# Metric tensor on unit sphere

1. Jul 20, 2006

### quasar987

Wiki says it's

1 0
0 $sin^2\theta$

My book says it's

$cos²\theta$ 0
0 1

I calculate

1 0
0 $cos^2\theta$

?! which is it?

2. Jul 20, 2006

### yenchin

Which angle is your theta? Mathematician's and physicist's notations of spherical coordinates seem to be opposite of each other.

3. Jul 20, 2006

### HallsofIvy

Staff Emeritus
Exactly what coordinate system are you using on the surface of the sphere?

A standard choice (I would have thought) would be spherical coordinates, with $\theta$ as "longitudinal" angle, $\phi$ as "co-latitude" angle (measured from the positive z-axis), with $\rho$ set equal to 1. Using those I get
$$\left( \begin{array}{cc}sin^2(\phi) & 0 & 0 & 1\end{array} \right)$$
similar to what Wikpedia gives but with $\theta$ as latitude and reversing the order of the components.

I can't see how you could possibly get $cos(\theta)$ without a square. If, as yenchin said, you use "latitude" instead of "co-latitude", you would get $cos^2(\theta)$. Is it possible that your text has
$$\left(\begin{array}{cc}cos^2(\theta) & 0 \\0 & 1 \end{array}\right)$$
instead of just the "cos" you give? In that case, the only difference between that and what you give is choice of order of components. The only difference between Wikpedia's result and yours is choice of "co-latitude" rather than "latitude

4. Jul 20, 2006

### quasar987

I see now how g_11 and g_22 would be inverted depending on how you order the coordinates, i.e.

$$g_{11}(\theta,\phi) = g_{22}(\phi, \theta)$$

I used the same meaning of polar angles as the author of my book, i.e. theta as the "latitude" angle (measured from the projection of $\vec{r}$ on the Oxy plane) and and phi as the "longitudinal" angle (measured from the positive x-axis to the projection of $\vec{r}$ on the Oxy plane)

P.S. in my OP, the second matrix has cos² as its first term and not cos. Very sorry.