This note introduces basic properties of the electromagnetic field.
Maxwell’s Equations
The effect of fields on material objects: \[ \begin{aligned} \nabla \times \mathbf{H} - \frac{1}{c}\frac{\partial \mathbf{D}}{\partial t} &= \frac{4\pi}{c}\mathbf{j},\\ \nabla \times \mathbf{E} + \frac{1}{c}\frac{\partial \mathbf{B}}{\partial t} &= \mathbf{0}, \end{aligned} \tag{1}\] Here, \(\mathbf{H}\) is the magnetic vector, \(\mathbf{D}\) is the electric displacement, and \(\mathbf{j}\) is the current density.
They are supplemented by two scalar relations: \[ \begin{aligned} \nabla \cdot \mathbf{D} &= 4\pi\rho,\\ \nabla \cdot \mathbf{B} &= 0. \end{aligned} \tag{2}\] The first equation relates the electric displacement to the charge density, while the second implies that no free magnetic poles exist.
From Equation 1 it follows that \[ \begin{aligned} \nabla \cdot (\nabla \times \mathbf{H}) - \nabla \cdot (\frac{1}{c}\frac{\partial \mathbf{D}}{\partial t}) &= \nabla \cdot (\frac{4\pi}{c}\mathbf{j}) \\ - \frac{1}{4 \pi} \nabla \cdot \frac{\partial \mathbf{D}}{\partial t} =\nabla \cdot \mathbf{j} \end{aligned} \] using Equation 2, \[ \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{j} = 0, \tag{3}\] The continuity equation is analogous to the equation encountered in hydrodynamics. It expresses the fact that charge is conserved in the neighbourhood of any point. Indeed, if one integrates Equation 3 over any region of space, one obtains, with the help of Gauss’s theorem, \[ \frac{d}{dt} \int \rho dV + \int \mathbf{j} \cdot \mathbf{n} dS = 0, \] This equation implies that the total charge: \[ e = \int \rho dV \tag{4}\]
In optical fields, the field vectors vary rapidly with time. The word stationary is often used in a wider sense. \[ \begin{array}{c|c|c} \textbf{Field type} & \textbf{Time dependence} & \textbf{Current density} \\[4pt] \hline \text{Static field} & \displaystyle \frac{\partial \mathbf{E}}{\partial t}=\frac{\partial \mathbf{D}}{\partial t}=\frac{\partial \mathbf{B}}{\partial t}=\frac{\partial \mathbf{H}}{\partial t}=\mathbf{0} & \mathbf{j}=\mathbf{0} \\[10pt] \text{Stationary field} & \displaystyle \frac{\partial \mathbf{E}}{\partial t}=\frac{\partial \mathbf{D}}{\partial t}=\frac{\partial \mathbf{B}}{\partial t}=\frac{\partial \mathbf{H}}{\partial t}=\mathbf{0} & \mathbf{j}\neq\mathbf{0} \end{array} \]
Material equations
If the bodies are at rest or move very slowly relative to one another, and if the material is isotropic, the material equations usually take the following relatively simple form: \[ \begin{aligned} \mathbf{j} &= \sigma \mathbf{E}, \\ \mathbf{D} &= \epsilon \mathbf{E}, \\ \mathbf{B} &= \mu \mathbf{H}. \end{aligned} \tag{5}\] Here \(\sigma\) is called the specific conductivity, \(\epsilon\) is known as the dielectric constant (or permittivity), and \(\mu\) is called the magnetic permeability.
