# Is the Lagrangian density of electromagnetism half-blind?

The quantity you propose is a total derivative; specifically, $$\frac{1}{2} \epsilon_{abcd} F^{ab} F^{cd} = \partial^a \left( \epsilon_{abcd} A^b F^{cd} \right).$$ Since adding a total derivative to any Lagrangian doesn't change the classical equations of motion, it doesn't matter if this invariant is in the Lagrangian or not, and it's customary to just leave it out.

(At the quantum level there are interesting physically observable phenomena that can arise from total-derivative terms, but that's a separate question and one I'm not as qualified to answer.)

You can add this to the Lagrangian if you want, but it will have no effect whatsoever. Try running the Lagrangian with the extra term through the Euler-Lagrange equation; it's a bit tedious, but you'll see it has no effect on the equations of motion. The reason why is that this term can be written as a total derivative (see this question), and two Lagrangians differing by the total derivative of a function will describe the same physical system (i.e. will return the same equations of motion).

$$\boldsymbol{\S}$$ A. The function $$\,\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\,$$ as a total derivative

We'll prove at an elementary level that for the electromagnetic field the Lorentz invariant scalar function $$\,\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\,$$ is the 4-divergence of a 4-dimensional vector function. So adding this scalar to the Lagrangian density of the field doesn't change the equations of motion, that is the Maxwell equations.

From the expressions of $$\,\mathbf E,\mathbf B\,$$ in terms of the scalar and vector potentials $$\phi,\mathbf A\,$$ \begin{align} \mathbf E & \boldsymbol{=}\boldsymbol{-}\boldsymbol{\nabla}\phi\boldsymbol{-}\dfrac{\partial \mathbf A}{\partial t} \tag{01a}\label{01a}\\ \mathbf B & \boldsymbol{=} \boldsymbol{\nabla}\boldsymbol{\times}\mathbf A \tag{01b}\label{01b} \end{align} we have $$$$\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\boldsymbol{=}\boldsymbol{-}\underbrace{\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\left(\dfrac{\partial \mathbf A}{\partial t}\right)}}_{\boldsymbol{\boxed{1}}}\boldsymbol{-}\underbrace{\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)}_{\boldsymbol{\boxed{2}}} \tag{02}\label{02}$$$$ Our target would be to find, if there exist, a real scalar function $$\,\eta\,$$ and a real 3-vector function $$\,\boldsymbol{\xi}\boldsymbol{=}\left(\xi^1,\xi^2,\xi^3\right)\,$$ , that is a 4-dimensional vector function $$\,\boldsymbol{\Xi}\boldsymbol{=}\left(\xi^1,\xi^2,\xi^3,\eta\right)\,$$ such that it yields the equality $$$$\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\boldsymbol{=}\dfrac{\partial \eta}{c\partial t}\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\boldsymbol{\xi}\boldsymbol{=}\partial_{\mu}\Xi^{\mu} \tag{03}\label{03}$$$$ In the following we'll make use of the identity $$$$\boxed{\:\:\boldsymbol{\nabla}\boldsymbol{\cdot}\left(\mathbf a\boldsymbol{\times}\mathbf b\right)\boldsymbol{=}\mathbf b\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf a\right)\boldsymbol{-}\mathbf a\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf b\right)\vphantom{\dfrac{a}{b}}\:\:} \tag{04}\label{04}$$$$ Using the identity \eqref{04} with $$\,\mathbf a\boldsymbol{\equiv}\boldsymbol{\nabla}\phi\,$$ and $$\,\mathbf b\boldsymbol{\equiv}\mathbf A\,$$ we have $$$$\boldsymbol{\boxed{1}}\boldsymbol{=}\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\left(\dfrac{\partial \mathbf A}{\partial t}\right)}\boldsymbol{=}\mathbf A\boldsymbol{\cdot}\underbrace{\left[\boldsymbol{\nabla}\boldsymbol{\times} \left(\boldsymbol{\nabla}\phi\right)\vphantom{\dfrac{a}{b}}\right]}_{\boldsymbol{0}}\boldsymbol{-}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\times}\mathbf A \vphantom{\dfrac{a}{b}}\right] \tag{05}\label{05}$$$$ that is $$$$\boldsymbol{\boxed{1}}\boldsymbol{=}\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\left(\dfrac{\partial \mathbf A}{\partial t}\right)}\boldsymbol{=}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\boldsymbol{-}\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\times}\mathbf A \vphantom{\dfrac{a}{b}}\right] \tag{06}\label{06}$$$$ Now $$$$\dfrac{\partial }{\partial t}\left[\mathbf A\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{=}\underbrace{\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)}_{\boldsymbol{\boxed{2}}}\boldsymbol{+}\mathbf A\boldsymbol{\cdot}\left[\boldsymbol{\nabla}\boldsymbol{\times}\left(\dfrac{\partial \mathbf A}{\partial t}\right)\right] \tag{07}\label{07}$$$$ From identity \eqref{04} with $$\,\mathbf a\boldsymbol{\equiv}\left(\dfrac{\partial \mathbf A}{\partial t}\right)\,$$ and $$\,\mathbf b\boldsymbol{\equiv}\mathbf A\,$$ $$$$\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\times}\mathbf A\right]\boldsymbol{=}\mathbf A\boldsymbol{\cdot}\left[\boldsymbol{\nabla}\boldsymbol{\times}\left(\dfrac{\partial \mathbf A}{\partial t}\right)\right]\boldsymbol{-}\underbrace{\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)}_{\boldsymbol{\boxed{2}}} \tag{08}\label{08}$$$$ Subtracting equations \eqref{07},\eqref{08} side-by-side yields $$$$\boldsymbol{\boxed{2}}\boldsymbol{=}\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\boldsymbol{=}\dfrac{\partial }{\partial t}\left[\frac12\mathbf A\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{-}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\frac12\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\times}\mathbf A\right] \tag{09}\label{09}$$$$ that is $$$$\boldsymbol{\boxed{2}}\boldsymbol{=}\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\boldsymbol{=}\dfrac{\partial }{c\partial t}\left[\frac12 c \mathbf A\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{-}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\frac12\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\times}\mathbf A\right] \tag{10}\label{10}$$$$ From equations \eqref{02},\eqref{06} and \eqref{10} we have $$$$\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right) \boldsymbol{=}\boldsymbol{-}\underbrace{\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\left(\dfrac{\partial \mathbf A}{\partial t}\right)}}_{\boldsymbol{\boxed{1}}}\boldsymbol{-}\underbrace{\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)}_{\boldsymbol{\boxed{2}}}\qquad \boldsymbol{\Longrightarrow} \nonumber$$$$ $$$$\boxed{\:\:\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\boldsymbol{=} \dfrac{\partial }{c\partial t}\left[\boldsymbol{-}\frac12 c \mathbf A\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]\boldsymbol{+}\boldsymbol{\nabla}\boldsymbol{\cdot}\left[\frac12\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\times}\mathbf A\boldsymbol{+}\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\times}\mathbf A\right]\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{11}\label{11}$$$$

