Dirac equation in cosmology with scale factor

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pleasehelpmeno
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Hi i am trying to derive the Dirac equation of the form:
[itex][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[/itex] where a is the scale factor for expnasion of the universe.


I understand that the matter action is [itex]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)][/itex] but i don't understand firstly why there is a vierbein and not a [itex]\sqrt{-g}[/itex] term.

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

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

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.
 
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.
 
yeah thanks, i have a method to work on now.
I know that one can relate the spin connection to the gamma matrices by: [itex]\Gamma_{\mu}[/itex] to [itex]\gamma[/itex] by [itex][\Gamma_{\mu},\gamma^{\nu}][/itex] but is this simply a standard commutator relationship or is it something more because wouldn't [itex]\Gamma_{1}\gamma^{1} - \gamma^{1}\Gamma_{1} =0[/itex] for example?