I have a problem deriving the conservation of energy from time translation invariance. The invariance of the Lagrangian under infinitesimal time displacements $t \rightarrow t' = t + \epsilon$ can be written as \begin{equation} \delta L = L\left( q(t),\frac{dq(t)}{dt},t\right) - L\left( q(t+ \epsilon),\frac{dq(t+ \epsilon)}{dt},t+\epsilon \right) = 0. \end{equation} Using Taylor series, keeping only first order terms this gives \begin{equation}\rightarrow \delta L =- \frac{\partial L }{\partial q} \frac{\partial q}{\partial t} \epsilon- \frac{\partial L }{\partial \dot{q}} \frac{\partial \dot{q}}{\partial t} \epsilon - \frac{\partial L }{\partial t} \epsilon = 0. \end{equation} Using the Euler-Lagrange equation and assuming that the Lagrangian does not depend explicitly on time we get \begin{equation}\rightarrow \delta L =- \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \right) \frac{\partial q}{\partial t} \epsilon- \frac{\partial L }{\partial \dot{q}} \frac{\partial \dot{q}}{\partial t} \epsilon =0. \end{equation} Which we can write as \begin{equation}\rightarrow \delta L = - \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \frac{\partial q}{\partial t} \right) \epsilon = - \frac{d}{d t} \left(p \frac{\partial q}{\partial t} \right) \epsilon = 0. \end{equation} But unfortunatly this is not the Hamiltonian. This computation should yield \begin{equation} \rightarrow \frac{d}{dt} \left( p \dot{q} - L \right) = 0. \end{equation} But I can't find no reason why and how the the extra $-L$ should emerge. I can see that this term can be written at the place where it is written because we have $\delta L = - \frac{d L}{dt } \epsilon$ and therefore \begin{equation} \rightarrow \delta L = - \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \frac{\partial q}{\partial t} \right) \epsilon = - \frac{d L}{dt } \epsilon. \end{equation} And then the desired equation would only say $0-0=0$. Any idea where i did a mistake would be much appreciated.

  • $\begingroup$ I was currently reading this pdf staff.science.uu.nl/~aruty101/CFT.pdf $\endgroup$
    – jak
    Commented Jan 19, 2014 at 18:41
  • $\begingroup$ I would recommend the following pfd: physics.uc.edu/~vaz/lectures/mec.pdf . Section 10.4; it's quite detailed. $\endgroup$
    – Hunter
    Commented Jan 19, 2014 at 18:56
  • $\begingroup$ Thanks for your reading tip. On Page 229 it is simply stated that $\dot H = 0$ (as in most other sources) and i can prove this, too. My problem is that i can't see how this notion appears from time translation invariance, just as the momentum appears naturally from spatial translation invariance or the angular momentum from rotational invariance. $\endgroup$
    – jak
    Commented Jan 19, 2014 at 19:08
  • 1
    $\begingroup$ Hi JakobH & @Hunter: Both your links are now dead. $\endgroup$
    – Qmechanic
    Commented Dec 6, 2015 at 14:01

6 Answers 6


Reiterating pppqqq's answer, your error is right at the beginning where you set $\delta L = 0$. The Lagrangian is not a constant of motion, so this equation is fallacious.

Instead, you want

$\frac{dL}{dt} = \frac{\partial L}{\partial q} \dot{q} + \frac{\partial L}{\partial \dot{q}}\ddot{q}$

which assumes $\frac{\partial L}{\partial t} = 0$.

When you apply the Euler-Lagrange equation, you get

$\frac{dL}{dt} = \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}} \dot{q})$

which is just a short algebra step from showing that the Hamiltonian is conserved.

Your original derivation simply shows that if the Lagrangian is time-independent and if also it is a constant of motion, then $p \dot{q}$ is also a constant of motion.

  • $\begingroup$ By far the best answer! So clean and no need to do integrals. Highlights the core point, which is that what is essential is to use somewhere in the proof that L does not depend explicitly on time. $\endgroup$
    – Econ
    Commented Mar 25, 2022 at 20:05

I) Firstly, we mention that Noether's Theorem (in its original form) concerns a symmetry of the action $S$, not necessarily the Lagrangian $L$. The relevant notion for the Lagrangian is quasi-symmetry, cf. this Phys.SE answer.

