gain is finite. In case of strong external filed we say it is the field who makes the work on increasing the oscillator "internal" energy. Our mechanical equations are sufficient to calculate it. We do not calculate the external wave weakening as redundant.

 

In case of one photon that disappears (a weak external field) we see the energy-momentum transformations: the field gets weakened and the oscillator gets excited.

 

If the source of the wave is far away and neutral (no Coulomb tail exists), the processes of radiation and absortion are very separated in time so it is not a direct interaction between charges but emmission and absorption of real photons. In other words, the source loses the certain energy forever and it does not depend on whether the real photons are absorbed or not and where and when.

 

Unfortunately in perturbative calculations they have admitted such a sloppines (emission and absorpion by itself, self-action) that confuses the real processes and the loose language of perturbative corrections. I am sure we can do without this language.

Edited December 1, 2009 by Bob_for_short

[ajb]

I think I will have to disagree with this. It is virtual as it does correspond (to tree level) to an internal line in a Feynman diagram, as part of the Sun and eye system.

 

I expect, as Severian has stated that in practice this photon is very near on-shell. So by real, we mean very very near on-shell. If the photon were on shell it would be an energy eigenstate and could not disappear and be detected by your eye (or whatever detector you are using).

[Bob_for_short]

I think I will have to disagree with this. It is virtual as it does correspond (to tree level) to an internal line in a Feynman diagram, as part of the Sun and eye system.

 

I expect, as Severian has stated that in practice this photon is very near on-shell. So by real, we mean very very near on-shell. If the photon were on shell it would be an energy eigenstate and could not disappear and be detected by your eye (or whatever detector you are using).

 

I propose you to compare the photon's being near shell due to finite Earth-Sun distance and due to finite wave packet width. You will see that the latter will dominate. So the photon is real.

 

Speaking of the tree-level calculation, consider an electron-proton scattering in QED. The energy momentum remains within the system e+p. Nothing is radiated in this approximation. It is quite different from radiation and absorption. Why to invent some "carriers of interactions" if they are not independent and do not exist? Besides, the real photon cannot be entirely absorbed with a free charge.

 

So calling the Coulomb interaction a "virtual photon" is a bad habit. Calling the real photon a "virtual photon" is also a bad habit. I find this lagnuage extremely misleading. It is not admissible in exact sciences.

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Merged post follows:

Consecutive posts merged

One more argument in favour of photon's being real. While calculating we can have on the paper the "source" of the wave and the "absorber". So we can first find the radiated wave, its energy-momentum which are the source dependent things. Then we inject this solution into the absorber equation.

 

In experiment we may not have an idea what the photon source is. We describe the measured photon flux as a free wave (which is a good approximation to the source-dependent wave solution at far distances). So we disconnect the source and absorber. This makes the radiated field real and propagating. It carries the energy-momentum itself. We can manage it: reflect, guide, irradiate our target with it.

 

It is much more practical to think of it as of real field, don't you find it so?

Edited December 1, 2009 by Bob_for_short
Consecutive posts merged.

[ajb]

I propose you to compare the photon's being near shell due to finite Earth-Sun distance and due to finite wave packet width. You will see that the latter will dominate. So the photon is real.

 

Try it with the lamp on your desk. It makes no difference.

 

As you said yourself dominate. So we are not really in disagreement I think.

 

It is much more practical to think of it as of real field, don't you find it so?

 

Maybe true, but again it does not really make what I or Severian said any less correct.

 

Anyway, there is no reason for us to completely agree on such issues. However, I am pretty sure that most particle theorists will consider any photons observed to be virtual.

 

If virtual is a good name or not is a separate matter.

[Bob_for_short]

The misleading part is not "virtual" but "particle" in the term a "virtual particle" (if we speak of the photon propagator).

 

Well, I think we all expressed our positions. No need to repeat them.

[ajb]

The misleading part is not "virtual" but "particle" in the term a "virtual particle".

 

That I can agree on. The notion of particle in QFT is not obvious and very unclear outside perturbation theory. When doing calculations one is dealing with fields and not particles. The association of fields, states and particles is really a perturbative notion.

 

It seems quite sloppy at times, one may say "electron" but really we mean "Dirac field" etc. I am not sure how much confusion this really causes, especially when dealing with fields on Minkowski space-time.