Attractive force from gauge particle exchange

1. Dec 5, 2013

tommybee

Photons are referred to as the field quanta for the electromagnetic field.

I would like to understand how the exchange of field quanta such as photons can create an attractive force.

For example, I can see that an electron could repel another electron by hitting it with photons. (radiation pressure ?)

How can the exchange of photons create an attractive force, though ?

For example, how does the exchange of photons cause the attraction of an electron to a proton ? If a proton emits photons that hit an electron, then Newton's third law would suggest that the electron and the proton should repel each other.

Please forgive my naivete if this type of simplistic "billiard ball" physics does no apply to elementary particles.

2. Dec 5, 2013

3. Dec 5, 2013

dauto

These are exchange of virtual particles. Virtual particles are allowed to bend some rules which real particles must follow. You can for instance have a photon moving in one direction with a momentum in a different direction.

4. Dec 5, 2013

tommybee

Thanks dauto,

I had read that these field quanta are virtual particles - distinct from standard photons, etc. I would appreciate a link or other reference that explains virtual particles without oversimplification. In this case the oversimplification of some popular physics articles has lead to misconceptions such as in my original question above.

Have these virtual photons been identified (observed) experimentally ?

5. Dec 5, 2013

dauto

Virtual particles cannot be observed directly but their effects can. For instance the lamb shift ix explainable as a consequence of the emission and subsequent absorption of virtual photons.

6. Dec 8, 2013

tom.stoer

A simple explanation for the attractive force in QED works w/o virtual particles. A sketch of the general idea is

fix the gauge

$A_0=0$

solve Gauß law constraint

$\nabla E = \rho$

and replace the term

$A_0\,\rho$

in the Hamiltonian with the solution for the Gauß law:

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

This shows that the Coulomb potential does not depend on "virtual particles.

7. Dec 13, 2013

lpetrich

Maybe not explicitly, but it seems to me that the 1/|x-y| term is essentially the propagator of a virtual photon.

8. Dec 13, 2013

tom.stoer

If you like you can interpret it that way, but it's not necessary; all physical (transversal) photon d.o.f. contribute to other terms in the Hamiltonian

9. Dec 13, 2013

Chenkb

I think A.Zee's book "Quantum field theory in a nutshell", gives a good explanation, just in the first chapter.