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At a stationary symmetry-breaking vacuum of a gauge-invariant scalar potential, the angular field has zero potential curvature. Its free kinetic term is $(\partial b{)}^2/2$, while the displayed gauge-fixing term adds $-\xi g^2{v}^2{b}^2/2$. HenceThis is the would-be Goldstone boson, whose gauge-dependent pole accompanies the longitudinal pole in i; it is not an additional physical scalar particle of that mass. For $M_A\ne 0$, fixed finite $p$ and $\xi \to +\infty$,The angular field becomes infinitely heavy in this gauge and is absent from the unitary gauge propagating description. The qualification about fixed momentum matters: loop integrations probe arbitrarily large $p$, so one cannot automatically interchange the $\xi \to \infty$ limit with ultraviolet integration. If $gv=0$, there is no massive Higgs phase and the asserted heavy-field limit does not follow. A separate radial Higgs boson may remain; its mass depends on the unspecified potential and is not determined by this gauge-fixing term.