II) Secondly, we make the assumption that

$$ \text{The Lagrangian } L=L(q,\dot{q}) \text{ has no }{\it explicit} \text{ time dependence.} \tag{1} $$

We would like to use Noether's theorem to prove that the energy function$^1$

$$ h~:=~p_i\dot{q}^i-L,\qquad p_i ~:=~\frac{\partial L}{\partial \dot{q}^i },\tag{2} $$

is then conserved on-shell

$$ \frac{dh}{dt}~\approx~0.\tag{3} $$ Hence we should identify the relevant symmetry. (Here the $\approx$ symbol means equality modulo eom. Observe btw that we will not use eom for the remainder of this answer. This is because the assumptions of Noether's theorem demand that the symmetry holds also for virtual off-shell configurations which violate eom.)

III) It is apparent from OP's first equation that he is considering an infinitesimal pure time translation

$$ t^{\prime} - t ~=:~\delta t ~=~-\varepsilon, \qquad \text{(horizontal variation)}\tag{A}$$ $$ q^{\prime i}(t) - q^i(t)~=:~\delta_0 q^i ~=~0, \qquad \text{(no vertical variation)}\tag{B}$$ $$ q^{\prime i}(t^{\prime}) - q^i(t)~=:~\delta q^i ~=~-\varepsilon\dot{q}^i. \qquad \text{(full variation)}\tag{C}$$

(The words horizontal and vertical refer to translation in the $t$ direction and the $q^i$ directions, respectively). Also note that we have changed the sign in front of $\varepsilon$ for later convenience. A pure time translation (A) is in general not a symmetry of the Lagrangian

$$ \delta L ~=~ \frac{dL}{dt}\delta t ~=~ -\varepsilon \frac{dL}{dt}~\neq~0.\tag{D} $$

The full explanation why the pure horizontal transformation (A)-(C) cannot be used to prove energy conservation is given in Section VI below. But first we show two other transformations that do work in the next Sections IV and V.

IV) If we change time (A), the values of $q^{i}$ and $\dot{q}^{i}$ will in general also change. In other words, we must introduce a compensating vertical variation (B'), so that the full variation (C') of the generalized positions are zero:

$$ t^{\prime} - t ~=:~\delta t ~=~-\varepsilon, \qquad \text{(horizontal variation)}\tag{A'}$$ $$ q^{\prime i}(t) - q^i(t)~=:~\delta_0 q^i ~=~\varepsilon\dot{q}^i, \qquad \text{(vertical variation)}\tag{B'}$$ $$ q^{\prime i}(t^{\prime}) - q^i(t)~=:~\delta q^i ~=~0. \qquad \text{(full variation)}\tag{C'}$$

The transformation (A') - (C') is a symmetry of the Lagrangian:

$$\begin{align} \delta L ~=~&\frac{\partial L}{\partial q^i }\delta_0 q^i + \frac{\partial L}{\partial \dot{q}^i }\delta_0 \dot{q}^i + \frac{dL}{dt}\delta t \cr ~=~&-\varepsilon\frac{\partial L}{\partial t }~=~0,\end{align} \tag{D'} $$

where we in the last equality used that the Lagrangian $L$ has no explicit time dependence.

Using the standard formula mentioned on Wikipedia, the (bare) Noether current (multiplied with $\varepsilon$) becomes the energy (multiplied with $\varepsilon$)

$$ \varepsilon j ~:=~ p_i \delta_0 q^i + L \delta t~=~ p_i \delta q^i - h \delta t~=~ \varepsilon h ,\tag{E'}$$

as we wanted to show.

