Maxwell equation summary

22 Sep 2020 - tsp

This blog post is just a short summary of the Maxwell equations, how one can show some basic properties of electromagnetic waves and two of the most commonly used Gauges and their rationals.

[ \newcommand{\curl}{\mathop{\rm rot}\nolimits} \newcommand{\div}{\mathop{\rm div}\nolimits} \newcommand{\grad}{\mathop{\rm grad}\nolimits} ]

Differential operators used

In the following blog post a few differential operators are used that are worth looking at in the first place.

To reduce some amount of typing and writing I’ll use a shorthand notation for differential such that $\partial_x = \frac{\partial}{\partial x}$.

The first one is the Nabla operator that’s just a vector containing the corresponding differentials for each coordinate axis:

[ \vec{\nabla} = \left(\begin{matrix}\partial_x \\ \partial_y \\ \partial_z \end{matrix}\right) ]

There are three ways this operator will mainly be used in this article - namely the divergence $\div$, the curl $\curl$ and the gradient $\grad$:

[ \div \vec{u} = \vec{\nabla} \vec{u} = \partial_x * u_x + \partial_y * u_y + \partial_z * u_z \\ \curl \vec{u} = \vec{\nabla} \times \vec{u} = \left(\begin{matrix}\partial_y u_z - \partial_z u_y \\ -\partial_x u_z + \partial_z u_x \\ \partial_x u_y - \partial_z u_x \end{matrix}\right) \\ \grad f = \vec{\nabla} f = \left(\begin{matrix}\partial_x f \\ \partial_y f \\ \partial_z f \end{matrix}\right) ]

As one can see the divergence yields a scalar field that describes the degree of a vector fields source terms at each given position. The curl describes the infinitesimal rotation and yields - again - a vector field. The direction of the rotor is the rotational axis of the whole field at any given point, the magnitude of course the magnitude of the rotation (such as angular speed). The gradient is an operator that’s applied to a scalar field that yields a vector field. It describes the gradient, i.e. the variation, of the scalar field at every point.

There are some basic rules that are interesting in the context of Maxwell equations. First one can look at the repeated application of curl and divergence / gradient:

[ \vec{\nabla} (\vec{\nabla} \phi) = \div (\grad \phi) = \Delta \phi \\ \vec{\nabla} (\vec{\nabla} \times \vec{A}) = \div (\curl \vec{A}) = 0 \\ \vec{\nabla} \times (\vec{\nabla} f) = \curl (\grad f) = 0 \\ \vec{\nabla} \times (\vec{\nabla} \times \vec{A}) = \curl (\curl \vec{A}) = \grad (\div \vec{A}) - \Delta \vec{A} ]

Maxwell equations

So first let’s start with stating the Maxwell equations:

Electromagnetic waves in vacuum

To reason about electromagnetic waves it’s really easy to simply apply another curl operator on the third or fourth equation and assume that in vacuum the charge carrier density $\rho(\vec{r},t)=0$ and the current thus also $\vec{j}(\vec{r},t)=0$. To make equations more readable I drop the arguments $(\vec{r},t)$ at all fields $\vec{E}(\vec{r},t)$, $\vec{B}(\vec{r},t)$, $\rho(\vec{r},t)$, $\vec{j}(\vec{r},t)$.

[ \curl \vec{E} = - \partial_t \vec{B} \\ \curl \curl \vec{E} = - \curl \partial_t \vec{B} \\ \grad \underbrace{(\div \vec{E})}_{=\frac{\rho}{\epsilon_0}=0} - \Delta \vec{E} = - \partial_t \curl \vec{B} \\ \to -\Delta \vec{E} = -\partial_t (\underbrace{\mu_0 \vec{j}}_{=0} + \mu_0 \epsilon_0 \partial_t E) \\ \to -\Delta \vec{E} = - \mu_0 \epsilon_0 \partial^2_t \vec{E} \\ \nabla^2 \vec{E} = \mu_0 \epsilon_0 \partial^2_t \vec{E} ]

As one can see this is a classical second order wave equation. The same calculation can be done for the magnetic field:

