# A paradox in electromagnetic theory

1. Aug 31, 2007

### lugita15

In Volume II Chapter 28 of his Lectures on Physics, Feynman describes a fundamental inconsistency in classical electromagnetic theory, concerning electromagnetic mass:
Consider an electron in which all of the charge q is uniformly distributed on the surface of a sphere of radius a. The magnitude E of the electric field at a distance r from the center of the electron is $$E=\frac{q}{4\pi\epsilon_{0}r^{2}}$$and the density u of its field energy is given by
$$u=\frac{\epsilon_{0}E^{2}}{2}=\frac{q^{2}}{32\pi^{2}\epsilon_{0}r^{4}}$$
In order to find the total energy in the electric field produced by the electron, we must integrate this density over all space:
$$U=\int^{\infty}_{a}\frac{q^{2}}{8\pi\epsilon_{0}r^{2}}dr=\frac{1}{2}\frac{q^{2}}{4\pi\epsilon_{0}}\frac{1}{a}=\frac{1}{2}\frac{e^{2}}{a}$$
where $$e^{2}=\frac{q^{2}}{4\pi\epsilon_{0}}$$
Now according to the theory of relativity, the energy U of a particle is always $$mc^{2}$$. So according to relativity, the mass of an electron due to its own electric field is
$$m=\frac{1}{2}\frac{e^{2}}{ac^{2}}$$
However, the mass of an electron due to its own field can also be calculated another way. Consider an electron moving at a speed v<<c. The momentum density $$\vec{g}$$ of its electric field is given by
$$\vec{g}=\epsilon_{0}\vec{E}\times\vec{B}$$ where $$\vec{B}=\vec{v}\times\vec{E}/c^{2}$$.
In order to find the total momentum in the electric field produced by the electron, we must integrate this density over all of space. This integration is quite messy, but the result is $$\vec{p}=\frac{2}{3}\frac{e^{2}}{ac^{2}}\vec{v}$$.
At low velocities, $$\vec{p}=m\vec{v}$$. So according to electromagnetic theory, the mass of an electron due to its own electric field is
$$m=\frac{2}{3}\frac{e^{2}}{ac^{2}}$$
So electromagnetic theory and special relativity give contradictory results for the mass of an electron due to its own electron field. What is the resolution to this apparent paradox?

Any help would be greatly appreciated.

Last edited: Aug 31, 2007
2. Aug 31, 2007

### meopemuk

There are lot of paradoxes in classical electromagnetic theory. You can find dozens of references (including those relevant to the "electromagnetic mass" controversy that you mentioned) in

G. Spavieri, G. T. Gillies, "Fundamental tests of electrodynamic theories: Conceptual investigations of the Trouton-Noble and hidden momentum effects", Nuovo Cim. B 118 (2003), 205.

In my personal opinion, many of these paradoxes simply cannot be resolved within Maxwell's theory. We need a better theory of electromagnetism.

Eugene.

Last edited: Aug 31, 2007
3. Aug 31, 2007

### lugita15

Is that paper online? If so, what is the link?

4. Aug 31, 2007

### meopemuk

5. Aug 31, 2007

### cesiumfrog

A large number of these "paradoxes" have turned out to be subtle coordinate transformation errors.

6. Aug 31, 2007

### lugita15

Is there any error in coordinate transformations in the paradox discussed in my post? For that matter, were there any coordinate transformations in my post?

7. Aug 31, 2007

### lugita15

8. Aug 31, 2007

### cesiumfrog

I'm not claiming to resolve this specific paradox, just commenting on other examples I've studied (here, for example, is one that Griffiths published earlier work on - having obtained a remarkably similar difference in leading factor). In your post you needed to transform the electric and magnetic fields into a moving frame of coordinates.

9. Aug 31, 2007

### meopemuk

So bad. You can try to search for other articles of these authors on Google Scholar. Perhaps, some of them are more readily available, e.g., as preprints on the arXiv. I read this paper in the physics library. I am lucky to have a good research university nearby.

I've seen this "electromagnetic mass" controversy mentioned in many places. However, I don't think anybody has a logical resolution of this paradox within classical electrodynamics.

My other point was that there are many more electromagnetic paradoxes not mentioned in most textbooks. There are also recent experiments, which seem to contradict Maxwell's electrodynamics:

A. L. Kholmetskii, et al. "Experimental evidence of non-applicability of the standard retardation condition to bound magnetic fields and on new generalized Biot-Savart law",
J. Appl. Phys. 101 (2007), 023532; http://www.arxiv.org/abs/physics/0601084

N. Graneau, T. Phipps Jr., D. Roscoe, "An experimental confirmation of longitudinal electromagnetic forces" Eur. Phys. J. D 15 (2001), 87

Eugene.

