Blog #31, Worksheet #10.1, Problem #2, Circumference Ratios

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 dφ = 0 because a circle encloses a two-dimensional surface. For the flat case, this part is just \[{ds_{2D}}^2 = d r^2 + r^2 d \theta^2\] The circumference is found by fixing the radial coordinate (r = R and dr = 0) and both sides of the equation (note that θ is integrated from 0 to 2π). 
The radius is found by fixing the angular coordinate (θ, dθ = 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. 

We first set r = R and dr = 0 to find the circumference. We get \[ds^2 = R^2 d \theta^2 \]We then take the root of both sides to get \[ds = R d \theta\]and when we integrate both sides we get \[\int_{0}^{s} ds = \int_{0}^{2 \pi} R d \theta\]Which gives us \[S_{circ} = 2 \pi R\]Then we solve for the radius by setting \(d \theta = 0 \) and get \[S_{rad} = \int_{0}^{R} dr = R \]Finally we take the ratio of our S factors to get the desired ratio \[\frac{S_{circ}}{S_{rad}} = \frac{2 \pi R}{R} = 2 \pi \]

(b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in Problem (1). This can be written as: \[{ds_{2D}}^2 = d \xi^2 + sin \xi^2 d \theta^2\]  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? (You can try some arbitrary values of ξ to get some examples.) 

We repeat a similar process compared to part a) although now our \(S_{circ}\) comes out to be \[S = \int_{0}^{2 \pi} sin X d \theta = 2 \pi sin X\]As in part a), our radius integral simply comes out to \[S_{rad} = X\]Then we place the same ratio as in part a) to get  \[\frac{S_{circ}}{S_{rad}} = \frac{2 \pi sin X}{X}\]We know that the function \(\frac{sin(X)}{X}\) can never be greater than one (see plot below). Thus the flat ratio will always be larger than the closed case, except at X = 0.

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

The open case is identical to the to the closed case except where the closed case has a sin term, the open case has a sinh term. This is due to the \(\frac{dr^2}{1- kr^2}\) that reduces to sinusoidal terms for the closed case, yet remains in the open system. Thus our ratio for the the open case becomes \[\frac{S_{circ}}{S_{rad}} = \frac{2 \pi sinh X}{X}\]However the plot of \(\frac{sinh(X)}{X}\) is always rising and must always be larger than the flat case of \(2 \pi\) (see graph below).

(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? 

As the above graphs show, the only time that both cases approach the flat ratio of \(2 \pi\) is time 0. When we estimate the universe as taking different spacetime geometries, we still approximate all estimates as flat because the universe is so large and crosses such a expansive time, that any approximations we make are based on a very small, short time period and can be approximated as 0.