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
    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
    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
    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
    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
    May 20, 2020 at 15:28

2 Answers 2


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.