# 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... 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
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. Apr 8, 2014 at 16:34
• @Jim No, because if you keep r constant you do not find a geodesic of that spacetime. 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! 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.) Apr 8, 2014 at 18:01