- Reduced version
- Vacuum equations
- Covariant vacuum equations
- Electromagnetic waves
- Example: Polarised light

### Reduced version

The full set of *eight* Maxwell's equations are

$$\begin{matrix}\nabla \times \vec{E} &= -\frac{\partial \vec{B}}{\partial t}&\nabla \times \vec{B} &= \mu_0 \varepsilon_0\frac{\partial \vec{E}}{\partial t} + \mu_0 \vec{J}\\\nabla \cdot \vec{E} &= \frac{\rho_0}{\varepsilon_0}&\nabla \cdot \vec{B} &= 0\end{matrix}$$ where

- $\vec{E}, \vec{B}$ are the electric and magnetic fields
- $\rho_0$ is called the
**charge density**(we will define this later) - $\vec{J}$ is called the
**current density**(we will define this later) - $\varepsilon_0$ is a constant called the
*permittivity of free space* - $\mu_0$ is a constant called the
*permeability of free space*

The speed of light, $c$, was calculated from these equations using the formula $$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$$ but since we are setting $c \equiv 1$, we can eliminate one of these constants, $$\mu_0 = \frac{1}{\varepsilon_0}$$

The equations simplify somewhat to $$\begin{matrix}\nabla \times \vec{E} &= -\frac{\partial \vec{B}}{\partial t}&\nabla \times \vec{B} &= \frac{\partial \vec{E}}{\partial t} + \frac{1}{\varepsilon_0} \vec{J}\\\nabla \cdot \vec{E} &= \frac{\rho_0}{\varepsilon_0}&\nabla \cdot \vec{B} &= 0\end{matrix}$$

Professor Susskind then subsumes the other constant into the charge and current densities, $$\begin{align*}\rho &= \frac{\rho_0}{\varepsilon_0}\\\vec{j} &= \frac{1}{\varepsilon_0} \vec{J}\end{align*}$$

We now have what we might call the *reduced* version of Maxwell's equations, $$\begin{matrix}\nabla \times \vec{E} &= -\frac{\partial \vec{B}}{\partial t}&\nabla \times \vec{B} &= \frac{\partial \vec{E}}{\partial t} + \vec{j}\\\nabla \cdot \vec{E} &= \rho&\nabla \cdot \vec{B} &= 0\end{matrix}$$

### Vacuum equations

The simplest situation to consider first is when there are no charges or currents present so that $$\begin{align*}\rho &= 0\\\vec{j} &= 0\end{align*}$$ leading to **Maxwell's vacuum equations**, $$\begin{matrix}\nabla \times \vec{E} &= -\frac{\partial \vec{B}}{\partial t}&\nabla \times \vec{B} &= \frac{\partial \vec{E}}{\partial t}\\\nabla \cdot \vec{E} &= 0&\nabla \cdot \vec{B} &= 0\end{matrix}$$

An observer moving along the same axis as a light-ray might expect to see it moving at a different speed than a stationary observer, but these equations describe light-rays *always moving with the same speed* regardless of the speed of the observer.

The speed of light *is* invariant, according to the vacuum equations. But, the equations themselves aren't invariant as they are written above. We would like to recast them in terms of tensors.

### Covariant vacuum equations

Recall the Faraday tensor, $$F^{\mu \nu} = \begin{bmatrix}0 & -E_x & -E_y & -E_z\\E_x & 0 & -B_z & B_y\\E_y & B_z & 0 & -B_x\\E_z & -B_y & B_x & 0\end{bmatrix}$$ from the Lorentz force equation $$\frac{\mathrm{d} P^\mu}{\mathrm{d} \tau} = q F^{\mu \nu} U_\nu$$

There is another matrix of components of the electric and magnetic fields which also transforms correctly as a tensor. To derive it, notice that that the electric and magnetic fields appear in the vacuum equations anti-symmetrically. That is, we could substitute one for the other in the vacuum equations and they would look exactly the same, except that we have to include a minus-sign, $$\begin{align*}\vec{E} &\rightarrow -\vec{B}\\\vec{B} &\rightarrow \vec{E}\end{align*}$$

Equivalently, we can exchange the components of the Faraday tensor, $$\begin{align*}E_m &\rightarrow -B_m\\B_m &\rightarrow E_m\end{align*}$$ to give $$\tilde{F}^{\mu \nu} = \begin{bmatrix}0 & B_x & B_y & B_z\\-B_x & 0 & -E_z & E_y\\-B_y & E_z & 0 & -E_x\\-B_z & -E_y & E_x & 0\end{bmatrix}$$

We want to write vacuum equations in terms of these tensors. Notice that they are all first-order differential equations, so we might consider covariantly differentiating once each of these tensors. But, just doing that would actually lead to many more than eight equations.

