0
$\begingroup$

The spin angular momentum of a gyro is represented, in special and general relativity, by a spin four-vector $S^{\mu}$. In the rest frame of the gyro, the spin four-vector takes the form $S^{\mu}=(0,S^{i})$, where $S^{i}$ is the ordinary three-vector angular momentum and the time component $S^{0}$ is zero. Why does the spin four-vector have vanishing time component in its rest frame? The four-velocity of a particle does not follow this pattern -- it has a non-zero time component in the rest frame of the particle. Is there a way to see why $S^{0}=0$?

$\endgroup$
  • 1
    $\begingroup$ E.g. the Wikipedia article details the definition and calculation of that vector, and it turns out to not have a time component in that frame. Can you be more specific what you want to know about that? $\endgroup$ – ACuriousMind Nov 20 '15 at 16:39
  • $\begingroup$ Where a lot of vectors generalize to 4 vectors, spin and angular momentum (and magnetic fields) generalize to a 4x4 matrix. $\endgroup$ – David Elm Oct 7 '17 at 3:07
1
$\begingroup$

So in a four-dimensional orientable space we have a $[0\;4]$ orientation tensor $$\epsilon_{\alpha\beta\gamma\delta} = - \epsilon_{\beta\alpha\gamma\delta} = - \epsilon_{\gamma\beta\alpha\delta} = - \epsilon_{\delta\beta\gamma\alpha}$$Usually with respect to some basis we choose $\epsilon_{0123} = 1$ or so to finish off the specification of the whole tensor. Since we could just as easily get the same orientable space by choosing $\epsilon_{0123} = -1$ these are sometimes called "pseudovectors" and "pseudotensors", but you can flip that perspective around and just say "this space just has this orientation tensor by definition" and reflections will simply change the sign of the components of the orientation tensor, leading to a consistent calculation with no "pseudos" to speak of.

Now suppose you have a bunch of worldlines of interacting particles: due to Poincaré symmetry and Noether's theorem there are some conserved quantities $P^\mu$ and $J^{\mu\nu},$ and while there is no perfect definition of "center of mass", we are free to choose the frame where $P^\mu$ points purely in the time-direction as the "center of mass frame". Regardless there is a 4-covector called the Pauli-Lubanski spin pseudovector defined by the orientation tensor as: $$ S_\alpha = \frac12~\epsilon_{\alpha\beta\gamma\delta}~P^\beta~J^{\gamma\delta}$$ However in this frame in particular it has no time component, and it's for a super-simple reason.

Let $T^\alpha$ be the unit 4-vector in the time direction for this particular center-of-mass-frame. In this frame, $P^\mu = m c^2 T^\mu$ for some effective mass $m$. This means that the time component of the spin vector is therefore $$S_\alpha T^\alpha = \frac1{2mc^2}~\epsilon_{\alpha\beta\gamma\delta}~P^\alpha ~P^\beta~J^{\gamma\delta}$$However, $A_{\gamma\delta} = \epsilon_{\alpha\beta\gamma\delta} V^\alpha V^\beta$ is $0_{\gamma\delta}$ for any $V^\alpha$ because of antisymmetry: the first equality in the first equation in this answer says that it must be equal to $-\epsilon_{\beta\alpha\gamma\delta} V^\alpha V^\beta$ which under the $\alpha \leftrightarrow \beta$ relabeling isomorphism is just $-A_{\gamma\delta};$ the only tensor which is its own negative is the zero tensor.

Therefore, the component of the spin 4-vector in the center-of-mass frame is 0.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.