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Deriving the 4d continuity equation

  1. May 9, 2006 #1
    Well we start out with

    [tex]-\frac {d} {dt} \int_{V}^{} \sigma dV = \int_{\Pi}^{} \vec{J} \cdot d\vec{\Pi}[/tex]

    Using the Gauss theorem

    [tex]\int_{V}^{} (\frac{ \partial {\sigma}}{ \partial {t}} + div \vec{J}) dV = 0[/tex]


    [tex]\frac{ \partial {\sigma}}{ \partial {t}} + div \vec{J} = 0[/tex]

    and written in 4D..

    [tex]\frac{ \partial {J^n}}{ \partial {x^n}} = 0 \qquad\quad\ (n = 0,1,2,3)[/tex]

    I can't seem to get my head around that last step. How does it expand out?
    Last edited: May 9, 2006
  2. jcsd
  3. May 9, 2006 #2
    Ok done editing.
  4. May 9, 2006 #3
    This field is really confusing. Should I just gauss the answer?
  5. May 10, 2006 #4
  6. May 10, 2006 #5


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    Are you familiar with four-vector notation and so on? Then it is simply a question of definition that the last expression is [itex] \partial_\mu J_\mu =0[/itex] where Einstein's summation convention is used.
  7. May 11, 2006 #6
    Yeah well I'm familiar with it, but I needed the definition spelled out. I eventually worked it out.
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