# Lagrangian for a free Dirac field equal zero?

The Lagrangian (density) for a free Dirac field is given as $${\mathcal L}_\mathrm{Dirac} = \bar{\psi} \left( i \gamma^{\mu} \partial_{\mu} - m \right) \psi,$$ but given that $\psi$ obeys the Dirac equation, $$\left( i \gamma^{\mu} \partial_{\mu} - m \right) \psi = 0$$ Doesn't this mean the Lagrangian (density) is zero?

Since eoms are found by stabilising the action $S = \int \mathcal{L}$, considering only the solution of the eoms in place of a generic field configuration is like considering a function $f(x)$ only at its stationary point; in general it is not sufficient because you need to know how the funciton behaves in a whole neighbourhood of such points.
• The Lagrangian (density) for the KG field, ${\mathcal L}_\mathrm{KG} = \frac{1}{2}(\partial_{\mu} \phi)(\partial^{\mu} \phi) - \frac{1}{2} m^2 \phi^2$, doesn't vanish? – jim Jan 1 '17 at 13:34
• It does, in the sense that your $\mathcal{L}_\text{KG}$ is equivalent to $\mathcal{L}' = -\frac{1}{2} \phi (\Box + m^2)\phi$ up to a 4-divergence. But even if it did not vanish, it would have been just a constant (with respect to the fields), and you can redefine the lagrangian subtracting it so that it vanishes. – yoric Jan 1 '17 at 14:14
• @yoric your answer is reeeeally helpful. I don't understand though how, calculating the energy momentum tensor $T_{\mu \nu}$ for a dirac spinor field I can remove the term with the lagrangian (based on the considerations above), while when calculating the one for a scalar field i need to consider it in order to get the right tetramomentum vector – user129511 Jun 4 '17 at 15:07