# If all conserved quantities of a system are known, can they be explained by symmetries?

If a system has $N$ degrees of freedom (DOF) and therefore $N$ independent1 conserved quantities integrals of motion, can continuous symmetries with a total of $N$ parameters be found that deliver these conserved quantities by means of Noether's theorem? I think this is not exactly the opposite of Noether's theorem since I don't ask whether for each conserved quantity a symmetry can be retrieved, I ask about a connection between the whole set of conserved quantities and symmetries.

1) or $2N-1$, or $N$, depending on definition and details that are irrelevant here. But let me expand on it anyway... I consider the number of DOFs equal to the number of initial conditions required to fully describe a system in Classical Mechanics. That means, velocities (or momenta) are considered individual DOFs, and not that each pair of coordinate + velocity make up only one DOF. Time is no DOF however, it is a parameter. Please discuss this in this question if you disagree.

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There seems to be a definition of "conserved quantities" propagating in this forums which is not shared by the rest of humanity. A conserved quantity is some combination of the dynamical variables which does not depend on time. The existence of those depend crucially on Noether theorem. The quantities you are referring to are combinations of the dynamical variables AND TIME, which characterize the initial conditions of the system. These are unrelated to symmetries, are "conserved" in only the most trivial way, and do not give you any information about the system. – user566 Apr 18 '11 at 14:54
The question itself is answered correctly by Marek, but the premise to the question (that any system has 2N or 2N-1 conserved quantities) is not correct. A generic system does not have any conserved quantities, and the most you can have is N, when the system is completely integrable. You seem to be following Vladimir in confusing conserved quantities with integration constants. – user566 Apr 18 '11 at 15:14
Dear @Hiatus. 1) The definition that a quantity $F=F(q,p,t)$ is a constant of motion (c.o.m.) is just that the total time derivative vanishes $\frac{dF}{dt}=0$. The definition has nothing to do with explicit time dependence, which may anyway change under coordinate transformations in phase space. 2) For finite many d.o.f. $N$ there are at most $2N$ independent c.o.m. The are not infinitely many independent c.o.m. 3) In the example of a free 1D particle in my comment above, it is possible to construct a symmetry that via Noether Thm leads to $F:=q-\frac{pt}{m}$ as a conserved Noether current. – Qmechanic Apr 18 '11 at 20:59
@Qmechanic: I agree with everything you say about “constants of motion”, which is what Landau and Lifshitz call “integrals of motion”. I am just pointing out that these are not the same thing as what is conventionally called “conserved quantities” which are related to symmetries and are used e.g. to simplify the equations of motion. The latter have general relation to symmetries via the Noether theorem, the former are just formal quantities that don’t help you solve anything or characterize the system (e.g. in general their form depends not only on time but on the specific trajectory). – user566 Apr 18 '11 at 21:33
@Hiatus I appreciate your input, it clarifies exactly what is being discussed here. You discussion of "integrals of motion" vs "conserved quantities" was helpful. I hope you don't stop posting here. – John Apr 21 '11 at 0:49

There are already several good answers. However, the off-shell aspect related to Noether Theorem has not been addressed so far. (The words on-shell and off-shell refer to whether the equations of motion (e.o.m.) are satisfied or not.) Let me rephrase the problem as follows.

Consider a (not necessarily isolated) Hamiltonian system with $N$ degrees of freedom (d.o.f.). The phase space has $2N$ coordinates, which we denote $(z^1, \ldots, z^{2N})$.

(We shall have nothing to say about the corresponding Lagrangian problem.)

1) Symplectic structure. Usually, we work in Darboux coordinates $(q^1, \ldots, q^N; p_1, \ldots, p_N)$, with the canonical symplectic potential one-form

$$\vartheta=\sum_{i=1}^N p_i dq^i.$$

However, it turns out to be more efficient in later calculations, if we instead from the beginning consider general coordinates $(z^1, \ldots, z^{2N})$ and a general (globally defined) symplectic potential one-form

$$\vartheta=\sum_{I=1}^{2N} \vartheta_I(z;t) dz^I,$$

with non-degenerate (=invertible) symplectic two-form

$$\omega = \frac{1}{2}\sum_{I,J=1}^{2N} \omega_{IJ} \ dz^I \wedge dz^J = d\vartheta,\qquad\omega_{IJ} =\partial_{[I}\vartheta_{J]}=\partial_{I}\vartheta_{J}-\partial_{J}\vartheta_{I}.$$

The corresponding Poisson bracket is

$$\{f,g\} = \sum_{I,J=1}^{2N} (\partial_I f) \omega^{IJ} (\partial_J g), \qquad \sum_{J=1}^{2N} \omega_{IJ}\omega^{JK}= \delta_I^K.$$

