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.


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 Answers 2


Comments to the post (v3):

  1. OP is considering point mechanics. Noether's theorem also holds in field theory.

  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.

  3. Boundary conditions (BCs) are not allowed in the definition of a quasi-symmetry, cf. e.g. this Phys.SE post.

OP's treatment seems correct except where they contemplate imposing BCs, cf. pt. 3.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.