Electromagnetism

Electrostatics

Electrostatics is the physics of source charge distributions that are constant in time.

Coulomb's Law

Coulomb's law is the empirical equation for the force $\vec{F}$ on a point electric charge $q$ at position $\r$ by a source point electric charge $q'$ at source position $\r'$ $$ \vec{F} = \q{1}{4 \pi \ep_0} \q{\uR}{R^2} q q' . $$ Coulomb's law is linear and thus the forces of source charges obey superposition $$ \vec{F} = q \bb{\q{1}{4 \pi \ep_0} \sum_i \q{\uR_i}{R_i^2}q'_i} . $$ The bracketed factor is a vector field independent of $q$. Define the electric field as $$ \boxed{\vec{F} = q \E} $$ $$ \E(\r) = \q{1}{4 \pi \ep_0} \sum_i \q{\uR_i}{R_i^2} q'_i . $$ Charges that are small and numerous can be modeled as a continuum. Summation over point charges becomes integration of charge density over a geometric domain $$ \boxed{\E(\r) = \q{1}{4 \pi \ep_0} \int \q{\uR}{R^2} dq'} $$ $$ \boxed{dq' = \lambda(\r') dL' = \sigma(\r') dS' = \rho(\r') dV' .} $$ In particular, the electric field of a volume charge density over volume $V$ is $$ \E(\r) = \q{1}{4 \pi \ep_0} \int_V \q{\uR}{R^2} \rho(\r') dV' . $$

Divergence and Curl

Helmholtz decomposition implies a vector field is specified by its divergence and curl.

Consider the divergence of $\E$ for an electric volume density. Apply divergence, exchange the order of divergence and integration, and apply product rule (3) $$ \div \E = \q{1}{4 \pi \ep_0} \int_V \div \bb{\q{\uR}{R^2} \rho} dV' = \q{1}{4 \pi \ep_0} \int_V \bb{\grad \rho \dot \q{\uR}{R^2} + \rho \div \q{\uR}{R^2}} dV' . $$ $\grad \rho(\r') = \vec{0}$ because $\rho$ is not a function of $\r$. Care must be taken to evaluate the second term with the identity $\div [\uR / R^2] = 4 \pi \delta(\R)$ (see Vector Calculus page) $$ \div \E = \q{1}{4 \pi \ep_0} \int_V [0 + 4 \pi \delta(\R) \rho(\r')] dV' = \q{\rho(\r)}{\ep_0} . $$ The result is Gauss's law in differential form $$ \boxed{\div \E = \q{\rho}{\ep_0} .} $$ Integrate this over a volume $$ \int_V \div \E dV = \q{1}{\ep_0} \int_V \rho dV $$ define the total enclosed charge as $$ \boxed{Q \equiv \int_V \rho(\r) dV} $$ and apply the divergence theorem to obtain Gauss's law in integral form $$ \boxed{\oint_S \E \dot \dS = \q{Q}{\ep_0} .} $$

Compute the curl of $\E$ by first considering a single source electric point charge at $\r' = \vec{0}$ and form a line integral in spherical coordinates $$ \int_L \E \dot \dL = \int_L \bb{\q{1}{4 \pi \ep_0} \q{q'}{r^2} \ur} \dot [\ur dr + \utheta r d\theta + \uphi r \sin\theta d\phi] = \q{q'}{4 \pi \ep_0} \int_{r_a}^{r_b} \q{dr}{r^2} = -\q{1}{4 \pi \ep_0} \q{q'}{r} \bigg|_{r=r_a}^{r_b} . $$ The line integral only depends on the end points. A closed loop evaluates to zero $$ \boxed{\oint_L \E \dot \dL = 0} $$ and the curl theorem implies $$ \boxed{\curl \E = \vec{0} .} $$

Scalar Potential

Line integrals of electrostatic fields only depend on the end points, not the path. Thus a scalar potential function can be defined relative to some reference point $\r_0$ $$ \boxed{V(\r) \equiv - \int_{\r_0}^{\r} \E(\r') \dot \dL' .} $$ Subtract $V$ evaluated at $\r_a$ from $\r_b$, swap limits of integration, and combine integrals $$ V(\r_b) - V(\r_a) = -\int_{\r_0}^{\r_b} \E \dot \dL' + \int_{\r_0}^{\r_a} \E \dot \dL' = -\int_{\r_a}^{\r_b} \E \dot \dL' . $$ Compare this to the gradient theorem to identify $$ \boxed{\E = - \grad V .} $$ Insert this into Gauss's law to derive Poisson's equation $$ \boxed{\del^2 V = - \q{\rho}{\ep_0} .} $$ The scalar potential relative to infinity of a source electric point charge at $\r' = \vec{0}$ is $$ V(\r) = - \int_\infty^r \bb{\q{1}{4 \pi \ep_0} \q{q'}{r'^2} \ur} \dot [\ur dr'] = - \q{q'}{4 \pi \ep_0} \int_\infty^r \q{dr'}{r'^2} %= \q{1}{4 \pi \ep_0} \q{q'}{r'} \bigg|_{r'=\infty}^r = \q{1}{4 \pi \ep_0} \q{q'}{r} . $$ Generalize this to arbitrary source positions $$ V(\r) = \q{1}{4 \pi \ep_0} \q{q'}{R} $$ then to an arbitrary number of source charges $$ V(\r) = \q{1}{4 \pi \ep_0} \sum_i \q{q'_i}{R_i} $$ then to a continuous charge distribution $$ \boxed{V(\r) = \q{1}{4 \pi \ep_0} \int \q{dq'}{R} .} $$ In particular, the scalar potential for a volume charge density is $$ V(\r) = \q{1}{4 \pi \ep_0} \int_V \q{\rho(\r')}{R} dV' . $$

