How can I use vector identities to prove and explain the Ponyting theorem?

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Homework Statement



Prove and explain the Ponyting theorem

Homework 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]

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]tbrowsing 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
 
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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]
 
[itex]\frac{2}{2} H . \frac{\partial H} {\partial t}[/itex]

not sure why you need to chain rule though?
 
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.
 
G01 said:
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.

apologies. Was being dumb for a moment there. I believe I have it now. thanks