1. Dec 8, 2014

### Ozgen Eren

Hey guys,

This is my fist post and I am curious enough about this to get a new account. Go on wikipedia for the formula and derivation of the abraham lorentz force: http://en.wikipedia.org/wiki/Abraham–Lorentz_force

Anyway my question is as follows:

Lets consider a charge that I am pulling with a constant force F.
Then obviously by F = ma it has a constant acceleration.
Thus it is radiating some energy.
But according to Abraham-Lorentz force there is no recoil force on the particle, so I spend no energy for radiation.

Because if this was the case I would just take a charge, pull it with a constant acceleration, harvest all the radiating energy, and then absorb the kinetic energy I just applied, and do it again and again. It would be my pretty little infinite energy source. Doesnt that imply any harmonic motion spreads infinite energy?

What is wrong, is it the formula or is it my reasoning?

2. Dec 8, 2014

### Vagn

How are you keeping the (resultant) force constant? If the particle emits a photon then it will recoil, and the resultant force will change over a very short period, giving a non-zero jerk. The Abraham-Lorentz force is a result of the emission, not the cause of it.

3. Dec 8, 2014

### Ozgen Eren

I didn't mean Abraham Lorentz force causes emission, I know its a result of our action upon the charge. I try to think of it as an "emission friction" if thats the right way to put it.

So you suggest that instead of creating an opposite force to balance emission energy, it creates a non-zero jerk which corresponds to a derivative of force on F=ma. That actually makes sense but still the original formula don't seem like supporting it.

If we just apply a constant force to a charge, the derivative of acceleration will be zero as long as F=ma is valid. So no Abraham-Lorentz force, but emission. Thats what Im trying to understand.

4. Dec 8, 2014

### Ehsan Farooq

If you are moving the charge with uniform acceleration ( though inverse square fields produce non uniform acceleration), you are doing some work as dictated by work energy theorm, that is appearing as radiation energy.. I see no paradox

5. Dec 8, 2014

### Vagn

The paper below discusses how to resolve the problem.
http://dx.doi.org/10.1016/0003-4916(60)90105-6 [Broken]

Last edited by a moderator: May 7, 2017
6. Dec 8, 2014

### Ozgen Eren

No the work is done upon the kinetic energy, not upon the radiation. So what you applied = 1/2*m*V*V, what seems to exist : 1/2*m*V*V + Eradiation.

I will take a look on that one, thanks.

7. Dec 8, 2014

### Ozgen Eren

Well I didn't buy the original article but found a free one which refers to it and restates the explanation. Here is the link: