Prove that Gauss's Law Holds Under Translations

Question

We know that Gauss's Law says $\nabla \cdot E(\textbf{x}) = \frac{\rho(\textbf{x})}{\varepsilon}$, and we also know that this relationship should be true regardless of where you're located in three space. Here we'll take $\textbf{x} \in \mathbb{R}^3$.

Therefore, it should be possible that if $E$ is a solution to Gauss's Law with charge density $\rho$, then for any $p$ a constant vector in $\mathbb{R}^3$, it should follow that $$\nabla \cdot E(\textbf{x} - p) = \frac{\rho(\textbf{x} - p)}{\varepsilon}$$

However, I am having trouble proving this statement. I understand that it should be a straightforward computation - simply take the divergence of $E(\textbf{x}-p)$.

Yet, I am having difficulty actually applying the chain rule.

Clearly we require $E(\textbf{x} - p) = (E_1(\textbf{x} - p), E_2(\textbf{x} - p), E_3(\textbf{x} - p))$ however I am not sure how to find the divergence of this to show the relation is true. The reason I'm confused is because we are taking the partial derivative of a scalar function and must apply the chain rule when the component function is itself a vector.

Thus, I would really appreciate it if someone could show how to explicitly find the divergence of $\nabla \cdot E(\textbf{x} - p)$ and show how the chain rule yields what we want.

Answer 1

Let $\textbf{g}=\textbf{x}-a$

If

$E(\textbf{x} - p) = (E_1(\textbf{x} - p), E_2(\textbf{x} - p), E_3(\textbf{x} - p))$

then

$\nabla \cdot E(\textbf{x} - p) = \partial_x E_1(\textbf{x} - p)+ \partial_y E_2(\textbf{x} - p)+ \partial_z E_3(\textbf{x} - p)= \nabla \cdot E(\textbf{g})$

Because $p$ is a constant.

and

$\nabla \cdot E(\textbf{g}) = \frac{\rho(\textbf{g})}{\epsilon}$

The chain rule is trivial. $\frac{\partial f(x-a)}{\partial x} =\frac{\partial f(g)}{\partial (g)} \frac{d(x-a)}{dx} =\frac{\partial f(g)}{\partial (g)}$ where $g=x-a$. Apply this to each component of the electric field.

Nothing interesting or unusual is going to happen with a vector $\textbf{x}=(x,y,z)$ because the $y$ and $z$ terms will just drop out when you take the derivative with respect to x.

Instead of thinking of $E_1$ as a function of a vector $\textbf{x}$, think of it as an ordinary function of $x$,$y$ and $z$. Because it is.

So rotate your coordinates so that $p$ lies on the x axis and calculate

$\partial_x E_1(x - p,y,z)+\partial_y E_2 (x - p, y ,z)+\partial_z E_3(x - p,y,z)$