# Geodesic devation on a two sphere

So the geodesic deviation equation gives the relative acceleration between two geodesics in motion. But given a pair of geodesic (let's say on the two sphere) that start at the equator, separated by some distance. Is there a way to compute their separation as a function of time without using the geodesic equation? Let's say they're moving at northward toward the pole along a line of constant longitude at unit velocity.

• I do not understand, what is time on the 2-sphere? That is a Riemannian manifold, not a Lorentzian one... Commented Apr 8, 2014 at 16:24
• @V.Moretti $ds^2=-dt^2+dr^2+r^2d\Omega_2^2$ where $r$ is constant, thus $dr^2=0$. Does that answer your question?
– Jim
Commented Apr 8, 2014 at 16:33
• @V.Moretti Well, let's say they're moving at northward toward the pole along a line of constant longitude at unit velocity. Commented Apr 8, 2014 at 16:34
• @Jim No, because if you keep r constant you do not find a geodesic of that spacetime. Commented Apr 8, 2014 at 17:25
• @user44056 The geodesic deviation is usually understood for a congruence of (for instance timelike) geodesics in a given spacetime and the parameter along the geodesics is the proper time. The sphere is not a Lorenzian manifold, so there is nothing like proper time. In a Riemannian manifold the natural affine parameter along geodesics is the length parameter on the curves. In this case however there is nothing like an acceleration! Commented Apr 8, 2014 at 17:29

Re your edited question, this is just simple spherical geometry. If the initial separation is $d$ then the separation at time $t$ is $d \cos(vt/r)$, where $r$ is the radius of the sphere, $v$ is the vehicle speed and $t$ is time.

The diagram shows a cross section through the poles. The vehicle is driving north at a velocity v, so the distance it drives in a time $t$ is just $s = vt$, so the angle $\theta$ is:

$$\theta = \frac{vt}{2\pi r} 2\pi = \frac{vt}{r}$$

Suppose the vehicles start out at a separation $d$. The angular separation along the equator $\Delta\phi$ is:

$$\Delta\phi = \frac{d}{2\pi r} 2\pi = \frac{d}{r}$$

As the two vehicles drive north the angular separation $\phi$ doesn't change, so we just need to calculate the circumference of the line of latitude at the angle $\theta$, $C_\theta$, and the separation will be $C_\theta\tfrac{\Delta\phi}{2\pi}$.

$$C_\theta = 2\pi r \cos\theta$$

So the separation $s$ is:

\begin{align} s &= 2\pi r \cos\theta \frac{\Delta\phi}{2\pi} \\ &= 2\pi r \cos\left(\frac{vt}{r}\right) \frac{d/r}{2\pi} \\ &= d \cos\left(\frac{vt}{r} \right) \end{align}

• A downvote? Did I get something wrong? (I answered in a hurry so it's entirely possible.) Commented Apr 8, 2014 at 18:01