# Light from non accelerating charge?

1. Oct 9, 2009

### gonchenshi

A point charge moving at constant velocity causes unconstant changes in the electric field. Won't that generate an electromagnetic wave?

2. Oct 10, 2009

### Count Iblis

Due to Lorentz invariance. The laws of physics work in the same way when you transform to a moving frame. A point charge in rest does not radiate electromagnetic waves, therefore a moving point charge cannot radiate electromagnetic waves either.

3. Oct 10, 2009

### Bob_for_short

It generates (is a source of) a variable electromagnetic field indeed but it is "attached" to the charge if the charge moves in vacuum.

When a charge moves in a transparent media, the resulting field may include radiated field if the charge velocity exceeds the light velocity c/n. Such a radiation is called Cherenkov's one. It is a collective effect.

Last edited: Oct 10, 2009
4. Oct 10, 2009

### gonchenshi

Well, what I am thinking is that this may be a problem for the principle of relativity, so is there an explanation that doesn't use it?

5. Oct 10, 2009

### gonchenshi

Can you explain the attached part in more detail, please? (Maths is ok)

6. Oct 10, 2009

### Count Iblis

Electromagnetism is intimately related to special relativity. Now, it is true that the computation from first principles of the radiation emitted by a point charge is a highly non-trivial problem. The derivation given in text-books (e.g. the book by Jackson) is heuristic at best.

To see the problem, just think about where the energy and momentum of the electromagnetic radiation comes from. Of course, it must come from the point charge. So, the momentum of the charge changes due to momentum that is carried away in the form of electromagnetic radiation. The rate of momentum change of a charge is the force and that is always given by the Lorentz force if there are only electromagnetic interactions involved. In this case there are no external electromagnetic fields, all the fields are generated by the charge itself andin the Lorentz force formula we usually only include the external fields.

So, the conclusion has to be that the usual Lorentz force formula is wrong and that this has to be corrected by somehow including the interaction of the electromagnetic fields of the charge itself. Many attempts were made to do this in a consistent way, but until recently all the approaches were problematic.

The formula given in most textbooks, the so-called "Abraham–Lorentz-Dirac force" is known to be fundamentally flawed, but it can be used to do computations if you put in some ad-hoc prescriptions (discard runaway solutions and ignore pre-acceleration effects).

A fully rigorous solution was obtained only this year, see here:

http://arxiv.org/abs/0905.2391

From this result you can see that the Abraham–Lorentz-Dirac force formula (plus ad hoc prescriptions) is not an exact solution.

7. Oct 10, 2009

### Phrak

This may depend on what we mean by electromagnetic waves. As you say, take the static solution, and transform coordinate systems where there are both changing electric and magnetic fields. Now take Fourier transforms in chosen coordinates. We'll need solutions for E = E(x-ct) and B=(x-ct) if we want propagating waves. Take the parts where E and B are in the right proportions, and throw out the rest. Does Jackson have anything to say about this?

8. Oct 11, 2009

### Bob_for_short

When you have only one charge, it is difficult to separate the "proper" field from the radiated because they both come in superposition. It is much easier to separate them if there are two charges of the opposite signs. Then at large distances the dipole field fades out quickly and the radiated becomes dominant. If you take a constant dipole and look at it from a moving RF, it will not have radiated tail, only the transformed dipole EM filed that is rather weak.

Last edited: Oct 11, 2009