Neglecting terms in a Lagrangian

[Kyleric]

Say we have a Lagrangian $$\mathcal{L}=\bar{u}i\kern+0.15em /\kern-0.65em Du+\bar{d}i\kern+0.15em /\kern-0.65em Dd-m_u\bar{u}u-m_d\bar{d}d,$$
where u and d are fermions. In Peskin&Schroeder p. 667 it says that if $$m_u$$ and $$m_d$$ are very small, we can neglect the last two terms of the Lagrangian.

I'd like to know a somewhat rigorous reason for this.

 

[Kyleric]

Ok nevermind, it's just from the Dirac equation.

 

[Polyrhythmic]

In physics, quantities which are small compared to others are simply set to zero as an approximation. This is exactly what happens here.

 

[tom.stoer]

The problem is that setting the quark masses to zero changes the symmetry structure of the theory. Therefore for some effects there is no smooth limit to m_q → 0.

 

[Kyleric]

In physics, quantities which are small compared to others are simply set to zero as an approximation. This is exactly what happens here.

True for numbers, but the components of the Lagrangian are operators.

The problem is that setting the quark masses to zero changes the symmetry structure of the theory. Therefore for some effects there is no smooth limit to m_q → 0.

Yes that's what I was thinking also. But they took the approximation as a step towards a discussion about spontaneous symmetry breaking in QCD, so it was very well motivated.

 

[tom.stoer]

In QCD the chiral symmetry breaking results in massless Goldstone bosons (pions) and non-zero quark masses add small explicit symmetry breaking terms; what you get is something like

$$m_\pi^2 = (m_u + m_d)\frac{M^2}{f_\pi}$$

where M is related to the quark condensate which acts as order parameter