# Tensorial Proof that $\det(F^{\mu\nu})=(\vec{E}\cdot\vec{B})^2$?

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.

• 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 – Triatticus Sep 30 '18 at 21:17
• By factor of $3$ you mean factor of $8/3$ – J.G. Sep 30 '18 at 21:28
• @J.G. true, I just meant I can imagine how factors of 2 might pop up, but not factors of 3. – WillG Sep 30 '18 at 21:36
• – Emilio Pisanty Oct 9 '18 at 11:35

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.

• 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. – WillG Sep 30 '18 at 22:29
• @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 $\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}$. – J.G. Oct 1 '18 at 5:34

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.

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.

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.