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

  • $\begingroup$ Which textbook? Which page? $\endgroup$
    – Qmechanic
    Commented May 20, 2020 at 12:00
  • $\begingroup$ 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) $\endgroup$
    – Lyders
    Commented May 20, 2020 at 12:16
  • $\begingroup$ Possible duplicate: How is it possible to vary time without affect the coordinates or their derivatives? $\endgroup$
    – Qmechanic
    Commented May 20, 2020 at 13:24
  • $\begingroup$ $\dfrac{\partial L}{\partial t} = 0$ is only if L is not a function of t $\endgroup$
    – Eli
    Commented May 20, 2020 at 14:45
  • $\begingroup$ 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) $\endgroup$
    – Lyders
    Commented May 20, 2020 at 15:28

2 Answers 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.


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.


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