# Prove that Gauss's Law Holds Under Translations

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.

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

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