# Partial derivative of Dirac Lagrangian with respect to derivatives of fields

Why is $$\frac{\partial\mathcal{L}}{\partial(\partial_\nu \bar{\psi})} = 0$$, for the Dirac Lagrangian $$\mathcal{L} = \bar{\psi}(i \gamma^\mu \partial_\mu - m)\psi$$?

This comes up in deriving the Noether current for $$\psi \rightarrow e^{i\alpha}\psi$$ for example.

My confusion comes from the fact that we can write the following term in the Lagrangian $$i\bar{\psi}\gamma^\mu\partial_\mu\psi = -i(\partial_\mu \bar{\psi})\gamma^\mu\psi$$ by integrating by parts which makes it look like $$\frac{\partial\mathcal{L}}{\partial(\partial_\nu \bar{\psi})} = -i \gamma^\mu \psi$$. In fact, this is how we get the equations of motion for $$\bar{\psi}$$.

• Are you asking why $\frac{\partial\mathcal{L}}{\partial(\partial_\nu \bar{\psi})} = -i \gamma^\mu \psi$? Oct 30, 2020 at 0:25
• How can you integrate by parts without an integral? Oct 30, 2020 at 1:01

1. $$\psi$$ and $$\bar \psi$$ are thought as two independent variables in the Lagrangian.
2. If you write a Lagrangian as $$\mathcal{L}_1 =\bar\psi(...)\psi$$, you should use it to calculate the Noether current or equation of motion. If you have the other one, $$\mathcal{L}_2 =\psi(...)\bar\psi$$, you have to perform the derivatives based on this one.