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  Conservative forces and self-energy

+ 1 like - 0 dislike

In electrostatics, we can define the assembly energy of a distribution of charge since the work required is path independent. Quantitatively, since the $E$ field is conservative, the infinitesimal work needed to bring in an infinitesimal distribution $\delta\rho$ on top of an already assembled distribution $\rho$ is just

$\delta W \thicksim \int dx^3dx'^3 \delta\rho(\mathbf{x}) \frac{1}{|\mathbf{x}-\mathbf{x'}|}\rho(\mathbf{x'}) = \delta \int dx^3dx'^3 \frac{1}{2}\rho(\mathbf{x}) \frac{1}{|\mathbf{x}-\mathbf{x'}|}\rho(\mathbf{x'})$

Quite a few things come together for this to happen: the field is conservative, the potential is linear in the charge density, etc. My question is: are there any counter examples of a conservative field which does not have a well-defined self-energy? If not, is there a well-established result that every conservative field has self-energy?

asked Dec 26, 2016 in Theoretical Physics by Ivan (5 points) [ revision history ]

There is solely a well defined notion of the interaction energy. Self-energy is a failed notion by definition: something acts on itself under condition that this action changes nothing. It is meaningless physically and has no certain numerical value. When it appears in calculations, it is subtracted (discarded). Your $\delta\rho$ is thus not contained in your $\rho$, it is an extra charge, so you may not write $\rho\delta\rho=\delta\frac{1}{2}\rho^2$.

I am a bit confused about your argument. I could easily define a family of charge distributions controlled by some random parameter, i.e. $\rho(x,\alpha)$. Then the differential $\delta$ is a differential over $\alpha$. Another way of thinking about the r.h.s. of the equation above is as the integral of the functional derivative times a small variation $\delta\rho$. Anyway, regardless of the math notation, I can still ask the question: is the work required to assemble a field path independent? For the electrostatic field, the answer is yes, but are there any obvious counter examples? The more I think about it, the more I see this might be related to differentiability of functions in infinite dimensional spaces, something in the lines of the Cauchy relations for complex analytic functions.

Instead of $\alpha$ you have to write the interaction energy like this: $W_{12}=\int d^3xd^3x' \rho_1({\bf{x}})\rho_2({\bf{x}}')/|{\bf{x}}-{\bf{x}}'|$, where one charge cloud $\rho_1$ is not contained in the other cloud $\rho_2$.

1 Answer

+ 1 like - 0 dislike

That's an interesting question, but I can only think of an answer that renders the question trivial.

Before that, however, I would like to remark that you should not call it "self-energy". Your expression for $W$ (after integration over $\delta W$) is simply the energy contained in the electric field $\mathbf{E}$ generated by the charge density $\rho(x)$. The term "self-energy" specifically refers to the question of how the energy is to be interpreted for an isolated point particle, say an electron at position $x_0$, which corresponds to a charge density $\rho(x) = (-e) \delta(x-x_0)$. Does the electron "feel" its own electric field, thus contributing to the energy $W$? This seems impossible, because the integral is infinite in this case. Fortunately, in the case of *continuous* (or just *bounded*) charge distributions, the integral is well-defined and the "self-energy" problem is moot.

Now, concerning the question proper. There are two effects at play here: The electric field is generated ("emanates from") the charge distribution $\rho(x)$. This is Gauss' law. On the other hand, it also acts exerts a conservative force on the charge, i.e. the Lorentz law $\mathbf{F} = \rho\mathbf{E}$. These seem to be independent laws, changing either one of them will yield a different expression for the energy density. Your question is whether we can (at least formally) change these laws and obtain something that prohibits the definition of a total energy.

However, there is also the Lagrangian formulation of electrodynamics. Typically, a conservative system always has a Lagrangian formulation. (And this slight overgeneralization is what renders the question trivial, because you were only asking about the force being conservative, not the combined system force + charge, whatever that is exactly.) There, the interaction between electrons and field is mediated by an integral over a simple product $\int d^3x dt \, A_\mu j^\mu$. In particular, the contribution of the electric potential $\int d^3xdt\, A_0 \rho$. This term encodes *both* the force that the electric field puts on the charge *and* how the electric field is generated by the charge, as can be seen by functional differentiation with respect to $\rho(x)$ and $A_0$, respectively. So, regardless of how the other terms of the Lagrangian look like, this interaction energy will stay the same, and you will always get a well-defined energy for a charge density after solving for $A_0$.

Unfortunately, this answer seems circular, because the result we got out — the ability to define an energy density for the charge $\rho(x)$ — is something we already put in — the assumption that the physics are described by an interaction energy  $\int d^3xdt\, A_0 \rho$ in the first place.

answered Dec 28, 2016 by Greg Graviton (775 points) [ no revision ]

When we deal with plasma theory or something with continuous charge distribution $\rho({\bf{x}},t)$, there is no infinite contributions due to point-like charges. The self-energy integral, i.e., with equal arguments in both $\rho$, vanishes.

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