State space of a pendulum as $S^{1} \times \Bbb{R}$?

Question

I was reading about the State space of a Dynamical System from Scholarpedia article, I understood that a state space is a set of all possible sets of the Dynamical system.

But I was not able to understand how in the case of a pendulum where the degrees of freedom is 2 comprising of angle and velocity, then how does it represent $S^{1} \times \Bbb{R}$ like why a "cross" between them but I can visualize that $S^{1} \times \Bbb{R}$ is a cylinder.

The values your two-dimensional state can take belong to the Cartesian product of $S^1$ and $\mathbb{R}$. — Calculon, May 07 '18 at 16:00

score 1 · Accepted Answer · answered May 07 '18 at 15:59

1

These are pairs of points. The angle parametrizes $S^1$, and the velocity is taken as a scalar in $\mathbb R$. So the tuple is $(\theta,v) \in S^1 \times \mathbb R$.

answered May 07 '18 at 15:59

Andres Mejia

20,977

Let me know if this is what you are looking for, it's actually a bit difficult for me to see the confusion. – Andres Mejia May 07 '18 at 15:59
Yes, that is it, I was a bit late identifying $(\theta , v)$ with $S^{1} \times \Bbb{R}$! – BAYMAX May 07 '18 at 16:01
ah okay. Glad it worked out. – Andres Mejia May 07 '18 at 16:07

score 1 · Answer 2 · answered May 07 '18 at 15:59

The angle can be represented by some point on a circle, which is $S^1$. The velocity is some real number, i.e. $\mathbb{R}$. So, the whole system is represented by a pair $(\theta, v)$ where $\theta \in S^1$ and $v \in \mathbb{R}$, and hence is represented by the cross $S^1 \times \mathbb{R}$.

State space of a pendulum as $S^{1} \times \Bbb{R}$?

2 Answers2