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Say you have your kitchen sink full of water and you pull the plug. Then a nice little vortex will form where the water disappears into the drain.

My question is:

Is this an example of a mathematical singularity occurring in real life? And if it is, what is the function of which it is a singularity of?

Edit

A singular point of a $C^1$-map $f: M \to N$ where $M$ is a smooth $m$-manifold and $N$ a smooth $n$ manifold is any point $x_0$ where $f'(x_0) = 0$.

a student
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1 Answers1

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It all depends on how you define a singularity.

Wikipedia says:

In mathematics, a singularity is in general a point at which a given mathematical object is not defined, or a point of an exceptional set where it fails to be well-behaved in some particular way, such as differentiability.

A vortex in the kitchen sink can be described as an irrotational vortex. A simple example of an irrational vortex is given by the velocity field $$v(x,y)=\frac{\alpha}{x^2+y^2}\cdot(-y,x).$$ $v(x,y)$ is the velocity of the fluid at a given point $(x,y)$ (we forgot about the 3rd dimension for simplicity). At the center of the vortex $(x,y)=(0,0)$, the magnitude of the velocity is divergent $|v(0,0)|\rightarrow\infty$. The direction of the velocity is given by the angle $$\theta(x,y)=\arctan (-y/x),$$ which is not defined in the center of the vortex $x=y=0$. Therefore, a vortex in a kitchen sink is a singular point in the velocity field of the water, since the velocity diverges and is not differentiable at the center of the vortex. Actually, it is a topological singularity, since there is no continuous transformation of the velocity field which removes the singular point of an irrotational vortex.

PS: One has to assume that the water is a continuous medium, i.e., one has to forget the fact that water or any other fluid is an ensemble of many microscopical molecules. In fact, at a microscopical level, there cannot be any singularity in a kitchen sink.

sintetico
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  • What about the third dimension? – Bernhard Jul 08 '15 at 15:03
  • I think it is not relevant. Just consider $v(x,y,z)=\frac{v_0}{x^2+y^2+z^2}⋅(−y,x,0)$. – sintetico Jul 08 '15 at 15:19
  • My point is that in the vortex in the sink, there typically is a vertical velocity component. – Bernhard Jul 08 '15 at 16:05
  • Your physical dimensions do not make sense: $v_0$ can't be a velocity - just rewrite your vortex as $\mathbf{v} = v_0 \mathbf{e}\phi/r$ with $ \mathbf{e}\phi$ the angular unit vector. So $v_0$ must be something like vortex strength. In addition, the modulus (aka speed) $v=v_0/r$ is divergent, hence indeed singular, at $r=0$! – Tom Heinzl Jul 08 '15 at 18:56
  • Thank you for the comment, I think I forgot something in the equation...I will fix that – sintetico Jul 08 '15 at 18:58
  • No worries - note that the Wikipedia page discusses two types of vortices, a non-singular "rigid body" type and a singular one (same as yours), which they call "irrotational". – Tom Heinzl Jul 08 '15 at 19:09
  • @TomHeinzl You probably know this, but it's not just Wikipedia that calls Sintetico's vortex "irrotational": it's pretty universal because the curl of this velocity field vanishes everywhere (aside from at the singularity, where it's undefined). This is the classic $\Omega=\log z$ irrotational, inviscid, incompressible velocity complex potential. – Selene Routley Jul 09 '15 at 00:12
  • @TomHeinzl can you check whether dimensions (and other things) do make sense now? – sintetico Jul 09 '15 at 06:02
  • Looks good to me: $\mathrm{d}_z (i,\frac{\Gamma}{2,\pi},\log z)^* = i,\Gamma/z^* = i,\frac{\Gamma}{2,\pi},z/|z|^2 = \frac{\Gamma}{2,\pi},(-y+i,x)/(x^2+y^2)$, so your vortex is the velocity complex potential $\Omega = i,\frac{\Gamma}{2,\pi},\log z$, where $\Gamma$ has the dimensions of velocity times length: it is the circulation of the vortex. – Selene Routley Jul 09 '15 at 06:57
  • I'm happy with the amended answer and the comments by @WetSavannaAnimal. Thanks! – Tom Heinzl Jul 09 '15 at 11:08