 Solution

In a supersymmetric Higgs mechanism, an internal gauge symmetry breaks while supersymmetry can remain unbroken. A massless vector multiplet absorbs a chiral multiplet and becomes a massive multiplet with spins $(1,\frac{1}{2},\frac{1}{2},0)$: the gauge boson acquires its longitudinal particle polarization from the Goldstone mode. In the super-Higgs mechanism, local supersymmetry itself breaks: the spin-$3/2$ gravitino absorbs the spin-$1/2$ goldstino, gaining two longitudinal polarizations and becoming massive. With only global supersymmetry breaking, the goldstino instead remains a physical massless fermion.