The Maxwell Equations

Differential Form

In cgs/gaussian units:

\[\begin{aligned} \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{E}(\boldsymbol{r}, t) &= 4 \pi \rho(\boldsymbol{r},t) \\ \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{B}(\boldsymbol{r},t) &= 0 \\ \boldsymbol{\nabla} \boldsymbol{\times} \boldsymbol{E}(\boldsymbol{r},t) &= - \frac{1}{c} \frac{\partial }{\partial t} \boldsymbol{B}(\boldsymbol{r},t) \\ \boldsymbol{\nabla} \boldsymbol{\times} \boldsymbol{B}(\boldsymbol{r},t) &= \frac{1}{c} \biggl(4 \pi \boldsymbol{j}(\boldsymbol{r},t) + \frac{\partial }{\partial t} \boldsymbol{E}(\boldsymbol{r},t)\biggr) \end{aligned}\]

Homogeneous vs. Inhomogeneous

\[\begin{aligned} \boldsymbol{\nabla \cdot} \boldsymbol{B}(\boldsymbol{r},t) &= 0 \\ \boldsymbol{\nabla \times} \boldsymbol{E}(\boldsymbol{r},t) + \frac{1}{c} \frac{\partial }{\partial t} \boldsymbol{B}(\boldsymbol{r},t) &= 0 \\ \boldsymbol{\nabla \cdot} \boldsymbol{E}(\boldsymbol{r}, t) &= 4 \pi \rho(\boldsymbol{r},t) \\ \boldsymbol{\nabla \times} \boldsymbol{B}(\boldsymbol{r},t) - \frac{1}{c} \frac{\partial}{\partial t} \boldsymbol{E}(\boldsymbol{r},t) &= \frac{4 \pi}{c} \boldsymbol{j}(\boldsymbol{r},t) \end{aligned}\]

Integral Form

Gauss Theorem:

\[\begin{aligned} \int\limits_{V} \mathrm{d} V \ \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{F} = \oint\limits_{\partial V} \mathrm{d}S \ \boldsymbol{\hat{n}}_{\perp \partial V} \boldsymbol{\cdot} \boldsymbol{F} \end{aligned}\]

Stokes Theorem:

\[\begin{aligned} \int\limits_{S} \mathrm{d}S \ \boldsymbol{\hat{n}}_{\perp S} \boldsymbol{\cdot} (\boldsymbol{\nabla} \boldsymbol{\times} \boldsymbol{F}) = \oint\limits_{\partial S} \mathrm{d}\ell \ \boldsymbol{\hat{n}}_{\parallel \partial S} \boldsymbol{\cdot} \boldsymbol{F} \end{aligned}\]

Therefore:

Gauss’s Law for Electricity:

\[\begin{aligned} \int\limits_{V} \mathrm{d} V \ \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{E}(\boldsymbol{r},t) &= \oint\limits_{\partial V} \mathrm{d}S \ \boldsymbol{\hat{n}}_{\perp \partial V} \boldsymbol{\cdot} \boldsymbol{E}(\boldsymbol{r},t) = 4\pi Q_{\text{enc}} = 4 \pi \int\limits_{V} \mathrm{d}V \ \rho(\boldsymbol{r},t) \end{aligned}\]

Gauss’s Law for Magnetism:

\[\begin{aligned} \int\limits_{V} \mathrm{d} V \ \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{B}(\boldsymbol{r},t) &= \oint\limits_{\partial V} \mathrm{d}S \ \boldsymbol{\hat{n}}_{\perp \partial V} \boldsymbol{\cdot} \boldsymbol{B}(\boldsymbol{r},t) = 0 \end{aligned}\]