So the Lorentz invariant scalar function $$\,\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\,$$ is the 4-divergence of the following 4-dimensional vector function $$$$\boxed{\:\:\boldsymbol{\Xi} \boldsymbol{=}\left(\boldsymbol{\xi},\eta \right)\boldsymbol{=} \Biggl(\left[\frac12\left(\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\times}\mathbf A\boldsymbol{+}\left(\boldsymbol{\nabla}\phi\right)\boldsymbol{\times}\mathbf A\right],\left[\boldsymbol{-}\frac12 c \mathbf A\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)\vphantom{\dfrac{a}{b}}\right]\Biggr)\:\:} \tag{12}\label{12}$$$$

$$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$$

$$\boldsymbol{\S}$$ B. The function $$\,\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\,$$ as Lagrange density producing identically zero equations of motion

Motivated by John Dumancic's answer I give the proof of the above conclusion

So, consider that the Lagrangian density $$\,\mathcal{L}$$ is only this function $$$$\mathcal{L}\boldsymbol{=}\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\boldsymbol{=}\left(\boldsymbol{-}\boldsymbol{\nabla}\phi\boldsymbol{-}\dfrac{\partial \mathbf A}{\partial t}\right)\boldsymbol{\cdot}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right) \tag{C-01}\label{C-01}$$$$

We must consider this density as function of the four ''field coordinates'', the components of the electromagnetic 4-vector $$$$\mathcal A\boldsymbol{=}\left(A_0,A_1,A_2,A_3\right)\boldsymbol{=}\left(\phi,\mathbf A\right) \tag{C-02}\label{C-02}$$$$ and their time and space 1rst order derivatives so that $$$$\mathcal{L}\left(A_{\jmath}, \dfrac{\partial A_{\jmath}}{\partial t}, \dfrac{\partial A_{\jmath}}{\partial x_k}\right) \qquad \left(\jmath=0,1,2,3\right) \qquad \left(k=1,2,3\right) \tag{C-03}\label{C-03}$$$$ We express the Lagrangian density of equation \eqref{C-01} in terms of these coordinates \begin{align} \mathcal{L}\boldsymbol{=}&\boldsymbol{-}\left(\dfrac{\partial \phi}{\partial x_1}\boldsymbol{+}\dfrac{\partial A_1}{\partial t}\right)\left(\dfrac{\partial A_3}{\partial x_2}\boldsymbol{-}\dfrac{\partial A_2}{\partial x_3}\right)\boldsymbol{-} \left(\dfrac{\partial \phi}{\partial x_2}\boldsymbol{+}\dfrac{\partial A_2}{\partial t}\right)\left(\dfrac{\partial A_1}{\partial x_3}\boldsymbol{-}\dfrac{\partial A_3}{\partial x_1}\right) \nonumber\\ &\boldsymbol{-} \left(\dfrac{\partial \phi}{\partial x_3}\boldsymbol{+}\dfrac{\partial A_3}{\partial t}\right)\left(\dfrac{\partial A_2}{\partial x_1}\boldsymbol{-}\dfrac{\partial A_1}{\partial x_2}\right) \tag{C-04}\label{C-04} \end{align}

