Maxwell's Equations with Differential Forms

Today, I want to take Maxwell's equations and write them in the language of differential forms. The resulting equations are clearly covariant (i.e. they look the same after you apply a Lorentz transformation), and look a lot simpler than Maxwell's equations in vector notation. This is one of my favorite examples of how differential forms can make life easier.

Maxwell's Equations

In CGS units, Maxwell's equations are given by

  1. $\nabla \cdot E = 4 \pi \rho$
  2. $\nabla \cdot B = 0$
  3. $\nabla \times E = -\frac 1 c \frac{\partial B}{\partial t}$
  4. $\nabla \times B = \frac{4\pi}c J + \frac 1 c \frac{\partial E}{\partial t}$

$E$ is the electric field, $B$ is the magnetic field, $J$ is the electric current density, and $\rho$ is the electric charge density. $E, B$ and $J$ are vector fields and $\rho$ is a scalar field.

If we want to write Maxwell's equations with differential forms, we need to decide what type of forms will represent $E, B, J$ and $\rho$. Following Stern et al we will decide this based on how these fields are used in various equations.

First, we consider Faraday's law \[ \oint_C E \cdot d\ell = -\frac d {dt} \int_S B \cdot dA \]

$E$ is integrated over a curve, so $E$ naturally corresponds to a 1-form (Since one-forms are objects that you can integrate along curves). We will write the 1-form associated to $E$ as $\eta = E^\flat$. Meanwhile $B$ is integrated over a surface, so $B$ naturally corresponds to a 2-form. We will write this 2-form as $\beta = \star B^\flat$ as the 2-form associated to $B$. $\beta$ can be thought of as a function that measures the flux of $B$ through oriented parallelograms. $J$, the current density, is integrated over surfaces to find the current passing through the surface, so $J$ is naturally a 2-form. We will write this forms as $\mathscr J = \star J^\flat$. Finally, $\rho$ is integrated over volumes to find the enclosed charge, so $\rho$ is naturally a 3-form, and we will call the 3-form $\rho$.

We recall the following rules for translating from vector calculus to differential forms: \[\begin{aligned} (\nabla \cdot v)^\flat &= \star d \star v^\flat\\ (\nabla \times v)^\flat &= \star d v^\flat \end{aligned}\] We can use these rules to write Maxwell's equations in terms of $\eta, \beta, \mathscr J,$ and $\rho$.

We will start with the first equation. On the left hand side, $\nabla \cdot E$ becomes $\star d \star \eta$, which is a 0-form. We want to set it equal to the 3-form $4\pi \rho$. So we use the Hodge star to turn $\star d \star \eta$ into a 3 form and find that $\star \star d \star \eta = 4\pi \rho$. In 3D, $\star\star = 1$, so our equation is just

1'.  $d\star \eta = 4\pi \rho$

Now, we move on to the second equation. $\nabla \cdot B$ becomes $\star d \star (\star \beta)$. Since $\star\star = 1$, this just becomes $\star d \beta$. So our equation is $\star d \beta = 0$. Applying $\star$ to both sides gives $d\beta = 0$.

2'.  $d\beta = 0$

For the third equation, our substitutions gives us $\star d \eta = -\frac 1 c \frac{\partial\star \beta} {\partial t}$. Applying $\star$ to both sides yields $d\eta = -\frac 1 c \star \frac{\partial\star \beta} {\partial t}$. But we can pull the $\star$ inside the derivative to get

3'.  $d \eta = -\frac 1 c \frac{\partial \beta}{\partial t}$

Finally, we translate the last equation. Our substitution rules give us

4'.  $\star d \star \beta = \frac {4\pi} c \star \mathscr J + \frac 1 c \frac{\partial \eta}{\partial t}$

Putting all of the equations together, we have
1'.$d \star \eta = 4 \pi \rho$
2'.$d \beta = 0$
3'.$d \eta = - \frac 1 c \frac{\partial \beta}{\partial t}$
4'.$\star d \star \beta = \frac{4\pi}{c} \star \mathscr J + \frac 1 c \frac {\partial \eta}{\partial t}$

Now, we have written the equations using differential forms. But the equations still don't look very relativistic yet - we still have a big distinction between space and time derivatives. For our next step, we will stop thinking about forms on space that change over time, and instead think about forms on $3+1$-dimensional spacetime (i.e. spacetime with 3 spatial dimensions and 1 time dimension). In spacetime, we have both the spatial exterior derivative, the spatial Hodge star, the spacetime exterior derivative, and the spacetime Hodge star. We will denote the spatial operators $d_s, \star_s$ respectively, and we will denote the spacetime operators $d$ and $\star$.

