Compact expression of Maxwell's equations: is there a missing minus sign?

Question

The compact form of Maxwell's equations:

$$\boxed{\square\, \boldsymbol{\mathsf{F}}=\mu_0 \boldsymbol{\mathcal{J}}} \tag{1}$$ where the current density quadrivector is given by the relation $\boldsymbol{\mathcal{J}}=(\bar J, ic\rho)$. The tensor of the electromagnetic field is given by $$F_{\mu\nu}:=\frac{\partial \mathcal{A}_{\nu}}{\partial X_{\mu}}-\frac{\partial \mathcal{A}_{\mu}}{\partial X_\nu} \tag{2}$$ calculated using the four-potential $\boldsymbol{\mathcal{A}}=\left(\bar{A}, \dfrac ic \varphi\right)$.

It is known $F_{\mu\nu}=-F_{\nu\mu}$ and with several steps (which I am not reporting) I have proved that: \begin{equation} \sum^4_{\nu=1}\frac{\partial F_{\mu\nu}}{\partial X_\nu}=\mu_0\mathcal{J}_\mu,\quad \mu=1,2,3,4. \tag{3} \end{equation} Hence $(3) \iff (1)$. Obviously if I exchange the subscripts of the tensor of the electromagnetic field I easily get the minus sign. Infact being $$F_{\mu\nu}=-F_{\nu\mu} \tag{4}:$$ \begin{equation} \sum^4_{\mu=1}\frac{\partial F_{\nu\mu}}{\partial X_\mu}=-\mu_0\mathcal{J}_\nu,\quad \nu=1,2,3,4. \tag{5} \end{equation} or $$\square \,\boldsymbol{\mathsf{F}} =-\mu_0 \boldsymbol{\mathcal{J}} \tag{6}$$

The (6) and the (1) are the same things or with the minus sign it has another meaning?

Answer 1

Starting from your equation (3) we have \begin{equation} \mu_0\mathcal{J}_\mu = \sum^4_{\mu=1}\frac{\partial F_{\mu\nu}}{\partial X_\nu} = -\sum^4_{\mu=1}\frac{\partial F_{\nu\mu}}{\partial X_\nu} \end{equation} therefore your equation (5) should read \begin{equation} \sum^4_{\mu=1}\frac{\partial F_{\nu\mu}}{\partial X_\nu} = - \mu_0\mathcal{J}_\mu \end{equation} which can also be written (after renaming indices) \begin{equation} \sum^4_{\mu=1}\frac{\partial F_{\mu\nu}}{\partial X_\mu} = - \mu_0\mathcal{J}_\nu \end{equation} therefore your equation (5) has a sign error.

But perhaps what you meant to say is that you do not obtain (5) from (3) but rather you just assert that (5) is what you would get if you started out from a different equation in the first place.

I think the problem here may be that you are not aware of an assumption built into the index-free notation you have adopted in (1), concerning on which index of $F$ the differential operator is being employed. There must be a convention, otherwise the equation is ambiguous because $F$ is not symmetric.

The general point is that you cannot expect $\partial_\mu F^{\mu b}$ to be equal to $\partial_\mu F^{b \mu}$ when $F^{ab} \ne F^{ba}$, so any equation involving such quantities has to pay attention to whether the derivative is taken on the first or the second index.

Answer 2

In order to derive the inhomogeneous Maxwell equations, you do not need to use the antisymmetry of the field strength tensor. They follow from the equations of motion for $A_\mu$. If you start with $$ \mathscr{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e J^\mu A_\mu\;,$$ the Euler-Lagrange equations give you $$ \partial_\mu F^{\mu\nu}=e J^\nu$$ up to a total derivative term that can be gauged or argued away. If you relabel $\mu\leftrightarrow\nu$, you get $$ \partial_\nu F^{\nu\mu}=e J^\mu\;.$$ It might very well be that our conventions differ because I do not work with these $\mu_0$ oddities. That is, all the above assumes natural units, in which $c=1$. In any case, you do not get a sign when renaming the summation indices.