# Converse of Noether's Theorem

1. Aug 23, 2014

### Boorglar

A free rigid body (no forces/torques acting on it) has a constant angular momentum. And yet, I am puzzled because there seems not to be a corresponding rotational symmetry in the Lagrangian, in this case.

While studying the equations of motion for a free rigid body, I decided to work out the Lagrangian in terms of the Euler angles as generalized coordinates. I got the following expression for the kinetic energy (rotational, ignoring the center of mass movement), which is also the Lagrangian:
$$L = \frac{1}{2}\left(I_1ω_1^2 + I_2ω_2^2 + I_3ω_3^2\right) = \frac{1}{2}\left[I_1\left(\dot{\phi}\sinθ\sinψ + \dot{θ}\cosψ\right)^2 + I_2\left(\dot{\phi}\sinθ\cosψ-\dot{θ}\sinψ\right)^2 + I_3\left(\dot{ψ}+\dot{\phi}\cosθ\right)^2\right]$$
The $ω_i$ are the components of the angular velocity in the body frame, and can be written in terms of the euler angles ($\phi,\ θ,\ ψ$ for precession, nutation and spin respectively of the body frame). The $I_i$ are the principal moments of inertia.

Now Noether's Theorem implies that if the Lagrangian is rotationally symmetric, angular momentum is conserved. But in this case, the Lagrangian does not seem to be rotationally symmetric. Indeed, a rotation would correspond to a change in one of the Euler angles. And yet $\frac{∂L}{∂θ},\ \frac{∂L}{∂ψ} ≠ 0$.

Does this mean that not every conservation law arises from symmetries of the Lagrangian, or have I misunderstood what the rotational symmetries should be (i.e. a change in one Euler angle is not actually a rotation)?

2. Aug 24, 2014

### ShayanJ

The point is the symmetries are w.r.t. angles around the body coordinates, not the fixed coordinates and so the ignorable coordinates are not Euler angles but the angles around body coordinates.

3. Aug 24, 2014

### Boorglar

Hello Shyan, and thanks for the answer.
Can you explain what are the "angles around body coordinates"? I thought those were the Euler angles, but now I am not sure...

4. Aug 24, 2014

### ShayanJ

Euler angles are expressing the orientation of the body through giving the inclination of a coordinate system natural to it w.r.t. a fixed coordinate system. So Euler angles are angles around the fixed axes.

5. Aug 25, 2014

### vanhees71

Noether's theorem goes in both directions! If there is a one-parameter symmetry Lie group the corresponding "generator" is a conserved quantity and any conserved quantity defines a one-parameter symmetry group.

As a closed system the free top has 10 such conserved quantities corresponding to the natural 10 one-parameter subgroups of the Galilei group

homogeneity of time <-> energy
homogeneity of space <-> momentum
isotropy of space <-> angular momentum
invariance under Galilei boosts <-> center of mass velocity

In your example, only the Euler angle $\phi$ which refers to the rotation around the fixed 3-axis in the lab frame (which is tacitly choosen as an inertial frame of course) reflects one such one-parameter group, and thus only this coordinate is cyclic and the corresponding conserved quantity, the associated canonical momenum, is the three-component of angular momentum. Since you know that all three components of angular momentum is conserved, it is wise to choose it in direction of the lab-frame 3-axis when introducing the Euler angles.

The other two angles are rotations around axes referring to non-inertial reference frames. E.g., $\psi$ is the rotation around a axis fixed in the body (the 3-axis in the body system). It becomes cyclic for the symmetric top, i.e., for the case $I_1=I_2$, because then the rotation of the body around this axis is a symmetry of the system.

6. Aug 26, 2014

### samalkhaiat

This is not correct. The logical implication in Noether theorem goes as follow:

