$$\require{cancel}$$

# S04. Curvature - SOLUTIONS

Box $$\PageIndex{1}$$

Exercise 4.1.1: Use Equation 4.2 to show that the distance along the path with constant $$\phi$$ that goes from the origin to $$r = r_1$$ is simply given by $$r_1$$. (You've already done this, or something very similar, in Chapter 1. If you like, you can just go back now and review what you did. Or you can just do it again, it should be quick.)

Solution $$\PageIndex{1.1}$$

\begin{equation*} \begin{aligned} ds^2 = dr^2 + r^2 d\phi^2 \end{aligned} \end{equation*}

"constant $$\phi$$", so $$d\phi = 0$$, and then $$ds = dr$$. Therefore, the distance is

\begin{equation*} \begin{aligned} \int ds = \int_{0}^{r_1} dr = r_1 \end{aligned} \end{equation*}

Box $$\PageIndex{2}$$

Exercise 4.2.1: Consider the set of points all at $$r = r_1$$. These define a circle because they are the same distance from a common point (in this case the origin). Assuming that $$r, \phi$$ and $$r, \phi+2\pi$$ are the same point, use Equation 4.2 to show that the circumference of this circle is $$2\pi r_1$$.  (You've already done this, or something very similar, in Chapter 1. If you like, you can just go back now and review what you did. Or you can just do it again, it should be quick.)

Solution $$\PageIndex{2.1}$$

"all at $$r = r_1$$", so now $$dr = 0$$, and then $$ds = r_1 d\phi$$.

$$\phi$$ chaning by $$2\pi$$ takes us once around the circle so the circumference is

\begin{equation*} \begin{aligned} \int ds = \int_{0}^{2\pi} r_1 d\phi = 2\pi r_1 \end{aligned} \end{equation*}

Box $$\PageIndex{3}$$

Exercise 4.3.1: How long would a path be that stretches from $$r=0$$ to $$r=r_1$$ at constant $$\phi$$? Call the length $$\ell$$ and express it as an integral that depends on $$r_1$$ and $$k$$. Assume that $$r_1$$ is much less than $$\sqrt{1/k}$$. Make a Taylor expansion that approximates the integrand so that it contains the first order corrections due to $$k \ne 0$$. After this approximation, do the integral.

Solution $$\PageIndex{3.1}$$

\begin{equation*} \begin{aligned} d\phi = 0 \quad \Longrightarrow \quad \ell = \int ds = \int_{0}^{r_1} \frac{1}{\sqrt{1 - kr^2}}dr \end{aligned} \end{equation*}

If$$\quad r \ll 1/\sqrt{k} \quad \Longrightarrow \quad kr_1^2 \ll 1 \quad \Longrightarrow \quad kr^2 \ll 1 \quad$$for$$\quad r < r_1$$.

Therefore the integrand $$\frac{1}{\sqrt{1 - kr^2}} \simeq 1 + \frac{1}{2}kr^2$$ by the Taylor expansion $$(1 + \epsilon)^n \simeq 1 + n\epsilon$$, where $$n = -\frac{1}{2}$$.

Now we have

\begin{equation*} \begin{aligned} \ell \simeq \int_{0}^{r_1} \Big(1 + \frac{1}{2}kr^2\Big)dr = r_1 + \frac{1}{6}kr_1^3 = r_1\Big(1 + \frac{1}{6}kr_1^2\Big) \end{aligned} \end{equation*}

Box $$\PageIndex{4}$$

Exercise 4.4.1: Consider the set of points all at $$r= r_1$$ with all values of $$\phi$$. Is this a circle? What is the circumference of this object? Again, assume that $$r,\phi$$ and $$r,\phi+2\pi$$ are the same point. First find the circumference as a function of $$r_1$$ and then use your result from the previous problem to express it as a function of $$k$$ and $$\ell$$.

Solution $$\PageIndex{4.1}$$

Yes it is a circle. The distance from the center to $$r_1$$ does not depend on the value of $$\phi$$, so all these points are at the same distance from the center.

The angular part of the invariant distance is unchanged from what we did in Exercise 4.2.1, so the circumference is

\begin{equation*} \begin{aligned} \int ds = \int_{0}^{2\pi} r_1 d\phi = 2\pi r_1 \end{aligned} \end{equation*}

Therefore $$C = 2\pi r_1$$, while the radius $$= \ell = r_1\big(1 + \frac{1}{6}kr_1^2\big)$$ from the result we obtained in Exercise 4.3.1.

