Maxwell equations: exterior derivative vs partial derivative

Question

In the language of differential forms, Maxwell's equations are

$$dF=0,\quad d\star F= J,\quad dJ=0.$$

If we write the dual field as $$\star F=\epsilon_{\alpha\beta\mu\nu}F^{\alpha\beta}\,dx^\mu\wedge dx^\nu,$$ by the Bianchi identity one can see that $$\partial_\mu\,(\star F)^{\mu\nu}=0,$$ such that $(\star F)^{\mu\nu}$ is a conserved quantity. Please correct me if the indices on this are wrong however

Some authors seem to equate this equation to $d\star F=0$, however this doesn't seem right to me as then the second Maxwell equation listed would always be equal to zero (I don't see where I have made assumptions of the absence of sources so far), and also if we explicitly write it out we have

$$d\star F=\frac{1}{2}\epsilon_{\alpha\beta\mu\nu}\frac{\partial}{\partial x^{\lambda}}F^{\alpha\beta}\,dx^\lambda\wedge dx^\mu\wedge dx^\nu,$$

which is not the same. Furthermore, the index of the derivative is not present in the levi civita and so the Bianchi identity cannot be used. So my question is if there is a relation between the exterior and partial derivative here, or if in any situation the two equations

$$\partial_\mu\,(\star F)^{\mu\nu}=0,\quad d\star F=0,$$ mean the same thing.

Another confusion I have here however is that the result for this computation on Wikipedia is the following

$$d\star F=\frac{1}{6}\epsilon_{\beta\gamma\delta\eta}\frac{\partial}{\partial x^{\alpha}}F^{\alpha\beta}\,dx^\gamma\wedge dx^\delta\wedge dx^\eta,$$

so I am unsure if I am doing the calculation correct.

Answer 1

In standard notation, $F=\frac{1}{2}F_{ab}dx^a \wedge dx^b$ and $\star F =\frac{1}{4}F^{ab}\epsilon_{abcd}dx^c\wedge dx^d$.

Then $d \star F = \frac{1}{4}\partial_a(F^{bc}\epsilon_{bcde})dx^a \wedge dx^d \wedge dx^e$. This is what we have called $J$. (Note that sometimes what we have defined to be $J$ is defined as $\star J$ instead).

$\partial_\mu(\star F)^{\mu\nu}$ is identically zero as you mention, which ultimately arises from $F=dA$.

$d \star F$ is $\bf{not}$ equivalent to $\partial_\mu(\star F)^{\mu\nu}$. But obviously in the absence of any source, they happen to take the same value of zero.

As for what you have found on wikipedia, I believe you have missed out a factor of $\sqrt{-g}$, which then gives the corresponding result for EM in curved spacetime ($g$ is the metric).