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3

Let us first replace the Minkowski metric tensor $$\eta~=~\eta_{\mu\nu}~\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}$$ with a more general constant metric tensor $$g~=~g_{\mu\nu}~\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}.$$ Note that the raised EM tensor $$F^{\mu\nu}~:=~g^{\mu\lambda} F_{\lambda\kappa}g^{\kappa\nu}$$ depends on the (inverse) metric. The Maxwell ...

4

There is an additional condition coming from the third term on the left hand side of your transformed equation, where you have used what seemed to be the chain rule $$\frac{\partial x^\lambda}{\partial x^{\mu'}}\frac{\partial x^{\mu'}}{\partial x^\mu}=\delta^\lambda_\mu$$ In reality however, the nontrivial positioning of the covariant and contra variant ...

0

Once $^*$ is the Hodge star operator, its definition in non-flat spacetime $(\star \eta)_{i_1,i_2,\ldots,i_{n-k}} = \frac{1}{(k)!} \eta^{j_1,\ldots,j_k}\,\sqrt {|\det g|} \,\epsilon_{j_1,\ldots,j_k,i_1,\ldots,i_{n-k}}$ say that we only need the determinant of the metric tensor $\bf{g}$ to change from $\bf{F}$ to $\bf{^*F}$. Thanks for @ACuriousMind and ...

1

As ACuriousMind said in a comment above, such an equation is manifestly covariant. Suppose the homogeneous Maxwell equations hold in the Lorentz transformed coordinates, i.e. that $$\bar\partial_\alpha \bar F_{\beta\gamma} + \bar\partial_\gamma \bar F_{\alpha\beta} + \bar \partial_\beta \bar F_{\gamma\alpha} = 0$$ Lorentz transformation are global and thus ...

0

I assume what is meant by Faraday's law of induction is what Griffiths refers to as the "universal flux rule", the statement of which can be found in this question. This covers both cases 1) and 2), even though in 1) it is justified by the third Maxwell equation1 and in 2) by the Lorentz force law. The universal flux rule is a consequence of the third ...

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