**Calculate** $V^{\alpha}_{;\alpha}$

**Solution:**
$$
{\partial \vec V\over\partial x^\beta}={\partial\over\partial x^{\beta}}(V^\alpha\vec e_{\alpha})={\partial V^\alpha\over\partial x^\beta}\vec e_\alpha+{\partial\vec e_\alpha\over\partial x^\beta}V^\alpha\\=V^\alpha_{,\beta}\vec e_\alpha+V^\mu\Gamma^\alpha_{\mu\beta}\vec e_{\alpha}
$$

$$
V^\alpha_{;\beta}=V^\alpha_{;\beta}+\Gamma^{\alpha}_{\mu\beta}V^{\mu}
$$

$$
V^{\alpha}_{;\alpha}=V^\alpha_{,\alpha}+\Gamma^{\alpha}_{\mu\alpha}V^{\mu}
$$

$$
V^\alpha_{,\alpha}={\partial V^r\over\partial r}+{\partial V^{\theta}\over\partial\theta}
$$

$$
\Gamma^{\alpha}_{\mu\alpha}V^{\mu}=\Gamma^r_{\mu r}V^\mu+\Gamma^\theta_{\mu\theta}V^\mu\\=\Gamma^r_{rr}V^r+\Gamma^r_{\theta r}V^\theta+\Gamma^{\theta}_{r\theta}V^r+\Gamma^{\theta}_{\theta\theta}V^{\theta}
$$

$$
{\partial \vec e_\alpha\over\partial x^\beta}=\Gamma^{\mu}_{\alpha\beta}\vec e_{\mu}
$$

$$
\vec e_r=\cos\theta \vec e_x+\sin\theta\vec e_y\\\vec e_\theta=-{1\over r}\sin\theta\vec e_x+{1\over r}\cos\theta \vec e_y
$$

$$
{\partial \vec e_r\over\partial r}=0\\{\partial \vec e_r\over\partial \theta}={1\over r}\vec e_\theta\\{\partial \vec e_\theta\over\partial r}=-{1\over r}\vec e_\theta\\{\partial \vec e_\theta\over\partial \theta}=-{1\over r}\vec e_r
$$

$$
\begin{array}
{ll}\Gamma^ r_{rr}=0,&\Gamma^\theta_{rr}=0\\
\Gamma^{\theta}_{r\theta}={1\over r},&\Gamma^r_{r\theta}=0\\\Gamma^\theta_{\theta r}=-{1\over r},&\Gamma^r_{\theta r}=0\\
\Gamma^r_{\theta\theta}=-{1\over r},&\Gamma^\theta_{\theta\theta}=0
\end{array}
$$

$$
V^\alpha_{;\alpha}={\partial V^r\over\partial r}+{\partial V^{\theta}\over\partial\theta}+{1\over r}V^r
$$