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First the problem in detail should be,

"Starting with 1-form potential $\mathbf{A}$, derive the relation $F_{\mu\nu}^{\;\;\;,\nu}=A_{\nu,\mu}^{\;\;\;\;,\nu}-A_{\mu,\nu}^{\;\;\;\;,\nu}=4\pi J_\mu$, using relations $\mathbf{F}=\mathbf{dA}$ and $\mathbf{d\ast F}=4\pi\ast\mathbf{J}$."

I was reading chapter 4 of Gravitation by Charles W. Misner. There I stumble upon this not as an exercise but as a statement that said $\mathbf{\ast d\ast F}=\mathbf{\ast d\ast dA}=4\pi\mathbf{J}$ reduce to the expression in the question in index notation. So I wanted to check for practice. But first let me list all relevent formulas that I know, if any one of them is wrong please let me know.

So if $\mathbf{B}$ is a $p$-form then,

$$\mathbf{B}=\frac{1}{p!}B_{\alpha_1\alpha_1\cdots\alpha_p}\mathbf{d}x^{\alpha_1}\wedge\mathbf{d}x^{\alpha_2}\wedge\cdots\wedge\mathbf{d}x^{\alpha_p}$$

and the exterior derivative of $\mathbf{B}$ is,

$$\mathbf{dB}=\frac{1}{p!}\frac{\partial B_{\alpha_1\alpha_1\cdots\alpha_p}}{\partial x^\beta}\mathbf{d}x^\beta\wedge\mathbf{d}x^{\alpha_1}\wedge\mathbf{d}x^{\alpha_2}\wedge\cdots\wedge\mathbf{d}x^{\alpha_p}.$$

Which is a (p+1)-form.

Again for $\mathbf{B}=B_{\alpha_1\alpha_1\cdots\alpha_p}\mathbf{d}x^{\alpha_1}\otimes\mathbf{d}x^{\alpha_2}\otimes\cdots\otimes\mathbf{d}x^{\alpha_p}$, the Hodge Dual is (assuming $p\leq n$)

$$\ast\mathbf{B}=\frac{1}{p!}B^{\alpha_1\alpha_1\cdots\alpha_p}\epsilon_{\alpha_1\alpha_1\cdots\alpha_p\beta_{p+1}\beta_{p+2}\cdots\beta_n}\mathbf{d}x^{\beta_{p+1}}\otimes\mathbf{d}x^{\beta_{p+2}}\otimes\cdots\otimes\mathbf{d}x^{\beta_n}.$$

Also $$\delta^{\alpha_{p+1}\cdots\alpha_n}_{\;\;\;\;\;\;\;\;\;\;\beta_{p+1}\cdots\beta_n}=-\frac{1}{p!}\epsilon^{\gamma_1\cdots\gamma_p\alpha_{p+1}\cdots\alpha_n}\epsilon_{\gamma_1\cdots\gamma_p\beta_{p+1}\cdots\beta_n}$$

And $$p!B_{\beta_1\cdots\beta_p}=B_{\alpha_1\cdots\alpha_p}\delta^{\alpha_1\cdots\alpha_p}_{\;\;\;\;\;\;\;\;\beta_1\cdots\beta_p}$$

So now I begin working on the problem. Given 1-form $\mathbf{A}$ I assumed that $$\mathbf{A}=A_\mu\mathbf{d}x^\mu$$ Taking exterior derivative we get, $$\mathbf{dA}=\frac{\partial A_\mu}{\partial x^\nu}\mathbf{d}x^\nu\wedge\mathbf{d}x^\mu$$ Expressing in terms of tensor products $$\mathbf{F}=\mathbf{dA}=2\frac{\partial A_\mu}{\partial x^\nu}\mathbf{d}x^\nu\otimes\mathbf{d}x^\mu$$

Now taking the Hodge Dual,

$$\ast F=\ast\mathbf{dA}=\frac{1}{2}(\mathbf{dA})_{\nu\mu}g^{\nu\alpha}g^{\mu\beta}\epsilon_{\alpha\beta\gamma\lambda}\mathbf{d}x^\gamma\otimes\mathbf{d}x^\lambda$$

Substituting $\mathbf{dA}$, $$\ast F=\ast\mathbf{dA}=\frac{1}{2}2\frac{\partial A_\mu}{\partial x^\nu}g^{\nu\alpha}g^{\mu\beta}\epsilon_{\alpha\beta\gamma\lambda}\mathbf{d}x^\gamma\otimes\mathbf{d}x^\lambda$$

$$\Rightarrow\ast F=\ast\mathbf{dA}=\frac{\partial A_\mu}{\partial x^\nu}\epsilon^{\nu\mu}_{\;\;\;\gamma\lambda}\mathbf{d}x^\gamma\otimes\mathbf{d}x^\lambda$$

Reexpressing in terms of wedge products,

$$\ast F=\ast\mathbf{dA}=\frac{1}{2}\frac{\partial A_\mu}{\partial x^\nu}\epsilon^{\nu\mu}_{\;\;\;\gamma\lambda}\mathbf{d}x^\gamma\wedge\mathbf{d}x^\lambda$$

