Cosmology AA 2002/20031 The concept of Curvature Guido Chincarini This lecture is after the Luminosity Function and before the Metric_3D The idea is to.

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Presentation transcript:

Cosmology AA 2002/20031 The concept of Curvature Guido Chincarini This lecture is after the Luminosity Function and before the Metric_3D The idea is to give a good feeling for the concept of curvature in a 3 dimension space. We refer to the Gauss curvature that will be later connected to the metric of Robertson Walker.

Cosmology AA 2002/20032 The Circumference R  a/R By definition I set a center in O and I draw All the rays of size a. o o a Cir=2  a Cir= 2  x = 2  R sin(a/R) Sin(x) = x - x 3 /3! + x 5 /5! + … Cir= 2  R (a/R-a 3 /6R 3 + ….= 2  a (1 – a 2 /6R 2 + …) If I define the curvature of a sphere as K=1/R 2 and write from above 1/R 2 = 3/  a-Cir)/a 3

Cosmology AA 2002/20033 Circumference And I can write: K = 3/  lim a->0 (2  a- Cir)/a 3 ·That is I measure the curvature by estimating how much the circumference I measure on my surface about a small area around the observer differs from the circumference as measured on a flat surface. ·I can do this for any surface

Cosmology AA 2002/20034 Curve in Space

Cosmology AA 2002/20035 Equation of a Sphere

Cosmology AA 2002/20036 Arc of a curve in Space

Cosmology AA 2002/20037 Spherical Surface

Cosmology AA 2002/20038 Curvatures Geodesic equations: Two points can be joined by the shortest curve. Mean Curvature:½ (1/R 1 +1/R 2 ) Gauss Curvature:K = 1/(R 1 R 2 ) Gauss Curvature: See Theorema Egregium – See Berry & The book on Mathematics by Gellert Kunstner ….. The question:How can one draw conclusions about the spatial form of a surface from measurements on the surfece itself. The intrinsic geometry is completely governed by the element of arc on the surface. Indeed the element of arc determines the intrinsic geometry of the surface completely. Gauss showed that he Gaussian curvature is not only invariant under motions and parameter transformation, but also under bending. That is why he called this striking and unexpected result Theorema Egregium. Bending means deforming isometrically, that is: Two surfaces are isometric if and only if one can find parametric representation of them for which the elements of arc coincides.

Cosmology AA 2002/20039 Here only notation – see elsewhere details

Cosmology AA 2002/ Curvature See also Bronshtein & Semendyayev R 1 = a R 2 = Inf H=2a, K=0 R 1 = R R 2 = R H=1/R, K=1/R 2 R 1 > 0R 2 < 0

Cosmology AA 2002/ Metric 1.Now the best way to go is to use the metric since this characterize the surface. 2.We check if the Pythagorean theorem is satisfied. 3.We must look for a transformation of coordinates such that we can write the distance in Cartesian coordinates.

Cosmology AA 2002/ & in Polar Coordinates? How do I find out whether it is a flat surface? I must find out whether it is possible to find a transformation of coordinates which allow to write a distance as in cartesian coordinates. r  P  rr  x,  y

Cosmology AA 2002/ Differentiate That is the transformation shows that we have A Flat Euclidean Space

Cosmology AA 2002/ Cylinder We have the same for a Cylinder. Obviously the form and the topology differs but the geometry remains that of a flat sppace.

Cosmology AA 2002/ Sphere r     R Sin  r  ss

Cosmology AA 2002/ Element of arc We are unable to find a transformation that is capable to give us in cartesian coordinates a flat space. On the other hand if the space is not flat we must measure the curvature and we know that the Gauss curvature is not a function of the coordinates used. It must be an invariant.

Cosmology AA 2002/ And since we know that a metric could be written as a diagonal matrix with g 12 = g 21, we have:

Cosmology AA 2002/ For the sphere

Cosmology AA 2002/200319