This collection is inspired on my EM II class. Further, it showcases formulas at the appropriate level for tests such as the PhysGRE.
Since these were originally conceived as a personal aid for my EM II class, a lot of electrostatics and magnetostatics are omitted, and what's included has been consolidated into a single section. The order follows that of my class and Griffiths's book. A lot of results and applications can be seen in my class page when I was the TA for EM I.
Constants.
Vector Calculus Identities.
Electrostatics.
Standard electrostatic configurations.
Magnetostatics.
Standard magnetostatics configurations.
Electromotive force.
Electromagnetic induction.
Maxwell Equations. (+ in medium)
Conservation laws.
Classical waves.
EM Waves (Vacuum).
EM Waves (in matter).
EM Waves (in conductors).
EM Waves (dispersion).
EM Waves (Wave Guides).
Potentials.
Potentials in continuous distributions.
Potentials and point charges.
Permittivity of free space
\[\epsilon _0 =8.85\times 10^{-12} \frac{\text{C} ^2 }{\text{N} \cdot \text{m} ^2 }\]Permeability of free space:
\[\mu _0 =4\pi \times 10^{-7}\frac{\text{N} }{\text{A} ^2 } \]Fundamental theorem for Gradients
\[\int_{\mathbf{a}}^{\mathbf{b}} (\nabla T)\cdot d\mathbf{l}=T(\mathbf{b})-T(\mathbf{a}) \]Fundamental theorem for Divergences (Gauss's/Green's/divergence theorem)
\[\int _{\mathcal{V}}(\nabla \cdot \mathbf{v})d\tau =\oint _{\mathcal{S}}\mathbf{v}\cdot d\mathbf{a} \]Fundamental theorem for Curls (Stokes' theorem)
\[\int _{\mathcal{S}}(\nabla \times \mathbf{v})\cdot d\mathbf{a}=\oint _{\mathcal{P}}\mathbf{v}\cdot d\mathbf{l} \]Dot/cross triple products
\begin{align*} \mathbf{A}\cdot (\mathbf{B}\times \mathbf{C}) &= \mathbf{B}\cdot (\mathbf{C}\times \mathbf{A})=\mathbf{C}\cdot (\mathbf{A}\times \mathbf{B})\tag{Cyclic permutations} \\ \mathbf{A}\times (\mathbf{B}\times \mathbf{C}) &= \mathbf{B}(\mathbf{A}\cdot \mathbf{C})-\mathbf{C}(\mathbf{A}\cdot \mathbf{B}) \end{align*}Gradient, Divergence, and Curl product rules:
\begin{align*} \nabla (fg) &= f(\nabla g)+g(\nabla f) \\ \nabla (\mathbf{A}\cdot \mathbf{B}) &= \mathbf{A}\times (\nabla \times \mathbf{B})+\mathbf{B}\times (\nabla \times \mathbf{A})+(\mathbf{A}\cdot \nabla )\mathbf{B}+(\mathbf{B}\cdot \nabla )\mathbf{A} \\ \nabla \cdot (f\mathbf{A}) &= f(\nabla \cdot \mathbf{A})+\mathbf{A}\cdot (\nabla f) \\ \nabla \cdot (\mathbf{A}\times \mathbf{B}) &= \mathbf{B}\cdot (\nabla \times \mathbf{A})-\mathbf{A}\cdot (\nabla \times \mathbf{B}) \\ \nabla \times (f\mathbf{A}) &= f(\nabla \times \mathbf{A})-\mathbf{A}\times (\nabla f) \\ \nabla \times (\mathbf{A}\times \mathbf{B}) &= (\mathbf{B}\cdot \nabla )\mathbf{A}-(\mathbf{A}\cdot \nabla )\mathbf{B}+\mathbf{A}(\nabla \cdot \mathbf{B})-\mathbf{B}(\nabla \cdot \mathbf{A}) \end{align*}Second derivatives:
\begin{align*} \nabla \cdot (\nabla \times \mathbf{A}) &= 0 \\ \nabla \times (\nabla f) &= \mathbf{0} \\ \nabla \times (\nabla \times \mathbf{A}) &= \nabla (\nabla \cdot \mathbf{A})-\nabla ^2 \mathbf{A} \end{align*}Spherical line and volume elements:
\begin{align*} d\mathbf{l} &= dr\hat{\mathbf{r}}+rd\theta \hat{\theta }+r\sin \theta d\phi \hat{\phi } \\ d\tau &= r^2 \sin \theta drd\theta d\phi \end{align*}Cylindrical line and volume elements:
\begin{align*} d\mathbf{l} &= ds\hat{\mathbf{s}}+sd\phi \hat{\phi }+dz\hat{z} \\ d\tau &= sdsd\phi dz \end{align*}Separation vector:
\[\mathscr{r}=\mathbf{r}-\mathbf{r}' \]Coulomb's law:
\[\mathbf{F}= \frac{1}{4\pi \epsilon _0 }\frac{qQ}{\mathscr{r}^2 }\hat{\mathscr{r}}\]Electric field for a line, surface, volume charge:
\begin{align*} \mathbf{E}(\mathbf{r}) &= \frac{1}{4\pi \epsilon _0 }\int \frac{\lambda (\mathbf{r}')}{\mathscr{r}^2 }\hat{\mathscr{r}}dl' \\ \mathbf{E}(\mathbf{r}) &= \frac{1}{4\pi \epsilon _0 }\int \frac{\sigma (\mathbf{r}')}{\mathscr{r}^2 }\hat{\mathscr{r}}da' \\ \mathbf{E}(\mathbf{r}) &= \frac{1}{4\pi \epsilon _0 }\int \frac{\rho (\mathbf{r}')}{\mathscr{r}^2 }\hat{\mathscr{r}}d\tau ' \end{align*}Electric flux:
\[\Phi _E=\int _{\mathcal{S}}\mathbf{E}\cdot d\mathbf{a} \]Gauss's law:
\[\oint \mathbf{E}\cdot d\mathbf{a}=\frac{1}{\epsilon _0 }Q_{\text{enc} } \]Gauss's law in differential form:
\[\nabla \cdot \mathbf{E}=\frac{1}{\epsilon _0 }\rho \]3 useful symmetries for Gauss's law:
In the case of electrostatics:
\[\nabla \times \mathbf{E}=\mathbf{0} \]Electric potential $V $:
\[\mathbf{E}=-\nabla V \]Inversely:
\[V=-\int \mathbf{E}\cdot d\mathbf{l} \]Poisson's equation:
\[\nabla ^2 V=-\frac{\rho }{\epsilon _0 } \]General solution to Poisson's equation:
\[V(\mathbf{r})=\frac{1}{4\pi \epsilon _0 }\int \frac{\rho (\mathbf{r}')}{\mathscr{r}}d\tau ' \]Laplace's equation:
\[\nabla ^2 V=0 \]General solution to Laplace's equation in spherical coordinates with azimuthal symmetry:
\[V(r,\theta )=\sum_{l=0}^{\infty } \left( A_lr^{l}+\frac{B_l}{r^{l+1}} \right) P_l(\cos \theta ) \]List of Legendre polynomials:
\begin{align*} P_{0}(x) &= 1 \\ P_{1}(x) &= x \\ P_{2}(x) &= \frac{3x^2 -1}{2} \\ P_{3}(x) &= \frac{5x^3 -3x}{2} \\ P_{4}(x) &= \frac{35x^4 -30x^2 +3}{8} \\ P_{5}(x) &= \frac{63x^5 -70x^3 +15x}{8} \end{align*}General solution to Laplace's equation in cylindrical coordinates with cylindrical symmetry:
\[V(s,\phi )=a_0 +b_0 \ln s + \sum_{k=1}^{\infty } \bigg[ s^k (a_k \cos \left( k\phi \right) +b_k \sin \left( k\phi \right) )+s^{-k}(c_k \cos \left( k\phi \right) +d_k \sin \left( k\phi \right) )\bigg] \]Uniqueness theorems:
Boundary conditions for the $\mathbf{E} $ field:
\begin{align*} E_{\text{above} }^{\perp}-E_{\text{below} }^{\perp} &= \frac{1}{\epsilon _0 }\sigma \\ \mathbf{E}^{\parallel}_{\text{above} } &= \mathbf{E}^{\parallel}_{\text{below} } \end{align*}Boundary conditions for the $V $ field:
\begin{align*} V_{\text{above} } &= V_{\text{below} } \\ \frac{\partial V_{\text{above} }}{\partial n}-\frac{\partial V_{\text{below} }}{\partial n} &= -\frac{1}{\epsilon _0 }\sigma \end{align*}Energy of a continuous charge distribution:
\begin{align*} W &= \frac{1}{2}\int \rho Vd\tau \\ &= \frac{\epsilon _0 }{2}\int_{\text{all space} } E^2 d\tau \end{align*}Ideal conductors are such that
Capacitors:
Multipole expansion:
\[V(\mathbf{r})=\frac{1}{4\pi \epsilon _0 }\sum_{n=0}^{\infty } \frac{1}{r^{(n+1)}}\int (r')^n P_n (\cos \left( \alpha \right) )\rho (\mathbf{r}')d\tau ' \]Monopole term:
\[V_{\text{mon} }(\mathbf{r})=\frac{1}{4\pi \epsilon _0 }\frac{Q}{r} \]Dipole moment:
\[\mathbf{p}\equiv \int \mathbf{r}'\rho (\mathbf{r}')d\tau ' \]Dipole term:
\[V_{\text{dip} }(\mathbf{r})=\frac{1}{4\pi \epsilon _0 }\frac{\mathbf{p}\cdot \hat{\mathbf{r}}}{r^2 } \]Physical dipole
\[\mathbf{p}=q(q\mathbf{r}'_+-q\mathbf{r}'_-)=q\mathbf{d} \] \begin{align*} V_{\text{dip} }(r,\theta ) &= \frac{p\cos \theta }{4\pi \epsilon _0 r^2 } \\ \mathbf{E}_{\text{dip} }(r,\theta ) &= \frac{p}{4\pi \epsilon _0 r^3 }\bigg(2\cos \theta \hat{\mathbf{r}}+\sin \theta \hat{\theta }) \end{align*}Torque experienced by an ideal dipole due to an external field:
\[\mathbf{n}=\mathbf{p}\times \mathbf{E} \]Energy of an ideal dipole in an electric field:
\[U=-\mathbf{p}\cdot \mathbf{E} \]Polarization
\[\mathbf{P}\equiv \text{dipole moment per unit volume} \]Surface and volume bound charges
\begin{align*} \sigma _b &= \mathbf{P}\cdot \hat{n} \\ \rho _b &= -\nabla \cdot \mathbf{P} \end{align*}Electric Displacement:
\[\mathbf{D}=\epsilon _0 \mathbf{E}+\mathbf{P} \]Gauss's law for the electric displacement $\mathbf{D} $:
\[\nabla \cdot \mathbf{D}=\rho _f \;\Leftrightarrow \; \oint \mathbf{D}\cdot d\mathbf{a}=Q_{f_{\text{enc} }} \]Boundary conditions for the electric displacement $\mathbf{D} $:
\begin{align*} D^{\perp}_{\text{above} }-D^{\perp}_{\text{below} } &= \sigma _f \\ \mathbf{D}^{\parallel}_{\text{above} }-\mathbf{D}^{\parallel}_{\text{below} } &= \mathbf{P}^{\parallel}_{\text{above} }-\mathbf{P}^{\parallel}_{\text{below} } \end{align*}In a linear dielectric:
\[\mathbf{P}=\epsilon _0 \chi _e\mathbf{E} \]Where $\chi _e $ is the electric susceptibility.
With the permittivity of a material:
\[\epsilon \equiv \epsilon _0 (1+\chi _e) \]The electric displacement is:
\[\mathbf{D}=\epsilon \mathbf{E} \]Relative permittivity/Dielectric constant:
\[\epsilon _r\equiv 1+\chi _e=\frac{\epsilon }{\epsilon _0 } \]$\mathbf{E} $ field a distance $z $ above the midpoint of a straight line segment of length $2L $ and (uniform) line charge density $\lambda $.
\[\mathbf{E}=\frac{1}{4\pi \epsilon _0 }\frac{2\lambda L}{z\sqrt{z^2 +L^2 }}\hat{\mathbf{z}} \]$\mathbf{E} $ field a distance $z $ above the center of a square loop of side $a $ with uniform line charge density $\lambda $.
\[\mathbf{E}=\frac{1}{4\pi \epsilon _0 }\frac{4\lambda az}{(z^2 +\frac{a^2 }{4})\sqrt{z^2 +\frac{a^2 }{2}}} \]$\mathbf{E} $ field a distance $z $ above the center of a circular loop of radius $r $ with uniform line charge density $\lambda $.
\[\mathbf{E}=\frac{1}{4\pi \epsilon _0 }\frac{\lambda (2\pi r)z}{(r^2 +z^2 )^{3/2}}\hat{\mathbf{z}} \]$\mathbf{E} $ field a distance $z $ above the center of a flat circular disk of radius $R $ and uniform surface charge density $\sigma $.
\[\mathbf{E}=\frac{1}{4\pi \epsilon _0 }2\pi \sigma z \left[ \frac{1}{z}-\frac{1}{\sqrt{R^2 +z^2 }} \right] \hat{\mathbf{z}} \]$E $ field of a spherical surface of radius $R $ with uniform surface charge density $\sigma $ (total charge $Q $).