Boundary conditions at a surface of discontinuity
Magnetic Induction and Electric Displacement
\[ \int \nabla \cdot \mathbf{B} dV = \int \mathbf{B} \cdot \mathbf{n} dS = 0 \tag{6}\] Since the areas \(\delta A_1\) and \(\delta A_2\) are assumed to be small, \(\mathbf{B}\) may be considered to have constant values \(\mathbf{B}^{(1)}\) and \(\mathbf{B}^{(2)}\) on \(\delta A_1\) and \(\delta A_2\), and \((12)\) may then be replaced by \[ \mathbf{B}^{(1)} \cdot \mathbf{n}_1 \delta A_1 + \mathbf{B}^{(2)} \cdot \mathbf{n}_2 \delta A_2 + \text{contribution from walls} = 0 \] When the cylinder shrinks to zero, \[ (\mathbf{B}^{(1)} \cdot \mathbf{n}_1 + \mathbf{B}^{(2)} \cdot \mathbf{n}_2) \delta A = 0 \] \[ \mathbf{n}_{12} \cdot (\mathbf{B}^{(2)} - \mathbf{B}^{(1)}) = 0 \tag{7}\] The normal component of the magnetic induction is continuous across the boundary.
From Equation 2, we have the electric displacement \(\mathbf{D}\): \[ \int \nabla \cdot \mathbf{D} dV = \int \mathbf{D} \cdot \mathbf{n} dS = 4\pi \int \rho dV \tag{8}\] When the contribution from the walls tends to zero, the derivation is identical to that of Equation 7. \[ \mathbf{n}_{12} \cdot (\mathbf{D}^{(2)} - \mathbf{D}^{(1)}) = 4 \pi \hat{\rho} \tag{9}\] The normal component of the electric displacement is discontinuous across the boundary.
Surface Density
As \(\delta A_1\) and \(\delta A_2\) shrink together, the total charge remains finite, so that the volume density becomes infinite. Instead of the volume charge density \(\rho\), the concept of surface charge density \(\hat{\rho}\) must be used. \[ \lim_{\delta h \to 0} \int \rho dV = \int \hat{\rho} dA \tag{10}\] The surface current density \(\hat{\mathbf{j}}\) is defined in a similar way: \[ \lim_{\delta h \to 0} \int \mathbf{j} dV = \int \hat{\mathbf{j}} dA \tag{11}\]
Electric Vector and Magnetic Vector
\(\mathbf{b}\) is the unit vector perpendicular to the plane of the rectangle. It follows from Equation 1 and from Stokes’ theorem that \[ \int (\nabla \times \mathbf{E}) \cdot \mathbf{b} dS = \int \mathbf{E} \cdot d \mathbf{r} = - \frac{1}{c} \int \frac{\partial \mathbf{B}}{\partial t} \cdot \mathbf{b} dS \] If the lengths \(P_1 Q_1 = \delta s_1\) and \(P_2 Q_2 = \delta s_2\) are small: \[ \mathbf{E}^{(1)} \cdot \mathbf{t}_1 \delta s_1 + \mathbf{E}^{(2)} \cdot \mathbf{t}_2 \delta s_2 + \text{contribution from ends} = -\frac{1}{c} \frac{\partial \mathbf{B}}{\partial t} \cdot \mathbf{b} \delta s \delta h \] where \(\delta s\) is the line element along which the rectangle intersects the surface. \[ (\mathbf{E}^{(1)} \cdot \mathbf{t}_1 + \mathbf{E}^{(2)} \cdot \mathbf{t}_2) \delta s = 0 \] \[ \mathbf{b} \cdot [\mathbf{n}_{12} \times (\mathbf{E}^{(2)} - \mathbf{E}^{(1)})] = 0 \] \[ \mathbf{n}_{12} \times (\mathbf{E}^{(2)} - \mathbf{E}^{(1)}) = 0 \tag{12}\] The tangential component of the electric vector is continuous across the boundary.
Consider the behaviour of the tangential component of the magnetic vector. The analysis is similar, but there is an additional term if currents are present. \[ \mathbf{H}^{(1)} \cdot \mathbf{t}_1 \delta s_1 + \mathbf{H}^{(2)} \cdot \mathbf{t}_2 \delta s_2 + \text{contribution from ends} = \frac{1}{c} \frac{\partial \mathbf{D}}{\partial t} \cdot \mathbf{b} \delta s \delta h + \frac{4 \pi}{c} \hat{\mathbf{j}} \cdot \mathbf{b} \delta s \] When \(\delta h \to 0\), the derivation is identical to that of Equation 12. \[ \mathbf{n}_{12} \times (\mathbf{H}^{(2)} - \mathbf{H}^{(1)}) = \frac{4 \pi}{c} \hat{\mathbf{j}} \tag{13}\] The tangential component of the magnetic vector is discontinuous across the boundary.
