Which of these two textbook equations of geodesic deviation is correct?

Question

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.

Answer 1

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

$$\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}$$

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.

Answer 2

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.