Why does a system whose equations of movement are $\lambda^2U^{\alpha} + \partial_{\mu}F^{\mu \alpha} = 0$ have three degrees of freedom? I'm trying to understand the solution of a problem where I have to study a field ($U^\mu$) which Lagrangian is:
$$\mathscr{L} = - \frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{1}{2} \lambda^2 U_{\mu} U^{\mu}$$
Where $ F_{\mu \nu}=\partial_{\mu} U_{\nu} - \partial_{\nu} U_{\mu}$. I got the equations of movement:
$$\lambda^2U^{\alpha} + \partial_{\mu}F^{\mu \alpha} = 0$$
But here my professor told us that if $\lambda \neq 0$ then we have three degrees of freedom in the problem. Why is that so? I've been able to grasp it so far.
 A: Because if $\lambda \neq 0$ you are treating the massive Vector Field, namely you're talking about vector particles (spin $1$) with mass: $W^+$, $W^-$ and $Z^0$ bosons.
Indeed by using the identity you wrote; $F_{\mu \nu}=\partial_{\mu} U_{\nu} - \partial_{\nu} U_{\mu}$, you can write your Lagrangian in the form
$$\mathcal{L} = -\frac{1}{2}\partial_{\mu}U_{\nu}\partial^{\nu}U^{\mu} + \frac{1}{2}\left(\partial_{\mu}U^{\mu}\right)^2  - \frac{1}{2}\lambda^2 U_{\mu}U^{\mu}$$
which is similar to the Klein-Goron Lagrangian but with one term more. Varying $U_{\mu}$ you obtain the equation of motion (after writing the action $\mathcal{S} = \int\ \mathcal{L}\ dt$) getting
$$\partial^{\mu}\partial^{\nu}F_{\mu\nu} - \lambda^2 \partial^{\nu}U_{\nu} = 0$$
and for $\lambda\neq 0$ you have the constraint 
$$\partial^{\mu}U_{\nu} = 0$$
Using those facts, the equations of motion can be written in the form:
$$(\Box - \lambda^2)U_{\mu} = 0$$
$$\partial^{\mu}U_{\mu} = 0$$
This is what tells us that of the four $U_{\mu}$ fields, only three of them are independent fields, and they describe in a covariant way the three associated polarizations of a spin $1$ particle.