2) Action. The Hamiltonian action $S$ reads

$$S[z]= \int dt\ L_H(z^1, \ldots, z^{2N};\dot{z}^1, \ldots, \dot{z}^{2N};t),$$

where

$$L_H(z;\dot{z};t)= \sum_{I=1}^{2N} \vartheta_I(z;t) \dot{z}^I- H(z;t)$$

is the Hamiltonian Lagrangian. By infinitesimal variation $$\delta S = \int dt\sum_{I=1}^{2N}\delta z^I \left( \sum_{J=1}^{2N}\omega_{IJ} \dot{z}^J-\partial_I H - \partial_0\vartheta_I\right)+ \int dt \frac{d}{dt}\sum_{I=1}^{2N}\vartheta_I \delta z^I, \qquad \partial_0 \equiv\frac{\partial }{\partial t},$$

of the action $S$, we find the Hamilton e.o.m.

$$\dot{z}^I \approx \sum_{J=1}^{2N}\omega^{IJ}\left(\partial_J H + \partial_0\vartheta_J\right) = \{z^I,H\} + \sum_{J=1}^{2N}\omega^{IJ}\partial_0\vartheta_J.$$

(We will use the $\approx$ sign to stress that an equation is an on-shell equation.)

3) Constants of motion. The solution

$$z^I = Z^I(a^1, \ldots, a^{2N};t)$$

to the first-order Hamilton e.o.m. depends on $2N$ constants of integration $(a^1, \ldots, a^{2N})$. Assuming appropriate regularity conditions, it is in principle possible to invert locally this relation such that the constants of integration

$$a^I=A^I(z^1, \ldots, z^{2N};t)$$

are expressed in terms of the $(z^1, \ldots, z^{2N})$ variables and time $t$. These functions $A^I$ are $2N$ constants of motion (c.o.m.), i.e., constant in time $\frac{dA^I}{dt}\approx0$. Any function $B(A^1, \ldots, A^{2N})$ of the $A$'s, but without explicit time dependence, will again be a c.o.m. In particular, we may express the initial values $(z^1_0, \ldots, z^{2N}_0)$ at time $t=0$ as functions

$$Z^J_0(z;t)=Z^J(A^1(z;t), \ldots, A^{2N}(z;t); t=0)$$

of the $A$'s, so that $Z^J_0$ become c.o.m.

Now, let

$$b^I=B^I(z^1, \ldots, z^{2N};t)$$

be $2N$ independent c.o.m., which we have argued above must exist. The question is if there exist $2N$ off-shell symmetries of the action $S$, such that the corresponding Noether currents are on-shell c.o.m.?

Remark. It should be stressed that an on-shell symmetry is a vacuous notion, because if we vary the action $\delta S$ and apply e.o.m., then $\delta S\approx 0$ vanishes by definition (modulo boundary terms), independent of what the variation $\delta$ consists of. For this reason we often just shorten off-shell symmetry into symmetry. On the other hand, when speaking of c.o.m., we always assume e.o.m.

4) Change of coordinates. Since the action $S$ is invariant under change of coordinates, we may simply change coordinates $z\to b = B(z;t)$ to the $2N$ c.o.m., and use the $b$'s as coordinates (which we will just call $z$ from now on). Then the e.o.m. in these coordinates are just

$$\frac{dz^I}{dt}\approx0,$$

so we conclude that in these coordinates, we have

$$\partial_J H + \partial_0 \vartheta_J=0$$

as an off-shell equation. [An aside: This implies that the symplectic matrix $\omega_{IJ}$ does not depend explicitly on time,

$$\partial_0\omega_{IJ} =\partial_0\partial_{[I}\vartheta_{J]}=\partial_{[I} \partial_0\vartheta_{J]}=-\partial_{[I}\partial_{J]} H=0.$$

Hence the Poisson matrix $\{z^I,z^J\}=\omega^{IJ}$ does not depend explicitly on time. By Darboux Theorem, we may locally find Darboux coordinates $(q^1, \ldots, q^N; p_1, \ldots, p_N)$, which are also c.o.m.]

5) Variation. We now perform an infinitesimal variation $\delta= \varepsilon\{z^{I_0}, \cdot \}$,

$$\delta z^J = \varepsilon\{z^{I_0}, z^J\}=\varepsilon \omega^{I_0 J},$$

with Hamiltonian generator $z^{I_0}$, where $I_0\in\{1, \ldots, 2N\}$. It is straightforward to check that the infinitesimal variation $\delta= \varepsilon\{z^{I_0}, \cdot \}$ is an off-shell symmetry of the action (modulo boundary terms)

$$\delta S = \varepsilon\int dt \frac{d f^0}{dt},$$

where

$$f^0 = z^{I_0}+ \sum_{J=1}^{2N}\omega^{I_0J}\vartheta_J.$$ The bare Noether current is

$$j^0 = \sum_{J=1}^{2N}\frac{\partial L_H}{\partial \dot{z}^J} \omega^{I_0 J}=\sum_{J=1}^{2N}\omega^{I_0J}\vartheta_J,$$

so that the full Noether current

$$J^0=j^0-f^0=-z^{I_0}$$

becomes just (minus) the Hamiltonian generator $z^{I_0}$, which is conserved on-shell $\frac{dJ^0}{dt}\approx 0$ by definition.

