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Surface with Ricci scalar equal to two

  1. Nov 25, 2007 #1
    A two-dimensional Rienmannian manifold has a metric given by
    ds^2=e^f dr^2 + r^2 dTHETA^2
    where f=f(r) is a function of the coordinate r

    Eventually I calculated that Ricci scalar is R=-1/r* d(e^-f)/dr

    if e^-f=1-r^2 what this surface is?

    In this case R comes to be equal to 2
    I've read on wikipedia that Ricci scalar of a sphere with radius r is equal to 2/r^2

    So, is this surface a sphere or radius r=1?
  2. jcsd
  3. Nov 26, 2007 #2


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    Science Advisor
    Homework Helper

    Yes, it is. If you want an explicit coordinate representation in 3-space, it's x=r*cos(theta), y=r*sin(theta), z=sqrt(1-r^2).
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