When considering whether or not the total energy of a system varies over time, we consider the time dependence of the Lagrangian, namely by looking at $$\frac{\partial L}{\partial t} \, .$$
Why is it that we only consider explicit time dependence in this case, and not the dependence of each variable within the Lagrangian? My notes on this topic are fairly vague and I just wondered if there was a simple reason.
Short answer: it can be shown that the Hamiltonian satisfies $$\frac{dH}{dt}=\frac{\partial H}{\partial t}=-\frac{\partial L}{\partial t}.$$Long answer: since $H=\dot{\mathbf{q}}\cdot\mathbf{p}-L$ we have $\partial_t H=-\partial_t L$, and by Hamilton's equations $$\frac{dH}{dt}-\frac{\partial H}{\partial t}=\frac{dH}{d\mathbf{q}}\cdot\dot{\mathbf{q}}+\frac{dH}{d\mathbf{p}}\cdot\dot{\mathbf{p}}=\frac{dH}{d\mathbf{q}}\cdot\frac{dH}{d\mathbf{p}}-\frac{dH}{d\mathbf{p}}\cdot\frac{dH}{d\mathbf{q}}=0.$$
\begin{align} \frac{\mathrm dE}{\mathrm dt} &\equiv \frac{\mathrm d}{\mathrm dt} \left(\frac{\partial L}{\partial \dot q}\dot q - L\right) \\&= \left(\frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot q} \right) \dot q + \frac{\partial L}{\partial \dot q} \ddot q - \frac{\partial L}{\partial q} \dot q - \frac{\partial L}{\partial \dot q} \ddot q - \frac{\partial L}{\partial t} \\&= \left( \frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} \right) \dot q - \frac{\partial L}{\partial t} \\&\overset{\text{eom}}= - \frac{\partial L}{\partial t} \end{align}
That being said, I'd quite like to know the physical reason behind this!
– arcturus7 Apr 09 '17 at 16:31