The Spin Term of the Angular Momentum Tensor in Relativistic Quantum Mechanics In Relativistic Quantum Mechanics, an Angular Momentum Tensor is defined ($J_{\nu \mu}$).  This tensor is divided into two terms, one responsible for the spin angular momentum ($S_{\nu \mu}$) and one responsible for the orbital angular momentum ($L_{\nu \mu}$). In essence:
$J_{\nu \mu}$ = $L_{\nu \mu}$ + $S_{\nu \mu}$
I read that the zeroth-components (the time components) of $S_{\nu \mu}$ is zero.  In other words, I read that:
$S_{00}$ = $S_{01}$ = $S_{02}$ = $S_{03}$ = 0.
Why is this the case?
PS:


*

*This question arose when I was concerned about the equality between equation (9.36) and equation (9.37) on page 81 of the following set of notes (http://www-pnp.physics.ox.ac.uk/~tseng/teaching/b2/b2-lectures-2018.pdf).

*A possible explanation I have: the zeroth-components of the Angular Momentum Tensor corresponds to boosts and there's no boosts when dealing with spin, so there's no zeroth-components.  Am I correct?

 A: If you want $S^{i0}$ to be zero, then the spin  $S^{\mu\nu}$ is the angular momentum of the body about its energy centroid in the lab frame. 
Let's see how this comes about.
Suppose that the we have a conserved and symmetric energy-momentum tensor:
$$
\partial_\mu T^{\mu\nu}=0, \quad T^{\mu\nu}=T^{\nu\mu}
$$
which is non-zero only within the body of interest.
Let $x_{\rm A}^\mu$ be a space-time event, $\Sigma$ a spacelike surface, and define the angular momentum about $x_{\rm A}$ by
$$
M^{\mu\nu}_{\rm A} = \int_\Sigma\left\{(x^\mu-x_{\rm A}^\mu)T^{\gamma\nu}- (x^\nu-x_{\rm A}^\nu)T^{\gamma\mu}\right\}d\Sigma_\gamma
$$
Then  $M^{\mu\nu}_{\rm A} $ is a tensor and independent of the choice of  the choice of  $\Sigma$.  
We now choose a lab  frame and define the mass-centroid $X^i_{\rm L}$ in that frame by
$$
\left\{\int_{t=\rm const.} T^{00}d^3x \right\}\,X^i_{\rm L}= \int_{t=\rm const.} x^iT^{00}d^3x.
$$
Note that 
$$
\partial_t  \int T^{00}d^3x =  \int \partial_0 T^{00}d^3x = - \int \partial_j T^{j0}d^3x=0,
$$
and 
$$
\partial_t  \int x^i T^{00}d^3x =  \int  x^i\,\partial_0 T^{00}d^3x =   -\int  x^i\,\partial_j T^{j0}d^3x=  \int \delta^i_j \,T^{j0}d^3x= p^i.
$$
So, differentiating its definition with respect to $t$, we read off that the ordinary three-velocity of the centroid is 
$$
\dot {\bf X}_{\rm L}= {\bf p}/E.
$$
Here
$$
E= \int T^{00}d^3x, \quad p^i =   \int  T^{0i}d^3x.
$$
Now take  $\Sigma$ to be the lab-frame  surface $t=$const with $x^\mu_{\rm A}$ a point in that surface. Then
$$
M^{i0}_{\rm A} = \int_\Sigma\left\{(x^i-x_{\rm A}^i)T^{00}- (x^0-x_{\rm A}^0)T^{0i}\right\}d^3x\\
= (X^i_{\rm L}-x^i_{\rm A})E.
$$
(The second term  is zero because $x^0- x^0_{\rm A}$  is zero everywhere in the integral.)
Thus $M^{i0}_A$ is zero when ${\rm A}$ is the centroid in the lab frame.  If we replace the lab frame with a frame having four-velocity $v^\mu$ we have that 
$M^{\mu\nu}_{\rm A} v_\nu=0$ if and only if ${\rm A}$ is the mass centroid in that frame.
Define the centre of mass $X^i_{\rm CM}$ to be the mass-centroid in the frame where the three momentum $p^i=0$,  then  the spin  or intrinsic angular momentum $S^{\mu\nu}$ is usually defined to be the angular momentum about the centre of mass.  Thus a more natural condition is $S^{\mu\nu}p_\nu=0$. 
The total angular momentum about an arbitrary point $x_A$ is then 
$$
M^{\mu\nu}= (x_{\rm CM}-x_A)^\mu p^\nu- (x_{\rm CM}-x_a)^\nu p^\mu + S^{\mu\nu}\\
= L^{\mu\nu}_A+S^{\mu\nu}.
$$
A rather useful combination is the totally antisymmetric Pauli-Lubanski tensor (invented by Myron Matthisson)
$$
S^{\lambda\mu\nu}= p^\lambda S^{\mu\nu}+ p^\mu S^{\nu\lambda}+p^\nu S^{\lambda\mu}.
$$
This  object turns out to be independendent of the choice of $x_A$ and is also sometimes called the intrinsic spin.