[ \curl \vec{B} = \underbrace{\mu_0 \vec{j}}_{=0} + \mu_0 \epsilon_0 \partial_t \vec{E} \\ \curl \curl \vec{B} = \mu_0 \epsilon_0 \partial_t \curl \vec{E} \\ \grad \underbrace{\div \vec{B}}_{=0} - \Delta \vec{B} = - \mu_0 \epsilon_0 \partial_t^2 \vec{B} \\ \nabla^2 \vec{B} = \mu_0 \epsilon_0 \partial_t^2 \vec{B} ]

When one recalls how one could solve such wave equations in one dimension using the ansatz $f(x,t) = c_2 * e^{- i \omega t + i k x}$:

[ \partial_x^2 f(x,t) = c_1 \partial_t f(x,t) \\ k^2 f(x,t) = c_1 \omega^2 f(x,t) \\ \frac{k^2}{\omega ^2} = c_1 \\ \frac{\omega^2}{k^2} = \frac{1}{c_1} \\ \frac{\omega}{k} = \frac{1}{\sqrt{c_1}} ]

and one also recalls that the phase velocity of a plane wave has been defined as

[ \phi = kx - \omega t \\ v_p = \frac{dx}{dt} = \frac{\partial_t \phi}{\partial_x \phi} = - \frac{i \omega}{i k} = -\frac{\omega}{k} \\ v_p^2 = \frac{\omega^2}{k^2} ]

one can identify the constant $\frac{1}{\sqrt{\epsilon_0 \mu_0}}$ as the phase velocity of the electromagnetic wave, i.e. the speed of light:

[ c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} ]

Note that the phase velocity is of course not the velocity by which information can be transferred. This is described by the group velocity $v_g = \frac{\partial \omega}{\partial k}$

Electromagnetic (EM) waves in vacuum are traversal

How can one show that electromagnetic waves in vacuum have to be traversal, i.e. they do not have a longitudinal component oscillating in the direction of propagation? This is easily done by inserting the ansatz into the divergence equations in vacuum:

[ \nabla \vec{E} = \nabla E_0 e^{- i \omega t + i \vec{k} \vec{r}} = 0 ]

Assuming that the wave travels in $z$ direction all other components vanish and one can simply look at the z component of the divergence:

[ \nabla \vec{E_0} e^{- i \omega t + i k_z z} = 0 \\ k_z * E_0z e^{-i \omega t + i \vec{k} \vec{r}} = 0 \\ \to E_0z = 0 \\ \to B_0z = 0 ]

From this one can see that all electromagnetic waves in vacuum only have traversal components, i.e. they oscillate perpendicular to the direction of propagation.

Vector and scalar potential, Gauge invariance

As with any field theory one might be tempted to use a potential describing the field structure. It turns out that this works for time independent Maxwell equations for electric fields by defining

[ \vec{E}(\vec{r}) = - \grad \phi(\vec{r}) ]

Unfortunately this does not work any more for time dependent fields as can be seen easily when inserting into Maxwell equations. To solve that problem one can first define a vector potential $\vec{A}$ such that:

[ \vec{B} = \curl \vec{A} ]

This directly ensures the satisfaction of the Gaussian magnetic law independent of the selected vector field $\vec{A}$ since the divergence of a curl vanishes:

[ \vec{\nabla} \vec{B} = 0 \\ \vec{\nabla} (\vec{\nabla} \times \vec{A}) = 0 \\ \div (\curl \vec{A}) = 0 ]

One can now insert this vector potential into the induction law:

[ \vec{\nabla} \times \vec{E} = \frac{\partial}{\partial t} \vec{B} \\ \vec{\nabla} \times \vec{E} = \frac{\partial}{\partial t} \vec{\nabla} \times \vec{A} \\ \vec{\nabla} \times \vec{E} = \vec{\nabla} \times \frac{\partial}{\partial t} \vec{A} \\ \vec{\nabla} \times \vec{E} - \vec{\nabla} \times \frac{\partial}{\partial t} \vec{A} = 0 \\ \vec{\nabla} \times \left(\vec{E} - \partial_t \vec{A} \right) = 0 ]