V) Alternatively, as is done in Example 1 on Wikipedia, we can consider a purely vertical infinitesimal transformation

$$ t^{\prime} - t ~=:~\delta t ~=~0, \qquad \text{(no horizontal variation)}\tag{A''}$$ $$ q^{\prime i}(t) - q^i(t)~=:~\delta_0 q^i ~=~\varepsilon\dot{q}^i, \qquad \text{(vertical variation)}\tag{B''}$$ $$ q^{\prime i}(t^{\prime}) - q^i(t)~=:~\delta q^i ~=~\varepsilon\dot{q}^i. \qquad \text{(full variation)}\tag{C''}$$

The transformation (A'') - (C'') is a quasi-symmetry of the Lagrangian:

$$\begin{align} \delta L ~=~&\frac{\partial L}{\partial q^i }\delta_0 q^i + \frac{\partial L}{\partial \dot{q}^i }\delta_0 \dot{q}^i \cr ~=~&\varepsilon\frac{\partial L}{\partial q^i }\dot{q}^i + \varepsilon\frac{\partial L}{\partial \dot{q}^i } \ddot{q}^i\cr ~=~& \varepsilon\frac{dL}{dt}-\varepsilon\frac{\partial L}{\partial t}~=~ \varepsilon\frac{dL}{dt}, \end{align} \tag{D''}$$

where we in the last equality used that the Lagrangian $L$ has no explicit time dependence.

The (bare) Noether current (multiplied with $\varepsilon$) becomes

$$ \varepsilon j ~:=~ p_i \delta_0 q^i + L \delta t~=~ \varepsilon p_i\dot{q}^i.\tag{E''}$$

The Noether current must be corrected because of the appearance of the total time derivative in eq. (D''). The full Noether current becomes the energy function

$$ J~=~j-L~=~p_i\dot{q}^i-L~=~h,\tag{F''}$$

as we wanted to show.

VI) Finally, let us return to OP's pure horizontal transformation (A)-(C). While not a symmetry, it is still a quasi-symmetry of the Lagrangian $L$, cf. eq. (D). The (bare) Noether current (multiplied with $\varepsilon$) becomes

$$ \varepsilon j ~:=~ p_i \delta_0 q^i + L \delta t~=~ -\varepsilon L .\tag{E}$$

The Noether current must be corrected because of the appearance of the total time derivative in eq. (D). The full Noether current becomes zero:

$$ J~=~j-(-L)~=~-L+L~=~0.\tag{F}$$

In other words, the corresponding conservation law is a triviality! This is because we never used in eq. (D) the non-trivial fact (1) that the Lagrangian $L$ has no explicit time dependence.


$^1$ The energy function $h(q,\dot{q},t)$ in the Lagrangian formalism corresponds to the Hamiltonian $H(q,p,t)$ in the Hamiltonian formalism.

  • 1
    $\begingroup$ When you say vertical and horizontal do you refer to the way a vector bundle can be split up into vertical and horizontal components?en.wikipedia.org/wiki/Horizontal_bundle $\endgroup$ Commented Jul 24, 2015 at 9:39
  • 1
    $\begingroup$ The terminology is inspired by that. $\endgroup$
    – Qmechanic
    Commented Jul 24, 2015 at 10:13

Here's the right way to understand this (not that I'm biased or anything). Let me begin that I agree with others who point out that $\delta L \neq 0$ in this case, but I'd like to demonstrate why in a convincing manner. Hopefully the way I present the resolution will be clear. I'll be mathematically precise, but I won't worry about certain technical assumptions such as degrees of differentiability of the functions involved.


So that we can be absolutely sure that there is no confusion, let me review some notation and definitions.

Let a path $q:[t_a, t_b]\to \mathbb R$ in configuration space be given. Let $\hat q:[t_a, t_b]\times (\epsilon_a, \epsilon_b)\to \mathbb R$ be a one-parameter deformation of $q$ with $\epsilon_a<0<\epsilon_b$. We define the variation of $q$ and its derivative $\dot q$ with respect to this deformation as follows: \begin{align} \delta q(t) = \frac{\partial \hat q}{\partial\epsilon}(t,0) , \qquad \delta\dot q(t) = \frac{\partial^2\hat q}{\partial \epsilon\partial t}(t,0) \end{align} By the way, to get some intuition for this (and especially my notation), you might find the following post useful:

Lagrangian Mechanics - Commutativity Rule $\frac{d}{dt}\delta q=\delta \frac{dq}{dt} $

