Maxwell's Equations

This is given by the divergence of an inverse square field, specifically for an electric field which is the same as in :

\begin{align*} \vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_{0}} \end{align*}

The divergence of the magnetic field is the same as in magnetostatics:

\begin{align*} \vec{\nabla} \cdot \vec{B} = 0 \end{align*}

The magnetostatic magnetic field is given by the Bio-Savart Law:

\begin{align*} \vec{B} = \frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{J} \times \hat{r}}{r^{2}}d\tau \end{align*}

Now the curl of this field is given by magnetostatics:

\begin{align*} \vec{\nabla} \times \vec{B} = \mu_{0}\vec{J} \end{align*}

However, if you take the divergence of this equation, the left hand side reduces to zero by the definition of the del operator, but the right hand side does not always:

\begin{align*} \vec{\nabla} \cdot \mu_{0}\vec{J} = \mu_{0} (\vec{\nabla} \cdot \vec{J}) \neq 0 \end{align*}

Given this problem, a correction is given via the continuity equation:

\begin{align*} \vec{\nabla} \cdot \vec{J} = -\frac{\partial \rho}{\partial t} \\ \epsilon_{0}(\vec{\nabla} \cdot \vec{E}) = \rho \\ \vec{\nabla} \cdot \vec{J} = -\epsilon_{0}\vec{\nabla} \cdot \frac{\partial\vec{E}}{\partial t} \end{align*}

So therefore when we account for the fact that \(\vec{\nabla} \cdot \vec{\nabla} \times \vec{B} = 0\)

\begin{align*} \vec{\nabla} \times \vec{B} = \mu_{0}\vec{J} + \mu_{0}\epsilon_{0}\frac{\partial\vec{E}}{\partial t} \end{align*}

By definition of electromagnetic induction (and to make Ampere's law consistent with relativity):

\begin{align*} \vec{\nabla} \times \vec{E} = - \frac{\partial\vec{B}}{\partial t} \end{align*}

Instead of assuming induction as an axiom, it is possible to fix Ampere's equation with the continuity equation first, and then assume Lorentz Covariance. This explanation is a work in progress. Though, a possible explanation is through the fact that this is the simplest way for Maxwell's equation to describe waves in a vacuum.

The continuity equation has a Lorentz Invariant counterpart, as predicted by . We take a look at Maxwell's equations (specifically Gauss' law and Ampere's law) in the uncondensed form:

\begin{align} \partial_{x}E_{x} + \partial_{y}E_{y} + \partial_{z}E_{z} = \frac{\rho}{\epsilon_{0}} \\ \partial_{x}E_{x} + \partial_{y}E_{y} + \partial_{z}E_{z} = \mu_{0}c^{2}\rho \\ \frac{1}{c}(\partial_{x}E_{x} + \partial_{y}E_{y} + \partial_{z}E_{z}) = \mu_{0}j_{t} \end{align}

Where \(j_{t} = c\rho\). Once we formulate a covariant form of the continuity equation, this will become more clear. And now the unexpanded version of Ampere's law:

\begin{align} \partial_{y}E_{z} - \partial_{z}E_{y} = \mu_{0}j_{x} + \frac{1}{c}\partial_{t}E_{x} \\ \partial_{z}E_{x} - \partial_{x}E_{z} = \mu_{0}j_{y} + \frac{1}{c}\partial_{t}E_{y} \\ \partial_{x}E_{y} - \partial_{y}E_{x} = \mu_{0}j_{z} + \frac{1}{c}\partial_{t}E_{z} \end{align}

(where \(\partial_{t} = \frac{1}{c}\frac{\partial}{\partial t}\)). We can now construct a 4-dimensional rank-2 tensor equation for these equations:

\begin{align} D \cdot \begin{pmatrix} 0 && \frac{1}{c}E_{x} && \frac{1}{c}E_{y} && \frac{1}{c}E_{z} \\ -\frac{1}{c}E_{x} && 0 && E_{z} && -E_{y} \\ -\frac{1}{c}E_{y} && -E_{z} && 0 && E_{x} \\ -\frac{1}{c}E_{z} && E_{y} && -E_{x} && 0 \\ \end{pmatrix} = \mu_{0} \begin{pmatrix} j_{t} \\ j_{x} \\ j_{y} \\ j_{z} \end{pmatrix} \end{align}

where the right hand side is an emergent four-vector \((c\rho, j)\), and \(D\) is an operator that takes the t, x, y, and z derivative of each respective column and then sums the rows to make a vector. We can reformulate the continuity equation in terms of this four-vector:

\begin{align*} \vec{\nabla} \cdot \vec{j} = 0 \end{align*}

Where \(\vec{\nabla}\) in this case represents the del operator but with a forth time dimension. If we condense the notation, we can write:

\begin{align*} D \cdot M = \vec{j} \end{align*}

By the same logic, we can use the remaining two equations (the divergence and curl of \(B\)) to construct this tensor:

\begin{align} D \cdot \begin{pmatrix} 0 && -B_{x} && -B_{y} && -B_{z} \\ B_{x} && 0 && -B_{z} && B_{y} \\ B_{y} && B_{z} && 0 && -B_{x} \\ B_{z} && -B_{y} && B_{x} && 0 \\ \end{pmatrix} = \vec{0} \end{align}

we can write this in compact form:

\begin{align*} D \cdot M' = \vec{0} \end{align*}

\(M'\) is the dual tensor of \(M\), where all the \(E_{n}\) are swapped with \(B_{n}\), and it is negative. Note that all these tensors are antisymmetric; they are symmetric under reflection across their diagonals and by flipping the signs.

The covariant form can be reached by gauge-fixing. Gauge fixing is a process wherein we utilize gauge invariance (invariance to a scalar addition to a potential) in order to fix a single scalar and remove a redundancy. Remember that in electrostatics:

\begin{align} \vec{E} = -\vec{\nabla}V \end{align}

and in magnetostatics:

\begin{align} \vec{B} = \vec{\nabla} \times \vec{A} \end{align}

but in electrodynamics:

\begin{align} \vec{E} = -\vec{\nabla}V - \frac{\partial \vec{A}}{\partial t} \\ \vec{B} = \vec{\nabla} \times \vec{A} \end{align}

because any field added to \(\vec{A}\) preserves the properties of \(\vec{B}\) when you take the curl. You can verify that if we take it for granted that \((\frac{V}{c}, A)\) is covariant, we can formulate the above two tensor equations in terms of a single equation:

\begin{align} D \cdot (M + M') = \mu_{0}\vec{j} \\ \Box A = \mu_{0}\vec{j} \end{align}

where \(\Box = \partial^{2}_{t} - \vec{\nabla}\). This single equation is the covariant Maxwell's equation. Truly a beautiful sight.