 Solution

Use $n_{\mu }=-\alpha {\nabla }_{\mu }t$. Differentiating, contracting with $n^{\rho }$, and using $n^{\rho }{\nabla }_{\rho }t=1/\alpha$ gives a gradient of $\log \alpha$ plus a component parallel to $n_{\mu }$. The acceleration is orthogonal to $n^{\mu }$, so spatial projection removes the parallel part: