# Disappearing term in proof of Noether's theorem

[I know this is a topic in mathematical physics, but I feel like my question is more on the mathematics than the physics. At any rate, feel free to migrate it to Physics.SE if necessary.]

In my notes, one of the implications of Noether's theorem (in the Lagrangian formalism) is stated briefly as such:

Theorem. If the Euler-Lagrange equations hold, and an infinitesimal transformation is invariant, then there exists a conserved quantity.

An infinitesimal transformation is one that takes the form $Q = q + \delta q$ and, for simplicity, we assume that it does not modify the time coordinate. An invariant transformation is one such that the Euler-Lagrange equations take the same functional form, i.e.: if the Lagrangian function is $L_q(q,\dot q,t)$ in the old coordinates and $L_Q(Q,\dot Q ,t)$ in the new, then the latter also lies in the kernel of the Euler-Lagrange operator $$\frac{\mathfrak D}{\mathfrak D q^k} = \frac{\partial}{\partial q^k} - \frac d {dt} \frac{\partial}{\partial \dot q^k}$$ expressed in the old coordinates. We have seen a lemma that states that such a transformation must be such that the new and old Lagrangians are connected by the formula $$L_Q(Q,\dot Q,t) = L_q(Q,\dot Q,t) + \frac{d\Lambda}{dt}(Q,t) \tag1$$ for some arbitrary regular function $\Lambda$ (actually, we have only proven that if it satisfies $(1)$, the transformation is invariant, but I trust that the other implication holds).

The demonstration goes like this: consider the infinitesimal transformation $Q = q + \delta q$ and suppose that it is invariant, so that it satisfies $(1)$. The Lagrangian is uniquely defined (its value is not dependent on the chosen chart), so $L_q(q,\dot q,t) = L_Q(Q,\dot Q,t)$ and we may revert to the old Lagrangian on the RHS of $(1)$. Substituting the formula for $Q$, we get $$L_q(q,\dot q,t) = L_q(q+\delta q, \dot q + \delta \dot q,t) + \frac{d\Lambda}{dt}(q+\delta q,t)$$ We assume that the Lagrangian and the $\Lambda$ are sufficiently regular, so that we may expand both to first order and obtain $$\frac{\partial L_q}{\partial q^k}\delta q^k + \frac{\partial L_q}{\partial \dot q^k}\delta \dot q^k + \frac d {dt} \left(\Lambda(q,t) + \frac{\partial \Lambda}{\partial q^k}\delta q^k \right) = 0$$ and, by a reverse Leibniz rule, $$\frac{\partial L_q}{\partial q^k}\delta q^k + \frac{d}{dt}\left(\frac{\partial L_q}{\partial \dot q^k}\delta q^k\right) - \frac d{dt}\left( \frac{\partial L_q}{\partial \dot q^k}\right) \delta q^k + \frac d {dt} \left(\Lambda(q,t) + \frac{\partial \Lambda}{\partial q^k}\delta q^k \right) = 0.$$ Since by hypothesis the E-L equations hold in the old variables $\forall k$, the first and third terms disappear and we are left with the total time derivative $$\frac {d}{dt} \left(\frac{\partial L_q}{\partial \dot q^k}\delta q^k + \Lambda + \frac{\partial \Lambda}{\partial q^k}\delta q^k \right) = 0$$ which implies that the quantity in parenthesis is conserved. However, in my notes, somewhere along the way the $\Lambda$ term disappears, without justification! And indeed the correct conserved quantity is $$\frac{\partial L_q}{\partial \dot q^k}\delta q^k + \delta \Lambda = \text{const}.$$

Why does this happen? Is $\Lambda (q,t)$ itself zero for some reason? Is its total time derivative zero? And how can either option be justified thoroughly?

I don't think the last term $\frac{d\Lambda}{dt}(Q,t)$ of (1) is correct. It needs to vanish when $\delta q=0$, but in this circumstance, it gives $\frac{d\Lambda}{dt}(q,t)$.
Let us consider an example, and denote $L[q]=L(q, \dot q,t)$. Let $L=\dot q^2/2$, and $Q=q+vt$ (Galilean boost). Then $L[Q]=\dot q^2+2v\dot q+v^2=L[q]+\frac{d}{dt}(2vq+v^2t)$. Up to a constant, we have $\Lambda=2vq+v^2t$ (I will add a minus sign to $\Lambda$ here). Since $\delta q=vt$, we can identify $v$ as the small parameter. Observe then that $\Lambda=0$ when $v=0$.
I think a more correct way to write (1) is like this: $$L[Q]=L[q]+\frac{d}{dt}\left(\delta\Lambda[q]+O(\delta^2)\right).$$ In the above example, we would have $\delta\Lambda[q]+O(\delta^2)=2vq+v^2t$, or $\delta\Lambda[q]=2vq$. Expanding this to first order then yields: $$\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial \dot q}\delta\dot q=\frac{d}{dt}\delta\Lambda[q].$$ And from here we can finish Noether: $$\delta q\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot q}\right)=\frac{d}{dt}\left(-\delta q\frac{\partial L}{\partial \dot q}+\delta\Lambda[q]\right).$$
• So would it be correct to say that $\Lambda$ must be chosen so that it vanishes for the identity transformation? Is this also true for general invariant transformation or just specific to Noether’s theorem? – giobrach Jul 25 '18 at 7:42
• I believe yes to both questions, because otherwise the $O(1)$ term does not vanish when you expand to first order in $\delta$. So I would say this is true for invariant continuous transformations. – user254433 Jul 25 '18 at 8:07
• So this is a huge stretch, but: would it be correct to say that if we are given a family of invariant transformations that depend continuously on a parameter $\alpha$ and forms a group with respect to said parameter, then the set of “gauges” $d\Lambda/dt$ corresponding to those transformations also forms a group that is isomorphic to the former through the map $\text{transf}(\alpha) \mapsto \text{gauge}(\alpha)$, so that the identity must be preserved? – giobrach Jul 25 '18 at 8:35
• Well, there are examples where gauge$(\alpha)\equiv 0$, like $\delta q=1$ in the above example, so we would hope for a homomorphism. But I don't think the target space has a group law. If $X=a(t,q)\partial/\partial q$ is the infinitesimal generator, then we have gauge$(\alpha)=[\exp(\alpha X)-1]L[q]$. One try is $g(\alpha)+g(\beta)=g(\alpha+\beta)$, but the right hand side gives $(e_\alpha+e_\beta-2)L$. Another try is $\exp(\beta X)g(\alpha)=g(\alpha+\beta)$, but the RHS is $[\exp(\alpha+\beta)X-\exp(\beta)]L=g(\alpha+\beta)-g(\beta)$. – user254433 Jul 25 '18 at 9:42