When proving that the homogeneity of time leads to the conservation of energy,
(This is the proof from Landau for the case when there is no field present.)
(Uses the Einstein's summation convention)
$$\frac{dL}{dt} = \frac{\partial L}{\partial q_i} \dot{q_i} + \frac{\partial L}{\partial \dot{q_i}} \ddot{q_i} + \frac{\partial L}{\partial t}$$
But, $\partial L / \partial t = 0$ assuming the homogeneity of time and so,
$$\frac{dL}{dt} = \frac{\partial L}{\partial q_i} \dot{q_i} $$
Using the Lagrange-Euler equation, (summing over all possible $i$'s)
$$\frac{\partial L}{ \partial q_i} = \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{q_i}} \Big)$$
and so,
$$\frac{dL}{dt} = \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{q_i}} \Big) \dot{q_i} $$
$$0 = \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{q_i}} \dot{q_i} – L\Big) $$
which gives,
$$\frac{d(E)}{dt} = 0$$
One assumption not mentioned above, that I believe is important, is that $\ddot{q_i} = 0$ which makes sense because in the lack of a field, a particle won't accelerate which again ties back to the statement that $L$ does not depend on time as then $\dot{q_i}$ would be constant and so $L = T – U = T-0 = T$ where $T$ is a quadratic function of $\dot{q_i}$ which is a constant would lead to the proof as shown above.
However, when there is a field $U = U(q_i)$ present, even when it is constant and does not depend on time, I'm not sure how $L$ should also be independent of time. Because now that a field is present, $\dot{q}$ would depend on time and so $\ddot{q} \not = 0$ and so I'm confused as to how to proceed with the proof. The situation is further complicated by the fact that because $L$ depends on $\dot{q}$ and because $\dot{q}$ depends on time, $L$ should depend on time and so the assumption that $\partial L / \partial t = 0$ wouldn't hold true.
Any pointers would be really appreciated.
Best Answer
Well I don't think you necessarily need that assumption.
Begin as you did with
$$\frac{dL}{dt}=\frac{\partial L}{\partial q}\dot q +\frac{\partial L}{\partial \dot q}\ddot q +\frac{\partial L}{\partial t}$$
And with $\frac{\partial L}{\partial t}=0$ by assumption.
Now using the Euler Lagrange replace $\frac{\partial L}{\partial q}$ by $\frac{d}{dt}\frac{\partial L}{\partial \dot q}$. The we have:
$$\frac{dL}{dt}=\frac{d}{dt}\left (\frac{\partial L}{\partial \dot q}\right)\dot q +\frac{\partial L}{\partial \dot q}\ddot q$$
And we recognise the Rhs by the product rule to be $\frac{d}{dt}\left (\frac{\partial L}{\partial \dot q}\dot q\right)$
Which implies that
$$\frac{d}{dt}\left (\frac{\partial L}{\partial \dot q}\dot q -L\right)=0$$