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
\[ds^2_{2d}=dr^2 +r^2 d\theta^2\]
The circumference is found by fixing the radial coordinate (r = R and dr = 0) and integrating 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.
(b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in Problem (1). This can be written as:
\[ds^2_{2d}=d\xi^2+sin\xi^2 d\theta^2\]
Repeat the same calculation above and derive the ratio for the closed geometry.
The circumference is found by fixing the radial coordinate (r = R and dr = 0) and integrating 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.
(b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in Problem (1). This can be written as:
\[ds^2_{2d}=d\xi^2+sin\xi^2 d\theta^2\]
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.)
(c) Repeat the same analyses for the open geometry, and comparing to 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?
(c) Repeat the same analyses for the open geometry, and comparing to 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?
Let's begin.
(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
\[ds^2_{2d}=dr^2 +r^2 d\theta^2\]
The circumference is found by fixing the radial coordinate (r = R and dr = 0) and integrating 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.
The circumference is found by fixing the radial coordinate (r = R and dr = 0) and integrating 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.
Let's start with the given equation:
\[ds^2_{2d}=dr^2 +r^2 d\theta^2\]
So, to calculate C, we let r=R and dr=0.
\[ds^2_{2d}=R^2 d\theta^2\]
\[ds_{2d}=R d\theta\]
Integrating, we get:
\[\int ds_{2d}=\int^{2\pi}_0 R d\theta\]
\[C=2\pi R\]
This is as expected.
Now, we solve for radius.
\[ds^2_{2d}=dr^2 +r^2 d\theta^2\]
\[ds^2_{2d}=dr^2\]
Integrating, we get:
\[\int ds_{2d}=\int^R_0 dr\]
\[R=R\]
Now to find our ratio:
\[\frac{C}{R}=\frac{2\pi R}{R}=\boxed{2\pi}\]
This is obvious, but it sets up a useful frame work for the rest of the calculations.
(b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in Problem (1). This can be written as:
\[ds^2_{2d}=d\xi^2+sin\xi^2 d\theta^2\]
Repeat the same calculation above and derive the ratio for the closed geometry.
\[ds^2_{2d}=d\xi^2+sin\xi^2 d\theta^2\]
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.)
Let's solve for C.
\[ds^2_{2d}=d\xi^2+sin\xi^2 d\theta^2\]
\[ds^2_{2d}=sin\xi^2 d\theta^2\]
\[ds_{2d}=sin\xi d\theta\]
\[\int ds_{2d}=\int^{2\pi}_{0}sin\xi d\theta\]
\[C=2\pi sin\xi\]
Solving for R:
\[ds^2_{2d}=d\xi^2+sin\xi^2 d\theta^2\]
\[ds^2_{2d}=d\xi^2\]
\[ds_{2d}=d\xi\]
\[\int ds_{2d}=\int^{\xi}_0 d\xi\]
\[R=\xi\]
Now for the ratio:
\[\frac{C}{R}=\boxed{\frac{2\pi sin\xi}{\xi}}\]
We will compare everything is part (c).
(c) Repeat the same analyses for the open geometry, and comparing to the flat case.
\[ds_{2d}^2=\frac{dr^2}{1+r^2}+r^2 d\theta^2\]
Solving for C:
\[ds_{2d}^2=\frac{dr^2}{1+r^2}+r^2 d\theta^2\]
\[ds_{2d}^2=R^2 d\theta^2\]
\[ds_{2d}=R d\theta\]
\[\int ds_{2d}=\int^{2\pi}_0 R d\theta\]
\[C=2\pi R\]
Solving for R:
\[ds_{2d}^2=\frac{dr^2}{1+r^2}\]
\[ds_{2d}=\frac{dr}{(1+r^2)^{1/2}}\]
\[\int ds_{2d}=\int^R_0 \frac{dr}{(1+r^2)^{1/2}}\]
Using fancy hyperbolic trigonometry identities:
\[R=sinh^{-1}(R)\]
Now for the ratio:
\[\frac{C}{R}=\frac{2\pi R}{sinh^{-1}R}=\boxed{\frac{2\pi sinh\xi}{\xi}}\]
(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?
First let's compare our 3 values.
\[\frac{2\pi sin(\xi)}{\xi}\le 2\pi R \le \frac{2\pi sinh(\xi)}{\xi}\]
Now, as we talk the limit as \(\xi\) goes to 0, a remarkable thing happens: they all become \(2\pi R\).
This makes sense mathematically, but more importantly, it makes sense physically. For example, we live on the curved surface of the Earth, but when we are close to it, it looks flat. The same applies for curved 3D space, in small portions it still looks flat.
I worked with B. Brzycki, G. Grell, and N. James on this problem.
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