# Why does a cyclic coordinate reduce the dimension of the cotangent manifold by 2?

Our professor's notes read, "In general, in Hamiltonian dynamics a constant of motion will reduce the dimension of the phase space by two dimensions, not just one as it does in Lagrangian dynamics." To demonstrate this, he uses the central force Hamiltonian,

$$H=\frac{P_r^2}{2m}+\frac{p_{\theta}^2}{2mr^2}+ \frac{p_{\phi}}{2mr^2 sin^2 \theta} + V(r).$$

By Hamilton's equation $\dot{p_{\phi}}=0$, this is a constant of the motion. As a result, specifying $p_{\phi}=\mu$ gives us a 5 dimensional manifold. The notes go on to state, "Furthermore, on each invariant submanifold the Hamiltonian can be written:

$$H=\frac{P_r^2}{2m}+\frac{p_{\theta}^2}{2mr^2}+ \frac{\mu}{2mr^2 sin^2 \theta} + V(r),$$

which is a Hamiltonian involving only two freedoms $r$ and $\theta$. Therefore the motion actually occurs on a 4-dimensional submanifold of the 5-dimensional submanifold of $T^*Q$ . . ."

However, to me it looks like we still have five degrees of freedom: $p_{\theta},$ $p_r,$ $r,$ $\theta,$ and $\phi$. Thus, I'm not sure what he means when he says that the presence of a constant of motion reduces the dimension of the cotangent manifold by 2. Is he saying that if we specify a numerical value for H, then the dimension is reduced from 5 to 4? Or does just the presence of a cyclic coordinate reduce the dimension from 6 to 4?

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Recall that $$\dot{\mathbf{p}}=-\frac{\partial H}{\partial\mathbf{q}}$$ Since $H$ does not actually depend on $\phi$, then $$\dot{p}_\phi=0=-\frac{\partial H}{\partial\phi}$$

This will eliminate $p_\phi$ and $\phi$ from your coordinates: $H(p_r,p_\theta,p_\phi,r,\theta,\phi)\to H(p_r,p_\theta,r,\theta)$.

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But $\phi$ is not constant, right? I guess I'm a little confused -- is the cotangent manifold just the domain of the Hamiltonian? The way I'm imagining it is that its the configuration space AND the momentum space combined. I suspect this is not correct. –  Dargscisyhp Dec 10 '13 at 20:47
Sure $\phi$ can vary, but the key is that moving around on $\phi$ produces no change in your Hamiltonian. –  Kyle Kanos Dec 10 '13 at 20:50
@KyleKanos It's not clear to me how you've shown that the existence of a cyclic coordinate somehow "reduces" the dimension of the phase space by two. You've argued that it causes the Hamiltonian to "depend on" two less coordinates in some sense, but how exactly does this tell us something about the reduction of phase space dimension? See my answer below for more reasons for which I'm skeptical about these statements; I'd be curious to get your input. –  joshphysics Dec 10 '13 at 22:35
I think it would be helpful if we could get a definition of the cotangent manifold with regards to Hamiltonian dynamics. According to wikipedia $T^*Q$ "can be thought of as the set of possible positions and momenta." In that case, even though the Hamiltonian is independent of $\phi, \phi$ itself changes. So if the cotangent manifold is just the domain of your Hamiltonian then I agree that the domain has dropped by two dimensions, but if it is the collection of generalized coordinates and momenta, then it still seems 5-dimensional. –  Dargscisyhp Dec 10 '13 at 22:55

In my opinion, your professor is being liberal with terminology in a confusing way, and I think you've essentially already pointed out why in your comment on Kyle's answer.

Let's examine a simple example. Consider the free particle moving in three spatial dimensions. The configuration space of the free particle is $\mathbb R^3$ and its momentum space is also $\mathbb R^3$ since it can have any triple of numbers $(p_x, p_y,p_z)$ as its momenta. So the whole phase space is $\mathbb R^6$; it's six-dimensional.

Now, recall that the free particle hamiltonian is \begin{align} H(p_x, p_y,p_z, x,y,z) = \frac{1}{2m}(p_x^2+p_y^2+p_z^2) \end{align} Each of $x$, $y$, and $z$ is a cyclic coordinate, so along any solution to the equations of motion, $p_x$, $p_y$, and $p_z$ are constant. This means that as the particle moves, it stays at its initial point $(p_x(t_0), p_y(t_0),p_z(t_0))$ in its momentum space. Hamilton's equations also show that the particle will move on straight lines in configuration space.

But would we say that the phase space of the free particle is zero-dimensional because there are three cyclic coordinates and $6-2-2-2=0$? That would be extremely non-standard terminology.

The professor's language indicates (as far as I can tell) that he is trying to say something about invariant submanifolds of the Hamiltonian, namely submanifolds of phase space that are mapped into themselves under Hamiltonian time-evolution. Even so, I'm not sure one can make much sense out of this business of each cyclic coordinate reducing the phase space dimension by two.

In the free particle case, for example, the manifold with coordinates $\{(0,0,0,x,y,z)\,|\, x,y,z\in\mathbb R^3\}$, namely all of configuration space, is an invariant submanifold of the Hamiltonian evolution which is three-dimensional (not zero-dimensional).

There are other situations in which it would be standard to say that there is a reduction in the phase space dimension, namely situations in which there is a constraint on the system as opposed to a conserved quantity.

Take, for example, the free particle moving on the plane subject to the constraint $y=y_0$ for some constant $y_0$. In this case, the canonical momentum in the $y$-direction will be restricted to vanish, $p_y = 0$, because otherwise the particle would move off of the $x$-axis. In this case, even though the particle is moving on phase space $\mathbb R^4$, it motion is always restricted to lie on a particular two-dimensional submanifold $\{(x,0,p_x,0)\,|\,\mathbb R^2\}$ of $\mathbb R^4$ that is isomorphic to $\mathbb R^2$.

Addendum. The standard terminology is that for a given system, the configuration space $Q$ is precisely the set of all possible positions of the system. If there are $N$ particles in the system, then the configuration space will be a subset of $\mathbb R^{3N}$, and is often a smooth manifold. If it is, then phase space is precisely the cotangent bundle $T^*Q$ of $Q$, a standard mathematical object whose definition is precise. The Hamiltonian is then a function on $T^*Q$. Notice, in particular, that if there is a cyclic coordinate in the Hamiltonian, then strictly speaking, phase space doesn't change, namely the domain of $H$ remains $T^*Q$, but the Hamiltonian simply does not depend on one of the coordinates. To use the terminology that phase space has changed dimension in this case is, in my opinion, a bit of an abuse.

As an analogy, suppose I had a real-valued function of two real variables $f$ defined as follows: \begin{align} f(x,y) = x^2 \end{align} Is the domain of the function $\mathbb R$ instead of $\mathbb R^2$ simply because it doesn't depend on $y$? No; it's domain is still $\mathbb R^2$.

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Thank you for your comment. This is exactly what is confusing me. I think it would be helpful to have a definition of what a cotangent bundle actually is, because I wonder if the problem is just terminology. –  Dargscisyhp Dec 10 '13 at 22:58
@user35268 I think it is essentially a problem of terminology. Regarding the tangent bundle, see my addendum. I'm not even sure there is much to be gained by trying to determine exactly how your professor is using the terminology as long as you understand which statements about the dynamics of the system are true, and which are false. –  joshphysics Dec 11 '13 at 1:51