Why is energy (or in more general terms,the Hamiltonian) not conserved when the Lagrangian has an explicit time dependence?
I know that we can derive the identity:
$\frac{d \mathcal{H}}{d t} = - {\partial \mathcal{L}\over \partial t}$
but is there a more physical and intuitive explanation for the conservation?
I) If a system (and hence the Lagrangian) depends explicitly on time, it can often be physically interpreted as that the system is not an isolated system. In other words, that it interacts with the environment, and hence there is no reason that the energy of the system should be conserved.
Example: A non-relativistic 1D point particle with Lagrangian $$\tag{1} L~=~\frac{1}{2}m \dot{q}^2 -V(q,t),$$ where the potential
$$\tag{2} V(q,t)~=~V_0(t) - F(t) q +\frac{1}{2}k(t)q^2$$
is quadratic in $q$ and the three coefficients $V_0(t)$, $- F(t)$ and $k(t)$ depend explicitly on time. We may interpret:
$V_0(t)$ as a fluctuating (choice of) zero-point energy. This is a total derivative term in the Lagrangian, and does not affect the equations of motion.
$F(t)$ as an external force.
$k(t)$ as a changing spring constant.
II) On the other hand, if the Lagrangian does not depend explicitly on time, so that time translation is a symmetry of the Lagrangian, then Noether's (first) theorem yields that the corresponding Noether charge (=the energy) is conserved.
The time dependent hamiltonian also implies, that the volume element in the phase space is not conserved under time evolution.
