Euler-Lagrange equations using $\vec{E}$ and $\vec{B}$ instead of $A^\mu$

Question

We all know that the lagrangian for the free electromagnetic field is given by $$ \mathscr{L} = -\frac{1}{4}F^{\mu \nu}F_{\mu \nu} $$ where $ F^{\mu \nu} = \partial^\mu A^\nu -\partial^\nu A^\mu $ is the electromagnetic field tensor. But also we know that

Lets consider $c=1$ for simplicity. Then, doing the math, the lagrangian can be written as $$ \mathscr{L} = \frac{1}{2} (|\vec{E}|^2 - |\vec{B}|^2) $$

By applying Euler-Lagrange, i.e. $$ \partial_\mu \left( \frac{\partial \mathscr{L}}{\partial (\partial_\mu \phi_i)} \right) - \frac{\partial \mathscr{L}}{\partial \phi_i} = 0 $$ where $\phi_i$ is each of the components of each field, I find $$ \vec{E} = 0 $$ and $$ \vec{B} = 0 $$ but not the Maxwell equations... What is going on?

Answer 1

$\vec{E}(x,t)$ and $\vec{B}(x,t)$ are not totally independent variables.

---

I'm not familar with field theory, but have simpler example with LC circuit, a zero dimensional field theory:

$$ H=T+V=\frac{L}{2}I^2+\frac{1}{2C}Q^2 \quad \sim \quad \frac{1}{2}(|\vec{B}|^2+|\vec{E}|^2) $$

$$ L=T-V=\frac{L}{2}\dot{Q}^2-\frac{1}{2C}Q^2 \quad \sim \quad \frac{1}{2}(|\vec{B}|^2-|\vec{E}|^2) $$

---

Charge $Q$ is the dynamic variable. $|\vec{E}|$ is propotional to $Q$, $|\vec{B}|$ is propotional to $\dot{Q}=I$,

You can't take current $I$ as independent from $Q$. This is the constraint of the system.

---

---

If you insist to take $I$ and $Q$ as independent variable, then you get a different system: a capacitor and a inductor seperately