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

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!

• Isn't that conservation of charge? Divergence theorem is applied there. Look up any book on electromagnetism. – Weltschmerz Oct 6 '10 at 21:57
• But where is the divergence? And the flux? please explain a little bit, thanks! =) – Abramo Oct 6 '10 at 22:06
• The flux is in the surface integral, the divergence is in the volume integral. Recall an important Maxwell equation. Again, this is all thoroughly explained in many books on electrodynamics, see for example the Landau and Lifschitz book. – Weltschmerz Oct 6 '10 at 22:11
• I know the divergence theorem, but simply i can't see any surface here! The dS is not a surface element, but -dx-dy-dz+cdt – Abramo Oct 6 '10 at 22:17

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.