# Clearing up some simple details of the types of symmetries involved in Noether's theorem

I would just like to ensure that I have fully understood the content of Noether's theorem and a few of its details. The generic statement of Noether's theorem is relatively straight forward however there are subtleties associated with what exactly constitutes a symmetry and the consequences of working on- and off-shell.

Symmetry

I discuss two notions of symmetry here, they are:

Quasi-symmetry: In which, to first order, the action changes by a boundary term and/or the Lagrangian changes by a total derivative: $$\delta S=[B(q)]_{t_0}^{t_1},\quad \delta L=\frac{dF(q)}{dt} \tag{1.a}.$$

Symmetry: In which, to first order, both quantities are invariant: $$\delta S=0,\quad \delta L=0. \tag{1.b}$$

On-shell vs. off shell

By on-shell we mean the subset of curves through configuration space, $$Q\cong R^N$$, that solve the Euler-Lagrange equations.

By off-shell we mean the more general set of curves that do not necessarily solve the Euler-Lagrange equations.

Noether's Theorem

Now we discuss the actual content of Nother's theorem, that an off-shell symmetry (or more generally quasi-symmetry) of the action implies the existence of a conserved quantity on-shell. In other words, a generic transformation on the domain of the action functional implies a conserved quantity along the subset of the domain that solve the Euler-Lagrange equations.

A generic infinitesimal change to the action can be written:

$$\delta S[q(t)]=\int_{t_0}^{t_1} dt\left( \frac{\partial L}{\partial q}-\frac{d}{dt}\left( \frac{\partial L}{\partial \dot q} \right)\right)\delta q+\left[\frac{\partial L}{\partial \dot q}\delta q\right]_{t_0}^{t_1}, \tag{2}$$

If we now work on-shell, in which $$q(t)$$ solves the E-L equations and the integrand vanishes, we are left with a few possibilities:

1. $$\delta q$$ satisfies the boundary condition $$\delta q(t_0)=\delta q(t_1)=0$$, in which case the boundary term vanishes and this is simply the statement that all first-order transformations of the action along the equations of motion vanish.

2. $$\delta q$$ does not satisfy the boundary conditions, in which case we are left with two further possibilities, either $$\delta q$$ is a symmetry or $$\delta q$$ is a quasi-symmetry (assuming of course it is a symmetry at all).

If $$\delta q$$ is a symmetry, then the following is true:

$$\left[\frac{\partial L}{\partial \dot q}\delta q\right]_{t_0}^{t_1}=0\quad \Rightarrow \quad \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\delta q\right)=0, \tag{3}$$ and we obtain our conserved "Noether charge".

If, however $$\delta q$$ is a quasi-symmetry then we find by (1.a):

$$\left[\frac{\partial L}{\partial \dot q}\delta q\right]_{t_0}^{t_1}=\left[B(q(t))\right]_{t_0}^{t_1}\quad \Rightarrow \quad \frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\delta q-B(q(t))\right)=0, \tag{4}$$

and we obtain a slightly different Noether charge. We can then finally (cf. this post) relate the quasi-symmetry of the action and the quasi-symmetry of the Lagrangian by:

$$B(q(t))=F(q(t)) \tag{5},$$

which closes the possible outcomes.

My question is are the above statements correct if we confine our attention to quasi-symmetries, and if not, where in my definitions and/or derivation have I made mistakes?

I understand that this is a bit of an open ended question, but this topic is discussed at length on this site, and after fairly extensive reading of related questions I think this summary question serves an ok role. But apologies if it is against the rules.

2. OP is considering infinitesimal vertical transformation $$\delta q$$ only with no infinitesimal horizontal transformation $$\delta t=0$$. Noether's theorem holds more generally for combinations of vertical & horizontal infinitesimal quasi-symmetries.
If it is not in the field theory context, the Noether charge is really the generalized momentum projected to the direction of $$\delta q$$. So the conclusion that you get is something not necessarily a function of $$q(t)$$. Take the example, where $$B(q(t))=0$$, and your Lagrangian is about a point mass with translational symmetry, what you should get is $$\delta q$$ being the identity, and $$p=\rm{Const}$$ for linear momentum conservation.
With the quasi-symmetry, what you should get is simply $$\frac{\partial L}{\partial\dot{q}}\delta q - B(q(t))=\rm{Const}$$. That is, the momentum projected to the direction of symmetry has a difference from $$B(q(t))$$ that is a constant, assuming $$q(t)$$ satisfies the Euler-Lagrange equations.