Lorentz transformation of the dual tensor

Question

I am trying to lorentz-transform the dual electromagnetic tensor $G^{\mu \nu}:= \frac{1}{2} \epsilon ^{\mu \nu \alpha \beta} F_{\alpha \beta}$ and also show (perhaps by using that last result) that $G^{\mu \nu}F_{\mu \nu} = - \frac{4}{c} \vec{E}\vec{B}$ is really (or not really) invariant under Lorentz transformation.

So far i have looked at it in different ways:

1. Idea: Not contracting $\epsilon$ \begin{equation*} {G^{\mu \nu }}' = \Lambda ^\mu _\sigma \Lambda ^\nu _\tau G^{\sigma \tau} = \Lambda ^\mu _\sigma \Lambda ^\nu _\tau \frac{1}{2} \epsilon ^{\sigma \tau \alpha \beta} F_{\alpha \beta} \end{equation*} this seems plausible but what about the levi civita pseudotensor? Shouldn't I Lorentz-transform it fully? As in

2. Idea: Contracting every tensor within \begin{equation*} {G^{\mu \nu}}' = (\frac{1}{2} \epsilon ^{\mu \nu \alpha \beta})' (F_{\alpha \beta})' =\Lambda ^\mu _\sigma \Lambda ^\nu _\tau \Lambda ^\alpha _\rho \Lambda ^\beta _\gamma ( \frac{1}{2} \epsilon ^{\sigma \tau \rho \gamma} )\Lambda _\alpha ^\xi \Lambda _\beta ^\omega (F_{\xi \omega}) \end{equation*} I apologize for the many indices. But is this the general way to go about the lorentz transformation? It seems to take me to the same last result as in 1.idea if i do the following: \begin{equation*} {G^{\mu \nu}}' = \Lambda ^\mu _\sigma \Lambda ^\nu _\tau \Lambda ^\alpha _\rho \Lambda ^\beta _\gamma ( \frac{1}{2} \epsilon ^{\sigma \tau \rho \gamma} )\Lambda _\alpha ^\xi \Lambda _\beta ^\omega (F_{\xi \omega}) = \Lambda ^\mu _\sigma \Lambda ^\nu _\tau \delta ^\xi _\rho \delta ^\omega _\gamma \frac{1}{2} \epsilon ^{\sigma \tau \rho \gamma} F_{\xi \omega} = \\ \Lambda ^\mu _\sigma \Lambda ^\nu _\tau \frac{1}{2} \epsilon ^{\sigma \tau \rho \gamma} F_{\rho \gamma} = \Lambda ^\mu _\sigma \Lambda ^\nu _\tau G^{\sigma \tau} \end{equation*} here i used the identity $\Lambda ^\alpha _\beta \Lambda ^\beta _\gamma = \delta ^\alpha _\gamma$ (is this true for all lorentz transformations?)

So i get the two ways and it seems they're equal, but what if I see it this way:

3. Idea: seeing the Lorentz-transformation of levi-civita as Determinant \begin{equation*} {G^{\mu \nu}}' = \Lambda ^\mu _\sigma \Lambda ^\nu _\tau \Lambda ^\alpha _\rho \Lambda ^\beta _\gamma ( \frac{1}{2} \epsilon ^{\sigma \tau \rho \gamma} )\Lambda _\alpha ^\xi \Lambda _\beta ^\omega (F_{\xi \omega}) =\frac{1}{2} det(\Lambda) \epsilon ^{\mu \nu \alpha \beta} \Lambda _\alpha ^\xi \Lambda _\beta ^\omega F_{\xi \omega} = \\ \pm \frac{1}{2}\epsilon ^{\mu \nu \alpha \beta} \Lambda _\alpha ^\xi \Lambda _\beta ^\omega F_{\xi \omega} = \pm \frac{1}{2}\epsilon ^{\mu \nu \alpha \beta} (F_{\alpha \beta})' \end{equation*} here i used $Det(\Lambda) = \pm 1$ so what is going on here? Where did i do a mistake? Does this have anything to do with non-symmetrical lorentz transformations, as in rotations? If i now solve for the lorentz transformation of $G^{\mu \nu} F_{\mu \nu}$ i get in one instance that they are invariant, on the other instance, they are not; the $\pm$ that comes from the determinant contradicts the equality.

I am inclined to my 2. Idea as being the way to go about it. It seems that one chooses often out of the determinant the +sign, but that is only for proper lorentz transformations, I would really appreciate learning if $G^{\mu \nu} F_{\mu \nu}$ is invariant under Lorentz transformations in general.

