Blog #31: WS10.1, #2

2. Ratio of circumference to radius. Let's continue to study the difference between closed, flat, and open geometries by computing the ratio between the circumference and radius of a circle. 
a) To compute the radius and circumference of a circle, we look at the spatial part of the metric and concentrate on the two-dimensional part by setting $d\phi=0$ because a circle encloses a two-dimensional surface. For the flat case, this part is just  $$ds^2_{2d}=dr^2+r^2d\theta^2.$$ The circumference is found by fixing the radial coordinate (r = R and dr = 0) and both sides of the equation (note that $\theta$ is integrated from 0 to 2π). 
The radius is found by fixing the angular coordinate ($\theta, d\theta=0$) and integrating both sides (note that dr is integrated from 0 to R). 
Compute the circumference and radius to reproduce the famous Euclidean ratio 2π. 

Since dr = 0 and r = R, our equation becomes $ds^2_{2d}=R^2d\theta^2$.
$ds_{2d}=Rd\theta$
We can then integrate. Since we want the equation for circumference, we integrate ds from 0 to the full circumference.
$\int^{circ}_0ds_{2d}=R\int^{2\pi}_0d\theta$
$circumference=2\pi R$, as we know.
The ratio of circumference to radius would therefore be $\frac{2\pi R}{R}=2\pi$.

b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in problem 1. This can be written as: $$ds^2_{2d}=d\xi^2+sin^2\xi d\theta^2.$$ Repeat the same calculation above and derive the ratio for the closed geometry. 
Compare your results to the flat (Euclidean) case; which ratio is larger? 

Since $\xi$ essentially represents the radius of the circle in this geometry, we can again fix it such that $d\xi=0$. The process is then very similar to that of part a:
$ds^2_{2d}=sin^2\xi d\theta^2$
$ds_{2d}=sin\xi d\theta$
$\int^{circ}_0ds_{2s}=\int^{2\pi}_0sin\xi d\theta$
$circumference=2\pi sin\xi$
The ratio of circumference to radius would be $\frac{2\pi sin\xi}{\xi}$. This would be smaller than the Euclidean ratio, since $\xi$, the denominator, would increase faster than $\sin\xi$, the numerator, would.

c) Repeat the same analyses for the open geometry, and comparing to the flat case.

Since the simplified metric is not provided, we have to start from the FRW metric. Since this is an open, hyperbolic geometry, we know that k = -1, $r=sinh\xi$, and therefore $dr=cosh\xi$. Since this is the 2-dimensional case, we can again set $d\phi=0$.
$ds^2_{3d}=\frac{dr^2}{1-kr^2}+r^2(d\theta^2+sin^2\theta d\phi^2)$
$ds^2_{2d}=\frac{cosh^2\xi}{1+sinh^2\xi}d\xi^2+sinh^2\xi d\theta^2$
$ds^2_{2d}=d\xi^2+sinh^2\xi d\theta^2$
Since we are working with a constant radius, $d\xi=0$.
$ds^2_{2d}=sinh^2\xi d\theta^2$

$\int^{circ}_0ds_{2d}=\int^{2\pi}_0sinh\xi d\theta$

$circumference=2\pi sinh\xi$ 

The ratio of circumference to radius would be $\frac{2\pi sinh\xi}{\xi}$. This would be larger than the Euclidean case. 

d) You may have noticed that, except for the flat case, this ratio is not a constant value. However, in both the open and closed case, there is a limit where the ratio approaches the flat case. Which limit is that? 

At small radii, both the closed and open geometries will have ratios close to that of the flat geometry because $\frac{sin(x)}{x}$ and $\frac{sinh(x)}{x}$ both tend towards 1 as x approaches 0. This makes sense, because a circle drawn on a sphere, if much smaller than the sphere, will appear to be flat. The Earth is a good example--if you draw a circle on the floor, it will look flat even though it's technically on a sphere.