# Problem with Poisson's Theorem

Gold Member
So, Poisson's theorem states that if 2 variables, u and v, are constants of the motion, then one can find a third constant of the motion {u,v} where {u,v} is the Poisson bracket. This is a consequence of Jacobi's identity and the fact that:

$$\frac{du}{dt}=[u,H]+\frac{\partial u}{\partial t}$$

Here [] are Poisson brackets (The {} symbol goes away in latex...I'm not sure how to put that in there).
And then we can find that {Lx,Ly}=Lz where L is the angular momentum. Therefore, if Lx and Ly are constants of the motion, so is Lz. This theorem seems quite general to me.

But this is very counter-intuitive for me, since it makes no sense for me why I can't just apply a torque in the z-direction. In fact, why can't I label whichever direction I apply the torque in, as the z-direction, and therefore only change Lz, but not Lx and Ly.

As a simple example, consider the pendulum. The pendulum oscillates in the x-y plane. Thus, the L vector is only in the z plane, and it oscillates from positive to negative as the pendulum bobs back and forth. In this example, only Lz is changing, Lx and Ly are identically 0 for all time.

So dLx/dt=0, dLy/dt=0, but dLz/dt=torque_z.

So it would seem contradictory with Poisson's Theorem.

I can't find the source of this contradiction, and it's bothering me. Please someone enlighten me.

Gold Member
I'm sure someone has an answer to this...=[

lalbatros
It depends on the boundaries of the system.
If you spin a top with your fingers, then the hamiltonian for a top in a gravity field cannot represent what happens. In this case, you would need to include your full personal hamiltonian into the theory!
And the theorem would still hold!
The z-spin transferred to the top would be balanced by the opposite z-spin tranferred to your body.

Obvioulsy the Poisson theorem does not apply to this situation.
It applies to a well defined class of system.
It happens that this class is wide enough and covers almost everything one needs in physics.
But don't touch a top with your fingers!
Hamiltonians have no fingers.

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Gold Member
But the pendulum is just in a gravitational field, I'm not touching it. Are you saying that for this case, I would have to include the angular momentum of the Earth? In that case...isn't this just a statement of conservation of angular momentum again? <_<

Gold Member
What I'm saying is, even in a conservative force-field, I can always define my coordinate system such that the torque is directed along only 1 axis can't I? In that case, why can't I only change just 1 component of the angular momentum?

Pavlov
Hi, I know that topic is old but I was looking for the solution of the same problem and after a long time I've found this: http://www.physics.sc.edu/~yar/Phys701_2009/homework/hw9_solutions.pdf [Broken] (last paragraph).

In Section 9.7, it is shown that if any two components of the angular momentum are conserved, then the total angular momentum is conserved. It would then seem that if two of the components are identically zero and constant, the third must be conserved. From this it would appear to follow that in any motion confined to a plane, so that the components of the angular momentum in the plane are zero, the total angular momentum is constant.
There appear to be a number of obvious contradictions to this prediction; for example, the angular momentum of an oscillating spring in a watch, or the angular momentum of a plane disk rolling down an inclined plane all in the same vertical plane. Discuss the force of these objections and whether the statement of the theorem requires any restrictions.
Solution: Here the point is that having Lx and Ly being equal to zero and having them conserved are two different statements. Conservation means that I can set Lx and Ly to any values and they will not change in time. Clearly in all the examples given in the problem it is not possible to set Lx and Ly to any values other then zero. These projection are constant not because they are conserved, but because there are reaction torques of the environment that keep them at zero values. We conclude that the theorem of Sec.
9.7 is correct, but was incorrectly applied in the problem.

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