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I am trying to show that the time-invariance of the Lagrangian of a simple one-particle system implies energy conservation for that system. The first step is, well, to show that the Lagrangian is time-invariant, but I'm getting stuck at a certain part, and it is confusing me.

Consider a small first-order step in time $t$, $\tau$. It is first order, so $\tau^n,\,n=2,3,\cdots$ is negligibly small. A time displacement of the Lagrangian can be written as

\begin{align*} T_t(\tau)L(q,\dot{q},t)&=e^{\tau\frac{d}{dt}}L(q,\dot{q},t)\\ &=\left(1+\left(\tau\frac{d}{dt}\right)+\frac{1}{2}\left(\tau\frac{d}{dt}\right)^2+\cdots\right)L(q,\dot{q},t)\\ &\approx L + \tau\frac{d}{dt}L(q,\dot{q},t)\\ &=L + \tau\frac{d}{dt}\left(\frac{1}{2}m\dot{q}^2-V(q)\right)\\ &=L + \tau\left(m\dot{q}\ddot{q}-\frac{d\,V(q)}{dq}\dot{q}\right)\\ &=L + \tau\dot{q}\left(m\ddot{q}-\frac{d\,V(q)}{dq}\right)\\ &= ??? \end{align*}

My plan was to use the relation $m\ddot{q}=-\frac{d\,V(q)}{dq}$ from the E-L equation, which is satisfied as a particular time $t$, to show that the Lagrangian doesn't change under time-displacement of first order $T_t(\tau)$ (hence E-L equation is satisfied for all time $t$), but that doesn't quite fit in my last line.

Where have I gone wrong? Is my logic screwy?

Arturo don Juan
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  • This question (v1) is essentially a duplicate of http://physics.stackexchange.com/q/94381/2451 and links therein. – Qmechanic Mar 19 '15 at 20:53
  • I didn't quite see the exact similarity at first, but now I do. I posted a comment on @pppqqq's answer in the question you linked regarding the problem both the question-er in the link you posted and I myself had. Maybe you might be able to address it? (I have not gotten a response yet) – Arturo don Juan Mar 19 '15 at 21:58

1 Answers1

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We have such equation:

$$H = \frac{\partial L}{\partial \dot{q}} \dot{q} - L$$

You can show by calculation, that it holds in your special case, too.

Now we use chain rule, and Euler-Lagrange equation: $$\frac{dL}{dt} = \frac{\partial L}{\partial q} \dot{q} + \frac{\partial L}{\partial \dot{q}} \ddot{q} =\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}} \right) \dot{q}+\frac{\partial L}{\partial \dot{q}} \ddot{q} = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \dot{q} \right)$$

So: $$\frac{dH}{dt} = \frac{dL}{dt} - \frac{dL}{dt} =0.$$ As you can see, this works only for lagrangian without explicit dependence on $t$. Otherwise we would have some additional terms in $\frac{dL}{dt}$.

Ageeine
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