I am trying to prove the following.
Suppose we have a scalar function $\phi$ (sufficiently differentiable), the metric tensor $g_{ij} = \dfrac{\partial y^\alpha}{\partial x^i}\dfrac{\partial y^\alpha}{\partial x^j}$, and the operation of covariant differentiation $D_j(V_i)$ of a covariant vector $V_i$ where $ D_j(V_i) = \dfrac{\partial V_i}{\partial x^j} – \Gamma^\alpha_{ij}V_\alpha$. Then we have the scalar
$$g^{ij}D_{ij}(\phi) = \dfrac{1}{\sqrt{|g|}}\dfrac{\partial}{\partial x^i}\left(\sqrt{|g|}g^{ij}\dfrac{\partial\phi}{\partial x^j} \right)$$
where $D_{ij}(\phi) = D_j(D_i(\phi))$ and $|g|$ indicates the determinant of the metric tensor. I've obtained $g^{ij}D_{ij}(\phi) = g^{ij}\left[\dfrac{\partial}{\partial x^j} \left(\dfrac{\phi}{\partial x^i}\right) – \Gamma^\alpha_{ij}\dfrac{\partial\phi}{\partial x^\alpha} \right]$ (covariant derivative of a scalar being the same as the ordinary derivative) from the definition of covariant differentiation, but I can't seem to obtain the conclusion.
I have tried employing the fact that $\dfrac{\partial}{\partial x^i}\log{\sqrt{|g|}} = \Gamma^\alpha_{\alpha i}$, but nothing so far.
Best Answer
Using your identity
$$\tag{1} \Gamma_{r\,\,i}^{\,\,i}=\frac{\partial }{\partial x^i}\ln(\sqrt g)=\frac{1}{\sqrt{g}}\frac{\partial }{\partial x^i}(\sqrt{g})$$
We can establish an important intermediate result. Lets study the contraction of the covariant derivative for a contravariant tensor $T^i$ (ie the divergence of $T^i$)
$$\tag{2} D_iT^i=\frac{\partial T^i}{\partial x^i}+\Gamma_{r\,\,i}^{\,\,i}T^r$$
Now using $(1)$ and the normal product rule backwards we have (rename dummy r to i)
$$\tag{3}D_iT^i=\frac{\partial T^i}{\partial x^i}+\frac{1}{\sqrt g}\frac{\partial }{\partial x^r}(\sqrt g)T^r=\frac{1}{\sqrt g}\frac{\partial}{\partial x^i}\left(\sqrt g T^i\right)$$
This important intermediate result is called the Voss-Weyl formula. Finally, with $T^i=g^{ij}\frac{\partial \phi}{\partial x^j}$ equation $(3)$ is
$$D_i(g^{ij}D_j\phi)=\frac{1}{\sqrt g}\frac{\partial }{x^i}\left (\sqrt g g^{ij} \frac{\partial \phi}{\partial x^j}\right)$$