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In which frame does the unit 2-sphere look locally flat?

  1. Sep 24, 2006 #1
    Can someone please tell me, the coordinate transformation that will make the following metric:

    [tex]ds^2=dr^2 + r^2d\theta^2[/tex]

    look locally flat? Many thanks.
     
  2. jcsd
  3. Sep 24, 2006 #2

    George Jones

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    I think that the transformation is already very familiar to you. :smile:
     
  4. Sep 24, 2006 #3
    Heh, fair enough. So I did sit down and think about it for a bit, and is it this one?

    [tex]\left( \begin{array}{clrr} r' \\ \theta' \end{array} \right) =
    \left( \begin{array}{clrr} 1 & 0 \\ 0 & \mbox{$\frac{1}{r}$} \end{array} \right)
    \left( \begin{array}{clrr} r \\ \theta \end{array} \right)[/tex]

    So that:

    [tex]dx'^\mu =
    \left( \begin{array}{clrr} dr' \\ d\theta' \end{array} \right) =
    \left( \begin{array}{clrr} dr \\ \mbox{$\frac{d\theta}{r}$} \end{array} \right)[/tex]

    So that:

    [tex]ds^2 = dr^2 + d\theta ^2[/tex]
     
    Last edited: Sep 24, 2006
  5. Sep 24, 2006 #4

    George Jones

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    Careful, if, e.g.,

    [tex]\theta' = \frac{\theta}{r},[/tex]

    then

    [tex] d \theta' = \frac{\partial \theta'}{\partial r} d r + \frac{\partial \theta'}{\partial \theta} d \theta.[/tex]

    Also, I'm now a bit confused. When, I first responded, I didn't really read the title of the thread. The metric you give in the original post seems not to be the metric for a 2-sphere. It looks more like the metric 2-dimensional plane written in terms of polar coordinates, and hence my first reply.
     
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