Tricks for evaluating tensor contractions with Levi-Civita symbol

Question

I am trying to evaluate the Lorentz invariant $\epsilon^{\alpha\beta\gamma\delta}F_{\alpha\beta}F_{\gamma\delta}$, where $F_{\mu\nu}$ is the electromagnetic field tensor, $$ F_{\mu\nu} = \begin{bmatrix}0 & E^1 & E^2 & E^3\\-E^1 & 0 & -B^3 & B^2\\-E^2 & B^3 & 0 & -B^1\\-E^3 & -B^2 & B^1 & 0\end{bmatrix} $$ and $\epsilon^{\alpha\beta\gamma\delta}$ is the Levi-Civita symbol. I am wondering if there is an efficient way of evaluating this contraction. I know that only permutations of ${1,2,3,4}$ are nonzero because of the Levi-Civita symbol (which means that only the non-zero entries of $F_{\mu\nu}$ will appear), but from there I am left with writing out the 24 permutations of those four numbers, giving me 24 (very simple) terms to evaluate. I have no problem with performing this algebra, but is there a shortcut?

I suppose that since $F$ is anti-symmetric, I could cut the number of terms in half.

Answer 1

Don't use indices. Use a sophisticated and efficient notation for tensor manipulations: through clifford algebra.

Let $\gamma^0, \gamma^1, \gamma^2, \gamma^3$ be basis vectors. Under the clifford product operation, they obey the following:

$$\gamma^\mu \gamma^\nu = \begin{cases}\eta^{\mu \nu} & \mu = \nu \\ -\gamma^\mu \gamma^\nu & \mu \neq \nu\end{cases}$$

The product operation is also associative.

Write $E$ and $B$ in terms of $\gamma^1, \gamma^2, \gamma^3$ as you would write any vectors: $E = E_x \gamma^1 + \ldots$ for instance. Then $F$ can be written

$$F = \gamma^0 E - \gamma^1 \gamma^2 \gamma^3 B$$

Define $\epsilon = -\gamma^0 \gamma^1 \gamma^2 \gamma^3$. Then the tensor equation $\epsilon^{\alpha \beta \gamma \delta} F_{\alpha \beta} F_{\gamma \delta}$ can be written

$$\epsilon^{\alpha \beta \gamma \delta} F_{\alpha \beta} F_{\gamma \delta} = \langle \epsilon FF\rangle_0$$

The brackets denote that we're looking for the scalar part of this expression. This is equivalent to $\epsilon \langle FF \rangle_4 = \epsilon (F \wedge F)$ in the common shorthand.

Now we can use the associativity of the product to our advantage. For brevity, let $i = \gamma^1 \gamma^2 \gamma^3$, such that $\gamma^0 i = \epsilon$, and see that

$$\langle \epsilon FF \rangle_0 = \langle \epsilon (\gamma^0 E - iB)(\gamma^0 E - iB) \rangle_0$$

Multiply out the terms:

$$\langle \epsilon FF \rangle_0 = \langle \epsilon (\gamma^0 E \gamma^0 E -\gamma^0 E iB - iB \gamma^0 E + iBiB )\rangle_0$$

Use the properties of the product to deduce that $E\gamma^0 = -\gamma^0 E$, $\gamma^0 B = -B \gamma^0$, $Ei = iE$ and $iB = Bi$, and $ii = 1$. The result reduces to

$$\langle \epsilon FF \rangle_0 = \langle \epsilon (-EE - 2 \epsilon EB + BB) \rangle_0$$

Finally, deduce that $\epsilon \epsilon = -1$, $EE = -|E|^2$ and $BB = -|B|^2$ to get

$$\langle \epsilon FF \rangle_0 = \langle \epsilon (|E|^2 - |B|^2) + 2 E \cdot B \rangle_0$$

The only term that survives the angled brackets is the $2 E \cdot B$ term. Of course, staring us right in the face is the other well-known invariant, $E^2 - B^2$. That is one advantage of this method: both invariants can be derived from the same approach, and at the same time. Moreover, we never have to break into components to evaluate the expression. We merely use the commutation properties of basis elements in the clifford algebra--an algebra already widely used in quantum mechanics.

