Christoffel Symbol on a Spherical Metric

This Christoffel symbol is given by, computing:

$\begin{aligned} \Gamma^{\phi}_{\phi \theta} &= \frac12 g^{\phi \lambda} (\partial_{\phi} g_{\lambda \theta} + \partial_{\theta} g_{\phi \lambda} - \partial_{\lambda} g_{\phi \theta}) \\ &= \frac12 g^{\phi \phi} \partial_{\theta} g_{\phi \phi} \\ &= \frac12 (1+r^2) \frac{1}{\sin^2 \theta} \frac{1}{1+r^2} 2\sin \theta \cos \theta \\ &= \cot \theta \end{aligned}$

The invariant interval allows us to determine the covariant components of the metric ( $g_{\mu\nu}$ ):

$ds^2 = r^2d\theta^2 + \frac{1}{1 + r^2}\sin^2(\theta)d\phi^2 = \begin{pmatrix} d\theta & d\phi \end{pmatrix} \begin{pmatrix} r^2 & 0\\ 0 & \frac{1}{1 + r^2}\sin^2\theta \end{pmatrix} \begin{pmatrix} d\theta\\ d\phi \end{pmatrix},$

where the $2 \times 2$ matrix is $g_{\mu\nu}$ . It only has $\theta\text{-}$ and $\phi\text{-}$ related components as we are on the surface of the sphere, where $r$ is constant. The contravariant components of the metric (i.e. the inverse metric components, $g^{\mu\nu}$ ) can be determined, since

$g^{\mu\nu}g_{\mu\nu} = \delta^{\mu}_{\nu} \Rightarrow \begin{pmatrix} g^{\theta \theta} & g^{\theta \phi}\\ g^{\phi \theta} & g^{\phi \phi} \end{pmatrix} \begin{pmatrix} r^2 & 0\\ 0 & \frac{1}{1 + r^2}\sin^2\theta \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \text{\,\, (eq. 1)}.$

From this, one finds that

$g^{\theta\phi} = g^{\phi\theta} = 0, g^{\theta\theta} = \frac{1}{r^2} \text{\,\,\, and \,\,\,} g^{\phi\phi} = \frac{1 + r^2}{\sin^2\theta}.$

(It's actually a general consequence of eq. 1 that, for orthogonal metrics such as this, the leading diagonal terms of the contravariant metric and the leading diagonal terms of the covariant metric are the multiplicative inverses of each other, i.e. $g^{\mu\mu} = 1/g_{\mu\mu}$ , and all other terms are 0.)

Using the given formula for the Christoffel symbols,

$\Gamma^{\phi}_{\phi\theta} = \frac{1}{2}g^{\phi\sigma}\left(\partial_{\phi}g_{\sigma\theta} + \partial_{\theta}g_{\phi\sigma} - \partial_{\sigma}g_{\phi\theta} \right) \text{\,\, (eq. 2)},$

where $\sigma$ is a dummy index that sums over the coordinates ( $\theta$ and $\phi$ in this case).

However, as I stated above, it's a consequence of eq. 1 that $g_{\mu\nu} = g^{\mu\nu} = 0$ when $\mu \neq \nu$ $\text{\,\, (eq. 3)}$ .

Eq. 3 means that the last term in the bracket in eq. 2 will be zero no matter what $\sigma$ is, so we can discard that term. Doing this, expanding the bracket and summing over the index $\sigma$ explicitly, we get

$\Gamma^{\phi}_{\phi\theta} = \frac{1}{2}\left[g^{\phi\sigma}\partial_{\phi}g_{\sigma\theta} + g^{\phi\sigma}\partial_{\theta}g_{\phi\sigma}\right]$

$= \frac{1}{2}\left[\left(g^{\phi\theta}\partial_{\phi}g_{\theta\theta} + g^{\phi\phi}\partial_{\phi}g_{\phi\theta}\right) + \left(g^{\phi\theta}\partial_{\theta}g_{\phi\theta} + g^{\phi\phi}\partial_{\theta}g_{\phi\phi}\right)\right]$

$\Rightarrow \Gamma^{\phi}_{\phi\theta} = \frac{1}{2}g^{\phi\phi}\partial_{\theta}g_{\phi\phi} \text{\,\, (eq. 4)},$

where eq. 3 has been used to simplify the expression. Working out the partial derivative is as follows:

$\partial_{\theta}g_{\phi\phi} = \partial_{\theta}\left(\frac{\sin^2\theta}{1 + r^2}\right) = \frac{1}{1 + r^2}\partial_{\theta}\sin^2\theta = \frac{1}{1 + r^2}\cdot 2\sin\theta\cos\theta.$

Plugging our now-known values into eq. 4, we end up with

$\Gamma^{\phi}_{\phi\theta} = \frac{1}{2}\left(\frac{1 + r^2}{\sin^2\theta}\right)\left(\frac{1}{1 + r^2}\cdot 2\sin\theta\cos\theta\right) = \frac{1}{\sin^2\theta}\cdot \sin\theta\cos\theta = \frac{\cos\theta}{\sin\theta} = \frac{1}{\tan\theta} = \cot\theta,$

which is the answer :)

I sincerely hope this was helpful!