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@cabraham
I recently posted a short derivation of Poynting's theorem in https://www.physicsforums.com/threads/work-done-by-magnetic-field.825806/
It reminded me of our conversations about Poynting's theorem in a motor. As we discussed earlier, the work term ##E \cdot J## includes all the work done on matter, including but not limited to the Ohmic losses. So I wanted to see if any further analysis of that term could be done to pull out the Ohmic losses.
If a piece of material is moving at velocity ##v<<c## then we can make the following transformations:
##E' = E + v \times B##
##J' = J - \rho v##
where the primes indicate values in the reference frame where the material is at rest.
Substituting those in we have:
##E \cdot J = (E' - v \times B) \cdot (J' + \rho v)##
##=E' \cdot J' + E' \cdot \rho v - (v \times B) \cdot J' - (v \times B) \cdot \rho v##
##=E' \cdot J' + \rho v \cdot (E+ v \times B) - (J-\rho v) \cdot (v \times B)##
##=E' \cdot J' + \rho v \cdot E - J \cdot (v \times B)##
So finally we end with
##E \cdot J=E' \cdot J' + v \cdot (\rho E + J \times B)##
Since ##E'## and ##J'## are in the reference frame of the material, I would interpret the term ##E' \cdot J'## as being Ohmic losses. Then ##E \cdot J## contains not only the Ohmic losses, but also the material velocity times the Lorentz force density, which I would call the Lorentz power density.
We had discussed earlier how ##E \cdot J## depends on ##B##, so I think that this calculation makes that dependency more explicit, and the final result is somewhat obvious in retrospect.
I don't have a reference for this, so I don't know if someone else has already do this derivation, but I thought you might like to see it.
I recently posted a short derivation of Poynting's theorem in https://www.physicsforums.com/threads/work-done-by-magnetic-field.825806/
It reminded me of our conversations about Poynting's theorem in a motor. As we discussed earlier, the work term ##E \cdot J## includes all the work done on matter, including but not limited to the Ohmic losses. So I wanted to see if any further analysis of that term could be done to pull out the Ohmic losses.
If a piece of material is moving at velocity ##v<<c## then we can make the following transformations:
##E' = E + v \times B##
##J' = J - \rho v##
where the primes indicate values in the reference frame where the material is at rest.
Substituting those in we have:
##E \cdot J = (E' - v \times B) \cdot (J' + \rho v)##
##=E' \cdot J' + E' \cdot \rho v - (v \times B) \cdot J' - (v \times B) \cdot \rho v##
##=E' \cdot J' + \rho v \cdot (E+ v \times B) - (J-\rho v) \cdot (v \times B)##
##=E' \cdot J' + \rho v \cdot E - J \cdot (v \times B)##
So finally we end with
##E \cdot J=E' \cdot J' + v \cdot (\rho E + J \times B)##
Since ##E'## and ##J'## are in the reference frame of the material, I would interpret the term ##E' \cdot J'## as being Ohmic losses. Then ##E \cdot J## contains not only the Ohmic losses, but also the material velocity times the Lorentz force density, which I would call the Lorentz power density.
We had discussed earlier how ##E \cdot J## depends on ##B##, so I think that this calculation makes that dependency more explicit, and the final result is somewhat obvious in retrospect.
I don't have a reference for this, so I don't know if someone else has already do this derivation, but I thought you might like to see it.
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