$\require{cancel}$ $\require{physics}$ $\require{mathtools}$

# E&M Notes

## Maxwell’s Equations and Potentials

\begin{gather*} \nabla\cdot \vb{E} = \rho/\epsilon_0 \qquad \nabla\cdot \vb{B} = 0 \qquad \nabla\times \vb{E} = -\pdv{\vb{B}}{t} \qquad \nabla\times \vb{B} = \mu_0 \vb{J} + \mu_0\epsilon_0 \pdv{\vb{E}}{t}\\ \vb{E} = -\nabla V - \pdv{\vb{A}}{t} \qquad \vb{B} = \nabla\times\vb{A} \qquad \nabla\cdot\vb{J} = - \pdv{\rho}{t} \qquad \vb{J} = \sigma\vb{E} \end{gather*}

Can get MEs in potential form by subbing in pretty easily.

Gauge-dependent equations In the Lorenz gauge ($$\nabla\cdot\vb{A} + \mu_0\epsilon_0 V = 0$$) and in terms of four-vectors $$A^\mu = (V/c,\,\vb{A})$$ and $$J^\mu = (c\rho,\,\vb{J})$$

$$\qty(\epsilon_0\mu_0\pdv[2]{t}-\nabla^2) A^\mu = \mu_0 J^\mu$$

in magnetostatics, we can pick the coulomb gauge ($$\nabla\cdot\vb{A} = 0$$) to get biot savart:

$$\vb{A}(x) = \frac{\mu_0}{4\pi}\int \frac{\vb{J}}{|x-y|}\dd[3]{y}$$

Common BCs (from integrating gauss/ampere):

\begin{gather} \hat{n}\cdot\qty[\vb{D}_{\text{out}} -\vb{D}_\text{in}] = \sigma_f \qquad \hat{n}\times\qty[\vb{E}_{\text{out}} - \vb{E}_\text{in}] = 0\\ \hat{n}\cdot\qty[\vb{B}_{\text{out}} - \vb{B}_\text{in}] = 0 \qquad \hat{n}\times\qty[\vb{H}_{\text{out}} - \vb{H}_\text{in}] = \vb{K}_f \end{gather}

Plane waves in a vacuum

$$\vb{E} = \vb{E}_0e^{i(k\cdot r-\omega t)}\qquad \vb{B} = \vb{B}_0e^{i(k\cdot r-\omega t)}$$

satisfy maxwell iff

$$\vb{E}_0\perp \vb{B}_0 \perp \vb{k} \implies \vb{B}_0 = \frac{1}{\omega}\vb{k} \times \vb{E}_0$$

## Energy, Momentum, etc.

• The energy per unit volume stored in the EM fields is
$$u = \frac{1}{2}\qty(\epsilon_0\vb{E}^2 + \frac{1}{\mu_0}\vb{B}^2)$$
the integral gives the work needed to set up a given charge/current distrib
• The energy per unit time per unit area transported by the fields is
$$\vb{S} = \frac{1}{\mu_0}\vb{E}\times\vb{B}$$
the surface integral gives the energy flux
• The rate at which work is done on the charges per unit volume is
$$\dv{W}{t} = \vb{E}\cdot\vb{J} = -\dv{u}{t} - \nabla\cdot \vb{S}$$
this gives a continuity equation in the case of no charges/currents
• The force per unit volume done on charges is
$$\vb{f} = \dv{\vb{p}_\text{mech}}{t} = \rho \vb{E} + \vb{J} \times \vb{B} = \nabla\cdot\vb*{\sigma} - \epsilon_0\mu_0\pdv{\vb{S}}{t}$$
for $\sigma_{ij}$ the stress tensor, which is the force done in the $i$ direction on a surface oriented in the $j$ direction
$$\sigma_{ij} = \epsilon_0 E_iE_j + \frac{1}{\mu_0}B_iB_j - u\delta_{ij}$$
and the last term representing the momentum density stored in the EM fields,
$$\mu_0\epsilon_0\vb{S} = \epsilon_0\vb{E}\times\vb{B}$$
• The angular momentum stored in the fields is the definition you'd expect:
$$\vb*{\ell} = \vb{r}\times\mu_0\epsilon_0\vb{S} = \epsilon_0\vb{r}\times(\vb{E}\times\vb{B})$$
Note that static field scan still carry linear/angular momenta!

