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I Gravitation vs Curvilinear Coordinates

  1. Jun 16, 2017 #1
    In Weinberg's book, it is said that a given metric ##g_{\mu \nu}## could be describing a true gravitational field or can be just the metric ##\eta_{\alpha \beta}## of special relativity written in curvilinear coordinates. Then it is said that in the latter case, there will be a set of Minkowskian coordinates ##\xi^\alpha (x)## such that
    $$\eta^{\alpha \beta} = g^{\mu \nu}\frac{\partial \xi^{\alpha}(x)}{\partial x^\mu}\frac{\partial \xi^{\beta}(x)}{\partial x^\nu} \ (1)$$ everywhere, not just at every point ##X## and its infinitesimal neighborhood as the equivalence principle states.

    Now what is the difference of everywhere to every point ##X##? Since if we pick up every point ##X## we will end up with all points that make the space we are considering?

    Also, it is said in the book that for the metric which has coefficients
    $$g_{rr} = 1; \ g_{\theta \theta} = r^2; \ g_{\varphi \varphi} = r^2 sin^2 \theta; \ g_{tt} = -1$$ it's possible to find a set of Minkowskian coordinates
    $$\xi^1 = rsin \theta cos \varphi; \ \xi^2 = rsin \theta sin \varphi; \ \xi^3 = rcos \theta; \ \xi^4 = t$$ that satisfies (1) above.

    I don't understand, since by the above reasoning a sphere would'nt be a curved space, for it's possible to find that set of Minkowskian coordinates above that satisfy (1) at every point on the sphere. But we know that the sphere is a intrinsically curved space. Could the Riemann curvature tensor vanish for the sphere when we use spherical coordinates to evaluate it?
     
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  3. Jun 16, 2017 #2

    PeterDonis

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    If the spacetime is actually flat Minkowski spacetime, then you will be able to find one single coordinate chart in which the metric is ##\eta_{\alpha \beta}## everywhere.

    If the spacetime is curved, then you will be able to find a coordinate chart in which the metric is ##\eta_{\alpha \beta}## at some chosen point ##X##, but if you then pick a different point ##X'##, the chart in which the metric is ##\eta_{\alpha \beta}## at ##X'## will be a different chart from the one in which the metric is ##\eta_{\alpha \beta}## at ##X##

    Are you sure? Try it!

    No.
     
  4. Jun 16, 2017 #3
    I will try it. However, in Weinberg's book it is said that it's possible, by using $$\xi^1 = rsin \theta cos \varphi; \ \xi^2 = rsin \theta sin \varphi; \ \xi^3 = rcos \theta; \ \xi^4 = t$$ Maybe the author made a mistake?

    (Quote)
    ##"##For example, given the metric coefficients $$g_{rr} = 1; \ g_{\theta \theta} = r^2; \ g_{\varphi \varphi} = r^2 sin^2 \theta; \ g_{tt} = -1 \ \text{(6.4.2)}$$ we know that there is a set of ##\xi##s satisfying $$\eta^{\alpha \beta} = g^{\mu \nu}\frac{\partial \xi^{\alpha}(x)}{\partial x^\mu}\frac{\partial \xi^{\beta}(x)}{\partial x^\nu} \ $$ that is, $$\xi^1 = rsin \theta cos \varphi; \ \xi^2 = rsin \theta sin \varphi; \ \xi^3 = rcos \theta; \ \xi^4 = t$$ but how could we have told that (6.4.2) was really equivalent to the Minkowski metric ##\eta_{\alpha \beta}##, if we weren't clever enough to have recognized it as simply ##\eta_{\alpha \beta}## in spherical polar coordinates? Or, on the other hand, if we change ##g_{rr}## to an arbritary function of r, how can we tell that this really represents a gravitational field, that is, how can we tell that Eqs. (6.4.2) now have no solution?##"##
     
  5. Jun 16, 2017 #4

    PeterDonis

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    It's possible in a space where the metric coefficients are as you gave them. But is a space where the metric coefficients are as you gave them a sphere? Just because the coordinates are "spherical coordinates" does not mean the geometry is that of a sphere. You can have spherical coordinates on flat Euclidean space.

    To decide whether the metric is that of a sphere, or more generally whether it is even curved rather than flat, you need to compute its Riemann tensor. Have you tried that?
     
  6. Jun 16, 2017 #5
    But spherical coordinates automatically gives points that "live" on the surface of a sphere (for each r constant). So it automatically forms a sphere. How can we get a set of points forming a flat surface when using spherical coordinate system?
    I don't understand it. Since spherical coordinate system automatically gives points that form a sphere, it seems to me that the Riemann tensor will never vanish when we use spherical coordinates.
    Yes. I have computed all of the components a time ago and I remember that some of them were not zero.
     
  7. Jun 16, 2017 #6

    PeterDonis

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    But ##r## is not constant; it's one of the coordinates.

    Try computing the Riemann tensor for the metric Weinberg gives with ##r## as one of the coordinates, instead of being just a constant.
     
  8. Jun 16, 2017 #7
    Ohhh, so this is the key thing
    It would be great to do that and see the magic happening. I will perform the calculations later.
     
  9. Jun 16, 2017 #8

    Nugatory

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    You might try dropping down one dimension and considering whether using polar coordinates (##r##, ##\theta## where ##x=r\cos\theta## and ##y=r\sin\theta##) does anything to change the Euclidean flatness of the two-dimensional x-y plane.
     
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