# Non-relativistic limit of Dirac bilinear

1. Jun 10, 2015

### Andrea M.

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. Jun 10, 2015

### ChrisVer

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).

3. Jun 10, 2015

### ChrisVer

You can also have a look in Peskin, Ch4.8 Coulomb potential, where he gives $\bar{u} \gamma_0 u \approx 2m \xi^\dagger \xi$ 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 $p$.

4. Jun 10, 2015

### Andrea M.

Thank you so much, I'm having little hard time trying to understand the physics behind the direct detection techniques.