The energy law of the electromagnetic field
From Equation 1 it follows that \[ \mathbf{E} \cdot (\nabla \times \mathbf{H}) - \mathbf{H} \cdot (\nabla \times \mathbf{E}) = \frac{4 \pi}{c} \mathbf{j} \cdot \mathbf{E} + \frac{1}{c} \mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t} + \frac{1}{c} \mathbf{H} \cdot \frac{\partial \mathbf{B}}{\partial t} \] From \[ \begin{aligned} \\ \nabla \cdot (\mathbf A \times \mathbf B) &= \epsilon_{ijk} \partial_k (A_i B_j) \\ &= \epsilon_{ijk} [(\partial_k A_i) B_j + A_i (\partial_k B_j)] \\ &= B_j (\epsilon_{ijk} \partial_k A_i) + A_i (\epsilon_{ijk} \partial_k B_j) \\ &= B_j (\epsilon_{kij} \partial_k A_i) + A_i (-\epsilon_{kji} \partial_k B_j) \\ &= \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B}) \end{aligned} \] We have that \[ \mathbf{E} \cdot (\nabla \times \mathbf{H}) - \mathbf{H} \cdot (\nabla \times \mathbf{E}) = - \nabla \cdot (\mathbf{E} \times \mathbf{H}) \] \[ \frac{1}{c} (\mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t} + \mathbf{H} \cdot \frac{\partial \mathbf{B}}{\partial t}) + \frac{4 \pi}{c} \mathbf{j} \cdot \mathbf{E} + \nabla \cdot (\mathbf{E} \times \mathbf{H}) = 0 \] Integrating over an arbitrary volume and applying Gauss’s theorem, we obtain \[ \frac{1}{4 \pi} \int (\mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t} + \mathbf{H} \cdot \frac{\partial \mathbf{B}}{\partial t}) dV + \int \mathbf{j} \cdot \mathbf{E} dV + \frac{c}{4 \pi} \int (\mathbf{E} \times \mathbf{H}) \cdot \mathbf{n} dS = 0 \tag{14}\]
On using the material equations Equation 5, we have \[ \left. \begin{aligned} \frac{1}{4 \pi} (\mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t}) &= \frac{1}{4 \pi} \mathbf{E} \cdot \frac{\partial}{\partial t}(\epsilon \mathbf{E}) = \frac{1}{8 \pi} \frac{\partial}{\partial t}(\epsilon \mathbf{E}^2) = \frac{1}{8 \pi} \frac{\partial}{\partial t}(\mathbf{E} \cdot \mathbf{D}) \\ \frac{1}{4 \pi} (\mathbf{H} \cdot \frac{\partial \mathbf{H}}{\partial t}) &= \frac{1}{4 \pi} \mathbf{H} \cdot \frac{\partial}{\partial t}(\mu \mathbf{H}) = \frac{1}{8 \pi} \frac{\partial}{\partial t}(\mu \mathbf{H}^2) = \frac{1}{8 \pi} \frac{\partial}{\partial t}(\mathbf{H} \cdot \mathbf{B}) \end{aligned} \right\} \] Setting \[ w_e = \frac{1}{8 \pi} \mathbf{E} \cdot \mathbf{D},\qquad w_m = \frac{1}{8 \pi} \mathbf{H} \cdot \mathbf{B} \tag{15}\] and \[ W=\int (w_e + w_m) dV \tag{16}\] Equation 14 becomes \[ \frac{d W}{d t} + \int \mathbf{j} \cdot \mathbf{E} dV + \frac{c}{4 \pi} \int (\mathbf{E} \times \mathbf{H}) \cdot \mathbf{n} dS = 0 \tag{17}\] \(W\) represents the total energy contained within the volume, \(w_e\) may be identified with the electric energy density and \(w_m\) with the magnetic energy density of the field.