If $G$ is a Lie group of transformations, then
$$\{ G: \ \delta S[ \phi ] = 0 \} \Leftrightarrow \ \{ \frac{ \delta S }{ \delta \phi_{ a } } \delta \phi_{ a } + \partial_{ \mu } J^{ \mu } = 0 \} .$$
So by itself, current conservation does not implie that the associated charge, $Q = \int d^{ 3 } x J^{ 0 } ( x )$, generates a symmetry of the action. Indeed, the theory may allow you to construct various conserved currents (and charges) that may not correspond to any symmetry of the action. For example, from the tensor field $T_{ \mu \nu \rho } = - T_{ \nu \mu \rho }$, we can form the following identically conserved currents
$$J_{ \mu \rho } = \partial^{ \nu } T_{ \mu \nu \rho } , \ \Rightarrow \ \partial^{ \mu } J_{ \mu \rho } = 0 ,$$
$$J_{ \mu } = \partial^{ \nu } \partial^{ \rho } T_{ \mu \nu \rho } , \ \Rightarrow \ \partial^{ \mu } J_{ \mu } = 0 .$$
The well-known examples of non-symmetry currents are the so-called topological currents and charges (topological mapping index or winding number)
$$K_{ \mu } = \epsilon_{ \mu \nu \rho \sigma } \epsilon^{ a b c } \ \partial^{ \nu } \phi_{ a } \ \partial^{ \rho } \phi_{ b } \ \partial^{ \sigma } \phi_{ c } ,$$
$$Q = \int_{ S^{ 2 } } d^{ 2 } \sigma^{ i } \ \epsilon_{ i j k } \epsilon^{ a b c } \ \phi_{ a } \ \partial^{ j } \ \phi_{ b } \ \partial^{ k } \phi_{ c } .$$

Sam

7. Aug 27, 2014

### vanhees71

Thanks for pointing out this caveat.

I was referring to Noether's theorem in classical mechanics. If you have a conserved quantity $A$, which means its Poisson bracket with the Hamiltonian vanishes, it defines a one-parameter symmetry group with group elements
$$g(\lambda)=\exp(\lambda \mathcal{L}_A),$$
where
$$\mathcal{L}_A f=\{f,a \}_{\text{pb}}$$
is the Lie derivative on the symplectic phase space.

Of course, this is only true, if the exponential of the Lie derivative converges. Are there examples in mechanics, where this is not the case?

8. Aug 30, 2014

### samalkhaiat

1. Noether Theorem in the Lagrangian Formalism

Classical mechanics does not change the logic in Noether theorem. That is, the action is invariant if and only if the following (Noether) identity holds
$$\frac{ \delta S }{ \delta q_{ a } } \delta q_{ a } + \frac{ d C }{ d t } \equiv 0 , \ \ \ \ \ \ (1.1)$$
where
$$C( q , \dot{ q } ) = \frac{ \partial L }{ \partial \dot{ q }_{ a } } \delta q_{ a } + Q ( q , \dot{ q } , t ) , \ \ \ \ \ \ \ (1.2)$$
and
$$\frac{ d Q }{ d t } = \delta L . \ \ \ \ \ \ (1.3)$$
It is clear from (1.1) that $C( q , \dot{ q } , t )$ is conserved provided that its arguments, $(q , \dot{ q })$, satisfy the Euler-Lagrange equations
$$\frac{ \delta S }{ \delta q_{ a } } = 0 . \ \ \ \ \ \ (1.4)$$
A quantity that is conserved only on actual (extremal) paths is called constant of motion. So, the conserved Noether charge, $C$, is a constant of motion. However, quantities that are conserved for all paths are not constants of motion. In other words, if some $F(q , \dot{q} , t)$ exists such that $d F / d t = 0$ does not follow from (1.4), then this $F$ is not the conserved Noether charge (not constant of motion). While the conserved Noether charge (constant of motion) generates a one-parameter symmetry group of the action, no symmetry can be associated with the conserved quantity $F(q , \dot{q} , t)$.
Now, I would like to point out to the remarkable, peculiar and subtle differences between the Lagrangian and the Hamiltonian formalisms:
a) The Noether identity (1.1) does not point to the fact that the conserved Noether charge is a generator of symmetry transformations, i.e., the identity (1.1) contains no trace of the symmetry transformations
$$\delta_{ \epsilon} q_{ a } = \{ \ q_{ a } , C \ \} = \frac{ \partial C }{ \partial p^{ a } } , \ \ \ \ \ \ (1.5a)$$
or even
$$\delta_{ \epsilon } p^{ a } = \{ \ p^{ a } , C \ \} = - \frac{ \partial C }{ \partial q_{ a } } . \ \ \ \ \ \ (1.5b)$$
Indeed an extra work is needed to derive these transformations. Compare this with the Hamiltonian version of the identity to be discussed below.
b) Here comes one bizarre thing. The form of $C$, i.e., EQ(2), and the three fundamental Poisson brackets between $q_{ a }$ and $p^{ b }$ are sufficient to show that
$$\delta_{ \epsilon} q_{ a } = \{ \ q_{ a } , C \ \}, \ \ \ \ \ \ \ (1.6a)$$
and
$$\delta_{ \epsilon } p^{ a } = \{ \ p^{ a } , C \ \} + f(t) \frac{ \delta S }{ \delta q_{ a } } . \ \ \ \ \ \ (1.6b)$$
It is remarkable that these equations can be derived without any symmetry considerations (i.e., $C$ needs not be conserved) and without reference to the dynamics (i.e., the equations of motion need not be satisfied).
The presence of the equation of motion in (1.6b) points to the fact that $p^{ a }$ is not a dynamical variables in the Lagrangian formalism.
The important point to notice here is the fact that equations $(1.6a)$ and $(1.6b)$ do not imply that $C$ is a symmetry generator, i.e., they don’t imply $d C / d t = 0$. Again, compare this to the Hamiltonian case below.

