home | section main page


Maxwell's Equations

Table of Contents

1. Introduction

Maxwell's Equations are a set of four differential equations in multiple dimensions that produce a complete classical theory of electromagnetic phenomena.

2. Derivations

These are the derivations of all four laws in their differential forms based on the inverse square law for (Coulomb's Law), the Lorentz Force, as well as the continuity equation and electromagnetic induction (which are just special cases of ) as initial assumptions:

2.1. Gauss' Law

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*}

2.2. Divergence of Magnetic Field

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

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

2.3. Ampere's Law with Modifications

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*}

2.4. Faraday's Law of Induction

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.

3. Implications

Maxwell's Equations can be used to calculate all electromagnetic phenomena on the macro scale all the way down to the atom. In practice, solving Maxwell's Equations can be analytically impossible, so several simplifying assumptions are often made. To recap, these are the four equations:

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

4. Speed of Light

Maxwell's Equations can be shown to reproduce the speed of light in a vacuum, where:

\begin{align*} \mu_{0}\epsilon_{0} = \frac{1}{c^{2}} \end{align*}

5. Relativity

It is known that Maxwell's Equations are consistent with and can be expressed in terms of curved spacetime. In fact, if relativity is taken as an axiom, it can be proven that the electric and magnetic fields are descriptions of the same phenomena; this can be taken as a specific example of a duality. The result is a Lorentz Invariant theory of Electromagnetism.

5.1. Gauss-Ampere Equation

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*}

5.2. Gauss-Faraday Equation

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.

5.3. Covariant Form

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.

Copyright © 2024 Preston Pan