9.3 Energy conservation

Energy conservation in electromagnetism

Energy conservation in electromagnetism governs how energy is stored in fields, flows through space, and converts into other forms like heat or kinetic energy. Poynting's theorem provides the mathematical statement of this principle, connecting the time rate of change of field energy to the energy flux and the work done on charges.

Derivation of Poynting's theorem

The derivation starts from two of Maxwell's equations and uses a vector identity to combine them into a single energy-balance equation. Here are the steps:

1. Begin with Faraday's law: $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$

2. Begin with Ampère's law (with Maxwell's correction): $\nabla \times \vec{H} = \vec{J} + \frac{\partial \vec{D}}{\partial t}$

3. Dot $\vec{H}$ into Faraday's law and $\vec{E}$ into Ampère's law. This gives you expressions involving $\vec{H} \cdot (\nabla \times \vec{E})$ and $\vec{E} \cdot (\nabla \times \vec{H})$.

4. Subtract the two results. The left-hand side becomes $\vec{H} \cdot (\nabla \times \vec{E}) - \vec{E} \cdot (\nabla \times \vec{H})$, which equals $-\nabla \cdot (\vec{E} \times \vec{H})$ by the vector identity for the divergence of a cross product.

5. On the right-hand side, you get time derivatives of the field energy densities and a $\vec{J} \cdot \vec{E}$ term representing work done on charges.

6. The result is Poynting's theorem in differential form:

$-\nabla \cdot (\vec{E} \times \vec{H}) = \vec{J} \cdot \vec{E} + \frac{\partial}{\partial t}\left(\frac{1}{2}\varepsilon_0 E^2 + \frac{1}{2\mu_0}B^2\right)$

In integral form (applying the divergence theorem over a volume $V$ bounded by surface $\mathcal{S}$):

$-\oint_{\mathcal{S}} (\vec{E} \times \vec{H}) \cdot d\vec{a} = \int_V \vec{J} \cdot \vec{E}\, dV + \frac{d}{dt}\int_V \left(\frac{1}{2}\varepsilon_0 E^2 + \frac{1}{2\mu_0}B^2\right) dV$

Physical interpretation

Each term in Poynting's theorem has a direct physical meaning:

• $-\nabla \cdot \vec{S}$ (or the surface integral of $\vec{S}$): the net electromagnetic power flowing into the volume through its boundary.
• $\vec{J} \cdot \vec{E}$: the rate per unit volume at which the fields do work on free charges. This is where field energy converts to other forms (Joule heating in a resistor, kinetic energy of accelerated particles, etc.).
• $\frac{\partial u}{\partial t}$: the rate at which the local electromagnetic energy density changes.

The theorem says: the power flowing into a region either increases the stored field energy or does work on the charges inside. Nothing is created or destroyed.

Electromagnetic energy density

The total energy stored per unit volume in the electromagnetic field is the sum of the electric and magnetic contributions:

$u = u_E + u_B = \frac{1}{2}\varepsilon_0 E^2 + \frac{1}{2\mu_0}B^2$

Electric field energy density

$u_E = \frac{1}{2}\varepsilon_0 E^2$

Here $\varepsilon_0$ is the permittivity of free space and $E$ is the magnitude of the electric field. Because the energy density scales as $E^2$, doubling the field strength quadruples the stored energy. In a linear dielectric, replace $\varepsilon_0$ with $\varepsilon$ (or equivalently write $u_E = \frac{1}{2}\vec{D}\cdot\vec{E}$).

Magnetic field energy density

$u_B = \frac{1}{2\mu_0}B^2$

Here $\mu_0$ is the permeability of free space and $B$ is the magnitude of the magnetic field. The same quadratic scaling applies. In a linear magnetic medium, this generalizes to $u_B = \frac{1}{2}\vec{B}\cdot\vec{H}$.

Electromagnetic energy flux

Poynting vector

The Poynting vector gives the energy flux density (power per unit area) of the electromagnetic field:

$\vec{S} = \vec{E} \times \vec{H}$

Its units are $\text{W/m}^2$. The direction of $\vec{S}$ is the direction energy flows, and its magnitude is the instantaneous power crossing a unit area perpendicular to that direction.

Direction and magnitude of energy flow

Because $\vec{S}$ is a cross product, it's perpendicular to both $\vec{E}$ and $\vec{H}$. For a plane wave propagating in the $+\hat{z}$ direction with $\vec{E}$ along $\hat{x}$ and $\vec{H}$ along $\hat{y}$, the Poynting vector points along $+\hat{z}$, which is the propagation direction. This is the typical situation: energy flows in the direction the wave travels.

Be careful with the word "always" here. In waveguides or near reactive loads, the Poynting vector can have components that don't align with the nominal propagation direction, or it can even point backward locally.

Energy conservation in electromagnetic waves

Energy density of electromagnetic waves

For a plane wave in vacuum, the electric and magnetic energy densities are equal at every instant:

$u_E = u_B$

This follows from the plane-wave relation $B = E/c$, which gives $\frac{1}{2\mu_0}B^2 = \frac{1}{2}\varepsilon_0 E^2$. The total instantaneous energy density is therefore:

$u = \varepsilon_0 E^2 = \frac{B^2}{\mu_0}$

Both $u_E$ and $u_B$ oscillate in time and space (they go as $\cos^2$), so the energy density is not uniform along the wave.

