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I am trying to understand why $$\det(F^{\mu\nu})=(\vec{E}\cdot\vec{B})^2\tag{1}.$$ Of course one can just calculate the determinant of $F^{\mu\nu}$ expressed as a matrix with components given in terms of $E_x, E_y, ...$, etc., but I am looking for something a bit more insightful. In particular, I understand that $\vec{E}\cdot\vec{B}$ can be written as

$$\vec{E}\cdot\vec{B}=-\frac{1}{8}\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}.\tag{2}$$

This seems promising because then

$$(\vec{E}\cdot\vec{B})^2=\frac{1}{64}\epsilon_{\mu\nu\rho\sigma}\epsilon_{\alpha\beta\gamma\delta}F^{\mu\nu}F^{\rho\sigma}F^{\alpha\beta}F^{\gamma\delta},\tag{3}$$

which looks awfully like the expression for the determinant of $F$ in terms of Levi-Civita tensors,

$$\det(F^{\mu\nu})=\frac{1}{4!}\epsilon_{\mu\nu\rho\sigma}\epsilon_{\alpha\beta\gamma\delta}F^{\mu\alpha}F^{\nu\beta}F^{\rho\gamma}F^{\sigma\delta}.\tag{4}$$

But I can't figure out how to connect these two expressions. Since $F$ is antisymmetric, one can flip the two indices of any single $F$ at the cost of a minus sign, but I need a way to permute indices between different $F$'s. I'm also not sure how a factor of 3 could possibly enter in to change the 1/64 into a 1/24.

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  • $\begingroup$ You might be able to get away with using symmetrization/antisymmettization, i.e. that you can try and use $$F^{\mu \nu}F^{\alpha \beta} = 2! (F^{\mu [\nu}F^{\alpha] \beta} + F^{\mu \alpha}F^{\nu \beta})$$ and perhaps using the epsilons to get rid of certain parts. I haven't totally looked this over but the idea popped up in my head $\endgroup$ – Triatticus Sep 30 '18 at 21:17
  • $\begingroup$ By factor of $3$ you mean factor of $8/3$ $\endgroup$ – J.G. Sep 30 '18 at 21:28
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    $\begingroup$ @J.G. true, I just meant I can imagine how factors of 2 might pop up, but not factors of 3. $\endgroup$ – WillG Sep 30 '18 at 21:36
  • $\begingroup$ Related: Fundamental invariants of the electromagnetic field. $\endgroup$ – Emilio Pisanty Oct 9 '18 at 11:35
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The two expressions are only superficially similar; they have to be by dimensional analysis. The index structure is completely different, and I don't think you can convert one to the other without essentially undoing everything you did to get to the first result, then inserting the standard proof of the second result. Such a proof would be unenlightening since it'd just be unnecessarily complicated.

So let me just give a simple derivation of the result you want, directly. The determinant is $$\det F^{\mu\nu} = \epsilon_{\mu\nu\rho\sigma} F^{\mu 0} F^{\nu 1} F^{\rho 2} F^{\sigma 3}$$ There are naively $4!$ terms, but the only nonzero terms are those where $\mu \neq 0$, $\nu \neq 1$, $\rho \neq 2$, $\sigma \neq 3$. That is, we need to count the number of derangements of a 4-element set. It is easy to show by casework there are $9$.

Each of these $9$ terms is quadratic in $\mathbf{E}$, since there are two indices equal to zero, and hence quadratic in $\mathbf{B}$. Moreover, the sum of all $9$ terms is a tensorial invariant. There are only two independent invariants: $E^2 - B^2$ and $\mathbf{E} \cdot \mathbf{B}$. We can't use the first one, because otherwise our terms wouldn't all have the same degree in $E$ and $B$. Then the answer must be proportional to $(\mathbf{E} \cdot \mathbf{B})^2$, but since $(\mathbf{E} \cdot \mathbf{B})^2$ has $9$ terms, they must simply be equal.

You might complain I used components, but I had to because your expression is not tensorial. The fields $\mathbf{E}$ and $\mathbf{B}$ are not Lorentz tensors, but rather a way of writing components of $F_{\mu\nu}$. You can't expect to prove a statement about components without expanding in components.

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  • $\begingroup$ ok thanks, I'm fine to accept the component-based argument. However, could you expand on / justify the fact that $E^2-B^2$ and $\vec{E}\cdot\vec{B}$ are the only two possible invariants? This fact is not obvious to me. $\endgroup$ – WillG Sep 30 '18 at 22:29
  • $\begingroup$ @WillG It's well-known these are the only linearly independent scalars obtainable from $F_{\mu\nu}$, and are respectively proportional to $F_{\mu\nu}F^{\mu\nu}$ and $\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}$. $\endgroup$ – J.G. Oct 1 '18 at 5:34
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OP's eq. (2) is (minus) the Pfaffian ${\rm Pf}(F)$, and OP's eq. (4) is the determinant ${\rm Det}(F)$. The sought-for relation (1) follows because the square of the Pfaffian is the determinant:

$$ {\rm Pf}(F)^2~=~{\rm Det}(F). \tag{A}$$

A proof of eq. (A) is given in my Math.SE answer here.

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The determinant on the left is invariant under (proper) Lorentz transformations, so you can easily calculate it in a frame of reference where $E^2=E^3=0$, it equals $(E^1 H^1)^2=(\vec{E}\vec{H})^2 $. As $\vec{E}\vec{H} $ is an invariant under proper Lorentz transformations, this proves your formula in a general case.

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Quoting myself from an old PhysicsForums post

If I am not mistaken, (someone check my math) you can calculate the principal invariants of a tensor: $$F^a{}_a,\quad F^a{}_{[a} F^b{}_{b]},\quad F^a{}_{[a} F^b{}_{b}F^c{}_{c]},\quad F^a{}_{[a} F^b{}_{b} F^c{}_{c}F^d{}_{d]}$$ in 4-dimensions. You'd get the trace (sum of the eigenvalues), sum of products-of-pairs of eigenvalues, sum of products-of-triples, and finally, the product-of-the-4-eigenvalues (the determinant).

For a real antisymmetric matrix, the eigenvalues are imaginary. So, only those with sums of products of even-numbers-of-eigenvalues will be nonzero.

(Higher combinations like $F^a{}_{[a} F^b{}_{b} F^c{}_{c}F^d{}_{d}F^e{}_{e}F^f{}_{f]}$ vanish since it would have more than 4 factors being antisymmetrized). So, all but two are identically zero.

At some point in your calculation, you would probably have to make use of the epsilon-delta identities [sometimes written as determinants with delta-entries] (like this contracted combination $\epsilon_{abmn}\epsilon^{cdmn}=-4\delta_a{}^{[c}\delta_b{}^{d]}$ (up to sign conventions, based on signature and dimensionality)). Together with the antisymmetry of $F_{ab}=F_{[ab]}$, I think the index-gynmastics will work out.

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