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Proof of Ponytings theorem

  1. Jan 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove and explain the Ponyting theorem

    2. Relevant equations

    S = E x H [1]
    [itex]\nabla[/itex].S = [itex]\nabla[/itex].(E x H) [2]

    [itex]\nabla[/itex] x E = -[itex]\partial[/itex]B/[itex]\partial[/itex]t [3]

    [itex]\nabla[/itex] x H = [itex]\partial[/itex] D/[itex]\partial[/itex]t [4]

    D = [itex]\epsilon[/itex]E + P [5]

    B = [itex]\mu[/itex]H + [itex]\mu[/itex]M [6]

    3. The attempt at a solution

    I understand I am to use the vector identity to obtain [2]

    [itex]\nabla[/itex] . ( E x H ) = ([itex]\nabla[/itex] E).H - ([itex]\nabla[/itex] x H ). E

    I then substitute [3] and [4] into [2]

    I then use the definition from [5] and [6] and sub them into my equation

    I have:

    [itex]\nabla[/itex] . S = -[itex]\partial[/itex][[itex]\mu[/itex]H + [itex]\mu[/itex]M].H/[itex]\partial[/itex]t - [itex]\partial[/itex][[itex]\epsilon[/itex]E + P].E /[itex]\partial[/itex]t

    i can then expand out the bracket but I'm not sure what to do next

    The result i'm aiming for is

    [itex]\nabla[/itex].S = -[itex]\partial[/itex]/dt ( 1/2 [itex]\epsilon[/itex]E^2 + 1/2[itex]\mu[/itex] H^2) + E.[itex]\partial[/itex]P/[itex]\partial[/itex]t + [itex]\mu[/itex]H.[itex]\partial[/itex]M/[itex]\partial[/itex]t


    browsing google i have found this. 8.87 and 8.88 will give me the answer I want but I'm unfamiliar with them. could anyone shed some light on this
     
    Last edited: Jan 15, 2013
  2. jcsd
  3. Jan 15, 2013 #2

    G01

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    Your almost there. Here' a hint to shed light on those equations:

    Use the chain rule to expand

    [tex]\frac{1}{2}\frac{\partial H^2}{\partial t} = ?[/tex]
     
  4. Jan 15, 2013 #3
    [itex]\frac{2}{2} H . \frac{\partial H} {\partial t}[/itex]

    not sure why you need to chain rule though?
     
  5. Jan 15, 2013 #4

    G01

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    The chain rule is where the justification for EQ 8.87 and 8.88 comes from. You should be able to use the chain rule to get the E^2 and H^2 terms in Poynting's theorem from terms like E*dE/dt.
     
  6. Jan 15, 2013 #5
    apologies. Was being dumb for a moment there. I believe I have it now. thanks
     
  7. Jan 16, 2013 #6

    G01

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    No problem! :)
     
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