Blog Post 31, Worksheet 10.1, Problem 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 \[ ds_{2d}^2 = dr^2 + r^2 d\theta^2.\] The circumference is found by fixing the radial coordinate (r = R and dr = 0) and both sides of the equation (note that \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 ration 2π. 

As stated in the question, we are looking only at the spatial part of the metric: \[ ds_{2d}^2 = dr^2 + r^2 d\theta^2.\] To find the circumference of the circle, we need to set the radius equal to a constant, r = R, with dr = 0. Making these substitutions, we are left with: \[ds_{2d}^2 = R^2 d\theta^2\] After we take the square root of both sides, we have: \[ds_{2d} = R d\theta\] Now, we integrate both sides. We integrate \(\theta\) from 0 to 2π, and we integrate s from 0 to some arbitrary constant C: \[\int_0^C ds_{2d} = \int_0^{2π} Rd\theta\] This gives us our circumference, \[C = 2πR\]

To find the radius, we instead set \(\theta\) and \(d\theta\) equal to zero, as we are integrate outward from the center of the circle: \[ds_{2d}^2 = dr^2\] \[ds_{2d} = dr\] Now, we integrate r from 0 to R, and we integrate s from 0 to some arbitrary constant r: \[\int_0^r ds = int_0^R dr\] \[r = \text{radius} = R\] Now that we have the circumference and radius, we can calculate our circumference to radius ratio: \[\frac{\text{circumference}}{\text{radius}} = \frac{2πR}{R} = 2π\]

(b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in Problem (1). This can be written as: \[ds_{2d}^2 = d\xi^2 + \text{sin}\: \xi^2 d\theta^2.\] 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? 

To find the circumference for a closed geometry, we now set \(d\xi = 0\). This is analogous to setting \(dr = 0\) in the above problem, but now we are working in the hyperspherical coordinate system. \[ds_{2d}^2 = d\xi^2 + \text{sin}\: \xi^2 d\theta^2\] \[ds_{2d}^2 = \text{sin}\: \xi^2 d\theta^2\] \[ds_{2d} = \text{sin}\: \xi d\theta\] Now we integrate \(\theta\) from 0 to 2π, and we integrate s from 0 to some arbitrary constant C: \[\int_0^C ds_{2d} = \int_0^{2π} \text{sin}\: \xi d\theta\] \[C = \text{circumference} = 2π\text{sin}\: (\xi)\] To calculate the radius, we set \(d\theta = 0\), leaving us with: \[ds_{2d}^2 = d\xi^2\] \[ds = d\xi\] Now we integrate s from 0 to some radius r, and \(\xi\) from 0 to \(\xi\): \[\int_0^r ds_{2d} = \int_0^{\xi} d\xi\] \[r = \xi\] circumference to radius ratio is thus: \[\frac{\text{circumference}}{\text{radius}} = \frac{2π \text{sin }(\xi)}{\xi}\] The value of \(\text{sin }(\xi)\), will always be between -1 and 1, and thus this ratio will always be less than the ratio for the flat geometry, 2π. 

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

For the open geometry (k  = -1), the relevant equation becomes \[ds_{2d}^2 = d\xi^2 + \text{sinh }^2 d\theta^2\] Using the same logic as above, we derive the circumference by setting \(d\xi = 0\) and integrating \(\theta\) from 0 to 2π. To derive the radius, we set \(d\theta = 0\) and integrate \(\xi\) from 0 to \(\xi\). Our ratio thus becomes: \[\frac{\text{circumference}}{\text{radius}} = \frac{2π \text{sinh }(\xi)}{\xi}\] The value of \(\frac{\text{sinh }\xi }{\xi}\) will always be greater than one, so the ratio for the open geometry will always be greater than that of the flat 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?

Both the open and closed case approach the flat case ratio, 2π, in the limit as \(\xi\) goes to 0.