So the answer is yes in the Hamiltonian case.

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Does this answer have a reference for further reading? In particular to the 'bare Noether current'. Many thanks! :) – AngusTheMan Oct 28 '15 at 22:18

Yes, this is the opposite of Noether's theorem. So let's call our conserved quantity $A$ (we will consider just one conserved quantity for starters) and begin with $\left \{H, A \right \} = 0$ law for conservation. Because of the connection between Poisson bracket with flows on the phase space this tells you both that $\mathcal{L}_{V_H} A$ = 0 ($A$ is conserved in time evolution) and $\mathcal{L}_{V_A} H = 0$ (Hamiltonian is conserved under symmetry that has fundamental field $V_A$) as long as the vectors fields associated to the functions on phase space as $V_X = ({\rm d} X)^{\sharp}$ are non-degenerate in certain sense. Note that raising $(\cdot)^{\sharp}$ operator is defined obviously using the inverse symplectic form. This e.g. means that for example genuine constants (which are certainly also constants of motion) won't work because ${\rm d} C = 0$ and we obtain zero vector field.

On the other hand, as long as all the conserved quantities are non-degenerate we can always find the associated symmetry flows by integration of the above mentioned vector fields. But note that what we receive in the end are symmetries of the phase space. Whether these are also directly linked to some symmetry of the underlying position space (if there is such a space, that is; in general there need not be one but in usual applications we take the phase space of the position manifold $M$ as the cotangent bundle $T^*M$) is a question for further investigation. I'll try looking into it later, if I find some time.

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thanks for your answer. Unfortunately I'm not familiar with the notation $\mathcal{L}_{V_H}$ and $(dX)^\sharp$, what do they mean? (Pointing to some reference or wikipedia entry would suffice) – Tobias Kienzler Apr 18 '11 at 9:13
@Tobias: sorry about that. They are all basic notions in differential geometry; Lie derivative ${\mathcal L}_V$ with respect to vector field $V$, exterior derivative ${\rm d} \mathbb \omega$ of differential form $\omega$ and musical isomorphism $(\cdot)^{\sharp}$ which maps forms to vectors. E.g. if you had a metric, this is usual index acrobatics $V^i = g^{ij}V_j$ and so on... – Marek Apr 18 '11 at 10:17
@Tobias: likewise, I am sorry for being overly formal. It is apparently detrimental to the communication. Anyway, I think we get each other now. And indeed, if we have such a family of constants, we can think of them as coordinates of the whole phase space (if we add also time as the $2N^{\rm th}$ coordinate, that is). Therefore there will indeed always be $2N$ symmetries of the phase space. So having this out of the way, the gist of your question is when such a symmetry comes from a symmetry of a position space. Which I still intend looking into :) – Marek Apr 18 '11 at 15:54
@Tobias: although having read Moshe's last comment, I see I had gotten ahead of myself :) It might indeed not be possible to parametrize the phase space by $2N$ coordinates. E.g. thinking about chaotic systems, if you perturb initial condition a little (which corresponds to local change of coordinates) you obtain a family of paths that walk wildly all over the phase space. Obviously these can't really serve as global coordinates. Similar remarks are probably also an origin of there being at most $N$ conserved quantities -- a statement I am familiar with but haven't really looked into.. – Marek Apr 18 '11 at 16:05
@Marek, the gist of your answer is correct: the conserved charge generates a flow on the phase space which is a symmetry as long as the charge is conserved. – user566 Apr 18 '11 at 16:21

I'll start with some terminological subtleties, that have to be accounted for, when it comes to "conserved quantities" or "integrals of motion".

First of all it is important to state on which variables the quantities can depend. In the field of differential equations, Hamiltonian dynamics and dynamical systems, a conserved quantity by definition does not explicitly depends on time -- it is usually a function, defined on a space (a manifold), that fully describes the state of your system (uses all the degrees of freedom as you have defined them).

Another thing to be taken into account -- It seems to be possible to take a function of a constant of motion, therefore arriving at another quantity that is constant, therefore producing any amount of constants of motion. So there is an implicit condition for functional independence between those $N$ constants of motion. Which can be formulated as the vanishing of the Poisson brackets between any pair of the quantities. Here is the reference about it on Wikipedia, also I've asked a question about it some time ago...

So I think that your implicit statement:

If a system has N degrees of freedom (DOF) and therefore N conserved independent quantities

Is not totally correct, if one takes into account what was said about the notion of "conserved quantity".