Answer 2

Let $u^{\mu}$ be the 4-velocity of some observer. Then $F_{\alpha\beta} = 2E_{[\alpha}u_{\beta]} + \epsilon_{\alpha\beta\gamma\delta}u^{\gamma}B^{\delta}$ where $E^{\alpha}, B^{\alpha}$ are the electric and magnetic fields relative to this observer and brackets denote antisymmetrization. By definition, $u^{\alpha}E_{\alpha} = u^{\alpha}B_{\alpha} = 0$.

Thus, $$\epsilon^{\alpha\beta\gamma\delta}F_{\alpha\beta}F_{\gamma\delta} = \epsilon^{\alpha\beta\gamma\delta}(2E_{[\alpha}u_{\beta]} + \epsilon_{\alpha\beta\mu\nu}u^{\mu}B^{\nu})(2E_{[\gamma}u_{\delta]} + \epsilon_{\gamma\delta\sigma\rho}u^{\sigma}B^{\rho}) \\ = 2\epsilon^{\alpha\beta\gamma\delta}\epsilon_{\alpha\beta\mu\nu}u^{\mu}B^{\nu}E_{\gamma}u_{\delta} + 2\epsilon^{\alpha\beta\gamma\delta}\epsilon_{\gamma\delta\sigma\rho}u^{\sigma}B^{\rho}E_{\alpha}u_{\beta} + \epsilon^{\alpha\beta\gamma\delta}\epsilon_{\gamma\delta\sigma\rho}\epsilon_{\alpha\beta\mu\nu}u^{\sigma}B^{\rho}u^{\mu}B^{\nu} \\ = 4\epsilon^{\alpha\beta\gamma\delta}\epsilon_{\alpha\beta\mu\nu}u^{\mu}B^{\nu}E_{\gamma}u_{\delta} + \epsilon^{\alpha\beta\gamma\delta}\epsilon_{\gamma\delta\sigma\rho}\epsilon_{\alpha\beta\mu\nu}u^{\sigma}B^{\rho}u^{\mu}B^{\nu} \\ = -16u^{\mu}B^{\nu}E_{[\mu}u_{\nu]} -4\epsilon_{\gamma\delta\sigma\rho}u^{\sigma}B^{\rho}u^{\gamma}B^{\delta} \\ = -8E^{\mu}B_{\mu} = -8\vec{E}\cdot \vec{B}$$

Answer 3

Here's a systematic way to do this: first recall that the electromagnetic tensor is given by

$$F^{\alpha\beta}=\partial^\alpha A^\beta-\partial^\beta A^\alpha$$

where

$$F_{\alpha \beta}=\eta^{\mu \nu} F^{\alpha \beta}\eta^{\eta \nu}$$

and

$$\partial^\mu=-\nabla,\frac{1}{c}\frac{\partial}{\partial t}$$

$A$ is the magnetic vector potential and $A^0$ is the scalar potential. $\eta^{\mu \nu}$ is the Minkowski space-time metric. I look at $F^{\alpha \beta}$ rather than your $F_{\alpha \beta}$ because this is what I've been working with in my notes, and I don't want to make a sign error and confuse you (or me!) anymore than is needed.

We ultimately find that

\begin{align} \epsilon^{\alpha \beta\gamma \delta}F^{\alpha \beta}F^{\gamma \delta}&=\epsilon^{\alpha \beta \gamma \delta}\left(\partial^\alpha A^\beta-\partial^\beta A^\alpha\right)\left(\partial^\gamma A^\delta-\partial^\delta A^\gamma\right) \end{align}

You can now derive each quantity in the parentheses via \begin{align} E=-\nabla A-\frac{\partial A}{\partial t},\qquad B=\nabla\times A \end{align}

For example,

\begin{align} \partial^0 A^1-\partial^1 A^0=\frac{\partial A_x}{\partial t}+\frac{\partial A_0}{\partial x}=-E_x \end{align}

And thus

\begin{align} \epsilon^{0123}\left(\partial^0 A^1-\partial^1A^0 \right)\left(\partial^2 A^3-\partial^3 A^2\right)=(-E_x)(-B_x)=E_xB_x \end{align}

I leave the rest of the components (along with a transformation via the Minkowski metric) up to you--I don't see any faster way of doing this other than Meng Cheng's method, though I do find this way to be quite easy to follow and systematic.