Using Lorentz force law and Maxwell's equations, one can derive the following relations:

\begin{align} \dfrac{dW_{\text{mech}}}{dt}& =-\dfrac{\partial}{\partial t}\bigg(\dfrac{\epsilon_0 E^2}{2}+\dfrac{B^2}{2\mu_0}\bigg)-\nabla \cdot \bigg(\dfrac{1}{\mu_0}(\vec{E}\times \vec{B})\bigg) \tag{1} \\ \dfrac{d\vec{p_{\text{mech}}}}{dt}& =-\dfrac{\partial}{\partial t}(\epsilon_0\vec{E}\times \vec{B})+\nabla \cdot \textbf{T} \tag{2} \end{align} where $$T_{ij}=\epsilon _{0}\left ( E_{i}E_{j}\frac{1}{2}\delta _{ij}E^{2} \right )+\frac{1}{\mu_{0}}\left ( B_{i}B_{j}-\frac{1}{2}\delta _{ij}B^{2} \right ) $$ and $\vec{p}_\mathrm{mech}$ is the mechanical momentum of matter and $W_\mathrm{mech}$ is the work done on matter by the electromagnetic field.

This motivates us to define $\dfrac{\epsilon_0 E^2}{2}+\dfrac{B^2}{2\mu_0}$ as the energy density and $\epsilon_0\vec{E}\times \vec{B}$ as the momentum density of electromagnetic fields. But as Feynman has argued in his famous CalTech undergrad lectures, these are just the possible candidates for these densities. Because, in principle, you can possibly play around with the relations $(1)$ and $(2)$ and rewrite them in such a way that they are in the form of local conservation laws for energy and momentum but the expressions on which $\dfrac{\partial}{\partial t}$ operates are different from $\dfrac{\epsilon_0 E^2}{2}+\dfrac{B^2}{2\mu_0}$ and $\epsilon_0\vec{E}\times \vec{B}$ respectively (with correspondingly adjusted expressions on which $\nabla$ operates).

If this can be done then we will have some new expressions for candidate electromagnetic energy density and candidate electromagnetic momentum density. Feynman says that this has been done and it is an open problem as to which ones are the correct ones. I am curious as to whether this issue has been resolved so far or not.

  • $\begingroup$ Please don't use MathJax for emphasis in titles. There's a design reason why the Markdown *emphasis* doesn't work there. $\endgroup$ Oct 13, 2017 at 13:10

1 Answer 1


This is actually not an issue, because the value of energy of a system, or density of energy at any point, by itself, is a technical artifact that does not matter in physics.

For example, consider kinetic energy of a point particle. Usually, it is assumed that this energy is given by

$$ \frac{1}{2}mv^2. $$

How do we know this is the correct formula, and not, say, $$ \frac{1}{2}mv^2 + 5~\text{Joule} $$ or $$ \frac{1}{2}mv^2 + \frac{1}{2}mc^2 ? $$ These alternative formulae are just as good for stating the conservation law or the work-energy theorem, which are the reasons we bother to talk about energy at all.

The original expression $\frac{1}{2}mv^2$ is used rather that some other because of simplicity. Kinetic energy of point particle is defined to be given $\frac{1}{2}mv^2$. It is not as if people somehow 'found' or 'proved' that kinetic energy has to have particular value; they chose one.

The same holds for electromagnetic energy. It does not matter what the value of EM energy inside some definite region of space is. In macroscopic EM theory, it makes sense to define it as

$$ \int_V \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0} B^2 \,dV, $$

because of the Poynting theorem. But to say that value of EM energy in that volume region has some definite value to be found is in contradiction to how energy is defined and used in physics; that is as a conserved quantity that we are free to redefine by addition of a constant.

Except, well, for General Relativity.

In this theory, 'energy' is sometimes understood as something that ought to be a function of measurable manifestations of gravity field. More technically, Einstein equations relate total stress-energy tensor (whose term $T^{00}$ is said to be total energy density) to the Ricci tensor(characterizing gravity). So, in this theory, it is in principle possible to find the stress-energy tensor from the measurements of properties of spacetime.

However, this has no import on the question, because such determination will only fix a particular term that figures in the Einstein equations, but will hardly change the concept of energy as used in the rest of physics (a quantity subject to conservation laws, defined up to a constant).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.