Is the converse of Noether's first theorem true: Every conservation law has a symmetry?

Question

Noether's (first) theorem states that any differentiable symmetry of the action of a physical system has a corresponding conservation law.

Is the converse true: Any conservation law of a physical system has a differentiable symmetry of its action?

Answer 1

I) For a mathematical precise treatment of an inverse Noether's Theorem, one should consult e.g. Olver's book (Ref. 1, Thm. 5.58), as user orbifold also writes in his answer (v2). Here we would like give a heuristic and less technical discussion, to convey the heart of the matter, and try to avoid the language of jets and prolongations as much as possible.

In popular terms, we would like to formulate an "inverse Noether machine"

$$ \text{Input: Lagrangian system with known conservation laws} $$ $$ \Downarrow $$ $$ \text{[inverse Noether machine]} $$ $$ \Downarrow $$ $$ \text{Output: (quasi)symmetries of action functional} $$

Since this "machine" is supposed to be a mathematical theorem that should succeed everytime without exceptions (else it is by definition not a theorem!), we might have to narrow down the set/class/category of inputs that we allow into the machine in order not to have halting errors/breakdowns in the machinery.

II) Let us make the following non-necessary restrictions for simplicity:

1. Let us focus on point mechanics with a local action functional $$ S[q] ~=~ \int\! dt~ L\left(q(t), \frac{dq(t)}{dt}, \ldots,\frac{d^Nq(t)}{dt^N} ;t\right), \tag{1} $$ where $N\in\mathbb{N}_0$ is some finite order. Generalization to classical local field theory is straightforward.

2. Let us restrict to only vertical transformations $\delta q^i$, i.e., any horizontal transformation $\delta t=0$ vanishes. (Olver essentially calls these evolutionary vector fields, and he mentions that it is effectively enough to study those (Ref. 1, Prop. 5.52).)

3. Let us assume, as Olver also does, that the Lagrangian $L$ and the transformations are real analytic$^{\dagger}$.

The following technical restrictions/extensions are absolutely necessary:

1. The notion of symmetry $\delta S=0$ should be relaxed to quasisymmetry (QS). By definition a QS of the action $S$ only has to hold modulo boundary terms. (NB: Olver uses a different terminology: He calls a symmetry for a strict symmetry, and a quasisymmetry for a symmetry.)

2. The notion of QS transformations might only make sense infinitesimally/as a vector field/Lie algebra. There might not exist corresponding finite QS transformations/Lie group. In particular, the QS transformations are allowed to depend on the velocities $\dot{q}$. (Olver refers to this as generalized vector fields (Ref. 1, Def. 5.1).)

III) Noether's Theorem provides a canonical recipe of how to turn a QS of the action $S$ into a conservation law (CL),

$$ \frac{dQ}{dt}~\approx~0,\tag{2}$$

where $Q$ is the full Noether charge. (Here the $\approx$ symbol means equality on-shell, i.e. modulo the equations of motion (eom).)

Remark 1: Apart from time $t$, the QS transformations are only allowed to act on the variables $q^i$ that actively participate in the action principle. If there are passive external parameters, say, coupling constants, etc, the fact that they are constant in the model are just trivial CLs, which should obviously not count as genuine CLs. In particular, $\frac{d1}{dt}=0$ is just a trivial CL.

Remark 2: A CL should by definition hold for all solutions, not just for a particular solution.

Remark 3: A QS of the action $S$ is always implicitly assumed to hold off-shell. (It should be stressed that an on-shell QS of the action

$$ \delta S \approx \text{boundary terms}\tag{3} $$

is a vacuous notion, as the Euler-Lagrange equations remove any bulk term on-shell.)

Remark 4: It should be emphasized that a symmetry of eoms does not always lead to a QS of the Lagrangian, cf. e.g. Ref. 2, Example 1 below, and this Phys.SE post. Hence it is important to trace the off-shell aspects of Noether's Theorem.

Example 1: A symmetry of the eoms is not necessarily a QS of the Lagrangian. Let the Lagrangian be $L=\frac{1}{2}\sum_{i=1}^n \dot{q}^i g_{ij} \dot{q}^j$, where $g_{ij}$ is a constant non-degenerate metric. The eoms $\ddot{q}^i\approx 0$ have a $gl(n,\mathbb{R})$ symmetry $\delta q^{i}=\epsilon^i{}_j~q^{i}$, but only an $o(n,\mathbb{R})$ Lie subalgebra of the $gl(n,\mathbb{R})$ Lie algebra is a QS of the Lagrangian.

IV) Without further assumptions, there is a priori no guarantee that the Noether recipe will turn a QS into a non-trivial CL.

Example 2: Let the Lagrangian $L(q)=0$ be the trivial Lagrangian. The variable $q$ is pure gauge. Then the local gauge symmetry $\delta q(t)=\epsilon(t)$ is a symmetry, although the corresponding CL is trivial.

