# If electromagnetic fields give charge to particles, do photons carry charge?

As I understand these two statements:

1. An electromagnetic field gives particles charge
2. A photon is a quantum of electromagnetic field

It must mean that a photon carries charge. But I guess it isn't true. Why not?

• Your assumption 1) is backwards. It is the charge that creates the electric field around it. Charge is an intrinsic property of matter, like mass it can be defined in certain units; correspondingly units are derived for the fields created around a charge which depend on the definition of charge units. – anna v Jul 7 '12 at 4:00
• No the photon doesn't carry charge. It doesn't emit an electromagnetic field. Why would you find this surprising? The qualitative statements you give do not imply the statement you say they imply. – Ron Maimon Jul 7 '12 at 6:03

Remember, electromagnetic field is a distribution of electromagnetic force, not charge. Photon bosons are quantum of this field. So, they are force carriers.. not charge carriers. Only force is exchanged with these messenger particles. Based on this interaction, we determine charge of electrons etc involved. That's it!

The electromagentic field is the fundamental entity.

Charges (electrons, positrons, nuclei) are accompanied by (''emit'') an electromagnetic field - a soft virtual photon cloud in terms of QED.

Photons are elementary excitations of the quantum electromagnetic field. They do not carry charge.

It's actually quite simple. If the electromagnetic field (i.e. photons) were charged, that would imply non-linear self-interactions such as those occurring for the gluon octet that mediates the strong force. The gluonic Lagrangian takes the form $$\mathcal{L}_\text{SU(3)} = -\frac{1}{2} \, \mathrm{tr}(F^2) = -\frac{1}{4} \, \sum_{a = 1}^8 F_{\mu\nu}^a \, F_a^{\mu\nu},$$ where $$F_{\mu\nu}^a = \partial_\mu A\nu^a - \partial_\nu A\mu^a - g \, f^{a}{}_{bc} \, [A_\mu^b,A_\nu^c].$$ Here, $A_\mu^a$ denotes the gluon fields, $F_{\mu\nu}^a$ their field-strength tensors, $\mu$, $\nu$ are spacetime indices, and $a$, $b$, $c$ color indices. $g$ is the gauge coupling and $f^{abc}$ the structure constants of the $su(3)$ Lie algebra associated with the $SU(3)$ gauge group.

Looking at $F^2$, we see immediately that it contains quartic self-interactions among the gluon fields due to the commutator in the field strength. And here comes the crucial point: Since $U(1)$ (the gauge group of electromagnetism), unlike $SU(3)$, is abelian, i.e. has trivial group structure with all commutators vanishing, the structure constants of the Lie algebra associated with $U(1)$ are all zero. This suppresses the quartic interactions in the Lagrangian of electromagnetism and prohibits electrically charged photons.

The previous answers are based on classical electromagnetism. If we consider this from a quantum-electrodynamic (QED) standpoint, it's not so simple. In QED, the force between charged fermions is exclusively conveyed by an uncharged boson quanta, photons. This changes the classical problem to "how do the charges know which electromagnetic force to generate, an attracting or repelling one?" I may be wrong about this, but the QED argument I believe is much more direct and leaves the original question open. In fact, I have contributed to this question, because for many years I have been unable to answer it.

• "the force between charged fermions is exclusively conveyed by an uncharged boson quanta, photons". Not exclusively. In QED the photon goes into virtual loops of e+ e- ( and higher order virtual loops). The crossection and trajectory for the two cases, is different. This is an experimental fact : like charges repel, unlike attract. This experimental fact is included in the QFTheoretical formulation, which gives a different crossection and trajectory for e- nearing e+ than another e- – anna v Nov 16 '12 at 19:54