Form of the geodesic equation I have been studying the basics of general relativity with Hartle's Gravity. He presents the geodesic equation as
$$
\frac{d^2x^{{\mu}}}{ds^2}=-{\Gamma}^{{\mu}}_{{\alpha}{\beta}}\frac{dx^{{\alpha}}}{ds}\frac{dx^{{\beta}}}{ds}
$$
However, in reading Padmanabhan's Gravitation, he says that the equation
$$
0=k^{{\alpha}}{\nabla}_{{\alpha}}k_{{\beta}}
$$
is the geodesic equation for a wave vector $k$ defined as $k_{{\alpha}}={\nabla}_{{\alpha}}{\psi}$, where ${\psi}$ is just a scalar function.
How do these two definitions of the geodesic equation represent the same thing? They do not look at all alike. In fact, if I try to work it out, I get
$$
0=k^{{\alpha}}\,\bigg(\frac{{\partial}k_{{\beta}}}{{\partial}x^{{\alpha}}}-{\Gamma}^{{\delta}}_{{\beta}{\gamma}}k_{{\delta}}\bigg)
$$
$$
0=k^{{\alpha}}\,\frac{{\partial}k_{{\beta}}}{{\partial}x^{{\alpha}}}-{\Gamma}^{{\delta}}_{{\beta}{\gamma}}k_{{\delta}}k^{{\alpha}}
$$
For one thing, the $x^{{\alpha}}$ is not supposed to be on the bottom of the derivative! If $k_{{\alpha}}$ is supposed to satisfy the geodesic equation, I expected this to look like the first equation I wrote.
 A: In short, the first equation you wrote is in component form where curves are parametrized as $x^{\mu}(s)$, while the second equation is more general. Now let's flesh out the details. 
What does it mean for a curve to be a geodesic? Intuitively it has to be straightest curve possible in curved spacetime. The way you do that is to propagate the tangent vector $T^{\alpha}$ of a curve C along itself! That yields the coordinate independent geodesic equation: $$T^{\alpha}\nabla_{\alpha}T^{\beta} = 0$$
Then you can write this tensor equation in component form, by choosing a coordinate system and its associated Christoffel symbol. The covariant derivative is expressed as:
$$\nabla_{\alpha}T^{\beta} = \partial_{\alpha}T^{\beta}+\Gamma^{\beta}_{\alpha\gamma}T^{\gamma}$$ 
Now if you plug this equation into the coordinate independent geodesic equation, you get:
$$ T^{\alpha}\partial_{\alpha}T^{\beta}+T^{\alpha}\Gamma^{\beta}_{\alpha\gamma}T^{\gamma} = 0$$
Now if you parametrize the curve C as $x^{\mu}(s)$, then the tangent vector field becomes:$T^{\mu} = \frac{dx^{\mu}}{ds}$, and $T^{\alpha}\partial_{\alpha}$ represents the derivative operator $\frac{d}{ds}$.
Therefore, the previous equation turns into:
$$\frac{d}{ds}\frac{dx^{\beta}}{ds}+\Gamma^{\beta}_{\alpha\gamma}\frac{dx^{\alpha}}{ds}\frac{dx^{\gamma}}{ds} = 0$$
$$\frac{d^2x^{\beta}}{ds^2}+\Gamma^{\beta}_{\alpha\gamma}\frac{dx^{\alpha}}{ds}\frac{dx^{\gamma}}{ds} = 0$$
Which is precisely the equation that you started with. 