Faraday’s Law of Induction:

\[\begin{aligned} \int\limits_{S} \mathrm{d}S \ \boldsymbol{\hat{n}}_{\perp S} \boldsymbol{\cdot} (\boldsymbol{\nabla} \boldsymbol{\times} \boldsymbol{E}(\boldsymbol{r},t)) &= \oint\limits_{\partial S} \mathrm{d}\ell \ \boldsymbol{\hat{n}}_{\parallel \partial S} \boldsymbol{\cdot} \boldsymbol{E}(\boldsymbol{r},t) \\ &= -\frac{1}{c} \frac{\partial}{\partial t} \int\limits_{S} \mathrm{d}S \ \boldsymbol{\hat{n}_{\perp S}} \boldsymbol{\cdot} \boldsymbol{B}(\boldsymbol{r},t) \end{aligned}\]

Ampère–Maxwell Law:

\[\begin{aligned} \int\limits_{S} \mathrm{d}S \ \boldsymbol{\hat{n}}_{\perp S} \boldsymbol{\cdot} (\boldsymbol{\nabla} \boldsymbol{\times} \boldsymbol{B}(\boldsymbol{r},t)) &= \oint\limits_{\partial S} \mathrm{d}\ell \ \boldsymbol{\hat{n}}_{\parallel \partial S} \boldsymbol{\cdot} \boldsymbol{B}(\boldsymbol{r},t) \\ &= \frac{1}{c} \left[ \int\limits_{S} \mathrm{d}S \ \boldsymbol{\hat{n}}_{\perp S} \boldsymbol{\cdot} \left(4 \pi \boldsymbol{j}(\boldsymbol{r},t) + \frac{\partial}{\partial t} \boldsymbol{E}(\boldsymbol{r},t) \right) \right] \end{aligned}\]

Potential Formalism

First we need identities from vector calculus.

\[\begin{aligned} \boldsymbol{\nabla} \boldsymbol{\times} \boldsymbol{\nabla} \phi &= 0 \\ \boldsymbol{\nabla} \boldsymbol{\cdot} (\boldsymbol{\nabla} \boldsymbol{\times} \boldsymbol{A}) &= 0 \\ \boldsymbol{\nabla} \boldsymbol{\times} (\boldsymbol{\nabla} \boldsymbol{\times} \boldsymbol{A}) &= \boldsymbol{\nabla} (\boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{A}) - \boldsymbol{\nabla}^{2} \boldsymbol{A} \end{aligned}\]

With VCII, we can rewrite MWII as \(\begin{aligned} \boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{B}(\boldsymbol{r},t) &= 0 \ \Longrightarrow \boldsymbol{B}(\boldsymbol{r},t) =: \boldsymbol{\nabla} \boldsymbol{\times} \boldsymbol{A}(\boldsymbol{r},t). \end{aligned}\) Expressing our magnetic field by the curl of a vector potential and inserting this into MWIII leads to \(\begin{aligned} \boldsymbol{\nabla} \boldsymbol{\times} \boldsymbol{E}(\boldsymbol{r},t) = - \frac{1}{c} \frac{\partial}{\partial t} (\boldsymbol{\nabla} \boldsymbol{\times} \boldsymbol{A}(\boldsymbol{r},t)). \end{aligned}\) Since the curl operation and the time dervitative commute, \(\begin{aligned} \boldsymbol{\nabla} \boldsymbol{\times} \biggl[\boldsymbol{E}(\boldsymbol{r},t) + \frac{1}{c} \frac{\partial}{\partial t} \boldsymbol{A}(\boldsymbol{r},t) \biggr] &= 0 \\ \Longrightarrow \boldsymbol{E}(\boldsymbol{r},t) + \frac{1}{c} \frac{\partial}{\partial t} \boldsymbol{A}(\boldsymbol{r},t) &=: -\boldsymbol{\nabla} \phi(\boldsymbol{r},t) \\ \Longrightarrow \boldsymbol{E}(\boldsymbol{r},t) &= - \boldsymbol{\nabla}\phi(\boldsymbol{r},t) - \frac{1}{c} \frac{\partial}{\partial t} \boldsymbol{A}(\boldsymbol{r},t). \end{aligned}\)