Polarization

Taylor expand $R^{-1}$ about the origin (see Vector Calculus page) $$ \q{1}{R} = \q{1}{r} \sum_{n=0}^{\infty} \bb{\q{r'}{r}}^n P_n(\cos \theta) $$ and insert this series into the equation for scalar potential of a volume charge density $$ \boxed{V(\r) = \q{1}{4 \pi \ep_0} \sum_{n=0}^{\infty} \q{1}{r^{[n+1]}} \int_V [r']^n P_n(\cos \theta) \rho(\r') dV' .} $$ Expand the first few terms $$ V(\r) = \q{1}{4 \pi \ep_0} \bb{ \q{1}{r} \int_V \rho dV' + \q{1}{r^2} \int_V r' \cos \theta \rho dV' %+ \q{1}{r^3} \int_V [r']^2 \q{1}{2} [-1 + 3 \cos^2 \theta] \rho dV' + \ldots} . $$ The first term is the potential of an electric monopole moment at the origin $$ V_{mono}(\r) = \q{1}{4 \pi \ep_0} \q{Q}{r} . $$ The second term is the potential of an electric dipole moment at the origin $$ V_{di}(\r) = \q{1}{4 \pi \ep_0} \q{1}{r^2} \int_V r' \cos \theta \rho dV' . $$ This equation can be cast into vector form by observing $r' \cos \theta = \ur \dot \r'$ $$ V_{di}(\r) = \q{1}{4 \pi \ep_0} \q{1}{r^2} \int_V \ur \dot \r' \rho dV' = \q{1}{4 \pi \ep_0} \q{1}{r^2} \ur \dot \int_V \r' \rho dV' = \q{1}{4 \pi \ep_0} \q{\ur}{r^2} \dot \p $$ and the electric dipole moment is defined as $$ \boxed{\p \equiv \int_V \r' \rho(\r') dV' .} $$ Generalize the dipole potential to a point dipole at an arbitrary source position $$ V(\r) = \q{1}{4 \pi \ep_0} \q{\uR}{R^2} \dot \p $$ then to a continuous electric dipole moment volume density with $d \p(\r) = \P(\r) dV$ $$ V(\r) = \q{1}{4 \pi \ep_0} \int_V \q{\uR}{R^2}\dot \P(\r') dV' . $$ where $\P$ is polarization density $$ \P(\r) = \lim_{\Delta V \rightarrow 0} \q{\Delta \p}{\Delta V} $$ Use vector identity $\grad R^n = n R^{n-1} \uR$ to rewrite $\uR / R^2 = \grad' R^{-1}$ $$ V(\r) = \q{1}{4 \pi \ep_0} \int_V [\grad' R^{-1}] \dot \P(\r') dV' $$ and integrate by parts by using product rule (3) and the divergence theorem $$ V(\r) = \q{1}{4 \pi \ep_0} \int_V [\del' \dot [R^{-1} \P] - R^{-1} \del' \dot \P] dV' $$ $$ V(\r) = \q{1}{4 \pi \ep_0} \oint_S \q{\P \dot \dS'}{R} - \q{1}{4 \pi \ep_0} \int_V \q{\del' \dot \P}{R} dV' . $$ Compare these integrals to the scalar potential of a continuous charge distribution $$ V(\r) = \q{1}{4 \pi \ep_0} \oint_S \q{\sigma_b}{R} dS' + \q{1}{4 \pi \ep_0} \int_V \q{\rho_b}{R} dV' $$ and define bound electric surface and volume charge densities as $$ \boxed{\sigma_b = \P \dot \un} $$ $$ \boxed{\rho_b = -\div \P .} $$ Electric charge can be split into bound charge and free (not bound) charge $$ \boxed{\rho = \rho_f + \rho_b .} $$ Insert this into Gauss's law $$ \div \E = \q{\rho}{\ep_0} = \q{1}{\ep_0}[\rho_f + \rho_b] = \q{1}{\ep_0} [\rho_f - \div \P] $$ and isolate the free charge $$ \div [\ep_0 \E + \P] = \rho_f . $$ Define the $\D$ field as $$ \boxed{\D \equiv \ep_0 \E + \P .} $$ Gauss's law in matter in differential form is $$ \boxed{\div \D = \rho_f} $$ and in integral form is $$ \boxed{\oint_S \D \dot \dS = Q_f .} $$

Magnetostatics

Continuity equation $$ \boxed{\oint_S \J \dot \dS = - \pdt{Q}} $$ $$ \int_V \div \J dV = -\pdt{} \int_V \rho dV = -\int_V \pdt{\rho} dV $$ $$ \boxed{\div \J = - \pdt{\rho}} $$ Magnetostatics $$ \boxed{\div \J = 0} $$

