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.

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    $\begingroup$ I do not understand, what is time on the 2-sphere? That is a Riemannian manifold, not a Lorentzian one... $\endgroup$ Apr 8, 2014 at 16:24
  • $\begingroup$ @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? $\endgroup$
    – Jim
    Apr 8, 2014 at 16:33
  • $\begingroup$ @V.Moretti Well, let's say they're moving at northward toward the pole along a line of constant longitude at unit velocity. $\endgroup$
    – user44056
    Apr 8, 2014 at 16:34
  • $\begingroup$ @Jim No, because if you keep r constant you do not find a geodesic of that spacetime. $\endgroup$ Apr 8, 2014 at 17:25
  • $\begingroup$ @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! $\endgroup$ Apr 8, 2014 at 17:29

1 Answer 1


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:

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

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

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