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A The Fierz-Pauli action takes the form of the Maxwell action

  1. Apr 27, 2016 #1
    I read in a paper arXiv:1002.3877 a Higgs mechanism for massive gravity can be constructed by introducing 4 scalar fields (they play the role of the Higgs fields) [itex]\phi^A[/itex] with [itex]A=0,1,2,3[/itex] and then considering the Fierz-Pauli action [itex]S_{\rm{FP}}=\frac{m^2}{2}\int d^4x\sqrt{-g}(\bar{h}^2-\bar{h}^A{}_B\bar{h}^B{}_A)[/itex], where [itex]\bar{h}^{AB} =H^{AB}-\eta^{AB}[/itex], with [itex]H^{AB}=g^{\mu\nu}\partial_\mu\phi^A\partial_\nu\phi^B[/itex] and [itex]\eta_{AB}[/itex] is the Minkowski metric, with respect to which the indices [itex]A,B[/itex] are raised and lowered, and [itex]\bar{h}=\bar{h}^{AB}\eta_{AB}=\bar{h}^A{}_A[/itex].

    The paper says the way to see how, to the linear approximation in field perturbation, four degrees of freedom for the scalar fields could disappear giving only three extra degrees of freedom to the graviton is to figure out that the Fierz-Pauli action takes the form of the Maxwell action in the linear perturbation. When the fields are expanded around the vacuum Minkowski solution [itex]<g_{\mu\nu}>=\eta_{\mu\nu}, <\phi^A>=x^A[/itex], we have [itex]\phi^A=x^A+\chi^A, g^{\mu\nu}=\eta^{\mu\nu}+h^{\mu\nu}[/itex]. Setting [itex]h^{\mu\nu}=0[/itex], to the linear order in perturbation [itex]\chi^A[/itex] we have [itex]\bar{h}^A{}_B=\partial_B\chi^A+\partial^A\chi_B[/itex], which gives [itex]S_{\rm{FP}}=m^2\int d^4x[(\partial_A\chi^A)^2-(\partial_A\chi^B)(\partial^A\chi_B)][/itex], which the paper says takes the form of the Maxwell action for the 4-vecror potential [itex]\chi^A[/itex].

    My problems are that I can't see how [itex]\bar{h}^A{}_B=\partial_B\chi^A+\partial^A\chi_B[/itex] gives [itex]S_{\rm{FP}}=m^2\int d^4x[(\partial_A\chi^A)^2-(\partial_A\chi^B)(\partial^A\chi_B)][/itex] and that how the latter takes the form of the Maxwell action for the 4-vecror potential [itex]\chi^A[/itex]. I think the two problems lies in a common point: that the above two results hold entails [itex](\partial_A\chi^A)^2-(\partial_B\chi^A)(\partial_A\chi^B)=0[/itex], but I can't derive it and don't think that's the case generically. For the first problem, I think [tex]\frac{1}{2}[\bar{h}^2-\bar{h}^A{}_B\bar{h}^B{}_A] \\
    =[(\partial_A\chi^A)^2-(\partial_A\chi^B)(\partial^A\chi_B)]+[(\partial_A\chi^A)^2-(\partial_B\chi^A)(\partial_A\chi^B)].[/tex] Thus I think what the paper claims holds iff what in the second square bracket in the second line above vanishes, that is, [itex](\partial_A\chi^A)^2-(\partial_B\chi^A)(\partial_A\chi^B)=0[/itex]. For the second problem, I think the Maxwell action with the vector potential [itex]\chi^A[/itex] is [tex]\int d^4xF_{AB}F^{AB}=\int d^4x(\partial_A\chi_B-\partial_B\chi_A)(\partial ^A\chi^B-\partial ^B\chi^A) \\
    =\int d^4x 2[(\partial_A\chi_B)(\partial ^A\chi^B)-(\partial_A\chi_B)(\partial ^B\chi^A)] \\
    =\int d^4x 2[(\partial_A\chi^B)(\partial^A\chi_B)-(\partial_B\chi^A)(\partial_A\chi^B)],[/tex] which I think is equal to [itex]-2\int d^4x[(\partial_A\chi^A)^2-(\partial_A\chi^B)(\partial^A\chi_B)][/itex] iff [itex](\partial_A\chi^A)^2-(\partial_B\chi^A)(\partial_A\chi^B)=0[/itex].

    I think it's impossible that [itex](\partial_A\chi^A)^2-(\partial_B\chi^A)(\partial_A\chi^B)=0[/itex] holds generically. So how are the two results arrived at?
     
    Last edited: Apr 27, 2016
  2. jcsd
  3. May 2, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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