How to prove time translation invariance of Lagrangian for a free particle?

Question

In my textbook, the author deduce the expression of the lagrangian $L(q_i(t), \dot q_i(t), t )$ of a free particle only using classical physical symmetries where the $q_i(t)$ are independent coordinates and the $\dot q_i(t)$ their derivatives by time. To simplify let's just pretend $q_i(t) = x(t)$, and the lagrangian becomes $L(x(t), \dot x(t), t)$.

The proof begins with explaining that "the homogeneity of time implies that the Lagrangian cannot contain explicitly the time $t$".

I would like the mathematical proof of this statement but I am struggling. I have to prove that $\dfrac{\partial L}{\partial t} = 0$.

Let $t'$ be the image of $t$ by a translation $t' = t + dt$ with $dt$ an infinitesimal time duration. This should induce $x$ and $\dot x$ variations:

$$ L(x(t'), \dot x(t'), t') = L(x(t) + \delta x, \dot x(t) + \delta \dot x, t + d\tau) $$

Assuming lagrangian is time translation invariant, $L(x(t'), \dot x(t'), t') = L(x(t), \dot x(t), t) \quad (*)$.

Using first order Taylor's expansion: $$ L(x(t'), \dot x(t'), t') = L(x(t), \dot x(t), t) + \dfrac{\partial L}{\partial x}\delta x + \dfrac{\partial L}{\partial \dot x}\delta \dot x + \dfrac{\partial L}{\partial t} dt + \mathcal{o}(\lvert\lvert (\delta x, \delta \dot x, dt)\rvert\rvert) $$

by neglecting higher order terms, $(*)$ becomes :

$$ L(x(t), \dot x(t), t) + \dfrac{\partial L}{\partial x}\delta x + \dfrac{\partial L}{\partial \dot x}\delta \dot x + \dfrac{\partial L}{\partial t} dt = L(x(t), \dot x(t), t) $$

$$ \dfrac{\partial L}{\partial x}\delta x + \dfrac{\partial L}{\partial \dot x}\delta \dot x + \dfrac{\partial L}{\partial t} dt = 0 $$

How can I deduct that $\dfrac{\partial L}{\partial t} = 0$ ?

Answer 1

Eq.(*) means that the Lagrangian is invariant under both time and space translation. Since you are assuming that the Lagrangian is only time translation invariant, that equation should've been \begin{equation} \label{eq:time} L(x(t'),\dot x(t'),t')=L(x(t'),\dot x(t'),t) \end{equation} which implies that $$L(x(t'),\dot x(t'),t)+{\partial L \over \partial t}dt=L(x(t'),\dot x(t'),t)$$ And immediate calculation proves the quoted statement.

Answer 2

Remember the definition of the partial derivative. $$\frac{\partial}{\partial t}L(x,y,t)=\lim_{h\rightarrow 0}\frac{L(x,y,t +h)-L(x,y,t)}{h}$$ Since the Lagrangian doesn't depend on $t$ explicitly we have that $$L(x,y,t')=L(x,y,t)\quad\text{ for all }t'$$ meaning the numerator of this limit is zero and consequently $\frac{\partial L}{\partial t}=0$. In this part of classical mechanics it is hard to keep track of all the derivatives, but the partial derivative is really simple in this regard.

Side note: I used $y$ in this expression to emphasize that the Lagrangian is just a function with three arguments. $L: \mathbb R^3\rightarrow\mathbb R$. The third argument is used for $t$ but nothing prevents you from also using $t$ in the other arguments. You could have something like $L(e^t,t^2-3t,t)$. But when you calculate the partial derivative you only use one of these arguments. The total derivative uses all the arguments.