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The two-dimensional Stokes theorem, written in vector notation, reads $$\oint_\mathcal{C} \vec{F} \cdot d \vec{r} = \int_\mathcal{S} (\vec{\nabla} \times\vec{F})\cdot d \vec{S},$$ where the closed curve $\mathcal{C}$ is the boundary of the surface $\mathcal{S}$, $\mathcal{C} = \partial \mathcal{S}$.

Now, the usual definition of a closed plane curve $\gamma$ is $$\gamma:[a,b] \to \mathbb{R}^2,$$ with $\gamma(a) = \gamma(b)$. However, wouldn't it make more sense to define a closed curve as $$\gamma:[a,b\rangle \to \mathbb{R}^2,$$ with the condition $$\lim_{t \to b} \gamma(t) = \gamma(a)?$$ In this way, no point on the curve is parametrized twice (assume $\gamma$ is injective), and this is consistent with, e.g., the polar coordinate $\varphi$ which has the domain $\varphi \in [0,2\pi\rangle$.

My question: would the change in the definition of a closed curve alter Stokes theorem in any way?

Fizikus
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  • I don't think this improves things much. Admittedly you don't parameterise anything twice, as you put it, but you are still taking the interval $[a,b]$, bending it around and joining up the ends, so to speak. A better approach might be to define $\gamma'$ on the real line and wind the real line around and around, identifying any point $t\in [a, b]$ with $t+2\pi$, $t+4\pi$ and so on so that $\gamma$ on the interval $[a,b]$ is just the restriction of $\gamma'$. Now any properties of $\gamma'$, such as differentiability say, should I think hold for $\gamma$, too. – James Smith Jan 07 '18 at 17:21
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    Better still is to define a closed curve to be a continuous function $\gamma \colon S^1 \to \mathbb{R}^2$. This way you don't have to make an arbitrary choice of start / end point, and it makes it clear that you can use whatever coordinate system for the circle that you want. Of course, this still isn't quite the right language for Stokes theorem, which is still true and useful in cases where the boundary is not connected. – Paul Siegel Jan 07 '18 at 17:44
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    In order for Stokes's theorem to hold on surfaces with merely continuous boundary, you need to have vanishing boundary capacity, so continuity is not enough; see Sauvigny, Partial Differential Equations, volume 1, p. 38. – Ben McKay Jan 07 '18 at 18:00
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    The question doesn't make sense, as the curve $\gamma$ is in the plane but the boundary $C$ is in the surface $S$, which is not assumed to be in the plane. – Ben McKay Jan 07 '18 at 18:01

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