Hamilton principle and classical Electromagnetic field Lagrangian

Question

Hamilton's principle states that the Lagrangian for a system is $L=T-V$ where $T$ is kinetic energy and $U$ is potential energy. Lagrangian can also be defined as $L=\int \mathcal{L}\ d^3{x}$ where $\mathcal{L}$ is the Lagrangian density. In many text books I have seen that the Lagrangian density for the classical Electromagnetic fields is stated as \begin{equation} \mathcal{L}=-\rho(x,t)\phi(x,t)+J(x,t)\cdot A(x,t)+\dfrac{\varepsilon_{0}}{2}E^2(x,t)-\dfrac{1}{2\mu_{0}}B^2(x,t).\tag{1} \end{equation} I understand the first two terms, but why does the third term has positive sign? It should be negative according to the Hamilton principle, Shouldn't it? Because fields energy density is $$u=\dfrac{\varepsilon_{0}}{2}E^2(x,t)+\dfrac{1}{2\mu_{0}}B^2(x,t).\tag{2}$$

Answer 1

1. The positive sign of the $\frac{\varepsilon_{0}}{2}\vec{E}^2$ term in the Lagrangian density (1) is correct because it (among other things) contains the kinetic energy term $\frac{\varepsilon_{0}}{2}\dot{\vec{A}}^2$, which should be positive, cf. e.g. this Phys.SE post.

   In contrast, the $\frac{1}{2\mu_{0}}\vec{B}^2$ term is a potential term, so it should come with a minus sign in the Lagrangian density (1).

2. In the Hamiltonian formulation the electric field $\vec{E}$ becomes the corresponding momentum, so the $\frac{\varepsilon_{0}}{2}\vec{E}^2$ term should also be positive in the Hamiltonian density (2).