A spacetime contains a one-parameter family of geodesics $x^{a}=x^{a}(\lambda, \mu)$, where $\lambda$ is a parameter along each geodesic, and $\mu$ labels the geodesics. The tangent to the geodesics is $T^{a}=\partial x^{a} / \partial \lambda$, and $N^{a}=\partial x^{a} / \partial \mu$ is a connecting vector. Prove that

$$\nabla_{\mu} T^{a}=\nabla_{\lambda} N^{a}$$

and hence derive the equation of geodesic deviation:

$$\nabla_{\lambda}^{2} N^{a}+R_{b c d}^{a} T^{b} N^{c} T^{d}=0$$

[You may assume $R_{b c d}^{a}=-R_{b d c}^{a}$ and the Ricci identity in the form

$$\left.\left(\nabla_{\lambda} \nabla_{\mu}-\nabla_{\mu} \nabla_{\lambda}\right) T^{a}=R_{b c d}^{a} T^{b} T^{c} N^{d}\right]$$

Consider the two-dimensional space consisting of the sphere of radius $r$ with line element

$$d s^{2}=r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right) .$$

Show that one may choose $T^{a}=(1,0), N^{a}=(0,1)$, and that

$$\nabla_{\theta} N^{a}=\cot \theta N^{a} .$$

Hence show that $R=2 / r^{2}$, using the geodesic deviation equation and the identity in any two-dimensional space

$$R_{b c d}^{a}=\frac{1}{2} R\left(\delta_{c}^{a} g_{b d}-\delta_{d}^{a} g_{b c}\right)$$

where $R$ is the Ricci scalar.

Verify your answer by direct computation of $R$.

[You may assume that the only non-zero connection components are

$$\Gamma_{\phi \theta}^{\phi}=\Gamma_{\theta \phi}^{\phi}=\cot \theta$$

and

$$\Gamma_{\phi \phi}^{\theta}=-\sin \theta \cos \theta .$$

You may also use the definition

$$\left.R_{b c d}^{a}=\Gamma_{b d, c}^{a}-\Gamma_{b c, d}^{a}+\Gamma_{e c}^{a} \Gamma_{b d}^{e}-\Gamma_{e d}^{a} \Gamma_{b c}^{e}\right]$$