# 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. – anna v Jan 29 '17 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. – MaxB Jan 29 '17 at 20:10
• more spheres=more surface – user126422 Jan 29 '17 at 20:10
• Not sure high viscosity allows you to neglect inertia? – JMLCarter Jan 29 '17 at 23:04
• @robertbristow-johnson wouldn't it be possible that it collapses and spins? No : angular momentum conservation – MaxB Jan 30 '17 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.
• 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 '17 at 7:25