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Flattening metric tensor

  1. Jun 16, 2009 #1

    I'm trying to learn some geometry for general relativity, and I am having a bit of trouble understanding how to tell flat and curved spaces apart. Specifically, I heard that a space is flat if you can "flatten" the metric by finding a coordinate system where [tex]ds^2 = dx^2 + dy^2 + ...[/tex]. Unfortunately, if this is true, then I can show that the surface of a sphere is flat.

    Let [tex]\theta[/tex] be the latitude and [tex]\phi[/tex] be the longitude on a unit sphere. Then the metric is

    [tex]ds^2 = d\theta^2 + \sin^2(\theta)d\phi^2[/tex]

    Now let coordinates x and y be

    [tex]x = \theta[/tex]
    [tex]y = \phi \sin(\theta)[/tex]

    Then we get

    [tex]d\theta = dx[/tex]
    [tex]d\phi = \frac{dy}{\sin(\theta)}[/tex]

    Plugging into the metric,

    [tex]ds^2 = d\theta^2 + \sin^2(\theta)d\phi^2 = dx^2 + dy^2\frac{\sin^2(\theta)}{\sin^2(\theta)} = dx^2 + dy^2[/tex]

    This is in the "flat" form.

    I'm sure I've done something wrong. Maybe putting the metric in this form does not actually imply that the space is flat, or maybe I did something bad with the coordinate change?

  2. jcsd
  3. Jun 17, 2009 #2


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    Well, the unit sphere in two dimensions isn't curved, so you might be right.

    Curvature in 2d is a tricky concept. For example, a cone is not curved except at the "tip", but most smooth objects in 2d are intrinsically flat. They just look curved to us, when embedded in 3d space.
  4. Jun 17, 2009 #3


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    You differentiated incorrectly.

    [tex]dy = \phi \, d\!\sin(\theta) + \sin(\theta) \, d\phi[/tex]

    or alternatively,

    [tex]d\phi = d\left( \frac{y}{\sin(\theta)} \right) = \frac{ \sin(\theta)\, dy - y\, d\!\sin(\theta) }{(\sin\theta)^2}[/tex]

    (Of course, [itex]d\!\sin(\theta) = \cos(\theta) \, d\theta[/itex])
  5. Jun 17, 2009 #4


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    I'm under the impression that there aren't any flat metrics on the sphere. (Of course, all metrics on the circle are flat)
  6. Jun 17, 2009 #5


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    Good point. Forget what I said? :blush:
  7. Jun 19, 2009 #6


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    Basically, how I would define flat manifolds is to show that the Riemanntensor is 0. This means that if I would transport vectors on the manifold along arbitrary loops, then the orientation of the vector would stay the same. This is true for the cirkel [itex]S^{1}[/itex]. But also for the cylinder or the doughnut; try it yourself. After this trying you could set up a coordinate system for these manifolds, labourously calculate the connection coeficients etc and construct the Riemann-tensor. Because it's a tensor (a multilinear mapping from tangent spaces and cotangent spaces to the real line), if it vanishes in one coordinate system, it will vanish in every coordinate system.

    Hope this helps :)
  8. Jun 26, 2009 #7
    Nor on any compact surface in R^3.
  9. Aug 11, 2009 #8
    Just because the Riemann curvature tensor is identically zero does not mean that parallel transport of arbitrary vectors around arbitrary loops bring the vector back to itself. What about parallel translation on the flat Klein bottle? In three dimensions there is a manifold with zero Riemann tensor in which no vector transports back to itself (I think).

    In fact the only manifolds with zero Riemann tensor where all vectors transport around all loops back to themselves are flat tori, cylinders, and Euclidean space.
  10. Aug 11, 2009 #9


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    No. You didn't differentiate the "[itex]sin(\theta)[/itex]" which you have to do. [itex]\theta[/itex] is not a constant! [itex]dy= d\phi sin(\theta)+ \phi cos(\theta)d\theta[/itex]. Now you can say that [itex]d\theta= dx[/itex] but you still have that sine and cosine.

    [tex]d\phi= \frac{dy- \phi cos(\theta) dx}{sin(\theta)}[/tex]
    Now, [itex]cos(\theta)= cos(x)[/itex], [itex]sin(\theta)= sin(x)[/itex] and [itex]\phi= y/(sin(\theta)= y/sin(x)[/itex] so
    [tex]d\phi= \frac{dy- y cot(x) dx}{sin(x)}[/itex]

    [tex]ds^2= dx^2+ x^2\frac{(dy- y cot(x) dx)^2}{sin^2(x)}[/tex]
    [tex]= \frac{x^2(1+ y^2cot^2(x))}{sin^2(x)}dx^2- \frac{2x^2ycot(x)}{sin^2(x)}dxdy+ \fra{x^2}{sin^2(x)}dy^2[/tex]
    far from "flat".

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