Let's consider the classical Lagrangian density for a real vector field $V^{\mu}$ of mass $M$: $$L_{V} = -\frac{1}{4} V_{\mu\nu}V^{\mu\nu} + \frac{1}{2} M^{2} V_{\mu}V^{\mu} $$ The Eulero-Lagrange equation reads as: $$\boxed{(\Box + M^{2})V^{\mu} - \partial^{\mu} \partial_{\nu}V^{\nu}=0}$$ that is the Proca equation. It can be reduced to a Klein-Gordon equation with a constrain: $$(\Box + M^{2})V^{\mu}=0$$ $$ \partial_{\mu}V^{\mu}=0$$
Now arises my question: i'm trying to demonstate that one has: $$\boxed{\pi_{i}(\partial_{i}V_{0}) = \partial_{i}(\pi_{i}V_{0}) - M^{2}V_{0}^{2}} (1) $$ by using the Eulero-Lagrange equation. In the above expression: $\pi^{\mu} = - V^{0\mu}$ is the conjugate momentum of the field $V^{\mu}$ and for the skew-symmetry of the tensor $V^{\mu\nu}$ one has $\pi^{0} = - V^{00} = 0$: this means that $V^{0}$ is not a dynamical variable, and so the only non-zero conjugate momenta are $\pi^{i}=-V^{0i}$. I tried to demonstrate this by computing: $$\boxed{\partial_{i}(\pi_{i}V_{0}) = \pi_{i}(\partial_{i}V_{0})+ (\partial_{i}\pi_{i})V_{0}} (2)$$ Doing (1) + (2): $$(\partial_{i}\pi_{i})V_{0} = M^2 V_{0}^{2} $$ but i'm not able yet to demonstrate the last equality. Please, can anyone help me?
