Geodesic equation as centripetal acceleration? I was looking at the geodesic equation 
$$ \frac{d^2 x^\lambda}{dt^2}+\Gamma^{\lambda}_{\mu\nu}\frac{d x^\mu}{dt}\frac{d x^\nu}{dt}=0$$
and it occurred to me that it looks like the centripetal force equation $a=v^2/r$. I mean, the first term is a second time derivative and the other term is a product of velocities.
I feel that this interpretation should be valid somehow, in the sense that when a particle is moving freely on a curved surface, it must feel a centripetal acceleration in order to keep up with the curvature.
On the other hand, when we do differential geometry we are supposed to use intrinsic coordinates, i.e. make no reference to an ambient space where our manifold would be embedded. In that case, it makes no sense to talk about a "normal direction" to the surface, and this is the direction in which a centripetal acceleration would have to point.
So, the question is: can I interpret the geodesic equation as a centripetal force equation? 
Also, books on differential geometry, even for physicist, tend to be very abstract with all the connections and covariant derivatives. I would appreciate any hints about books that have simple concrete examples of this stuff.
 A: Consider Minkowski space from the perspective of an observer rotating with constant angular frequency. First, pick Minkowski space in cylindrical coordinates : 
$$ds^2 = -dt^2 + dz^2 + d\rho^2 + \rho^2 d\phi^2$$
The orthonormal frame of a rotating observer at distance $R$ is given by the Langevin observer, which is a "boosted" stationary observer along the angular coordinates, with a speed of $\omega R$ : 
\begin{eqnarray}
e_0 &=& \frac{1}{\sqrt{1 - \omega^2 R^2}} \partial_t + \frac{\omega R}{\sqrt{1 - \omega^2 R^2}} \frac{1}{R} \partial_\phi\\
e_1 &=& \partial_z\\
e_2 &=& \partial_\rho\\
e_3 &=& \frac{1}{\sqrt{1 - \omega^2 R^2}} \frac{1}{R} \partial_\phi + \frac{\omega R}{\sqrt{1 - \omega^2 R^2}} \partial_t
\end{eqnarray}
Then, using the coordinat transformation $\phi' = \phi - \omega t$, we get
$$ds^{2}=-\left(1-\omega^{2}\rho^{2}\right)dt^{2}+2\omega \rho^{2}dtd\phi' +dz^{2}+d\rho^{2}+\rho^{2}d\phi'^{2}$$
Those are the Born coordinates, in which the Langevin observers are straight lines. 
Now, consider the geodesic equation. First, we compute the Christoffel symbols
$${\Gamma^\sigma}_{\mu\nu} = \frac{1}{2} g^{\sigma\tau}(\partial_{\mu}g_{\nu\tau} + \partial_{\nu}g_{\mu\tau} - \partial_{\tau}g_{\mu\nu})$$
I'll only consider the $r$ component of the symbols, since this is what we need to demonstrate the centrifugal force (if you wish to show the effect of the Coriolis force or Euler force, please expand the problem fully). This will be : 
\begin{eqnarray}
{\Gamma^\rho}_{\mu\nu} &=& \frac{1}{2} (\partial_{\mu}g_{\nu\rho} + \partial_{\nu}g_{\mu\rho} - \partial_{\rho}g_{\mu\nu})
\end{eqnarray}
For $\sigma = \rho$, $\mu, \nu \neq \rho$ (more than one $\rho$ is zero here), this leaves 
$${\Gamma^\rho}_{\mu\nu} = -\frac{1}{2} \partial_{\rho}g_{\mu\nu}$$
Meaning that 
\begin{eqnarray}
{\Gamma^\rho}_{tt} &=& - \omega^{2} \rho\\ 
{\Gamma^\rho}_{\phi\phi} &=& - \rho\\
{\Gamma^\rho}_{t\phi} &=& {\Gamma^\rho}_{\phi t} = - \omega \rho \\
\end{eqnarray}
The geodesic equation then becomes : 
\begin{eqnarray}
\ddot{\rho} - \omega^{2} \rho \dot{t}\dot{t} - \omega \rho \dot{t}\dot{\phi} - \rho \dot{\phi} \dot{\phi} = 0
\end{eqnarray}
The last term isn't terribly interesting (it's a term that only appears due to the choice of coordinates). Let's ignore it for a bit. If we multiply the rest by $m$ : 
$$m\ddot{\rho} = m  \omega^{2} \rho \dot{t}\dot{t} + m \omega \rho \dot{t}\dot{\phi}$$
If we assume classical mechanics, we have that $\dot t = \gamma \approx 1$
$$m\ddot{\rho} =  m \omega^{2} \rho + m \omega \rho \dot{\phi}$$
The first term is indeed what corresponds to the centrifugal term, once we work out the coordinates. I think the second term is part of the Coriolis force.
The same kind of reasoning applies to a variety of spacetime, by considering various rotating frame fields.
A: 
can I interpret the geodesic equation as a centripetal force equation?

From a formal point of view both the geodesic deviation and the centripedal force equation are about acceleration - as you said. The physical meaning is not comparable however, so you can't interpret one as the other. The main differences concern the curvature of space-time and the force an object feels.  
The Geodesic deviation describes the relative acceleration between objects in free fall in curved spacetime. Whereas the centripedal force equation describes a force on an object in flat space-time.  

I feel that this interpretation should be valid somehow, in the sense that when a particle is moving freely on a curved surface, it must feel a centripetal acceleration in order to keep up with the curvature.

If in flat Minkowski space-time a particle moves freely it's path is a straight line, a geodesic. It's path is a curve and not a geodesic if it accelerates and thus feels a force.  
