I have a homework question asking me to prove that $\frac{\partial \mathcal{H}}{\partial t}=-\frac{\partial\mathcal{L}}{\partial t}$, and the hint is to first consider a system with one degree of freedom where the equation for the Hamiltonian simplifies to $\mathcal{H}(q,p,t)=p\dot{q}(q,p,t)-\mathcal{L}(q,\dot{q}(q,p,t),t)$. Now I tried to take the partial derivative with respect to time but I'm not sure that I'm doing it right, for my first step I get $$\frac{\partial\mathcal{H}}{\partial t}=p\frac{\partial \dot{q}(q,p,t)}{\partial t}\frac{\partial q}{\partial t}-\frac{\partial \mathcal{L}(q,\dot{q}(q,p,t),t)}{\partial t}\frac{\partial q}{\partial t}\frac{\partial \dot{q}(q,p,t)}{\partial t}\frac{\partial q}{\partial t}$$
I'll spare you the details of simplifying, but I get stuck at $$\frac{\partial\mathcal{H}}{\partial t}=-\frac{\partial \mathcal{L}}{\partial t}\Big(\frac{\partial ^2q}{\partial t^2}\frac{\partial\dot{q}}{\partial t}-\frac{\partial q}{\partial t}\Big)$$
So in theory, the stuff in the parenthesis should equal one if I'm doing this right, but I'm unsure if I am. Any help would be greatly appreciated!
1 Answers
Your opening non inline equation is also incorrect on dimensional grounds, the first factor by $LT^{-1}$ and the second by $L^3 T^{-4}$.
For a one dimensional system as you are considering here, $H=H(q,p,t)$, and so one has $$dH = \frac{\partial H}{\partial q}dq + \frac{\partial H}{\partial p}dp + \frac{\partial H}{\partial t}dt\,\,\,\,\,\,(1)$$ generically.
From the explicit realisation of the hamiltonian function as a legendre transform of the lagrangian, as you have written in your question, we also find $$dH = p \,d \dot{q} + dp \, \dot q - \left(\frac{\partial L}{\partial q}dq + \frac{\partial L}{\partial \dot q}d \dot q + \frac{\partial L}{\partial t}dt\right) $$
This can be simplified in favour of writing the single generalised momenta in terms of the generalised velocity $p = \partial L/\partial \dot q$ to obtain $$dH = dp \, \dot q - \frac{\partial L}{\partial q}dq - \frac{\partial L}{\partial t}dt\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$
Comparing $(1)$ and $(2)$ will give you your desired relation, in addition to two other equations which collectively comprise Hamilton's equations.
You can also proceed more directly using your legendre transform. Since there is explicit and implicit instances of $t$ in the legendre transform, it is fruitful to extract a total derivative of $H$ with respect to $t$. So, $$\frac{d H}{dt} = p \frac{d \dot q}{dt} - \frac{dL}{dt} = p \frac{d \dot q}{dt} - \left(\frac{\partial L}{\partial \dot q} \frac{d \dot q}{dt} + \frac{\partial L}{\partial t} \right) = -\frac{\partial L}{\partial t}.$$
Using $(1)$ together with the Hamilton's equations, you find in fact $dH/dt = \partial H/\partial t$ obtaining again $$\frac{\partial H}{\partial t} = -\frac{\partial L}{\partial t}$$
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