Energy flux of electromagnetic waves

For practical purposes you usually care about the time-averaged Poynting vector, since the instantaneous value oscillates at twice the wave frequency. For a sinusoidal plane wave with peak electric field amplitude $E_0$:

$\langle S \rangle = \frac{1}{2}\frac{E_0^2}{\mu_0 c} = \frac{1}{2}\sqrt{\frac{\varepsilon_0}{\mu_0}}\,E_0^2 = \frac{E_0^2}{2\eta_0}$

where $\eta_0 = \sqrt{\mu_0/\varepsilon_0} \approx 377\,\Omega$ is the impedance of free space and $c = 1/\sqrt{\mu_0\varepsilon_0}$. This quantity is often called the intensity $I$ of the wave.

In a lossless medium, $\langle S \rangle$ stays constant along the propagation direction. In a dissipative medium (finite conductivity or complex permittivity), the fields decay exponentially and so does the energy flux.

Applications of energy conservation

Waveguides and transmission lines

These structures channel electromagnetic energy from one location to another. Energy conservation requires:

$P_{\text{in}} = P_{\text{out}} + P_{\text{loss}}$

Losses come from ohmic dissipation in the conducting walls and dielectric losses in the filling material. You can compute the power flowing at any cross-section by integrating the Poynting vector over that cross-section.

Antennas and radiation

An antenna converts guided-wave energy into free-space radiation (or the reverse on receive). The total radiated power is found by integrating the Poynting vector over a closed surface surrounding the antenna:

$P_{\text{rad}} = \oint_{\mathcal{S}} \vec{S} \cdot d\vec{a}$

Energy conservation gives: $P_{\text{input}} = P_{\text{rad}} + P_{\text{ohmic}} + P_{\text{other losses}}$. The ratio $P_{\text{rad}}/P_{\text{input}}$ defines the antenna's radiation efficiency.

Electromagnetic shielding

When an electromagnetic wave hits a shield, the incident energy splits into three parts:

• Reflected energy (bounced back)
• Absorbed energy (converted to heat inside the shield)
• Transmitted energy (passed through)

Energy conservation requires $P_{\text{incident}} = P_{\text{reflected}} + P_{\text{absorbed}} + P_{\text{transmitted}}$. Effective shields maximize reflection and absorption so that very little energy is transmitted.

Energy dissipation in electromagnetic systems

Ohmic losses

When current flows through a conductor with finite conductivity, electrons scatter off the lattice and convert field energy into heat. The local power dissipation density is:

$p = \vec{J} \cdot \vec{E} = \sigma E^2$

where $\sigma$ is the conductivity. For a lumped element carrying current $I$ through resistance $R$, this integrates to the familiar $P = I^2 R$. This is exactly the $\vec{J} \cdot \vec{E}$ term in Poynting's theorem.

Radiative losses

Energy can leak out of a system unintentionally through radiation. Common causes include bends, discontinuities, or gaps in waveguides and transmission lines. These losses reduce the power delivered to the load and can also create electromagnetic interference with nearby systems. Radiative losses are typically minimized through proper shielding and smooth geometric transitions.

Energy storage in electromagnetic fields

Capacitive energy storage

A capacitor stores energy in the electric field between its plates:

$W = \frac{1}{2}CV^2$

You can derive this from the field perspective: integrate the electric energy density $u_E = \frac{1}{2}\varepsilon_0 E^2$ over the volume between the plates. For a parallel-plate capacitor with plate area $A$, separation $d$, and uniform field $E = V/d$, you get $W = \frac{1}{2}\varepsilon_0 (V/d)^2 \cdot Ad = \frac{1}{2}CV^2$, confirming consistency between the circuit and field descriptions.

Inductive energy storage

An inductor stores energy in the magnetic field generated by the current through it:

$W = \frac{1}{2}LI^2$

Similarly, this equals the integral of $u_B = \frac{1}{2\mu_0}B^2$ over the volume where the magnetic field exists (e.g., inside a solenoid). The circuit formula and the field integral give the same answer, which is a nice consistency check on the energy density expressions.

Boundary conditions and energy conservation

Continuity of energy flux at boundaries

At an interface between two media (assuming no surface currents or charges that could store energy), the normal component of the Poynting vector must be continuous:

$\vec{S}_1 \cdot \hat{n} = \vec{S}_2 \cdot \hat{n}$

This ensures that energy doesn't pile up or vanish at the boundary. A discontinuity in the normal component of $\vec{S}$ would imply energy appearing or disappearing at the surface, which is non-physical (unless there's a resistive surface layer doing work).

Energy reflection and transmission at interfaces

When a plane wave hits an interface between two media, some fraction of the energy reflects and the rest transmits. The reflectance $R$ and transmittance $T$ satisfy:

$R + T = 1$

(for a lossless interface with no absorption). These depend on the impedances of the two media and the angle of incidence. For normal incidence on a boundary between media with impedances $\eta_1$ and $\eta_2$:

$R = \left(\frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}\right)^2, \qquad T = 1 - R$

Note that $R$ and $T$ here refer to power (intensity) ratios, not field amplitude ratios. The Fresnel coefficients for the field amplitudes are related but distinct.