\begin{align*} \mathbf{E}_{\text{in} } &= \mathbf{0} \\ \mathbf{E}_{\text{out} } &= \frac{1}{4\pi \epsilon _0 }\frac{q}{z^2 \hat{z}} \end{align*}$\mathbf{E} $ field of a solid sphere of radius $R $ with uniform volume charge density $\rho $ (total charge $Q $).
\begin{align*} \mathbf{E}_{\text{in} } &= \frac{1}{4\pi \epsilon _0 }\frac{Q}{R^3 }r\hat{\mathbf{r}} \\ \mathbf{E}_{\text{out} } &= \frac{1}{4\pi \epsilon _0 }\frac{Q}{r^2 }\hat{r} \end{align*}$\mathbf{E} $ field of an infinite plane with uniform surface charge density $\sigma $.
\[\mathbf{E}=\frac{\sigma }{2\epsilon _0 }\hat{n} \]$\mathbf{E} $ field of an infinitely long straight wire with uniform line charge density $\lambda $:
\[\mathbf{E}=\frac{\lambda }{2\pi \epsilon _0 s}\hat{\mathbf{s}} \]$V $ field a distance $z $ above the midpoint of a straight line segment of length $2L $ and (uniform) line charge density $\lambda $.
\[V(z)=\frac{\lambda }{2\pi \epsilon _0 }\ln \left( \frac{L+\sqrt{z^2 +L^2 }}{z} \right) \]$V $ field a distance $z $ above the center of a flat circular disk of radius $R $ and uniform surface charge density $\sigma $.
\[ V(z)=\frac{\sigma }{2\epsilon _0 } \left( \sqrt{R^2 +z^2 }-z \right) \]$V $ field inside a solid uniformly charged sphere of radius $R $ and total charge $Q $.
\[V(r)=\frac{q}{8\pi \epsilon _0 R} \left( 3-\frac{r^2 }{R^2 } \right) \]Lorentz Force law:
\[\mathbf{F}=Q \left[ \mathbf{E}+(\mathbf{v}\times \mathbf{B}) \right] \]In regards to work:
\[\text{Magnetic forces do no work.} \]Unit of amperes (current being charge per unit time):
\[1\;\text{A} =1\;\frac{\text{C} }{\text{s} } \]Magnetic force on a segment of current-carrying wire:
\[\mathbf{F}_{\text{mag} }=I\int d\mathbf{l}\times \mathbf{B} \]Surface current density:
\begin{align*} \mathbf{K} &= \frac{d\mathbf{I}}{dl_{\perp}} \\ &= \sigma \mathbf{v} \end{align*}Volume current density:
\begin{align*} \mathbf{J} &= \frac{d\mathbf{I}}{da_{\perp}} \\ &= \rho \mathbf{v} \end{align*}Magnetic force on a volume current:
\begin{align*} \mathbf{F}_{\text{mag} } &= \int (\mathbf{v}\times \mathbf{B})\rho d\tau \\ &= \int (\mathbf{J}\times \mathbf{B})d\tau \end{align*}Total current crossing a surface $\mathcal{S} $:
\[I=\int _{\mathcal{S}}J\;da_{\perp} \]Continuity equation:
\[\nabla \cdot \mathbf{J}=-\frac{\partial \rho }{\partial t} \]Magnetostatics regime:
\[\frac{\partial \mathbf{J}}{\partial t}=\mathbf{0} \]Unit of teslas:
\[1\;\text{T} =1\frac{\text{N} }{\text{A} \cdot \text{m} } \]Biot-Savart law:
\begin{align*} \mathbf{B}(\mathbf{r}) &= \frac{\mu _0 }{4\pi }\int \frac{\mathbf{I}\times \hat{\mathscr{r}}}{\mathscr{r}^2 } \\ &= \frac{\mu _0 }{4\pi }I\int \frac{d\mathbf{l}'\times \hat{\mathscr{r}}}{\mathscr{r}^2 } \end{align*}Divergence of $\mathbf{B} $:
\[\nabla \cdot \mathbf{B}=0 \]Curl of $\mathbf{B} $:
\[\nabla \times \mathbf{B}=\mu _0 \mathbf{J} \]In integral form, it becomes Ampere's law:
\[\oint \mathbf{B}\cdot d\mathbf{l}=\mu _0 I_{\text{enc} } \]Where
\[I_{\text{enc} }=\int \mathbf{J}\cdot d\mathbf{a} \]Since $\nabla \cdot \mathbf{B}=0 $, we can introduce a vector potential
\[\mathbf{B}=\nabla \times \mathbf{A} \]We set the divergence of $\mathbf{A} $:
\[\nabla \cdot \mathbf{A}=0 \]And we get (three) Poisson's equation(s):
\[\nabla ^2 \mathbf{A}=-\mu _0 \mathbf{J} \]Cyclotron motion
\begin{align*} QvB &= m\frac{v^2 }{R} \\ \omega &= \frac{QB}{m} \end{align*}Magnetic field of an infinite wire with current $I $:
\[B(s)=\frac{\mu _0 }{2\pi s}I \]Magnetic field a distance $z $ above the center of a circular loop of radius $R $ with steady current $I $.
\[B(z)=\frac{\mu _0 I }{2} \frac{R^2 }{(R^2 +z^2 )^{3/2}} \]Magnetic field of an infinitely-long solenoid with $n $-turns per unit length carrying a current $I $:
\[B =\begin{cases} \begin{matrix} \mu _0 nI & \text{inside} \\ 0 & \text{outside} \\ \end{matrix} \end{cases} \]Magnetic field of an infinite surface current $\mathbf{K}=k\hat{x} $ on the $xy $-plane:
\[\mathbf{B}=\begin{cases} \begin{matrix} (\mu _0 /2)K\hat{\mathbf{y}} & z < 0 \\ -(\mu _0 /2) K \hat{\mathbf{y}} & z > 0 \\ \end{matrix} \end{cases} \]Magnetic field on a toroidal wire with $N $ total turns, carrying a current $I $:
\[\mathbf{B}(\mathbf{r})=\begin{cases} \begin{matrix} \frac{\mu _0 NI}{2\pi s}\hat{\phi } & \text{inside the coil} \\ \mathbf{0} & \text{outside} \\ \end{matrix} \end{cases} \]Current density $\mathbf{J} $ and force per unit charge $\mathbf{f} $ are related through conductivity $\sigma $:
\[\mathbf{J}=\sigma \mathbf{f} \]Resistivity:
\[\rho =\frac{1}{\sigma } \]In terms of resistivity,
\begin{align*} \text{Perfect conductors} \Rightarrow \;\; \sigma &= \infty \\ \text{Insulators} \Rightarrow \;\; \sigma &= 0 \end{align*}In the limit of small velocity, $\mathbf{J}=\sigma (\mathbf{E}+\cancel{\mathbf{v}\times \mathbf{B}}) $ giving Ohm's law:
\[\mathbf{J}=\sigma \mathbf{E} \]In general,
\[V=IR \]Where $R $ is the resistance (units of Ohms).
