I am planning to start self studying GR and, before choosing a better book to use, I have been flipping through an old copy of Weinberg's book (Gravitation and Cosmology) and came upon something that is not making sense to me. I assume that I am missing something obvious.(adsbygoogle = window.adsbygoogle || []).push({});

When he talks about Thomas precession he uses the fact that ##\frac{dS^{\alpha}}{d\tau}=0## in a torque free frame (if it exists) and since we were already considering a free falling frame, ##\frac{d\vec{x}}{dt}=0##. He then states that this implies for all inertial frames ##\frac{dS^{\alpha}}{d\tau}=\Theta U^{\alpha}## for some constant ##\Theta##. I don't see why. I'm sure that I'm just missing something obvious. Below are my thoughts, but I might be over thinking it.

First, ##\frac{dS^{\alpha}}{d\tau}## is not a vector. If we transform to another (inertial) frame we find

##\frac{dS^{\beta}}{d\tau}=\frac{\partial x^{\beta}}{\partial x^{\alpha}}\frac{dS^{\alpha}}{d\tau}+\frac{d}{d\tau}(\frac{\partial x^{\beta}}{\partial x^{\alpha}})S^{\alpha}##

Therefore

##

\frac{d}{d\tau}(\frac{\partial x^{\beta}}{\partial x^{\alpha}})S^{\alpha}=\Theta \frac{\partial x^{\beta}}{\partial x^0}

##

since ##U^0=1## and ##U^i=0##.

I think we can rewrite this as

##

S^{\alpha}\frac{\partial}{\partial x^{\alpha}}(\frac{\partial x^{\beta}}{\partial\tau})=\Theta\frac{\partial x^{\beta}}{\partial\tau}

##

where ##t=\tau## since the ##t## being considered was the time in the rest frame (of whatever particle we are considering) and therefore the proper time (maybe?). If this is a legitimate manipulation, this seems plausible, but I do not understand why it must be true.

Anyway, I will not be using this text as a self study text. I bought a used copy because the book was so inexpensive, but I think I would rather Carroll for self study.

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# Thomas Precession in Weinberg

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