Answer 1

Recall that the Lorentz group $O(1,3)$ consists of all $4\times 4$ matrices $\Lambda$ satisfying $\Lambda^T\eta\Lambda=\eta$. This gives, in particular, $\det(\Lambda)=\pm1$. The subgroup $SO(1,3)$ consists of all transformations with determinant $+1$.

$\vec{E}\cdot\vec{B}$ is invariant under $SO(1,3)$ but not under all of $O(1,3)$ - it's a pseudoscalar. This is because it's a product of the vector $\vec{E}$ and the pseudovector $\vec{B}$. Here's an easy way of seeing that $\vec{B}$ transforms as a pseudovector:

Consider the Lorentz force law $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$. Now apply a reflection: \begin{equation}R:(t,\vec{x})\mapsto(t,-\vec{x}).\end{equation} We have: \begin{equation}\vec{F}\mapsto-\vec{F} \qquad \text{and} \qquad\vec{E}\mapsto-\vec{E},\end{equation} so that $\vec{v}\times\vec{B}\mapsto-\vec{v}\times\vec{B}$.

But the velocity transforms as $\vec{v}\mapsto-\vec{v}$ so we have the transformation of $\vec{B}$ under $R$:\begin{equation}\vec{B}\mapsto\vec{B}=(-1)(-\vec{B})=\det(R)(R\cdot\vec{B}).\end{equation}

Answer 2

When physicists refer to Lorentz symmetry, it is usually understood to refer to just the "proper" Lorentz symmetry. After all, no one would claim the standard model violates Lorentz symmetry, although it does not have parity symmetry. So to avoid confusion, I'll break the coordinate transformations of the full $O(1,3)$ (which is what I assume you mean by "Lorentz transformations in general") into proper Lorentz transformations, parity transformation, and time reversal transformation. Breaking it up this way also makes discussing the value in various coordinate systems easier to discuss.

You are asking several things here, but with all the approaches it sounds like the main issue is how to determine if something like $G^{\mu\nu}F_{\mu\nu}$ is invariant to Lorentz transformations and spatial or time reflections, and also how to figure out its value.

First, while a boring tautology, you cannot write out the transformations unless you know how the objects transform. $F_{\mu\nu}$ is a tensor, and as you mention, the Levi-Civita symbol $\epsilon^{\alpha\beta\mu\nu}$ is actually a pseudo-tensor. A pseudo-tensor will transform like a tensor except transformations that contain parity changes ("handedness" changes) and time flips (reversal of the time coordinate), in which case there will be a sign change according to $det(\Lambda^\mu{}_\nu)$.

Since this is the only thing that distinguishes the transformation of these objects from tensors, we don't need to restrict ourselves to Lorentz transformations. We can consider general transformations and break them into four classes based on whether there is a spatial or time reflection. Restricting ourselves to 'proper' transformations, everything will transform like a tensor, and then we can calculate the results in 'improper' transformations just by considering its parity of time reversal action.

The great thing about Einstein notation is once we are working with tensors, we can just read off whether something is coordinate independent. All the indices are contracted, so this is independent of coordinate transformation (within the restricted class we are considering at the moment). Transformation of an "upper" (contravariant) index will cancel transformation of the "lower" (covariant) index and so on. This means $G^{\mu\nu}F_{\mu\nu}$ is invariant to all proper transformations. For improper transformations, the value will be the same except for a possible sign change. Change handedness -> change sign. Change time direction -> change sign. Change both handedness and time direction -> the two sign changes cancel.

As for seeing what $G^{\mu\nu}F_{\mu\nu}$ equals, just choose a coordinate system that is convenient, then you can easily see what the value is in other coordinate systems. So I'd suggest just using an inertial coordinate system, looking at how $\mathbf{E}$ and $\mathbf{B}$ are defined as components of the electromagnetic tensor $F_{\mu\nu}$ and then calculating $G^{\mu\nu}$ (which is just permuting components using Levi-Civita when viewed as a matrix of components). Then calculating $G^{\mu\nu}F_{\mu\nu}$ is easy and can be seen to be proportional to $\mathbf{E} \cdot \mathbf{B}$.

Then, applying a coordinate transformation with a spatial reflection or time reversal (so the sign of $G^{\mu\nu}$ changes), we see that under these "improper" transformations $G^{\mu\nu}F_{\mu\nu}$ will get a negative sign. So, yes this is invariant for the usual proper Lorentz transformations, but not for improper Lorentz transformations (although the absolute value would be completely invariant to coordinate transformation).