Joule heating law:
\[P=VI=I^2 R \]Electromotive force (emf), for a source force $\mathbf{f}_s $:
\[\mathcal{E}=\oint \mathbf{f}_S\cdot d\mathbf{l} \]Where $\mathcal{E} $ is "work done per unit charge."
Flux rule for motional emf:
\[\mathcal{E}=-\frac{d\Phi }{dt} \]A changing magnetic field induces an electric field.
We therefore get Faraday's law (integral form):
\[\oint \mathbf{E}\cdot d\mathbf{l}=-\int \frac{\partial \mathbf{B}}{\partial t}\cdot d\mathbf{a} \]Equivalently, Faraday's law (differential form):
\[\nabla \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t} \]Universal flux rule: Whenever the magnetic flux through a loop changes, an emf will appear in the loop:
\[\mathcal{E}=-\frac{d\Phi }{dt} \]Lenz's law:
\[\text{Nature abhors a change in flux.} \]For a pure Faraday field,
\begin{align*} \nabla \cdot \mathbf{E}_{\text{far} } &= 0 \\ \nabla \times \mathbf{E}_{\text{far} } &= -\frac{\partial \mathbf{B}}{\partial t} \end{align*}This similarity with Magnetostatics
\[-\frac{\partial \mathbf{B}}{\partial t}\Leftrightarrow \mu _0 \mathbf{J} \]gives rise to a Biot-Savart-like law for the $\mathbf{E} $ (Faradyan) field:
\[\mathbf{E}=-\frac{1}{4\pi }\frac{\partial }{\partial t}\int \frac{\mathbf{B}\times \hat{\mathscr{r}}}{\mathscr{r}^2 }d\tau \]And Ampere's law in integral form is just Faraday's law in integral form, with
\[-\frac{d\Phi }{dt}\Leftrightarrow \mu _0 I_{\text{enc} } \]Mutual inductance is the constant $M_{21} $ such that
\[\Phi _2 =M_{21}I_1 \]Where
Neumann formula
\[M_{21}=\frac{\mu _0 }{4\pi }\oint \oint \frac{d\mathbf{l}_1 \cdot d\mathbf{l}_2 }{\mathscr{r}} \]Which showcases the symmetric nature of the mutual inductances:
\[M_{21}=M_{12} \]Self inductance $L $ (units of Henries $\text{H} $) relates the current of a loop and its flux through it:
\[\Phi =LI \]The loop generates an emf on itself:
\[\mathcal{E}=-L\frac{dI}{dt} \]Energy stored in a magnetic field:
\begin{align*} W &= \frac{1}{2}LI^2 \\ &= \frac{1}{2\mu _0 }\int _{\text{all space} }B^2 d\tau \end{align*}We have four Maxwell equations:
\begin{align*} \nabla \cdot \mathbf{E} &= \frac{1}{\epsilon _0 }\rho \tag{Gauss's law} \\ \nabla \cdot \mathbf{B} &= 0 \\ \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t}\tag{Faraday's law}\\ \nabla \times \mathbf{B} &= \mu _0 \mathbf{J} + \mu _0 \epsilon _0 \frac{\partial \mathbf{E}}{\partial t}\tag{(corrected) Ampere's law} \end{align*}Where the displacement "current":
\[\mathbf{J}_d\equiv \epsilon _0 \frac{\partial \mathbf{E}}{\partial t} \]The theoretical content of electrodynamics is summarized in Maxwell's equations and the force law:
\[\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B}) \]In matter, we have the familiar bound charge density (for polarization $\mathbf{P} $):
\begin{align*} \rho &= \rho _f+\rho _b \\ &= \rho _f - \nabla \cdot \mathbf{P} \end{align*}And current density has an additional polarization current term (besides the magnetization bound current):
\begin{align*} \mathbf{J} &= \mathbf{J}_f+\mathbf{J}_b+\mathbf{J}_p \\ &= \mathbf{J}_f+\nabla \times \mathbf{M}+\frac{\partial \mathbf{P}}{\partial t} \end{align*}By defining the fields:
\begin{align*} \mathbf{D} &= \epsilon _0 \mathbf{E}+\mathbf{P} \\ \mathbf{H} &= \frac{1}{\mu _0 }\mathbf{B}-\mathbf{M} \end{align*}We have the four Maxwell equations:
\begin{align*} \nabla \cdot \mathbf{D} &= \rho _f \\ \nabla \cdot \mathbf{B} &= 0 \\ \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t}\\ \nabla \times \mathbf{H} &= \mathbf{J}_f+\frac{\partial \mathbf{D}}{\partial t} \end{align*}Where the displacement current:
\[\mathbf{J}_d\equiv \frac{\partial \mathbf{D}}{\partial t} \]For linear media, we have:
\begin{align*} \mathbf{P} &= \epsilon _0 \chi _e\mathbf{E} \\ \Rightarrow \;\mathbf{D} &= \epsilon \mathbf{E}\tag{$\epsilon \equiv \epsilon _0 (1+\chi _e) $} \\ \mathbf{M} &= \chi _m \mathbf{H}\\ \Rightarrow \; \mathbf{H} &= \frac{1}{\mu }\mathbf{B}\tag{$\mu \equiv \mu _0 (1+\chi _m) $} \end{align*}There are four boundary conditions:
\begin{align*} D_1 ^{\perp }-D_2 ^{\perp } &= \sigma _f \\ B_1 ^{\perp }-B_2 ^{\perp } &= 0 \\ \mathbf{E}_1 ^{\parallel}-\mathbf{E}_2 ^{\parallel} &= \mathbf{0} \\ \mathbf{H}_1 ^{\parallel}-\mathbf{H}_2 ^{\parallel} &= \mathbf{K}_f\times \hat{\mathbf{n}} \end{align*}In particular for linear media:
