- #1
Apashanka
- 429
- 15
If considering a 2D surface (plane) having polar coordinate r,θ (where r is the distance from the origin and θ is the anticlockwise angle from the base line as usual)
The metric is now actually ds2=dr2+r2dθ2
If now this 2D surface is given a positive curvature of +1 (equivalent to the surface of a sphere) in which the the azimuthal angle is θ and the radial coordinate r is measured from the north pole of the sphere to the point on the surface along a great circle and R is the radius of the sphere.
In that case the metric of a 2D surface having a positive curvature becomes
ds2=dr2+R2sin2(r/R)dθ2
and for large radius of curvature R>>r the metric matches with the metric of a plane 2D surface(e.g without curvature).
Hence for a plane 2D surface being given a positive curvature of +1 it's actually equivalent to the curved surface of a 3D sphere.
My question is if a 3D surface (say for exm sphere) is given a positive or negative curvature then whether it's metric will be in some 4D minkowski space??
Or equivalent to some 3d metric in 4D minkowski space as in the case of a 2D plane.
Thanks
The metric is now actually ds2=dr2+r2dθ2
If now this 2D surface is given a positive curvature of +1 (equivalent to the surface of a sphere) in which the the azimuthal angle is θ and the radial coordinate r is measured from the north pole of the sphere to the point on the surface along a great circle and R is the radius of the sphere.
In that case the metric of a 2D surface having a positive curvature becomes
ds2=dr2+R2sin2(r/R)dθ2
and for large radius of curvature R>>r the metric matches with the metric of a plane 2D surface(e.g without curvature).
Hence for a plane 2D surface being given a positive curvature of +1 it's actually equivalent to the curved surface of a 3D sphere.
My question is if a 3D surface (say for exm sphere) is given a positive or negative curvature then whether it's metric will be in some 4D minkowski space??
Or equivalent to some 3d metric in 4D minkowski space as in the case of a 2D plane.
Thanks