2. The Hamiltonian Version of Noether Theorem

Given the action
$$S = \int_{ 1 }^{ 2 } d t \ \left( p^{ a } q_{ a } - H( q , p ) \right) , \ \ \ \ \ \ (2.1)$$
and the phase space transformations
$$q_{ a } \rightarrow q_{ a } + \delta q_{ a } , \ \ p^{ a } \rightarrow p^{ a } + \delta p^{ a } , \ \ \ \ \ (2.2)$$
one can show that the necessary and sufficient condition for the invariance of the action is given by the following “Noether” identity
$$\left( \dot{ q }_{ a } - \frac{ \partial H }{ \partial p^{ a } } \right) \delta p^{ a } - \left( \dot{ p }^{ a } + \frac{ \partial H }{ \partial q_{ a } } \right) \delta q_{ a } + \frac{ d C }{ d t } \equiv 0 , \ \ \ \ \ (2.3a)$$
where
$$C( q , p ) = p^{ a } \delta q_{ a } + Q( q , p ) , \ \ \ \ \ (2.4)$$
and the change in the value of $H$ at a fixed point of phase space is given by
$$\delta H = - \frac{ d Q }{ d t } . \ \ \ \ \ \ \ (2.5)$$
I encourage the reader to derive the identity (2.3a), as it requires good understanding of the principle of least action.
As in the Lagrangian formalism, the “Noether” identity (2.3a) show that the conservation of $C$ follows from the equations of motion (i.e., the Hamilton equations). However, unlike the Lagrangian version, implicit in (2.3a) is the fact that $C$ is the generator of the canonical symmetry transformations (2.2). To see this, we use
$$\frac{ d C }{ d t } = \frac{ \partial C }{ \partial q_{ a } } \dot{ q }_{ a } + \frac{ \partial C }{ \partial p^{ a } } \dot{ p }^{ a } + \frac{ \partial C }{ \partial t } ,$$
and rewrite (2.3a) in the form
$$\dot{ q }_{ a } \left( \delta p^{ a } + \frac{ \partial C }{ \partial q_{ a } } \right) - \dot{ p }^{ a } \left( \delta q_{ a } - \frac{ \partial C }{ \partial p^{ a } } \right) + \frac{ \partial C }{ \partial t } - \frac{ \partial H }{ \partial p^{ a } } \delta p^{ a } - \frac{ \partial H }{ \partial q_{ a } } \delta q_{ a } = 0 . \ \ (2.3b)$$
From this identity, it follows that
$$\delta q_{ a } = \frac{ \partial C }{ \partial p^{ a } } \equiv \{ \ q_{ a } , C \ \} , \ \ \ \ \ \ \ (2.6a)$$
$$\delta p^{ a } = - \frac{ \partial C }{ \partial q_{ a } } \equiv \{ \ p^{ a } , C \ \} , \ \ \ \ \ (2.6b)$$
and the “change” in the form of the Hamiltonian will then be given by
$$\bar{ \delta } H = \frac{ \partial C }{ \partial t } + \{ \ C , H \ \} = \frac{ d C }{ d t } = 0 . \ \ \ \ \ (2.7)$$
Thus, the statement of Noether theorem in the Hamiltonian formalism becomes that of form invariance of the Hamiltonian function.
So, if the conservation of some quantity $F( q , p )$ does not follow from the Hamilton equations (i.e., the phase space curve $( q(t) , p(t) )$ is not extremal), we can at best write
$$0 = \frac{ d F }{ d t } = \frac{ \partial F }{ \partial q_{ a } } \dot{ q }_{ a } + \frac{ \partial F }{ \partial p^{ a } } \dot{ p }^{ a } .$$
Since we can not use the Hamilton equations, we cannot even form the Poisson bracket. So, in general, it is not correct to say that $d F / d t = 0$ implies $\{ \ F , H \ \} = 0$. However, if $F$ is a constant of motion, i.e., it is conserved only on-shell, then it is true that
$$d F / d t = 0 \ \Leftrightarrow \ \{ \ C , H \ \} = 0 \ \Leftrightarrow \ C \ \mbox{is a generator}.$$
Indeed, some Hamiltonian systems in fluid dynamics have infinitely many conserved quantities but possess only 2 or 3 constants of motion.

