# A circle in a non-euclidean geometry

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1. Oct 19, 2015

### June_cosmo

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:
$$r=r_0$$
$$\theta=\pi/2$$
(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. Oct 19, 2015

### andrewkirk

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

The length is
$$\int_C\sqrt{ds^2}$$
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 $\theta$?

3. Oct 19, 2015

### June_cosmo

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