Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spin connection problem

  1. Mar 21, 2013 #1
    In trying to derive the Dirac equation in space-time [itex] (1,-a^{2},-a^{2},-a^{2})[/itex], I have read that the Dirac equation is [itex](i\bar{\gamma}^{\mu}(\partial_{\mu}+\Gamma_{\mu})-m)\psi=0 [/itex] where,

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

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

    With my metric choice [itex]\gamma_{\mu}[/itex] should equal [itex]\frac{3}{2}\frac{\dot{a}}{a}[/itex].

    I don't see how with this [itex]\sum[/itex] term that this is possible, have I made a mistake and can anyone help?
     
  2. jcsd
  3. Mar 21, 2013 #2

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.

    Well, [itex]\gamma_{\mu}[/itex] (or even [itex]\Gamma_{\mu}[/itex]) 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}##.
     
  4. Mar 21, 2013 #3
    Sorry I think I may have made a mistake isn't: [itex]e_{\alpha}^{\mbox{ }\nu}[/itex] equal to [itex](1,1/a,1/a,1/a)[/itex] and so on. I understand these e terms but I can't see how to deal with [itex] \sum ^{\alpha\beta} [/itex] because surely it would give a gamma matrix but it shouldn't be there
     
  5. Mar 21, 2013 #4

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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 [itex] \sum ^{\alpha\beta} [/itex] 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.
     
  6. Mar 21, 2013 #5
    I know that [itex] \Gamma_{\mu}=\frac{3\dot{a}}{2a} [/itex] so how can one get rid of the matrix?
     
  7. Mar 21, 2013 #6

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  8. Mar 21, 2013 #7
    thx I am still struggling with this sum term.
    Surely when [itex]\alpha = 1 [/itex] then [itex] \nu =1 [/itex] then [itex]\beta =1 [/itex].
    and the same for [itex]\beta =[/itex] to 2 and 3.

    So when [itex]\beta= 1 [/itex] wouldn't [itex] \sum ^{11} =0 [/itex] and the same for beta =2,3 thus all these sum terms would become zero?
     
  9. Mar 21, 2013 #8

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  10. Mar 21, 2013 #9
    So [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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Spin connection problem
  1. Connected Spaces (Replies: 5)

  2. Connected Sets (Replies: 7)

  3. Connections and forms (Replies: 12)

Loading...