Curved Dirac equation, Spin connection

1. Mar 22, 2013

$(1,a^2,a^2,a^2)$) from the action; $\mathcal{S}_{D}[\phi,\psi,e^{\alpha}_{\mu}] = \int d^4 x \det(e^{\alpha}_{\mu}) \left[ \mathcal{L}_{KG} + i\bar{\psi}\bar{\gamma}^{\mu}D_{\mu}\psi - (m_{\psi} + g\phi)\bar{\psi}\psi \right]$ I can show that, $i\bar{\gamma}^{\mu}D_{\mu}\psi - (m_{\psi}+g\phi)\psi =0$ by varying the action. I know that $D_{\mu}=\partial_{\mu}+\frac{1}{4}\gamma_{\alpha\beta}\omega^{\alpha\beta}_{\mu}$ and I know how to relate $\bar{\gamma}^{\mu}$to the flat space-time gamma matrices $\gamma$, I am just stuck trying to prove that $\frac{1}{4}\gamma_{\alpha\beta}\omega^{\alpha\beta}_{\mu}=\frac{3}{2}\frac{\dot{a}}{a}$ I think this term is equal to $\frac{1}{4}\left( \gamma^{\alpha}\gamma^{\beta} - \gamma^{\beta}\gamma^{\alpha}\right) \left( e_{\alpha}^{\nu}(\frac{\partial}{\partial x^{\mu}})e_{\beta\nu}+e_{\alpha\nu}e_{\beta}^{\sigma}\Gamma^{\nu}_{\sigma \mu}\right)$ but if it is I can't get the above result, can anyone help?

2. Mar 22, 2013

fzero

This is a repost of the question you asked in https://www.physicsforums.com/showthread.php?p=4316692#post4316692. You are still somehow under the impression that a matrix should equal a scalar. You also seem to want people to do algebra for you, since you haven't bothered to show that you've done any additional work on the calculation since I tried to help you.

Please tell us what result you get when you try to compute

$$\frac{1}{4}\left( \gamma^{\alpha}\gamma^{\beta} - \gamma^{\beta}\gamma^{\alpha}\right) \left( e_{\alpha}^{\nu}(\frac{\partial}{\partial x^{\mu}})e_{\beta\nu}+e_{\alpha\nu}e_{\beta}^{\sigma}\Gamma^{\nu}_{\sigma \mu}\right).$$

3. Mar 23, 2013

I didn't wamt to post a long answer but here goes,

So $\Gamma^{0}_{ij}=\dot{a}a$ and $\Gamma^{i}_{0j}=\frac{\dot{a}}{a}$
$e^{\nu}_{\alpha}=(1,1/a,1/a,1/a)$
$e_{\beta\nu}=(1,a,a,a)$
When $\alpha$ or $\beta$ equals zero then $\gamma^{0}\gamma^{\rho}-\gamma^{\rho}\gamma^{0}=0$ so this isn't allowed.

If $\alpha=\beta$ then the gamma matrices also go to zero, so $\alpha \neq \beta$ to contribute.
So if $\nu\neq\alpha$or$\beta$ the $e^{\nu}_{\alpha}$or $e_{\beta\nu}$ will equal zero or so this term can immediately be discarded as it will always equal zero since $\alpha \neq \beta$. $\Gamma^{\nu}_{\sigma\mu}$ will only be non zero if $\nu=0$ but then $\alpha=0$ so this is not possible or $\sigma =0$ and $\nu=\mu$ but then $\beta$ is zero so this isn't possible. Thus ther eis no contribution to the component obviusly wrong. Since the tetrads are added together can they be relabelled so: $( e_{\alpha}^{\nu}(\frac{\partial}{\partial x^{\mu}})e_{\beta\nu}+e_{\alpha\nu}e_{\beta}^{\sigma}\Gamma^{\nu}_{\sigma\mu})$ otherwise I can't see why this would be nonzero, unless maybe i have made a mistake with the gamma matrices

4. Mar 23, 2013

fzero

This is not true. For example, in the Dirac basis

$$\gamma^0 = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix} ,~~~~\gamma^i \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix},$$

you find

$$[\gamma^0,\gamma^i] = -2 \begin{pmatrix} \sigma^i & 0 \\ 0 & \sigma^i \end{pmatrix}.$$

I think that the ${\Gamma^0}_{ij}$ term is the only one that contributes, but you should check. I agree that this is fairly tricky to sort out.

5. Mar 23, 2013

Am I correct in thinking that the beta's and alpha's are linked?

6. Mar 23, 2013

fzero

They do not have to be equal. For example, $e_{00} {e_1}^1 {\Gamma^0}_{11}$ seems to be a non-vanishing contribution to the spin connection.

7. Mar 23, 2013

$\frac{1}{8}( \gamma^{\alpha}\gamma^{\beta} - \gamma^{\beta}\gamma^{\alpha}) ( e_{\alpha}^{\nu}(\frac{\partial}{\partial x^{\mu}})e_{\beta\nu}+e_{\alpha\nu}e_{\beta}^{\sigma}\Gamma^{\nu}_{\sigma\mu})$
The trouble is I get $-\frac{1}{4}\frac{1}{a}[\gamma^{1}\begin{pmatrix} \sigma^{1} & 0\\ 0&\sigma^{1} \end{pmatrix}+\gamma^{2}\begin{pmatrix} \sigma^{2} & 0\\ 0&\sigma^{2} \end{pmatrix}+\gamma^{3}\begin{pmatrix} \sigma^{3} & 0\\ 0&\sigma^{3} \end{pmatrix}](a\frac{\dot{a}}{a})$ (and another identical version for beta = 0.
Which when combined gives $\frac{3}{2}(\frac{\dot{a}}{a})$ It is very close to the expected result of $\frac{3}{2}[\gamma^{0}](\frac{\dot{a}}{a})$ but I can't see any obvius mistake, The only way to get a $\gamma^0$ term present is by choosing [itex]\mu=0[\itex] but that isn't allowed because then whole term would equal zero anyway? Can you think of my mistake or how to get this gamma zero term?