2
$\begingroup$

In Weinberg's Gravitation and Cosmology he has a pretty concise derivation of the Thomas precession formula (Eq. 5.1.13). But I don't get the first step...

A particle with intrinsic spin is under the influence of a force $f^\mu$ which does not torque the spin. We are in an inertial frame where the particle is at rest (at some instant).

In this frame the spin vector is constant since there's no torque: $$\frac{d\mathbf{S}}{dt}=0,$$ where $\mathbf{S}$ is the spatial part of the spin 4-vector which is defined earlier (in particular, in the rest frame of the particle the time component $S^0=0$).

Also, since we're in a comoving frame the 3-velocity of the particle is also zero: $d\mathbf{x}/dt=0$.

But then comes the claim that in this particular coordinate system we can write the zero-torque condition as: $$\frac{d S^\alpha}{d \tau} \propto U^\alpha,$$ where $\tau$ is proper time and $U^\alpha$ is the 4-velocity.

I do not get this. First, what about the $\alpha=0$ component? Isn't this $0$ on the left side of the equation and $1$ on the right?

Second, it feels like the argument is something like "both of these vectors are $0$ so therefore they are proportional to each other." But this seems weird. You can also get a Lorentz invariant equation by just setting a 4-vector equal to 0? I get the feeling that there's some very useful trick being applied here that will help in other situations too to derive covariant physical laws.

$\endgroup$
  • $\begingroup$ It might be best to post the second question as a separate question, so that people who'd like to answer one but not the other can respond to it. $\endgroup$ – Display Name Oct 12 '18 at 2:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.