Now, suppose that a lagrangian $L$ that is local in $q$ and $\dot q$ is given, then for a given path $q$ we define its variation with respect to the deformation $\hat q$ as follows: \begin{align} \delta L(q(t), \dot q(t), t) = \frac{\partial}{\partial\epsilon}L\left(\hat q(t,\epsilon), \frac{\partial\hat q}{\partial t}(t,\epsilon), t\right)\Big|_{\epsilon=0} \end{align} From these two definitions, we find the following expression for the variation of the Lagrangian (where we suppress the arguments of functions for notational compactness) \begin{align} \delta L = \frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot q}\delta\dot q \end{align} We call a given deformation a symmetry of $L$ provided there exists a function $F$ that is local in paths $q$ such that \begin{align} \delta L(q(t), \dot q(t), t) = \frac{dF_q}{dt}(t) \end{align} for any $q$. In other words, a symmetry is a deformation that, to first order in the deformation parameter $\epsilon$, only changes the Lagrangian by at most a total time derivative. These definitions allows us to compactly write the following Lagrangian version of Noether's theorem

For every symmetry of the Lagrangian, the quantity \begin{align} Q_q(t) = \frac{\partial L}{\partial \dot q}(q(t), \dot q(t), t) \delta q(t) - F_q(t) \end{align} is conserved for all $q$ satisfying the Euler-Lagrange equations.

Time translation symmetry.

We consider the deformation \begin{align} \hat q(t,\epsilon) = q(t+\epsilon). \end{align} which, of course, we call time translation. Now, a short computation shows that under this deformation, one has the following variations: \begin{align} \delta q(t) = \dot q(t), \qquad \delta \dot q(t) = \ddot q(t) \end{align} It follows that for any Lagrangian (not just one that has time-translation symmetry) a short computation gives \begin{align} \delta L(q(t), \dot q(t) t) = \frac{d}{dt}L(q(t), \dot q(t), t) - \frac{\partial L}{\partial t}(q(t), \dot q(t), t), \end{align} and we immediately get the following result:

If $\partial L/\partial t = 0$, then time-translation is a symmetry of $L$ where the function $F$ is simply given by the Lagrangian itself.

Noether's theorem then tells us that there is a conserved charge; \begin{align} Q_q(t) = \frac{\partial L}{\partial \dot q}(q(t), \dot q(t), t)\dot q(t) - L(q(t), \dot q(t), t) \end{align} which is precisely the Hamiltonian.

  • $\begingroup$ The variation of the Lagrangian that you derive for $\partial L/\partial t=0$ is $$ \delta L(q(t), \dot q(t) t) =\frac{d}{dt}L(q(t), \dot q(t), t) , $$ i.e. the total derivative of the Lagrangian itself. However, above you mention that we only get a conserved quantity if the variation is a total derivative of a function $F_q(t)$ that does not depend on $\dot q(t)$. However, the Lagrangian depends, in general, on $\dot q(t)$ and therefore I don't understand why the time translations are still a symmetry $\endgroup$
    – jak
    Commented Jul 4, 2017 at 9:32
  • $\begingroup$ @JakobH You're misunderstanding what $F_q$ denotes. $F$ is a local function of the path $q$ (the path $q$ is itself a function -- a parameterized curve in configuration space). This means, for example, that there could be a function $f:\mathbb R^{n+1} \to \mathbb R$ for which $F_q(t) = f(q(t), \dot q(t), \ddot q(t), \dots, q^{(n)}(t))$. In other words, the expression for $F_q(t)$ can contain on derivatives of $q$ evaluated at a time $t$. $\endgroup$ Commented Jul 4, 2017 at 17:22

I think the problem is in the first line: invariance for finite time displacement is $$L(q,\dot q ,t+h)-L(q,\dot q ,t)=0.$$ In the infinitesimal case this should become: $$L (q,\dot q, t+h)-L (q,\dot q,t)=O(h^2) \iff \partial _t L(q,\dot q ,t)=0$$ (note that $q$ and $\dot q $ here are not functions of time). With this and Lagrange's equation of motion, you should be able to prove that $H=p\dot q-L$ is conserved along solutions.