Biot-Savart Law

$$ \vec{F}_m = q \vec{v} \cross \B $$ $$ \boxed{\B(\r) = \q{\mu_0}{4 \pi} \int \q{d\I' \cross \uR}{R^2}} $$ $$ \boxed{d\I' = \I(\r') dL' = \K(\r') dS' = \J(\r') dV'} $$ $$ \B(\r) = \q{\mu_0}{4 \pi} \int_V \q{\J(\r') \cross \uR}{R^2} dV' $$

Divergence and Curl

Divergence of B $$ \div \B = \q{\mu_0}{4 \pi} \int_V \div \bb{\J \cross \q{\uR}{R^2}} dV' = \q{\mu_0}{4 \pi} \int_V \bb{\q{\uR}{R^2} \dot \curl \J - \J \dot \curl \q{\uR}{R^2}} dV' $$ $\div [\vec{A} \cross \vec{B}] = \vec{B} \dot \curl \vec{A} - \vec{A} \dot \curl \vec{B}$ $\curl \J(\r') = \vec{0}$ $\curl [\uR / R^2] = \vec{0}$ (see vector calc review) $$ \boxed{\div \B = 0} $$ $$ \int_V \div \B dV = \int_V 0 dV $$ $$ \boxed{\oint_S \B \dot \dS = 0} $$

Curl of B $$ \curl \B = \q{\mu_0}{4 \pi} \int_V \curl \bb{\J \cross \q{\uR}{R^2}} dV' = \q{\mu_0}{4 \pi} \int_V \bb{\J \bb{\div \q{\uR}{R^2}} - [\J \dot \del]\q{\uR}{R^2}} dV' $$ $\curl [\vec{A} \cross \vec{B}] = \vec{A} [\div \vec{B}] - \vec{B} [\div \vec{A}] - [\vec{A} \dot \del] \vec{B} + [\vec{B} \dot \del] \vec{A}$ $\div [\uR / R^2] = 4 \pi \delta(\R)$ $$ -[\J \dot \del] \q{\uR}{R^2} = [\J \dot \del'] \q{\uR}{R^2} $$ $$ \ux \int_V [\J \dot \del'] \q{R_x}{R^3} dV' = \ux \int_V \bb{\del' \dot \bb{\J \q{R_x}{R^3}} - \q{R_x}{R^3} \del' \dot \J} dV' = \ux \oint_S \J \q{R_x}{R^3} \dot \dS' = \vec{0} $$ $\div [f \vec{A}] = \grad f \dot \vec{A} + f \div \vec{A}$ $$ \int_V [\J \dot \del] \q{\uR}{R^2} dV' = \vec{0} $$ $$ \curl \B = \q{\mu_0}{4 \pi} \int_V \J(\r') [4 \pi \delta(\R)] dV' = \mu_0 \J(\r) $$ $$ \boxed{\curl \B = \mu_0 \J} $$ $$ \int_S [\curl \B] \dot \dS = \mu_0 \int_S \J \dot \dS $$ $$ \boxed{I = \int_S \J \dot \dS} $$ $$ \boxed{\oint_L \B \dot \dL = \mu_0 I} $$

Vector Potential

$$ \boxed{\B = \curl \A} $$ There is no restriction on divergence of A, and for now it's nice to get rid of it. Suppose $\div \A_o \neq 0$. Add a gradient (doesn't change the curl) $\A = \A_0 + \grad \lambda$ $$ \div \A = \div \A_0 + \del^2 \lambda $$ $$ \del^2 \lambda = - \div \A_0 $$ $$ \lambda = \q{1}{4 \pi} \int_V \q{\div \A_0}{R} dV' $$ $$ \boxed{\div \A = 0} $$ Poisson's equation x3 $$ \curl [\curl \A] = \grad \div \A - \del^2 \A = \mu_0 \J $$ $$ \boxed{\del^2 \A = -\mu_0 \J} $$ vector potential of current distribution $$ \boxed{\A(\r) = \q{\mu_0}{4 \pi} \int_V \q{\J(\r')}{R} dV'} $$ $$ \q{1}{R} = \q{1}{r} \sum_{n=0}^{\infty} \bb{\q{r'}{r}}^n P_n(\cos \theta) $$ $$ \A(\r) = \q{\mu_0 I}{4 \pi} \oint_L \q{\dL'}{R} $$