Example 3: Let the Lagrangian be $L=\frac{1}{2}\sum_{i=1}^3(q^i)^2-q^1q^2q^3$. The eom are $q_1\approx q_2q_3$ and cyclic permutations. It follows that the positions $q^i\in\{ 0,\pm 1\}$ are constant. (Only $1+1+3=5$ out of the $3^3=27$ branches are consistent.) Any function $Q=Q(q)$ is a conserved quantity. The transformation $\delta q^i=\epsilon \dot{q}^i$ is a QS of the action $S$.

If we want to formulate a bijection between QSs and CLs, we must consider equivalence classes of QSs and CLs modulo trivial QSs and CLs, respectively.

• A QS transformation $\delta q^i$ is called trivial if it vanishes on-shell (Ref. 1, p.292).

A CL is called

1. trivial of first kind if the Noether current $Q$ vanishes on-shell.

2. trivial of second kind if CL vanishes off-shell.

3. trivial if it is a linear combination of CLs of first and second kinds (Ref. 1, p.264-265).

V) The most crucial assumption is that the eoms are assumed to be (totally) non-degenerate. Olver writes (Ref. 1, Def. 2.83.): A system of differential equations is called totally non-degenerate if it and all its prolongations are both of maximal rank and locally solvable$^{\ddagger}$.

The non-degeneracy assumption exclude that the action $S$ has a local gauge symmetry. If $N=1$, i.e. $L=L(q,\dot{q},t)$, the non-degeneracy assumption means that the Legendre transformation is regular, so that we may easily construct a corresponding Hamiltonian formulation $H=H(q,p,t)$. The Hamiltonian Lagrangian reads

$$ L_H~=~p_i \dot{q}^i-H.\tag{4} $$

VI) For a Hamiltonian action functional $S_H[p,q] = \int\! dt~ L_H$, there is a canonical way to define an inverse map from a conserved quantity $Q=Q(q,p,t)$ to a transformation of $q^i$ and $p_i$ by using the Noether charge $Q$ as Hamiltonian generator for the transformations, as also explained in e.g. my Phys.SE answer here. Here we briefly recall the proof. The on-shell CL (2) implies

$$ \{Q,H\}+\frac{\partial Q}{\partial t}~=~0\tag{5} $$

off-shell, cf. Remark 2 and this Phys.SE post. The corresponding transformation

$$ \delta q^i~=~ \{q^i,Q\}\epsilon~=~\frac{\partial Q}{\partial p_i}\epsilon\qquad \text{and}\qquad \delta p_i~=~ \{p_i,Q\}\epsilon~=~-\frac{\partial Q}{\partial q^i}\epsilon\tag{6} $$

is a QS of the Hamiltonian Lagrangian

$$\begin{align} \delta L_H ~\stackrel{(4)}{=}~&\dot{q}^i \delta p_i -\dot{p}_i \delta q^i -\delta H+\frac{d}{dt}(p_i \delta q^i)\cr ~\stackrel{(6)+(8)}{=}& -\dot{q}^i\frac{\partial Q}{\partial q^i}\epsilon -\dot{p}_i\frac{\partial Q}{\partial p_i}\epsilon -\{H,Q\}\epsilon + \epsilon \frac{d Q^0}{dt}\cr ~\stackrel{(5)}{=}~& \epsilon \frac{d (Q^0-Q)}{dt}~\stackrel{(9)}{=}~ \epsilon \frac{d f^0}{dt},\end{align}\tag{7} $$

because $\delta L_H$ is a total time derivative. Here $Q^0$ is the bare Noether charge

$$ Q^0~=~ \frac{\partial L_H}{\partial \dot{q}^i} \{q^i,Q\} + \frac{\partial L_H}{\partial \dot{p}_i} \{p_i,Q\} ~=~ p_i \frac{\partial Q}{\partial p_i},\tag{8} $$

and

$$ f^0~=~ Q^0-Q .\tag{9} $$

Hence the corresponding full Noether charge

$$ Q~=~Q^0-f^0\tag{10} $$

is precisely the conserved quantity $Q$ that we began with. Therefore the inverse map works in the Hamiltonian case.

Example 4: The non-relativistic free particle $L_H=p\dot{q}-\frac{p^2}{2m}$ has e.g. the two conserved charges $Q_1=p$ and $Q_2=q-\frac{pt}{m}$.

The inverse Noether Theorem for non-degenerate systems (Ref. 1, Thm. 5.58) can intuitively be understood from the fact, that:

1. Firstly, there exists an underlying Hamiltonian system $S_H[p,q]$, where the bijective correspondence between QS and CL is evident.

2. Secondly, by integrating out the momenta $p_i$ we may argue that the same bijective correspondence holds for the original Lagrangian system.

VII) Finally, Ref. 3 lists KdV and sine-Gordon as counterexamples to an inverse Noether Theorem. KdV and sine-Gordon are integrable systems with infinitely many conserved charges $Q_n$, and one can introduce infinitely many corresponding commuting Hamiltonians $\hat{H}_n$ and times $t_n$. According to Olver, KdV and sine-Gordon are not really counterexamples, but just a result of a failure to properly distinguishing between non-trivial and trivial CL. See also Ref. 4.

