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I am concerned about generators of the Poincare Lie Group that is responsible for boosts and rotations (often labelled by $J_{\nu \mu}$).

$J_{\nu \mu}$ is often divided into a term responsible for the spin angular momentum ($S_{\nu \mu}$) and another term responsible for the orbital angular momentum ($L_{\nu \mu}$). Therefore,

$J_{\nu \mu}$ = $L_{\nu \mu}$ + $S_{\nu \mu}$

I am wondering why is the $S_{0 \mu}$ components equal to zero?

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  • $\begingroup$ This depends on a particular choice of coordinate that you are using (time is the $0$th coordinate). Then it all boils down to the fact that $O(1,3)$ contains the orthogonal subgroup $O(3)$ fixing the $0$th coordinate vector pointwise, hence, the 1st column vector of the corresponding group elements is $(1,0,....))$. On the level of the Lie algebra, this becomes $(0,...,0)$. $\endgroup$ – Moishe Kohan Dec 6 '19 at 13:24
  • $\begingroup$ @MoisheKohan Can you elaborate a bit? You said that "This depends" so do you mean the 𝑆_(0𝜇) components need not be zero? $\endgroup$ – The First StyleBender Dec 6 '19 at 14:32
  • $\begingroup$ Yes, for instance, if you, say, swap the 1st and 4th coordinates making the Lorentzian form $x_1^2+x_2^2+x_3^2 -x_4^2$. $\endgroup$ – Moishe Kohan Dec 6 '19 at 15:06

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