2013-09-07

# Curvature Guesswork(I.5.5, p. 79)*

# Contents

## Exercise

Here we are asked to compute the curvature at the origin for the metric $$ds^2=dr^2+ (r h(r))^2d \theta^2, $$ where $h(0)=1$. For me this is guesswork, for two reasons:

- The curvature is not defined at this point in the book. All we I have is a seemingly ad-hoc formula $$ R = \lim_{\rho\to0}\frac{6}{\rho^2}(1-\frac{\zeta}{2\pi\rho}) \qquad \text{(p. 6)}\label{R}\tag{1} $$ where $\rho$ is the radius of a small circle and $\zeta$ its circumference.
- I seem to need the information that the coordinates used are polar-like, otherwise I would not know how to compute radius and circumference. But this has to be guessed from the naming convention $(r, \theta)$.

## Solution

So I deduce that $\rho=r$ and $\zeta=r h(r)$. Inserting into \ref{R} we get: \begin{align} R &= \lim_{r\to0}\frac{6}{r^2}(1-\frac{2 \pi r\,h(r)}{2 \pi r})\\ &=\lim_{r\to0}\frac{6}{r^2}(1-h(r))\\ &=\lim_{r\to0}6\frac{1-h(r)}{r^2} \end{align} Since both nominator and denominator tend to zero, we may invoke l’Hospital's rule and differentiate both separately without changing the result and get \begin{align} R&= \lim_{r \to 0} 6 \frac{-h'(r)}{2 r}\,. \end{align} Now, if $h'(r)$ tends to anything but zero for $r\to 0$, the resulting $R$ would be plus or minus infinity, so assume that it tends to zero and apply L'Hospital's rule again. \begin{align} R &= \lim_{r \to0} -3h''(r) \end{align} If $h$ curves upwards at the origin, envisage for example a parabola, the 2nd derivative is positive and $R$ becomes negative, as we are asked to prove in the exercise. The exercise further asks to compute $R$ for $h(r)=\sin(r)/r$ and for $h(r)=\sinh(r)/r$. In both cases it is quicker to apply L'Hospital's rule as often as needed, than to compute the 2nd derivative. For the sine we get: \begin{align} R&=\lim_{r\to0} 6\frac{1-\sin(r)/r}{r^2}\\ &=\lim_{r\to0} 6\frac{r-\sin(r)}{r^3}\\ &=\lim_{r\to0} 6\frac{1-\cos(r)}{3r^2}\\ &=\lim_{r\to0} 6\frac{\sin(r)}{6r}\\ &=\lim_{r\to0} \cos(r) =1 \end{align} For the hyperbolic sine, the derivation is just the same, except for one less change of sign, so that the result is $-1$, as given in the solution section of the book.