Magnetization

$$ \boxed{\A(\r) = \q{\mu_0 I}{4 \pi} \sum_{n=0}^{\infty} \q{1}{r^{[n+1]}} \oint_L [r']^n P_n(\cos \theta) \dL'} $$ $$ \A(\r) = \q{\mu_0 I}{4 \pi} \bb{\q{1}{r} \oint_L \dL' + \q{1}{r^2} \oint_L r' \cos\theta \dL' + \q{1}{r^3} \oint_L \q{r'^2}{2}[3 \cos^2 \theta - 1] \dL' + \ldots} $$ $$ \oint_L \dL' = \vec{0} $$ $$ \A_{mono}(\r) = \vec{0} $$ $$ \A_{di}(\r) = \q{\mu_0}{4 \pi} \q{I}{r^2} \oint_L r' \cos\theta \dL' $$ $\ur \dot \r' = r' \cos\theta$ $\oint_L \vec{c} \dot \r \dL = \int_S \dS \cross \vec{c}$ $$ \A_{di}(\r) = \q{\mu_0}{4 \pi} \q{I}{r^2} \oint_L \ur \dot \r' \dL' = \q{\mu_0}{4 \pi} \q{I}{r^2} \bb{\int_S \dS'} \cross \ur $$ $$ \A_{di}(\r) = \q{\mu_0}{4 \pi} \q{\m \cross \ur}{r^2} $$ $$ \boxed{\m \equiv I \int_S \dS} $$ $$ \A(\r) = \q{\mu_0}{4 \pi} \q{\m \cross \uR}{R^2} $$ $d\m' = \M(\r') dV'$ $$ \A(\r) = \q{\mu_0}{4 \pi} \int_V \q{\M(\r') \cross \uR}{R^2} dV' $$ $[\uR / R^2] = \del' R^{-1}$ $$ \A(\r) = \q{\mu_0}{4 \pi} \int_V \M \cross \del' \q{1}{R} dV' $$ $$ \A(\r) = \q{\mu_0}{4 \pi} \int_V \bb{\q{1}{R} \del' \cross \M - \del' \cross \q{\M}{R}} dV' $$ $\curl [f \vec{A}] = \grad f \cross \vec{A} + f \curl \vec{A}$ $\int_V \curl \vec{A} dV = -\oint_S \vec{A} \cross \dS$ (see Vector Calculus page) $$ \A(\r) = \q{\mu_0}{4 \pi} \int_V \q{\del' \cross \M}{R} dV' + \q{\mu_0}{4 \pi} \oint_S \q{\M \cross \dS'}{R} $$ $$ \A(\r) = \q{\mu_0}{4 \pi} \int_V \q{\J_b}{R} dV' + \q{\mu_0}{4 \pi} \oint_S \q{\K_b}{R} dS' $$ $$ \boxed{\J_b = \curl \M} $$ $$ \boxed{\K_b = \M \cross \un} $$ $$ \J = \J_f + \J_b $$ $$ \curl \B = \mu_0 [\J_f + \J_b] = \mu_0 [\J_f + \curl \M] $$ $$ \curl \bb{\q{1}{\mu_0} \B - \M} = \J_f $$ $$ \boxed{\curl \H = \J_f} $$ $$ \boxed{\H \equiv \q{1}{\mu_0} \B - \M} $$

Electrodynamics

Define emf as the closed line integral of force per charge $$ \mathcal{E} \equiv \oint_L \vec{f} \dot \dL $$ $$ \Phi_B \equiv \int_S \B \dot \dS $$ $$ d\Phi_B = \Phi_B(t + dt) - \Phi_B(t) = \int_{dS} \B \dot \dS $$ $$ \dS = [\vec{v} dt] \cross \dL $$ $$ \odt{\Phi_B} = \oint_L \B \dot [\vec{v} \cross \dL] $$ $$ \odt{\Phi_B} = -\oint_L [\vec{v} \cross \B] \dot \dL = -\oint_L \vec{f}_{mag} \dot \dL = -\mathcal{E} $$ $$ \mathcal{E} = - \odt{\Phi_B} $$ Faraday hypothesized a change in magnetic flux induces an electric field emf $$ \boxed{\oint_L \E \dot \dL = - \odt{\Phi_B} .} $$ This is Faraday's law in integral form. Expand the definition for magnetic flux $$ \oint_L \E \dot \dL = - \odt{} \int_S \B \dot \dS $$ apply the curl theorem and swap the order of differentiation and integration $$ \int_S [\curl \E] \dot \dS = - \int_S \pdt{\B} \dot \dS $$ This equation has to hold for any surface $S$, so the integrands must be equal $$ \boxed{\curl \E = -\pdt{\B} .} $$ This is the differential form of Faraday's law.

Maxwell recognized in electrodynamics the divergence of Ampere's law doesn't necessarily equal zero because their are no constraints on $\div \J$ $$ \div [\curl \B] = \mu_0 \div \J $$ $$ \vec{0} \neq \mu_0 \div \J $$ Maxwell used the continuity equation and Gauss's law $$ \div \J = -\pdt{\rho} = - \pdt{} \div [\ep_0 \E] $$ to propose a consistent modification to Ampere's law on theoretical grounds $$ \boxed{\curl \B = \mu_0 \J + \ep_0 \mu_0 \pdt{\E} .} $$ This modification explains observation and couples the electric and magnetic fields.

Maxwell's Equations

The differential form of Maxwell's equations is $$ \boxed{ \begin{cases} \curl \E = -\pdt{\B} \\ \curl \B = \mu_0 \J + \ep_0 \mu_0 \pdt{\E} \\ \div \E = \q{\rho}{\ep_0} \\ \div \B = 0 \end{cases} \quad\quad\text{BCs:} \begin{cases} \un \cross \Delta \E = \vec{0} \\ \un \cross \Delta \B = \mu_0 \K \\ \un \dot \Delta \E = \q{\sigma}{\ep_0} \\ \un \dot \Delta \B = 0 \end{cases} } $$ and the electromagnetic (Lorentz) force law is $$ \boxed{\vec{f} = \E + \vec{v} \cross \B .} $$

Maxwell's Equations in matter

Electrodynamics in matter requires considering current of changing polarization density $$ \boxed{\J_p \equiv \pdt{\P}} $$ $$ \J = \J_f + \J_b + \J_p = \J_f + \curl \M + \pdt{\P} $$ $$ \curl \B = \mu_0 \bb{\J_f + \curl \M + \pdt{\P}} + \ep_0 \mu_0 \pdt{\E} $$ $$ \curl \bb{\q{1}{\mu_0}\B - \M} = \J_f + \pdt{} [\ep_0 \E + \P] $$ $$ \curl \H = \J_f + \pdt{\D} $$ $$ \rho = \rho_f + \rho_b = \rho_f - \div \P $$ $$ \div \E = \q{1}{\ep_0} [\rho_f - \div \P] $$ $$ \div [\ep_0 \E + \P] = \rho_f $$ $$ \div \D = \rho_f $$ $$ \boxed{ \begin{cases} \curl \E = -\pdt{\B} \\ \curl \H = \J_f + \pdt{\D} \\ \div \D = \rho_f \\ \div \B = 0 \end{cases} \quad\quad\text{BCs:} \begin{cases} \un \cross \Delta \E = \vec{0} \\ \un \cross \Delta \H = \K_f \\ \un \dot \Delta \D = \sigma_f \\ \un \dot \Delta \B = 0 \end{cases} } $$ Auxiliary fields $$ \boxed{\D \equiv \ep_0 \E + \P} $$ $$ \boxed{\H \equiv \q{1}{\mu_0} \B - \M} $$

Symmetric Maxwell's Equations in Matter

see forward model $$ \boxed{ \begin{cases} \curl \E = -\J_{mf} - \pdt{\B} \\ \curl \H = \J_{ef} + \pdt{\D} \\ \div \D = \rho_{ef} \\ \div \B = \rho_{mf} \end{cases} \quad\quad\text{BCs:} \begin{cases} \un \cross \Delta \E = - \K_{mf} \\ \un \cross \Delta \H = \K_{ef} \\ \un \dot \Delta \D = \sigma_{ef} \\ \un \dot \Delta \B = \sigma_{mf} \end{cases} } $$ in (nonmagentoelectric) linear matter $$ \boxed{ \begin{cases} \curl \E = -\J_{mf} - \pdt{}[\mu \H] \\ \curl \H = \J_{ef} + \pdt{}[\ep \E] \\ \div [\ep \E] = \rho_{ef} \\ \div [\mu \H] = \rho_{mf} \end{cases} \quad\quad\text{BCs:} \begin{cases} \un \cross \Delta \E = - \K_{mf} \\ \un \cross \Delta \H = \K_{ef} \\ \un \dot \Delta [\ep \E] = \sigma_{ef} \\ \un \dot \Delta [\mu \H] = \sigma_{mf} \end{cases} } $$

Time Harmonic Fields

see forward model $$ \boxed{ \begin{cases} \curl \z{\E} = -\z{\J}_{mi} - j \w \z{\mu} \z{\H} \\ \curl \z{\H} = \z{\J}_{ei} + j \w \z{\ep} \z{\E} \\ \div [\z{\ep} \z{\E}] = \z{\rho}_{ei} \\ \div [\z{\mu} \z{\H}] = \z{\rho}_{mi} \\ \end{cases} \quad\quad\text{BCs:} \begin{cases} \un \cross \Delta \z{\E} = - \z{\K}_{mi} \\ \un \cross \Delta \z{\H} = \z{\K}_{ei} \\ \un \dot \Delta [\z{\ep} \z{\E}] = \z{\sigma}_{ei} \\ \un \dot \Delta [\z{\mu} \z{\H}] = \z{\sigma}_{mi} \end{cases} } $$ note in em review that in time harmonic formalism, dot product is no longer an inner product, but an instruction of how to combine harmonic quantities. dot products (and cross products) are rarely applied. carefully consider meaning every time dot or cross product is applied. grad, div, and curl are real operators applied to complex values, so act normally even on time harmonic quantities. when in doubt, remember the physical situation is to take the real part of all fields and then applying operations.

Theorems

Uniqueness

Time Harmonic Fields in lossy materials (take limit in lossless case) $$ \begin{cases} \curl \E_1 = -\J_{mi} - j \w \mu \H_1 \\ \curl \H_1 = \J_{ei} + j \w \ep \E_1 \end{cases} \quad\quad\quad \begin{cases} \curl \E_2 = -\J_{mi} - j \w \mu \H_2 \\ \curl \H_2 = \J_{ei} + j \w \ep \E_2 \end{cases} $$ $$ \begin{cases} \curl \Delta \E = -j \w \mu \Delta \H \\ \curl \Delta \H = j \w \ep \Delta \E \end{cases} $$ $$ \div [\Delta \E \cross \conj{\Delta \H}] = -j \w \conj{\mu} \nn{\Delta \H}^2 + j \w \ep \nn{\Delta \E}^2 $$ $$ \int_V \Re(\div [\Delta \E \cross \conj{\Delta \H}]) dV = \int_V \Re(-j \w \conj{\mu} \nn{\Delta \H}^2 + j \w \ep \nn{\Delta \E}^2) dV $$ $\z{\ep} \equiv \ep + \kappa_e/[j \w]$ and $\z{\mu} \equiv \mu + \kappa_m/[j \w]$ $$ \oint_S \Re(\Delta \E \cross \conj{\Delta \H}) \dot \dS = -\int_V [\kappa_m \nn{\Delta \H}^2 + \kappa_e \nn{\Delta \E}^2] dV $$ i feel like if i took the $\Im$ instead this would always hold even in lossless media...

