My previous question Textbook disagreement on geodesic deviation on a 2-sphere got shot down as “off topic”, so I'm having a second stab at it.

Misner et al's Gravitation (p34) gives the geodesic deviation equation as$$\frac{D^{2}\xi^{\alpha}}{D\tau^{2}}+R_{\phantom{\mu}\beta\gamma\delta}^{\alpha}\frac{dx^{\beta}}{d\tau}\xi^{\gamma}\frac{dx^{\delta}}{d\tau}=0,$$ with the right-hand side $\xi$ index $\gamma$ equal to the second lower index on the Riemann tensor. Lambourne's Relativity, Gravitation and Cosmology (p185), on the other hand, gives $$\frac{D^{2}\xi^{\mu}}{D\lambda^{2}}+R_{\phantom{\mu}\alpha\beta\gamma}^{\mu}\xi^{\alpha}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=0,$$ with the right-hand side $\xi$ index $\alpha$ equal to the first lower index on the Riemann tensor.

My question is, which of these two equations is correct?

I tried to answer this question myself by using the two equations to calculate the geodesic deviation on the surface of a unit 2-sphere. With Misner's equation (substituting $\lambda$ for $\tau$) I got $$\frac{D^{2}\xi^{\theta}}{D\lambda^{2}}=\left(\sin^{2}\theta\right)\left(u^{\phi}u^{\theta}\right)\xi^{\phi}-\left(\sin^{2}\theta\right)\left(u^{\phi}u^{\phi}\right)\xi^{\theta}$$ and $$\frac{D^{2}\xi^{\phi}}{D\lambda^{2}}=\xi^{\theta}\left(u^{\theta}u^{\phi}\right)-\xi^{\phi}\left(u^{\theta}u^{\theta}\right).$$

You can see my calculation on my previous question Textbook disagreement on geodesic deviation on a 2-sphere With Lambourne's equation I got

$$\frac{D^{2}\xi^{\theta}}{D\lambda^{2}}=0$$ and $$\frac{D^{2}\xi^{\phi}}{D\lambda^{2}}=0.$$ This didn't seem right to me so I concluded that Lambourne's equation is incorrect.

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    $\begingroup$ The proper method of dealing with a closed question is to edit the closed question, not write it out again (albeit slightly modified). If you merge this question with your other, closed one, I will be willing to nominate for reopening. $\endgroup$
    – Kyle Kanos
    Commented Aug 28, 2014 at 14:38
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    $\begingroup$ There are two different conventions for the Riemann tensor, and the tensors can differ by a minus sign between the two conventions. Without telling us which version the two books are using, this question, as written, is unanswerable, since the two expressions you give also differ by a minus sign. $\endgroup$ Commented Aug 28, 2014 at 14:51
  • $\begingroup$ @JerrySchirmer: Lambourne uses +--- and MTW use -+++ $\endgroup$ Commented Aug 28, 2014 at 15:17
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    $\begingroup$ @JohnRennie: yes, but there is an additional convention on the Riemann tensor, depending on whether $\nabla_{[a}\nabla_{b]}\omega_{c} = R_{abc}{}^{d}\omega_{d}$ or whether $\nabla_{[a}\nabla_{b]}\omega_{c} = R^{d}{}_{abc}{}\omega_{d}$. But looking at these two expressions again, the second one is wrong, as it is always identically zero. $\endgroup$ Commented Aug 28, 2014 at 15:19
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    $\begingroup$ @JerrySchirmer: oops, yes, sorry I didn't read your comment carefully enough. I have both books but don't know them well enough to find the relevant sections without a prolonged search. I guess I'll admit defeat :-) $\endgroup$ Commented Aug 28, 2014 at 15:23

2 Answers 2


Despite my comment, on second look your second equation, attributed to Lambourne, is always identically zero. This is because you multiply the symmetric tensor


against $R^{\mu}{}_{\nu\beta\gamma}$, and the riemann tensor is antisymmetric on those last two indices, and tracing a symmetric tensor against an antisymmetric tensor always gives zero.

