# The magnetic field of a photon

1. Nov 22, 2009

### Identity

The magnetic field of a moving charge can be found using the Biot-Savart Law, but what about the magnetic field of a photon?
Photons are 'electromagnetic radiation', so technically they should have a magnetic field, right? How would you find it? (And as an additional question, how do you find the electric field?)
Thanks :)

2. Nov 22, 2009

### Phrak

The electromagnetic fields associated with a photon in vacuum are a solution to the Laplace equation in Minkowski space, including boudry conditions,

$$\frac{\partial^2 E}{\partial x^2} + \frac{\partial^2 E}{\partial y^2} + \frac{\partial^2 E}{\partial z^2} - \frac{\partial^2 E}{\partial (ct)^2} =0$$

for all coordinates;

$$E = E(x,y,z,t) \ .$$

The same pertains to the magnetic field.

Last edited: Nov 22, 2009
3. Nov 22, 2009

### Identity

Thanks Phrak, I don't really understand how to interpret that equation but I guess I just need to do advanced physics first

4. Nov 23, 2009

### diazona

Here's the thing: a photon doesn't have a magnetic field in the same way that something like an electron "has" a magnetic field. The electron produces a magnetic field because it's a charged particle that is moving, but a photon is not a charged particle, so it doesn't produce any magnetic field of its own.

In fact, the photon is a magnetic field - well, maybe more like a "piece" of a magnetic field. When that electron (or any charged particle) moves around in an erratic fashion, it produces waves in the magnetic field that move away at the speed of light, kind of like how a duck splashing around in the water produces ripples/waves that move away from it. The waves are the photons, roughly speaking. (At least that's the best explanation I can think of without invoking quantum electrodynamics)

5. Nov 23, 2009

### Identity

Ah ok, that's fascinating :)

6. Nov 23, 2009

### hiyok

I think it is impossible to talk about the magnetic field of a photon without taking into account QED (quantum electrodynamics). Actually, a photon does not have a definite magnetic (and electric) field because of a broad uncertainty principle. This principle states that, one cannot have simultaneously a definite electromagnetic field and a single photon with definite frequency. Formally, this goes under the name 'number-phase duality'. Due to this, the mean electromagnetic field of a photon turns out to be zero! But the field fluctuates and gives energy to the photon.

Basically, this uncertainty concerning light should be related to the conventional uncertainty concerning the position and momentum of a particle, as the electromagnetic field is sourced by a charge, whose motion determines the former.

Pls see Heisenberg's book for details.

7. Nov 23, 2009

### Staff: Mentor

The "field of a photon" is not the classical electromagnetic field described by E and B. It's the 4-vector potential $A_{\mu}$ whose components are the electric potential $\phi$ and the magnetic vector potential $\vec A$. When you quantize this field, you get photons.

8. Nov 23, 2009

### Phrak

I swear, this thread was moved overnite to the quantum physics folder, wasn't it?

In the original folder it would be appropriate to argue that the field of a photon would be equally well described by A or F (where F=dA), or even d*F. But, by slight of hand, and folder relocation the laws of physics have changed!

9. Nov 23, 2009

### Staff: Mentor

Well, photons are quintessentially quantum objects, aren't they?

10. Nov 23, 2009

### Phrak

But shouldn't you mean quantum waves? :uhh:

11. Nov 23, 2009

### Staff: Mentor

I prefer to dodge the "wave or particle" question.

12. Nov 23, 2009

### Neo_Anderson

A photon is a transverse wave consisting of a magnetic component (H) and an electric component (E). A light polarizer will remove one or the other component, but not both. So when you say, "The magnetic field of a photon," what you're really referring to is the magnetic component of the transverse wave.

13. Nov 23, 2009

### hiyok

NOOOOOOOP. pls see jtbell's posts for explanation

14. Nov 24, 2009

### Frame Dragger

Exactly. You can't talk about the quanta of the magnetic field without first adressing the basic theory in which the photon's properties are described. Unlike other gauge bosons, the photon is massless (and chargless).

15. Nov 24, 2009

### Phrak

Well, beyond the theoretical interpretations, classically, a propagating electric field will not exist without an accompanying propagating magnetic field. A polarizer acts to select one direction for the electric field. The magnetic field is not 'stripped away', which is what you seem to be saying. I appologize if I've misinterpretted.

16. Nov 24, 2009

### Neo_Anderson

Well some guys here, for reasons beyond my considerable comprehension, throw themselves into a frenzy when I said that the photon is a transverse EM wave with an E-component and a perpendicular H-component, in-line with Maxwell's equations. For some reason, they--like the ape in its cage after being prodded by the branch--threw a fit, saying, "The photon is quantized! The photon is quantized!" I don't care if it's quantized or not (it is, of course). What I do care about is the fact that the wave (not particle) component obeys Maxwell's equations. And Maxwell's equations explicitly address both an E-component to light as well as the H-component, both anti-parallel to each other, as the wave propagates at speed c.
I was referring to the E-ray and the O-ray, which should clarify any misunderstanding in post #12.

Last edited: Nov 24, 2009
17. Nov 25, 2009

### Frame Dragger

For my part, I focused on your conclusion that there is an equivalence between the OP's question and your answer that the magnetic componant of a propogating light wave IS the magnetic field of the photon. Remember, he was asking explicitly if a photon's magnetic field could be found. An answer that goes on to describe a photon as having a magnetic componant could be misleading. After all, the question isn't if the photon has a magnetic componant, but if it has a measurable magnetic field.