The correct equations require us to *differentiate and then contract on the covariant index*, which leads to the **covariant vacuum equations**, $$\begin{align*}\partial_\mu F^{\mu \nu} &= 0\\\partial_\mu \tilde{F}^{\mu \nu} &= 0\end{align*}$$

By definition, these equations are invariant under Lorentz transformations and we can recover each of the classical equations by choosing a value for the index $\nu$.

Nb. Unlike the derivation for the covariant Lorentz force law, we needed to consider the limiting condition, where the speeds involved were low compared to the speed of light, here, we recover the classical equations exactly.

Consider first $\nu = x$, $$\begin{align*} & \partial_\mu F^{\mu x} = 0\\\Rightarrow & \frac{\partial F^{t x}}{\partial t} + \frac{\partial F^{x x}}{\partial x} + \frac{\partial F^{y x}}{\partial y} + \frac{\partial F^{z x}}{\partial z} = 0\\\Rightarrow & \frac{\partial (-E_x)}{\partial t} + \frac{\partial (0)}{\partial x} + \frac{\partial (B_z)}{\partial y} + \frac{\partial (-B_y)}{\partial z} = 0\\\Rightarrow & -\frac{\partial E_x}{\partial t} + \frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z} = 0\\\Rightarrow & \left[\nabla \times \vec{B}\right]_x = \left[\frac{\partial \vec{E}}{\partial t}\right]_x\end{align*}$$ and this pattern is repeated for the other spatial components, $m = x,y,z$, so that $$\nabla \times \vec{B} = \frac{\partial \vec{E}}{\partial t}$$

The time component gives $$\begin{align*} & \partial_\mu F^{\mu t} = 0\\\Rightarrow & \frac{\partial F^{t t}}{\partial t} + \frac{\partial F^{x t}}{\partial x} + \frac{\partial F^{y t}}{\partial y} + \frac{\partial F^{z t}}{\partial z} = 0\\\Rightarrow & \frac{\partial (0)}{\partial t} + \frac{\partial (E_x)}{\partial x} + \frac{\partial (E_y)}{\partial y} + \frac{\partial (E_z)}{\partial z} = 0\\\Rightarrow & \nabla \cdot \vec{E} = 0\end{align*}$$

The other four equations follow directly from the second tensor equation, thus we can write $$\begin{align*}\partial_\mu F^{\mu \nu} = 0 &\Rightarrow \left\{\begin{matrix}\nabla \times \vec{B} = \frac{\partial \vec{E}}{\partial t}\\\nabla \cdot \vec{E} = 0\end{matrix}\right.\\\partial_\mu \tilde{F}^{\mu \nu} = 0 &\Rightarrow \left\{\begin{matrix}\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}\\\nabla \cdot \vec{B} = 0\end{matrix}\right.\end{align*}$$

### Electromagnetic waves

Staying with the vacuum form for now, we would to find out how electromagnetic waves stem from Maxwell's equations.

A wave equation has $2^\text{nd}$ order derivatives, whereas Maxwell's equations are all $1^\text{st}$ order. However, they depend on two fields, $\vec{E}, \vec{B}$, so they are equivalent to $2^\text{nd}$ order equations in either of the fields separately.

Consider, we can differentiate the equation, $$\nabla \times \vec{B} = \frac{\partial \vec{E}}{\partial t}$$ with respect to time, to give $$\nabla \times \frac{\partial \vec{B}}{\partial t} = \frac{\partial^2 \vec{E}}{\partial t^2}$$

But $$\frac{\partial \vec{B}}{\partial t} = -\nabla \times \vec{E}$$ which we can substitute in to the above expression (and rearrange) to give $$\nabla \times \left(\nabla \times \vec{E}\right) = -\frac{\partial^2 \vec{E}}{\partial t^2}$$ which is now a $2^\text{nd}$ order equation in the $\vec{E}$-field only.

It is a straightforward exercise to show that $$\left[\nabla \times \left(\nabla \times \vec{E}\right)\right]_m = -\nabla^2 E_m + \partial_m \left(\nabla \cdot \vec{E}\right)$$ where $$\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$$

But we also know that $$\nabla \cdot \vec{E} = 0$$ thus we end up with three wave equations where the field in each case is one of the three components of $\vec{E}$, $$\frac{\partial^2 E_m}{\partial t^2} - \frac{\partial^2 E_m}{\partial x^2} - \frac{\partial^2 E_m}{\partial y^2} - \frac{\partial^2 E_m}{\partial z^2} = 0$$

A similar argument shows that the components of the magnetic field $\vec{B}$ also satisfy wave equations, $$\frac{\partial^2 B_m}{\partial t^2} - \frac{\partial^2 B_m}{\partial x^2} - \frac{\partial^2 B_m}{\partial y^2} - \frac{\partial^2 B_m}{\partial z^2} = 0$$

These waves can move at the speed of light in any direction, and we can show (mathematically) that these waves are made up of sums of sinusoidal functions, each of which can be considered as a particular solution.

