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.
One way to see the difference is to look at the operators in a 3-d notation:
The boost components:
$$ K^i = S^{0i}, \quad i=1,2,3$$
The angular momentum operators
$$J^i = \epsilon_{jk}^i S^{jk}, \quad i, j, k=1,2,3$$
The Pauli-Lubanski vector time component:
$$W^0 = \mathbf{J}\cdot \mathbf{P}$$
($\mathbf{P}$ is the momentum operator)
and the spatial components
$$\mathbf{W} = \mathbf{P}\times \mathbf{K} + E \mathbf{J}$$
($E = P^0$, the energy). From this representation, we can see that the spatial Pauli-Lubanski components coincide in the rest frame with the angular momentum at the rest frame, i.e., the spin.
But the existence of a rest frame is exclusive to a massive particle; and this interpretaion is false in the massless case. In this case, there is no rest frame. Furthermore, the constraint $W^{\mu}P_{\mu} = 0$, together with the maslessness condition $P^{\mu}P_{\mu} = 0$, imply that the Pauli-Lubanski is necessarily proportional to the momentum vector:
$$W^{\mu} = \lambda P^{\mu}$$
Looking at the zero component, we get:
$$\lambda = \frac{\mathbf{J}\cdot \mathbf{P}}{E}$$
The helicity.
Therefore for a massless particle the Pauli Lubansky vector does not coincide and is not proportional to the spin.
Best Answer
The time component of the Pauli-Lubanski vector is equal to the helicity times the (three) momentum magnitude:
$w^0 = \lambda ||\mathbf{p}|| =\mathbf{j}.\mathbf{p}$
Where $\lambda$ is the helicity, $\mathbf{j}$ is the (total) angular momentum and $\mathbf{p}$ is the three momentum. Please see the following article by Carineña, Garcia-Bondía, Lizzi, Marmo and Vitale (the second formula of section 2). Please, see also, the next formula where the transformation of the spatial and time components of the Pauli-Lubanski vector under a general boost is written:
$ w^0 \rightarrow cosh(\xi)w^0 + sinh(\xi) \mathbf{n}.\mathbf{w}$.
$ \mathbf{w } \rightarrow \mathbf{w} - sinh(\xi)w^0 \mathbf{n} + (cosh(\xi)-1) (\mathbf{n}.\mathbf{w}) \mathbf{w}$.
Where $\mathbf{w}$ are the spatial components of the Pauli-Lubanski vector. $\xi$ is the rapidity, $\mathbf{n}$ is the boost direction
Now it is easy to deduce the properties of the time component of the Pauli-Lubanski by inspection:
1) For a spinless particle, this component is identically zero in all reference frames:
2) For a massless particle, and a Lorentz transformation which preserves the momentum. The angular momentum rotates around the momentum vector (Wigner rotation) such that the helicity is conserved. This is because for a lightlike 4-momentum, the Pauli-Lubanski vector must be proportional to the momentum vector, therefore its time component does not change under a momentum preserving Lorentz transformation.
Update
The reason is as follows: For a massless particle, the Pauli-Lubanskii 4-vector is light-like. Taking into accout that it is always orthogonal to the momentum 4-vector (which is also light-like in this case), the two vectors must be proportional (two orthogonal light-like vectors must be proportional). The proportionality factor is just the ratio between the helicity (time component of the of the Pauli-Lubanski vector) and the energy (time component of the 4-momentum). This suggets that when the kinetic energy of a particle is much larger than its rest mass, the Pauli-Lubanski and the momentum vectors tend to be aligned. In order to see that more explicitely, one can use the expression of the Pauli-Lubanski spatial components in terms of the spin and momentum vectors for a massive particle:
$\mathbf{w} = m \mathbf{s} + \frac{ \mathbf{p}.\mathbf{s}}{p_0+m}\mathbf{p}$.
From this formula it is clear that when the particle speed becomes large, the second term dominates and the pauli-Lubanski spatial components 3-vector becomes almost aligned with the momentum spatial components 3-vector.