The Euler-Lagrange equations of motion are $$$$\frac{\partial }{\partial t}\left[\frac{\partial \mathcal{L}}{\partial \left(\dfrac{\partial A_{\jmath}}{\partial t}\right)}\right]\boldsymbol{+}\sum_{k\boldsymbol{=}1}^{k\boldsymbol{=}3}\frac{\partial }{\partial x_{k}}\left[\frac{\partial \mathcal{L}}{\partial \left(\dfrac{\partial A_{\jmath}}{\partial x_{k}}\right)}\right]\boldsymbol{-} \frac{\partial \mathcal{L}}{\partial A_{\jmath}}\boldsymbol{=}0 \qquad \left(\jmath=0,1,2,3\right) \tag{C-05}\label{C-05}$$$$

For $$\jmath\boldsymbol{=}0\:( A_0\boldsymbol{=}\phi)$$ we have $$$$\frac{\partial }{\partial t}\underbrace{\left(\frac{\partial \mathcal{L}}{\partial \overset{\:\,_{\boldsymbol \cdot}}{\phi}}\right)}_{0}\boldsymbol{+}\underbrace{\boldsymbol{\nabla}\boldsymbol{\cdot}\overbrace{\left(\frac{\partial \mathcal{L}}{\partial \boldsymbol{\nabla}\phi\vphantom{\overset{\:\,_{\boldsymbol \cdot}}{\phi}}}\right)}^{\boldsymbol{-}\left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf A\right)}}_{0}\boldsymbol{-} \underbrace{\frac{\partial \mathcal{L}}{\partial \phi\vphantom{\overset{\:\,_{\boldsymbol \cdot}}{\phi}}}}_{0}\boldsymbol{=}0 \tag{C-06}\label{C-06}$$$$ that is the lhs is an identically zero term. This happens for the rest three equations, for example for $$\jmath\boldsymbol{=}1$$ we have $$$$\underbrace{\frac{\partial }{\partial t}\overbrace{\left(\frac{\partial \mathcal{L}}{\partial \overset{\:\,_{\boldsymbol \cdot}}{A}_1}\right)}^{\dfrac{\partial A_2}{\partial x_3}\boldsymbol{-}\dfrac{\partial A_3}{\partial x_2}}}_{\dfrac{\partial^2 A_2}{\partial t\partial x_3}\boldsymbol{-}\dfrac{\partial^2 A_3}{\partial t\partial x_2}}\boldsymbol{+}\underbrace{\boldsymbol{\nabla}\boldsymbol{\cdot}\overbrace{\left(\frac{\partial \mathcal{L}}{\partial \boldsymbol{\nabla}A_1\vphantom{\overset{\:\,_{\boldsymbol \cdot}}{\phi}}}\right)}^{ \begin{bmatrix} 0\\ \dfrac{\partial \phi}{\partial x_3}\boldsymbol{+}\dfrac{\partial A_3}{\partial t}\\ \boldsymbol{-}\dfrac{\partial \phi}{\partial x_2}\boldsymbol{-}\dfrac{\partial A_2}{\partial t} \end{bmatrix} }}_{\dfrac{\partial^2 A_3}{\partial x_2\partial t}\boldsymbol{-}\dfrac{\partial^2 A_2}{\partial x_3\partial t}}\boldsymbol{-} \underbrace{\frac{\partial \mathcal{L}}{\partial A_1\vphantom{\overset{\:\,_{\boldsymbol \cdot}}{\phi}}}}_{0}\boldsymbol{=}0 \tag{C-07}\label{C-07}$$$$ that is a lhs identically zero too. Similarly for $$\jmath\boldsymbol{=}2,3$$.

Conclusion : The function $$\,\left(\mathbf E\boldsymbol{\cdot}\mathbf B\right)\,$$ as Lagrangian density alone produces identically zero equations of motion. So, adding it to any Lagrangian density of the electromagnetic field has no effect to the equations of motion.