Deriving the 4d continuity equation

[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?

 

[Thrice]

Ok done editing.

 

[Thrice]

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

 

[Thrice]

Am I not explaining it properly or..

http://scienceworld.wolfram.com/physics/ContinuityEquation.html

Well that's the last bump.

 

[nrqed]

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?

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.

 

[Thrice]

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