# How to get the formula of the energy of EM waves?

I am trying to get the formula for energy of EM waves:

$$W = \frac{E^2 + B^2}{2}$$ calculating the work done on a test charge by the force: $$\mathbf F = q(\mathbf E + v \times \mathbf B)$$ $$\mathbf E$$ and $$\mathbf B$$ are vectors of the type $$\mathbf F(u)$$, $$u = (\mathbf {k.x} - \omega t)$$ and $$\omega = \frac{c}{|k|}$$ solutions of Maxwell wave equation. It seems go well until I get $$\frac{\partial E_v}{\partial t} = \mathbf {j.E}$$ where the left side is power per unit of volume and $$\mathbf j$$ is density of current.

But if I try to get rid of $$\mathbf j$$, using the Maxwell equation:

$$\mathbf j = \nabla \times \mathbf B - \frac{\partial \mathbf E}{\partial t}$$ the right side vanishes. And it is not a surprise, because the wave equation, from which $$\mathbf E$$ and $$\mathbf B$$ are solutions requires no charges or currents.

Searching the web, the energy formula comes from circuits, inductors and capacitors storing energy. Energy of EM waves simply use that results.

The other approach is from Lagrangian, but in this case, as I understand, it is the opposite way: the expression for the energy is postulated, and Maxwell equations are derived from it.

Is it possible to derive the quadratic energy expression from the wave equation and Lorentz force?

• In your first formula you’ve used $E$ to mean two entirely different things. – G. Smith May 10 '20 at 19:27
• Yes, I changed it to W. – Claudio Saspinski May 10 '20 at 19:33
• en.wikipedia.org/wiki/Poynting%27s_theorem – d_b May 10 '20 at 19:36
• @d_b Wikipedia article adds a term: flux of energy to the work done per time. I assumed that the work done by the field is what we call energy. Anyway the quadratic formula is not derived there, it is an input. – Claudio Saspinski May 10 '20 at 19:52
• See sjsu.edu/faculty/watkins/fieldenergy.htm for derivation. Note that this derivation does not require the fields to have wave form or any other form. – G. Smith May 10 '20 at 20:30

The usual approach is to calculate the power due to Joule's heat, given by $$\mathbf{j}\cdot\mathbf{E}$$. Starting with the Maxwell equations $$\nabla \times\mathbf{B}=\frac{4\pi}{c}\mathbf{j} + \frac{1}{c}\frac{\partial\mathbf{E}}{\partial t},\\ \nabla \times\mathbf{E}=- \frac{1}{c}\frac{\partial\mathbf{B}}{\partial t}.$$ Multiplying the first equation by $$\mathbf{E}$$ we can express the Joule's heat as $$\frac{4\pi}{c}\mathbf{j}\cdot\mathbf{E} = -\frac{1}{c}\mathbf{E}\cdot\frac{\partial\mathbf{E}}{\partial t} + \mathbf{E}\cdot(\nabla \times\mathbf{B}).$$ Further $$-\frac{1}{c}\mathbf{E}\cdot\frac{\partial\mathbf{E}}{\partial t} = -\frac{1}{2c}\frac{\partial\mathbf{E}^2}{\partial t},$$ whereas $$\mathbf{E}\cdot\nabla \times\mathbf{B} = \nabla (\mathbf{B}\times\mathbf{E}) + \mathbf{B}\cdot(\nabla\times\mathbf{E}).$$ Here the last term is transformed using the other Maxwell equation, as it was done previously for $$\mathbf{E}$$, whereas the first term produces the Pointing vector.
• Using a finite sized charge, it is possible to get $W = qvE$ for the work done per time by the plane wave on the charge, and $W_v = \rho vE = jE$ for work per volume and time. – Claudio Saspinski May 11 '20 at 22:16