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Curved Dirac equation, Spin connection

  1. Mar 22, 2013 #1
    [itex](1,a^2,a^2,a^2)[/itex]) from the action; [itex] \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]
    [/itex] I can show that, [itex]i\bar{\gamma}^{\mu}D_{\mu}\psi - (m_{\psi}+g\phi)\psi =0 [/itex] by varying the action. I know that [itex]D_{\mu}=\partial_{\mu}+\frac{1}{4}\gamma_{\alpha\beta}\omega^{\alpha\beta}_{\mu}[/itex] and I know how to relate [itex]\bar{\gamma}^{\mu}[/itex]to the flat space-time gamma matrices [itex]\gamma[/itex], I am just stuck trying to prove that [itex]\frac{1}{4}\gamma_{\alpha\beta}\omega^{\alpha\beta}_{\mu}=\frac{3}{2}\frac{\dot{a}}{a}[/itex] I think this term is equal to [itex] \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)[/itex] but if it is I can't get the above result, can anyone help?
  2. jcsd
  3. Mar 22, 2013 #2


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    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

    $$\frac{1}{4}\left( \gamma^{\alpha}\gamma^{\beta} - \gamma^{\beta}\gamma^{\alpha}\right) \left( e_{\alpha}^{\nu}(\frac{\partial}{\partial x^{\mu}})e_{\beta\nu}+e_{\alpha\nu}e_{\beta}^{\sigma}\Gamma^{\nu}_{\sigma \mu}\right).$$
  4. Mar 23, 2013 #3
    I didn't wamt to post a long answer but here goes,

    So [itex]\Gamma^{0}_{ij}=\dot{a}a[/itex] and [itex]\Gamma^{i}_{0j}=\frac{\dot{a}}{a}[/itex]
    When [itex]\alpha[/itex] or [itex]\beta[/itex] equals zero then [itex]\gamma^{0}\gamma^{\rho}-\gamma^{\rho}\gamma^{0}=0[/itex] so this isn't allowed.

    If [itex]\alpha=\beta[/itex] then the gamma matrices also go to zero, so [itex]\alpha \neq \beta [/itex] to contribute.
    So if [itex]\nu\neq\alpha[/itex]or[itex]\beta[/itex] the [itex]e^{\nu}_{\alpha} [/itex]or [itex]e_{\beta\nu}[/itex] will equal zero or so this term can immediately be discarded as it will always equal zero since [itex]\alpha \neq \beta [/itex]. [itex]\Gamma^{\nu}_{\sigma\mu}[/itex] will only be non zero if [itex]\nu=0[/itex] but then [itex]\alpha=0[/itex] so this is not possible or [itex]\sigma =0[/itex] and [itex]\nu=\mu[/itex] but then [itex]\beta[/itex] 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: [itex]( e_{\alpha}^{\nu}(\frac{\partial}{\partial x^{\mu}})e_{\beta\nu}+e_{\alpha\nu}e_{\beta}^{\sigma}\Gamma^{\nu}_{\sigma\mu}) [/itex] otherwise I can't see why this would be nonzero, unless maybe i have made a mistake with the gamma matrices
  5. Mar 23, 2013 #4


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    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.
  6. Mar 23, 2013 #5
    Am I correct in thinking that the beta's and alpha's are linked?
  7. Mar 23, 2013 #6


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    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.
  8. Mar 23, 2013 #7
    I have almost cracked it, I think it should be should there also be
    [itex] \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})[/itex]

    The trouble is I get [itex]-\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}) [/itex] (and another identical version for beta = 0.

    Which when combined gives [itex]\frac{3}{2}(\frac{\dot{a}}{a}) [/itex] It is very close to the expected result of [itex]\frac{3}{2}[\gamma^{0}](\frac{\dot{a}}{a}) [/itex] but I can't see any obvius mistake, The only way to get a [itex] \gamma^0[/itex] 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?
    Last edited: Mar 23, 2013
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