- #1
davidge
- 554
- 21
Hi. I was trying to translate the divergence theorem and the Green's theorem to tensor notation that we use in Relativity. For the divergence theorem, it was easy (please tell me if I'm wrong in the below derivation). I'm using the standard electromagnetic tensor ##F_{\mu \nu}## in place of the individual ##E## and ##B## vector fields. And I haven't written the various integral signs in the multiple integrals below. Since ##\vec{\nabla} \cdot \vec{E} \to \partial_{\mu}E^{\mu}## for a vector field ##E##,
$$ \int F_{\mu \nu}dx^{\mu}dx^{\nu} = \int \partial_{\rho}F_{\mu \nu} dx^{\mu}dx^{\nu}dx^{\rho} = \int \partial F_{\mu \nu} dx^{\mu}dx^{\nu}$$
The solution is evident.
Now, I'm having trouble in deriving the Green's theorem. I'm not sure if the correct translation for the "curl" of ##F_{\mu \nu}## is $$\epsilon^{\mu \kappa}{}_{\sigma} \epsilon^{\nu \lambda}{}_{\rho} \partial_{\lambda} F_{\mu \nu} = J_{\sigma \rho}$$ (##\epsilon## = 1, for indices 123, 1 and -1 for even and odd permutation of them and 0 otherwise.) It will become evident below why I inserted that "don't matching" ##\kappa## - index in the previous expression.
An attempt to the Green's equality then would be
$$ \int \epsilon^{\nu \lambda}{}_{\rho} dx^{\rho} \int \epsilon^{\mu \kappa}{}_{\sigma} \partial_{\lambda} F_{\mu \nu} dx^{\sigma} = \int F_{\rho \sigma}dx^{\rho}dx^{\sigma}dx^{\kappa}$$
Now, for this to be true, we must have
$$\epsilon^{\nu \lambda}{}_{\rho} \epsilon^{\mu \kappa}{}_{\sigma} \partial_{\lambda} F_{\mu \nu} = \int F_{\rho \sigma} dx^{\kappa}$$
Is this correct?
$$ \int F_{\mu \nu}dx^{\mu}dx^{\nu} = \int \partial_{\rho}F_{\mu \nu} dx^{\mu}dx^{\nu}dx^{\rho} = \int \partial F_{\mu \nu} dx^{\mu}dx^{\nu}$$
The solution is evident.
Now, I'm having trouble in deriving the Green's theorem. I'm not sure if the correct translation for the "curl" of ##F_{\mu \nu}## is $$\epsilon^{\mu \kappa}{}_{\sigma} \epsilon^{\nu \lambda}{}_{\rho} \partial_{\lambda} F_{\mu \nu} = J_{\sigma \rho}$$ (##\epsilon## = 1, for indices 123, 1 and -1 for even and odd permutation of them and 0 otherwise.) It will become evident below why I inserted that "don't matching" ##\kappa## - index in the previous expression.
An attempt to the Green's equality then would be
$$ \int \epsilon^{\nu \lambda}{}_{\rho} dx^{\rho} \int \epsilon^{\mu \kappa}{}_{\sigma} \partial_{\lambda} F_{\mu \nu} dx^{\sigma} = \int F_{\rho \sigma}dx^{\rho}dx^{\sigma}dx^{\kappa}$$
Now, for this to be true, we must have
$$\epsilon^{\nu \lambda}{}_{\rho} \epsilon^{\mu \kappa}{}_{\sigma} \partial_{\lambda} F_{\mu \nu} = \int F_{\rho \sigma} dx^{\kappa}$$
Is this correct?
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