1. $\un \cross \E$ is specified on $S$, i.e. $\un \cross \Delta \E = \vec{0}$ on $S$
2. $\un \cross \H$ is specified on $S$, i.e. $\un \cross \Delta \H = \vec{0}$ on $S$
3. $S$ is partitioned into $S_i$ each of which has either $\un \cross \E$ or $\un \cross \H$ specified

Note, Maxwell's equations are not linearly independent, and neither are the BCs. By knowing the tangential components of the fields on a surface, the normal components can be deduced (how? continuity equation?).

Duality

Electromagnetic equations are symmetric in electric and magnetic quantities. Given any electromagnetic equation, a dual equation is derived by making the substitutions

Reciprocity

see forward model

note, need to first consider integration volume that encloses both current distributions. then it can be broken into two disjoint integration volumes over currents if desired (and as currently presented).

Image Theory

for PEC/PMC planes $$ \r' = \r - 2 \un [\un \dot \r] $$ $$ \J_e(\r) = \mp \J_e(\r') \pm 2 \un [\un \dot \J_e(\r')] $$ $$ \J_m(\r) = \pm \J_m(\r') \mp 2 \un [\un \dot \J_m(\r')] $$ for $\r \dot \un < 0$. this is a type of equivalence theorem where the PEC/PMC is replaced by the background medium and the BCs at the interface are satisfied.

Equivalence

see forward model

Radiation

This appendix derives the fields radiated by arbitrary source current distributions in a homogenous linear isotropic medium.

Symmetric Potentials

Assume \underline{linear media}. $$ \boxed{\E = \E_e + \E_m} $$ $$ \boxed{\H = \H_e + \H_m} $$ Assume isotropic homogeneous nonmagnetoelectric media. Assume underline{time harmonic formalism.

No magnetic sources $$ \begin{cases} \curl \E_e = - j \w \mu \H_e \\ \curl \H_e = \J_{ei} + j \w \ep \E_e \\ \div [\ep \E_e] = \rho_{ei} \\ \div [\mu \H_e] = 0 \end{cases} $$ $$ \mu \H_e = \curl \A_e $$ $$ \curl [\E_e + j \w \A_e] = \vec{0} $$ $$ \E_e + j \w \A_e = - \grad \phi_e $$ $$ \E_e = - \grad \phi_e - j \w \A_e $$ Impose Lorenz gauge $$ \div \A_e + j \w \ep \mu \phi_e = 0 $$ $$ \phi_e = - \q{1}{j \w \ep \mu} \div \A_e $$ $$ \del^2 \A_e + k^2 \A_e = - \mu \J_{ei} $$ $$ \del^2 \phi_e + k^2 \phi_e = - \q{\rho_{ei}}{\ep} $$

No electric sources $$ \begin{cases} \curl \E_m = - \J_{mi} - j \w \mu \H_m \\ \curl \H_m = j \w \ep \E_m \\ \div [\ep \E_m] = 0 \\ \div [\mu \H_m] = \rho_{mi} \end{cases} $$ $$ \ep \E_m = - \curl \A_m $$ $$ \curl [\H_m + j \w \A_m] = 0 $$ $$ \H_m + j \w \A_m = - \grad \phi_m $$ $$ \H_m = - \grad \phi_m - j \w \A_m $$ Impose Lorenz gauge $$ \div \A_m + j \w \ep \mu \phi_m = 0 $$ $$ \del^2 \A_m + k^2 \A_m = - \ep \J_{mi} $$ $$ \del^2 \phi_m + k^2 \phi_m = - \q{\rho_{mi}}{\mu} $$

