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Spinors in curved space time

  1. Jan 15, 2013 #1
    I have to compute the square of the Dirac operator, D=γaeμaDμ , in curved space time (DμΨ=∂μΨ+AabμΣab is the covariant derivative of the spinor field and Σab the Lorentz generators involving gamma matrices). Dirac equation for the massless fermion is γaeμaDμΨ=0. In particular I have to show that Dirac spinors obey the following equation:
    (−DμDμ+14R)Ψ=0 (1)
    where R is (I guess) the Ricci scalar. Appling to the Dirac eq, the operator γνDν and decomposing the product γμγν in symmetric and antisymmetric part I found:
    DμDμΨ+14[γμν][Dμ,Dν]Ψ=0
    Now I have troubles to show that this last object is related with the Ricci scalar. Can somebody help me or suggest me the right way to solve Eq. (1)?
     
  2. jcsd
  3. Jan 15, 2013 #2
    In the last Equation is 1/4 and not 14.
     
  4. Jan 15, 2013 #3

    dextercioby

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    The commutator of <curved> covariant derivatives is proportional to a term involving the curvature tensor (in terms of vierbeins) and to another term involving the torsion tensor (again written in terms of vierbeins). This is a standard result which you should check by yourself or look it up in a book. You should really compute this term:

    [tex] [\gamma^{\mu},\gamma^{\nu}][D_{\mu},D_{\nu}] \Psi [/tex]
     
  5. Jan 16, 2013 #4
    Thanks. The problem is that I'm not able to compute that term. I guess someting like:

    μρ] Rμρ ψ

    Moreover I do not know if this is the right way to solve Equation (1).
     
  6. Jan 17, 2013 #5

    dextercioby

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    Try Nakahara's book for a useful reference on the vierbein - spin connection formalism. We can't perform the calculations for you.
     
  7. Jan 18, 2013 #6
    @gauge86:you might be aware of that covariant derivatives don't commute.In GR you might have seen that this difference amounts to curvature tensor.So that commutator must likely to be ricci curvature tensor or something equivalent.
     
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