Maxwell's equations from differential forms

Question

I found the following in some lecture notes I took some time ago:

$$ \mathbf{E}=-\text{grad}\Phi-\partial_t\mathbf{A}\\ \mathbf{B}=\mathrm{rot}\mathbf{A} $$

These are the electromagnetic fields expressed in the potential form $A^{\mu}=(\Phi,\mathbf{A})$.

Now I'd like to derive Maxwell's equations using differential forms. The potential is a $1$-form, the electromagnetic field tensor $F_{\mu\nu}:=(dA)_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ is a $2$-form. Because $d^2=0$, $df=0$, so the homogeneous Maxwell equations are automatically satisfied. The inhomogeneous ones are

$$ \partial_\nu F^{\nu\mu}=j^\mu $$

Distinguishing between $\mu=0$ and $\mu=i$, this translates to

$$ \begin{align} \rho = j^0 &= \partial_\nu(\partial^\nu A^0-\partial^0 A^\nu)\\ &= \partial_j\partial^j A^0- \partial_0\partial^j A^j\\ & \\ j^i &= \partial_\nu(\partial^\nu A^i-\partial^i A^\nu)\\ &= \partial_0\partial^0 A^i- \partial_0\partial^i A^0 + \partial_j\partial^j A^i- \partial_j\partial^i A^j\\ &= \partial_0(\partial^0 A^i-\partial^i A^0) + \partial_j\partial^j A^i- \partial^i\partial_j A^j \\ \end{align} $$

so we have

$$ \begin{align} \rho &=\nabla(\color{red}{+}\nabla\Phi-\partial_t\mathbf{A}) \\ \mathbf{j} &=\partial_t(\partial^t\mathbf{A}-\nabla\Phi) + (\nabla^2)\mathbf{A}-\nabla(\nabla\cdot\mathbf{A})\\ &=\partial_t(\color{red}{-}\nabla\Phi+\partial^t\mathbf{A}) \color{red}{-}\nabla\times(\nabla\times\mathbf{A}) \end{align} $$

But the inhomogeneous Maxwell equations are

$$ \begin{align} \rho =\nabla\mathbf{E} &= \nabla(\color{red}{-}\nabla\Phi-\partial_t\mathbf{A}) \\ \mathbf{j} =-\partial_t\mathbf{E}+\mathrm{rot}\mathbf{B} & =\partial_t(\color{red}{+}\nabla\Phi+\partial^t\mathbf{A}) \color{red}{+} \nabla\times(\nabla\times\mathbf{A}) \end{align} $$

So I got some signs wrong, but I cannot see why. I thought that maybe I got some wrong signs due to mixing some upper and lower indices (because of the Minkowski metric), but since the wrong signs are mostly with the $\Phi$, I doubt that's the reason.

Any hints?

Answer 1

Hint: OP's sign convention in Maxwell's equations $$\partial_{\nu} F^{\nu\mu}~=~+ j^\mu$$ implicitly implies that the sign convention for the Minkowski metric is $(+,-,-,-)$, cf. e.g. this Phys.SE post. This implies that $\partial^i=-\partial_i$ for spatial indices, which seems to be absent from OP's posts (v3).

Answer 2

You're right, the reason is that you have to use the Minkowski metric to pull indices up and down. You are using a "mostly minus" convention, so $$ \partial^0 = \partial_0 = \frac{\partial}{\partial t} \;, \quad \partial^i = -\partial_i = -\frac{\partial}{\partial x^i} $$ This minus explains all your troubles: Whenever you translate $\partial^i$ into $\nabla$, you get an extra sign.