Energy Conservation

We have seen that the energy density of an electric field is given by [see Equation (2.84)]

$$U_E = \frac{\epsilon_0\,E^2}{2},$$ (2.546)

whereas the energy density of a magnetic field takes the form [see Equation (2.377)]

$$U_B = \frac{B^2}{2\,\mu_0}.$$ (2.547)

This suggests that the energy density of a general electromagnetic field is

$$U= \frac{\epsilon_0\,E^2}{2}+\frac{B^2}{2\,\mu_0}.$$ (2.548)

Let us now demonstrate that Maxwell's equations conserve energy. We have already come across one conservation law in electromagnetism; namely,

$$\frac{\partial\rho}{\partial t} +\nabla\cdot {\bf j} = 0.$$ (2.549)

(See Equation (2.473).] The previous expression described the conservation of electric charge. Thus, integrating over some volume $V$, bounded by a surface $S$, and making use of the divergence theorem (see Section A.20), we obtain

$$-\frac{\partial}{\partial t}\!\int_V \rho\,d V = \oint_S {\bf j}\cdot d{\bf S}.$$ (2.550)

In other words, the rate of decrease of the electric charge contained in volume $V$ is equal to the net flux of charge out of surface $S$. This suggests that an energy conservation law for electromagnetic fields should have the form

$$-\frac{\partial}{\partial t}\!\int_V U\,d V = \oint_S {\bf u}\cdot d{\bf S}.$$ (2.551)

Here, $U$ is the energy density of the electromagnetic field, and ${\bf u}$ is the flux of electromagnetic energy (i.e., energy $\vert{\bf u}\vert$ per unit time, per unit cross-sectional area, passes a given point in the direction of ${\bf u}$). According to the previous equation, the rate of decrease of the electromagnetic energy in volume $V$ is equal to the net flux of electromagnetic energy out of surface $S$.

Equation (2.551) is actually incomplete, because electromagnetic fields can gain or lose energy by interacting with matter. We need to incorporate this fact into our analysis. We saw earlier (see Section 2.1.11) that the rate of heat dissipation per unit volume in a conductor (the so-called ohmic heating rate) is ${\bf E}\cdot{\bf j}$. This energy is extracted from electromagnetic fields, so the rate of energy loss of the fields in a volume $V$ due to interaction with matter is $\int_V {\bf E}\cdot{\bf j}\,dV$. Thus, Equation (2.551) generalizes to give

$$-\frac{\partial}{\partial t} \!\int_V U\,dV= \oint_S {\bf u}\cdot d{\bf S} + \int_V {\bf E}\cdot{\bf j}\,dV.$$ (2.552)

From the divergence theorem (see Section A.20), the previous equation is equivalent to

$$\frac{\partial U}{\partial t} +\nabla\cdot {\bf u} = - {\bf E} \cdot {\bf j}.$$ (2.553)

Let us now see if we can derive an expression of this form from Maxwell's equations.

We start from the differential form of Ampère's law (including the displacement current), (2.487):

$$\nabla\times{\bf B} = \mu_0\, {\bf j} +\epsilon_0\,\mu_0 \frac{\partial {\bf E}} {\partial t}.$$ (2.554)

The scalar product of the electric field with this equation yields

$$- {\bf E} \cdot {\bf j} = - \frac{ {\bf E}\cdot \nabla\times{\bf B}}{\mu_0} +\epsilon_0 \,{\bf E}\cdot \frac{\partial {\bf E}}{\partial t}.$$ (2.555)

The previous expression can be rewritten

$$- {\bf E} \cdot {\bf j} = - \frac{ {\bf E}\cdot \nabla\times{\bf B}}{\mu_0} +\frac{\partial}{\partial t}\!\left(\frac{\epsilon_0 \,E^2}{2}\right).$$ (2.556)

However (see Section A.24),

$$\nabla\cdot({\bf E}\times{\bf B}) \equiv {\bf B}\cdot\nabla\times{\bf E} - {\bf E} \cdot \nabla\times{\bf B},$$ (2.557)

so

$$- {\bf E} \cdot {\bf j} = \nabla \!\cdot\! \left(\frac{{\bf E}\ti... ...{\mu_0} + \frac{\partial}{\partial t}\!\left(\frac{\epsilon_0\, E^2}{2}\right).$$ (2.558)

The differential form of Faraday's law, (2.486), yields

$$\nabla\times{\bf E} = - \frac{\partial {\bf B}}{\partial t},$$ (2.559)

so

$$- {\bf E} \cdot {\bf j} = \nabla \!\cdot\! \left(\frac{{\bf E}\ti... ...rtial t} + \frac{\partial}{\partial t}\!\left(\frac{\epsilon_0\,E^2}{2}\right),$$ (2.560)

which can be rewritten as

$$- {\bf E} \cdot {\bf j} = \nabla\! \cdot \!\left(\frac{{\bf E}\ti... ...l}{\partial t}\!\left( \frac{\epsilon_0\,E^2}{2} +\frac{B^2}{2\,\mu_0} \right).$$ (2.561)

Thus, we obtain the desired conservation law,

$$\frac{\partial U}{\partial t} + \nabla\cdot{\bf u} = -{\bf E}\cdot {\bf j},$$ (2.562)

where

$$U = \frac{\epsilon_0\,E^2}{2} + \frac{B^2}{2\,\mu_0}$$ (2.563)

is the electromagnetic energy density, and

$${\bf u} = \frac{{\bf E}\times{\bf B}}{\mu_0}$$ (2.564)

is the electromagnetic energy flux. The latter quantity is usually called the Poynting flux, after John Poynting who derived it in 1884.

Let us see whether our expression for the electromagnetic energy flux makes physical sense. We know that if we stand in the Sun then we get hot. This occurs because we absorb electromagnetic radiation emitted by the Sun. So, radiation must transport energy. The electric and magnetic fields in electromagnetic radiation are mutually perpendicular, and are also perpendicular to the direction of propagation, $\hat{\bf k}$ (which is a unit vector). Furthermore, $B=E/c$. (See Section 2.4.4.) Equation (2.521) can easily be transformed into the following relation between the electric and magnetic fields of an electromagnetic wave:

$${\bf E} \times{\bf B} = \frac{E^2}{c}\, \hat{\bf k}.$$ (2.565)

Thus, the Poynting flux for electromagnetic radiation is

$${\bf u} = \frac{E^2}{\mu_0\, c} \,\hat{\bf k} = \epsilon_0 \,c \,E^2\, \hat{\bf k}.$$ (2.566)

The previous expression states that electromagnetic waves transport energy along their direction of propagation, which seems to make sense.

The energy density of electromagnetic radiation is

$$U = \frac{\epsilon_0\,E^2}{2} + \frac{B^2}{2\,\mu_0} = \frac{\epsilon_0\,E^2}{2} + \frac{E^2}{2\,\mu_0 \,c^2} = \epsilon_0 \,E^2,$$ (2.567)

where use has been made of $B=E/c$, and $c=1/\!\sqrt{\epsilon_0\,\mu_0}$. Note that the electric and magnetic components of an electromagnetic wave have equal energy densities. Because electromagnetic waves travel at the speed of light in vacuum, we would expect the energy flux through one square meter in one second to equal the energy contained in a volume of length $c$, and unit cross-sectional area; that is, $c$ multiplied by the electromagnetic energy density. Thus,

$$\vert{\bf u}\vert = c\,U = \epsilon_0\, c\, E^2,$$ (2.568)

which is in accordance with the previous two equations.