How do you get a flat surface of water?

October 11, 2006 1:54 PM

Does gravity make small pools of water curve slightly (I’m thinking a small lake size) or is surface tension enough to flatten it out? More importantly - how would I calculate whether the resistance from the surface tension is enough to keep it flat and when/what size that breaks off?

Is there any way to create a perfectly, or very near perfectly flat surface of water?

posted by Smedleyman to science & nature (19 comments total)

Gravity will make the water tend towards the gravitational equipotential. They're all curved from that.

On a smaller scale surface tension acts to make the surface curve even more. Look closely at a glass of water.

It'll never be flat that I can tell.
posted by edd at 2:06 PM on October 11, 2006

Is there any way to create a perfectly, or very near perfectly flat surface of water?

First, empty the universe. Then make an infinite plane made out of something nice and dense, like lead. Then pour an infinite amount of water on top of the infinite plane. Voila -- one perfectly flat surface of water.
posted by chrismear at 2:12 PM on October 11, 2006

rotate the body of water just enough to overcome the equapotential.
posted by hortense at 2:20 PM on October 11, 2006 [1 favorite]

Water isn't flat, but pretty close.

If you had a kilometer-wide pool of water on Earth, the far side would be about 8 centimeters lower than it should be if it were precisely flat.

That's pretty flat. Google "float glass".
posted by jellicle at 2:26 PM on October 11, 2006

would a giant glob of water in space form a sphere or would there be flat spots? there is your answer
posted by jockc at 2:49 PM on October 11, 2006

“If you had a kilometer-wide pool of water on Earth, the far side would be about 8 centimeters lower than it should be if it were precisely flat.”

How did you calculate that? And what is the affect of surface tension on that?
posted by Smedleyman at 2:51 PM on October 11, 2006

Hortense has the right answer. And if you spin it even more, you can get negative curvature, where the center is lower than the edges.

There are several telescopes which do exactly that with mercury, to create large cheap mirrors that have the substantial drawback that they cannot be tilted.
posted by Steven C. Den Beste at 2:55 PM on October 11, 2006

Smedleyman, the calculation is based on the curvature of the earth. And surface tension would have no effect on it. (Think about the shape of the Pacific Ocean for a larger example of that effect.)
posted by Steven C. Den Beste at 2:57 PM on October 11, 2006

rotate the body of water just enough to overcome the equapotential.
posted by hortense at 2:20 PM PST

Maybe I'm dense but can someone explain how a spherical gravitational field and a hyperbolic rotation curve can combine to create a perfectly flat surface?
posted by vacapinta at 3:26 PM on October 11, 2006 [1 favorite]

Is there any way to create a perfectly, or very near perfectly flat surface of water?

Perfectly? A monolayer (or very thin layer) of water molecules.
posted by Krrrlson at 3:46 PM on October 11, 2006

Krrrlson, that can be done with oil but not with water. Water would bead up, unless there was alcohol or detergent in it.
posted by Steven C. Den Beste at 3:56 PM on October 11, 2006

"...spherical gravitational field and a hyperbolic rotation curve can combine to create a perfectly flat surface?"

Kinda where I was headed but my brain started smoking.
posted by Smedleyman at 3:58 PM on October 11, 2006

what vacapinta said, except that i think the shape is a paraboloid, not hyperboloid.

also, talking about something molecularly flat isn't really correct when you consider gas-liquid interfaces. what appears to be a smooth surface is actually a steep gradient in the density of water molecules, on the order of 20 or so molecule-lengths wide.
posted by sergeant sandwich at 4:49 PM on October 11, 2006

Does gravity make small pools of water curve slightly (I’m thinking a small lake size)

Excuse my ignorance, but why should it? If the whole body is subject to the same gravity, it should all experience the same pull (relative to the curvature of the earth, of course), no?
posted by wackybrit at 5:07 PM on October 11, 2006

If you really wanted an exact answer, you would need to compute the energy of the system, and minimize it with respect to the shape of the free surface, with the restriction that the volume of the system is fixed. You can integrate the gravitational potential energy, GMρ/r over the volume of the water, and the surface energy is just the surface tension multiplied by the free surface area. I think assuming a spherical surface of unknown radius would tell you whether the balance favors surface tension or gravity, even if that isn't the correct shape.

An easier answer, by way of handwaving, is to look at the Bond Number, ρga²/σ which represents the ratio of gravitational to surface forces for a system with characteristic length a. For anything larger than 1 cm, the Bond number is quite large, indicating gravitational dominance. The trick is figuring out what the relevant length scale is in this case.
posted by yarmond at 5:10 PM on October 11, 2006

Hortense's answer is inspired!

How I wish I'd had the wit to think of that!

A rotating body of water, neglecting surface tension, will assume a parabolic shape in a uniform (magnitude and direction, remember) gravitational field, but I haven't tried to calculate what the shape would be here on earth where the magnitude of the field on the surface of a small body of water is almost perfectly uniform but the directions of the vectors are not, since they must all point to the center of the earth.

Vacapinta, the compensation by spinning can be made almost perfect anyway, because a very small circular section of any sphere is almost exactly coincident with the apex of a paraboloid the focal length of which is half the radius of the sphere. In other words, a small circular piece of any sphere is a paraboloid to a very good approximation-- a fact made good use of in optics labs and solar furnaces.
posted by jamjam at 5:39 PM on October 11, 2006 [1 favorite]

You could get pretty close by putting your pond in a spaceship undergoing constant acceleration, and playing with the water-affinity of its wall lining to make the meniscus flatten out.
posted by flabdablet at 6:21 PM on October 11, 2006

Krrrlson, that can be done with oil but not with water. Water would bead up, unless there was alcohol or detergent in it.

Ah, but what about an H20 monolayer on a perfectly flat surface of hydrophilic molecules?
posted by Krrrlson at 9:37 PM on October 11, 2006

Another way to make the surface, use a mirrored membrane like mylar, closing a cylinder filled with water, draw out water to adjust the surface, then you could point SDB's telescope any direction you please. air would work also.
posted by hortense at 1:08 AM on October 14, 2006