Potentials due to Sources

$$ \del^2 \A_e + k^2 \A_e = - \mu \J_{ei} $$ $$ \J_{ei} = \uz J_{e0} \delta(\r) $$ $$ \del^2 A_{ez} + k^2 A_{ez} = -\mu J_{e0} \delta(\r) $$ $$ \del^2 = \q{1}{r^2} \pd{}{r} r^2 \pd{}{r} $$ $$ A_{ez} = \q{f(r)}{r} $$ $$ \begin{split} \del^2 \bb{\q{f}{r}} &= \q{1}{r^2} \od{}{r} r^2 \od{}{r} \bb{\q{f}{r}} \\ &= \q{1}{r^2} \od{}{r} r^2 \bb{\q{1}{r} \od{f}{r} - \q{f}{r^2}} \\ &= \q{1}{r^2} \od{}{r} \bb{r \od{f}{r} - f} \\ &= \q{1}{r^2} \bb{\od{f}{r} + r \odn{f}{r}{2} - \od{f}{r}} \\ &= \q{1}{r} \odn{f}{r}{2} \end{split} $$ $$ \q{1}{r}\odn{f}{r}{2} + k^2 \q{f}{r} = - \mu J_{e0} \delta(\r) $$ $$ \odn{f}{r}{2} + k^2 f = - \mu J_{e0} r \delta(\r) $$ $$ f(r) = C_1 e^{-j k r} + C_2 e^{j k r} $$ apply radiation BCs $$ f(r) = C_1 e^{-j k r} $$ $$ A_{ez}(r) = C_1 \q{1}{r} e^{-j k r} $$ $$ \div \grad A_{ez} + k^2 A_{ez} = - \mu J_{e0} \delta(\r) $$ $$ \oint_S \grad A_{ez} \dot \dS + \int_V k^2 A_{ez} dV = - \int_V \mu J_{e0} \delta(\r) dV $$ $$ \grad A_{ez} = \ur \od{}{r} \bb{C_1 \q{1}{r} e^{-j k r}} = \ur C_1 \bb{-\q{1}{r^2} - \q{j k}{r}} e^{-j k r} $$ $$ \begin{split} \int_S \grad A_{ez} \dot \dS &= \int_0^{2 \pi} \int_0^\pi \bb{\ur C_1 \bb{-\q{1}{a^2} - \q{j k}{a}} e^{-j k a}} \dot [\ur a^2 \sin\theta d\theta d\phi] \\ &= \int_0^{2 \pi} \int_0^\pi C_1 [-1 - j k a] e^{-j k a} \sin\theta d\theta d\phi \\ &= C_1[-1 - j k a] e^{-j k a} \int_0^\pi \sin\theta d\theta \int_0^{2 \pi} d\phi \\ &= C_1[-1 - j k a] e^{-j k a} [2] [2 \pi] \end{split} $$ $$ \begin{split} \int_V k^2 A_{ez} dV &= \int_0^{2 \pi} \int_0^\pi \int_0^a k^2 \bb{C_1 \q{1}{r} e^{-j k r}} [r^2 \sin\theta dr d\theta d\phi] \\ &= \int_0^{2 \pi} \int_0^\pi \int_0^a k^2 C_1 r e^{-j k r} \sin\theta dr d\theta d\phi \\ &= C_1 k^2 \int_0^a r e^{-j k r} dr \int_0^\pi \sin\theta d\theta \int_0^{2 \pi} d\phi \\ &= C_1 k^2 \int_0^a r e^{-j k r} dr [2] [2 \pi] \end{split} $$ $$ -\int_V \mu J_{e0} \delta(\r) dV = - \mu J_{e0} $$ $$ C_1 4 \pi [-1 - j k a] e^{-j k a} + C_1 4 \pi k^2 \int_0^a r e^{-j k r} dr = - \mu J_{e0} $$ $$ \lim_{a \rightarrow 0} \bb{C_1 4 \pi [-1 - j k a] e^{-j k a} + C_1 4 \pi k^2 \int_0^a r e^{-j k r} dr = - \mu J_{e0}} $$ $$ - C_1 4 \pi = - \mu J_{e0} $$ $$ C_1 = \q{\mu J_{e0}}{4 \pi} $$ $$ A_{ez} = \mu J_{e0} \q{1}{4 \pi r} e^{-j k r} $$ $$ \A_e = \mu \J_{e0} \q{1}{4 \pi r} e^{-j k r} $$ $$ \A_e = \mu \J_{e0} \q{1}{4 \pi R} e^{-j k R} $$ $$ \A_e = \mu \int_V \q{1}{4 \pi R} e^{-j k R} \J_{e} dV' $$ $$ \A_e = \mu \int_V G(\r, \r') \J_e(\r') dV' $$ $$ G(\r, \r') = \q{1}{4 \pi R} e^{-j k R} $$ $$ \A_m = \ep \int_V G(\r, \r')_{3D} \J_m(\r') dV' $$

