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I have to prove that $$\vec{F}=\frac{\vec{r}}{r^2}$$ is a conservative vector field without finding it's potential function.

However, $\vec{F}$ has a hole in its domain, at $(x,y,z)=(0,0,0)$, so I can't just check it's curl.

How do I go about proving this?

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Assuming that $r$ refers as usual to the radial spherical coordinate, for this example it still suffices to show that $\operatorname{curl} \vec{\bf F} = 0$. It's true that the domain of $\vec {\bf F}$ has a hole, but the domain is still simply connected: Any loop in the domain, say, based at a point $\vec{\textbf{x}}_0$ can be continuously deformed to the constant path at $\vec{\textbf{x}}_0$.

Alternatively, in spherical coordinates $(r, \theta, \varphi)$, the gradient is $$\operatorname{grad} f = f_r \vec{\bf r} + \frac{1}{r} f_{\theta} \vec{\bf \unicode[Arial]{x3B8}} + \frac{1}{r \sin \theta} f_{\varphi} \vec{\bf \unicode[Arial]{x3C6}}.$$ In particular, the gradient of a radial function $f(r)$ is $$\operatorname{grad} (f(r)) = f_r(r) \vec{\bf r} ,$$ so the Fundamental Theorem of Calculus guarantees that any continuous radial vector field (not defined at the origin, where spherical coordinates behave badly) is conservative.


For a vector field $\vec{\bf G}$ whose domain is not simply connected, $\operatorname{curl} \vec{\bf G} = 0$ is inconclusive. For example, both $$\vec{\bf R} := \frac{x}{x^2 + y^2} \vec{\bf x} + \frac{y}{x^2 + y^2} {\vec{\bf y}}$$ and $$\vec{\bf V} := \frac{-y}{x^2 + y^2} \vec{\bf x} + \frac{x}{x^2 + y^2} {\vec{\bf y}}$$ have domain $\{(x, y, z) : (x, y) \neq (0, 0)\}$ and have zero curl, but $\vec{\bf R}$ is conservative and $\vec{\bf V}$ is not. (In cylindrical coordinates $(R, \theta, z)$, $\vec{\bf G} = \frac{1}{R} \vec{\bf R}$ and $\vec{\bf V} = \vec{\bf \unicode[Arial]{x3B8}}$.)

In this particular case, the domain has a single hole that prevents it from being simply connected (more precisely, its fundamental group is a free with a single generator), so to check whether the vector fields are conservative it suffices to check whether the integral along a single suitable curve $\gamma$ vanishes. Here, suitable just means that any loop $\alpha$ is continuously deformable to the curve $n \gamma$ for some $n$, where $n \gamma$ denotes tracing out the loop $n$ times (or $-n$ times backward if $n < 0$); informally, $n$ just counts the (signed) number of times $\alpha$ wraps around the hole. Then the vanishing of the curl guarantees that $$\int_{\alpha} \vec{\bf G} \cdot d\vec{\bf s} = \int_{n \gamma} \vec{\bf G} \cdot d\vec{\bf s} = n \int_{\gamma} \vec{\bf G} \cdot d\vec{\bf s} .$$

So, $\vec{\bf G}$ is conservative, that is, $\int_{\alpha} \vec{\bf G} \cdot d\vec{\bf s}$ vanishes for all loops $\alpha$, iff $\int_{\gamma} \vec{\bf G} \cdot d\vec{\bf s}$. In our example, we can take the loop $\gamma(t) := (\cos t, \sin t, 0)$, $0 \leq t \leq 2 \pi$, and computing gives $$\int_{\alpha} \vec{\bf R} \cdot d\vec{\bf s} = 0 \qquad \textrm{but} \qquad \int_{\alpha} \vec{\bf V} \cdot d\vec{\bf s} = 2\pi ,$$ and we conclude that $\vec{\bf R}$ is conservative but $\vec{\bf V}$ is not, as claimed.

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  • $\begingroup$ Perhaps the downvoter would explain their objection? $\endgroup$ – Travis Oct 21 '18 at 18:21
  • $\begingroup$ The second part is beyond me, but for domains in $\mathbb{R}^2$ and $\mathbb{R}^3$ where I can visualize a closed curve. Is it true that domains in $\mathbb{R}^2$ or $\mathbb{R}^3$ are simply connect iff there is no hole going all the way through the domain? And finally, if the curl is 0 and the domain is not simply connected, can I just pick any closed curve $\gamma$ which I can visualize as being deformable into any and all other closed curves, and check to see if the circulation is $0$? I don't know how to prove mathematically that $\gamma$ is deformable into any other $\gamma$, but if it $\endgroup$ – DWade64 Oct 21 '18 at 18:50
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    $\begingroup$ It's hard to give precise answers more of the language of topology available. In any case: The plane $\Bbb R^2$ is different from higher-dimensional Euclidean spaces in the sense that removing a single point gives a space $\Bbb R^2 - \{\ast\}$ that is not simply connected. A loop that goes around $\ast$ once, for example, cannot be continuously deformed into a constant loop. (But this is not true for $\Bbb R^n$, $n \geq 3$.) "No hole going all the way through the domain" is a good way to think informally about the condition on holes in $\Bbb R^3$. $\endgroup$ – Travis Oct 21 '18 at 19:00
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    $\begingroup$ In the example, it's not that we can pick a loop $\gamma$ that is deformable into all other loops (this is only possible when the space is simply connected in the first place), but rather that we can pick a loop $\gamma$ such that for every curve $\alpha$, there is some integer $n$ such that $n \gamma$---the curve that wraps around gamma $n$ times---is deformable into $\alpha$. $\endgroup$ – Travis Oct 21 '18 at 19:02
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    $\begingroup$ If we can find such a curve $\gamma$, then like you say, it suffices to check whether the circulation along $\gamma$ is zero, but for some more complicated domains there will be no such curve gamma---instead, we'd need to pick curves $\gamma_1, \ldots, \gamma_k$ such that any curve $\alpha$ can be deformed into some curve $\gamma_{a_1}^{b_1} \cdots \gamma_{a_r}^{b_r}$ for some indices $a_1, \ldots, a_k$ and integers $b_i$ and then check the circulation of the vector field over each $\gamma_i$. In general it is not easy to determine visually whether a single curve $\gamma$ suffices. $\endgroup$ – Travis Oct 21 '18 at 19:09

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