For a spin 1 field described by a 4-vector, the condition [tex]m \partial_\mu A^\mu=0[/tex] can be derived from the equations of motion. This condition reduces the degrees of freedom from 4 to 3 for a massive particle (when m is not equal to zero). However, for the photon, [tex] \partial_\mu A^\mu=0[/tex] has to be imposed by hand since the above equation is trivially zero on the LHS for m=0. So shouldn't the photon still have 3 degrees of freedom once you add the Lorentz gauge condition [tex] \partial_\mu A^\mu=0[/tex] ? But I read that it's the gauge condition that reduces the degrees of freedom from 3 to 2. But it seems it should reduce it from 4 to 3, the same as imposing the Lorentz gauge condition on a massive vector field.