How many field components are there in vector-spinor field? I am trying to find out the degrees of freedom of the vector-spinor field ($s=3/2$). The degrees of freedom are given by $N=\frac{1}{2}\left(N_{F}-N_{C}\right)$ for this spin where $N_F$ is the number of (real) field components and $N_{C}$ the number of (real) constraints on the field that follow from the field equation.
I have derived the two constrains it has, thus $N_C=2$. I have to figure out what $N_F$ is for this kind of field to see what $N$ will be, but I am not able to reason how many components will it have although it was kind of intuitive to figure out for lower spin fields.
 A: Massless Rarita-Schwinger (=spin-vector) field in $4$ spacetime dimensions has $4$ gauge degrees of freedom (this symmetry becomes local supersymmetry once graviton is included), since the gauge parameter is a Weyl spinor. You subtract those from $16$ real components to get $12$ off-shell DOF.
Equations of motion then show that the time-component of the spin-vector $\psi_\mu$ is non-propagating, thus $-4$ DOF. There is another constraint from the equations of motion: $\partial^\mu\psi_\mu=0$ which further subtracts $4$ DOF. So far we have $12-8=4$ DOF which should be divided by two due to the field equations for the remaining spatial components of $\psi_\mu$ (Weyl/Dirac equations reduce the number of independent components of spinors by half).
Now for massive spin-vector the gauge freedom is lost, so off shell we have $16$ DOF, while on-shell $(16-4-4)/2=4$ independent DOF. The two extra degrees of freedom are longitudinal modes, similarly to a massive vector field having one longitudinal DOF. In the context of supersymmetry breaking, the two extra degrees of freedom of a massive gravitino are from a Goldstone (Weyl) fermion.
A: I might have solved my question after giving it a thought but please correct me if I am wrong. I am considering a vector-spinor field and thus from $\psi_{\mu \alpha}$, we will have 16 complex components as $\mu=0,1,2,3 ; \alpha=1,2,3,4$ and therefore, it will have 32 real components meaning that $N_F=32$.
Now for the case of a vector-field, we have 2 constrains, each having 4 complex components and thus 8 real entries meaning that in total, $N_C=2\cdot 8=16$.
Thus, we have that
$$
N=\frac{1}{2}\left(N_{F}-N_{C}\right)=\frac{1}{2}\left(32-16\right)=8
$$
