The Hamiltonian of a system is defined as $$H=\sum_ip_i\dot{q_i}-L$$ where $L=T-V$.

I have read thatin the cases where the Hamiltonian doesn't explicitly depend on time, we have $$H=E_{total}=T+V$$ But try as I might, I can't figure out how to prove it. Any help?

  • $\begingroup$ Look here: physics.stackexchange.com/questions/283165/… (By the way, the hamiltonian is not in general equivalent to energy when it is time independent, see for example the hamiltonian of a charged particle in an EM field: en.wikipedia.org/wiki/… $\endgroup$ Jan 17 '18 at 23:24
  • 2
    $\begingroup$ In classical non-relativistic mechanics, when $L = T - V$ and $T$ is a quadratic function of the velocities, $H = p_i \dot{q}_i - L = \frac{\partial L}{\partial \dot{q}^i} \dot{q}^i - (T - V) = 2T - (T - V) = T + V$. Noether's theorem says that $p_i \dot{q}_i - L = E$ is constant for time-independent Lagrangians, and the Legendre transform says $H = p_i \dot{q}_i - L$, so $H = E$. c.f. Landau Mechanics Sec. 6, 40. $\endgroup$
    – bolbteppa
    Jan 17 '18 at 23:39
  • $\begingroup$ Interesting. If it isn't too much to ask, how do you prove that $T$ is always a quadratic function of velocities? $\endgroup$ Jan 18 '18 at 0:57
  • 1
    $\begingroup$ It isn't in physics in general, but in some cases for a particle Landau gives some reasons why it should be quadratic using symmetries, check the first 5 sections of that book. $\endgroup$
    – bolbteppa
    Jan 18 '18 at 2:42