I was following the proof of Noether's theorem in Lemos - Analytical Mechanics, page 73.

He considers a full infinitesimal transformation: $$t'=t+\epsilon X(q(t),t),$$ $$q'(t')=q(t)+\epsilon\Psi(q(t),t),\tag{2.160}$$ whose change in the action is $$\Delta S=\int_{t_1'}^{t_2'}L\left(q'(t'),\frac{dq'(t')}{dt'},t'\right)dt'-\int_{t_1}^{t_2}L\left(q(t),\frac{dq(t)}{dt},t\right)dt.\tag{2.161}$$ Note that integration limits are changed in the first term of the RHS.

Then after plugging the transformation in $\Delta S$ he gets $$\Delta S=\int_{t_1}^{t_2}L(q+\epsilon\Psi,\dot q+\epsilon \xi,t+\epsilon X)(1+\epsilon\dot X)dt-\int_{t_1}^{t_2}L(q,\dot q,t)dt,\tag{2.166}$$ where $$\xi=\dot\Psi-\dot q\dot X.\tag{2.165}$$

  1. Why is the first integral above over $[t_1,t_2]$ instead of $[t_1',t_2']?$

  2. Is not there a term proportional to $\epsilon\left[L(q(t_2),\dot q(t_2),t_2)-L(q(t_1),\dot q(t_1),t_1)\right]$ being neglected in the $\Delta S$ above?

  1. It is customary to let the integration region flow with the so-called horizontal transformation (2.160a). Ref. 1 starts out very ambitious by declaring in eq. (2.160a) that the horizontal generator $X(q(t),t)$ is a function of $q(t)$, which is unusual. Later Ref. 1 seems to implicitly assume that $X(t)$ is only a function of time $t$, as is normally assumed.

  2. Theorem 2.7.1 on p. 74 in Ref. 1 only discusses the case when the action $S$ has a strict symmetry. In principle Noether's theorem also works if the action has a quasi-symmetry, i.e. if it is only invariant up to boundary terms, see p. 75 in Ref. 1.


  1. N.A. Lemos, Analytical Mechanics, 2018; Section 2.7.
  • $\begingroup$ 1. What does make this custom mathematically true? 2. The action will be quase-invariant if the difference is the integral of a total derivative of a function of $q$ and $t$ only. But in the present case, the difference is the integral of $dL/dt$, i.e. a total derivative of a function of $q$, $\dot q$ and $t$. $\endgroup$ – Diracology Apr 23 '19 at 17:16
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Apr 23 '19 at 17:50

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