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Homework Help: A circle in a non-euclidean geometry

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider a universe described by the Friedmann-Robertson-Walker metric which describes an open, closed, or
    at universe, depending on the value of k:
    $$ds^2=a^2(t)[\frac{dr^2}{1-kr^2}+r^2(d\theta^2+sin^2\theta d\phi^2)]$$
    This problem will involve only the geometry of space at some fixed time, so we can ignore thedependence of a on t, and think of it as a constant. Consider a circle described by the equations:
    (a) Find the circumference S of this circle. (Hint:break the circle into infinitesimal segments of angular size dphi,
    calculate the arc length of such a segment, and integrate.)
    b) Find the radius Rc of this circle. Note that Rc is the length of a line which runs from the origin to the circle (r = r0), along a trajectory of theta=pi/2 and phi= const. Consider the case of open and closed universes separately, and take k= 1 or k=-1 as discussed in lecture. (Hint: Break the line into infinitesimal segments of coordinate length dr, calculate the length of such a segment, and integrate.)

    2. Relevant equations

    3. The attempt at a solution
    I don't know how to derive dphi from the first equation provide?
  2. jcsd
  3. Oct 19, 2015 #2


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    Gold Member

    You don't need to derive it. You use it as your integration differential.

    The length is
    where ##C## is the circle around which you are integrating. Use the metric equation to convert the ##\sqrt{ds^2}## into a simple function of ##d\phi##. What happens to ##dr## and ##d\theta## given you've fixed ##r## and ##\theta##?
  4. Oct 19, 2015 #3
    Oh I get it! so dr and dtheta would be 0. so that
    $$ds^2=a^(t)*r^2sin^2\theta d\phi^2$$
    $$ds=a*rd\phi$$, the circumstance would be 2*pi*r0a.
    but I still don't quite get the second question. If we hold theta and phi constant this time, $$ds=a\sqrt{\frac{1}{1-kr^2}}dr$$,from where to where do we integrate?
    Last edited: Oct 19, 2015
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