Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
Related: physics.stackexchange.com/q/8626/2451 –  Qmechanic Apr 29 '12 at 19:02
1  
You should be slightly more clear about what you mean by "conservation law". If I define the auxiliary quantity $\mathrm{Blah}$ by requiring it to be equal to $1$ on the entire phase space, then trivially $\mathrm{Blah}$ is conserved. For a slightly less trivial analogue: if your phase space is not connected, you can have $\mathrm{Blah}$ be some piecewise constant function. This is the case for topological charge in field theories. –  Willie Wong Apr 30 '12 at 11:26
3  
@WillieWong: These are not counterexamples--- the constant functions are trivial conservation laws, and they correspond to the trivial symmetry which does nothing. The piecewise constant functions are also trivial symmetries (corresponding to a different phase rotation on the different disconnected pieces, or an independent shift in the action on the different peices) –  Ron Maimon Apr 30 '12 at 13:01
    
@Ron: if you allow the trivial symmetry, isn't it also necessarily a symmetry of any conservation law? This I think violates the "spirit" of the question. I interpret the question to mean more along the lines of "whether every conservation law can be written as the conservation law derived from Noether's theorem". –  Willie Wong May 1 '12 at 7:43
2  
@WillieWong: the symmetry that is the trivial symmetry for any value of the parameter S is uniquely associated with the constant functions. This is the motion generated by using a constant Hamiltonian. It does not violate the spirit--- constants correspond to trivial motions, and any locally nonconstant function generates an infinitesimal motion. –  Ron Maimon May 1 '12 at 12:51

5 Answers 5

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 $$ S[q] ~=~ \int\! dt~ L(q(t), \frac{dq(t)}{dt}, \ldots,\frac{d^Nq(t)}{dt^N} ;t), $$ 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),

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

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

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

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. Ref. 2 and Example 1. 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

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

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. this answer. Here we briefly recall the proof. The on-shell CL implies

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

off-shell, cf. Remark 2. The corresponding transformation

$$\delta q^i~=~ \{q^i,Q\}\epsilon\qquad \text{and}\qquad \delta p_i~=~ \{p_i,Q\}\epsilon$$

is a QS of the Hamiltonian Lagrangian

$$\delta L_H ~=~ \dot{q}^i \delta p_i -\dot{p}_i \delta q^i -\delta H -\frac{d }{dt}(p_i \delta q^i)$$ $$~=~ -\dot{q}^i\frac{\partial Q}{\partial q^i}\epsilon -\dot{p}_i\frac{\partial Q}{\partial p_i}\epsilon -\{H,Q\}\epsilon + \epsilon \frac{d Q^0}{dt} ~=~ \epsilon \frac{d f^0}{dt},$$

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

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

and

$$f^0~=~ Q^0-Q .$$

Hence the corresponding full Noether charge

$$ Q~=~Q^0-f^0 $$

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.


$^{\dagger}$ 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.

share|improve this answer
    
I don't like this answer--- it is unnecessary, intimidating nonsense. Here are specific objections: 1. The requirement of real analyticity is certainly unnecessary, you can have a spherically symmetric potential with a slope discontinuity, and angular momentum is still conserved. It is absolutely wrong to insist on this, no matter what Olver says. 2. The conserved quantities in KdV and Sine Gordon do not constitute counterexamples in any sense, I disagree with this--- the Lax conservation laws are real, and give nontrivial canonical motions. There is no way to call them counterexamples. –  Ron Maimon May 10 '12 at 23:55
    
I'm not competent enough to comment on the correctness of this answer, but +1 for the well crafted, elegant beauty of its structure and +1 for the interesting links and references. –  John McVirgo May 15 '12 at 0:25
1  
@Qmechanic: The answer is better now, +1, but quibbles: 1. The requirement of analyticity is still no good, you should not need it, and I don't see a counterexample which isn't forbidden by much less draconian conditions. 2. The statement that an on-shell conservation law is trivial is strange--- why is it trivial? It is a statement that a quantity is conserved along trajectories, and it still makes a canonical transformation. (I erased a previous comment you adressed with the edits). –  Ron Maimon May 15 '12 at 2:40

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.

share|improve this answer
    
In the triangular case, finding the third particle using the centre of mass and two particle coordinates is always trivially possible. Given the symmetry, only two of those three data are needed. –  Emilio Pisanty Apr 30 '12 at 16:44
    
@episanty: Of course you're right. I fixed it, thanks. –  Ron Maimon Apr 30 '12 at 23:34

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.

share|improve this answer
    
There is a reverse Noether theorem for conservation laws which are not special to a particular initial condition. The solitary wave equations have extra conservation laws, but conservation of "soliton shape" is not exactly the way to formulate it. –  Ron Maimon Apr 30 '12 at 4:37
1  
Before this answer is upvoted too much, what is the precise counterexample? The extra conservation laws in soliton equations also correspond to symmetries of the phase space. –  Ron Maimon May 1 '12 at 4:03
    
Well, can't argue with the master himself - Goldstein. But even so, I'm shocked. –  Larry Harson May 4 '12 at 21:10
    
This answer is wrong or at least imprecise. One should never accept arguments from authority. –  orbifold May 6 '12 at 10:53
    
-1: Especially when the argument from authority is incorrect. –  Ron Maimon May 6 '12 at 21:39

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.

share|improve this answer

Ah well, I'll take a chance and point out an interesting feature of differentiation itself: Smoothness, all by itself as a mathematical assertion about any system, seems unavoidably to imply some concept of conservation.

Here's what I mean by that: Why should a given line or space not be infinitely disconnected, as in the Cantor set or perhaps some more chaotic equivalent of it? Well, for one thing, it would require infinite information to fully specific such a space, so just from that alone, an infinitely disconnected space cannot exist as an actual, existing entity within our universe. Like the Mandelbrot space, we can only explore tiny pieces of such a space, and even then never to the infinite limit.

So, when we talk about differentiable spaces in mathematics, we do so because there is no other choice for creating structures that apply meaningfully to the limits that exist within our particular universe for quantities such as mass, energy, and information. The disconnectedness of the Cantor set seems a lot more intuitive as a starting assumption. It's then the conservation constraint of "sorry, but only get a limited set of resources with which to represent it" that causes more familiar concepts of smooth lines and limited detail at very fine scales to pop out.

So, apart even from the symmetry issue, I think it's important to think very carefully about how nominally "pure mathematics" concepts such as differentiation apply to the physics of our universe as we find it. We are very much creatures of that universe, and we tend to be subtly (and not-so-subtly, e.g. linear algebra) guided by its built-in "givens" to favor certain forms of mathematical analysis as superior or in some way more fundamental.

share|improve this answer
1  
Negative comments are welcome! I added this answer "at risk" knowing quite well it's an unusual spin, but I also find the issue of continuity as a given of physics quite fascinating and I think relevant. Human minds are built to accept everyday physics without much questioning, so I really do think some awareness is good. But tell me what I said wrong, please! –  Terry Bollinger May 9 '12 at 1:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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