Is the Spin Connection Problem Affecting My Derivation of the Dirac Equation?

[pleasehelpmeno]

In trying to derive the Dirac equation in space-time $(1,-a^{2},-a^{2},-a^{2})$, I have read that the Dirac equation is $(i\bar{\gamma}^{\mu}(\partial_{\mu}+\Gamma_{\mu})-m)\psi=0$ where,

$\Gamma_{\mu}=\frac{1}{2}\sum ^{\alpha \beta}e_{\alpha}^{\mbox{ }\nu}(\frac{\partial}{\partial x^{\mu}}e_{\beta \nu})$

Is it correct that $e_{\beta\nu}$ is equal to $(1,a^{2},a^{2},a^{2})$, $e_{\alpha}^{\mbox{ }\nu}$ equal to $(1,1/a^{2},1/a^{2},1/a^{2})$ and finally $\sum^{\alpha\beta}=\frac{1}{4}(\gamma^{\alpha}\gamma^{\beta} -\gamma^{\beta} \gamma^{\alpha})$?

With my metric choice $\gamma_{\mu}$ should equal $\frac{3}{2}\frac{\dot{a}}{a}$.

I don't see how with this $\sum$ term that this is possible, have I made a mistake and can anyone help?

 

[fzero]

In trying to derive the Dirac equation in space-time $(1,-a^{2},-a^{2},-a^{2})$, I have read that the Dirac equation is $(i\bar{\gamma}^{\mu}(\partial_{\mu}+\Gamma_{\mu})-m)\psi=0$ where,

$\Gamma_{\mu}=\frac{1}{2}\sum ^{\alpha \beta}e_{\alpha}^{\mbox{ }\nu}(\frac{\partial}{\partial x^{\mu}}e_{\beta \nu})$


Is it correct that $e_{\beta\nu}$ is equal to $(1,a^{2},a^{2},a^{2})$, $e_{\alpha}^{\mbox{ }\nu}$ equal to $(1,1/a^{2},1/a^{2},1/a^{2})$ and finally $\sum^{\alpha\beta}=\frac{1}{4}(\gamma^{\alpha}\gamma^{\beta} -\gamma^{\beta} \gamma^{\alpha})$?

These are both matrices, so it is more accurate to say that

$$e_{\beta\nu} = \mathrm{diag} (1,a^{2},a^{2},a^{2}),$$
$$e_{\alpha}^{\mbox{ }\nu}=\mathrm{diag} (1,1/a^{2},1/a^{2},1/a^{2}).$$

Note that if these were not diagonal, then we would have to be careful about which indices labeled the rows vs columns.

With my metric choice $\gamma_{\mu}$ should equal $\frac{3}{2}\frac{\dot{a}}{a}$.

I don't see how with this $\sum$ term that this is possible, have I made a mistake and can anyone help?

Well, $\gamma_{\mu}$ (or even $\Gamma_{\mu}$) is a collection of 4x4 matrices, so it cannot be equal to a scalar. However, it would appear that

$$e_{\alpha}^{\mbox{ }\nu}\left(\frac{\partial}{\partial t}e_{\beta \nu}\right) = 6\frac{\dot{a}}{a} \mathrm{diag} (0,1,1,1),$$

where this is a matrix in the frame indices $\alpha,\beta$. You can no doubt work out the other components and then multiply against $\Sigma^{\alpha\beta}$.

 

[pleasehelpmeno]

Sorry I think I may have made a mistake isn't: $e_{\alpha}^{\mbox{ }\nu}$ equal to $(1,1/a,1/a,1/a)$ and so on. I understand these e terms but I can't see how to deal with $\sum ^{\alpha\beta}$ because surely it would give a gamma matrix but it shouldn't be there

 

[fzero]

Sorry I think I may have made a mistake isn't: $e_{\alpha}^{\mbox{ }\nu}$ equal to $(1,1/a,1/a,1/a)$ and so on. I understand these e terms but I can't see how to deal with $\sum ^{\alpha\beta}$ because surely it would give a gamma matrix but it shouldn't be there

You're right about the powers of $a$, I missed that. So

$$e_{\alpha}^{\mbox{ }\nu}\left(\frac{\partial}{\partial t}e_{\beta \nu}\right) = 3\frac{\dot{a}}{a} \mathrm{diag} (0,1,1,1),$$

The components of $\sum ^{\alpha\beta}$ and $\Gamma_\mu$ are 4x4 matrices in spinor space, acting on the spinor $\psi$. I'm not sure what it is that you're implying.

 

[pleasehelpmeno]

I know that $\Gamma_{\mu}=\frac{3\dot{a}}{2a}$ so how can one get rid of the matrix?

 

[fzero]

I know that $\Gamma_{\mu}=\frac{3\dot{a}}{2a}$ so how can one get rid of the matrix?

I'm not sure what you mean by "get rid of the matrix," $\Gamma_\mu$ is a matrix in the spinor space. What you've written down is possibly the value of the nonzero components, but we can't change the fact that it's a matrix.

 

[pleasehelpmeno]

thx I am still struggling with this sum term.
Surely when $\alpha = 1$ then $\nu =1$ then $\beta =1$.
and the same for $\beta =$ to 2 and 3.

So when $\beta= 1$ wouldn't $\sum ^{11} =0$ and the same for beta =2,3 thus all these sum terms would become zero?

 

[fzero]

thx I am still struggling with this sum term.
Surely when $\alpha = 1$ then $\nu =1$ then $\beta =1$.
and the same for $\beta =$ to 2 and 3.

So when $\beta= 1$ wouldn't $\sum ^{11} =0$ and the same for beta =2,3 thus all these sum terms would become zero?

I should have looked it up earlier, but you appear to be missing a term in your spin connection, namely

$$\Gamma_{\mu}=\frac{1}{2}\sum ^{\alpha \beta} \left( e_{\alpha}^{\mbox{ }\nu}(\frac{\partial}{\partial x^{\mu}}e_{\beta \nu}) + {e_\alpha}_\nu {e_b}^\sigma {\Gamma^\nu}_{\sigma\mu}\right). $$

The first term might vanish by symmetry arguments (I could have also made a mistake), but presumably the 2nd term does not.

 

[pleasehelpmeno]

So $\frac{1}{8}( \gamma^{\alpha}\gamma^{\beta} - \gamma^{\beta}\gamma^{\alpha}) ( e_{\alpha}^{\nu}(\frac{\partial}{\partial x^{\mu}})e_{\beta\nu}+e_{\alpha\nu}e_{\beta}^{\sigma}\Gamma^{\nu}_{\sigma\mu})$