# Causality¶

Field theory shows a lot of casality conditions. Here is a collection of them.

Maxwell’s equations in vacuum are

$\begin{split}\nabla\cdot \vec E &= 4\pi \rho, \\ \nabla \cdot \vec B & = 0, \\ \nabla \times \vec E & = -\frac{1}{c}\partial_t \vec B, \\ \nabla \times \vec B & = \frac{1}{c}\partial_t E + \frac{4\pi}{c}\vec j.\end{split}$

To write down the wave equation, we could switch to the scalar potential $$\phi$$ and vector potential $$\vec A$$.

Divergence free means that we can always have

$\vec B = \nabla \times \vec A.$

By using the above relation, I could rewrite this to

$\begin{split}\nabla \times \vec E &= -\frac{1}{c} \partial_t (\nabla \times \vec A),\\ \nabla \times ( \vec E + \frac{1}{c} \partial_t \vec A ) &= 0.\end{split}$

This means I can write all inside divergance written as a gradient of a scalar function or a constant time the gradient of some scalar function.

$\vec E= -\nabla \phi - \frac{1}{c} \partial_t \vec A.$

With the definition of scalar and vector potentials, we could plug them in and find the wave equations. However, since the values of these potential are gauge dependent, I should choose a convinient gauge. Hereby, I use Lorenz gauge.

$\frac{1}{c} \partial_t \phi + \nabla \cdot \vec A = 0.$

The importance of this gauge is that it is Lorentz invariant.

Using this gauge the two other Maxwell’s equations, I have the wave equations,

$\begin{split}(\frac{1}{c^2} \partial_t^2 -\nabla^2) \phi &= 4\pi \rho ,\\ (\frac{1}{c^2} \partial_t^2 -\nabla^2) \vec A &= \frac{4\pi}{c} \vec j .\end{split}$

Solving these Helmholtz equations, I get the solution as a function of retarted time $$t_{ret}=t-\frac{R}{c}$$, where $$R=\vert \vec x - \vec x' \vert$$.

$\begin{split}\phi(t,\vec x) &= \int d^3x' \frac{\rho(t_{ret},\vec x')}{R}, \\ \vec A(t,\vec x) & = \int d^3 x' \frac{\vec j(t_{ret},\vec x')}{R}.\end{split}$

Here it clearly shows that the observation depends on the history $$R/c$$ ago. This is the signal propagation time.

## Response of Matter¶

$\vec P = \int \chi(t,t') \vec E(t') dt'.$