Sam

Last edited: Aug 30, 2014
9. Aug 30, 2014

### vanhees71

I'm a bit puzzled by the very last part of your posting. The math of the Noether theorem is, of course correct.

Now, for any field on phase space, $F(t,q,p)$, plugging in the trajectory, which satisfies the Hamilton canonical equations,
$$\dot{q}=\nabla_{p} H, \quad \dot{p}=-\nabla_q H,$$
you get
$$\frac{\mathrm{d} F}{\mathrm{d} t}=\partial_t F + \dot{q} \cdot \nabla_{q} F+\dot{p} \cdot \nabla_{p} F=\partial_t F + \nabla_{p} H \cdot \nabla_{q} F-\nabla_{q} H \nabla_{p} F=\partial_t F + \{F,H \}.$$
Now, if this vanishes this is precisely the condition for an infinitesimal symmetry transformation generated by $F$.

I'm also aware of the fact that gauge transformations, i.e., such transformations that keep the action invariant identically, i.e., also not only for the solutions of the equations of motion, then you have in fact a constraint system and you have to use Dirac's formalism to switch to the Hamiltonian formulation, substituting the Poisson Brackets by Dirac Brackets, and that then there is no conserved quantity corresponding to the gauge transformation.

Could you point me to a textbook or paper about the statement on fluid dynamics? That sounds very interesting.

10. Sep 1, 2014

### samalkhaiat

There is no puzzle. You have already mentioned one part of the problem. Time-independent constraints define conserved quantities. The constraint $\vec{ r } (t) \cdot \vec{ r } (t) = 1$ is a conserved quantity. What symmetry group does $r^{ 2 }$ generate?

This quantity is a generator of symmetry (constant of motion) because its conservation follows from the equation of motion. On-shell conservation generates symmetry of the action.

The other part of the problem is related to the existence of identically conserved quantities. This can be explained by the following simple example. Consider the action integral of vibrating one-dimensional spring
$$S = \frac{ 1 }{ 2 } \int d t \ d x \ \left( ( \frac{ \partial Y }{ \partial t } )^{ 2 } - ( \frac{ \partial Y }{ \partial x } )^{ 2 } \right) .$$
Treating this as (1+1)-dimensional field theory, we can define the following currents:

a) $J_{ \mu } = \partial_{ \mu } Y(t , x )$. Since the conservation of this current follows from the equation of motion, it must be symmetry current.

b)
$$j^{ \mu } = \epsilon^{ \mu \nu }\partial_{ \nu } Y( t , x ) . \ \ \ \ (1)$$
This current is identically conserved. Therefore, it is not related to any symmetry of the action.
The idea of infinitely many conserved currents follows (essentially) from this type of current by power expansion. Basically, the “field” $Y$ in the problem depends on some spectral parameter. So, expanding $Y(t , x ; \lambda )$ and $j^{ \mu } ( t , x ; \lambda )$ in power of $\lambda$ will generate an infinite number of conserved currents.

You can look up “infinite conservation laws” in Kadomtsev-Petviashvili equation (in fluid and plasma), Davey-Stewartson equations (which model the evolution of non-linear water waves of infinite depth), the KdV equation, the Burgers-Hopf equation and soliton solutions in non-linear evolution equations such as the Sine-Gordon equation.

Sam

11. Dec 16, 2014

### ShayanJ

I know this thread is pretty old but the quoted post seems very important to me(I can't remember considering one of Sam's posts as not important, thanks Sam!)and I wanna grasp it fully. But I can't. Things aren't clear to me. Can someone add some intermediate steps and explanations or introduce some references?
Thanks