# Thomas Precession in Weinberg

1. Jun 25, 2014

### DrewD

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.

2. Jun 25, 2014

### WannabeNewton

Of course $\frac{d}{d\tau}S^{\alpha}$ is a 4-vector; $\nabla_{u} = \frac{d}{d\tau}$ always maps a 4-vector to a 4-vector along the integral curve of $u$. Secondly what Weinberg says is $\frac{d\vec{S}}{d\hat{t}} = 0$ in a local inertial frame momentarily at rest with respect to the spinning particle; note that I'm using $\hat{t}$ as opposed to the $t$ Weinberg uses to emphasize that we are in a comoving local inertial frame.

Now, it is certainly not true that $\frac{dS^{\alpha}}{d\tau} = 0$ for a torque-free spinning particle or gyroscope unless the particle or gyroscope is freely falling (more precisely, it also has to be spherical and sufficiently small to eliminate tidal torques that deviate it from geodesic motion) and it is only true that $\frac{d\vec{S}}{d\tau} = 0$ if we are in a momentarily comoving local inertial frame because therein $\hat{t} = \tau$ by definition of a comoving local inertial frame, which is the situation we have above. Weinberg does not assume the particle is freely falling-he just uses a freely falling frame momentarily at rest with respect to the particle. The transport equation for a torque-free spinning particle or gyroscope is called Fermi-Walker transport, which is what Weinberg is deriving. Nowhere does he claim that $\frac{dS^{\alpha}}{d\tau} = 0$ when all the forces on the particle act on its center of mass.

Lastly, $\frac{dS^{\alpha}}{d\tau} = (\frac{dS^0}{d\hat{t}}, \frac{d\vec{S}}{d\hat{t}}) = (\frac{dS^0}{d \hat{t}}, 0) = \frac{dS^0}{d \hat{t}} u^{\alpha}$ since $u^{\alpha} = \delta^{\alpha}_{\hat{t}}$ in the comoving local inertial frame. What Weinberg then does is use $S_{\alpha}u^{\alpha} = 0$ to find a covariant expression for $\Theta \equiv \frac{dS^0}{d \hat{t}}$.

Last edited: Jun 25, 2014
3. Jun 25, 2014

### Bill_K

Wannabe, I believe his notation is $\nabla_{u} = \frac{D}{D\tau}$. When he writes $\frac{d}{d\tau}S^{\alpha}$ it means just ordinary derivatives, not covariant.

4. Jun 25, 2014

### WannabeNewton

Oh I see. My apologies then! Thankfully he's working in a local inertial frame :p

5. Jun 25, 2014

### DrewD

Yes, in the part in question (first few pages of ch. 5) he is talking about ordinary differentiation and not covariant.

Yes, Weinberg starts out by saying that we are considering a freely falling frame (not particle, sorry) and the instantaneous comoving frame that, in this case, has zero torque. I understand what he is deriving eventually, but unless I am incorrect, the part I am confused by is just SR (the section ends with Fermi-Walker transport)

I think this might be clearing things up. I was thinking that since

$S^{\alpha}=(S^0,\vec{S})=(0,\vec{S})$

then the $S^0$ component would always be zero. If this were true, $\frac{d S^0}{d\tau}$ would be identically zero and in this special case where $\frac{d \vec{S}}{d\tau}=0$, then the four "vector" would be zero in the comoving frame. This isn't true since $S^0$ in another frame will depend on $\vec{S}$.

I think it makes perfect sense (but I'm at work and may find more confusion when I think about it later), and if I am right, it was a silly mistake.

6. Jun 25, 2014

### WannabeNewton

Ah I see your confusion. Let's consider the following argument. We have $S^{\alpha}u_{\alpha} = 0$ all along the world-line; then $\frac{d}{d\tau}(S^{\alpha}u_{\alpha}) = 0$. Let us now consider a locally inertial frame momentarily comoving with the particle described by this world-line. I will again use hats above indices to indicate components with respect to this frame. Now in this frame $u_{\hat{\alpha}} = \eta_{\hat{\alpha}\hat{\beta}}u^{\hat{\beta}} = \eta_{\hat{\alpha}\hat{\beta}}\delta^{\hat{\beta}}_{\hat{0}} = -\delta ^{\hat{0}}_{\hat{\alpha}}$ so clearly then $\frac{d}{d\tau}(S^{\alpha}u_{\alpha}) = -\frac{d}{d\tau}S^0 = 0$.

But this is certainly wrong. Where did we go wrong? Well remember, we are in a momentarily comoving local inertial frame. This means the relation $u^{\hat{\alpha}} = \delta^{\hat{\alpha}}_{\hat{0}}$, and hence $u_{\hat{\alpha}} = -\delta ^{\hat{0}}_{\hat{\alpha}}$, is only valid at the event on the world-line at which this local inertial frame is momentarily comoving with the particle. But the expression $\frac{d}{d\tau}(S^{\alpha}u_{\alpha})$ involves a derivative of $u^{\alpha}$ and hence requires knowledge of $u^{\alpha}$ in some open interval surrounding the aforementioned event and not just at this event, meaning the relation $u_{\hat{\alpha}} = -\delta ^{\hat{0}}_{\hat{\alpha}}$ in the momentarily comoving frame cannot simply be substituted into $\frac{d}{d\tau}(S^{\alpha}u_{\alpha})=0$.

Indeed $0 = \frac{d}{d\tau}(S^{\alpha}u_{\alpha}) = u_{\alpha}\frac{dS^{\alpha}}{d\tau} + S^{\alpha}a_{\alpha}$ so in the momentarily comoving local inertial frame we have $\frac{dS^{0}}{d\tau} = \frac{dS^{0}}{d\hat{t}} = -S^{\hat{i}}a_{\hat{i}}$ and to calculate $a_{\hat{i}}$ we need more than just $u^{\hat{\alpha}} = \delta^{\hat{\alpha}}_{\hat{0}}$ in the momentarily comoving local inertial frame at a single event because the acceleration is the derivative of the velocity for which we need to know the latter on an entire open interval. In fact by incorrectly saying $\frac{d}{d\tau}(S^{\alpha}u_{\alpha}) = -\frac{d}{d\tau}S^0 = 0$ in the momentarily comoving local inertial frame what we've really done is $\frac{d}{d\tau}u^{\hat{\alpha}} = \frac{d}{d\tau}\delta^{\hat{\alpha}}_{\hat{0}} = 0$ which is certainly an invalid step. I hope that helps.

By the way, in the above I'm using $\frac{d}{d\tau}$ for the covariant derivative as opposed to Weinberg's $\frac{D}{D\tau}$; sorry for the confusion of notation but I'm just really used to the former.