\begin{align*} \epsilon _1 E_1 ^{\perp }-\epsilon _2 E_2 ^{\perp } &= \sigma _f \\ B^{\perp }_1 -B_2 ^{\perp } &= 0 \\ \mathbf{E}_1 ^{\parallel}-\mathbf{E}_2 ^{\parallel} &= \mathbf{0} \\ \frac{1}{\mu _1 }\mathbf{B}_1 ^{\parallel}-\frac{1}{\mu _2 }\mathbf{B}_2 ^{\parallel} &= \mathbf{K}_f\times \hat{n} \end{align*}Continuity equation (local conservation of charge):
\[\frac{\partial \rho }{\partial t}=-\nabla \cdot \mathbf{J} \]Energy density in electromagnetic fields:
\[u=\frac{1}{2} \left( \epsilon _0 E^2 +\frac{1}{\mu _0 }B^2 \right) \]Poynting vector (energy flux density):
\[ \mathbf{S}\equiv \frac{1}{\mu _0 }(\mathbf{E}\times \mathbf{B}) \]Poynting's theorem (work-energy theorem):
\[\frac{dW}{dt}=-\frac{d}{dt}\int _{\mathcal{V}}u d\tau -\oint _{\mathcal{S}}\mathbf{S}\cdot d\mathbf{a} \]"The work done on the charges is equal to the decrease in energy remaining in the fields (first term) minus the energy that flowed out through the surface (second term)"
Maxwell stress tensor (force per unit area):
\[T_{ij}\equiv \epsilon _0 \left( E_i E_j - \frac{1}{2}\delta _{ij}E^2 \right) +\frac{1}{\mu _0 } \left( B_i B_j-\frac{1}{2}\delta _{ij}B^2 \right) \]Force per unit volume:
\[\mathbf{f}=\nabla \cdot \mathbf{T}-\epsilon _0 \mu _0 \frac{\partial \mathbf{S}}{\partial t} \]Total electromagnetic force
\[\mathbf{F}=\oint _{\mathcal{S}}\mathbf{T}\cdot d\mathbf{a}-\epsilon _0 \mu _0 \frac{d}{dt}\int _{\mathcal{V}}\mathbf{S}d\tau \]Divergence of $\mathbf{T} $:
\[ \left( \nabla \cdot \mathbf{T} \right) _j=\epsilon _0 \left[ (\nabla \cdot \mathbf{E})E_{j}+(\mathbf{E}\cdot \nabla )E_j-\frac{1}{2}\nabla _j E^2 \right] +\frac{1}{\mu _0 } \left[ (\nabla \cdot \mathbf{B}) B_j + (\mathbf{B}\cdot \nabla )B_j-\frac{1}{2}\nabla _jB^2 \right] \]Conservation of momentum in electrodynamics:
\[\frac{d\mathbf{p}_{\text{mech} }}{dt}=-\epsilon _0 \mu _0 \frac{d}{dt}\int _{\mathcal{V}}\mathbf{S}d\tau +\oint_{\mathcal{S}}\mathbf{T}\cdot d\mathbf{a} \]Momentum density in the fields:
\[\mathbf{g}=\mu _0 \epsilon _0 \mathbf{S} \]Momentum flux transported by the fields:
\[-\mathbf{T} \]Electromagnetic fields carry angular momentum:
\[\ell = \mathbf{r}\times \mathbf{g} = \epsilon _{0} \left[ \mathbf{r}\times (\mathbf{E}\times \mathbf{B}) \right] \]Classical wave equation
\[\frac{\partial ^2 f}{\partial z^2 }=\frac{1}{v^2 }\frac{\partial ^2 f}{\partial t^2 } \]Where a "wave" is any function whose dependence on $x,t $ is of the form $z-vt $. This equation is linear in $f $.
Given a sinusoidal wave $f(z,t)=A\cos \left( k(z-vt)+\delta \right) $:
Given a complex wave $\tilde{f}(z,t)=\tilde{A}e^{i(kz-\omega t)} $:
Longitudinal waves have displacement along the direction of propagation.
Transverse waves have displacement perpendicular to the direction of propagation. Transverse waves are polarized by some polarization angle $\theta $:
\begin{align*} \tilde{\mathbf{f}}(z,t) &= \tilde{A}e^{i(kz-\omega t)}\hat{\mathbf{n}} \\ \hat{\mathbf{n}} &= \cos \left( \theta \right) \hat{\mathbf{x}}+\sin \left( \theta \right) \hat{\mathbf{y}} \end{align*}Wave equations for the $\mathbf{E} $ and $\mathbf{B} $ fields:
\begin{align*} \nabla ^2 \mathbf{E} &= \mu _0 \epsilon _0 \frac{\partial ^2 \mathbf{E}}{\partial t^2 } \\ \nabla ^2 \mathbf{B} &= \mu _0 \epsilon _0 \frac{\partial ^2 \mathbf{B}}{\partial t^2 } \end{align*}With the identity:
\[v=\frac{1}{\epsilon _0 \mu _0 }=c \]Monochromatic (single $\omega $) plane waves (moving in the $z $ direction, uniform across planes perpendicular to propagation direction) are given by (fields being the real part):
\begin{align*} \tilde{\mathbf{E}}(z,t) &= \tilde{\mathbf{E}}_0 e^{i(kz-\omega t)} \\ \tilde{\mathbf{B}}(z,t) &= \tilde{\mathbf{B}}_0 e^{i(kz-\omega t)} \end{align*}Where
\[\omega =ck \]And (EM waves are transverse)
\begin{align*} (\tilde{E}_0 )_z &= 0 \\ (\tilde{B}_0 )_z &= 0 \end{align*}As well as the orthogonality condition:
\begin{align*} \tilde{\mathbf{B}}_0 &= \frac{k}{\omega }(\hat{\mathbf{z}}\times \tilde{\mathbf{E}}_0 ) \\ \text{Real parts} \;\;\;B_0 &= \frac{k}{\omega }E_0 =\frac{1}{c}E_0 \end{align*}Polarization convention: The direction of $\mathbf{E} $ specifies the direction of polarization.
