Re: [math-fun] newton rotating liquid mirror
Situation 1: Non-rotating Earth. The child's circular pool (R=~3') is placed on a lazy susan and rotated at 1 revolution per day. Result: concave parabolic surface. This is the canonical Newton telescope, except for the extremely slow rotational rate. Situation 2: Rotating Earth; child's pool located at the South Pole. Pool still rotates at 1 revolution per day. Pool & water within pool can't tell the difference between situation 1 and situation 2. Result: same concave parabolic surface. If you rotate a child's pool, the water will rush towards the edges; the only thing keeping the water from flowing out is the pool wall, hence the pool wall will feel not only the static load of the water, but also an incremental force due to the centrifugal force of the rotating water within the pool. Thus, it is the _constraint_ of the pool wall which keeps the surface of the pool water in a more-or-less parabolic shape. At 06:27 AM 3/6/2015, Andy Latto wrote:
So the shape of the surface near the center of the pool is not determined by the local forces on the water at the center of the pool, but by some kind of "spooky action at a distance from the pool walls? And accurate measurement of the water at the center of the pool can determine the size of the pool? Can it also determine the shape of the pool? This sure sounds wrong to me.
Suppose the pool is 100 feet in diameter. Consider the circle 50 feet in diameter around the center of the pool. No water is flowing in or out of that circle. So if we introduce a new boundary there, there will be no change in the height of water inside or outside that circle.
Andy
On Fri, Mar 6, 2015 at 8:21 AM, Henry Baker <hbaker1@pipeline.com> wrote:
No, I wouldn't expect a lake at the south pole to have a concave area in the center, but I would expect a child's round swimming pool at the south pole to have a concave surface.
There will be forces against the side of the pool to keep the edge of the water higher than the middle. In the middle of a lake, these forces will be dissipated because the water will move away from the center.
At 09:40 PM 3/5/2015, meekerdb wrote: Would you expect a large lake at the south pole to have a concave area in the center, but match the geoid elsewhere? -- Andy.Latto@pobox.com
I assume people know that many modern liquid mercury mirrors have been made. The Canadians are fond of them, working at Laval University in Quebec City. The built a more sophisticated one in British Columbia. I oince looked into making a small (1 meter) one with an air bearing. It takes a lot of mercury and there are significant toxicity issues. http://en.wikipedia.org/wiki/Liquid_mirror_telescope
On Mar 6, 2015, at 6:43 AM, Henry Baker <hbaker1@pipeline.com> wrote:
Situation 1: Non-rotating Earth. The child's circular pool (R=~3') is placed on a lazy susan and rotated at 1 revolution per day. Result: concave parabolic surface. This is the canonical Newton telescope, except for the extremely slow rotational rate.
Situation 2: Rotating Earth; child's pool located at the South Pole. Pool still rotates at 1 revolution per day. Pool & water within pool can't tell the difference between situation 1 and situation 2. Result: same concave parabolic surface.
If you rotate a child's pool, the water will rush towards the edges; the only thing keeping the water from flowing out is the pool wall, hence the pool wall will feel not only the static load of the water, but also an incremental force due to the centrifugal force of the rotating water within the pool.
Thus, it is the _constraint_ of the pool wall which keeps the surface of the pool water in a more-or-less parabolic shape.
At 06:27 AM 3/6/2015, Andy Latto wrote:
So the shape of the surface near the center of the pool is not determined by the local forces on the water at the center of the pool, but by some kind of "spooky action at a distance from the pool walls? And accurate measurement of the water at the center of the pool can determine the size of the pool? Can it also determine the shape of the pool? This sure sounds wrong to me.
Suppose the pool is 100 feet in diameter. Consider the circle 50 feet in diameter around the center of the pool. No water is flowing in or out of that circle. So if we introduce a new boundary there, there will be no change in the height of water inside or outside that circle.
Andy
On Fri, Mar 6, 2015 at 8:21 AM, Henry Baker <hbaker1@pipeline.com> wrote:
No, I wouldn't expect a lake at the south pole to have a concave area in the center, but I would expect a child's round swimming pool at the south pole to have a concave surface.
There will be forces against the side of the pool to keep the edge of the water higher than the middle. In the middle of a lake, these forces will be dissipated because the water will move away from the center.
