most likely candidate arc of tractrix and circle that are equal (proofs??) #401 Correcting Math
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From: Archimedes Plutonium on 7 Feb 2010 02:45 gudi wrote: (snipped) > I have put in the following sketch. > > http://i50.tinypic.com/2bpmbk.jpg > If there is a line arc on the pseudosphere that matches a line arc on the sphere the most likely place to find it is in Narasimham's sketch of the cusp at the upper left where the circle and tractrix meet. Tractrix http://en.wikipedia.org/wiki/Tractrix Looking at the Wikipedia drawing, the most likely candidate of an arc that is equal for both the circle and the tractrix is again at the cusp. The arc at the cusp is tending to be the same as the circle arc. There is a good chance in all of this that there are no equal arcs that can be replaced from tractrix to circle and vice versa. But if there is one, I should suspect it is at the cusp region where the curvature is at its largest. Now maybe there is a proof that no such arc exists that can be cut and pasted out of the circle and put into the tractrix and vice versa. Such a proof would rely on the idea that the curvature of arc is so very much different from any two points on the circle versus the tractrix. So our commonsense would be persuaded without ever looking deeper that there cannot be such equal arcs. But a proof that is contrary would likely say that the tractrix has a variable arcs along its path and that one of them is identical to the circle of same radius. In this picture we have a tractrix of varying arcs and where the circle arc is represented as one of those varying arcs. And where the circle arcs are all of a constancy. Another proof that a tractrix or pseudosphere contains at least one arc identical to a circle arc is a proof that would frame a circle inside a square and where we have an additional nested frame of a tractrix that has been disassembled and reorganized to form a square like frame where the circle is enclosed. In this proof scheme, the only frame that has a point tangency is the "square" but that the hyperbolic geometry frame of a re-organized tractrix has to have a arc-line-segment tangency with the circle enclosed. So that this proof, if it is true and works, would prove that the tractrix must have a arc that matches a arc of the circle. Archimedes Plutonium www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies
From: gudi on 7 Feb 2010 15:05 Hi Archimedes Plutonium, Without differential geometry basics I am afraid you cannot progress much in obtaining a satisfying answers for your questions. Just as surfaces in flat Euclidean geometry are developable and can be made to contact everywhere, in doubly curved geometry we have the corresponding concept that is called " applicability". However there is only local isometry, not global isometry. It is referred to as being " in the small", or in local neighborhoods, and in contrast we have " in the large" etc. We have to be content with contact along lines at a single point only. " Geometry and Imagination " is a good book. Many Catenoid to Helicoid bendings Java simulations are available on the net.They appear to be two different shapes but can be bent from one shape to the other shape i.e., by deforming it. You can also practically take a thin plastic water container with narrow waist part cut out and get a helix by a single cut along meridian and then stretching out. During an "application" between helicoid/catenoid thin foils, Gauss curvature remains the same, at each corresponding point during the deformation. As to your question of where exactly the contact is established, in surface theory, we have for each point( x,y,z ) = ((f(u,v),g(u,v)) parameterization where the parameters u and v describe the grid lines making up the curved surface in a curvilinear coordinate system. The choice is entirely yours. You can choose an orthogonal system, geodesic polar coordinates, asymptotic lines of a Chebycheff Net etc. and go along any u- line or v- line of your choice. A special case of Gauss Egregium theorem is the Minding therem that states that for surfaces class >=3,in the small or locally, an isometry is possible between two surfaces of constant equal Gauss curvature. The earlier sketch is modified somewhat here: http://i46.tinypic.com/f0rm9u.jpg However, it serves no purpose in understanding of "applicability" of surfaces. In isometry the area remains same ( through E*G - F^2 determinant of first fundamental form) but in the converse, if your question is, the area is same and what straining of u- and v- lines is to be chosen to get back isometry of each du and dv element, one has to use Beltrami differential operators etc., I think it takes lot more effort. Narasimham
From: gudi on 7 Feb 2010 15:12
On Feb 8, 1:05 am, gudi <mathm...(a)hotmail.com> wrote: BTW, the pseudosphere and sphere have same surface area = 4 pi R^2, but pseudosphere has half the volume of the sphere on same axial length, viz., 2/3 pi R^3 only,so much for the "pseudo".. Narasimham |