# Proof of Noether's theorem: How to deal with transformations in time?

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. 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 $t$. Commented Apr 23, 2019 at 17:16