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Non-relativistic limit of Dirac bilinear

  1. Jun 10, 2015 #1
    Hi,
    I'm studying direct detection techniques for dark matter and in almost all the articles I read (e.g.
    Gondolo, P. (1996, May 13). Phenomenological Introduction to Direct Dark Matter Detection. arXiv.org.) the authors say that in the non-relativistic limit the vector and axial currents take the following forms: $$\bar{\nu}\gamma_{\mu}\nu\rightarrow\nu^{\dagger}\nu$$ $$\bar{\nu}\gamma_{\mu}\gamma_5\nu\rightarrow\bar{\nu}\vec{\gamma}\gamma_5\nu$$
    can anyone explain me why?
     
  2. jcsd
  3. Jun 10, 2015 #2

    ChrisVer

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    One fast answer, in the non-relativistic limit, the first two components of the Dirac spinor are large and the last two are small...
    That means that when you take the products of the type: ## \bar{\psi} \gamma_\mu \psi## the ##\gamma_i## which are off-diagonal will couple your small component to the large... while the ##\gamma_0## will couple your large to large and small to small. That's why the ##\bar{\psi} \gamma_i \psi## can be neglected versus the ##\bar{\psi} \gamma_0 \psi = \psi^\dagger \psi##.

    For the ##\gamma_5## case the thing is the opposite because gamma5 reverses them (the gamma0 becomes off-diagonal in block form, while the gamma_i become diagonal).
     
  4. Jun 10, 2015 #3

    ChrisVer

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    You can also have a look in Peskin, Ch4.8 Coulomb potential, where he gives [itex] \bar{u} \gamma_0 u \approx 2m \xi^\dagger \xi[/itex] while he also mentions that the other can be neglected for small momenta... The thing is that again the 1st two go with ##m## for ##p \rightarrow 0## while the other two go with ##p##.
     
  5. Jun 10, 2015 #4
    Thank you so much, I'm having little hard time trying to understand the physics behind the direct detection techniques.
     
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