How to prove that $dh/dt = ∂h/∂t$ in Lagrangian mechanics?

Question

We already know that the energy function $h(q,\dot{q},t)$ (not the Hamiltonian!) in classical mechanics follows the equation $$dh/dt = −∂L/∂t\tag{1}$$ but how can we show that $$dh/dt = ∂h/∂t\tag{2}$$ is also true?

Answer 1

In Lagrangian mechanics (as opposed to Hamiltonian mechanics), we calculate: $$ \frac{\partial h}{\partial t}-\frac{d h}{d t} ~\stackrel{\text{EL eqs.}}{\approx}~\frac{\partial (h+L)}{\partial t}~=~\frac{\partial (\dot{q}^ip_i)}{\partial t}~=~\dot{q}^i\frac{\partial^2 L}{\partial t~\partial \dot{q}^i}. $$ There is no generic reason why this should be zero.

Answer 2

This can simply follow from writing down the total derivative $\text{d}h/\text{d}t$, where $h=h(q,p,t)$, with the imposition of Hamilton’s equations.

See e.g Proving one of the canonical Hamilton equations for more elaborate discussion.

Answer 3

You write down the total derivative for $\dfrac{dH}{dt}$, then replace the partials via Hamilton's equations:

$$ \dfrac{dH}{dt} = \Sigma_i\Big(\dfrac{\partial H}{\partial p_i}\dfrac{dp_i}{dt}-\dfrac{\partial H}{\partial q_i}\dfrac{dq_i}{dt}\Big) - \dfrac{\partial H}{\partial t} $$ Insert Hamilton's equations: $\dot q_i = \partial H/ \partial p_i$, $ \dot p_i = -\partial H/\partial q_i$ $$ \dfrac{dH}{dt} = \Sigma_i\Big(\dot q_i\dot p_i - \dot p_i \dot q_i\Big) - \dfrac{\partial H}{\partial t} $$ $$ \dfrac{dH}{dt} = \dfrac{\partial H}{\partial t} $$