# Deriving the dirac equation

1. Jan 30, 2013

Hi i am trying to derive the Dirac equation of the form:
$[i\gamma^0 \partial_0 + i\frac{1}{a(t)}\gamma.\nabla +i\frac{3}{2}(\frac{\dot{a}}{a})\gamma^0 - (m+h\phi)]\psi$ where a is the scale factor for expnasion of the universe.

I understand that the matter action is $S=\int d^{4}x e [\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi - V(\phi) + i \bar{\psi}\bar{\gamma}^{\mu}\vec{D}_{\mu}\psi -(m+h\phi)\bar{\psi}\psi)]$ but i don't understand firstly why there is a vierbein and not a $\sqrt{-g}$ term.

I don't really understand why this is the case $D_{\mu}=\frac{1}{4}\bar{\psi}\bar{\gamma}^{\mu} \gamma_{\alpha \beta}\omega^{\alpha \beta}_{\mu}$ and why the arrow above the D is gone.

And lastly I don't understand why $\bar{\gamma}^{i}=\frac{1}{a(t)}\gamma^{i}$

I understand that one needs to vary the action and i can do that bit but I don't understand some of these conversions, thx. I would appareciate any help that anyone can offer in tis challenge.

2. Jan 31, 2013

### Bill_K

To deal with spinors in curved spacetimes (or even just curvilinear coordinates) you need to use a set of basis vectors. This is because the gamma matrices that obey {γμ, γν} = 2gμν aren't constant, so we use instead matrices referred to a basis, in which {γa, γb} = 2ηab.

The covariant derivative is Dμ = ∂μ - (1/4)σabωabμ where σab is the usual Dirac matrix, and ωabμ are the Ricci rotation coefficients associated with the vierbein.

I think the only reason there's an arrow over the D is to remind us that it acts on the spinor to its right.

3. Jan 31, 2013

I know that one can relate the spin connection to the gamma matrices by: $\Gamma_{\mu}$ to $\gamma$ by $[\Gamma_{\mu},\gamma^{\nu}]$ but is this simply a standard commutator relationship or is it something more because wouldn't $\Gamma_{1}\gamma^{1} - \gamma^{1}\Gamma_{1} =0$ for example?