Curved Dirac equation, Spin connection

[pleasehelpmeno]

(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]<br /> 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?

 

[fzero]

(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]<br /> 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?

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

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

 

[pleasehelpmeno]

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\alphaor\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

 

[fzero]

\gamma^{0}\gamma^{\rho}-\gamma^{\rho}\gamma^{0}=0 so this isn't allowed.

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.

 

[pleasehelpmeno]

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

 

[fzero]

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

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.

 

[pleasehelpmeno]

I have almost cracked it, I think it should be should there also be
\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} &amp; 0\\ 0&amp;\sigma^{1}<br /> \end{pmatrix}+\gamma^{2}\begin{pmatrix} \sigma^{2} &amp; 0\\ 0&amp;\sigma^{2}<br /> \end{pmatrix}+\gamma^{3}\begin{pmatrix} \sigma^{3} &amp; 0\\ 0&amp;\sigma^{3}<br /> \end{pmatrix}](a\frac{\dot{a}}{a}) (and another identical version for beta = 0.

Which when combined gives \frac{3}{2}<i>(\frac{\dot{a}}{a}) </i> 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 \mu=0[\itex] but that isn&#039;t allowed because then whole term would equal zero anyway? Can you think of my mistake or how to get this gamma zero term?