Let $x^{a}(\lambda)$ be a path $P$ with tangent vector $T^{a}=\frac{d}{d \lambda} x^{a}(\lambda)$. For vectors $X^{a}(x(\lambda))$ and $Y^{a}(x(\lambda))$ defined on $P$ let

$$\nabla_{T} X^{a}=\frac{d}{d \lambda} X^{a}+\Gamma_{b c}^{a}(x(\lambda)) X^{b} T^{c}$$

where $\Gamma_{b c}^{a}(x)$ is the metric connection for a metric $g_{a b}(x) . \nabla_{T} Y^{a}$ is defined similarly. Suppose $P$ is geodesic and $\lambda$ is an affine parameter. Explain why $\nabla_{T} T^{a}=0$. Show that if $\nabla_{T} X^{a}=\nabla_{T} Y^{a}=0$ then $g_{a b}(x(\lambda)) X^{a}(x(\lambda)) Y^{b}(x(\lambda))$ is constant along $P$.

If $x^{a}(\lambda, \mu)$ is a family of geodesics which depend on $\mu$, let $S^{a}=\frac{\partial}{\partial \mu} x^{a}$ and define

$$\nabla_{S} X^{a}=\frac{\partial}{\partial \mu} X^{a}+\Gamma_{b c}^{a}(x(\lambda)) X^{b} S^{c}$$

Show that $\nabla_{T} S^{a}=\nabla_{S} T^{a}$ and obtain

$$\nabla_{T}{ }^{2} S^{a} \equiv \nabla_{T}\left(\nabla_{T} S^{a}\right)=R_{b c d}^{a} T^{b} T^{c} S^{d}$$

What is the physical relevance of this equation in general relativity? Describe briefly how this is relevant for an observer moving under gravity.

[You may assume $\left[\nabla_{T}, \nabla_{S}\right] X^{a}=R_{b c d}^{a} X^{b} T^{c} S^{d}$.]