I'm not sure about what does the term “infinitesimal time displacement” mean. If $g^{\varepsilon}\colon M \to M$ is a one parameter transformation of the configuration space, then the condition $$\dfrac {\partial}{\partial \varepsilon} |_{\varepsilon =0}L(g^\varepsilon _*(\dot q),t)=0,$$ that I believe expresses symmetry under infinitesimal displacement, is different from $$L(g^\varepsilon _*(\dot q),t)=L(\dot q, t)$$ which is (according to Arnold) the usual definition of (finite displacement) symmetry. If we look at the special case where $g^\varepsilon$ (which involves a slight generalization of the precedent discourse) is the time translation, then it's obvious that the finite and infinitesimal displacement symmetry conditions are the same.

I'll try to answer the question “how can we see energy naturally emerge from time translation symmetry” in the only sense that I can understand it, that is, “can energy be seen as a Noether's charge?”. Alert: the proof is messy.

Recall the definition of the Noether's charge associated to a 1 parameter group of symmetries $g^{\varepsilon}$: $$I=\dfrac{\partial L}{\partial \dot q}\dfrac{\partial }{\partial \varepsilon}|_{\varepsilon = 0}(g^\varepsilon q).$$ Noether's theorem states that $I$ is conserved along solutions if $\partial _\varepsilon |_{\varepsilon =0}L(g_* ^\varepsilon \dot q)=0$.

As it is, the theorem is stated for an autonomous lagrangian, that is, not time dependent Lagrangian. In order to see the energy naturally emerge as a Noether's charge, one approach is indicated in Arnold's book and is as follows.

If $M$ is the configuration space and $L$ is the spurious (i.e. non autonomous) Lagrangian, define the generalized configuration space as $M'=M\times \mathbb R $. Define the Lagrangian on $TM'$: $$\tilde L(q,\dot q,\tau ,\dot \tau)=L(q,\frac{\dot q}{\dot \tau},\tau)\dot \tau.$$ If $q\colon \mathbb [\tau _1 ,\tau _2] \to M$ and $\tau \colon [t_1,t_2] \to [\tau _1,\tau _2] $, note that the action: $$\tilde S[q,\tau]=\int _{t_1} ^{t_2}\tilde L(q(\tau(t)),\dot q (\tau (t)),\tau (t),\dot \tau (t))\text d t=\int _{\tau _1}^{\tau _2}L(q(\tau),\dot q (\tau),\tau)\text d \tau=S[q]$$ doesn't depend on $\tau$. So if $q$ is an extremal of $S$, then $(q\circ \tau,\tau)$ is an extremal of $\tilde S$ and satisfies Euler-Lagrange equations.

So we can apply Noether's theorem to $\tilde L$. Note that $\partial _\tau \tilde L(q,\dot q ,\tau , \dot \tau)=\partial _\tau L (q,\dot q/\dot \tau,\tau) \dot \tau$, so $\tilde L$ admits time translations if $L$ does. Finally, Noethers charge related to the time translation is: $$\dfrac{\partial \tilde L}{\partial \dot \tau}=L-\dfrac{\partial L}{\partial \dot q}\frac{\dot q}{\dot \tau},$$ that ìis minus the energy.

  • $\begingroup$ Unfortunately I can't see how this is different from what i did. I worked with infinitesimal transformations and therefore neglecting quadratic terms. Can you give me another tip how the term -L should emerge? $\endgroup$
    – jak
    Commented Jan 19, 2014 at 18:43
  • $\begingroup$ The difference is that you are taking $q$ and $\dot q$ as functions of time, instead of simply a vector of $\mathbb R ^{2n}$ (or, better a tangent vector to $\mathbb R ^n$). I'm not sure about what does the technical term “infinitesimal time displacement” mean, since you can see that the two conditions I've wrote are indeed equivalent. However, if we take “$\partial _t L=0$“ as a definition of “time translational symmetry“, the proof of $\dot H = 0$ is really a simple calculation. $\endgroup$
    – pppqqq
    Commented Jan 19, 2014 at 18:57
  • $\begingroup$ thanks for your reply. I have no problem prooving that $\dot H = 0$, but my problem is seeing how this notion emerges naturally from time-translation invariance $\endgroup$
    – jak
    Commented Jan 19, 2014 at 19:05
  • 1
    $\begingroup$ pppqqq Is correct. Your procedure begins by assuming that for a given trajectory, the value of the Lagrangian is constant. I.E., you throw a ball through the air and over the course of its flight kinetic-potential energy stays constant. This is wrong, so of course the result you get is wrong. $\endgroup$ Commented Jan 19, 2014 at 19:58
  • 2
    $\begingroup$ @ArturoDonJuan the problem here is in the interpretation of "time translation invariance". The OP assumed that this meant $\frac{\text d}{\text{d}t} L=0$, $\text d / \text d t$ being the total derivative of the lagrangian along the classical path (see equation (1-2-3) of the OP). This is not correct. Instead, it is the partial derivative $\partial L /\partial t$ which is zero. Moreover, this is not a property of the path, as $\text d /\text{d} t = 0$ is, but is a property of the lagrangian. $\endgroup$
    – pppqqq
    Commented Mar 26, 2015 at 23:26

