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

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 $$\dfrac{\partial L}{\partial t} = 0$$ ?

• Which textbook? Which page? May 20, 2020 at 12:00
• I'm french, so it is a ENS course (phys.ens.fr/~hare/FIP/Meca_anal_Hare_2007.pdf section 3.1), but I found the same proof in english in mechanics by Landau and Lifshitz (page 5) May 20, 2020 at 12:16
• May 20, 2020 at 13:24
• $\dfrac{\partial L}{\partial t} = 0$ is only if L is not a function of t
– Eli
May 20, 2020 at 14:45
• yes but that's what I want to prove: one starts with the most general expression of the Lagrangian $L(x(t), \dot x(t), t)$ and considering a free particle and using the time translation invariance, one should deduct that $L$ does not explicitly depends on $t$ (renaissance.ucsd.edu/courses/mae207/mech.pdf - page 5 is the argumentation but not real proof) May 20, 2020 at 15:28

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.
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.