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 advanced_tools:manifold [2019/07/29 11:54]jakobadmin [Layman] advanced_tools:manifold [2020/01/11 02:46] (current)sebsonf Radius equation Both sides previous revision Previous revision 2020/01/11 02:46 sebsonf Radius equation2019/07/29 11:54 jakobadmin [Layman] 2018/12/19 10:01 jakobadmin ↷ Links adapted because of a move operation2018/05/04 07:53 jakobadmin ↷ Links adapted because of a move operation2018/04/08 15:14 georgefarr ↷ Links adapted because of a move operation2018/03/28 06:53 jakobadmin ↷ Links adapted because of a move operation2018/03/06 14:52 jakobadmin [Why is it interesting?] 2018/03/06 14:49 jakobadmin [Examples] 2018/03/06 14:47 jakobadmin [Examples] 2018/03/06 14:47 jakobadmin [Examples] 2018/03/06 14:45 jakobadmin created 2020/01/11 02:46 sebsonf Radius equation2019/07/29 11:54 jakobadmin [Layman] 2018/12/19 10:01 jakobadmin ↷ Links adapted because of a move operation2018/05/04 07:53 jakobadmin ↷ Links adapted because of a move operation2018/04/08 15:14 georgefarr ↷ Links adapted because of a move operation2018/03/28 06:53 jakobadmin ↷ Links adapted because of a move operation2018/03/06 14:52 jakobadmin [Why is it interesting?] 2018/03/06 14:49 jakobadmin [Examples] 2018/03/06 14:47 jakobadmin [Examples] 2018/03/06 14:47 jakobadmin [Examples] 2018/03/06 14:45 jakobadmin created Line 62: Line 62: {{ :​advanced_tools:​spheremanifoldexample2.png?​nolink&​400|}} {{ :​advanced_tools:​spheremanifoldexample2.png?​nolink&​400|}} - Another example of a manifold is the two-sphere. The two-sphere $S^2$ is defined as the set of points in $R^3$ for which $y^2+y^2+z^2=const$ holds,​ where $const$ is, of course, the radius of the sphere. The two-sphere is two-dimensional because the definition involves $3$ coordinates and one condition, which eliminates one degree of freedom. Therefore to see that the sphere is a manifold we need a map onto $R^2$. This map is given by the usual spherical coordinates. + Another example of a manifold is the two-sphere. The two-sphere $S^2$ is defined as the set of points in $R^3$ for which $x^2+y^2+z^2=const$ holds,​ where $const$ is, of course, the radius of the sphere. The two-sphere is two-dimensional because the definition involves $3$ coordinates and one condition, which eliminates one degree of freedom. Therefore to see that the sphere is a manifold we need a map onto $R^2$. This map is given by the usual spherical coordinates. Almost all points on the surface of the sphere can be identified unambiguously with a coordinate combination of the form $(\varphi, \theta)$. Almost all! Where is the pole $\varphi=0$ mapped to? There is no one-to-one identification possible because the pole is mapped to a whole line, as indicated in the image.Therefore this map does not work for the complete sphere and we need another map in the neighborhood of the pole to describe things there. A similar problem appears for the map on the semicircle $\theta=0$.  Each point can be mapped to in the $R^2$ to $\theta=0$ and $\theta=2 \pi$, which is again not a one-to-one map. This illustrates the fact that for manifolds there is in general not one coordinate system for all points of the manifold, only local coordinates,​ which are valid in some neighborhood. This is no problem because a manifold is defined to look locally like $R^n$. Almost all points on the surface of the sphere can be identified unambiguously with a coordinate combination of the form $(\varphi, \theta)$. Almost all! Where is the pole $\varphi=0$ mapped to? There is no one-to-one identification possible because the pole is mapped to a whole line, as indicated in the image.Therefore this map does not work for the complete sphere and we need another map in the neighborhood of the pole to describe things there. A similar problem appears for the map on the semicircle $\theta=0$.  Each point can be mapped to in the $R^2$ to $\theta=0$ and $\theta=2 \pi$, which is again not a one-to-one map. This illustrates the fact that for manifolds there is in general not one coordinate system for all points of the manifold, only local coordinates,​ which are valid in some neighborhood. This is no problem because a manifold is defined to look locally like $R^n$.