# Neglecting terms in a Lagrangian

1. Jan 27, 2012

### 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.

Last edited: Jan 27, 2012
2. Jan 27, 2012

### Kyleric

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

3. Jan 27, 2012

### Polyrhythmic

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

4. Jan 27, 2012

### 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 mq → 0.

5. Jan 27, 2012

### Kyleric

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

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.

6. Jan 27, 2012

### 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