How do I reconcile these two definitions of acceleration?

Question

How do I reconcile these two definitions of acceleration?

$$a=\frac{d\bar{v}}{dt}=(\frac{dv^k}{dt}+v^i v^j \Gamma^k_{ij})\bar{e}_k \tag{1}$$

and

$$a^k=v^{\small\beta} \nabla_{\small\beta} v^k.\tag{2}$$

I tried doing it for gravity in polar coordinates.

$v^r=gt$

$r=\frac{1}{2} gt^2$

This clearly works for the first formula because

$a=\frac{d\bar{v}}{dt}=(\frac{dv^r}{dt}+v^r v^r \Gamma^r_{rr})\bar{e}_r $

and

$\Gamma^r_{rr}=0$

Now, the second formula.

$a^r = v^r \nabla_r v^r $ because $v^\theta =0$

$a^r = v^r (\frac{\partial v^r}{\partial r}) $

?

$= \frac{\partial r}{\partial t} (\frac{\partial v^r}{\partial r})=\frac{\partial v}{\partial t} $

Is this correct?

If so, how should I do it in general?

Answer 1

In general,

\begin{align} v^\beta \nabla_\beta v^k = v^\beta (\partial_\beta v^k +\Gamma^k_{\beta \alpha}v^\alpha) = v^\beta \partial_\beta v^k + v^\beta v^\alpha \Gamma^k_{\beta \alpha}. \end{align} The only thing left to realise is that \begin{align} v^\beta \partial_\beta v^k = \frac{\partial x^\beta}{\partial t}\frac{\partial v^k}{\partial x^\beta} = \frac{\partial v^k}{\partial t}, \end{align} (if you're not happy with the last step, perhaps convince yourself that you'd be happy to go from the right hand side to the left hand side) and this is exactly what you have in the first expression (just contracted with a basis).