Deriving the 4d continuity equation

1. May 9, 2006

Thrice

Well we start out with

$$-\frac {d} {dt} \int_{V}^{} \sigma dV = \int_{\Pi}^{} \vec{J} \cdot d\vec{\Pi}$$

Using the Gauss theorem

$$\int_{V}^{} (\frac{ \partial {\sigma}}{ \partial {t}} + div \vec{J}) dV = 0$$

so

$$\frac{ \partial {\sigma}}{ \partial {t}} + div \vec{J} = 0$$

and written in 4D..

$$\frac{ \partial {J^n}}{ \partial {x^n}} = 0 \qquad\quad\ (n = 0,1,2,3)$$

I can't seem to get my head around that last step. How does it expand out?

Last edited: May 9, 2006
2. May 9, 2006

Thrice

Ok done editing.

3. May 9, 2006

Thrice

This field is really confusing. Should I just gauss the answer?

4. May 10, 2006

Thrice

5. May 10, 2006

nrqed

Are you familiar with four-vector notation and so on? Then it is simply a question of definition that the last expression is $\partial_\mu J_\mu =0$ where Einstein's summation convention is used.

6. May 11, 2006

Thrice

Yeah well I'm familiar with it, but I needed the definition spelled out. I eventually worked it out.