- #1
DrewD
- 529
- 28
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.
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.
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.