Does a moving charge have additional (magnetic) energy

My understanding so far:

A moving charge produces a magnetic field, B, in an analogous way to a current produces one.

A magnetic field has an energy density which is proportional to B squared.

My question is: Does a charged particle in motion have an additional energy associated with it's motion due to this magnetic field?

I suspect the answer is no, for a number of reasons, but want to check.

Additional work would then have to be done to accelerate a charged particle. This would be work against a force, and I can see no mechanism which could produce that force (other than just 'the increase in field energy').

If the charge is a point charge then the energy density would diverge close to that charge. (The volume integral would also diverge). So moving a charge would require infinite energy. I'm guessing this is similar to why you can't define the electric field energy of a single point charge, it doesn't interact with anything as it dwindles into infinity.

• Well you just point to the problem of classical electrodynamics, it has an inconsistency (the inifite energy densities of point charges, but non-point charges are disallowed by relativity). But indeed a charged particle requires additional force to be accelerated, the force is caused by the momentum carried by the emitted radiation. Commented May 18, 2015 at 16:59
• @SebastianRiese: Is this problem solved in quantum mechanics? Thank you. Commented May 18, 2015 at 18:26
• As far as I know not properly (the electron space eigenstates are still $\delta$-like). But I don't remember exactly, I know there is something on this in the Feynman lectures. Commented May 18, 2015 at 18:55

I will try to elaborate the self-action in the standard classical electrodynamic theory.

An accelerated charge radiates electromagnetic waves-that is energy and momentum. The radiated energy is removed from the the kinetic energy of the point charged particle. So under the influence of a particular force, a charged particle seems to accelerates lesser than a non-charged.

One can prove that the total radiated power is equal to: $$P={μ_0 q^2 a^2 \over 6πc}$$ where: a:acceleration,q:charge,c:velocity of light.

The energy the particle loses because of the self-force is: $$\int_{t_1}^{t_2} F_r \cdot v dt= -{μ_0 q^2 \over 6πc} \int_{t_1}^{t_2} a^2 dt$$ From here, one can prove that: $$F_r = {μ_0 q^2 \over 6πc } {da \over dt}$$ where $F_r$ is of course the self-force. Note that this is not a rigorous proof but rather a simplification.

A strange result is this: Assume no external forces on the particle. Then from the second law of Newton: $$F_r =ma--> a(t)=a_0 e^{t/τ}$$ where $τ={m_0 q^2 \over 6πmc}$ That means the acceleration is self-increacing. To solve this try and put $a_0=0$. One can prove then that bigger "problems" arise, more or less the solutions then give a pre-acceleration, an acceleration that is acted on the particle before a force is being applied to it. And the thing is, both of these issues cannot solved together.

I hope this helps you.

Note: As commented, this is a problem indeed for Classical electrodynamics, but I think the problem still arises in quantum mechanics and quantum field theory. Maybe someone with more knowledge on the subject of QFT could elaborate more.

• Commented May 18, 2015 at 19:46
• You mentioned radiated power, which is the time rate that energy is supplied to the radiation (acceleration) fields, but energy is also put into the Schott (velocity) fields and so both contribute to self-force. And field energy is quadratic therefore not additive. So there is no simple relationship between energy supplied to radiation and self force. Self force would have to balance all the energy and you are only looking at the energy that escapes to surfaces very very far away. If you wanted to introduce multiple self forces then what you write might work, but as written it is misleading. Commented Jun 3, 2015 at 18:11
• Hi @Timaeus and thanks for the comment. First, I didn't posted the entire mathematical proof. Beyond, that, from what I've read, the velocity fields doesn't contribute to the radiation, and my answer doesn't indeed contain the Lienard generalization for the Larmor equation. Please, if you can elaborate on your comment. It will be much appreciated. Thanks. Commented Jun 4, 2015 at 16:35
• The term self force sounds generic like a total self force or a unique self force. If you wanted to talk about the radiation reaction self force then it'd be more clear that it is one self force amongst possibly many self forces. The velocity field doesn't contribute to the radiation but it does take energy and momentum and field energy and field momentum are not additive you can not add radiation field energy and velocity field energy to get total field energy. When energy is supplied it goes to kinetic and to changing the total field energy not just changing the radiation fields. Commented Jun 4, 2015 at 19:37
• @Timaeus So, there are other self-forces and the velocity field could be one of them? Could you suggest some links or articles for further reading? Thank you. Commented Jun 4, 2015 at 19:44

Does a charged particle in motion have an additional energy associated with it's motion due to this magnetic field?

For extended particles with finite charge density, Poynting theorem is valid inside the particle and thus can be used to derive expression for EM energy. This energy is only partially in the particle, the rest is in the surrounding space.

Since some energy is in the particle as well, it makes sense to say the particle has additional contribution due to EM energy inside it.

Is this (positive) contribution of energy higher when the particle moves? I do not see any easy way to find out, other than directly comparing the two energies.

To find out these two numbers, you may do the following calculation. Calculate Poynting energy in the region occupied by a uniformly charged sphere at rest. Then calculate the same integral (with magnetic contribution) in other frame where the sphere moves with constant velocity. The particle will have shape of oblate ellipsoid and the fields will change according to the Lorentz transformation, so the answer is not immediately obvious and probably will require some calculations.

This path seems demanding but should be manageable, if not analytically then with help of computer. The disadvantage of this approach is that it depends on specific model of charged distribution. Another particle which does not have uniform charge distribution will need another calculation.

For point particles, Poynting theorem is not valid at points where the particles are and thus cannot be used to derive expression for EM energy. Another work-energy theorem can be derived for point particles that have no self-interaction. Theory of charged point particles has been published many times, for example:

J. Frenkel, Zur Elektrodynamik punktfoermiger Elektronen, Zeits. f. Phys., 32, (1925), p. 518-534. http://dx.doi.org/10.1007/BF01331692