Proving validity of expression for the electromagnetic tensor

I am trying to show that the following expression for the electromagnetic tensor, given in Geroch's lecture notes on GR, is valid:

$$F_{ab}=2\xi_{[a}E_{b]}-\epsilon_{abcd}\xi^cB^d \quad(*)$$

For that, the text suggests simple substitution of the electric ($E^a$) and magnetic ($B_a)$ fields into $(*)$ to get identities. The field vectors are:

$$E^a=F_{ab}\xi^b$$

$$B_a=\frac{1}{2}\epsilon_{abcd}\xi^bF^{cd}$$

However, I'm stuck and can't work out the math after plugging everything into $(*)$:

$$\xi_dg_{bd}F_{ab}\xi^b-\xi_bg_{ab}F_{bc}\xi^c-\frac{1}{2}\epsilon_{abcd}\xi^cg^{da}\epsilon_{abcd}\xi^bF^{cd}$$

Any help on how to get out of this tensor mess would be appreciated.

Thanks!

• Look in MTW for how the manage multiplication of Levi-Civita tensors. I also notice that in your last equation you have multiple appearances of indices, which is probably also a source of confusion. – Lawrence B. Crowell Sep 2 '16 at 10:40

Firstly, note that indices must match on both sides of an equation, so we get $E_a = F_{ab}\xi^b$ (or $E^a = F^{ab}\xi_b$). Secondly it is helpful to consider the corresponding contravariant equation for the magnetic field: $B^a = -\frac{1}{2}\epsilon^{abcd}\xi_{b}F_{cd}$ (notice the minus sign due to the pseudo-tensor quality of the Levi-Civita tensor). Making these substitutions we get the terms \begin{align} 2\xi_{[a}E_{b]} &= 2 \xi_{[a}F_{b]c}\xi^c, \\ \epsilon_{abcd}\xi^cB^d &= -\frac{1}{2} \epsilon_{abcd}\xi^c\epsilon^{d\alpha\beta\gamma}\xi_{\alpha}F_{\beta\gamma} \\ &= \frac{1}{2} \epsilon_{abcd}\epsilon^{\alpha\beta\gamma d}\xi^c\xi_\alpha F_{\beta\gamma} \\ &= \frac{3!}{2}\delta^{[\alpha}_a\delta^\beta_b\delta^{\gamma]}_c\xi^c\xi_\alpha F_{\beta\gamma} \\ &= \frac{1}{2}\left(\xi_aF_{bc}\xi^c + \xi_cF_{ab}\xi^c + \xi_bF_{ca}\xi^c - \xi_bF_{ac}\xi^c - \xi_cF_{ba}\xi^c - \xi_aF_{cb}\xi^c\right) \\ &= 2\xi_{[a}F_{b]c}\xi^c + \xi^c\xi_cF_{ab}. \end{align} Now, I assume $\xi^a$ is a timelike vector whence, depending on your sign convention, $\xi^c\xi_c = \pm 1$, so I get correspondingly $$\mp F_{ab} = 2\xi_{[a}E_{b]} - \epsilon_{abcd}\xi^cB^d,$$ suggesting you are using a spacelike convention, is this correct?
EDIT: Regarding the contraction of the Levi-Civita tensors, first recall that \begin{align} \epsilon_{abcd} &= \sqrt{|\det[g_{ij}]|}\varepsilon_{abcd}, \\ \epsilon^{abcd} &= \frac{1}{\sqrt{|\det[g_{ij}]|}}\varepsilon^{abcd}, \end{align} where $\varepsilon$ denotes the Levi-Civita symbol. Thus contraction of $\epsilon$ equates with contraction of $\varepsilon$. Then note that straight from the definition we find \begin{align} \varepsilon_{abcd}\varepsilon^{\alpha\beta\gamma\delta} &= \begin{cases} +1 & \text{if $(\alpha,\beta,\gamma,\delta)$ and $(a,b,c,d)$ are permutations of the same sign,} \\ -1 & \text{if they are permutations of different sign,} \\ 0 & \text{otherwise,} \end{cases} \end{align} so similarly it follows directly that \begin{align} \varepsilon_{abcd}\varepsilon^{\alpha\beta\gamma d} &= \begin{cases} +1 & \text{if $(\alpha,\beta,\gamma)$ and $(a,b,c)$ are permutations of the same sign,} \\ -1 & \text{if they are permutations of different sign}, \\ 0 & \text{otherwise.} \end{cases} \end{align} In the above I did not state what is permuted for brevity, but I think that should be obvious. As for $\delta^{[\alpha}_a\delta^\beta_b\delta^{\gamma]}_c$ observe that if either of $(\alpha,\beta,\gamma)$ or $(a,b,c)$ contain a repeated index the expression vanishes to zero by the anticommutation, and similarly if $\{\alpha,\beta,\gamma\} \neq \{a,b,c\}$ by the Kroenecker deltas. So far so good. Now if $(\alpha,\beta,\gamma)$ is an even permutation of $(a,b,c)$ then the only non-zero term in the expansion will be one with a $+$ from the anti-commutation, and if it is an odd permutation the only non-zero term will be one with $-$ from the anti-commutation. Thus we recover $$\varepsilon_{abcd}\varepsilon^{\alpha\beta\gamma d} = 3!\delta^{[\alpha}_a\delta^\beta_b\delta^{\gamma]}_c.$$
• May I ask where you got the expression for the product of the Levi-Civita tensors from? I have not seen this expression using commutator brackets; I just know the brute force method using the determinant of a matrix of Kronecker-Deltas. I mean in the end I get the same 6 terms but it is a lot messier to do: Calculating the determinant, equating the contracted index $d$... – N0va Sep 2 '16 at 15:07
• I'm using the $\xi_a\xi^a=-1$ convention. Also, I got the $E^a=F_{ab}\xi^b$ from the source text (end of page 41): books.google.com.br/… Are the lecture notes inconsistent? I've heard they have some typos... – Lsheep Sep 2 '16 at 17:59
• $\xi_a \xi^a=-1$ is typical for GR, since the Lorentzian metric convention $(-+++)$ is often used. $E^a=F_{ab}\xi^b$ is $100\%$ wrong: it makes no sence looking at the indieces: the right side has one contracted summation index $b$ and the lower index $a$: $E_a=F_{ab}\xi^b$ is correct. – N0va Sep 2 '16 at 18:14
• Could someone clarify the passage from the second to the third line in the magnetic field expression? What property do we use to "commute" $\epsilon^{\alpha\beta\gamma d}$ with $\xi^c$ and change sign? – Lsheep Sep 2 '16 at 20:34
• @Lsheep I added a "proof" for the Levi-Civita contraction. In the step where the sign changes what is actually happening is that we are changing the order of indices in one of the Levi-Civita tensor (the second one: $d$ is moved from the first index to the last). I also reordered the terms, but that is inconsequential. I apologize if it is confusing, but I wanted to bring the Levi-Civita tensors next to each other to clarify the contraction. – Erik Jörgenfelt Sep 2 '16 at 21:15