# Will a contiguous, low-Re, low-Ca, liquid body always become a sphere at zero gravity?

Let's assume zero gravity, zero initial speed everywhere, $Re \ll 1$ and $Ca \ll 1$

Will such a liquid body always become a sphere or will it sometimes split?

$$\lim_{viscosity -> \infty} \lim_{t -> \infty} ShapeAtTime(t)$$

(Sufficiently high viscosity will also limit $Ca$, even though it is not directly in the expression)

I think it helps to think about this kind of experiment, but with an hourglass-like shape: Will its neck widen or expand at zero gravity?

• I think if you add the "no vibrational or rotational energy" caveat it would remain a sphere. If it is vibrating or rotating depending on the cohesion of the atoms it might break up into smaller ones. Jan 29, 2017 at 20:07
• @annav Added 0 initial speed as an assumption (and we already have high viscosity), but please note that I'm not assuming that the initial shape is already a sphere.
– MWB
Jan 29, 2017 at 20:10
• more spheres=more surface
– user126422
Jan 29, 2017 at 20:10
• Not sure high viscosity allows you to neglect inertia? Jan 29, 2017 at 23:04
• @robertbristow-johnson wouldn't it be possible that it collapses and spins? No : angular momentum conservation
– MWB
Jan 30, 2017 at 6:20

If the initial fluid blob had symmetric dumb-bell shape, then fluid pressure will be higher at its waist, and there will be flow from waist region to the two bulging regions, resulting in breakup into (at least) two smaller droplets (read up Rayleigh-Plateau instability). In other words, even if velocity is zero everywhere initially at $t=0$, you can always set up a situation where pressure gradient is not zero everywhere inside the fluid, resulting in a flow for $t>0$ and thus possible breakup. You can always have a flow so far as viscosity is finite, no matter how high, and this alone cannot prevent breakup.

• you can always set up a situation ... so far as viscosity is finite Your argument is: "for any given viscosity > 0 we can choose a shape such that ...". In other words, you are talking about a limit (math) over shapes, which isn't in the question (the shape is finite, viscosity isn't)
– MWB
Jan 30, 2017 at 6:36
• Cant say I fully understand your comment. Dumb-bell shape is a shape, not a limit of some sequence of shapes (although it can be set up as such). The shape is finite, yes. As viscosity $\to\infty$, so does the time required for break up (as far as I can see).
– Deep
Jan 30, 2017 at 7:25
• It's futile to try to explain infinitesimals in a tweet-like comment, but let me try: Overly simply put, you don't get to choose a shape in response to my viscosity. I get to choose the viscosity in response to your shape -- That's the meaning of the first "limit" in the question.
– MWB
Jan 30, 2017 at 9:14
• @MaxB I don't think he's trying to set the limits of your question. He's telling you a situation that may occur and is laying out the assumptions. There is a condition that he is stating for this to occur. It is up to you to decide if it properly answers your question. I don't think he did anything wrong. In fact, you say he's talking about a limit and reference it to math over shapes. If anything his answer is specifically about real scenarios up to the limit where the math becomes more abstract than his answer (I. E. Literal infinite viscosity, at infinite time).
– JMac
Jan 30, 2017 at 20:06
• @MaxB The same dumbbell shape works for every finite viscosity. Jan 30, 2017 at 20:48

If the object is infinitely viscous (and non-volatile, so that it can't evaporate), it will never change shape.