This problem actually has three parts, and I have questions about each part. (What is in block quotes is the actual text of the problem.)
- Part (a):
Consider the vector field on the circle given by $\dot{\theta} = \cos \theta$. (Note: $\dot{\theta} = \frac{d \theta}{dt}$) I need to show that this system has a single-valued potential $V(\theta)$; i.e., for each point on the circle, there is a well-defined value of $V$ such that $\dot{\theta}=-\frac{dV}{d\theta}$. (Then it says something parenthetically that kind of confuses me, as you'll see why in my question regarding Part (b): "As usual, $\theta$ and $\theta + 2\pi k$ are to be regarded as the same point on the circle, for each integer $k$.")
For this part, essentially, we want to find a $V(\theta)$ such that $-\frac{dV}{d\theta} = \cos \theta$. So, by separation of variables, we obtain $V(\theta) = -\sin \theta + C$, where $C$ is a constant of integration.
Now, according to my understanding, "single-valued" means that each element of the function's domain maps to a single, well-defined element of its range, correct? As in, to call a function "single-valued" is kind of redundant, because by defintion of a bona fide function, it will be "single-valued" anyway.
So, my question for this part is, do I just want to show that if $\theta$ and $\theta + 2\pi k$ are considered the same point on the circle, then $V(\theta) = V(\theta + 2\pi k)$ here? Or, if not, how do I prove that $V(\theta)$ is single-valued?
- Part (b):
Now, consider $\dot{\theta} = 1$. Show that there is no single-valued potential $V(\theta)$ for this vector field on the circle.
So, again, I solved the differential equation $-\frac{dV}{d\theta} = 1$ as per the definition of a potential, and got the solution $V(\theta) = -\theta + C$.
In a solution I looked up, it said that there is no single-valued potential $V(\theta)$ for the vector field on the circle because $V(\theta)$ is not $2 \pi$-periodic - i.e., $V(\theta + 2 \pi k) = -(\theta + 2 \pi k) + C = - \theta - 2 \pi k + C \neq - \theta + C = V(\theta)$.
However, and this is where the statement "As usual, $\theta$ and $\theta + 2 \pi k$ are to be regarded as the same point on the circle, for each integer $k$" confuses me: I mean, if we're assuming that $\theta$ and $\theta+2 \pi k$ are the same point on the circle - i.e., that $\theta = \theta + 2 \pi k$, for all intents and purposes - and $V(\theta) = -\theta + C$ and $V(\theta + 2 \pi k) = -(\theta + 2\pi k)+C$, how does it make any sense that $V(\theta) \neq V(\theta + 2 \pi k)$?
- Part (c):
What's the general rule? When does $\dot{\theta} = f(\theta)$ have a single-valued potential?
Again, a solution I read says that in order for $\dot{\theta} = f(\theta)$ to have a single-valued potential, $V(\theta)$ must be $2\pi$-periodic.
So, I'm trying to show that this is true: Solving the equation $\dot{\theta} = -\frac{dV}{d\theta}$, I obtain $V(\theta) = -\int f(\theta)d\theta$.
So, let's choose a constant $C$ such that $V(\theta) = -\int_{0}^{\theta}f(\theta) d\theta$. Then, it seems to me that $2\pi$-periodicity would imply that $V(\theta) = V(\theta + 2\pi k)$ (which, to me, seems like the exact opposite of single-valued. If someone could explain to me why it's not, it would be extremely helpful!).
So, I'd need $- \int_{0}^{\theta}f(\theta) d\theta = -\int_{0}^{\theta + 2 \pi k}f(\theta+2\pi k) d(\theta +2\pi k)$, or $- \int_{0}^{\theta}f(\theta) d\theta + \int_{0}^{\theta + 2 \pi k}f(\theta+2\pi k) d(\theta +2\pi k) = 0$. Now, I understand that there are a lot of questions asked an answered on here about how to show this when $f$ itself is periodic, but we're not necessarily told that here - we only know about $V$, not $f$. So, how do I go about showing this?
I realize I'm asking a lot here, and I'm probably overthinking a great deal of it, but I really want to understand potentials! Thank you for your time and patience!