## Saturday, November 21, 2015

### 31 | Worksheet 10.1, Problem 2: Ratio of circumference to radius

In this problem, we explored the ratio of the circumference to radius in flat, open, and closed geometries. To do this, we examined 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 $$\mathrm{d}\phi = 0$$ because a circle encloses a two-dimensional surface. For the flat case, this part is just:
$\mathrm{d}s^2 = \mathrm{d}r^2 + r^2 \mathrm{d}\theta^2$
The circumference is found by fixing the radial coordinate ($$r = R$$ and $$\mathrm{d}r =0$$) and integrating both sides of the equation. The radius is found by fixing the angular coordinate ($$\theta$$,$$\mathrm{d}\theta = 0$$) and integrating both sides.

Compute the circumference and radius to reproduce the famous Euclidean ration $$2\pi$$.

We are told that the metric we are working with is given by:
$\mathrm{d}s^2 = \mathrm{d}r^2 + r^2 \mathrm{d}\theta^2$
Let's start by finding the circumference. First, we need to fix the radial coordinate such that $$r = R$$ and $$\mathrm{d}r = 0$$. Doing so gives us;
$\mathrm{d}s^2 = R^2 \mathrm{d}\theta^2$
This simplifies to:
$\mathrm{d}s = R\mathrm{d}\theta$
Integrating over the angles $$0 \rightarrow 2\pi$$,
$\int\! \mathrm{d}s = \int_0^{2\pi} R \mathrm{d}\theta$
This gives us:
$S = 2\pi R$
This means that the circumference of the circle is $$2\pi R$$. This is consistent with what we know about circles.

Now, let's find the radius. To do this, we set $$\theta$$ constant, such that $$\mathrm{d}\theta = 0$$. Our expression now simplifies to:
$\mathrm{d}s = \mathrm{d}r$
If we integrate both sides, we get:
$\int \mathrm{d}s = \int_0^R \mathrm{d}r$
$s = R$
This means that the radius of our circle is $$R$$, which makes perfect sense.

Now, to find the ratio, we simply take the circumference over the radius:
$\frac{\text{circumference}}{\text{radius}} = \frac{2\pi R}{R} = \boxed{2\pi}$

For closed geometry, we calculated the analogous two-dimensional part of the metric in Problem 1. This can be written as:
$\mathrm{d}s^2 = \mathrm{d}\xi^2 + \sin^2 \xi \mathrm{d}\theta^2$
Repeat the same calculation as above to derive the ratio for closed geometry.  Compare your results to the Euclidean case.

Here we have a very similar problem to what we did in (a), so we can use the same technique. Let's start by considering the circumference. To do so, we will set $$\xi$$ constant, such that $$\xi = \Xi, \mathrm{d}\xi = 0$$). Our expression becomes:
$\mathrm{d}s^2 = \sin^2 \Xi \mathrm{d}\theta^2$
$\mathrm{d}s = \sin \Xi \mathrm{d}\theta$
Again, we can integrate both sides:
$\int \mathrm{d}s = \int_0^{2\pi}\sin \Xi\, \mathrm{d}\theta$
Solving gives us that the circumference of the circle in closed geometry is:
$S = 2\pi \sin \Xi$
Now, let's find the radius by setting the angular coordinate constant $$\mathrm{d}\theta =0$$. Simplification yields:
$\mathrm{d}s = \mathrm{d}\xi$
Now we can integrate from 0 to $$\Xi$$:
$\int \mathrm{d}s =\int_0^{\Xi}\mathrm{d}\xi$
Thus, the radius is simply given by $$\Xi$$.

Using these two values we find that the ratio is:
$\frac{\text{circumference}}{\text{radius}} = \boxed{\frac{2\pi \sin \Xi}{\Xi}}$
We can see that this is different from the Euclidean case. In this example, the peak of the ratio is $$2\pi$$ (same as Euclidean), but this only occurs in the limit as $$\Xi \rightarrow 0$$. Otherwise, this ratio is smaller.

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

This is a very similar problem to (b) except that:
$\mathrm{d}s^2 = \mathrm{d}\xi^2 + \sinh^2 \xi\, \mathrm{d} \theta^2$
I won't do the math here, as it is exactly the same as (b) except that we replace the sine function with the hyperbolic sine function. We end up at the conclusion that:
$\frac{\text{circumference}}{\text{radius}} = \boxed{\frac{2\pi \sinh \Xi}{\Xi}}$
Like in the closed geometry case, this ratio is dependent on the radial coordinate. In fact, this ratio has a very similar form to that of the closed 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 consider the closed case where we found the ratio to be:
$\frac{2\pi \sin \Xi}{\Xi}$
As $$\Xi \rightarrow 0$$, we can use the small angle approximation that says $$\sin x \approx x$$ for small angles. For small values of $$\Xi$$, the ratio we found becomes:
$\frac{2\pi \sin \Xi}{\Xi} \approx \boxed{2\pi}$
Thus, the limit for closed and open geometries as $$\Xi \rightarrow 0$$ is identical to the ratio for the flat geometry.