To justify the interpretation of \(W\) as the total energy we can assume the material bodies to be so small that they can be regarded as point charges \(e_k (k = 1, 2, \dots)\). The force exerted by a field \((\mathbf{E}, \mathbf{B})\) on a charge \(e\) moving with velocity \(\mathbf{v}\) is given by the so-called Lorentz law \[ \mathbf{F} = e(\mathbf{E} + \frac{1}{c} \mathbf{v} \times \mathbf{B}) \] It follows that if all the charges \(e_k\) are displaced by \(\delta \mathbf{x}_k\) (\(k = 1, 2, \dots\)) during a time interval \(\delta t\), the total work done is \[ \begin{aligned} \delta A &= \sum_k \mathbf{F}_k \cdot \delta \mathbf{x}_k = \sum_k e_k(\mathbf{E}_k + \frac{1}{c}\mathbf{v}_k \times \mathbf{B}) \cdot \mathbf{v}_k \delta t\\ &= \sum_k e_k \mathbf{E}_k \cdot \mathbf{v}_k \delta t + \sum_k(\frac{1}{c}\mathbf{v}_k \times \mathbf{B}) \cdot \mathbf{v}_k \delta t = \sum_k e_k \mathbf{E}_k \cdot \mathbf{v}_k \delta t \end{aligned} \] \[ (\frac{1}{c}\mathbf{v}_k \times \mathbf{B}) \cdot \delta \mathbf{x}_k = (\frac{1}{c}\mathbf{v}_k \times \mathbf{B}) \cdot \mathbf{v}_k \delta t \] If the number of charged particles is large, we can consider the distribution to be continuous. We introduce the charge density \(\rho\): \[ \delta A = \delta t \int \rho \mathbf{v} \cdot \mathbf{E} dV \] The current density \(\mathbf{j}\) appearing in Maxwell’s equations can be split into two parts: \[ \mathbf{j} = \mathbf{j}_c + \mathbf{j}_\nu \] where \[ \begin{aligned} \mathbf{j}_c &= \sigma \mathbf{E} \\ \mathbf{j}_\nu &= \rho \mathbf{v} \end{aligned} \] Using Equation 17, \[ \begin{aligned} \frac{d W}{d t} + \int \mathbf{j} \cdot \mathbf{E} dV &+ \frac{c}{4 \pi} \int (\mathbf{E} \times \mathbf{H}) \cdot \mathbf{n} dS = 0 \\ \frac{d W}{d t} + \int \mathbf{j}_c \cdot \mathbf{E} dV + \int \mathbf{j}_\nu \cdot \mathbf{E} dV &+ \frac{c}{4 \pi} \int (\mathbf{E} \times \mathbf{H}) \cdot \mathbf{n} dS = 0 \\ \frac{d W}{d t} + \int \sigma \mathbf{E}^2 dV + \int \rho \mathbf{v} \cdot \mathbf{E} dV &+ \frac{c}{4 \pi} \int (\mathbf{E} \times \mathbf{H}) \cdot \mathbf{n} dS = 0 \\ \frac{d W}{d t} + Q + \frac{\delta A}{\delta t} &+ \int \mathbf{S} \cdot \mathbf{n} dS = 0 \end{aligned} \] \[ \frac{d W}{d t} = - \frac{\delta A}{\delta t} - Q - \int \mathbf{S} \cdot \mathbf{n} dS \tag{18}\] where the term \(Q\) represents Joule heating, and the vector \(\mathbf{S}\) is the Poynting vector.
In a nonconducting medium (\(\sigma = 0\)) where no mechanical work is done (\(A = 0\)), the energy equation becomes: \[ \frac{\partial w}{\partial t} + \nabla \cdot \mathbf{S} = 0 \]