# How are photons involved in electrical flow?

Lets say you have an extremely simple DC circuit with just a 9 volt ideal battery and a superconductive wire connecting one terminal to the other (don't do this on a real battery). Electrons flow from one atom to the next in the wire, gradually reducing the battery's charge. Confusingly, the force carrier for electromagnetism is the photon, not the electron. So, in a simple circuit, where do photons come in? In AP Physics C I learned how electricity creates magnetism and vice versa, but photons were never mentioned.

• You won't learn about how the photon mediates the electromagnetic force unless you take a physics graduate course in quantum field theory. Simply stated, electrons accelerate by emitting photons. – Andrew May 24 at 15:34
• @Andrew an electron flow through a circuit is not caused by photons. Its the other way around. When you accelerate electrons they will radiate photons. – Bill Alsept May 24 at 18:42
• @BillAlsept Does that imply circuits should faintly glow (beyond the glowing caused by resistance) – qazwsx May 24 at 22:29
• @qazwsx everything above absolute zero radiates, even you. It’s hard to stop those electrons. – Bill Alsept May 24 at 23:13

Circuit theory is best described by the Classical laws of people like Ohm, Faraday and Maxwell.

Photons of electromagnetism only come into the picture at the detailed quantum level when explaining things like the photoelectric effect, and even there the main message to take away is energy threshold levels.

Electrons from a battery are pushed along a wire by an excess of negatively-charged electrons behind them, and/or pulled along by an excess of positively-charged protons in front of them. These electrostatic forces are sometimes explained as an exchange of "virtual photons" of a peculiar kind; if energy is exchanged between two particles then the photons are treated as "real", though still peculiar. But that is a handwavy analogy rather than a mathematical model - nobody has been able to define to me a quantum of the electrostatic field (Somebody please do so if you can!). Quantum field theory boils down to using the classical equations in such circumstances as you ask about.

The question assumes an ideal battery. An ideal battery is a current source with fixed voltage ($$V$$) that always satisfies Ohm's Law ($$V=IR$$) by supplying the required current:

$$I = \frac V R$$

NO MATTER WHAT. That is what "Ideal" means.

The question goes on to state that $$R=0$$, which requires infinite current. At this point, any discussion of atoms and electrons breaks down, as the required set-up is non-physical.

Classically, the force on a charge is (Lorentz Force Law):

$$\vec F = q(\vec E+ \vec v \times \vec B)$$

so that the force carriers are the electric ($$\vec E$$) and magnetic ($$\vec B$$) fields, visible only to things with charge ($$q$$).

We know, thanks to Maxwell's equations, that the E and B fields support traveling radiation, EM waves. Moreover, quantum theory (and experiment) show that these waves are quantized into packets with a minimum energy ($$E$$) at fixed frequency ($$\omega$$) given by the Planck relation:

$$E = \hbar\omega$$

In term of momentum, that is written:

$$\vec p = \hbar \vec k$$

The dispersion relation of electromagnetic waves:

$$\omega = c|\vec k|$$

then shows that photons are massless via:

$$m = \frac 1 {c^2}\sqrt{E^2-c^2|p|^2} = 0$$

The calculation of scattering amplitudes of charged particles can be broken down into various terms that are interpreted as an exchange of virtual photons. Virtual photons conserve energy and momentum when exchanged between 2 particles, and as such, are not massless:

$$E^2 -c^2p^2 \lt 0$$

in scattering, while

$$E^2 > c^2p^2 \lt 0$$

in annihilation. They are not real photons, but thanks to the remarkable success of quantum electrodynamics, photons are universally known as the force carrier of electromagnetism. At the end of the day, they are made from electric and magnetic fields, so it is the field that carries force.

Since similar formalisms work for the weak force ($$W^{\pm},Z$$), the color force (gluons), the residual strong force $$(\pi, \omega, \rho, \cdots)$$, phonons in crystals, etc... it is very handy to refer to the particle as the force carrier, though the particle is a quantum of some underlying field (as all particle are, for that matter).

Obviously, that has nothing to do with circuits. (Though I would not be shocked if someone could derive Ohm's law from a quantized field theory).

Circuit theory is a simplification of Maxwell's equations which is a simplification of quantum electrodynamics. The photon is a concept of quantum electrodynamics. It is simplified away when you go from quantum electrodynamics to Maxwell's equations. It does not resurface when going from Maxwell's equations to circuit theory.

Unfortunately, there is a terrible tendency to bring in unnecessary concepts from a more complicated theory into a simpler theory. This temptation should be avoided. Any concept of "photon" that you could introduce in circuit theory would necessarily be a severely crippled version of actual photons. Such a crippled version of photons would not provide any future benefit for learning quantum electrodynamics (in fact it would need to be unlearned), nor would it provide any immediate benefit for learning circuit theory. It should therefore not be attempted.

If something is worth doing at all then it is worth doing right. It is not worth teaching photons right in a circuit theory class, so it is not worth doing at all.

In the simple circuit you describe, there is only an electrostatic field applied to a conductor, so there is no variation with time of the electric field or of the magnetic field.

In that case, according to Maxwell's equations, no electromagnetic field is emitted, therefore no photon intervenes.