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    $\begingroup$ @Peter4075: all that's going on with that is the fact that the Riemann tensor is defined by $\nabla_{a}\nabla_{b}\omega_{c} - \nabla_{b}\nabla_{a}\omega_{c}$. It's an arbitrary choice whether you put the fourth index that you contract on $\omega$ first or last, so this could be either equal to $R_{abc}{}^{d}\omega_{d}$ or $R^{d}{}_{abc}\omega_{d}$. Using the properties of the Riemann Tensor, you can juggle indices around, and show that these two definitions are equal to each other, except that they differ by a minus sign. $\endgroup$ Commented Aug 28, 2014 at 17:53
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    $\begingroup$ No, they can't! Both expressions are exactly the same 4-vector --- they are the spacetime co-ordinates of a particular point particle. $\endgroup$
    – gj255
    Commented Aug 29, 2014 at 9:13
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    $\begingroup$ @Peter4075 it's just because $\frac{dx^\beta}{d\lambda}\frac{dx^\gamma}{d\lambda}=\frac{dx^\gamma}{d\lambda} \frac{dx^\beta}{d\lambda}$ :) $\endgroup$
    – Danu
    Commented Aug 29, 2014 at 13:46
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    $\begingroup$ @Peter4075: they're two copies of the SAME set of four functions. $\endgroup$ Commented Aug 29, 2014 at 15:18
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    $\begingroup$ So if, for eg, we had $x,y,z$ coordinates, the two terms would multiply out to give a symmetric matrix $$\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=\left(\begin{array}{ccc} \frac{dx}{d\lambda}\frac{dx}{d\lambda} & \frac{dx}{d\lambda}\frac{dy}{d\lambda} & \frac{dx}{d\lambda}\frac{dz}{d\lambda}\\ \frac{dy}{d\lambda}\frac{dx}{d\lambda} & \frac{dy}{d\lambda}\frac{dy}{d\lambda} & \frac{dy}{d\lambda}\frac{dz}{d\lambda}\\ \frac{dz}{d\lambda}\frac{dx}{d\lambda} & \frac{dz}{d\lambda}\frac{dy}{d\lambda} & \frac{dz}{d\lambda}\frac{dz}{d\lambda} \end{array}\right)$$ aka a symmetric tensor. $\endgroup$
    – Peter4075
    Commented Aug 29, 2014 at 18:28

EDIT2: I've just seen Jerry Schirmer's answer, which confirms to me that Lambourne's equation is incorrect. I would mark his as the accepted answer. I'll leave the below for reference, though it is entirely false! Indeed, as pointed out by Peter in the comments, Lambourne uses the same convention as MTW for the Riemann tensor (see equation 3.35). Funnily, this doesn't appear in the August 2012 errata of the book --- perhaps somebody should email him.

In Misner, Thorne and Wheeler, the Riemann tensor is given by $$ R^\mu{}_{\alpha \beta \gamma} = \frac{\partial \Gamma^\mu{}_{\alpha \gamma}}{\partial x^\beta} - \frac{\partial \Gamma^\mu{}_{\alpha \beta}}{\partial x^\gamma} + \Gamma^\mu{}_{\sigma \beta} \Gamma^\sigma{}_{\gamma \alpha} - \Gamma^\mu{}_{\sigma \gamma} \Gamma^\sigma{}_{\beta \alpha} $$ I believe Lambourne defines $R^\mu{}_{\alpha \beta \gamma}$ to be precisely the negative of this quantity. Now the Riemann tensor has various symmetries. The ones relevant here are

$$R_{abcd} = - R_{bacd} $$ $$R_{abcd} = - R_{abdc} $$ $$R_{abcd} = + R_{cdab} $$

With these symmetries and the two conventions considered, you see that the two equations of geodesic deviation are identical.

EDIT1: I've just noticed that this isn't right, despite the up-votes. If we think of the four indices on the Riemann tensor in pairs (1 and 2, and 3 and 4), the index on the separation vector sits with the free index in Lambourne, whereas it sits with one of the $\dot{x}$ terms in MTW. This is a significant difference, and in fact I think Lambourne's equation might indeed be incorrect. I just found a PDF of the book and it appears you've reproduced the relevant equation (6.23) faithfully here.

  • $\begingroup$ I'm struggling to take this in, but Lambourne gives $$R_{\phantom{l}ijk}^{l}=\frac{\partial\Gamma_{ik}^{l}}{\partial x^{j}}-\frac{\partial\Gamma_{ij}^{l}}{\partial x^{k}}+\Gamma_{ik}^{m}\Gamma_{mj}^{l}-\Gamma_{ij}^{m}\Gamma_{mk}^{l}.$$ $\endgroup$
    – Peter4075
    Commented Aug 28, 2014 at 15:28

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