# Geodesic equation as centripetal acceleration?

[this was asked first here at physics.SE]

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 acceleration equation $a=v^2/r$: 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 use intrinsic coordinates, i.e. make no reference to an ambient space where our manifold would be embedded, 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 physicists, 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.

• Take a look at (free) my differential geometry text (linked in my profile). It deals only with surfaces in $\Bbb R^3$ and makes frequent reference to physical interpretations.) – Ted Shifrin May 23 '18 at 22:35
• @TedShifrin Thank you! – thedude May 24 '18 at 1:08

## 1 Answer

I think this is not the case, but there is a better candidate. Indeed, centripetal acceleration arises from confining a particle to a non-geodesic submanifold of the ambient space. The Christoffel symbols do represent a deviation from being geodesic, but this is to do with the component of acceleration parallel to the submanifold. Moreover, it is always possible to choose coordinates ("geodesic normal coordinates") so that the Christoffel symbols vanish at a point. I recommend you think of the Christoffel symbols as telling you how much the coordinate lines differ from geodesics of the submanifold itself, in a tangential direction: this is consistent with the ability to make them disappear at a point by taking appropriate coordinates, and them being definable only in terms of quantities on the submanifold. (Deviation from being a geodesic is called geodesic curvature, and is given by the length of the part of the derivative of the unit tangent vector that is parallel to the submanifold.)

Ah, so what about the normal component of acceleration, which should represent deviation from geodesics of the ambient space itself, and be related to the embedding of the submanifold? The story is simplest in the codimension-one case, which itself is essentially a simple generalisation of the 3D case, so let's just consider that. The appropriate object here is the second fundamental form, which is defined using the local deviation of a geodesic curve through a point in the surface from the surface's tangent plane at that point (Wikipedia actually has a decent picture of how this works at the top of the article). While it is not coordinate-independent, one can use it, along with the first fundamental form/metric on the surface, to write down quantities that are coordinate-independent, namely the principal curvatures $k_1,k_2$, and some more theory implies that there is an almost-unique canonical set of coordinates, based only on the extrinsic shape of the surface, so that the surface locally looks like the quadric $z=k_1 x^2+k_2 y^2$.

In particular, a geodesic on the surface has only normal curvature, so it has no acceleration tangential to the surface, while the normal acceleration depends on only the shape of the surface in the direction of the tangent, as specified by the second fundamental form.

In the extremely special case of a particle moving on the submanifold $x^2+y^2=r^2$ of two-dimensional space (which has constant normal curvature $1/r$) at speed $v(t)$, the tangential acceleration is given by $\dot{v}$, while the normal acceleration is $v^2/r$, consistent with the curvature of the circle being $1/r$.