Physical meaning of the invariance of the dot product of $\vec{E}$ and $\vec{B}$

Question

I recently learned about the fact that $\vec{E} \cdot \vec{B}$ is invariant under Lorentz transformations, which seems like a really nice and useful result. Is there a physical meaning similar to the Lorentz-invariance of the dot product of two 4-velocities? I would like to understand this better on a conceptual level.

Answer 1

You may already know that contractions of tensors produce Lorentz invariants (similar to how dot products are invariant under rotations in non relativistic mechanics). This result is natural from the definition of tensors.

In special relativity, the electric and magnetic fields are not Lorentz vectors (first rank tensors), however the electromagnetic tensor $$ F^{\mu\nu} = \begin{bmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{bmatrix},$$ is a good Lorentz tensor. Specifically if we defined the dual tensor $\bar{F}^{\alpha\beta}=\frac12\epsilon_{\alpha\beta\mu\nu}F^{\mu\nu}$, we have that the contraction $$F^{\mu\nu}\bar{F}_{\mu\nu}=\frac{4}{c^2}\mathbf B\cdot \mathbf E,$$ is a Lorentz invariant (more precisely a pseudo-scalar if you care about reflections). It is also related to the another invariant $\det(F^{\mu\nu})=(\mathbf B\cdot \mathbf E)^2/c^2$.

This invariance is telling us that the usual spatial angle between $\mathbf E$ and $\mathbf B$ is the same for all inertial observers. For example if $\mathbf E$ and $\mathbf B$ are orthogonal, the two fields remain orthogonal in any frame of reference. Thus electromagnetic waves in vacuum are transverse independently of the frame of reference. This result is somehow enforced by the special role of light in special relativity .

Just for completeness, the last important invariant related to $F^{\mu\nu}$ (aside from its trace) is its contraction with itself, $$F^{\mu\nu}F_{\mu\nu}=2\left(|\mathbf B|^2-\frac{1}{c^2}|\mathbf E|^2\right)$$ which just means that if $|\mathbf B|>\frac{1}{c^2}|\mathbf E|$ for some inertial observer, it will remain so for the rest of inertial observers, same if $|\mathbf B|\leq\frac{1}{c^2}|\mathbf E|$. Note that neither $|\mathbf E|$ or $|\mathbf B|$ are invariants on their own.

Answer 2

Scalar products are usually invariant under some transformation. In the case of four-vectors, this is Lorentz transformations.
This can be seen by directly applying the transformations to $$E'_\mu B'^\mu=({\Lambda_\mu}^\nu E_\nu)({\Lambda^\mu}_\rho B^\rho)=({\Lambda_\mu}^\nu{\Lambda^\mu}_\rho)E_\nu B^\rho=\\=\delta^\nu_\rho E_\nu B^\rho=E_\nu B^\nu$$ which means that in any frame you take it, the scalar product between $E$ and $B$ has the same numerical value.
Fundamental examples of this are and the invariant mass and the invariant interval in Minkowski space (valid for both signatures of the metric)$$\eta_{\mu\nu} dx^\mu dx^\nu=c^2dt^2-dx^x-dy^2-dz^2$$ which is indeed the scalar product $dx^\mu dx_\mu$, although not very instructive in this form.
NOTE: this is not true for arbitrary scalar products. As I have said, it is valid for four-vectors!