• We combine everybody together into the stress-energy tensor:

$$T^{\mu\nu} = \mqty(u & \sqrt{\epsilon_0\mu_0}\vb{S} \\ \sqrt{\epsilon_0\mu_0}\vb{S} & -\vb*{\sigma})$$

## Dipoles and Stuff

\begin{gather*} \vb{p}=\int \vb{r}\rho(r)\dd[3]{\vb{r}} \qquad U = -\vb{p}\cdot\vb{E} \qquad \vb{N} = \vb{p}\times\vb{E}\\ V_\text{dip} = \frac{1}{4\pi\epsilon_0}\frac{\hat{r}\cdot \vb{p}}{r^2} \qquad \vb{E}_\text{dip} = \frac{1}{4\pi\epsilon_0}\frac{1}{r^3}\qty[3(\vb{p}\cdot \vu{r})\hat{r} -\vb{p}] = \frac{p}{4\pi\epsilon_0 r^3}\qty[2\cos\theta\hat{r} + \sin\theta\hat{\theta}]\\ \vb{Q}_{ij} = \int\rho(\vb{r})\qty(3r_ir_j - r^2\delta_{ij})\dd[3]{r}\\ \vb{m}=\frac{1}{2}\int\vb{r}\times\vb{J}(r)\dd[3]{\vb{r}} \qquad U = -\vb{m}\cdot\vb{B} \qquad \vb{N} = \vb{m}\times\vb{B}\\ \vb{A}_\text{dip}= \frac{\mu_0}{4\pi}\frac{1}{r^2}\vb{m}\times\vu{r} \qquad \vb{B}_\text{dip} = \frac{\mu_0}{4\pi}\frac{1}{r^3}\qty[3(\vb{m}\cdot \vb{r})\hat{r} -\vb{m}]= \frac{\mu_0m}{4\pi r^3}\qty[2\cos\theta\hat{r} + \sin\theta\hat{\theta}]\\ \end{gather*}

## Fields in Matter

$$\nabla\cdot \vb{D} = \rho_f \qquad \nabla\cdot \vb{B}=0 \qquad \nabla\times \vb{E} = - \pdv{\vb{B}}{t} \qquad \nabla\times \vb{H} = \vb{J}_f + \pdv{\vb{D}}{t}$$

Constitutive relations relate $$\vb{D}\leftrightarrow\vb{E}$$, $$\vb{H} \leftrightarrow\vb{E}$$. In linear media,

\begin{alignat}{4} \vb{D} &= \epsilon\vb{E} &&= \epsilon_0(1+\chi_e)\vb{E} &&= \epsilon_0\vb{E} + \vb{P}\\ \vb{B} &= \mu\vb{H} &&= \mu_0(1+\chi_m)\vb{H} &&= \mu_0(\vb{H} + \vb{M}) \end{alignat}

We get surface and bound charges and currents due to $$\vb{P}$$ and $$\vb{M}$$:

\begin{align} \sigma_b &= \vb{P}\cdot\vu{n} & \rho_b &= - \nabla\cdot\vb{P}\\ \vb{K}_b &= \vb{M}\times\vu{n} & \vb{J}_b &= \nabla\times\vb{M}\\ \end{align}

If we have a material with conductivity $$\sigma\gg 1$$, maxwell + ohm gives

$$\epsilon\mu \partial_t^2 \vb{E} - \nabla^2 \vb{E} + \mu\sigma\partial_t\vb{E} = 0$$

and assuming a wave ansatz

$$E = \frac{1}{2}\qty(E_0e^{i(kz-\omega t)} + \text{c.c.}) \implies k^2 = \mu\omega\qty(\epsilon \omega + i\sigma)$$

and we have the dispersion relation for the material

### Plasma

The idea here is that instead of using Ohm to rewrite the current, we take $$\vb{J} = -e\nu\vb{v}$$ and use newton’s second law to say $$m_e\partial_t\vb{v} = -e\vb{E}$$. Taking this all together and making our wave ansatz, we get

$$\omega^2 = c^2k^2 + \omega_p^2 \qq{for}\omega_p^2 = \frac{e^2\nu}{\epsilon_0 m_e}$$

and we must have a wave with a frequency above the plasma frequency for it to not decay evanescently. We can rewrite the wave equation (after assuming waveform) as

$$-\nabla^2\vb{E} - \mu_0\underbrace{\epsilon_0\qty(1 - \frac{\omega_p^2}{\omega^2})}_{\epsilon(\omega)}\omega^2\vb{E}$$

### Waveguides and Cavities

We assume propagation in the $$z$$ direction along the waveguide $$C$$,

$$\vb{E}(\vb{r},\,t) = e^{i(kz-\omega t)}\qty[\vb{E}_\perp(\vb{r}_\perp) + E_z(\vb{r}_\perp)\vu{z}] \qquad \vb{H}(\vb{r},\,t) = e^{i(kz-\omega t)}\qty[\vb{H}_\perp(\vb{r}_\perp) + H_z(\vb{r}_\perp)\vu{z}]$$

where the fields are not necessarily transverse! We have nonzero $$z$$ components of the fields. Plugging into Maxwell and looking only inside the perfect conducting waveguide (where $$\rho = \vb{J} = 0$$) we can assume