References:

1. P.J. Olver, Applications of Lie Groups to Differential Equations, 1993.

2. V.I. Arnold, Mathematical methods of Classical Mechanics, 2nd eds., 1989, footnote 38 on p. 88.

3. H. Goldstein, Classical Mechanics; 2nd eds., 1980, p. 594; or 3rd eds., 2001, p. 596.

4. L.H. Ryder, Quantum Field Theory, 2nd eds., 1996, p. 395.

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$^{\dagger}$ Note that if one abandons real analyticity, say for $C^k$ differentiability instead, the analysis may become very technical and cumbersome. Even if one works with the category of smooth $C^\infty$ functions rather than the category of real analytic functions, one could encounter the Lewy phenomenon, where the equations of motion (eom) have no solutions at all! Such situation would render the notion of a conservation law (CL) a bit academic! However, even without solutions, a CL may formally still exists as a formal consequence of eoms. Finally, let us add that if one is only interested in a particular action functional $S$ (as opposed to all action functionals within some class) most often, much less differentiability is usually needed to ensure regularity.

$^{\ddagger}$ Maximal rank is crucial, while locally solvable may not be necessary, cf. previous footnote.

Answer 2

If the conservation law is general, meaning that it isn't specific to one motion, but conserved in a general configuration, then the answer is yes. This follows from the theory of canonical transformations in classical mechanics.

First, consider a perfectly triangular symmetric initial condition of three particles arranged on an equilateral triangle with velocities that are rotated by the appropriate angle (120 degrees, 240 degrees) to give a three-fold rotational symmetry. In this initial condition, for triangularly invariant force laws, there is a conservation of triangular symmetry, so that the configuration has the property that given the position of the center of mass and one of the particles, you can find the other two. This is the classical discrete symmetry, and it does not generalize to an arbitrary motion, so it has no symmetry associated with it.

But if you have a general conserved quantity Q(x,p) on the phase space which is conserved for all initial conditions x,p, then

$$[Q,H]_\mathrm{cl}=0$$

Where the bracket is the Poisson bracket. It follows that the motion on the phase space using Q as a Hamitonian

$$ {dx^i\over ds} = - {\partial Q\over \partial p_i}$$ $$ {dp_i\over ds} = {\partial Q\over \partial x^i} $$

makes a transformation of the phase space taking x,p to x(s),p(s), and this transformation commutes with the Hamiltonian time evolution, and defines a symmetry on the phase space whose Noether current gives the conservation of Q.

The same idea works in reverse, and in quantum mechanics, you just replace the classical Poisson bracket with the commutator, and use Q as a Hamiltonian to generate the wavefunction evolution:

$$ |\psi\rangle \rightarrow e^{isQ}\psi\rangle$$

and this gives you the symmetry. The nice thing in QM is that even discrete symmetries which are quantum mechanically exact give rise to conserved quantities, so that the triangular force law preserves the operator that does rotations by 120 degrees on the wavefunction, and one may classify the stationary states by their Z_3 discrete charge. The key difference is that any state in quantum mechanics may be written as a superposition of symmetric states, by superposing with rotated versions of itself with the appropriate phase.

Answer 3

I do not know how to prove the following but it should answer your question factually at least. The following I quote from the book 'Classical Mechanics' by Goldstein- "It should be remarked that while Noether's theorem proves that a continuous symmetry property o a Lagrangian density leads to a conservation condition, the converse is not true. There seems to be conservation conditions that cannot correspond to a symmetry property. The most prominent examples at the moment are the fields that have soliton solutions, e.g. , are described by th sine-Gordon equation or the Korteweg-deVries equation."

I hope this answers your question.

Answer 4

It is indeed true that there is a one-to-one correspondence between one-parameter groups of generalized variational symmetries of some functional and the conservation laws of its associated Euler-Lagrange equations. Precise statements and definitions can for example be found in chapter 5 of Olver, "Applications of Lie Groups to Differential Equations". Regardless of the question I highly recommend that book, if you are interested in such questions. In fact Noether already stated her theorem in this generality, but usually only the trivial aspects of it are discussed in physics courses.

A perhaps more interesting question is, which sets of differential equations can be the Euler-Lagrange equations of some variational problem, since it is at least conceivable that the description of some physical systems do not arise from variational problems. This was already studied by Helmholtz and also discussed in this book.

Ironically the Korteg-de-Vries equation admits an infinite number of such generalized symmetries, which is the reason, why it is "exactly solvable" and for the soliton solutions. So the accepted answer is not only wrong, but even the example given by the author is a good counterexample.

Answer 5

The real converse of the first theorem is the second one. Your formulation of the converse of the first theorem is too literal and thus valid as a particular case under additional conditions.