Metric of a cone I am trying to solve the following problem (Eric Poisson's The Mechanics of Black Hole Physics, Chapter 1, Problem 1):

The surface of a two-dimensional cone is embedded in three-dimensional flat space. The cone has an opening angle of $2 \alpha$. Points on the cone which all have the same distance $r$ from the apex define a circle, and $\phi$ is the angle that runs along the circle.
Write down the metric of the cone, in terms of the coordinates $r$ and $\phi$.

My attempt so far is
$$\mathrm{d}s^2 = r \ \mathrm{d}r^2 + r \sin^2(\phi) \ \mathrm{d}\phi^2, \quad 0 \le r < \infty, \quad 0 \le \phi \le 2 \pi.$$
Reasons why this should be correct:

*

*For $r=0$ we have a point;

*As $r$ increases the the radius of the circle increases;

*$\sin^2(\phi)$  is always positive;

*For a constant radius we have the usual perimeter equation for the circle.

Reasons why this is wrong:

*

*The $2 \alpha$ angle is not in the metric;

*I am not sure, but shouldn't this metric be singular at $r = 0$ due to a coordinate singularity?

I would be happy if the answer was not given, but a suggestion of what may be right or wrong.
 A: Your first reason to disagree with the metric you guess is quite good. In fact, notice that for $\alpha = \frac{\pi}{2}$ the cone reduces to a plane, meaning the answer should be flat in this situation. The second reason is correct, but your metric already satisfies it: notice that for $r = 0$ the metric is degenerate, and hence that is already a coordinate singularity. In fact, apart from the case $\alpha = \frac{\pi}{2}$, I believe this singularity is essential, not merely a coordinate singularity, since in the vicinity of the point of the cone one can't really find a diffeomorphism with an open set of $\mathbb{R}^2$ (in other words, you can't really chart the tip of the cone with a differentiable map).
Notice that a cone can be thought of as a submanifold of $\mathbb{R}^3$. Hence, a particularly simple way to approach the problem is to find out what is the induced metric on the cone. In other words, you can try writing the cone as a surface in $\mathbb{R}^3$ and then using the formulae that describe the cone in the expression for the Euclidean metric of $\mathbb{R}^3$ to figure out what is the metric for the cone.
Let me provide an example with a different situation: a sphere of radius $R$. I'll format it as a spoiler in case providing an answer to a different, but similar, problem is too detailed for your goals.

 The metric for Euclidean space in spherical coordinates is
 $$\mathrm{d}l^2 = \mathrm{d}r^2 + r^2 \ \mathrm{d}\theta^2 + r^2 \sin^2\theta \ \mathrm{d}\phi^2.$$
 Now, the equation for a sphere of radius $R$ in spherical coordinates is simply $r = R$. Hence, substituting this in the metric we just wrote we get
 $$\mathrm{d}s^2 = R^2 \ \mathrm{d}\theta^2 + R^2 \sin^2\theta \ \mathrm{d}\phi^2,$$
 where $\mathrm{d}R = 0$ because $R$ is a constant.

At last, let me point out a different problem with your metric: for constant $r$, it does not reduce to the metric of a circle. In fact, I've never seen one "guessing" the metric of a manifold. We usually obtain it by computing the induced metric when these methods are available. In Relativity, we'll often either postulate a metric or solve the Einstein Field Equations.