A generalized wave has propagation/wave vector $\mathbf{k} $:
\begin{align*} \tilde{\mathbf{E}}(\mathbf{r},t) &= \tilde{E}_0 e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}\hat{\mathbf{n}} \\ \tilde{\mathbf{B}}(\mathbf{r},t) &= \frac{1}{c}\hat{\mathbf{k}}\times \tilde{\mathbf{E}} \end{align*}Where $\hat{\mathbf{n}} $ is the polarization vector, giving the fields:
\begin{align*} \mathbf{E}(\mathbf{r},t) &= E_0 \cos \left( \mathbf{k}\cdot \mathbf{r}-\omega t+\delta \right) \hat{\mathbf{n}} \\ \mathbf{B}(\mathbf{r},t) &= \frac{1}{c}E_0 \cos \left( \mathbf{k}\cdot \mathbf{r}-\omega t+\delta \right) (\hat{\mathbf{k}}\times \hat{\mathbf{n}}) \end{align*}Energy per unit volume of a monochromatic
plane wave: \begin{align*} u &= \epsilon _0 E^2 \\ &= \epsilon _0 E_0 ^2 \cos ^2 (kz-\omega t+\delta ) \end{align*}Poynting vector (energy flux density (energy per unit area, per unit time)) of a monochromatic plane wave:
\[\mathbf{S}=cu\hat{\mathbf{z}} \]Momentum density for monochromatic plane waves:
\[\mathbf{g}=\frac{1}{c}u\hat{\mathbf{z}} \]Average of relevant quantities:
\begin{align*} \langle u \rangle &= \frac{1}{2}\epsilon _0 E_0 ^2 \\ \langle \mathbf{S} \rangle &= \frac{1}{2}c\epsilon _0 E_0 ^2 \hat{\mathbf{z}} \\ \langle \mathbf{g} \rangle &= \frac{1}{2c}\epsilon _0 E_0 ^2 \hat{\mathbf{z}} \end{align*}Where the average power per unit area being transported is known as intensity $I $:
\begin{align*} I &= \langle S \rangle \\ &= \frac{1}{2}c\epsilon _0 E_0 ^2 \end{align*}Radiation pressure:
\[P=\frac{I}{c} \]Maxwell's equations with no free charge, no free current, and in homogeneous linear media:
\begin{align*} \nabla \cdot \mathbf{E} &= 0 \\ \nabla \cdot \mathbf{B} &= 0 \\ \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t} \\ \nabla \times \mathbf{B} &= \mu \epsilon \frac{\partial \mathbf{E}}{\partial t} \end{align*}By defining the index of refraction $n $:
\[n\equiv \sqrt{\frac{\epsilon \mu }{\epsilon _0 \mu _0 }} \]Waves have propagation speed:
\[v=\frac{1}{\sqrt{\epsilon \mu }}=\frac{c}{n} \]For monochromatic waves,
\[\omega =kv \]The energy density is
\[u=\frac{1}{2} \left( \epsilon E^2 +\frac{1}{\mu B^2 } \right) \]Intensity
\[ I=\frac{1}{2}\epsilon vE_0 ^2 \]And $B\leftrightarrow E $ as follows:
\[B_0 =\frac{1}{v}E_0 \]Boundary conditions between media:
\begin{align*} \epsilon _1 E_1 ^{\perp} &= \epsilon _2 E_2 ^{\perp } \\ B_1 ^{\perp } &= B_2 ^{\perp }\\ \mathbf{E}_1 ^{\parallel} &= \mathbf{E}_2 ^{\parallel}\\ \frac{1}{\mu _1 }\mathbf{B}_1 ^{\parallel} &= \frac{1}{\mu _2 }\mathbf{B}_2 ^{\parallel} \end{align*}Normal Incidence.
Define $\beta $
\[\beta \equiv \frac{\mu _1 n_2 }{\mu _2 n_1 } \]For plane waves, the reflected and transmitted amplitudes are:
\begin{align*} \tilde{E}_{0_R} &= \frac{1-\beta }{1+\beta } \tilde{E}_{0_I}\\ \tilde{E}_{0_T} &= \frac{2}{1+\beta }\tilde{E}_{0_I} \end{align*}The real amplitudes are:
\begin{align*} E_{0_R} &= \bigg|\frac{n_1-n_2 }{n_1 +n_2 }\bigg| E_{0_I} \\ E_{0_T} &= \frac{2n_1 }{n_1 +n_2 }E_{0_I} \end{align*}For, $\mu _1 =\mu _2 =\mu _0 $, the reflection and transmission coefficeint:
\begin{align*} R &= \left( \frac{n_1 -n_2 }{n_1 +n_2 } \right) ^2 \\ T &= \frac{4n_1 n_2 }{(n_1 +n_2 )^2 } \end{align*}Oblique Incidence.
There are three fundamental laws of geometrical optics
First law: The incident, reflected, and transmitted wave vectors form a plane (plane of incidence) which also includes the normal to the surface.
Second law (- of reflection): The angle of incidence is equal to the angle of reflection:
\[\theta _I=\theta _R \]Third law (- of refraction, Snell's): The transmitted angle is related to the incident angle as follows:
\[\frac{\sin \left( \theta _T \right) }{\sin \left( \theta _I \right) }=\frac{n_1 }{n_2 } \]The reflected and transmitted amplitudes are (Fresnel's equations):
\begin{align*} \tilde{E}_{0_R} &= \frac{\alpha -\beta }{\alpha +\beta }\tilde{E}_{0_I} \\ \tilde{E}_{0_T} &= \frac{2}{\alpha +\beta } \tilde{E}_{0_I} \end{align*}Where $\beta =\frac{\mu _1 n_2 }{\mu _2 n_1 } $, as before, and
\[\alpha \equiv \frac{\cos \left( \theta _T \right) }{\cos \left( \theta _I \right) } \]Where $\alpha =1 $ for normal incidence.
Brewster's angle (no reflected wave):
\[\sin ^2 \theta _B=\frac{1-\beta ^2 }{ \left( \frac{n_1 }{n_2 } \right) ^2 -\beta ^2 } \]If $\mu _1 \approx \mu _2 $, then
\[\tan \theta _B\approx \frac{n_2 }{n_1 } \]Reflection ($I_R/I_I $) and transmission ($I_T/I_I $) coefficients:
\begin{align*} R &= \left( \frac{\alpha -\beta }{\alpha +\beta } \right) ^2 \\ T &= \alpha \beta \left( \frac{2}{\alpha +\beta } \right) ^2 \end{align*}In homogeneous linear media, $\rho _f $ evolves as
\[\rho _f(t)=e^{-(\sigma /\epsilon )t}\rho _f(0) \]Giving a characteristic dissipation time of
\[\tau =\frac{\epsilon }{\sigma } \]After free charge disappears, Maxwell's equations are:
\begin{align*} \nabla \cdot \mathbf{E} &= 0 \\ \nabla \cdot \mathbf{B} &= 0 \\ \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t} \\ \nabla \times \mathbf{B} &= \mu \epsilon \frac{\partial \mathbf{E}}{\partial t}+\color{blue}\mu \sigma \mathbf{E}\color{black} \end{align*}We get the usual plane-wave solutions
\begin{align*} \tilde{\mathbf{E}}(z,t) &= \tilde{\mathbf{E}}_0 e^{i(\tilde{k}z-\omega t)} \\ \tilde{\mathbf{B}}(z,t) &= \tilde{\mathbf{B}}_0 e^{i(\tilde{k}z-\omega t)} \end{align*}Now with a complex wave number
\[\tilde{k}^2 =\mu \epsilon \omega ^2 +i\mu \sigma \omega \]Or
\begin{align*} \tilde{k} &= k+i\kappa \\ k &= \omega \sqrt{\frac{\epsilon \mu }{2}} \left[ \sqrt{1+ \left( \frac{\sigma }{\epsilon \omega } \right) ^2 }+1 \right] ^{1/2}\\ \kappa &= \omega \sqrt{\frac{\epsilon \mu }{2}} \left[ \sqrt{1+ \left( \frac{\sigma }{\epsilon \omega } \right) ^2 }-1 \right] ^{1/2} \end{align*}Waves are attenuated. The skin depth quantifies the distance at which the amplitude gets reduced by a factor of $1/e $:
\[d=\frac{1}{\kappa } \]$k $ determines:
\begin{align*} \text{Wavelength} \;\lambda &= \frac{2\pi }{k} \\ \text{Propagation speed} \;v &= \frac{\omega }{k}\\ \text{Index of refraction} \;n &= \frac{ck}{\omega } \end{align*}The fields are transverse and mutually perpendicular.
