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Gravitation by Charles W. Misner, Kip Throne and John Wheeler page 120 Box 4.4 Duality Plus Exterior Differentiation, related exercise 3.13, 3.17

I want to calculate the component of $*d*F$ in index notation, the answer should be $F_{\mu\nu}^{,\nu}=A_{\nu,\mu}^{,\nu}-A_{\mu,\nu}^{,\nu} $ where $F$ was the usual electromagnetic tensor(Farady) $F$.

Here's what I've done and the issue I encountered.

Since $F_{\mu\nu}=A_{\nu,\mu}-A_{\mu,\nu}$ , $F^{\mu\nu}=A^{\nu,\mu}-A^{\mu,\nu}$.

The component of $*F$ was $*F_{\alpha \beta} =\frac{1}{2}F^{\mu\nu} \epsilon_{\mu\nu\alpha\beta} =\frac{1}{2}(A^{\nu,\mu}-A^{\mu,\nu})\epsilon_{\mu\nu\alpha\beta} $. [The dual permuted the index of elements, thus I suppose I should include $\delta_{\alpha\beta}^{\mu\nu}$(permutation tensor) somewhere. Thus, it equals to $(A^{\nu,\mu}-A^{\mu,\nu})\delta^{\alpha\beta}_{\mu\nu}$? ]

Thus the component of $d*F$ was $(d*F)_{\alpha\beta,\gamma}=(\frac{1}{2}F^{\mu\nu} \epsilon_{\mu\nu\alpha\beta})_{,\gamma}$. [Does this equal to $(\frac{1}{2}F^{\mu\nu})_{,\gamma} \epsilon_{\mu\nu\alpha\beta}$?]

The component of $*d*F=?$ [I thought about contract with $e^{\alpha\beta\gamma\delta}$, but the index at $\gamma$ didn't match.]

There's a relation I found useful the permutation tensor $\delta^{\alpha\beta}_{\mu\nu}=-\frac{1}{2} \epsilon^{\alpha\beta\lambda\rho}\epsilon_{\mu\nu\lambda\rho}$[eq.3.50 i], but I have a hard time to write and prove it formally in index notation.

Could you help me to walk through the calculation and show me how it was done?

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For this to make sense, we have to talk about $F$ as a 2-form, i.e.

$\mathbf{F}=\frac{1}{2}F_{\mu\nu} \: dx^\mu \wedge dx^\nu$

Then $\star\mathbf{F}$ is its dual. In 4d space:

$\star \left(dx^\mu\wedge dx^\nu\right) = \frac{1}{2}g^{\mu\kappa}g^{\nu\sigma}\epsilon_{\kappa\sigma\rho\phi} dx^\rho\wedge dx^\phi$

Thus:

$\star \mathbf{F} = \frac{1}{4}F_{\mu\nu} g^{\mu\kappa}g^{\nu\sigma}\epsilon_{\kappa\sigma\rho\phi} \: dx^\rho\wedge dx^\phi$

Applying the exterior derivative, which we can convert into partial derivative, or may keep as a covariant derivative (torsion-free; so that derivative of metric is zero, and so is the drivative of metric inverse):

$\mathbf{d}\star \mathbf{F} = \frac{1}{4} \nabla_\zeta g^{\mu\kappa}g^{\nu\sigma} \left(F_{\mu\nu} \epsilon_{\kappa\sigma\rho\phi}\right) \: dx^\zeta \wedge dx^\rho\wedge dx^\phi = \frac{1}{4} g^{\mu\kappa}g^{\nu\sigma} \nabla_\zeta \left(F_{\mu\nu} \delta^{0123}_{\kappa\sigma\rho\phi}\right) \: dx^\zeta \wedge dx^\rho\wedge dx^\phi$

Where we have used generalized Kroenecker deltas https://en.wikipedia.org/wiki/Kronecker_delta#Definitions_of_the_generalized_Kronecker_delta. They are more convenient here since delta-s are tensors, whilst Levi-Civitas are relative tensors (matters for differentiation).

Then:

$\mathbf{d}\star \mathbf{F} = \frac{1}{4} g^{\mu\kappa}g^{\nu\sigma} \left( \delta^{0123}_{\kappa\sigma\rho\phi} \nabla_\zeta F_{\mu\nu} + F_{\mu\nu}\Gamma^\alpha_{\alpha\zeta}\delta^{0123}_{\kappa\sigma\rho\phi}-F_{\mu\nu}\Gamma^\beta_{\kappa\zeta}\delta^{0123}_{\beta\sigma\rho\phi} - F_{\mu\nu}\Gamma^\beta_{\sigma\zeta}\delta^{0123}_{\kappa\beta\rho\phi}\right) \: dx^\zeta \wedge dx^\rho\wedge dx^\phi$

Thus:

