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Thread: A Physical System :particularly hard lol

  1. #1
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    A Physical System :particularly hard lol

    a really hard one here. would appreciate if you could start me off:


    a physical system is governed by the following:

    curl E = $\displaystyle -\frac{\partial B}{\partial t}$,
    div B = 0,
    curl B = J + $\displaystyle \frac{\partial E}{\partial t}$,
    div E = $\displaystyle \rho$
    where t = time, and time derivatives commute with $\displaystyle \nabla$

    ............................................
    how could you show that $\displaystyle \frac{\partial p}{\partial t}$ + div J = 0
    ............................................
    when $\displaystyle \rho = 0$ and J = 0 everywhere how can you show that:
    $\displaystyle \nabla^2E - \frac{\partial^2E}{\partial t^2}$ = 0
    and
    $\displaystyle \nabla^2B - \frac{\partial^2B}{\partial t^2}$ = 0
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  2. #2
    Super Member Rebesques's Avatar
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    Playing around with Maxwell's equations?

    Name the given relations (1)-(4). For the first question, we get from (4) that
    $\displaystyle
    \frac{\partial p}{\partial t}=\frac{\partial (div {\bf E})}{\partial t}=div\frac{\partial {\bf E}}{\partial t}
    $ and substitute J and curlB from (3) to get $\displaystyle
    \frac{\partial p}{\partial t}=div\frac{\partial {\bf E}}{\partial t}=div(curl {\bf B}-{\bf J})=-div{\bf J},
    $ as $\displaystyle divcurl=0$.

    For the second one, the identities $\displaystyle div{\bf B}=0, div{\bf E}=0$ mean the vector fields are solenoidal, so there exist scalar functions $\displaystyle B,E$ such that $\displaystyle {\bf B}=curlB, {\bf E}=curlE$. Use now (1) and (3), remembering that $\displaystyle curl(curl E)=\nabla(divE)-\nabla^2 E$.
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