Now assuming that one wants to describe the potential again with a scalar field again yields to the description of the electric field using the scalar potential $\phi$ and the vector potential $\vec{A}$:

[ \vec{E} + \partial_t \vec{A} = -\vec{\nabla} \phi \\ \vec{E} = -\vec{\nabla} \phi - \partial_t \vec{A} ]

As one can see this definition is working as expected for the static (time independent) case:

[ \vec{B} = \vec{\nabla} \times \vec{A} \\ \vec{E} = -\vec{\nabla} \phi - \underbrace{\partial_t \vec{A}}_{0} \\ \to \vec{E}_{Static} = -\vec{\nabla} \phi ]

This definition helps to see that the Maxwell equations are gauge invariant. What does gauge invariance mean? There is a degree of freedom - one can simply add an gradient field to the vector potential without any change to the resulting magnetic field:

[ \vec{A'} = \vec{A} + \vec{\nabla} f ]

To compensate the effect on the electric field one has to apply a gauge transformation to the scalar potential though. This can be seen by requiring that the gauge transformation does not change the electric field either:

[ \vec{E} = -\vec{\nabla} \phi - \partial_t \vec{A} \\ = -\vec{\nabla} \phi' - \partial_t \vec{A'} \\ = -\vec{\nabla} \phi' - \partial_t \vec{A} - \partial_t \vec{\nabla} f \\ = -\vec{\nabla} (\phi' + \partial_t f) - \partial_t \vec{A} \\ \to \phi = \phi' + \partial_t f \\ \to \phi' = \phi - \partial_t f ]

Why is this gauge invariance that important? As Noether showed in her famous Noether theorem every continuous symmetry of a physical system is associated with a conserved quantity - and that each conserved quantity also generates a symmetry group.

Usually - especially when doing calculations in general relativity - the vector potential and the scalar potential are grouped together as the four dimensional vector called four potential:

[ A^\alpha = \left(\begin{matrix}\frac{\phi}{c} \\ \vec{A} \end{matrix}\right) ]

In Laurenz gauge (see below) this will simplify the Maxwell equations as a single application of the d’Alembert operator $\Box$ and the four-current $\vec{J}^\alpha$:

[ \Box = \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \vec{\nabla}^2 \\ \Box A^\alpha = \mu_0 \vec{J}^\alpha ]

Since the Maxwell equations are gauge invariant one is usually free to select a proper gauge to perform calculations.

Coulomb gauge

The Coulomb gauge selects the vector potential in a way that the divergent (source) term of the vector potential vanishes.

[ \vec{\nabla} A = 0 \\ \div A = 0 ]

When one looks at the Fourier transformation of the Coulomb gauge and uses the ansatz that $\vec{A}(\vec{r},t) = \vec{A_0} e^{i * (\vec{k}\vec{r} - \omega t)}$ one can see again that this requires the vector potential to be a traversal wave without any longitudinal components:

[ \mathfrak{F} \div \vec{A}(\vec{r},t) \\ = \mathfrak{F} \div \vec{A_0} e^{i * (\vec{k}\vec{r} - \omega t)} \\ = \mathfrak{F} i \vec{k} \vec{A}(\vec{r},t) \\ = i \vec{k} \vec{\tilde{A}}(\vec{k},t) = 0 ]

Lorenz gauge

The Lorenz gauge is often used in relativistic approaches. The gauge is chosen such that

[ \Box A^\alpha = \mu_0 \vec{J}^\alpha ]

One can see that this corresponds to

[ \frac{1}{c^2} \frac{\partial}{\partial t} \phi(\vec{r}, t) + \vec{\nabla} \vec{A}(\vec{r},t) = 0 ]

The big advantage of this gauge is that it imposes Lorentz invariance - i.e. the potentials are then invariant under Lorentz transformations.