$\tilde{k} $ using a modulus and phase:
\begin{align*} \tilde{k} &= Ke^{i\phi } \\ K &= \sqrt{k^2 +\kappa ^2 }=\omega \sqrt{\epsilon \mu \sqrt{1+ \left( \frac{\sigma }{\epsilon \omega } \right) ^2 }}\\ \phi &= \tan ^{-1} \frac{\kappa }{k} \end{align*}The complex amplitudes ($\tilde{E}_0 =E_0 e^{i\delta _E} $) are related by
\begin{align*} B_0 e^{i\delta _B} &= \frac{Ke^{i\phi }}{\omega }E_0 e^{i\delta _E} \\ \text{Magnetic field lags behind} \;\delta _B-\delta _E &= \phi \end{align*}Real amplitudes by
\[\frac{B_0 }{E_0 }=\frac{K}{\omega }=\sqrt{\epsilon \mu \sqrt{1+ \left( \frac{\sigma }{\epsilon \omega } \right) ^2 }} \] If we set $\mathbf{E} $ in the $\hat{\mathbf{x}} $ direction, then \begin{align*} \mathbf{E}(z,t) &= E_0 e^{-\kappa z}\cos \left( kz-\omega t+\delta _E \right) \hat{\mathbf{x}} \\ \mathbf{B}(z,t) &= B_0 e^{-\kappa z}\cos \left( kz-\omega t+\delta _E+\phi \right) \hat{\mathbf{y}} \end{align*}When approaching a boundary, the wave obeys
\begin{align*} \epsilon _1 E_1 ^{\perp}-\epsilon _2 E_2 ^{\perp} &= \sigma _f \tag{Free surface charge $\sigma _f $}\\ B_1 ^{\perp}-B_2 ^{\perp} &= 0 \\ \mathbf{E}_{1}^{\parallel}-\mathbf{E}_2 ^{\parallel} &= \mathbf{0}\\ \frac{1}{\mu _1 }\mathbf{B}_1 ^{\parallel}-\frac{1}{\mu _2 }\mathbf{B}_2 ^{\parallel} &= \mathbf{k}_f\times \hat{\mathbf{n}}\tag{Normal to surface $\hat{\mathbf{n}} $, free surface current $\mathbf{K}_f $} \end{align*}Given an incident wave in a nonconducting medium moving towards a conductor:
\begin{align*} \tilde{\mathbf{E}}_I (z,t) &= \tilde{E}_{0_I}e^{i(k_1 z-\omega t)}\hat{\mathbf{x}} \\ \tilde{\mathbf{B}}_I (z,t) &= \frac{1}{v_1 }\tilde{E}_{0_I}e^{i(k_1 z-\omega t)}\hat{\mathbf{y}} \end{align*}Then we get a reflected wave:
\begin{align*} \tilde{\mathbf{E}}_R (z,t) &= \tilde{E}_{0_R}e^{i(-k_1 z-\omega t)}\hat{\mathbf{x}} \\ \tilde{\mathbf{B}}_I (z,t) &= -\frac{1}{v_1 }\tilde{E}_{0_R}e^{i(-k_1 z-\omega t)}\hat{\mathbf{y}} \end{align*}And a transmitted (which will be attenuated inside) wave:
\begin{align*} \tilde{\mathbf{E}}_T (z,t) &= \tilde{E}_{0_T}e^{i(\tilde{k}_2 z-\omega t)}\hat{\mathbf{x}} \\ \tilde{\mathbf{B}}_T (z,t) &= \frac{\tilde{k_2 }}{\omega }\tilde{E}_{0_T}e^{i(\tilde{k}_2 z-\omega t)}\hat{\mathbf{y}} \end{align*}Such that
\begin{align*} \tilde{E}_{0_R} &= \frac{1-\tilde{\beta }}{1+\tilde{\beta }}\tilde{E}_{0_I} \\ \tilde{E}_{0_T} &= \frac{2}{1+\tilde{\beta }}\tilde{E}_{0_I} \\ \tilde{\beta } &= \frac{\mu _1 v_1}{\mu _2 \omega }\tilde{k}_2 \end{align*}Special case: Perfect conductors have $\sigma =\infty $, so $k_2 =\infty $, so $\tilde{\beta }=\infty $, so $\tilde{E}_{0_R}=-\tilde{E}_{0_I} $ and no transmitted wave.
Dispersive mediums: Mediums where speed depends on frequency.
Wave (phase) velocity:
\[v=\frac{\omega }{k} \]Group velocity:
\[v_g=\frac{d\omega }{dk} \]Complex polarization and susceptibility:
\[\tilde{\mathbf{P}}=\epsilon _0 \tilde{\chi }_e\tilde{\mathbf{E}} \]Complex dielectric constant:
\[\tilde{\epsilon }_r=\frac{\tilde{\epsilon }}{\epsilon _0 } \]Plane wave solution:
\begin{align*} \tilde{\mathbf{E}}(z,t) &= \tilde{\mathbf{E}}_0 e^{i(\tilde{k}z-\omega t)} \\ \tilde{k} &= \sqrt{\tilde{\epsilon }\mu _0 }\omega \end{align*}For $\tilde{k}=k+i\kappa $, we have the absorption coefficient:
\[\alpha \equiv 2\kappa \]Index of refraction
\[n=\frac{k}{\omega } \]Cauchy's formula:
\[n=1+A \left( 1+\frac{B}{\lambda ^2 } \right) \]Where $A $ is the coefficient of refraction and $B $ the coefficient of dispersion.
A wave guide is a perfect conductor that can be thought of as a hollow pipe.
BC on the inner walls of the wave guide:
\begin{align*} \mathbf{E}^{\parallel} &= \mathbf{0} \\ B^{\perp} &= 0 \end{align*}Remark: Longitudinal components needed to satisfy the above.