10. Sep 1, 2007

### lugita15

Yes, but we're dealing with speed much smaller than the speed of light, so Lorentz transformations of E and B don't really matter.

11. Sep 1, 2007

### lugita15

I just wanted to give Feynman's derivation of $$\vec{p}$$ by integrating $$\vec{g}$$ over all of space, since I didn't include it in my original post:

12. Sep 1, 2007

### meopemuk

Suppose that the electron is first at rest. Suppose also that the electric field makes contribution $U/c^2$ to the electron's mass. Now, let the electron accelerate under the influence of some external force. If we accept that perturbations in the field propagate at the speed of light, then the shape of the field around the electron changes with time. Does it mean that electron's mass is changing? Does electron's mass depend on its acceleration? It doesn't make sense to me.

Eugene.

13. Sep 2, 2007

### lugita15

Although the electromagnetic energy density u will change as the electron accelerates, when you integrate the density over all of space, the total energy U in the electric field stays the same. So the electron's mass doesn't depend on its acceleration.

14. Sep 2, 2007

### meopemuk

Can you prove that?

Eugene.

15. Sep 2, 2007

### pervect

Staff Emeritus
I think the point probably is that it's inconsistent to view the electron as some sort of small, charged rigid sphere, which was right in the initial problem statement.

Rigid spheres don't exist in relativity, as is well known.

One would like to argue that if the sphere is small enough, one could avoid this difficulty. If it is a small sphere, deformations of this sphere shouldn't matter much. But the self-energy of a small sphere approaches infinity as the sphere shrinks to a point. This means that one is attempting to justify small changes to an infinite quantity, an argument that is very suspect.

That's just my take, I haven't read on this in detail. This concept apparently has a lot of history. If you happen to be able to find "Concepts of mass in classical and Modern Physics" by Max Jammer, he has a chapter devoted to the idea of "electromagnetic mass" which is related to this problem. So there is quite a lot of literature on this to plow through, something I haven't done.

You'll probably need to get access to a university library to get to the better papers. Usually the biggest problem is getting parking on campus :-). Printing the papers also might get expensive (you can ask the librarian about alternatives available if any, I know of at least one university that attempts to put USB ports on the library computers so you can put a flash drive onto the library computers).

16. Sep 2, 2007

### meopemuk

There is another well-known paradox associated with assigning such properties as energy and momentum to electromagnetic fields. Electrons have spin, therefore there is a magnetic field B even around a stationary electron. One can find the (Poynting) vector of the field's momentum by integrating the vector product $\mathbf{E} \times \mathbf{B}$ over entire space. This integral is non-zero. This means that the linear momentum of the field is non-zero. Shall we then conclude that the electron at rest has a non-zero momentum?

Eugene.

17. Sep 2, 2007

### cesiumfrog

That seems a bit of a red herring.

Rigidity is irrelevant to comparing a static charge distribution between stationary and constant-velocity reference frames.

Whether or not it models an electron is irrelevant to whether or not different methods for determining the invariant mass of a given static classical charge distribution should be expected to give the same results.

18. Sep 6, 2007

### pervect

Staff Emeritus
I don't think it's as much of a red herring as you might think, since the invariant mass of a system of nonzero volume is only invariant if the system is a closed system. If you take a small piece of a bigger system, that small piece does not in general have a 4-momentum that transforms as a 4-vector. The correct approach is to divide the system into pieces of infinitesimial volume, which will transform properly - this approach winds up giving you the stress-energy tensor (which however transforms as a rank 2 tensor, not a 4-vector).

We had a thread on the energy in the field not too long ago, in fact, where I pointed out the need to include the energy in the charges as well as the energy in the field:

where I said:

So I think the missing piece here is that one needs to consider more than just the field, and one will "find" the missing energy and momentum. When one has an isolated system, the total energy and momentum does transform as a 4-vector. The problem here, I think, is that the invariant "mass" of the field isn't coming out right because it isn't a 4-vector, and it isn't a 4-vector because a field with a source isn't a closed system unless you consider the energy in the source.

Last edited: Sep 6, 2007
19. Sep 9, 2007

### lugita15

Could you please explain, in simple terms, what the stress-energy tensor is?

20. Sep 9, 2007

### lugita15

Yes, a particle with spin has not only linear momentum, but angular momentum as well. But this isn't as counterintuitive as it may seem at first glance. In quantum mechanics, no particle is really "moving," in the sense that it is going on a continuous path from point A to point B. So momentum is no longer a property of "motion."