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

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: $$$$\sum^4_{\nu=1}\frac{\partial F_{\mu\nu}}{\partial X_\nu}=\mu_0\mathcal{J}_\mu,\quad \mu=1,2,3,4. \tag{3}$$$$ Hence $$(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}:$$ $$$$\sum^4_{\mu=1}\frac{\partial F_{\nu\mu}}{\partial X_\mu}=-\mu_0\mathcal{J}_\nu,\quad \nu=1,2,3,4. \tag{5}$$$$ or $$\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?

• If (3) is correct, then (5) must be wrong (for $\mu_0\mathcal{J}_\nu\ne0$) since the equations differ only by an exchange of the indices, i.e. a relabeling of $\mu\leftrightarrow\nu$ and a sign.
– user178876
Jul 11 '19 at 21:10
• I have successfully erased all memory of $\mu_0$ but equation (3) can be derived very simply from the equations of motion for $A_\mu$. And by exchanging $\mu$ and $\nu$ one gets (5) up to the sign, whose origin I do not understand. if you use the antisymmetry of $F_{\mu\nu}$, then you get an equation with a minus, but where the derivative acts on the first index of $F$.
– user178876
Jul 11 '19 at 21:22
• Oh, please, do not use the $x_4 = ict$. It is so 1909, and we are in 2019... Jul 11 '19 at 22:03
• @Sebastiano you have done - implicitly - everywhere, in your definitions of four-vectors, derivatives etc. The use of $i$'s in the time-like components (and incidentally, having it as $x_4$ rather than $x_0$) is generally considered obsolete in favour of the modern geometrical/metric point of view.
– fqq
Jul 12 '19 at 15:45
• Also, $\square$ usually denotes the D'Alembert operator, which does not map 2-tensors to vectors as would be required for the equation to make sense. I can guess what you mean by $\square \overleftrightarrow F$, but AFAIK it's not standard notation, and since you are dealing with a sign problem I guess definitions are important. 9
– fqq
Jul 12 '19 at 15:46

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

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.

• You used the lagrangian and the CGS :-(. I have understood something :-) but the minus sign how do I make it appear? Jul 11 '19 at 21:54
• @Sebastiano I do not know. There are a lot of convention-dependent signs, such as the sign of e (which is negative in my conventions) or the signature of the metric (I am using $(+,-,-,-)$). All I can say is that you cannot get it by relabelling, Buona notte!
– user178876
Jul 11 '19 at 21:57
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