$$\nabla\times\vb{E} = i\omega\mu\vb{H} \qq{and} \nabla\times\vb{H} = -i\omega\epsilon\vb{E}$$

and Maxwell tells us (the first is an automatic consequence)

\begin{align*} \nabla_\bot\cdot \vb{E}_\bot &= -ikE_z & \nabla_\bot\cdot \vb{H}_\bot &= -ikH_z\\ \nabla_\bot\times \vb{E}_\bot &= i\omega\mu H_z\vu{z} & \nabla_\bot\times \vb{H}_\bot &= -i\omega\epsilon E_z\vu{z} \\ i\omega\mu\vb{H}_\bot - ik\vu{z}\times\vb{E}_\bot &= -\vu{z}\times(\nabla_\bot E_z) & i\omega\epsilon\vb{E}_\bot + ik\vu{z}\times\vb{H}_\bot &= \vu{z}\times(\nabla_\bot H_z) \end{align*}

taking $$\vu{z}\,\times$$ the bottom equations allows us to determine the transverse $$\vb{F}_\bot$$ from the longitudinal components:

\begin{gather} \vb{H}_\bot = \frac{ik}{\gamma^2}\nabla_\bot H_z + \frac{i\omega\epsilon}{\gamma^2}\vu{z}\times\nabla_\bot E_z \\ \vb{E}_\bot = \frac{ik}{\gamma^2}\nabla_\bot E_z - \frac{i\omega\mu}{\gamma^2}\vu{z}\times\nabla_\bot H_z \end{gather}

for $$\gamma^2 = \mu\epsilon\omega^2 - k^2$$. We can show that both transverse fields individually satisfy

$$[\nabla_\bot^2 + \gamma^2]F_z = 0$$

Inside a conducting tube with simply connected base, assuming that $$E_z = 0$$ forces $$\vb{H}_\bot = 0$$ and vice versa. This means that if we assume both are zero, we have no wave propagation. However, other shapes do support these TEM modes.

In general, the strategy is to solve for the $$F_z$$ and then use this to determine the longitudinal components. Any solution can be written as a linear combination of the three cases.

TEM modes for these waves, $$E_z=H_z =0$$. Thus, our $$\vb{E}_\bot$$ and $$\vb{H}_\bot$$ are both harmonic functions of some potential that satisfies Laplace’s equation:

$$\nabla^2_\bot \phi = 0$$

where one of the fields is determined by the other from the last Maxwell equation.

TE modes in this case, $$E_z = 0$$. Thus,

$$[\nabla_\bot^2 + \gamma^2]H_z = 0 \qquad \vb{H}_\bot = \frac{ik}{\gamma^2}\nabla_\bot H_z \qquad \vb{E}_\bot = - \frac{i\omega\mu}{\gamma^2}\vu{z}\times\nabla_\bot H_z$$

Where our boundary condition is that $$\partial_{\vb{n}}\vb{H}_z$$ is zero on the boundary.

TM modes in this case, $$H_z = 0$$. Thus,

$$[\nabla_\bot^2 + \gamma^2]E_z = 0 \qquad \vb{H}_\bot = \frac{i\omega\epsilon}{\gamma^2}\vu{z}\times\nabla_\bot E_z \qquad \vb{E}_\bot = \frac{ik}{\gamma^2}\nabla_\bot E_z$$

Where our boundary condition is that $$\vb{E}_z$$ is zero on the boundary.

Cavities this is a waveguide that is truncated to be of finite length, so now we need to consider the BCs on the end plates, which read

$$\vb{H}_z = 0 \qquad \vb{E}_\bot = 0$$

to solve this, treat it as an infinite waveguide and take linear combinations to get standing-wave type solutions that are zero on the plates.

## Laplace’s Equation and Multipole Expansion

### 2D Laplace

In two dimensions, after separation of variables, we find that both $$X$$ and $$Y$$ satisfy harmonic oscillator equations with the property that $$k_x^2 + k_y^2 = 0$$:

$$X''(x) = k^2X \qquad Y''(y) = -k^2Y$$

this means in one dimension we’ll have trig solutions and in the other we’ll have exponentials. Pick whatever fits your BCs. You may need to add together solutions for a number of $$k$$s; often this comes from using a fourier series to match the trig dimension to some boundary potential – to find coefficients here, use the orthogonality of the trig functions:

$$\int_0^a \sin(n\pi y/a)\sin(m\pi y/a) \dd{y}=\frac{a}{2}\delta_{mn}$$

### 3D Laplace

Cartesian coords Separation of variables gives us

$$X'' = (k^2 + \ell^2)X \qquad Y'' = -k^2 Y \qquad Z'' = -\ell^2 Z$$

so we again get some mix of exponentials and trigs depending on what makes the most physical sense

Spherical coords

$$\nabla^2 V = 0 \implies V(\vb{r}) = Y_{\ell m}(\theta,\,\phi) \qty[A_{\ell m}r^{\ell} + B_{\ell m}r^{-(\ell + 1)}]$$

in the case of cylindrical symmetry, $$m=0$$ and the equation simplifies to

$$V = P_{\ell}(\cos\theta)\qty[A_{\ell}r^{\ell} + B_{\ell}r^{-(\ell + 1)}]$$

## Method of Images

Idea: replace a conductor by point charges that we know how to deal with; we go from a boundary value problem to an equivalent setup that matches the BCs (keep in mind that it is just that, though – a boundary value problem). You can do this with charges or with magnetic dipoles depending on what you need.

