Re: Visualizing a curved space
- From: Greg Egan <gregegan@xxxxxxxxxxxxxxx>
- Date: Sat, 15 Jul 2006 05:29:03 +0000 (UTC)
In article <LYSdnUfK_voKgyrZnZ2dnUVZ_qWdnZ2d@xxxxxxxxxxxx>, Alan
<info@xxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
I have a 2D Riemann metric tensor with coordinates (x,y) in a half-plane
-Infinity < x < Infinity, y > 0. I can calculate the geodesics.
I am looking for suggestions on algorithms for visualizations
of the geometry. (Ideally, something I can implement in Mathematica).
The space is somewhat similar to the Poincare hyperbolic
half-plane, in that the metric has the form
ds^2 = a(y) dx^2 + b(y) dy^2
I am a real novice in differential geometry, so the more elementary
the suggestions, the better.
Thanks!
alan
One thing you might like to try is looking for a surface in
three-dimensional Euclidean space, such that the induced metric on that
surface is your two-dimensional Riemannian metric.
Suppose you have a surface in R^3 which is parameterised by your two
coordinates x and y, with the point (x,y) on the surface having R^3
Cartesian coordinates:
(x,y) -> (F(x,y), G(x,y), H(x,y))
The R^3 Cartesian coordinates of the tangent vectors are then:
@_x -> (@F/@x, @G/@x, @H/@x)
@_y -> (@F/@y, @G/@y, @H/@y)
The metric induced on the surface by the Euclidean metric on R^3 can be
found just by taking the ordinary dot products of these vectors with
themselves, and with each other:
ds^2 = [(@F/@x)^2 + (@G/@x)^2 + (@H/@x)^2] dx^2
+ 2 [(@F/@x)(@F/@y) + (@G/@x)(@G/@y) + (@H/@x)(@H/@y)] dx dy
+ [(@F/@y)^2 + (@G/@y)^2 + (@H/@y)^2] dy^2
Now, in your case you want the metric:
ds^2 = a(y) dx^2 + b(y) dy^2
so you want to look for solutions F(x,y), G(x,y), H(x,y) to the partial
differential equations:
(@F/@x)^2 + (@G/@x)^2 + (@H/@x)^2 = a(y)
(@F/@x)(@F/@y) + (@G/@x)(@G/@y) + (@H/@x)(@H/@y) = 0
(@F/@y)^2 + (@G/@y)^2 + (@H/@y)^2 = b(y)
One simple example would be a(y)=R^2 (sin y)^2, b(y)=R^2. Then one
solution would be:
F(x,y) = R sin y cos x
G(x,y) = R sin y sin x
H(x,y) = R cos y
Obviously the details will depend on your own a(y) and b(y). If you
can't find closed-form solutions, you might be able to find solutions
numerically. (My own experience with Mathematica 4 numerically solving
PDEs is that you should take a 2-week holiday while it churns away, and
the final result is as likely as not to be an error message saying that
it couldn't achieve the required precision, but you might have a better
version or faster hardware ...)
.
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