$E_z $ and $B_z $ give $E_x,E_y,B_x,B_y $:
\begin{align*} E_x &= \frac{i}{ \left( \frac{\omega }{c} \right) ^2 -k^2 } \left( k\frac{\partial E_z}{\partial x}+\omega \frac{\partial B_z}{\partial y} \right) \\ E_y &= \frac{i}{ \left( \frac{\omega }{c} \right) ^2 -k^2 } \left( k\frac{\partial E_z}{\partial y}-\omega \frac{\partial B_z}{\partial x} \right)\\ B_x &= \frac{i}{ \left( \frac{\omega }{c} \right) ^2 -k^2 } \left( k\frac{\partial B_z}{\partial x}-\frac{\omega }{c^2 } \frac{\partial E_z}{\partial y} \right)\\ B_y &= \frac{i}{ \left( \frac{\omega }{c} \right) ^2 -k^2 } \left( k\frac{\partial B_z}{\partial y}+\frac{\omega }{c^2 } \frac{\partial E_z}{\partial x} \right)\\ \end{align*}Equations for $E_z $ and $B_z $:
\begin{align*} \left[ \frac{\partial ^2 }{\partial x^2 }+\frac{\partial ^2 }{\partial y^2 }+ \left( \frac{\omega }{c} \right) ^2 -k^2 \right] E_z &= 0 \\ \left[ \frac{\partial ^2 }{\partial x^2 }+\frac{\partial ^2 }{\partial y^2 }+ \left( \frac{\omega }{c} \right) ^2 -k^2 \right] B_z &= 0 \end{align*}Types of waves:
The rectangular wave guide:
The coaxial transmission line:
$\mathbf{E} $ and $\mathbf{B} $ in terms of the potentials:
\begin{align*} \mathbf{E} &= -\nabla V-\frac{\partial \mathbf{A}}{\partial t} \\ \mathbf{B} &= \nabla \times \mathbf{A} \end{align*}Maxwell's equations in terms of potentials:
\begin{align*} \nabla ^2 V+\frac{\partial }{\partial t}(\nabla \cdot \mathbf{A}) &= -\frac{1}{\epsilon _0 }\rho \\ \left( \nabla ^2 \mathbf{A}-\mu _0 \epsilon _0 \frac{\partial ^2 \mathbf{A}}{\partial t^2 } \right) -\nabla \left( \nabla \cdot \mathbf{A}+\mu _0 \epsilon _0 \frac{\partial V}{\partial t} \right) &= -\mu _0 \mathbf{J} \end{align*}Gauge transformations for the potentials:
\begin{align*} \mathbf{A}' &= \mathbf{A}+\nabla \lambda \\ V' &= V-\frac{\partial \lambda }{\partial t} \end{align*}Coulomb gauge:
Lorenz gauge:
The Lorentz Force law is:
\begin{align*} \mathbf{F}=\frac{d\mathbf{p}}{dt} &= q \left[ -\nabla V-\frac{\partial \mathbf{A}}{\partial t}+\mathbf{v}\times (\nabla \times \mathbf{A}) \right] \\ &= -q \left[ \frac{\partial \mathbf{A}}{\partial t}+(\mathbf{v}\cdot \nabla )\mathbf{A} + \nabla (V-\mathbf{v}\cdot \mathbf{A}) \right] \end{align*}Where the convective derivative $d\mathbf{A}/dt $ is a total derivative:
\[\frac{d\mathbf{A}}{dt}=\frac{\partial \mathbf{A}}{\partial t}+(\mathbf{v}\cdot \nabla )\mathbf{A} \]Representing time rate of change of $\mathbf{A} $ at the location of a moving particle.
Canonical momentum:
\[\mathbf{p}_{\text{canonical} }=\mathbf{p}+q\mathbf{A} \]The rate of change of the particle's energy, $\frac{dE}{dt}=\frac{\partial U_{\text{vel} }}{\partial t} $, where
\[U_{\text{vel} }=q(V-\mathbf{v}\cdot \mathbf{A}) \]$\mathbf{A}\equiv $ "potential moemntum per unit charge," $V\equiv $ "potential energy per unit charge."
Statically,
\[\nabla ^2 V=-\frac{1}{\epsilon _0 }\rho \;\;\Rightarrow \;\; V(\mathbf{r})=\frac{1}{4\pi \epsilon _0 }\int \frac{\rho (\mathbf{r}')}{\mathscr{r} }d\tau ' \]And
\[\nabla ^2 \mathbf{A}=-\mu _0 \mathbf{J} \;\;\Rightarrow \;\; \mathbf{A}(\mathbf{r})=\frac{\mu _0 }{4\pi }\int \frac{\mathbf{J}(\mathbf{r}')}{\mathscr{r} }d\tau ' \tag{Multiple copies of Poisson's}\] Non-statically:Jefimenko's Equations
\begin{align*} \mathbf{E}(\mathbf{r},t) &= \frac{1}{4\pi \epsilon _0 }\int \left[ \frac{\rho (\mathbf{r}',t_r)}{\mathscr{r}^2 }\hat{\mathscr{r}}+\frac{\dot{\rho }(\mathbf{r}',t_r)}{c\mathscr{r}}\hat{\mathscr{r}}-\frac{\dot{\mathbf{J}}(\mathbf{r}',t_r)}{c^2 \mathscr{r}} \right] d\tau ' \\ \mathbf{B}(\mathbf{r},t) &= \frac{\mu _0 }{4\pi }\int \left[ \frac{\mathbf{J}(\mathbf{r}',t_r)}{\mathscr{r}^2 }+\frac{\dot{\mathbf{J}}(\mathbf{r}',t_r)}{c\mathscr{r}} \right] \times \hat{\mathscr{r}} d\tau ' \end{align*}If a charge $q $ is moving with trajectory $\mathbf{w}(t) $, then the potentials are given by the Liénard-Wiechert potentials:
\begin{align*} V(\mathbf{r},t) &= \frac{1}{4\pi \epsilon _0 }\frac{qc}{(\mathscr{r}c-\mathscr{r}\cdot \mathbf{v})} \\ \mathbf{A}(\mathbf{r},t) &= \frac{\mathbf{v}}{c^2 }V(\mathbf{r},t) \end{align*}Where
\begin{align*} |\mathbf{r}-\mathbf{w}(t_r)| &= c(t-t_r) \\ \mathscr{r} &= \mathbf{r}-\mathbf{w}(t_r) \end{align*}By introducing the vector
\[\mathbf{u}\equiv c\hat{\mathscr{r}}-\mathbf{v} \]The $\mathbf{E} $ and $\mathbf{B} $ fields of a moving point particle are:
\begin{align*} \mathbf{E}(\mathbf{r},t) &= \frac{q}{4\pi \epsilon _0 }\frac{\mathscr{r}}{(\mathscr{r}\cdot \mathbf{u})^3 } \left[ (c^2 -v^2 )\mathbf{u}+\mathscr{r}\times (\mathbf{u}\times \mathbf{a}) \right] \\ \mathbf{B}(\mathbf{r},t) &= \frac{1}{c}\hat{\mathscr{r}}\times \mathbf{E}(\mathbf{r},t) \end{align*}