Geodetic Precession of a Gyroscope in Hartle's GR Book In 14.3, Hartle deduces the geodetic precession angle per orbit of a gyroscope in Schwarzschild geometry. Immediately after eq.(14.18), the book reasons that the angle deduced is physically measured by an observer comoving with the gyro. However, I have questions in this derivation.

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*Eq.(14.17) doesn't make sense to me, since technically $e_\hat{r}$ and $\frac{s(t)}{s_*}$ are not in the same tangent space. I guess instead of $e_\hat{r}$, the book intends to consider a vector at the same point as $\frac{s(t)}{s_*}$, with the same components as $e_\hat{r}$ in coordinate basis? If that is the case, we can use the inner product between the two to compute the precession angle, which is a spatial angle, simply because $e_\hat{r}^0=0$?


*I don't understand the argument that this angle is measured by a comoving observer. In particular, what does "a radial direction in the observer's frame" mean? To me, there seems to be only one meaningful radial direction, namely $\frac{\partial}{\partial r}$? But then this will trivally be (0,1,0,0) in Schwarzschild coordinate basis. Besides, in my understanding, Lorentz boost only relates locally inertial frames, but Schwarzchild basis frame is not locally inertial, so how can we use Lorentz boost to transform a vector in an observer's frame to the coordinate frame?
Any help will be appreciated!
 A: After some thoughts, I may have come up with an answer, which is now posted here for confirmation and future reference.

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*Given a tetrad $\{\hat{e}_\alpha\}$, the spatial angle between 4-vectors $u,v$ may be computed from the dot product (as opposed to the full scalar product) $u^i v_i$, where $\alpha=0,1,2,3$ and $i=1,2,3$. In particular, we can construct a tetrad from the Schwarzschild coordinate basis, which corresponds to a stationary observer. The spatial angle between $e_{\hat{r}}$ and $\frac{s(t)}{s_*}$ can therefore be computed in this tetrad from the full scalar product as in Eq.(14.17) since $e_{\hat{r}}^0=0$. Thus, Eq.(14.18) gives the precession angle measured by a stationary observer.


*Now we can meaningfully talk about Lorentz boost between this stationary observer  and the comoving observer. The full scalar product of $e_{\hat{r}}$ and $\frac{s(t)}{s_*}$ remains the same in the comoving observer's tetrad. The spatial angle measured by this comoving observer can be computed from the dot product in his tetrad, which is again equal to the full scalar product, since in the comoving observer's tetrad $e_{\hat{r}}^0=0$.
