Infinite Self-Energy of a Charged Point Particle

1. Feb 17, 2009

Himanshu

Recently I became acquainted with the problem of Infinite Self-Energy of a Charged Point Particle as described by both classical and quantum theory. Infinities of this kind certainly hint at the inconsistency and incompleteness of the theory itself. I was speculating that infinity is arising dew to the dimensionless nature of the particle. But in the context of string theory this issue could have been resolved as we now do no longer have to deal with point particles. I was wondering if this issue has been resolved by other theories.

2. Feb 17, 2009

MathematicalPhysicist

How are the singularities get resolved, I thought singularities are inherent in every mathematical theory of physical phenomena, is it not?

3. Feb 17, 2009

cesiumfrog

Himanshu, what references were you reading? I understood that particular classical infinity becomes zero in quantum mechanics.

4. Feb 17, 2009

Stingray

You can "renormalize" in classical physics just like you do in quantum physics. Of course, there are no problems at all with classical continuum mechanics.

5. Feb 19, 2009

Himanshu

I am refering to Griffiths- Introduction to Electrodynamics. The book says that the infinite energy of a point charge is a reccuring source of embarassment for electromagnetic theory afflicting quantum version as well as the classical. I have no Idea about how the self-energy becomes infinite in the context of QM, but in classical EM theory its a straightforward calculation.

6. Feb 26, 2009

cesiumfrog

Do you have any example of a problem that can be solved classically by renormalisation?

7. Feb 26, 2009

Stingray

The typical place where the self-energy of a point charge is considered is in deriving the motion of such an object. Roughly, you get an equation that looks like

$$m a = q E - (q^2 / r) a + \ldots$$

for a body with charge q and radius r. The first term is the ordinary force on a charged particle. The second is due to the inertia of its self-field. Rearranging,

$$(m + q^2/r) a = q E + \ldots$$

The left-hand side diverges if r is shrunk while keeping q and m fixed. So you say that only $m + q^2/r$ is observable. That's the effective or renormalized mass. You imagine that m is infinitely negative in order to get a finite sum.

This is closely analogous to things done in QFT. The difference is that the real classical theory has objects with finite r. The mass still shifts (so it may be said to "renormalize"), but there are no infinities that get swept away.

8. Mar 2, 2009

Bob S

Many years ago I had to solve the problem of a heavy electron (like the negative muon) in bound atomic orbits around nuclei. This is a two body problem, because the muon is inside (nearly) all the electron cloud. When the muon is within a few Fermi of the nucleus, the electric field is so strong that it can create virtual electron-positron pairs (vacuum polarization) (within the limits of the uncertainty principal), and modify the one-over-r^2 dependence of the electric field. This was solved using the Uehling integral (Phys Rev, about 1937). The Uehling integral has been validated with both muonic and pionic atoms, and (I think) electron scattering. The cloud of virtual electrons and positrons in vacuum polarization shield the bare (un-normalized) charge.

Last edited: Mar 2, 2009