The Exterior Derivative and Hodge Star in Spacetime

Before we can write Maxwell's equations in spacetime, we have to learn about how our spatial operators are related to our spacetime operators. We will use the convention that coordinates in spacetime are written like \[ (x^0, x^1, x^2, x^3) = (ct, x, y, z) \]

The Exterior Derivative

Now, we will look at the relationship between $d$ and $d_s$. Let $\omega = \sum_I \omega_I dx^I$ be a spatial differential form (i.e. no component of $\omega$ involves a $dx^0$). We can compute $d\omega$ as follows \[\begin{aligned} d\omega &= \sum_I d \omega_i \wedge dx^I\\ &= (dx^0 \wedge \partial_0 + d_s)\omega \end{aligned}\]

I skipped some of the computation to save space. See below for the full computation. It's not too long.
\[\begin{aligned} d\omega&=\sum_I d\omega_i \wedge dx^I\\ &=\sum_I \left[\left(\sum_{i=0}^3\partial_i\omega_Idx^i\right)\wedge dx^I\right]\\ &=\sum_I\left(\partial_0\omega_I dx^0\wedge dx^I + \sum_{i=1}^3\partial_i\omega_I dx^i\wedge dx^I\right)\\ &=dx^0\wedge\partial_0\omega+d_s\omega\\ &=(dx^0\wedge\partial_0+d_s)\omega \end{aligned}\]
So we see that $d = dx^0 \wedge \partial_0 + d_s$. The spacetime exterior derivative is just the spatial exterior derivative with an extra term related to the time derivative.

The Hodge Star of Spatial Forms

Now, we will relate $\star$ and $\star_s$. Let $\omega$ be a spatial $k$-form. The spacetime Hodge star is defined by the property that \[ \omega \wedge \star \omega = \left\langle \omega, \omega\right\rangle \mu \] Here $\mu = dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$ is the spacetime volume form. Let $\mu_s = dx^1 \wedge dx^2 \wedge dx^3$ be the spatial volume form. Clearly $\mu = dx^0 \wedge \mu_s$. Furthermore, we know that $\omega \wedge \star_s \omega = \left\langle \omega, \omega\right\rangle \mu_s$. Therefore, \[\begin{aligned} \omega \wedge \star \omega &= \langle \omega, \omega \rangle \mu\\ &= dx^0 \wedge \langle \omega, \omega \rangle \mu_s\\ &= dx^0 \wedge \omega \wedge \star_s\omega\\ &= \omega \wedge (-1)^k (dx^0 \wedge \star_s \omega) \end{aligned}\] So $\star \omega = (-1)^{k} \; dx^0 \wedge \star_s \omega$ when $\omega$ is purely a spatial $k$-form.

The Hodge Star of Forms with a Time Component

Now, suppose that $\omega = dx^0 \wedge \omega_s$ where $\omega_s$ is a spatial form. Then $\left\langle \omega, \omega\right\rangle = -\left\langle \omega_s,\omega_s\right\rangle$, so we need $\omega \wedge \star \omega = -\left\langle\omega_s,\omega_s\right\rangle\mu$. We know that $\omega_s \wedge \star_s \omega_s = \left\langle\omega_s,\omega_s\right\rangle \mu_s$. Thus, \[\omega \wedge \star_s \omega_s = dx^0 \wedge \omega_s \wedge \star_s \omega_s = dx^0 \wedge \mu_s = \mu \] So, $\star \omega = -\star_s \omega$ when $\omega$ is the wedge product of $dx^0$ and a spatial form.

Finally, we note for completeness that $\star dx^0 = -\mu_s$.