### Example: Polarised light

Recall, solutions of wave equations always have the form $$\phi = F\left(z - t\right)$$ where $F$ is any real function. We can consider a particular solution of the wave equation for the electric field, where the wave is moving in the $z$-direction, but pointing in the $x$-direction. For example, $$\begin{align*}E_x &= \cos\left(z - t\right)\\E_y &= 0\end{align*}$$

This is an example of polarised light - in this case, polarised along the $x$-axis.

Having set $E_y = 0$, we can calculate $E_z$ using $\nabla \cdot \vec{E} = 0$ since this says $$\begin{align*}0 &= \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z}\\&= \frac{\partial (0)}{\partial x} + \frac{\partial}{\partial y}\cos\left(z - t\right) + \frac{\partial E_z}{\partial z}\\&= \frac{\partial E_z}{\partial z}\end{align*}$$ which means that $E_z$ must be constant along the $z$-axis. Since a constant could never contribute to any wave motion we can disregard it, and set it to zero. So we end up with only an $x$-component of the electric field - what about the magnetic field?

Thus, $$\vec{E} = \cos\left(z - t\right) \hat{x}$$

We can compute the magnetic field using $$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$

The only non-zero component of $\nabla \times \vec{E}$ is in the $y$-direction, whic is the only one that involves $$\frac{\partial E_x}{\partial z} = \frac{\partial}{\partial z}\sin\left(z - t\right) = \sin\left(z - t\right)$$. Thus, $$\frac{\partial \vec{B}}{\partial t} = \sin\left(z - t\right) \hat{y}$$

We can integrate this function with respect to time, $$\vec{B} = \cos\left(z - t\right) \hat{y} + \vec{b}$$ where $\vec{b} = \vec{b}(x,y,z)$ is a function of space only and constant in time. But that means it doesn't contribute to any wave motion, hence we can disregard it (and set it to zero).

Thus we get the pair of *perpendicular* electromagnetic waves, $$\begin{align*}\vec{E} &= \cos\left(z - t\right) \hat{x}\\\vec{B} &= \cos\left(z - t\right) \hat{y}\end{align*}$$ which have exactly the same form and travel at exactly the same speed - the speed of light.

We can consider what this pair of waves look like to an observer travelling at speed $v$ along the direction of the waves.

That means calculating a boost in the $z$-direction, transforming the Faraday tensor and reading off the relevant components. Actually, I just rotated the indices $$z \leftarrow x \leftarrow y \leftarrow z$$ in the expressions we computed for a boost in the $x$-direction in lecture 7, to get $$\begin{align*}E_{x'} &= \gamma\left(E_x - vB_y\right)\\E_{y'} &= \gamma\left(E_y + vB_x\right)\\E_{z'} &= E_z\\B_{x'} &= \gamma\left(B_x + vE_y\right)\\B_{y'} &= \gamma\left(B_y - vE_x\right)\\B_{z'} &= B_z\end{align*}$$

Now, we can substitute the components $$E_x = B_y = \cos\left(z - t\right) \text{ and } E_y = E_z = B_x = B_z = 0$$ to give $$\begin{align*}E_{x'} &= \sqrt{\frac{1 - v}{1 + v}} \cos\left(z - t\right)\\B_{y'} &= \sqrt{\frac{1 - v}{1 + v}} \cos\left(z - t\right)\\E_{y'} &= E_{z'} = B_{x'} = B_{z'} = 0\end{align*}$$

We could write $$\begin{align*}\vec{E}' &= \alpha(v)\vec{E}\\\vec{B}' &= \alpha(v)\vec{B}\end{align*}$$ where $\alpha(v)$ corresponds to a change in the magnitude of the electric and magnetic waves, $$\begin{align*}\alpha(v) \lt 1 &\text{ if } v \gt 0\\\alpha(v) \gt 1 &\text{ if } v \lt 0\end{align*}$$

Thus, two observers, moving at relative speed $v$ along the axis of the direction of the waves, will see

- the waves moving at the same speed
- the electric part of the wave perpendicular to the magnetic part of the wave
- different wave amplitudes, depending on the relative speed, $v$