Concerning your question about finding the symmetry transformation, that corresponds to the quantity -- I've addressed the issue in this question.

Briefly -- given a "generator" $\delta G$ and some quantity $A$. The small transformation, generated by $\delta G$ is: $$A \to A+\delta A,\quad\quad \delta A = -\{\delta G, A\}$$

Putting $A = H$, and noting that, if $\delta G$ is a constant of motion, then $\{\delta G, H\} = 0$. One immediately arrives at the conclusion that the transformation, generated by $\delta G$ is the symmetry transformation. (Some examples are here.)

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Dear @Kostya: I agree that the question formulation probably should have used the word integrals of motion rather than conserved quantities, in order not to clash with traditional use. However, that does not change the fact that if one reads beyond semantics, the heart of the question is mathematically sound, interesting and non-trivial. – Qmechanic Apr 19 '11 at 9:50
Thanks for this link. This seems to answer my question via a constructive proof, right? Also (@Qmechanic, too) I modified my question to state integrals of motion instead of conserved quantities, both of you are right here that talking about conservation may be misleading to confusion with time-independence – Tobias Kienzler Apr 19 '11 at 10:17
Simple test example: Relativistic mass conservation, $\delta G := -\frac12 p^2\epsilon$ yields $\delta A = \epsilon(p^\mu\partial_\mu)A$, thus generating motion in space-time. – Tobias Kienzler Apr 19 '11 at 10:22
Dear @Tobias Kienzler: Well, for starters, Kostya's answer (v1) and Marek's answer (v1) assume an isolated system, and they do not address the analogous Lagrangian question. – Qmechanic Apr 19 '11 at 10:38
@Qmechanic: Thanks, I'm fine with isolated systems (conservation in the presence of external forces either violate actio=ratio or require e.g. an infinite mass or some other artificial mechanism anyway, right?). But the restriction to Hamiltonian systems is true, though I wonder if it were sufficient to prove the existence of a (weak?) Hamiltonian since $H$ is not explicitly required to construct the symmetry, but only to state the conservation, as Kostya showed. So maybe the statement $\delta A = -\{\delta G, A\}$ is valid in a more general context if the poisson bracket is well-defined – Tobias Kienzler Apr 19 '11 at 10:46

Dear Tobias, everything is mush simpler: the equations are constraints to possible system motion and they themselves provide conserved quantities. You may call these equation "constraints" or "symmetry requirement", if you like but they are necessary and sufficient conditions for having conserved quantities (=solutions).

Let us consider an ODE: $F(\dot{x},x,t) = 0$ with some initial data $x_0 (t_0)$. After solving this equation you obtain a solution that can be casted in an implicit form like this: $f(x(t), t, x_0)=0$. On the other hand, this implicit form of solution is a conservation law: a combination of the problem variable and the time does not depend on time.

Now I differentiate this implicit solution and obtain: $f'_x\cdot \dot{x} + \dot{f} = 0$. It is an expression of conserving something, isn't it? On the other hand, it should coincide with or be equivalent to the original equation $F=0$. Now you understand that equations are necessary and sufficient requirements to have conserved quantities. The latter are to a great extent synonymous to solutions.

EDIT: There is no need in any symmetry to have conserved quantities. Only equations. Symmetries simplify the look of conserved quantities (solutions). Consider a 1D motion of a particle in a time-dependent external force. The energy and momentum are not conserved but there are still two conservation "laws".

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Dear @Vladimir Kalitvianski: This answer (v2) does not address the main question, namely, if there exist $2N$ symmetries of the action $S$. – Qmechanic Apr 18 '11 at 12:08
See my comment above, a conserved quantity in classical mechanics is defined as combination of the dynamical variables (without explicit dependence on time) which is time independent. The existence of those relies on symmetries. If you allow time dependence for your "conserved" quantities, the definition becomes meaningless since any quantity, with any time-dependence, can be made to be "conserved". Now, this is basic classical mechanics, let's see just how long it takes to establish this simple and universally accepted fact. – user566 Apr 18 '11 at 14:58
The quantities you construct depend explicitly on time, and are therefore not what is referred to conventionally as “conserved quantities” (which are related to Noether theorem). Having your own definition of conventional terms is one excellent way to spread confusion. I have zero expectation that you’ll modify your behaviour, but I am hoping other users may get less confused now. – user566 Apr 18 '11 at 17:32
Whatever. When people ask about relation between symmetries and conserved quantities, they are using both those terms as they are defined conventionally. If you want to have a “better” definition use a different term, otherwise you are just spreading confusion. – user566 Apr 18 '11 at 17:40
OK, I tried my best here. You hang on to your confusions for dear life. Just too bad they seem to be spreading. – user566 Apr 18 '11 at 17:48