Thursday, December 20, 2018

general relativity - Which of these two textbook equations of geodesic deviation is correct?



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



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.



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