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Is the following statement true?

Let $T$ be diffeomorphic to the solid torus. Let $v$ be a vector field such that $v$ and $curl(v)$ are both tangent to $\partial T$ everywhere and $|v|$ is constant on $\partial T$. Then $v$ is zero vector field.

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  • $\begingroup$ What is $curl(v)$? What is $|v|$? $\endgroup$ – abx Apr 4 at 4:52
  • $\begingroup$ the metric is euclidean and $curl(v)= \nabla \times u$ $\endgroup$ – J. Doee Apr 4 at 12:19
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    $\begingroup$ the question does not seem to be well written. Since you made no hypothesis at all for $v$ in the interior of $T$, but only on the boundary, you cannot conclude in the interior either. For example, any smooth vector field identically zero on a small neighborhood of $\partial T$ but not identically zero in $T$ is a counterexample. $\endgroup$ – Gael Meigniez Apr 5 at 11:07
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The answer is no.

The vector field $\nabla\times \vec{v}$ is tangent to $\partial T$ iff for any region $\Sigma\subset\partial T$ bounded by the Jordan curve $\partial\Sigma$ the flux integral $\iint_{\Sigma}(\nabla\times \vec{v}).\vec{n}\,{\rm{d}}S$ is zero; this is because the tangency to $\partial T$ means orthogonality to the unit normal vector $\vec{n}$ at any point of $\partial T$. Stokes' Theorem now implies that $\oint_{\,\partial\Sigma}\vec{v}.{\rm{d}}\vec{r}$ is zero. Thus the line integral of $\vec{v}$ along any simple closed curve that encloses a region in $\partial T$ (i.e. is a boundary) must be zero. This provides a reformulation of the condition on $\nabla\times \vec{v}$: Denoting the differential $1$-form corresponding to $\vec{v}$ by $\omega$, the $2$-form ${\rm{d}}\omega|_{\partial T}=d\left(\omega|_{\partial T}\right)$ vanishes.

Now I construct a counter-example: Write the solid torus $T$ as $\overline{\Bbb{D}}\times\Bbb{S}^1$ where $\overline{\Bbb{D}}$ with $\Bbb{S}^1$ and are the unit closed disk and the unit circle in $\Bbb{R}^2$. Equip $T$ with the Euclidean norm induced from $\Bbb{R}^2\times\Bbb{R}^2$. The vector field $-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}$ on $\overline{\Bbb{D}}$ could be thought of as a vector field on $\overline{\Bbb{D}}\times\Bbb{S}^1$ which is independent of the third coordinate. I claim that it works as $\vec{v}$. On $\partial T=\Bbb{S}^1\times\Bbb{S}^1$ the vector field $\vec{v}$ is given by $$ -y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}|_{\Bbb{S}^1}=\frac{\partial }{\partial \theta} $$
which is tangent to the first component and is of length one. It remains to check the tangency condition imposed on $\nabla\times \vec{v}$. The $1$-form $\omega$ corresponding to $\vec{v}$ is the pullback of the $1$-form $-y{\rm{d}}x+x{\rm{d}}y$ on $\overline{\Bbb{D}}$ via the projection $p_1:T=\overline{\Bbb{D}}\times\Bbb{S}^1\rightarrow\overline{\Bbb{D}}$. Restricted to the boundary $\partial T=\partial\overline{\Bbb{D}}\times\Bbb{S}^1=\Bbb{S}^1\times\Bbb{S}^1$, $\omega|_{\partial T}$ is the pullback of the closed $1$-form ${\rm{d}}\theta$ via $p_1:\partial T=\Bbb{S}^1\times\Bbb{S}^1\rightarrow\Bbb{S}^1$. Consequently, $\omega|_{\partial T}$ is closed as well.

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  • $\begingroup$ Thank-you very much for your answer. If I understand correctly, your counterexample is the unit vector field in the poloidal direction. What if one furthermore requires that the flow of the vector field generate $T$, namely that evolving a crosssection of $T$ by the flow of $v$ sweeps out all of $T$? An example of such a field would be the toroidal vector field, $e_\phi$ whose curl is in the $\hat{z}$ direction and is not tangent to the torus except on two circles. $\endgroup$ – J. Doee Apr 4 at 18:02

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