Ok, so from your comments I understand that you already know how to derive Noether's theorem(?), which means that the Noether's current: $$ j = \left( L- \frac{\partial L}{\partial \dot{q}}\dot{q}(t) \right) \epsilon(t) + \frac{\partial L}{\partial \dot{q}} \delta q (t) \tag{1} $$ is conserved: $$ \frac{d j }{dt} = 0 $$ if the action of a given system is invariant under the following infinitesimal transformations: \begin{equation} t \rightarrow t' = t + \delta t = t + \epsilon (t) \end{equation} \begin{equation} q(t) \rightarrow q'(t')=q(t) + \delta q (t) \end{equation}

Now, note that the Hamiltonian is defined as: $$H = \frac{\partial L}{\partial \dot{q}}\dot{q} - L$$ which means that equation $(1)$ can be written as: $$j = - H \epsilon(t) + \frac{\partial L}{\partial \dot{q}} \delta q (t) $$

Now, let us consider a Lagrangian that does not explicitly depend on time, i.e. $L=L(q,\dot{q})$. Subsequently, we consider a time translation: $$t \rightarrow t' = t + \delta t = t + \epsilon$$ where $\epsilon$ is a constant (i.e. $\epsilon\neq \epsilon (t)$). If $S$ is invariant ($\delta S = 0$) under time translations, then the Noether current is given by: \begin{equation} j = -H\epsilon \end{equation} (because the path is not affected by a time tranlation, that is $\delta q (t)=0$) and so the Hamiltonian is a constant of motion.


But unfortunatly this is not the Hamiltonian. This computation should yield \begin{equation} \rightarrow \frac{d}{dt} \left( p \dot{q} - L \right) = 0. \end{equation} But I can't find no reason why and how the the extra $-L$ should emerge. I can see that this term can be written at the place where it is written because we have $\delta L = - \frac{d L}{dt } \epsilon$ and therefore \begin{equation} \rightarrow \delta L = - \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \frac{\partial q}{\partial t} \right) \epsilon = - \frac{d L}{dt } \epsilon. \end{equation} And then the desired equation would only say $0-0=0$. Any idea where i did a mistake would be much appreciated.

You didn't make a mistake. Take your final equation: \begin{equation} - \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \frac{\partial q}{\partial t} \right) \epsilon = - \frac{d L}{dt } \epsilon. \end{equation}

Use the definition of momentum: $$ p = \frac{\partial L}{\partial \dot p} $$

And find: $$ - \frac{d}{d t} \left(p \frac{\partial q}{\partial t} \right) \epsilon = - \frac{d L}{dt } \epsilon. $$

Your $\partial q/\partial t$ should be written as $\dot q$, so you have: $$ - \frac{d}{d t} \left(p \dot q \right) \epsilon = - \frac{d L}{dt } \epsilon. $$

Cancel epsilon from both sides: $$ - \frac{d}{d t} \left(p \dot q \right) = - \frac{d L}{dt } . $$

Move the LHS term to the RHS: $$ 0 = \frac{d}{d t}\left(p \dot q - L \right) $$

This says that $p \dot q - L$ is conserved. I.e., the Hamiltonian/energy is conserved, which is exactly what you were trying to prove.


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