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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!

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  • $\begingroup$ 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. $\endgroup$ – Lawrence B. Crowell Sep 2 '16 at 10:40
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I want to preface by saying that I would recommend going back and rehearsing tensor calculations in index notation. Not because you could not figure this out, but because your indices are all messed up in what you showed.

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 $$ \varepsilon_{abcd}\varepsilon^{\alpha\beta\gamma d} = 3!\delta^{[\alpha}_a\delta^\beta_b\delta^{\gamma]}_c. $$

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  • $\begingroup$ 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 $d$... $\endgroup$ – N0va Sep 2 '16 at 15:07
  • $\begingroup$ 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... $\endgroup$ – Lsheep Sep 2 '16 at 17:59
  • $\begingroup$ $\xi_a \xi^a=-1$ 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. $\endgroup$ – N0va Sep 2 '16 at 18:14
  • $\begingroup$ 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? $\endgroup$ – Lsheep Sep 2 '16 at 20:34
  • $\begingroup$ @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. $\endgroup$ – Erik Jörgenfelt Sep 2 '16 at 21:15

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