# Traveling EM waves with fixed energy density

• lar49

#### lar49

Hi,
I'm taking an undergrad course in Electromagnetism and Optics, and in the lecture notes it reads:
"Consider the formal equation ε0D⋅E = 1 ... that must be obeyed for waves traveling in different directions as defined by the wave vector k but with a given energy density."
Could anyone help me as to where this equation comes from?

Thanks! A:The equation $\epsilon_0 \mathbf{D}\cdot \mathbf{E} = 1$ is a manifestation of Gauss' law in the presence of time-dependent fields.Gauss' law states that$$\nabla \cdot \mathbf{D} = \rho$$where $\rho$ is the charge density. In the absence of charges, $\rho = 0$ and thus$$\nabla \cdot \mathbf{D} = 0.$$To work with this equation more easily, it is useful to introduce the electric field $\mathbf{E}$ such that$$\mathbf{D} = \epsilon_0 \mathbf{E}.$$Then, substituting this into the equation we get$$\epsilon_0 \nabla \cdot (\epsilon_0 \mathbf{E}) = 0.$$Using the vector identity$$\nabla \cdot (\epsilon_0 \mathbf{E}) = \epsilon_0 \nabla \cdot \mathbf{E} + \mathbf{E}\cdot \nabla \epsilon_0$$we can rewrite the equation as$$\epsilon_0 \nabla \cdot \mathbf{E} + \mathbf{E}\cdot \nabla \epsilon_0 = 0.$$The second term on the left-hand side is zero since $\epsilon_0$ is a constant, so we are left with$$\epsilon_0 \nabla \cdot \mathbf{E} = 0.$$Now, if we assume that the electric field is time-dependent, then we can use the wave equation$$\nabla^2 \mathbf{E} - \frac{1}{c^2}\frac{\partial^2 \mathbf{E}}{\partial t^2} = 0$$to rewrite the equation as\epsilon_0 \nabla \cdot \left(\frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}\right) =