Taking exterior derivative again, $$\mathbf{d\ast F}=\mathbf{d\ast dA}=\frac{\partial}{\partial x^\beta}(\frac{1}{2}\frac{\partial A_\mu}{\partial x^\nu}\epsilon^{\nu\mu}_{\;\;\;\gamma\lambda})\mathbf{d}x^\beta\wedge\mathbf{d}x^\gamma\wedge\mathbf{d}x^\lambda$$

$$\Rightarrow\mathbf{d\ast F}=\mathbf{d\ast dA}=\frac{1}{2}\frac{\partial^2 A_\mu}{\partial x^\beta\partial x^\nu}\epsilon^{\nu\mu}_{\;\;\;\gamma\lambda}\mathbf{d}x^\beta\wedge\mathbf{d}x^\gamma\wedge\mathbf{d}x^\lambda$$

Expressing in tensor products, $$\mathbf{d\ast F}=\mathbf{d\ast dA}=3!\frac{1}{2}\frac{\partial^2 A_\mu}{\partial x^\beta\partial x^\nu}\epsilon^{\nu\mu}_{\;\;\;\gamma\lambda}\mathbf{d}x^\beta\otimes\mathbf{d}x^\gamma\otimes\mathbf{d}x^\lambda$$

Taking Hodge Dual again, $$\mathbf{\ast d\ast F}=\mathbf{\ast d\ast dA}=\frac{1}{3!}(\mathbf{d\ast dA})_{\beta\gamma\lambda}g^{\beta\eta}g^{\gamma\rho}g^{\lambda\alpha}\epsilon_{\eta\rho\alpha\xi}\mathbf{d}x^\xi$$ Substituting,

$$\mathbf{\ast d\ast F}=\mathbf{\ast d\ast dA}=\frac{1}{3!}3\frac{\partial^2 A_\mu}{\partial x^\beta\partial x^\nu}\epsilon^{\nu\mu}_{\;\;\;\gamma\lambda}g^{\beta\eta}g^{\gamma\rho}g^{\lambda\alpha}\epsilon_{\eta\rho\alpha\xi}\mathbf{d}x^\xi$$

$$\Rightarrow\mathbf{\ast d\ast F}=\mathbf{\ast d\ast dA}=\frac{1}{2}\frac{\partial^2 A_\mu}{\partial x^\beta\partial x^\nu}\epsilon^{\nu\mu\rho\alpha}\epsilon_{\eta\rho\alpha\xi}g^{\beta\eta}\mathbf{d}x^\xi$$

$$\Rightarrow\mathbf{\ast d\ast F}=\mathbf{\ast d\ast dA}=\frac{1}{2}\frac{\partial^2 A_\mu}{\partial x^\beta\partial x^\nu}\epsilon^{\nu\mu\rho\alpha}\epsilon_{\eta\xi\rho\alpha}g^{\beta\eta}\mathbf{d}x^\xi$$

$$\Rightarrow\mathbf{\ast d\ast F}=\mathbf{\ast d\ast dA}=\frac{1}{2}\frac{\partial^2 A_\mu}{\partial x^\beta\partial x^\nu}(-2\delta^{\nu\mu}_{\;\;\;\eta\xi})g^{\beta\eta}\mathbf{d}x^\xi$$

$$\Rightarrow\mathbf{\ast d\ast F}=\mathbf{\ast d\ast dA}=-\frac{\partial^2 A_\mu}{\partial x^\beta\partial x^\nu}\delta^{\nu\mu}_{\;\;\;\eta\xi}g^{\beta\eta}\mathbf{d}x^\xi$$

$$\Rightarrow\mathbf{\ast d\ast F}=\mathbf{\ast d\ast dA}=-2\frac{\partial^2 A_\eta}{\partial x^\beta\partial x^\xi}g^{\beta\eta}\mathbf{d}x^\xi$$

From electromagnetism we know that $\mathbf{d\ast F}=4\pi\ast\mathbf{J}$

So $\mathbf{\ast d\ast F}=4\pi\ast\ast\mathbf{J}=4\pi \mathbf{J}$, since $\mathbf{J}$ is a 1-form. Therefore $$-2\frac{\partial^2 A_\eta}{\partial x^\beta\partial x^\xi}g^{\beta\eta}\mathbf{d}x^\xi=4\pi\mathbf{J}$$

But as I can see, this doesn't give the desired result in the problem. Taking inner product with $\frac{\partial}{\partial x^\mu}$ on both side,

$$-2\frac{\partial^2 A_\eta}{\partial x^\beta\partial x^\xi}g^{\beta\eta}\delta^\xi_{\;\mu}=4\pi J_\mu$$

$$\Rightarrow-2\frac{\partial^2 A_\eta}{\partial x^\beta\partial x^\mu}g^{\beta\eta}=4\pi J_\mu$$

$$\Rightarrow-2A_{\eta,\beta,\mu}g^{\beta\eta}=4\pi J_\mu$$

$$\Rightarrow-2A_{\eta,\mu}^{\;\;\;,\eta}=4\pi J_\mu$$

This is the result that I got. I don't know what I did wrong or know wrong. That's why I need help.

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    $\begingroup$ Why don't you use the fact that the divergence is defined as $\star \circ d \circ \star$? $\endgroup$ Aug 17, 2021 at 14:07
  • $\begingroup$ I am sorry, I didn't know that definition. But I just wanted to see if I can correctly use the formula for exterior derivative and hodge dual because there seems to be something that I am missing and I cannot figure out what. $\endgroup$ Aug 17, 2021 at 14:18
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    $\begingroup$ Don't be sorry for that, one cannot know everything at any instant! I don't have the time to see where exactly is the error, but I've done this same exercise in the past, and what helped me was to just seek for $d \star F\propto J^{(3)}$ (where $J^{(3)}$ is the source 3-form) without using $A$. You can find a bit of this in the Wikipedia article Divergence. $\endgroup$ Aug 17, 2021 at 14:26

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