Fields due to Potentials

$$ \grad R^n = n R^{n - 1} \uR $$ $$ \grad e^{-j k R} = e^{-j k R} \grad [-j k R] = -j k e^{-j k R} \uR $$ $$ \begin{split} \H_e &= \q{1}{\mu} \curl \A_e \\ &= \q{1}{\mu} \curl \bb{\mu \int_V G \J_e dV'} \\ &= \int_V \grad G \cross \J_e dV' \\ &= \int_V \grad \bb{\q{1}{4 \pi R} e^{-j k R}} \cross \J_e dV' \end{split} $$ $$ \begin{split} \grad [[4 \pi R]^{-1} e^{-j k R}] &= [4 \pi]^{-1} [[\grad R^{-1}] e^{-j k R} + R^{-1} [\grad e^{-j k R}]] \\ &= [4 \pi]^{-1} [[- R^{-2} \uR] e^{-j k R} + R^{-1} [-j k e^{-j k R} \uR]] \\ &= -[4 \pi]^{-1} [j k R^{-1} + R^{-2}] e^{-j k R} \uR \\ &= -G_0 \uR \end{split} $$ $$ \boxed{G_0 \equiv [4 \pi]^{-1} [j k R^{-1} + R^{-2}] e^{-j k R}} $$ $$ \boxed{\H_e(\r) = -\int_V G_0 \uR \cross \J_e(\r') dV'} $$ $$ \E_e = \q{1}{j \w \ep} [\curl \H_e - \J_e] $$ $$ \begin{split} \curl \H_e &= \curl \bb{-\int_V G_0 \uR \cross \J_e dV'} \\ &= -\int_V \curl [G_0 \uR \cross \J_e] dV' \\ &= -\int_V [[\grad [G_0 R^{-1}]] \cross [\R \cross \J_e] + G_0 R^{-1} \curl [\R \cross \J_e]] dV' \end{split} $$ $$ \begin{split} \grad [G_0 R^{-1}] &= \grad [ [4 \pi]^{-1} [j k R^{-2} + R^{-3}] e^{-j k R}] \\ &= [4 \pi]^{-1} \grad [[j k R^{-2} + R^{-3}] e^{-j k R}] \\ &= [4 \pi]^{-1} [[\grad [j k R^{-2} + R^{-3}]] e^{-j k R} + [j k R^{-2} + R^{-3}] [\grad e^{-j k R}]] \\ &= [4 \pi]^{-1} [[-j 2 k R^{-3} \uR - 3 R^{-4} \uR] e^{-j k R} + [j k R^{-2} + R^{-3}] [-j k e^{-j k R} \uR]] \\ &= [4 \pi]^{-1} [-j 2 k R^{-3} - 3 R^{-4} + k^2 R^{-2} - j k R^{-3}] e^{-j k R} \uR \\ &= [4 \pi]^{-1} [k^2 R^{-2} - j 3 k R^{-3} - 3 R^{-4}] e^{-j k R} \uR \\ &= -[4 \pi]^{-1} [-k^2 R^{-1} + j 3 k R^{-2} + 3 R^{-3}] e^{-j k R} R^{-1} \uR \\ &= -G_1 R^{-1} \uR \\ \end{split} $$ $$ \boxed{G_1 \equiv [4 \pi]^{-1} [- k^2 R^{-1} + j 3 k R^{-2} + 3 R^{-3}] e^{-j k R}} $$ $$ \begin{split} \curl [\R \cross \J] &= \R [\div \J] - \J [\div \R] - [\R \dot \del] \J + [\J \dot \del] \R \\ &= -\J [\div \R] + [\J \dot \del] \R \\ &= -\J [3] + [\J] \\ &= - 2 \J \end{split} $$ $$ \uR \cross [\uR \cross \J] = \uR [\uR \dot \J] - \J [\uR \dot \uR] = \uR [\uR \dot \J] - \J $$ $$ \begin{split} \curl \H_e &= -\int_V [[-G_1 R^{-1} \uR] \cross [\R \cross \J_e] + G_0 R^{-1} [-2 \J_e]] dV' \\ &= \int_V [G_1 \uR \cross [\uR \cross \J_e] + 2 G_0 R^{-1} \J_e] dV' \\ &= \int_V [G_1 [\uR[\uR \dot \J_e] - \J_e] + 2 G_0 R^{-1} \J_e] dV' \\ &= \int_V [[2 G_0 R^{-1} - G_1] \J_e + G_1 \uR [\uR \dot \J_e]] dV' \\ &= \int_V [G_2 \J_e + G_1 \uR [\uR \dot \J_e]] dV' \\ \end{split} $$ $$ \boxed{G_2 \equiv [4 \pi]^{-1} [k^2 R^{-1} - j k R^{-2} - R^{-3}] e^{-j k R}} $$ $$ \boxed{\E_e(\r) = \q{1}{j \w \ep} \int_V [G_2 \J_e(\r') + G_1 \uR [\uR \dot \J_e(\r')]] dV' - \q{1}{j \w \ep} \J_e(\r)} $$ $$ \begin{align} \E_e(\r) &= \q{1}{j \w \ep} \int_{V} [G_2 \J_e(\r') + G_1 \uR [\uR \dot \J_e(\r')]] dV' - \q{1}{j \w \ep} \J_e(\r) \\ \E_m(\r) &= \int_{V} G_0 \uR \cross \J_m(\r') dV' \\ \H_e(\r) &= - \int_{V} G_0 \uR \cross \J_e(\r') dV' \\ \H_m(\r) &= \q{1}{j \w \mu} \int_{V} [G_2 \J_m(\r') + G_1 \uR [\uR \dot \J_m(\r')]] dV' - \q{1}{j \w \mu} \J_m(\r) \end{align} $$ where $$ \begin{align} G_0 &\equiv [4 \pi]^{-1} [j k R^{-1} + R^{-2}] e^{-j k R} \\ G_1 &\equiv [4 \pi]^{-1} [-k^2 R^{-1} + j 3 k R^{-2} + 3 R^{-3}] e^{-j k R} \\ G_2 &\equiv [4 \pi]^{-1} [k^2 R^{-1} - j k R^{-2} - R^{-3}] e^{-j k R} \end{align} $$ Sommerfeld Radiation $$ \lim_{r \rightarrow \infty} r [\E + \eta \ur \cross \H] = 0 $$ $$ \lim_{r \rightarrow \infty} r [\H - \eta^{-1} \ur \cross \E] = 0 $$ where $\eta = \sqrt{\mu / \ep}$