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?


2 Answers 2


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).


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.


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