So let us solve for $$r_1$$, being sure to keep in mind that $$kr_1^2 \ll 1$$:

\begin{equation*} \begin{aligned} r_1 &= \frac{\ell}{\Big(1 + \frac{1}{6}kr_1^2\Big)} \\ \\ &\simeq \ell\Big[1 - \frac{1}{6}kr_1^2\Big] \; {\rm (Taylor} \; {\rm expansion)} \\ \\ &\simeq \ell\Big[1 - \frac{1}{6}k\ell^2[1 - \frac{1}{6}kr_1^2]^2\Big] \; {\rm (substitution)} \\ \\ &\simeq \ell\Big[1 - \frac{1}{6}k\ell^2(1 - \frac{1}{3}kr_1^2)\Big] \; {\rm (Taylor)} \\ \\ &\simeq \ell\Big[1 - \frac{1}{6}k\ell^2 + \frac{1}{18}(k\ell^2)(kr_1^2)\Big] \\ \\ &\simeq \ell\Big[1 - \frac{1}{6}k\ell^2\Big] \; {\rm (neglecting} \; 2^{nd} \; {\rm order} \; {\rm term)} \end{aligned} \end{equation*}

Therefore the circumference can be expressed as

\begin{equation*} \begin{aligned} C \simeq 2\pi\ell\Big[1 - \frac{1}{6}k\ell^2\Big] \end{aligned} \end{equation*}

Exercise 4.4.2: Discuss the result from Exercises 4.3.1 and 4.4.1 and what it means qualitatively for the circumference-radius relationship for circles in spaces with $$k < 0$$, $$k > 0$$ and $$k=0$$.

Solution $$\PageIndex{4.2}$$

It follows then from the previous results that:

$$k < 0$$: These circles have a bigger circumference than the Euclidean result.

$$k = 0$$: This is the Euclidean result.

$$k > 0$$: These circles have a smaller circumference than the Euclidean result.

Box $$\PageIndex{5}$$

Exercise 4.5.1: If you were a two-dimensional creature, and could travel around this space with a measuring tape, describe in a few sentences at least one way for measuring the value of $$k$$.

Solution $$\PageIndex{5.1}$$

1) Pin down one end of the measuring tape and mark the set of points that are all a distance $$\ell$$ from it.

2) Lay the measuring tape along this circle to measure its length. Call that $$C$$.

3) Check that $$C$$ is close to $$2\pi\ell$$ (so approximations we have used so far in above exercises are good).

4) Use $$C = 2\pi\ell\Big[1 - \frac{1}{6}k\ell^2\Big]$$ to solve for $$k$$.

$$\quad$$ or . . .

Use $$C = 2\pi r_1$$ to get $$r_1$$ (using $$C$$ from (2)) and use $$\ell = r_1\Big(1 + \frac{1}{6}kr_1^2\Big)$$ with $$\ell$$ from (1) to get $$k$$.

Box $$\PageIndex{6}$$

Exercise 4.6.1: Observe the circle at constant coordinate value $$r$$ in the embedding diagram. The distance from the origin (top of the sphere) to any point in the circle is $$\ell$$. Is the circumference of the circle greater than, equal to, or less than $$2\pi \ell$$? Compare to the relevant result in the boxes above.

Solution $$\PageIndex{6.1}$$

The circumference is less than $$2\pi \ell$$. This is consistent with the $$k > 0$$ case in Exercise 4.4.2.

Box $$\PageIndex{7}$$

Exercise 4.7.1: You know that in a Euclidean space the relationship between radius and area of a sphere is $$A = 4\pi r^2$$ with $$r$$ specifying the radius. Note that the angular ( $$d\phi$$ and $$d\theta$$ ) parts of the invariant distance equation are unchanged by having $$k \ne 0$$. Therefore we still have $$A = 4\pi r^2$$. I also claim that the relationship between sphere area and radius does depend on $$k$$. How can these statements both be true?

Solution $$\PageIndex{7.1}$$

The statements are both true because while we have $$A = 4\pi r^2$$ even when $$k \ne 0$$, whether or not $$r$$ is the radius does depend on $$k$$, since the radius is obtained by integrating not $$dr$$ but $$dr/\sqrt{1-kr^2}$$.