# Non-symmetry of a lagrangian

If a transformation $\Phi \rightarrow \Phi + \alpha \partial \Phi/ \partial \alpha$ is not a symmetry of the Lagrangian, then the Noether current is no longer conserved, but rather $\partial_{\mu}J^{\mu} = \partial L/ \partial \alpha$. This result shows that for a massive Dirac fermion the conservation of the chiral current is softly broken by the mass term: $\partial_{\mu} j^{5,\mu} = -2m\bar{\psi} \gamma^5 \psi$

I am just trying to understand these two sentences in my notes. If

$$\delta L = \frac{\partial L}{\partial \Phi} \delta \Phi + \frac{\partial L}{\partial(\partial_{\mu} \Phi)} \delta (\partial_{\mu}\Phi)$$

and I understand $\Phi \rightarrow \Phi + \alpha \partial \Phi/ \partial \alpha$ to not be a symmetry of the Lagrangian to mean that when this transformation is imposed on the fields the lagrangian is not invariant, i.e $L \rightarrow L' \neq L$

Can rewrite the equation above as $$\frac{\partial L}{\partial \Phi} \delta \Phi + \frac{\partial L}{\partial(\partial_{\mu} \Phi)} \partial_{\mu} (\delta \Phi) = \alpha \left( \frac{\partial L}{\partial \Phi} + \frac{\partial L}{\partial( \partial_{\mu}\Phi)} \partial_{\mu} \right) \frac{\partial \Phi}{\partial \alpha} = \alpha \left( \frac{\partial L}{\partial \alpha} + \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \partial_{\mu} \frac{\partial \Phi}{\partial \alpha}\right)$$

But it doesn't state whether the piece of the lagrangian after the transformation not making it invariant is a total derivative. If it was, then I could use the equations of motion since the action wouldn't change.

Many thanks

• Comments to the question (v1): 1. Which notes? 2. The very last equality in the question fails to properly account for the $\alpha$-dependence of the Lagrangian density $L$ via field-derivatives $\partial_{\mu} \Phi$. The last right-hand side should be just $\alpha \frac{dL}{d\alpha}$. – Qmechanic Oct 30 '15 at 10:29
• @Qmechanic: Did you mean that I failed to include the term that arises because of the dependence of $\alpha$ on $x$? So there is an additional term $$(\partial_{\mu} \alpha) \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \frac{\partial \Phi}{\partial \alpha}$$ With this correction, I still don't see how to proceed. Thanks. – CAF Oct 30 '15 at 11:14
• No, you probably implicitly assume that $\alpha$ is independent of $x^{\mu}$. – Qmechanic Oct 30 '15 at 12:34
• @Qmechanic: the notes are hand written from my lectures. – CAF Oct 30 '15 at 16:31