Intuition on $E^2-B^2$ as a Lagrangian?

Question

I recall reading that instead of $E$ and $B$ vectors, gradients of $E^2-B^2$ are fundamental. I understand that it is the Lorentz invariant Lagrangian for the EM field. The signs of the energy imply one is a kinetic energy term and the other a potential energy term.

My question is a wider understanding of how $B^2$ is a potential energy-like term and how is $E^2$ a kinetic energy-like term? This usually means one is a derivative in time and the other a function of space - is that the case here? Are they not both changing in time? Also for far-field EM waves, isn't $E^2=B^2$ so $E^2-B^2=0$.

Answer 1

The short answer is that when we rewrite $\vec{E}$ (and $\vec{B}$) in terms of the gauge field $A_{\mu}$, then $\vec{E}$ contains (and $\vec{B}$ does not contain) time-derivatives $\dot{A}_{\mu}$, as we would expect from a kinetic term $T$ (and a potential term $V$), respectively.

Similarly, the sum $T+V$ is indeed the energy density of the EM field.

Answer 2

It’s easier to see this using potentials. In an inertial frame, I’ll write the $4-$vector potential as $(V,A)$. This gives the field in the usual way: $$ E=-\partial_t A -\nabla V \\ B=\nabla\times A $$ so your Lagrangian density is: $$ \mathcal L=\frac{1}{2\mu_0}\left[(\partial_t A+\nabla V)^2-(\nabla\times A)^2 \right] $$ The first term is the square of the time derivative of the field $A$ so the analogy with kinetic energy is very transparent. The second term contains no time derivative, and can be seen as some coupled harmonic system’s potential energy.

Note that nowhere there is a time derivative pf $V$, this is because it is not a dynamic variable. You can view it as a Lagrange multiplier enforcing Gauss’ law.

Hope this helps.

Answer 3

Once you accept that (a) the electromagnetic field is decribed by a U(1) gauge field $A^\mu(x)$ (as field variable) and (b) that the (free) field equations should be linear in $A^\mu$, the corresponding action is (up to a normalization factor) uniquely determined by the requirements of Poincare symmetry and gauge invariance:

The field strength tensor $F_{\mu \nu}= \partial_\mu A_\nu-\partial_\nu A_\mu$ is clearly invariant under the U(1) gauge transformation $A_\mu \to A_\mu + \partial_\mu \Lambda$ and possible candidates (bilinear in $A^\mu$) for terms in the Lagrange density (Lagrangian) are the scalar $F_{\mu \nu}F^{\mu \nu} = 2 (\vec{B}\cdot \vec{B}-\vec{E}\cdot \vec{E})$ or the pseudoscalar $\epsilon_{\mu \nu \alpha \beta}F^{\mu \nu} F^{\alpha \beta} = 8 \vec{E} \cdot \vec{B}$. The second possibilty is ruled out by that fact that this term can be written as a 4-gradient $\partial_\mu( \ldots)$, which does not affect the equations of motion. So one finds the Poincare invariant action $S= k \int d^4 x \, F_{\mu \nu}(x) F^{\mu \nu}(x)$, with $k =-1/16 \pi c$ in the Gauss system (or $-1/4c$ in the Heaviside system).

Note that a "mass term" $A_\mu A^\mu$ is forbidden as it violates gauge invariance and terms like $(F_{\mu \nu} F^{\mu \nu})^2$ or $(\epsilon_{\mu \nu \alpha \beta} F^{\mu \nu} F^{\alpha \beta})^2$ are excluded by condition (b).

Thus the "intuition" for introducing $\vec{E}^2-\vec{B}^2$ in the Lagrangian is the simple fact that this is a Lorentz invariant and gauge invariant quantity. Interpreting $\vec{E}^2$ and/or $\vec{B}^2$ as "kinetic energy" or a "potential" term is completely misleading. The correct way to identify the energy density (and the momentum density) of the electromagnetic field is to study its energy-momentum tensor.