In S. Carroll Lecture Notes on General Relativity, chapter 6, pages 152-153 we have equation (6.62) $$\tag{6.62} \frac{\partial^2}{\partial t^2} S^\mu=\frac{1}{2} S^\sigma \frac{\partial^2}{\partial t^2} h^\mu_{\; \sigma}.$$

While trying to deduce that equation from the previous one $$\tag{6.58} \frac{D^2}{d \tau^2} S^\mu= R^\mu_{\; \nu\rho\sigma} U^\nu U^\rho S^\sigma,$$ where $U^\mu$ is the velocity vector I noticed that he neglected the term $$\tag{*}\frac{\partial}{\partial \tau}(U^\rho \Gamma^\mu_{\; \rho \sigma} S^\sigma)$$ on the left hand side.

Is this approximation justified in his setting? Or am I mistaken somewhere?


1 Answer 1


No, the ∗ term is incomplete. The two total $D$-derivatives on the lhs. of eq. (6.58) in Ref. 1 is what is causing the curvature on the rhs. in the first place, see e.g. Ref. 2 p. 146.

Carroll is on the lhs. eq. (6.62) changing notation for the two total $D$-derivatives to two $\partial$-derivatives, but they are still total derivatives.

In fluid-dynamical language, one may say that Carroll is going from an Eulerian to a Lagrangian picture.

He is considering linearized gravity, so the Riemann curvature tensor is proportional to $\epsilon$, and we can (to the order that we are calculating, namely to first order in $\epsilon$) interpret $S^{\sigma}$ on the rhs. as following the flow.

Note that Ref. 2 contains slightly more details than Ref. 1.


  1. Sean Carroll, Lecture Notes on General Relativity, Chapter 6. The pdf file is available here.

  2. Sean Carroll, Spacetime and Geometry: An Introduction to General Relativity, 2003; Chapter 7.

  • $\begingroup$ Would you please clarify more, where am I wrong exactly? $\endgroup$
    – user48900
    Jan 25, 2015 at 19:29
  • $\begingroup$ The $*$ term is incomplete. The complete term would just lead to the curvature term on the rhs. of eq. (6.58). $\endgroup$
    – Qmechanic
    Jan 25, 2015 at 19:38

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