In the context of Noether's theorem , the Hamiltonian is the constant of motion associated with the time-translational invariance of the Lagrangian. Time-translational invariance is equivalent to the Lagrangian not depending explicitly on time that is $$\dfrac{∂L}{∂t}=0 .$$

The reason they're equivalent is that for an infinitesimal time translation, we can approximate the Lagrangian as the first order expansion of its Taylor series, that is

$$δL ≡ L(q, \dot{q},t +\epsilon ) − L(q, \dot{q},t)= \dfrac{∂L}{∂t}\epsilon $$ $$\text{the right one}$$

But Shouldn't $t \mapsto t+ \epsilon$ induce $q(t) \mapsto q(t+ \epsilon)$ and $\dot{q}(t) \mapsto \dot{q}(t+ \epsilon)$ ? and if that is the case then $$δL ≡ L( q(t+ \epsilon), \dot{q}(t+ \epsilon),t +\epsilon ) − L(q, \dot{q},t)= \dfrac{∂L}{∂t}\epsilon +\dfrac{∂L}{∂q}\epsilon+\dfrac{∂L}{∂ \dot{q}}\epsilon $$ $$\text{ the wrong one}$$

So that a Lagrangian is time transational invariant if and only if it does not explicitly depend on $q$,$\dot{q}$ and $t$ which does not make sense. So How is it possible to vary time without affect the coordinates or their derivatives which are themselves functions of time?


Let $t\to t+\varepsilon$ be an explicit variation in the time variable, which in turn reflects onto $q\to q+\delta q$ and $v\to v+\delta v$, respectively (as you pointed out).

Let $L(q,v,t)$ be the Lagrangian function undergoing the variation $$ \delta L(q,v,t) = \frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial v}\delta v + \frac{\partial L}{\partial t}\delta t $$ under the aforementioned time transformation, by definition of variation.

On the solution of the equation of motion (and only there) one has $$ \delta v =d\,(\delta q) $$ namely the variation commutes with the time derivative; hence the above becomes $$ \delta L(q,v,t) = \frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial v}d\,(\delta q)+ \frac{\partial L}{\partial t}\delta t\ . $$ Applying the Leibniz rule to the second contribution and using the equation of motion (onto whose solution we decided to be) one ends up with $$ \delta L(q,v,t) = d\,\Big(\frac{\partial L}{\partial v}\,\delta q\Big)+ \frac{\partial L}{\partial t}\delta t\ . $$ By hypothesis the Lagrangian is such to have fixed boundaries under variations, thus the first contribution in the above vanishes and the remaining one proves the result.

The error in your computation was that you considered $\delta q, \delta v$ to be both $\varepsilon$, but they are not.

  • $\begingroup$ I followed all of your argument except for "By hypothesis the Lagrangian is such to have fixed boundaries under variations, thus the first contribution in the above vanishes and the remaining one proves the result". I don't understand this, can you explain? $\endgroup$ – Omar Nagib Dec 18 '15 at 14:32
  • $\begingroup$ @AngusTheMan After thinking about it for a while, I'm again unconvinced. So $\delta q(t)$ is $q(t+\epsilon)-q(t)$ such that $\delta q(t_0)=q(t_f)=0$. I agree that using E-L equation is equivalent to using Hamilton's principle. But how this implies $d\,\Big(\frac{\partial L}{\partial v}\,\delta q\Big)$=0, it seems to me the latter condition is fulfilled if and only if $\delta q(t)=0$ for any $t$ which is not the case. The latter term will only vanish at $\delta q(t_0)$ or $\delta q(t_f)$ but not for any arbitrary $t$. so can you clear up my confusion? $\endgroup$ – Omar Nagib Dec 18 '15 at 23:59
  • $\begingroup$ @AngusTheMan $$\begin{equation} \frac{\partial L}{\partial \dot q}\delta q \bigg|^2- \frac{\partial L}{\partial \dot q}\delta q \bigg|^1 =0 \end{equation}$$ But that this equation vanishes follows from the fact that $\delta q(1)= \delta q(2)=0$. hence this does not imply this term is a constant function of time, does it? $\endgroup$ – Omar Nagib Dec 19 '15 at 1:08
  • $\begingroup$ Applying Hamilton's principle is the same as using E-L, and since we used E-L equation it follows(From Hamilton's principle) that $\delta q(1)=\delta q(2)=0$, so I don't understand how you can assume $\delta q(i)$ to be arbitrary. Can you help me with that? $\endgroup$ – Omar Nagib Dec 19 '15 at 2:31

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