Why $ \int{\rho}dV = \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i $?

Question

I'd like to understand why $ \int{\rho}dV = \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i $ (the second equality), where

$j^i = \rho \frac{dx^i}{dt} $ is the current density 4-vector

$\mathbf{j} = \rho \mathbf{v}$ is the current density 3-vector

$ j^i = (c\rho, \mathbf{j}) $

$\rho$ is the charge density

$dS_i$ is the element $-dx-dy-dz+cdt$

Are you able to explain me this equality?

Thank you very much!

Answer 1

Following the advice of Weltschmerz, from 'The Classical Theory of Fields':

The total charge present in all of space is equal to the integral $\int \rho dV$ over all space. We can write this integral in four-dimensional form:

Then, you have the equality you wrote:

$$\displaystyle \int{\rho}dV = \displaystyle \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i$$

where the integral is taken over the entire four-dimensional hyperplane perpendicular to the $x^{0}$ axis (clearly this integration means integration over the whole three dimensional space). Generally, the integral

$$\displaystyle \frac{1}{c}\int j^{i}dS_{i}$$

over an arbitrary hypersurface is the sum of the charges whose world lines pass through this surface.

Then, $dS_{i}$ is not a surface in the conventional sense. Instead, is a hyper-surface.