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I Mapping between two spaces

  1. Jun 14, 2017 #1
    Consider the mapping ##f: \mathbb{D}^1 \longrightarrow [0,1] \times (0,2 \pi]## where ##\mathbb{D}^1## is the unit disk. This is the familiar polar coordinate system.

    The area element is ##dx \wedge dy## in ##\mathbb{D}^1## and ##r dr \wedge d\theta## in ##[0,1] \times (0,2 \pi]##*. Now at ##r = 0## this vanishes. But ##dx \wedge dy = r dr \wedge d\theta## and that means ##dx \wedge dy = 0## as well.

    We know that the polar system is singular at ##r = 0## because there cannot be a bijection: for every ##\theta \in (0,2 \pi]## we have the point ##(0,0)## in ##\mathbb{D}^1##.

    But what is the meaning of the vanishing area element ##dx \wedge dy##?

    *I know this is two dimensional and the area element as defined in vector calculus as the vector product of two vector requires a third vector, which means a third dimension. But I think this is not relevant here.
     
  2. jcsd
  3. Jun 14, 2017 #2

    fresh_42

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    It means that you have allowed the RHS to be multiplied by zero and the LHS not. The equations doesn't hold for ##r=0\,##.
     
  4. Jun 14, 2017 #3
    Does it mean we have to remove the point ##(0,0)## from the disk?
     
  5. Jun 14, 2017 #4

    fresh_42

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    In order to do what? I consider ##(dx,dy)## as Cartesian coordinates. If we give them coefficients, zero shouldn't be a problem. To me it is as if you asked, whether the origin of the plane should be removed, because the standard basis vectors aren't zero, whereas the radius can be zero.
     
  6. Jun 14, 2017 #5
    Yes, this is what I was asking you
    Me too
    In order not to have a inconsistent equation when ##r = 0##.
     
  7. Jun 14, 2017 #6

    fresh_42

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    Well, the relation ##(0,0) \longleftrightarrow \{0\} \times (0,2\pi]## is obviously not one-to-one which causes problems. To remove a point is problematic, as it directly affects the functions defined on ##\mathbb{D}^1##. I'd rather change the coordinate system depending on the task.
     
  8. Jun 14, 2017 #7
    Any examples of such coordinate systems?
    This is not a specific problem, I was just thinking about what I mentioned in post #1.
     
  9. Jun 14, 2017 #8

    fresh_42

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    I assume in most cases it doesn't make a difference. If so, I'd simply use the Cartesian version, but that's only because I'm no big fan of polar coordinates. It would be interesting to have an example, where it makes a difference to see, what exactly shouldn't work.
     
  10. Jun 14, 2017 #9

    WWGD

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    Basically, your coordinate change map is not defined for r=0:

    We have:
    ##x=rcos\theta ##
    ## y=rsin\theta ##
    But this holds for ## r \neq 0 ## only , as you said . Then the same is true for ## dx \wedge dy = (\frac {\partial}{\partial \theta}d\theta+ \frac{\partial}{\partial r}dr )( rcos\theta+ ysin\theta)##. The wedge is not defined at 0 because the coordinate expression for r=0 is not defined.
    EDIT: the one way I can see ##dx \wedge dy ## turned into 0 is if the determinant of the Jacobian of a change of coordinate is 0. This is what ##f(x)dx\wedge dy ## measures; how much f(x) stretches/shrinks the area element under a trasnformation.
     
    Last edited: Jun 14, 2017
  11. Jun 14, 2017 #10
    Which is not the case we are considering, because for this case we as you said, ##r=0## is not on the domain, isn't?
     
  12. Jun 14, 2017 #11

    WWGD

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    Right, we are not really transforming the space, we are expressing it in different terms, i.e., changing coordinate systems, so no squashing ( nor stretching) is happenning. EDIT: By ( many) definition(s) , actually , a change of variable is supposed to be a diffeomorphism, meaning the ais invertible,and therefore its determinant is non-zero. Which kind of makes sense, that you would want that for a coordinate change.
     
    Last edited: Jun 14, 2017
  13. Jun 15, 2017 #12

    Orodruin

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    Am I the only one bothered by the fact that the set ##[0,1]\times [0,2\pi)## is not open and so the described mapping is not a chart? Polar coordinates can never give you a chart at ##r=0## and so it makes no sense to try to interpret coordinate expressions for tensors in polar coordinates at the origin.
     
  14. Jun 15, 2017 #13

    WWGD

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    But I think the map is just intended to be a (local) change of coordinates. Is this equivalent to ##[0,1] \times [0,2\pi) ## being a chart?EDIT: I can see how a manifold is a way of assigning coordinates, but I had never seen things that way. Still, I agree, the coordinate change is a local diffeomorphism so it should preserve the topology of ##\mathbb D^1 ##.
     
  15. Jun 15, 2017 #14

    Orodruin

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    A chart is a homeomorphism from an open subset of the manifold to an open subset of ##\mathbb R^n##. It does not need to cover the entire manifold. Remove ##r=0##, ##r=1## and ##\theta = 0## and you have a chart that covers almost the entire disc.
     
  16. Jun 15, 2017 #15
    @Orodruin I've been considering ##[0,1] \times [0,2 \pi)## throughout this thread because that set covers the entire disk and I think it's the most familiar way of mapping points on the disk, although it's not homeomorphic to ##\mathbb{R}^2##.
    Is the area element and all other quantities defined only when we have charts? Is it what you mean here?
     
  17. Jun 15, 2017 #16

    Orodruin

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    If you want to use the coordinate expressions at a point, you must use a chart that is valid at that point. Writing ##r\, dr\wedge d\theta## makes no sense at ##r = 0## because your coordinate system does not define a chart there. There are some charts that cover all of the disc, your coordinate system with the given set is not one of them because it is not a chart. If you use any chart that covers the origin (such as Cartesian coordinates), you will find that the area element is non-zero there.
     
  18. Jun 15, 2017 #17
    I got it. Thanks.
     
  19. Jun 16, 2017 #18

    lavinia

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    This mapping cannot be continuous.

    In the other direction the map is continuous but singular.
     
    Last edited: Jun 16, 2017
  20. Jun 16, 2017 #19
    Where it is singular?
     
  21. Jun 16, 2017 #20

    lavinia

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    The map in the other direction maps the entire half open interval ##{0}×(0,2π]## onto the center of the unit disk.
     
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