I am studying Lagrangian mechanics and I have come across something that I do not understand. Basically the text I am reading skipped steps and I do not know how to get from point A to point B. I believe it is relatively simple.

The text is trying to show that the action is invariant under infinitesimal transformations of the following form.

$$ q\rightarrow q+\dot{q}\epsilon \\ t \rightarrow t+\epsilon $$

So the first thing is to write down the variation in the action up to first order.

$$ S'-S=\int_{t_1}^{t_2}\left[ \frac{\partial L}{\partial t}\delta t + \frac{\partial L}{\partial q_i}\delta q_i + \frac{\partial L}{\partial \dot{q_i}}\delta \dot{q_i}\right]dt $$

$$ S'-S=\int_{t_1}^{t_2}\left[ \frac{\partial L}{\partial t}\epsilon + \frac{\partial L}{\partial q_i}\dot{q}_i \epsilon + \frac{\partial L}{\partial \dot{q_i}}\frac{d}{dt}\left( \dot{q}_i \epsilon \right)\right]dt $$

This next part is what I do not understand. The next line writes the following.

$$ S'-S=\int_{t_1}^{t_2}\left[ \frac{\partial L}{\partial t}\epsilon + \left( \frac{\partial L}{\partial \dot{q}_i}\dot{q}_i \right)\frac{d\epsilon}{dt} \right]dt $$

So somehow the last two terms turned into the last term in the previous expression. I have tried all kinds of things (identities, integration by parts, ect..) to obtain this term and I simply cannot figure it out. I know it is probably simple but I am at a loss.

If anyone can help me out I would greatly appreciate it!

  • $\begingroup$ What text are you using? $\endgroup$
    – Javier
    Sep 7, 2015 at 0:54
  • $\begingroup$ I just realized that the word "text" could be misunderstood for a textbook. I tend to say "text" for anything I am reading. It is actually a write up from MIT. Here is the link. Page 3 web.mit.edu/edbert/GR/gr5.pdf $\endgroup$
    – user41178
    Sep 7, 2015 at 1:00
  • $\begingroup$ I am studying basic Lagrangian mechanics not what is presented in the bulk of that write up. However the content on page 3 is directly related to what I am reading in "Classical Dynamics, a Contemporary Approach." $\endgroup$
    – user41178
    Sep 7, 2015 at 1:03

1 Answer 1


If you take a closer look, you will see that the last line in fact says $dL/dt$, not $\partial L/ \partial t$. The derivation would be:

$$\frac{\partial L}{\partial t} \epsilon + \frac{\partial L }{\partial q_i} \dot{q_i} \epsilon + \frac{\partial L}{\partial \dot{q_i}} \frac{d}{dt}(\dot{q_i} \epsilon) \\ = \frac{\partial L}{\partial t} \epsilon + \frac{\partial L }{\partial q_i} \dot{q_i} \epsilon + \frac{\partial L}{\partial \dot{q}_i} \ddot{q_i} \epsilon + \frac{\partial L}{\partial \dot{q_i}} \dot{q_i} \frac{d\epsilon}{dt} \\ = \left( \frac{\partial L}{\partial t} + \frac{\partial L }{\partial q_i} \frac{dq_i}{dt} + \frac{\partial L}{\partial \dot{q}_i} \frac{d\dot{q_i}}{dt} \right) \epsilon + \frac{\partial L}{\partial \dot{q_i}} \dot{q_i} \frac{d\epsilon}{dt} \\ \\ = \frac{dL}{dt}\epsilon + \frac{\partial L}{\partial \dot{q_i}} \dot{q_i} \frac{d\epsilon}{dt} $$

  • $\begingroup$ Of course! Thanks! I very much appreciate your help! $\endgroup$
    – user41178
    Sep 7, 2015 at 1:48

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