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I Coordinate Charts

  1. May 31, 2017 #1
    I feel embarrassed to ask this, but I may have a misunderstanding in my understanding of some basics.

    I was told that ##\psi: U \rightarrow \psi(U)##, where ##U = (0, \infty) \times (0, \pi) \times (-\pi, \pi)## and ##\psi(\rho, \varphi, \theta) = (\rho\cos\theta\sin\varphi, \rho\sin\theta\sin\varphi, \rho\cos\varphi)##, is not a coordinate map, but that ##\psi^{-1}## is a coordinate map.

    But doesn't the co-domain of a coordinate map have to be an open subset of ##\mathbb{R}^n## with Cartesian coordinates? The co-domain of ##\psi^{-1}## is a subset of ##\mathbb{R}^3## with spherical coordinates so how could ##\psi^{-1}## be a coordinate map?

    EDIT: I should add that I am using the convention that the domain of my coordinate maps are open subsets of my manifold, and the person who corrected me on the ##\psi^{-1}## thing is aware of this.
     
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  3. Jun 1, 2017 #2

    Orodruin

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    The coordinate functions are maps from the manifold to an open set in ##\mathbb R^3##. In your case the manifold itself is ##\mathbb R^3## so it is bound to get a bit confusing about what is what. However, what is intended is to use the spherical coordinates as coordinates on ##\mathbb R^3##. This means ##\psi^{-1}## is the coordinate map in your case.
     
  4. Jun 5, 2017 #3
    Sorry for such a late reply, the last few days have been hectic. Would it be correct to say that the codomain of a coordinate map must be an open subset of ##\mathbb{R}^n## with the standard smooth structure (the smooth structure determined by the chart ##(\mathbb{R}^n, Id)##), and since spherical coordinate charts belong to the standard smooth structure then ##\psi^{-1}## is a coordinate map even though its codomain does not have Cartesian coordinates?
     
  5. Jun 6, 2017 #4

    WWGD

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    Just for the sake of preciseness, it must be an open n-ball, and homeomorphic to ##\mathbb R^n ## itself.
     
  6. Jun 7, 2017 #5

    mathwonk

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    In fact, rho, phi, and theta are cartesian coordinates on your U. and it is fine that your U is an open rectangle, rather than an open ball. It is still diffeomeorphic to R^3. It is true that some authors ask a coordinate nbdh to be a ball, but that is inconvenient in actual examples like this one. Indeed it is more convenient and natural for them to be rectangles in practice. polar coordinates, or rather their inverse parametrization (rcos(theta), rsin(theta)), e.g. carry rectangles to segments of annuli. but r and theta are still cartesian coordinates on the rectangle in (r, theta) space..
     
  7. Jun 7, 2017 #6

    mathwonk

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    a coordinate map sends points to num,bers, i.e. it assigns coordinates to the points of the manifold. a parametrization maps a domain of numbers ionto points of the manifold. of course it is confusing because the main way we have to describe a manifold is to take a subset of euclidean space, so its points already have numbers. but those numbers do not count as manifold coordinates because they usually do not range over an open set. e.g. the points of the unit sphere in R^3 have "coordinates" x,y,z, but these are not manifold coordinates on the unit sphere because they range only iovr the closed non open set of the sphere. the spherical coordinates phi, theta, however do range over an open set of the phi, theta plane. so they give sherical coordinates at lest on a large open dense subset of the sphere. back in the phi, theta plane however they are actually cartesian coordinates in that plane.

    i.e. the map sending a point of the sphere in x,y,z space to the correspoonding pair phi, theta, is called the spherical cordinate map on the sphere. but the map sending a point of the phi. theta plane to the pair phi, theta, is the cartesian coordinate map on that plane. thus the spherical cordinate map is the composition of the map from the sphere to the phi theta plane, followed by the cartesian coordinate maps of that plane to its axes.
     
  8. Jun 7, 2017 #7

    WWGD

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    You're right that it does not have to be an open ball, I just did not have a simple way of saying it had to be an open subset homeomorphic to ## \mathbb R^n ##.
     
  9. Jun 9, 2017 #8
    Thank you for the excellent explanation mathwonk - you addressed my confusion exactly.
     
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