# How to calculate and express $*d*F$ in index notation? [closed]

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?

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 $$g=det\left(diag\left(1,-1,-1,-1\right)\right)=-1$$

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