Monday, 16 July 2018

general relativity - Curvature of Weyl-rescaled metric from curvature of original metric


I'm trying to go from the low-energy effective action in the string-frame to the corresponding action in the Einstein frame. The action in the string frame has the form


S=1(2π)7l8sd10xγ[e2Φ(R[γμν]+4(Φ)2)12|Fp+2|2]


Here γμν is the string-frame metric, and R[γμν] is the usual Einstein curvature for this metric, which appears in the first (dilaton corrected) Einstein-Hilbert like term in the action.



The process is two fold:




  1. Introduce a field ϕ for fluctuations eΦ=gseϕ (Φ in this language is the dilaton).




  2. Weyl rescale the metric gμν=eϕ/2γμν.




The second step ought to yield the following relationship between the scalar curvatures:



R[γμν]=eϕ/2[R[gμν]922ϕ92(ϕ)2]


So my problem right now is to cleanly derive this curvature relationship, which is really a GR problem. One way to go about it is to start from the Christoffel symbols for g and show that they are equal to the Christoffel symbols for γ plus ϕ-dependent correction terms, then go the Riemann curvature tensor, and then finally to the Ricci tensor, in the usual canonical way.


However, I am hoping there's a faster way to to do this (vielbeins maybe? I can't think of a direct application) because the algebra in this standard procedure is quite tedious.


The general problem has a simple statement: how do you relate the curvature of a Weyl rescaled metric to its original curvature?



Answer



Let Ω2=expϕ/2 so that γ=Ω2g. Then we have the standard formula Rγ=Ω2(Rg2(n1)ΔlnΩ(n2)(n1)[lnΩ]2) This is proven in any number of General Relativity texts, but the one in R.M. Wald, General Relativity (1984), is particularly easy to follow and is proved in the form shown here. From above we get Ω=expϕ/4 so that lnΩ=ϕ/4. Furthermore, we have n=10 so that the coefficients of the last two terms are 29 and 89. From the lnΩs we get factors of 1/4 and 1/42 on the second and third terms, respectively. Thus we have 9/2 in front of those terms. Putting it all together, we obtain Rγ=exp(ϕ/2)(Rg92Δϕ92(ϕ)2) as was to be shown.


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