## Capacitance and Inductance

\begin{gather} C =\frac{Q}{\Delta V} = -\epsilon_0\int_{\partial\Omega}\dd{S}\cdot\nabla\phi \qquad \phi(x) = \begin{cases}1 & \text{in }\Omega \\ 0 & \text{at }\infty\end{cases}\\ U_\text{cap} = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C} \end{gather}

Strategy: assume a charge $$\pm Q$$ on the parts, calculate the potential difference or energy content, use this to get the capacitance.

A loop of current will generate a magnetic field. The flux of that field through a different loop is

$$\Phi_2 = \int_2 \vb{B}_1\cdot\vb{S}_2 = M_{21}I_1$$

where $$M_{21}$$ is the mutual inductance and is symmetric. You can also calculate self inductance, relating the flux through a loop to the current in the loop producing it.

$$\Phi = LI \qquad U = \frac{1}{2}LI^2$$

also dont forget maxwell tells us induced EMF:

$$\mathcal{E} = -L\dv{I}{t}$$

We solve ME in Lorenz gauge using the retarded greens function

$$\qty(\epsilon_0\mu_0\pdv[2]{t}-\nabla^2) A^\mu = \mu_0 J^\mu = \mu_0\, q\, u^\mu(t) \delta(\vb{x}-\vb{x}_0(t))$$

where $$u^\mu = (c,\,\vb{v})$$ is the four velocity of the particle. Then

$$A^\mu(\vb{x},\,t) = \frac{\mu_0 q}{4\pi} \frac{u^\mu(\tau)}{|\vb{x}-\vb{x}_0(\tau)|} \frac{1}{1-\vb*{\beta}(\tau)\cdot \vb{n}(\tau)} \qquad \tau = t - \frac{1}{c}|\vb{x}-\vb{x}_0(\tau)|$$

where $$\vb*{\beta} = \vb{v}/c$$ and $$\vb{n} = \frac{\vb{x}-\vb{x}_0(\tau)}{|\vb{x}-\vb{x}_0(\tau)|}$$ is the unit vector pointing from the source to the point of measurement. This is the Lienard-Wiechert potential. The exact expressions of the fields can be derived from this. We approximate:

1. $|\vb{x}|\gg 1$, meaning we can drop $1/|x-x_0|^2$s
2. $\beta \ll 1$, meaning we can drop $(n\cdot \beta)$

which gives fields

$$\vb{E}(\vb{x},\,t) = \frac{q}{4\pi\epsilon_0} \frac{\vb{n}\times(\vb{n}\times\dot{\vb*{\beta}}(\tau))}{c|\vb{x}|} \qq{and} \vb{B}(\vb{x},\,t) = \frac{1}{c} \vb{n}\times \vb{E}(\vb{x},\,t)$$

The Poynting vector is

$$\vb{S}(\vb{r}) = \frac{E^2}{\mu_0 c}\vb{n} = \frac{\mu_0q^2}{16\pi^2c} \frac{a^2(\tau)\sin^2\theta}{|\vb{r}|^2}\hat{r}$$

with the $$\theta$$ measured from the direction of acceleration. Integrating gives Larmor,

$$P = \frac{\mu_0q^2a^2(\tau)}{6\pi c}$$

### Multipole expansion

Define the radiation vector $$\vb*{\alpha}$$ by

$$\pdv{\vb{A}_\text{rad}(\vb{r},\,t)}{t} = \frac{\mu_0}{4\pi r}\vb*{\alpha}(\vb{r},\,t) \implies \vb*{\alpha}(\vb{r},\,t) = \dv{t}\int\dd[3]{\vb{r'}} \vb{J}\qty(\vb{r},\,t-\frac{r + \vu{r}\cdot \vb{r'}}{c})$$

then the fields and power are given by

The idea here is to expand $$\vb{J}$$:
$$\vb*{\alpha}(\vb{r},\,t) = \dv[2]{t}\vb{p}(t-r/c) + \frac{1}{c}\dv[2]{t}\vb{m}(t-r/c) + \frac{1}{c}\dv[3]{t}\vb{Q}(t-r/c)\cdot\vu{r} + \dots$$