# Massive Spin-1 Lagrangian: Removal of Spin-0 degree of freedom

I am currently reading Schwartz on QFT and the Standard Model and arrived now at Chapter 8.2.2, where he derives a Lagrangian for a massive Spin-1 field. The final Lagrangian looks like this: $$\mathcal{L}=\frac{1}{2} A_{\mu} \square A_{\mu}-\frac{1}{2} A_{\mu} \partial_{\mu} \partial_{\nu} A_{\nu}+\frac{1}{2} m^{2} A_{\mu}^{2}\tag{8.23}$$ The equations of motion are $$(\square +m^{2}) A_{\mu}=0$$ with a Lorentz-invariant constraint: $$\partial_{\mu} A_{\mu}=0.$$

The Lagrangian was constructed in a way, that $$A_{\mu}$$ is forced to transform like a 4-vector under Lorentz-transformations in order to keep the Lagrangian Lorentz-invariant.

Thus, we have a 4-dimensional representation of the Lorentz-group acting on the vector space, given by all the $$A_{\mu}$$ solving above equations.

From later in the book (Chapter 10) we know that, if we have a 4-dimensional representation of the Lorentz-group, then the corresponding representation of the Rotation-group is $$3 \oplus 1$$ -dimensional.

Now here is my question: The constraint $$\partial_{\mu} A_{\mu}=0$$ reduces the number of degrees of freedom of the vectors $$A_{\mu}$$ to 3. Schwartz then states:

Since $$\partial_{\mu} A_{\mu}=0$$ is a Lorentz-invariant condition, it has to remove a complete representation, which with one degree of freedom can only be the spin-0 component.

Can anyone explain to me, what he means by that/ why that is? In other words: Why does Lorentz-invariance of the constraint necessarily remove the Spin-0 d.o.f. and not one of the three Spin-1 degrees of freedom?

A Lorentz-invariant constraint is rotation-invariant, among other things. The three spin-1 degrees of freedom are mixed with each other by rotations, so a rotation-invariant constraint cannot remove one of them without removing all of them. But a spin-0 degree of freedom is not mixed with anything else by rotations, so it can be removed by a rotation-invariant constraint. Since $$\partial^\mu A_\mu=0$$ is only one constraint, it can only remove one degree of freedom, which must therefore be the spin-0 degree of freedom.