# Z and $\gamma$ bosons as mixtures of W and B: Part II

When it is said that the photon is ["a mixture of W and B"][1] ($$B$$ being a gauge field associated with the $$U(1)$$ hypercharge)

I have one question on this:

• Why there isn't a boson directly generated by this $$B$$ gauge?
• link to part I: physics.stackexchange.com/q/477283/955 – lurscher May 2 at 17:58
• because $SU(2)_L\times U(1)_Y$ is broken down to $U(1)_{em}$. When the former symmetry is restored (early universe) there will be massless B boson, and also 3 massless W bosons. – Kosm May 2 at 19:55

As Kosm suggests in his comment, there technically is a massless boson directly generated by the $$B_\mu$$ field. The reason we do not see it in (say) particle accelerators is because our universe is "too cold," and hence spontaneous symmetry breaking in the Higgs field generates the $$A_\mu$$ (photon), $$Z_\mu$$, and two $$W^\pm_\mu$$ fields.