Turns out Euclid was right about math after all

We are bounded in a nutshell of Infinite space: Blog Post #31, Worksheet # 10.1, Problem #2: Turns out Euclid was right about math after all

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 $$ds_{2d}^2= dr^2 + r^2d\theta^2.$$

The circumference is found by fixing the radial coordinate ($r=\theta$ and $dr=0$) and both sides of the equation (note that $\theta$ is integrated from 0 to $2\pi$ ).

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 $2\pi$.

(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.)

(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?

(a) From the Friedman-Robertson-Walker Metric, an equation we will be looking on later in in this problem, we can derive a few basic equations that describes the physical phenomenon in universes with different configurations. Thought of either flat, open, or closed, the universe’s configuration describes how we perceive reality, where circles are the same in any location and their circumference is always $2\pi R$, but this could be different for open or closed universes (even though astronomers observe the universe is most likely flat).

For a flat universe, we can prove the traditional Euclidean description of a circle with a few simple calculations: assuming $$d\phi = 0 ,$$ we get: $$ds_{2d}^2= dr^2 + r^2d\theta^2.$$    This is a version of the FRW Metric with the $d\phi$ set to 0.

Once we have this equation, we simply have to establish some parameters such as $r = R, dr= 0$, and so the equation becomes: $$ds_{2d} = Rd\theta,$$ , and by setting the integrals like the problem asks, we solve: $$\int_0^{Circumference}ds_{2d}=  \int_0^{2\pi}Rd\theta,$$ $$Circumference = 2\pi R,$$ which is exactly the traditional Euclidean case.

For the radius, we have to establish how $\theta, d\theta= 0$ and the equation now becomes: $$ds_{2d}^2= dr^2 + r^2d\theta^2,$$ and following the same process as before for the circumference: $$ds_{2d}^2= dr^2,$$   $$\int_0^rds_{2d}= \int_0^R dr ,$$ and so a true statement appears for the description of our universe: $$r= R.$$

Knowing the circumference and radius, we can use both of these to establish a ratio between them that becomes a standard ratio for comparison with other physical universe interpretations: $$\frac{Circumference}{Radius} = \frac{2\pi R}{R} = 2\pi,$$ which is the Euclidean model the problem asked for.

(b) For a closed universe, the equation is altered slightly because of the base FRW metric used, so we now have: $$ds_{2d}^2 = d\xi^2 + \sin\xi^2 d\theta^2,$$ and with the alterations and limits the problem is establishing in order to better describe a closed universe, with  $$d\xi = 0 ,$$ the equation now becomes $$ds_{2d} = \sin\xi d\theta.$$ Next, using the hints the problem gives, we set up the integration: $$\int_0^{Circumference}ds_{2d}=  \int_0^{2\pi}\sin\xi d\theta,$$ and now have $$Circumference = 2\pi \sin\xi.$$

Next we find the radius, with a process similar to the circumference, and the problem already set up how $\theta, d\theta= 0$, so the original equation now becomes: $$ds_{2d}^2 = d\xi^2 + \sin\xi^2 d\theta^2,$$    $$ds_{2d}^2= d\xi^2$$ and integrating with the correct parameters: $$\int_0^rds_{2d}= \int_0^{\xi} d\xi ,$$ we now have the radius: $$r= \xi.$$

Finally, establishing the ratio of circumference to ratio, we have: $$\frac{Circumference}{Radius} = \frac{2\pi \sin\xi}{\xi},$$ which is identical to the flat universe ratio multiplied by the factor of $$\frac{\sin\xi}{\xi}.$$

(c) Taking the FRW metric as the basis, we can derive how: $$ds^2 = \frac{dr^2}{1-kr^2} + r^2 (d\theta^2 + \sin^2\theta d\phi^2),$$ becomes $$ds^2 = \frac{dr^2}{1-kr^2} + r^2 d\theta^2,$$ when $d\phi = 0$. Then in the case of an open universe where k = -1, the equation becomes: $$ds^2 = \frac{dr^2}{1+r^2} + r^2 d\theta^2,$$ so, taking a recommendation from an earlier problem of establishing that $r = \sinh\xi$, the equation can now be solved to find a simpler solution which is similar to the other ratios. Now, the equation becomes: $$ds^2 = \frac{\cosh^2\xi \cdot d\xi^2}{1+\sinh^2\xi} + (\sinh\xi)^2 d\theta^2,$$ and we can do this since an identity of hyperbolic functions like $\sinh\xi$ establishes that its derivative is $\cosh\xi d\xi$. Next, we simply use another identity says $\cosh^2x – \sinh^2x = 1$, so the equation becomes: $$ds^2 = \frac{\cosh^2\xi \cdot d\xi^2}{\cosh^2\xi } + \sinh^2\xi d\theta^2,$$ $$ds^2 = d\xi^2 + \sinh^2\xi d\theta^2.$$ Using the exact same process as earlier in part b, we find that the ratio of circumference to radius is: $$\frac{Circumference}{Radius}=\frac{2\pi \sinh\xi}{\xi},$$ which is the same as the flat universe ratio multiplied by the factor of $$\frac{\sinh\xi}{\xi}.$$

(d) From these ratios, a clear patter can be discerned once compared to the original, flat universe, case. With all the ratios illustrated clearly: (flat, closed, and open, respectively) $$2\pi , ~ \frac{2\pi \sin\xi}{\xi} , ~ \frac{2\pi \sinh\xi}{\xi}$$ there is a clear change. Both open and closed universe models exhibit the trigonometric and hyperbolic function, which both happen to have a similar Taylor Expansion (a way to approximate for small numbers), which turns the ratios into: $$2\pi , ~ \frac{2\pi \xi}{\xi} , ~ \frac{2\pi \xi}{\xi}$$ $$2\pi , ~2\pi , ~ 2\pi.$$ Therefore, as $\xi$ becomes small, all universes look more and more like each other, as a result of the limiting case as $$\xi \to 0 .$$