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Neglecting terms in a Lagrangian

  1. Jan 27, 2012 #1
    Say we have a Lagrangian [tex]\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,[/tex]
    where u and d are fermions. In Peskin&Schroeder p. 667 it says that if [tex]m_u[/tex] and [tex]m_d[/tex] 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. jcsd
  3. Jan 27, 2012 #2
    Ok nevermind, it's just from the Dirac equation.
     
  4. Jan 27, 2012 #3
    In physics, quantities which are small compared to others are simply set to zero as an approximation. This is exactly what happens here.
     
  5. Jan 27, 2012 #4

    tom.stoer

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    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.
     
  6. Jan 27, 2012 #5
    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.
     
  7. Jan 27, 2012 #6

    tom.stoer

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    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

    [tex]m_\pi^2 = (m_u + m_d)\frac{M^2}{f_\pi}[/tex]

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