At 09:40 PM 3/5/2015, meekerdb wrote: Would you expect a large lake at the south pole to have a concave area in the center, but match the geoid elsewhere? -- Andy.Latto@pobox.com
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On Mar 6, 2015, at 6:43 AM, Henry Baker <hbaker1@pipeline.com> wrote:
Situation 1: Non-rotating Earth. The child's circular pool (R=~3') is placed on a lazy susan and rotated at 1 revolution per day. Result: concave parabolic surface. This is the canonical Newton telescope, except for the extremely slow rotational rate. Not quite: the surface of the pool would still be acted upon by a spherical gravitational field.
A fluid particle at rest with respect to the rotating frame sees an additional potential energy function -m(wr)^2/2, where m is the particle mass, w is the angular velocity, and r is the distance from the rotation axis. Let h(r) be the small extra height of the pool surface above the round earth. The equilibrium h(r) is the function for which -m(wr)^2/2 +mgh(r) is a constant, whose value is 0 if we take h(0)=0. Therefore h(r) = (wr)^2/(2g). What would h(r) have to be for the pool surface to be a flat plane? By simple geometry h(r) = r^2/(2R) where R>>r is the radius of the earth. So if you want to make a perfectly flat mirror the angular velocity would have to be w=sqrt(g/R) Comparing with the earth’s angular velocity w_E we get w/w_E = 17.05. I think I’ll use this on my next mechanics prelim!
Situation 2: Rotating Earth; child's pool located at the South Pole. Pool still rotates at 1 revolution per day. Pool & water within pool can't tell the difference between situation 1 and situation 2. Result: same concave parabolic surface.
If you rotate a child's pool, the water will rush towards the edges; the only thing keeping the water from flowing out is the pool wall, hence the pool wall will feel not only the static load of the water, but also an incremental force due to the centrifugal force of the rotating water within the pool.
Thus, it is the _constraint_ of the pool wall which keeps the surface of the pool water in a more-or-less parabolic shape.
On 3/6/2015 6:43 AM, Henry Baker wrote:
Situation 1: Non-rotating Earth. The child's circular pool (R=~3') is placed on a lazy susan and rotated at 1 revolution per day. Result: concave parabolic surface. This is the canonical Newton telescope, except for the extremely slow rotational rate.
Because the rotation is so slow the surface will move toward being concave, BUT it started out as slightly convex (due to the sphericity of the Earth and the divergence of the Earth's gravitational field) it will not become concave, just slightly less convex.
Situation 2: Rotating Earth; child's pool located at the South Pole. Pool still rotates at 1 revolution per day. Pool & water within pool can't tell the difference between situation 1 and situation 2. Result: same concave parabolic surface.
The Earth, on large scales, is fluid so it will become slightly less convex in the same way the pool does and they will both match the WGS84. At least that's the way I think it works. I guess an explicit calculation is needed. Brent
If you rotate a child's pool, the water will rush towards the edges; the only thing keeping the water from flowing out is the pool wall, hence the pool wall will feel not only the static load of the water, but also an incremental force due to the centrifugal force of the rotating water within the pool.
Thus, it is the _constraint_ of the pool wall which keeps the surface of the pool water in a more-or-less parabolic shape.
At 06:27 AM 3/6/2015, Andy Latto wrote:
So the shape of the surface near the center of the pool is not determined by the local forces on the water at the center of the pool, but by some kind of "spooky action at a distance from the pool walls? And accurate measurement of the water at the center of the pool can determine the size of the pool? Can it also determine the shape of the pool? This sure sounds wrong to me.
Suppose the pool is 100 feet in diameter. Consider the circle 50 feet in diameter around the center of the pool. No water is flowing in or out of that circle. So if we introduce a new boundary there, there will be no change in the height of water inside or outside that circle.
Andy
On Fri, Mar 6, 2015 at 8:21 AM, Henry Baker <hbaker1@pipeline.com> wrote:
No, I wouldn't expect a lake at the south pole to have a concave area in the center, but I would expect a child's round swimming pool at the south pole to have a concave surface.
There will be forces against the side of the pool to keep the edge of the water higher than the middle. In the middle of a lake, these forces will be dissipated because the water will move away from the center.
At 09:40 PM 3/5/2015, meekerdb wrote: Would you expect a large lake at the south pole to have a concave area in the center, but match the geoid elsewhere? -- Andy.Latto@pobox.com
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