If the opening in the jar is small enough... will water still leak out? The instability (whether you call it "Taylor" or "fingering" or whatever) exists in the absence of surface tension. However, in order for this instability to happen, the surface needs to bend away from flat, say with part pushing in and another part pushing out. This increases the surface area. This suggests that this instability becomes "stability" if the surface tension is high enough. Now let us attempt to turn this into a semiquantitative law. Let the hole be circular of radius r. Regard both the "water" and "air" as incompressible inviscid fluids of different densities. Assume the water will be perturbed vertically by epsilon along some smooth surface which differs from flat by a volume-amount of order epsilon*r^2. Its surface will thus increase from pi*r^2 to pi*r^2+C*r*epsilon^2 at leading order for some dimensionless positive constant C. (By the isoperimetric theorem there is a lower bound on C.) This costs energy C*T*epsilon^2 where T is surface tension. The water will finger downward and air upward, and if the water finger is narrower and longer than the air finger, then this gains gravitational energy proportional to g*epsilon^2*densitydiff*r^2. If the water finger is geometrically identical to the air finger, though, then no gravitational energy gain or loss. So anyhow, there clearly exists a finger shape which causes instability in the limit epsilon-->0+, and this is true no matter how large the air-water surface tension. But if T is large enough compared to g*densitydiff*r^2, then the surface energy term can be larger than the gravitational term when, say, epsilon=0.1*r. Therefore, I would predict that for small enough r, equivalently large enough T, the instability will be shut down at finite size (although it will remain for infinitesimal epsilon). This predicts (using epsilon=0.1*r for this purpose) that the critical r where the leakage will be stopped should be of order rcrit = 0.1*T/(g*densitydiff) which putting in numbers g = 9.8 meter/sec^2 densitydiff = 1000 kg/meter^3 T = 0.072 Joule/meter^2 yields the crude estimate rcrit = 0.86 millimeters. So I suggest the following experiment: 1. for all large enough holes, the water will come out. 2. but for a small enough hole (perhaps a good choice is diameter=0.5 mm), we predict the water will not leak out. 3. But if soap is in the water (which ought to cut T by about a factor of 10) then it will too leak out.
I didn't see Where hydrostatic pressure figures into this calculation. If the water column in the jar is high enough it should overcome surface tension for any fixed size hole. -- Gene
________________________________ From: Warren D Smith <warren.wds@gmail.com> To: math-fun@mailman.xmission.com Sent: Tuesday, August 12, 2014 2:37 PM Subject: [math-fun] Dumping water
If the opening in the jar is small enough... will water still leak out?
The instability (whether you call it "Taylor" or "fingering" or whatever) exists in the absence of surface tension. However, in order for this instability to happen, the surface needs to bend away from flat, say with part pushing in and another part pushing out. This increases the surface area. This suggests that this instability becomes "stability" if the surface tension is high enough.
Now let us attempt to turn this into a semiquantitative law. Let the hole be circular of radius r. Regard both the "water" and "air" as incompressible inviscid fluids of different densities. Assume the water will be perturbed vertically by epsilon along some smooth surface which differs from flat by a volume-amount of order epsilon*r^2. Its surface will thus increase from pi*r^2 to pi*r^2+C*r*epsilon^2 at leading order for some dimensionless positive constant C. (By the isoperimetric theorem there is a lower bound on C.) This costs energy C*T*epsilon^2 where T is surface tension. The water will finger downward and air upward, and if the water finger is narrower and longer than the air finger, then this gains gravitational energy proportional to g*epsilon^2*densitydiff*r^2. If the water finger is geometrically identical to the air finger, though, then no gravitational energy gain or loss.
So anyhow, there clearly exists a finger shape which causes instability in the limit epsilon-->0+, and this is true no matter how large the air-water surface tension.
But if T is large enough compared to g*densitydiff*r^2, then the surface energy term can be larger than the gravitational term when, say, epsilon=0.1*r. Therefore, I would predict that for small enough r, equivalently large enough T, the instability will be shut down at finite size (although it will remain for infinitesimal epsilon).
This predicts (using epsilon=0.1*r for this purpose) that the critical r where the leakage will be stopped should be of order rcrit = 0.1*T/(g*densitydiff) which putting in numbers g = 9.8 meter/sec^2 densitydiff = 1000 kg/meter^3 T = 0.072 Joule/meter^2 yields the crude estimate rcrit = 0.86 millimeters.
So I suggest the following experiment: 1. for all large enough holes, the water will come out. 2. but for a small enough hole (perhaps a good choice is diameter=0.5 mm), we predict the water will not leak out. 3. But if soap is in the water (which ought to cut T by about a factor of 10) then it will too leak out.
Suppose we have a set X with a 1) binary operation #, 2) an identity e, and 3) a unique two-sided inverse for each element. Let's call (X,#) an I-magma. (There are all kinds of names for structures similar to this, like quasigroups and loops, but I haven't yet found an official name for this exact concept.) If an I-magma were also associative, it would be a group. But suppose that instead of being associative we ask that an I-magma also be alternative: ----- For all x, y in X: x(xy) = (xx)y, (xy)y = x(yy), x(yx) = (xy)x ----- One class of alternative I-magmas is Moufang loops (see < http://www.encyclopediaofmath.org/index.php/Moufang_loop >). But I haven't seen anything implying that alternative I-magmas must be Moufang loops, so I suspect this is not true. QUESTION: What are the smallest alternative I-magmas??? A paper states: "There are 13 [Moufang loops of size <= 31]: one of order 12, five of order 16, one of order 20, five of order 24, and one of order 28. (See < http://www.ams.org/journals/tran/1974-188-00/S0002-9947-1974-0330336-3/S0002... >.) So each of these must be an alternative I-magma. But are there others <= size 31 as well? --Dan
Actually, a special case of Moufang loops are groups. The statement below refers to only the Moufang loops of size <= 31 that are *not* groups -- i.e., that are not associative. --Dan On Aug 12, 2014, at 5:05 PM, Dan Asimov <dasimov@earthlink.net> wrote:
A paper states: "There are 13 [Moufang loops of size <= 31]: one of order 12, five of order 16, one of order 20, five of order 24, and one of order 28.
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