I've calculated the correct answer to my problem, but don't understand one of the assumptions I made when doing so.

I used the geodesic deviation equation $$\frac{D^{2}\xi^{\mu}}{D\lambda^{2}}+R_{\phantom{\mu}\beta\alpha\gamma}^{\mu}\xi^{\alpha}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=0$$

to show that on the surface of a unit sphere two particles separated by initial distance $d$, starting from the equator and travelling north (ie on lines of constant $\phi$) will have a separation $s$ after time $t$ equal to $$s=\xi^{\phi}=d\sin\theta=d\cos\left(vt\right).$$ This is similar to Geodesic devation on a two sphere except that question was solved using simple spherical geometry.

The assumption I made was that the second absolute derivative wrt $t$ equals the second ordinary derivative, ie

$$\frac{D^{2}\xi^{\mu}}{dt{}^{2}}=\frac{d^{2}\xi^{\mu}}{dt{}^{2}}.$$ My question is, why am I allowed to make this assumption?

I've been told on another physics forum that the answer is because the problem is framed in terms of Riemann normal coordinate (because the distance the cars travel along their separate geodesics is a linear function of time $t$). I can only assume that in some way this makes the connection coefficients disappear in the absolute derivative equation$$\frac{DV^{\alpha}}{d\lambda}=\frac{dV^{\alpha}}{d\lambda}+V^{\gamma}\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\lambda},$$ but I can't see why this is. As I noted in a comment below, I understand it is possible to choose coordinates at a point where the connection coefficients vanish, but I used the ordinary polar coordinates $\phi$ and $\theta$ to calculate the correct answer. To use two different sets of coordinates like this seems like a case of "having your cake and eating it".

The calculation, by the way, is here (my answer to my question): Geodesic deviation on a unit sphere

  • 2
    $\begingroup$ Are you aware that you can, locally, always choose coordinates that make the derivatives of the metric (and hence the Christoffels) vanish? $\endgroup$
    – ACuriousMind
    Oct 17, 2014 at 15:19
  • $\begingroup$ I think so. Isn't that what Schutz calls the "local flatness" theorem? What I don't understand is that I correctly calculated the answer using the ordinary spherical $\theta,\phi$ coordinates. I can't see how the connection coefficients vanish using these. $\endgroup$
    – Peter4075
    Oct 17, 2014 at 15:56
  • $\begingroup$ That is true only for a point $p$. $\endgroup$
    – yess
    Oct 17, 2014 at 17:00

1 Answer 1


The first reason is that your "distance" between geodesics is measured by a parallely propagated direction $\partial/\partial \phi$. If you take a look at the sphere, the difference $\Delta \phi$ does not correspond to the distance between the points on the geodesics. The distance between them would be measured by arc-lengths of great circles. But you are using $\theta=const$ circles which are not the great circles unless $\theta=\pi/2$.

The second reason is you are working on a space of constant curvature. See below.

Say you take a vector $\zeta^\mu$ and propagate it along your geodesic to get $\zeta^\mu(\lambda)$, i.e. you solve $$\frac{D \zeta^\mu}{d \lambda} = 0$$ Thanks to that you now have by a straightforward application of the Leibniz rule $$\frac{D^2(\xi^\mu \zeta_\mu)}{d^2 \lambda} = \frac{D^2\xi^\mu}{d^2 \lambda} \zeta_\mu$$ But $\xi^\mu \zeta_\mu$ is a scalar which is propagated trivially, so you actually also get $D/d\lambda \to d/d\lambda$. Now you can project your equation of geodesic equation into $\zeta^\mu$ to get $$\frac{d^2(\xi^\mu \zeta_\mu)}{d^2 \lambda} + R^\mu_{\;\nu \kappa \lambda} \xi^\kappa u^\nu u^\lambda \zeta_\mu = 0 \; \; \; \;(*)$$

Now to use the fact you are on a space of constant curvature. In such a space, you can express the curvature tensor as $$R_{\mu \nu \kappa \lambda} = K(g_{\mu \kappa} g_{\nu \lambda} - g_{\mu \lambda} g_{\nu \kappa})$$ When you plug this in into the geodesic deviation equation, you get $$\frac{D^2 \xi^\alpha}{d\lambda^2} + u^2 K \xi^\alpha = 0$$ Where $u = dx/d\lambda$ and we choose an orthogonal $\xi$ to it (as is also your case). When you make the projection into the parallely propagated $\zeta_\alpha$, you get $$\frac{d^2 (\xi^\alpha \zeta_\alpha)}{d^2\lambda} + u^2 K \xi^\alpha \zeta_\alpha = 0$$ I.e., if you are investigating only $\Delta \phi = \xi^\alpha \zeta_\alpha$ where $\zeta = \partial/\partial \phi$, you can use even this strictly linear equation.

Note that without using the constant curvature of the space you would end up with equation $(*)$ which doesn't give you much of a clue why you should be able to find $\xi^\alpha \zeta_\alpha$ in the sum with the curvature tensor. So your space is special and your measure of deviation is special - both are necessary ingredients for the assumption.

  • $\begingroup$ So, can I assume that as only one of the coordinate curves (hope that's the correct term) is a geodesic (ie the $\phi=constant$ one), then the answer has nothing to do with Riemann normal coordinates? Also, I didn't realise that if $\frac{D\zeta^{\mu}}{d\lambda}=0$ then $\frac{D\zeta_{\mu}}{d\lambda} =0$. I've seen a similar problem to this online but with one of the particles moving along the equator ($\theta =\pi/2$), meaning the connection coefficients “naturally” disappear and the absolute 2nd derivative equals the ordinary 2nd derivative, which makes things a lot simpler. $\endgroup$
    – Peter4075
    Oct 20, 2014 at 14:21
  • $\begingroup$ Why is the right-hand side of your equation 0? Everybody seems to assume this, but I can't find anyone who can explain what the foundation of this assumption is. $\endgroup$
    – user32023
    Oct 26, 2018 at 0:59
  • $\begingroup$ @DonaldAirey There are six equations in my post. Which one do you mean? $\endgroup$
    – Void
    Oct 26, 2018 at 5:00
  • $\begingroup$ @Void - The fifth one. The one that leads off with "If you plug this into the Geodesic Deviation Equation"... $\endgroup$
    – user32023
    Oct 26, 2018 at 12:32

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