# Why parametrise a curve in this way (on the unit circle)?

I saw papers saying something like "let $\gamma:S^1 \times [0,T] \to \mathbb{R}^2$ parametrise a curve. The second interval above just makes it time dependent, but why parametrise (for fixed time) the curve on S^1, the unit circle? I think it's to make it closed but what is confusing is that in papers they write $\gamma$ a function of a real variable $u$ and $t$, so as $\gamma(u, t)$ so how can the domain of the first variable $u$ be the unit circle? Should it not be $\gamma(u, v, t)$ for $u^2 + v^2 = 1$?

Thanks

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As you observe, it forces the curve to be closed without them needing to say it. The variables in a function don't need to be standard coordinates on $\mathbb{R}^n$ either - a circle is $1$-dimensional, so a single coordinate $u$ is perfectly good. Maybe it would help to think of $u$ as the angle round the circle (remembering that $u=a$ and $u=a+2\pi$ define the same point, so $\gamma(a,t)=\gamma(a+2\pi,t)$ for all $t$ and $a$). This isn't strictly accurate because I've now redefined $u$ as being a real number (and $\gamma$ as being $2\pi$-periodic in its first variable), rather than $u$ being a point on the circle, but if you're used to functions taking real inputs it might be helpful.
Note: The value $t$ in Thomas's answer is precisely this angle in his parameterization of the circle. This is not the only way of parameterizing a circle by one variable though, which is why philosophically it's cleaner to just take $u$ to be a point in the circle.
People often think of $S^1$ as the image of $t \mapsto e^{it}$ and then identify this with $t$. Using the exponential that way one also often identifies $[0,2\pi]$ (sometimes $[0,1]$ then with different speed of course) with $S^1$ (identifying the boundary points) more or less explicitly and then works with parameters from the interval. As long as only a parameter for a closed curve is needed and geometric poperties of the source are not relevant this works fine, and it's easier to write down derivatives.