What is the physical significance of $\mathbf{E}^2-\mathbf{B}^2$ in E and M? Choosing nice units, i.e $c=1$, the electromagnetic energy density is:
$$u=\frac{1}{2}\left(\mathbf{E}^2+\mathbf{B}^2\right).$$
This is not Lorentz invariant, which makes sense since our nonrelativistic form of energy is not Lorentz invariant. However, the quantity:
$$\mathbf{E}^2-\mathbf{B}^2$$
is Lorentz invariant. Furthermore, the Yang-Mills Lagrangian for $G=U(1)$ is (up to some sign/constant depending on your induced metric convention etc.):
$$L_{YM}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}=\frac{1}{2}(\mathbf{E}^2-\mathbf{B}^2),$$
which also makes sense since $L_{YM}$ should be gauge invariant and Lorentz invariant. So I assume there should be some physical meaning to this quantity as classical Lagrangians are often viewed as kinetic-potential. I am more mathematically inclined, and have little to no physical intuition, so if someone could spell this out for me that would be greatly appreciated.
 A: $E^2-B^2$ is always zero for the radiation part of the retarded field of a single point charged particle, or for simple plane or spherical wave of EM radiation far from the source.
But it is not zero near the source where the electric field is dominated by the Coulomb term and magnetic field is dominated by the corresponding term proportional to velocity of the charged particle (and thus, for small speeds, magnetic square is dominated by the electric square). Also, if we have a superposition of plane waves with different directions (including a standing wave), the difference of squares stops being zero in general.
So one "meaning" could be that it shows how "far" from a simple plane wave the EM field is.
A: I don't have a ready answer to what you're saying about one system when you mention it has twice as much $E^2-c^2B^2$ as another, but there is an obvious relevance to the sign of $E^2-c^2B^2$. If it's positive, $E>cB$; if it's negative, $E<cB$; if it's zero, $E=cB$ in all frames.
The other Lorentz invariant is $E\cdot B\propto\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}$ (proportionality constant is an exercise). The fields are orthogonal iff it's $0$. If it is, you can choose a frame where the shorter vector is $0$; if not, you can't make either $0$.
A: I will describe a less known but truly amazing network theorem discovered by Tellegen [1] with the hope that this physics group will find it interesting. It specifically pertains to lumped element circuits, and I follow Meixner [2].
Denote the impedance (admittance) of 1-port (2-pole) RLC network by $Z(p)$,  $Y(p)=1/Z(p)$, and let the input voltage excitation be $u(t)$, such that $u(t) = 0$ for $t < 0$. Denote by $u_k(t)$ and $i_k(t)$ the $k^{th}$ branch voltage and current, resp. Then, using operator notation $p=\frac{d}{dt}$,$p'=\frac{d}{dt'}$, Tellegen proves that
$$\frac{1}{2}\frac{Y(p)-Y(p')}{p-p'}u(t)u(t') = \sum_k\frac{1}{2C_k}q_k(t)q_k(t')-\sum_k\sum_{\ell}\frac{1}{2}L_{k \ell}i_k(t)i_{\ell}(t') \tag{1}\label{1},$$
where $C_k$, $L_{kk}$ are the capacitance and inductance in the $k^{th}$ branch, and $L_{k\ell}$ is the mutual inductance of the $k^{th}$ and $\ell^{th}$ branches.
If you let $p'\to p$ $\eqref{1}$ turns in to
$$\lim_{t'\to t}\frac{1}{2}\frac{Y(p)-Y(p')}{p-p'}u(t)u(t') = \sum_k\frac{1}{2C_k}q_k^2(t)-\sum_k\sum_{\ell}\frac{1}{2}L_{k \ell}i_k(t)i_{\ell}(t)\\
=W_e-W_m \tag{2}\label{2}$$
The right side is the difference between the stored electric and magnetic energies, the left side is a complicated linear integro-differential operator based on the admittance operator $Y(p)$ and is applied to something that depends on the outside excitation.
In other words the difference $W_e-W_m$ depends only on the excitation $u(t)$ and on the admittance $Y(p)$ it drives and not on the specific structure of the network having that very admittance.
What is surprising in this result is that the impedance (admittance) while it determines the dissipation inside the network it does not determine the sum of the stored (reactive) energies $W_e+W_m$. The best known example is to take a resistor $R$ and a capacitor $C$ in series forming $Z_1=R+\frac{1}{pC}$, and take a resistor $R$ and an inductor $L$ in series forming $Z_2=R+pL$. Now connect $Z_1$ and $Z_2$ in parallel so that we have their impedance $Z=\frac{Z_1Z_2}{Z_1+Z_2}$.
Now let $L=R^2C$, in this special case we get $Z=\frac{(R+pR^2C)(R+1/pC)}{R+pR^2C+R+1/pC}$ which after simplification gives $Z=R$. Nothing special about this so far but if you let the same voltage drop on the resistor $R$ by itself as on the same impedance RLC you will get the same total current through both and the same dissipation but with very different reactive energies. In the case of the resistor the reactive energies are zero, obviously, while the 4-element RLC network of impedance $R$ will have nonzero $W_e$ and nonzero $W_m$, both of the same size, $W_e -W_m=0$ as required.
So here is Tellegen's lumped element invariance principle: all networks that have the same impedance will have the same $W_e-W_m$.
[1] Tellegen: A general network theorem with applications, Phillips Research  Reports, 1952, pp259-269
[2] Meixner: Network Theory and Its Relation to the Theory of Linear Systems, IRE Trans Ant. Prop.  December 1959 pp435-439
