# S02. Curvature - SOLUTIONS

- Page ID
- 3917

# Exercise 2.1.1

**Answer**-
\[\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*}\]

# Exercise 2.2.1

**Answer**-
"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*}\]

# Exercise 2.3.1

**Answer**-
\[\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*}\]

# Exercise 2.4.1

**Answer**-
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 2.4.2

**Answer**-
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.

# Exercise 2.5.1

**Answer**-
- Pin down one end of the measuring tape and mark the set of points that are all a distance \(\ell\) from it.
- Lay the measuring tape along this circle to measure its length. Call that \(C\).
- Check that \(C\) is close to \(2\pi\ell\) (so approximations we have used so far in above exercises are good).
- 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 \(k\).

# Exercise 2.6.1

**Answer**-
The circumference is less than \(2\pi \ell\). This is consistent with the \(k > 0\) case in Exercise 2.4.2.

# Exercise 2.7.1

**Answer**-
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 \(dr/\sqrt{1-kr^2}\).