Covariant formulation of Maxwell's equations

Finally, we've developed all of the tools we need to write Maxwell's equations in spacetime. We will begin with the homogeneous equations (equations two and three). Maxwell's second equation tells us that $d\beta = 0$ and Maxwell's third equation tells us that $d_s \eta + \frac 1 c \frac{\partial \beta}{\partial t} = 0$. Because, we write coordinates in spacetime as \[ (x^0, x^1, x^2, x^3) = (ct, x, y, z) \] it turns out that $\frac 1 c \frac{\partial \beta}{\partial t} = \frac{\partial \beta}{\partial x^0} =: \partial_0 \beta$. So we can write Maxwell's third equation as $d_s \eta + \partial_0 \beta = 0$

The second equation is an equation of 3-forms, and the third equation is an equation of 2-forms. We can make them both equations of 3-forms by wedging the third equation with $dx^0$. Then we have $d_s \beta = 0$ and $dx^0 \wedge(d_s \eta + \partial_0\beta) = 0$. We can add these together to get $d_s \beta + dx^0 \wedge \partial_0 \beta + dx^0 \wedge d_s\eta = 0$. We note that because we are adding together forms of different types, their sum is 0 if and only if the individual terms in the sum are 0. So this equation expresses both Maxwell's second law and his third law. Inspecting the sum, we see that the first two terms are our expression for $d\beta$! Furthermore, the last term is $d(\eta \wedge dx^0)$. \[\begin{aligned} d(\eta\wedge dx^0) &= d\eta \wedge dx^0\\ &= (d_s \eta + dx^0 \wedge \partial_0 c\eta) \wedge dx^0 \\ &= d_s \eta \wedge dx^0\\ &= dx^0 \wedge d_s \eta \end{aligned}\] Therefore, we can write Maxwell's second and third equations as $d\beta + d(\eta \wedge dx^0) = 0$, or $d(\beta + \eta \wedge dx^0) = 0$. To simplify even more, we call $F := \beta + \eta \wedge dx^0$ the Faraday tensor, and simply write $dF = 0$.

Now, we consider $d \star F$. We'll start by computing $\star F$. \[ \star F = \star(\beta + \eta \wedge dx^0) = \star \beta + \star(\eta \wedge dx^0) \] Since $\beta$ is a spatial 2-form, $\star \beta = dx^0 \wedge \star_s \beta$. Since $\eta$ is a spatial 1-form, \[ \star(\eta \wedge dx^0) = -\star(dx^0 \wedge \eta) = -(- \star_s \eta) = \star_s \eta\] Putting this together shows us that $\star F = dx^0 \wedge \star_s \beta + \star_s \eta$. Now, we can take the exterior derivative. \[\begin{aligned} d \star F &= d(dx^0 \wedge \star_s \beta + \star_s \eta)\\ &= d_s \star_s \eta + dx^0 \wedge (\partial_0 \star_s \eta - d_s \star_s d_s \beta) \end{aligned}\]

I skipped most of the computation to save space. See below for the full computation. It's not too long.
\[\begin{aligned} d \star F &= d(dx^0 \wedge \star_s \beta + \star_s \eta)\\ &= (d_s + dx^0 \wedge \partial_0)(dx^0 \wedge \star_s \beta + \star_s \eta)\\ &= -dx^0 \wedge d_s \star_s \beta + d_s \star_s \eta + dx^0 \wedge \partial_0 \star_s \eta\\ &= d_s \star_s \eta + dx^0 \wedge (\partial_0 \star_s \eta - d_s \star_s d_s \beta) \end{aligned}\]

Maxwell's first equation tells us that $d_s \star_s \eta = 4\pi \rho$. Maxwell's fourth equation tells us that $\partial_0\eta - \star_s d_s \star_s \beta = -\frac{4\pi} c \star_s \mathscr J$. Therefore, \[\begin{aligned} d\star F &= 4\pi \star \rho\;dx^0 -\frac {4\pi} c dx^0 \wedge \mathscr J\\ \end{aligned}\] We define $c\rho - dx^0 \wedge \mathscr J = \mathfrak J$ to be the four-current. Now, our equation reads $d \star F = \frac {4\pi} c \mathfrak J$. This lets us finally express Maxwell's equations (in cgs units) as \[\begin{aligned} dF = 0 \quad\text{and}\quad d \star F = \frac{4\pi} c \mathfrak J \end{aligned}\]

Final Thoughts

I think this form of Maxwell's equations is very pretty. Because $F$ and $\mathfrak J$ are coordinate-independent objects, you can tell these equations must also be Lorentz covariant. And the equations don't distinguish between space and time coordinates.

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