# Is a static EM field made of photons? Do they move at c?

1. May 12, 2013

### JohnRood

So i am taking the MOOC of Galitski et al on Exploring Quantum Physics. I like it.

But i keep running into some difficulty with my understanding of what it means to quantize a static EM (or E or M) field: Is the field made up of photons? If so, are they somehow moving at the speed of light? (!) If not, what does it mean that they are (much!) slower? A particular variant of this question which would be: What is light in the brain? [end] has particular interest to me.

So i have a suspicion we are somehow dealing with a set of (essentially harmonic?) oscillators here? So they have defined / definable photon energies etc?

With regard to the level of math used in any possible reply, i am a mathematician and will attempt to deal with any level of math you can throw at me.

Hmmm. This may be completely irrelevant, but i might regard answers which tell me something is infinite as being dubious. Infinity (even only countable infinity) is really large.

2. May 12, 2013

### tom.stoer

The concept of photon is partially gauge dependent.

A natural way to quantize the el.-mag. field is to separate two physical photons (transversal polarizations) and two unphysical gauge degrees of freedom. One of the latter can be gauged away. The second one can be integrated out, and what remains is a Coulomb interaction term between (fermionic) charges.

So in this picture the static part of the el.-mag. field does not depend on dynamical degrees of freedom, i.e. it does not consist of photons.

3. May 12, 2013

### JohnRood

So the "Coulombic Interaction" is mediated (?) by exchange of photons (force carrying bosons)? So somehow this is tantamount to saying: There is no such thing as a static [E / M / EM] field?

This is what i get out of your answer, but i am most definitely naive on this topic.

Sort of seems to have "defined the problem away". And, frankly, this kind of answer would not really surprise me. But still seems somehow unsatisfying ... ?

And, again, it is perfectly possible that i have not understood your answer (although it does look quite interesting).

4. May 12, 2013

### tom.stoer

No. The pure Coulomb interaction of static charges exists w/o photons; (virtual) photons are required only to describe corrections to the static scenario.

5. May 12, 2013

### JohnRood

Wow. This is "blowing my mind". IF something even moves, there must be exchange of force carriers, afaik. My feeling is that this IS what "quantization" ultimately buys you. So, i am now led to a weirder question: What good is a "static field", then?

Certainly it has (plenty of) energy stored in it. But it is NOT made up of any fancy stuff like oscillating photons etc. But as soon as it [= its energy, right?] acts on anything, this occurs by (virtual??) photons? How about real photons? This is getting complicated / interesting ... ?

6. May 12, 2013

### tom.stoer

Have a look at the derivation of the QED Hamiltonian and especially 2.117 and 2.118 in http://sophia.dtp.fmph.uniba.sk/~peterp/QED_A.pdf

The last term in 2.118 is the Coulomb interaction; the other terms contain physical i.e. transversal photons.

7. May 12, 2013

### VantagePoint72

This is not correct. The classical Coulomb potential is the effective potential describing the first order QED Feynman diagrams; i.e. the exchange of one virtual photon. See, e.g., section 4.8 of Peskin and Schroeder. There is no static interaction without photons.

8. May 12, 2013

### JohnRood

OK, very good, a meaty article link. Looks very nice. I will attempt to decipher / grok it. Tyvm.

9. May 12, 2013

### tom.stoer

This is wrong!

Peskin & Schroeder is not a very good reference for quantum field theory. The derivation of the physical QED (or even QCD) Hamiltonian w/o using perturbation theory (!) shows exactly that the Coulomb interaction does not depend on the physical photon d.o.f.

10. May 12, 2013

### JohnRood

Hmmm, very interesting. I have a copy of Peskin and Schroeder, which i will also look at. My (very naive) bias is with the answer of LastOne ... .

11. May 12, 2013

### JohnRood

Sorry "d. o. f." will be "degree of freedom" ?

12. May 12, 2013

### VantagePoint72

The entire concept of virtual photons is tied with perturbation theory. Saying "(virtual) photons are required only to describe corrections to the static scenario" and claiming you have arrived at this result without perturbation theory is incoherent.

That said, I'm happy to investigate this more. However, given Peskin and Schroeder is the standard reference around the world for teaching QFT, I am, for the moment, inclined to trust it over one person on an internet forum.

13. May 12, 2013

### tom.stoer

$$H_\text{Coul} \sim \int d^3x \int d^3y \frac{\rho(x)\,\rho(y)}{|x-y|}$$

This interaction, which results from integrating out the unphysical gauge degrees of freedom, does not contain the physical photon field (there are other interaction terms, of course).

14. May 12, 2013

### tom.stoer

The exact derivation does neither use perturbation theory nor virtual photons; it is valid non-perturbatively.

I do not say that.

15. May 12, 2013

### VantagePoint72

We are talking about virtual photons, not physical photons. Of course, the whole notion of 'virtual gauge bosons as force carriers' is questionable, as exact results like you are describing highlight. I'm firmly in the camp of Feynman diagrams being pure computational aids, and not reflecting actual processes.

However, you chose to invoke the language of virtual photons when you said, "The pure Coulomb interaction of static charges exists w/o photons; (virtual) photons are required only to describe corrections to the static scenario." This is a mixing of metaphors. If you want to work in the 'virtual photon exchange' picture, then the first-order Coulomb interaction is described by the exchange of one virtual photon. You cannot use half of one pedagogy and half off another and expect a sensible picture.

16. May 12, 2013

### tom.stoer

No. It is simply a different starting point.

The interaction term I have written down does not contain virtual photons b/c it is valid non-perturbatively, virtual photons are not necessary at this stage. Of course one can use it as a starting point for perturbation theory, but the Coulomb term will not generate virtual photons (propagators). They will be generated by other terms.

17. May 12, 2013

### VantagePoint72

Then rather than an off-hand dismissal, I welcome an explanation of what is wrong with Peskin and Schroeder's derivation. They begin with the QED Lagrangian too. The calculation implies that this:
is simply not correct. If you use it as a starting point for perturbation theory, it is absolutely explicit in P&S's calculation that the the Coulomb term generates the first order Feynman diagrams. Unless you can explain what is wrong with this calculation, your quoted statement above is wrong.

18. May 12, 2013

### tom.stoer

19. May 12, 2013

### JohnRood

Another nice link! (And i have a copy of Weinberg--all 3 ... not so easy to read ...)

20. May 12, 2013

### VantagePoint72

That does not answer the question. You are giving alternative derivations (that I am already familiar with), not explaining what is wrong with the one I am citing. There is no conflict from the fact that there are many way of arriving at the same conclusion. The problem, as I have said, is that the description in terms of virtual photons is inherently a perturbative one. As soon as you wish to start talking about virtual photons and perturbative corrections, then have committed to one particular picture: where the exact Coulomb interaction you are discussing is expanded via Wick's theorem. In that picture, the first order interaction corresponds to the classical EM Lagrangian.

As it stands, your statements are at odds with a standard reference in the field. Since you have repeatedly refused to respond directly to the standard reference, there is nothing further to say. If you decide to revisit this at some point when are more willing to say specifically why you believe Peskin and Schroeder's analysis is incorrect, I will be happy to read your thoughts. Until then, I will maintain my original position that you are mixing metaphors.