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

[EnigmaticField]

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?

 

[AndyW]

Thank you for bringing up your concerns about the paper arXiv:1002.3877 and the Higgs mechanism for massive gravity. I understand the importance of verifying and understanding results in scientific literature, and I will do my best to address your concerns.

Firstly, I would like to clarify that the paper is discussing the linearized Fierz-Pauli action, which is valid for small perturbations around the Minkowski vacuum solution. This means that the terms in the action will be expanded to linear order in the perturbation fields, and higher order terms will be neglected. This is a common approach in theoretical physics, as it allows us to simplify complicated equations and focus on the essential features of a theory.

Now, let's address your first concern about how \bar{h}^A{}_B=\partial_B\chi^A+\partial^A\chi_B leads to S_{\rm{FP}}=m^2\int d^4x[(\partial_A\chi^A)^2-(\partial_A\chi^B)(\partial^A\chi_B)]. To understand this, we need to expand the terms in the Fierz-Pauli action to linear order in the perturbation fields. This gives us:

\bar{h}^2=\partial_A\chi^A\partial_B\chi^B

\bar{h}^A{}_B=\partial_B\chi^A+\partial^A\chi_B

Plugging these into the Fierz-Pauli action, we get:

S_{\rm{FP}}=\frac{m^2}{2}\int d^4x\sqrt{-g}(\partial_A\chi^A\partial_B\chi^B-\partial_A\chi^A\partial_B\chi^B-\partial_A\chi^B\partial^A\chi_B)

S_{\rm{FP}}=m^2\int d^4x(\partial_A\chi^A)^2-m^2\int d^4x(\partial_A\chi^B)(\partial^A\chi_B)

As you can see, the term (\partial_A\chi^A)^2 cancels out, leaving us with S_{\rm{FP}}=m^2\int d^4x[(\partial_A\chi^A)^2-(\partial_A