Emergence of spin from special relativity I have pulled up and read as many answered questions as I can find here on why spin emerges as a consequence of making quantum mechanics compatible with special relativity- and still have problems understanding why this is so. The closest I can come is that Lorentz invariance is not the same as Galilean invariance (where time does not get mixed up with position at high velocities) and so rotations cannot be treated the same way in the Lorentz case as in the Galilean case- and that's why quantum mechanical spin pops up in the Lorentz case. I have a question the answer to which might help me grasp this, please correct me if it is ill-posed:
In the case of a quantum particle which possesses intrinsic spin, that spin gets labeled on the particle by imagining the particle has a vector running through it with its head end emerging at the particle's "north pole" and its tail emerging at its "south pole". This means that a description of that particle at some instant in time must specify not only its position in space but also the direction in which its "spin vector" is pointing. In the case where the spin vector is parallel to the particle's velocity, the spin vector also uniquely labels the "head end" and the "tail end" of the particle for us. Now my question:
Is the reason that Lorentz invariance introduces spin in any way related to the need to label and keep track of the head and tail ends of a relativistic particle- and that you don't need to keep track of that for slow-moving particles?
 A: Historically (Dirac equation thing mentioned in @Kasi's answer) and phenomenologically (e.g. scalar electrodynamics or QED?), one can see the connection between spin and special relativity where the latter is interpreted as "stuff moving fast".
But from a mathematical/logical perspective, I disagree.
First off, if you are just labelling your particle with an arrow in space, I don't see why you would need $SO(1,3)$ instead of the simple $SO(3)$.  As an interesting sidenote, the Poincaré group is given by $\mathcal{P} = SO(1,3) \ltimes \mathbb{R}^{1,3}$
while the Galilean group is given by $\mathcal{G} = (SO(3)\ltimes \mathbb{R}^3) \times (\mathbb{R}^1 \times \mathbb{R}^3)$. In both cases, you can find double covers such as $SO(1,3) \xleftarrow{2:1} SL(2,\mathbb{C})$ and $SO(3) \xleftarrow{2:1} SU(2)$. The double cover argument in key in determining that there can only be two distinct ways of permuting identical particles in 3D, i.e. fermions and bosons. So one can reach this conclusion also within Galilean transformations. If and when you assigned spins to fermions and bosons, you'd realise you need $SL(2,\mathbb{C})$ to represent spin-1/2 particles.
That digression aside, I think it all boils down to causality.
For your theory to be causal, the action of operators on states cannot be faster than the speed of light. This is especially important for space-like separated events. In these situations, you turn to the group of symmetries of spacetime. For non-curved (flat) geometries, this is the Poincaré group (which comprise also Lorentz transformations), and we are in the realm of special relativity. The representations of the Poincaré group, that is the things you need to do maths with, are uniquely labelled by mass and spin.
So, in conclusion, I think spin emerges from special relativity not "because it is moving fast", but because the latter is the framework governing transformations between different events (in space and time), which is needed to ensure causality. Causality requiring to be enforced is the reason quantum mechanics (and its relativistic correction attempts) need to dropped at some point in favour of quantum field theory.
A: 
Is the reason that Lorentz invariance introduces spin in any way related to the need to label and keep track of the head and tail ends of a relativistic particle- and that you don't need to keep track of that for slow-moving particles?

I think this is a slightly misleading statement. In non-relativistic quantum mechanics, the wave function of an electron in the position basis will be generally a function $\psi: \mathbb{R}^3 \to \mathbb{C}$. So here at each point in space, the wavefunction is scalar. Even in non-relativistic quantum mechanics if we assume that the wavefunction is not scalar and instead a vector then that particle will have spin (in this case the spin is 1). Within non-relativistic quantum mechanics, using the commutation properties of angular momentum we can show that the eigenvalues of spin are always integral multiples of $\frac{1}{2}$. For orbital angular momentum ${\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} }$ we can show that the eigenvalues are always integers. But ${\displaystyle \mathbf {L} }$ is only the generator of rotations which will assign $\vec{\psi}(x,y,z)$ to $\vec{\psi}(x',y',z')$ but it won't rotate the $\vec{\psi}(x,y,z)$ itself.
Read from page 324 in Principles of Quantum Mechanics by Ramamurti Shankar to understand spin in non-relativistic quantum mechanics. So even non-relativistic quantum mechanics says a lot about spin.
Relativistic quantum mechanics goes 1 step further and in it, if we assume that the equation is linear in time then the wavefunction cannot be scalar, it should be a spinor (intuitively these are like square roots of vectors). Lorentz invariance doesn't imply spin should exist. For example, the Klein–Gordon equation has spin-zero (that is the wavefunction is scalar). Just like in nonrelativistic quantum mechanics, in relativistic quantum mechanics, the eigenvalues of angular momentum are always integral multiples of $\frac{1}{2}$. Initially, the Klein–Gordon equation was assumed to be the relativistic equation for electrons (both Klein–Gordon and Dirac equations reduce to Schrödinger equation in the non-relativistic limit). But after detecting electron spin it was abandoned for the Dirac equation. The Dirac equation should be of the form
$$ {\displaystyle \left(\beta mc^{2}+c\sum _{n\mathop {=} 1}^{3}\alpha _{n}p_{n}\right)\psi (x,t)=i\hbar {\frac {\partial \psi (x,t)}{\partial t}}}$$
for it to satisfy ${\displaystyle E^{2}=m^{2}c^{4}+p^{2}c^{2}}$ and be linear in time. It can be shown that $\alpha_i$ and $\beta$ should be at least 4-dimensional matrices. If we take them as 4-dimensional matrices that implies that $\psi$ is a 4 component column matrix. The Klein–Gordon equation satisfies ${\displaystyle E^{2}=m^{2}c^{4}+p^{2}c^{2}}$ but is not linear in space-time coordinates.
The Pauli equation is the non-relativistic limit of the Dirac equation and it explains spin. If there are no electromagnetic fields then it will reduce to Schrödinger equation.
$${\displaystyle \left[{\frac {1}{2m}}({\boldsymbol {\sigma }}\cdot (\mathbf {p} -q\mathbf {A} ))^{2}+q\phi \right]|\psi \rangle =i\hbar {\frac {\partial }{\partial t}}|\psi \rangle }$$
A: You don't need special relativity to explain spin. The spin effects are observable at the non-relativistic limit. It is a consequence of linearlization of the quantum equations.
Dirac took the relativistic energy spectrum $E=\sqrt{m^2c^4+p^2c^2}$ and wrote a wave equation that is linear in space and time derivatives $p\to-i\hbar\nabla$ and $E\to i\hbar\partial_t$. He found that this could only be solved with a 4-component wave-functions (two-bispinors, the second one representing anti-particles).
You can do the same that Dirac did with the non-relativistic spectrum and linearize $E=p^2/2m$ and find Lévy-Leblond equations that are equivalent to the Pauli equation (Schrödinger equation for spin 1/2), which requires a two-valued spinor wavefunction. This happens because spin appears as one of the Casimirs of the Galilei algebra, which is non-relativistic. Spin is certainly a quantum effect, but not necessarily a relativistic one.
