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Let's take the pendulum equation $\ddot{x} = -\sin x$. Here $x \in \mathbb{T}^{1}$. Now rewrite it as a coupled first order system $$\dot{y} = -\sin x, \quad \dot{x}=y.$$

Intuitively we know that $y$ corresponds to velocity, the norm of which (i.e. speed) can be as large or small as we want, thus $y \in \mathbb{R}$. Hence the phase space of the pendulum is the cylinder $\mathbb{T}^{1} \times \mathbb{R}$.

However $x(t) = x(t+t_{0})$ for some period $t_{0}$ and by the definition $ y =\dot{x}$ we also expect $y(t)=y(t+t_{0})$, i.e. we can say $y \in \mathbb{T}^{1}$.

Is this a contradiction? Why do we define $y$ to be in $\mathbb{R}$ and not in $\mathbb{T}$?

Qmechanic
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Alex
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    That's...not how you write the harmonic oscillator. The equation for the harmonic oscillator is $m\ddot{x} = -kx$, not $\ddot{x} = -\sin(x)$! Also, $y=\dot{x}$ holds only on trajectories that are solutions to the equations of the (correct) equations of motion, not throughout the phase space ($\dot{x}$ doesn't even make sense without a trajectory). I'm not sure which of those exactly your problem is. – ACuriousMind Oct 09 '16 at 22:53
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    That's why I said "Simple pendulum", not a harmonic oscillator! Anyway, the context of my question focuses on the mathematical aspect, rather than the physics (I just used the term pendulum because it's a common example of a cylinder phase space for a dynamical system) @ACuriousMind – Alex Oct 09 '16 at 23:16
  • To me, the "simple" in "simple pendulum" means that we consider the approximation of small angles where it becomes simple. Anyway, it seems your underlying question is "Why does having a circle for the values of a generalized coordinate not force a circle for the values of its conjugate momentum?", right? – ACuriousMind Oct 09 '16 at 23:21
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    Essentially, yes. But I have a much more general question concerning an arbitrary dynamical system, without any relevance to physics: when we reduce the order of an ode by introducing a new variable, how do we determine in what space that new variable is? So suppose we had that ode I wrote above...without any reference to a pendulum. How would we decide if $y \in \mathbb{R}$, based on knowledge that $x \in \mathbb{T}$?... @ACuriousMind – Alex Oct 09 '16 at 23:24

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$x\in \mathbb{T}^1$ denotes the structure of the phase space itself, not the fact that the motion as a function of time is periodic. Any arbitrary motion of the pendulum can be represented in the phase space, not just the ones periodic (in time). We have $x\in \mathbb{T}^1$ because you can rotate the pendulum around the hinge for a full cycle and you end up with the same state. You cannot say the same for $y$.

pathintegral
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  • Ah, I see. Your comment gives me some valuable insights here. What about the general case, without any reference to physics. Suppose you have a 2nd order ode. You make a change of variables $y= \dot{x}$ to reduce the system to a first order coupled ode system. Suppose $x$ is in some space: say $x \in \mathbb{T}$. What can you say about $y$? Will it be in $\mathbb{T}$ or $\mathbb{R}$? @pathintegral – Alex Oct 09 '16 at 23:21
  • If you take out the physics I don't think you can say anything about $y$. In quantum mechanics, a situation when $y\in \mathbb{T}^1$ is when continuous translation is degraded to discrete ones, i.e., when your "universe" is a crystal lattice. But in QM, a state cannot be represented by a point in phase space because of Heisenberg's uncertainty principle. The size of the "pixel" in phase space is $\hbar$. – pathintegral Oct 09 '16 at 23:30
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  1. If the Lagrangian formulation has configuration space $M$, and the Legendre transformation is non-singular, then the corresponding phase space in the Hamiltonian formulation is the cotangent bundle $T^{\ast}M$. (For the pendulum, the configuration space $M\cong S^1$ is a circle.)

  2. For models with a 2-torus $S^1\times S^1$ as phase space, see this Phys.SE post.

Community
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Qmechanic
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Lagrangian mechanics is defined on a tangent bundle, the tangent bundle of the configuration space. For the pendulum the configuration space is $S^1$ the circle. Its tangent bundle is trivial, so it is $S^1 \times \mathbb R$.

You can pass to the Hamiltonian description, which lives on the cotangent bundle -- it too is trivial, so it is also $S^1 \times \mathbb R$. That is the space of initial conditions. The actual motion will be periodic in both variables, but that's something else.

(There is something called invariant tori, where the almost periodic motion traces out a torus in phase space. I'm not sure that applies to the pendulum.)

Robin Ekman
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Consider all possible motions of the pendulum, $x=x_I(t)\:, \dot{x}=\dot{x}_I(t)$ where $I=\{x_0, \dot{x}_0\}$ denotes the initial conditions of that solution of Hamilton equations. Varying $I$ you have all possible solutions.

Well, there is an evident asymmetry between the two Hamiltonian variables. $x$ can always be taken in a circle, as $-\pi \leq x \leq +\pi$ is enough to describe all motions of the pendulum independently from $I$. A larger interval would be redundant to describe the positions of the pendulum. Conversely, there is no sufficiently large interval $[-\dot{X}, \dot{X}]$ which may include all values of all functions $\mathbb R \ni t \mapsto \dot{x}_I(t)$ for every initial condition $I$. This fact holds even if every such curve $\mathbb R \ni t \mapsto \dot{x}_I(t)$ is periodic.

Valter Moretti
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