$\mathbf{d}\star \mathbf{F} = \frac{1}{4} g^{\mu\kappa}g^{\nu\sigma} \left( \epsilon_{\kappa\sigma\rho\phi} \nabla_\zeta F_{\mu\nu} + F_{\mu\nu}\cdot \frac{\partial_\zeta g}{2 g}\cdot\epsilon_{\kappa\sigma\rho\phi}-F_{\mu\nu}\Gamma^\beta_{\kappa\zeta}\epsilon_{\beta\sigma\rho\phi} - F_{\mu\nu}\Gamma^\beta_{\sigma\zeta}\epsilon_{\kappa\beta\rho\phi}\right) \: dx^\zeta \wedge dx^\rho\wedge dx^\phi$

Where $g=det\left(g_{\alpha\beta}\right)$ is the determinant of the metric. Assuming the connection ($\Gamma$) vanishes and metric determinant is constant, which would be true in flat space+Cartesian cooridnates:

$\mathbf{d}\star \mathbf{F} = \frac{1}{4} g^{\mu\kappa}g^{\nu\sigma} \epsilon_{\kappa\sigma\rho\phi} \,\partial_\zeta F_{\mu\nu} \: dx^\zeta \wedge dx^\rho\wedge dx^\phi$

Then:

$\star \left(dx^\zeta \wedge dx^\rho \wedge dx^\phi \right) = g^{\zeta\alpha}g^{\rho\beta}g^{\phi\gamma}\epsilon_{\alpha\beta\gamma\omega}dx^\omega$

So:

$\star \mathbf{d}\star\mathbf{F}= \frac{1}{4} g^{\mu\kappa}g^{\nu\sigma}\epsilon_{\kappa\sigma\rho\phi} \: \partial_\zeta F_{\mu\nu} \: g^{\zeta\alpha}g^{\rho\beta}g^{\phi\gamma}\epsilon_{\alpha\beta\gamma\omega}dx^\omega$

Next we want to convert one of the Levi-Civita into 'upstairs' form, here is a catch since, again, Levi-Civita is not a tensor. One can show that

$g^{\mu\kappa}g^{\nu\sigma}g^{\rho\beta}g^{\phi\gamma}\epsilon_{\kappa\sigma\rho\phi} = \epsilon^{\mu\nu\beta\gamma}/g$

It is easy to test this:

$\epsilon^{0123}=1=g \cdot g^{0\kappa}g^{1\sigma}g^{2\rho}g^{3\phi}\epsilon_{\kappa\sigma\rho\phi}=g\cdot det\left(g^{\alpha\beta}\right)=g\cdot \frac{1}{g}=1$

So:

$\star \mathbf{d}\star\mathbf{F}= \frac{1}{4} \partial_\zeta \left(F_{\mu\nu} \right) g^{\mu\kappa}g^{\nu\sigma}g^{\zeta\alpha}g^{\rho\beta}g^{\phi\gamma}\epsilon_{\kappa\sigma\rho\phi}\epsilon_{\alpha\beta\gamma\omega}dx^\omega = \frac{1}{4g} \partial_\zeta \left(F_{\mu\nu} \right) g^{\zeta\alpha}\epsilon^{\mu\nu\beta\gamma}\epsilon_{\alpha\beta\gamma\omega}dx^\omega = \frac{1}{4g} \partial_\zeta \left(F_{\mu\nu} \right) g^{\zeta\alpha}\delta^{\mu\nu\beta\gamma}_{\alpha\beta\gamma\omega}dx^\omega=\frac{1}{4g} \partial_\zeta \left(F_{\mu\nu} \right) g^{\zeta\alpha}\delta^{\mu\nu\beta\gamma}_{\alpha\omega\beta\gamma}dx^\omega=\frac{1}{4g} \partial_\zeta \left(F_{\mu\nu} \right) g^{\zeta\alpha}\,2\delta^{\mu\nu}_{\alpha\omega}\:dx^\omega$

Finally:

$\star \mathbf{d}\star\mathbf{F} = \frac{1}{2g} \partial^\alpha \left(F_{\mu\nu} \right) \delta^{\mu\nu}_{\alpha\omega}\:dx^\omega = \frac{1}{2g} \partial^\mu \left(F_{\mu\nu} \right) \: dx^\nu - \frac{1}{2g} \partial^\nu \left(F_{\mu\nu} \right) \: dx^\mu = \frac{1}{g}F^{,\mu}_{\mu\nu} \: dx^\nu$

The last step relies on the anti-symmetry of $F$. Compared to your answer there is a difference of -1 that comes from the fact that in flat spacetime $g=det\left(diag\left(1,-1,-1,-1\right)\right)=-1$

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  • $\begingroup$ Thank you. It helped a lot. Will the factor $(-1)$ come in when $e^{\mu\nu\beta\gamma}e_{\alpha\beta\gamma\omega}$ contract? $\endgroup$ – ShoutOutAndCalculate Oct 22 '19 at 6:38
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    $\begingroup$ Not really, it comes from the fact that Levi-Civita is not a tensor, but a relative tensor. Anyways, I fixed it. $\endgroup$ – Cryo Oct 23 '19 at 18:18

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