# 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.

• The centripetal force is a form of a force you get from the geodesic action, yes. It will be the force you get from considering for instance an observer using Born coordinates in general relativity. – Slereah May 16 '18 at 13:18
• There's actually a very natural interpretation of the Christoffel symbol term as a centrifugal force in a rotating coordinate system. To see this, write out the Minkowski metric $ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$ in terms of a set of rotating coordinates, and then find the Christoffel symbols in terms of these coordinates. – Michael Seifert May 16 '18 at 14:22
• @MichaelSeifert Do you have a reference for that calculation? I am not able to carry it out myself. – thedude May 16 '18 at 19:42
• This page looks like a fairly thorough working of the problem at a first glance, though I haven't vetted it carefully. – Michael Seifert May 16 '18 at 19:49
• @MichaelSeifert The text you linked is interesting, and seems to support my intuition that indeed the geodesic equation can be seen as a centripetal force equation. But using a rotating reference frame makes it a bit artificial, I would like to understand the question in the context of free motion on a manifold with static coordinates. – thedude May 16 '18 at 23:37

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.

• If you introduce a rotating coordinate system, then of course there will be a centrifugal force. I would like to understand the question in the context of free motion on a manifold with static coordinates. – thedude May 17 '18 at 13:04
• I understood your designation "free motion" in the sense of "free fall" which means a force free motion described by a geodesic. Did you mean that?-Tidal forces require geodesic deviation, I can't see any link to centripetal forces or I misunderstood you completely. – timm May 18 '18 at 9:35

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.