A quick justification

A simple rationale for choosing the Laurenz gauge despite the invariance can be found by plugging the definition of the vector potential into the induction law:

[ \curl \vec{E} = - \partial_t \vec{B} \\ \curl \vec{E} = - \partial_t \curl \vec{A} \\ \curl \vec{E} = - \curl \partial_t \vec{A} \\ \to \curl \left(\vec{E} + \partial_t \vec{A} \right) = 0 ]

Because the curl vanishes and the trivial solution of both fields being zero is non interesting one can assume that there exists a potential function $\psi$ such that the gradient yields the sum $\vec{E} + \partial_t \vec{A}$:

[ \curl \left(\vec{E} + \partial_t \vec{A} \right) = 0 \\ \to - \grad \psi = \vec{E} + \partial_t \vec{A} \\ \vec{E} = - \vec{\nabla} \psi - \partial_t \vec{A} ]

This definition is exactly the expression calculated above during the gauge transformation. Inserting this result into the magnetic induction law:

[ \curl \vec{B} = \mu_0 \vec{j} + \epsilon_0 \mu_0 \partial_t \vec{E} \\ \curl \vec{B} = \mu_0 \vec{j} + \epsilon_0 \mu_0 \partial_t \left(- \vec{\nabla} \psi - \partial_t \vec{A} \right) \\ \curl \curl \vec{A} = \mu_0 \vec{j} - \epsilon_0 \mu_0 \partial_t \vec{\nabla} \psi - \epsilon_0 \mu_0 \partial_t^2 \vec{A} \\ \grad (\div \vec{A}) - \Delta \vec{A} = \mu_0 \vec{j} - \vec{\nabla} \mu_0 \epsilon_0 \partial_t \psi - \epsilon_0 \mu_0 \partial_t^2 \vec{A} \\ \vec{\nabla} (\vec{\nabla} \vec{A} + \mu \epsilon_0 \partial_t \psi) = \mu_0 \vec{j} - \epsilon_0 \mu_0 \partial_t^2 \vec{A} + \Delta \vec{A} \\ \vec{\nabla} (\vec{\nabla} \vec{A} + \frac{1}{c^2} \partial_t \psi) = \mu_0 \vec{j} - \frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2} + \vec{\nabla}^2 \vec{A} ]

The fastest way to select the Laurenz gauge is to require the spatial and time derivative to be of the same order. This can simply be solved by requiring

[ \vec{\nabla} \vec{A} + \frac{1}{c^2} \partial_t \psi = 0 ]

This yields

[ 0 = \mu_0 \vec{j} - \frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2} + \vec{\nabla}^2 \vec{A} \\ \vec{\nabla}^2 \vec{A} - \frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2} = - \mu_0 \vec{j} \\ \underbrace{\left( \vec{\nabla}^2 - \frac{1}{c^2} \partial_t^2 \right)}_{\Box} \vec{A} = -\mu_0 \vec{j} ]

Here one can see the definition of the d’Alembert operator $\Box$.

To obtain an decoupled expression for the scalar potential one can use the Gauss law:

[ \vec{\nabla} \vec{E} = \frac{\rho}{\epsilon_0} \\ \vec{E} = - \vec{\nabla} \phi - \partial_t \vec{A} \\ \vec{\nabla}^2 \phi - \partial_t \vec{\nabla} \vec{A} = \frac{\rho}{\epsilon_0} ]

Now one can use Laurenz gauge

[ \vec{\nabla} \vec{A} + \frac{1}{c^2} \partial_t \phi = 0 \\ \vec{\nabla} \vec{A} = -\frac{1}{c^2} \partial_t \phi \\ \to \vec{\nabla}^2 \phi - \partial_t \underbrace{\frac{1}{c^2} \partial_t \phi}_{\div \vec{A}} = \frac{\rho}{\epsilon_0} \\ \underbrace{\left(\vec{\nabla}^2 - \frac{1}{c^2} \partial_t^2 \right)}_{\Box} \phi = -\frac{\rho}{\epsilon_0} ]

These two identities yielded to different equations for the vector potential and the scalar potential:

[ \Box \vec{A}(\vec{r}, t) = -\mu_0 \vec{j}(\vec{r}, t) \\ \Box \phi(\vec{r}, t) = -\frac{1}{\epsilon_0} \rho(\vec{r}, t) ]

This article is tagged: Physics, Tutorial, Electrodynamics


Dipl.-Ing. Thomas Spielauer, Wien (webcomplains389t48957@tspi.at)

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