# Wave operator applied to electromagnetic field tensor

I'm trying to understand an argument in "An introduction to general relativity" by Hughston and Todd (p37). Let $F_{ab}$ be the electromagnetic field tensor, I'm trying to show:

$$\Box F_{ab} = -4 \pi (\nabla_a J_b - \nabla_b J_a)$$

Now in the text they say that this follows as a consequence of contracting $\nabla_{[a}F_{bc]}$ with $\nabla^a$ however, I'm struggling to see how we could deal with terms of the form $\nabla^a \nabla_b F_{bc}$, I'd like to be able to use the other Maxwell equation $\nabla^a F_{ab}$ however this isn't immediately applicable without switching the order of the derivatives.

I could plausibly do this by switching the order of the derivatives and inserting a Riemann tensor however at this point of the book the Riemann tensor has not been introduced so I wondered if I am overlooking something?

• $\Box = \nabla_a \nabla^a$ – Wooster Jun 1 '16 at 15:43
• Oh, I've just realised something quite stupid. At this point in the text they are actually working in flat space so the derivatives do commute! Maybe this question could still be of interest if I ask if this still holds in curved space as well? – Wooster Jun 1 '16 at 15:49

It seems OP understands how to derive the equation in flat spacetime, we will show how to generalize it to the curved case. We list, for reference, the Maxwell equations $$\nabla_bF^{ab}=4\pi J^a$$ and $$\nabla_aF_{bc}+\nabla_c F_{ab}+\nabla_bF_{ac}=0.$$ The wave operator is $\Box=\nabla^a\nabla_a$, so by the homogeneous Maxwell equation we have $$\tag{1}\Box F_{bc}=-\nabla^a\nabla_c F_{ab}-\nabla^a\nabla_bF_{ac}=-\nabla_a\nabla_cF^a{}_b+\nabla_a\nabla_b F^a{}_c.$$ We need the following standard identity, see e.g. Wald (1984) for a proof: $$[\nabla_a,\nabla_b]T^c{}_d=R^c{}_{eab}T^e{}_d-R^e{}_{dab}T^c{}_e.$$ Using this, we can rework the first term in (1): \begin{align} \nabla_a\nabla_cF^a{}_b&=R^a{}_{eac}F^e{}_b-R^e{}_{bac}F^a{}_{e}+\nabla_c\nabla_a F^a{}_b\\ &=R^e{}_cF_{eb}-R_{ebac}F^{ae}-4\pi\nabla_cJ_b. \end{align} The second term is the same with $b\leftrightarrow c$: $$\nabla_a\nabla_bF^a{}_c=R^e{}_bF_{ec}-R_{ecab}F^{ae}-4\pi\nabla_bJ_c.$$ Putting it together, we have $$-\nabla_a\nabla_cF^a{}_b+\nabla_a\nabla_b F^a{}_c=-R^e{}_cF_{eb}+R^e{}_bF_{ec}+R_{ebac}F^{ae}-R_{ecab}F^{ae}+4\pi\nabla_cJ_b-4\pi\nabla_bJ_c.$$ Let us reorganize by putting all terms involving $F$ on the left side: $$\Box F_{bc}+(R^e{}_bF_{ce}-R^e{}_cF_{be})+(R_{ecab}-R_{ebac})F^{ae}=-4\pi(\nabla_bJ_c-\nabla_cJ_b).$$ The third term on the left can be simplified. The Riemann tensor is antisymmetric in its last two indices, so $R_{ebac}=-R_{ebca}$, so the term in the parentheses is $R_{ecab}+R_{ebca}$. But $bac$ and $cba$ are cyclic permutations of each other, so by the first Bianchi identity, $$R_{ecab}+R_{ebca}=-R_{eabc}=R_{aebc}=R_{bcae},$$ because the (totally covariant) Riemann tensor is antisymmetric in its first two indices and symmetric upon the interchanging of the first index pair with the second. Thus the generalized equation is $$\Box F_{ab}+(R^c{}_aF_{bc}-R^c{}_bF_{ac})+R_{abcd}F^{cd}=-4\pi(\nabla_aJ_b-\nabla_bJ_a).$$