Master's Thesis in Theoretical Physics - Universiteit Utrecht

Note that we have succeeded in removing the coupling of the Higgs field to gravity, but anadditional term has appeared. Let us simplify this term by making use of ¯∇ λ Ω = ∂ λ Ω, sinceΩ only contains a scalar field. Then we get[ ]−3MP√ 2 −ḡ ¯□lnΩ + ḡ ρλ (∂ ρ lnΩ)(∂ λ lnΩ)[ ( )= −3M 2 √P −ḡḡρλ 1¯∇ ρΩ ∂ λΩ + ( 1 Ω ∂ ρΩ)( 1 ]Ω ∂ λΩ)[= −3M 2 √P −ḡḡρλ− 1 Ω 2 ∂ ρ∂ λ Ω + 1 Ω ¯∇ ρ ∂ λ Ω + 1 ]Ω 2 ∂ ρΩ∂ λ Ω( √−ḡḡ= 3M 2 ¯∇ ρλ 1 )P ρ ∂ λ ΩΩ= −3 1 Ω 2 M2 P√−ḡḡ ρλ ∂ ρ Ω∂ λ Ω, (4.56)where in the last step I have used metric compatibility ¯∇ ρ ḡ µν = 0. Now I use the specificform of Ω 2 from Eq. (4.54) and I find−3 1 Ω 2 M2 P√−ḡḡ ρλ ∂ ρ Ω∂ λ Ω = −3 1 ξ 2 φ 2 √−ḡḡ ρλΩ 4 ∂ ρ φ∂ λ φ. (4.57)We see that this combines neatly with the kinetic term in the action (4.55) and we getS =∫d 4 x √ { 1−ḡ2 M2 ¯R P− 1 M 2 P Ω2 + 6ξ 2 φ 22 M 2 ḡ µν ∂ µ φ∂ ν φ − Ω −4 λP Ω4 4 (φ2 − v 2 ) 2} .We see that we have removed the coupling to gravity by doing the conformal transformation,but as a consequence we created a non-trivial kinetic term. We can actually write this againas a canonical kinetic term by defining a new field χ(φ), such that we haveM 2 P( ) dχ 2ḡ µν ∂ µ φ∂ ν φ. (4.58)− 1 2 ḡµν ∂ µ χ(φ)∂ ν χ(φ) = − 1 2dφWhen√ √√√dχdφ = M 2 P Ω2 + 6ξ 2 φ 2M 2 , (4.59)P Ω4we retrieve our non-trivial kinetic term. In terms of the new field χ the action becomes∫S = d 4 x √ 1−ḡ{2 M2 ¯R P− 1 2 ḡµν ∂ µ χ∂ ν χ − Ω −4 λ 4 (φ(χ)2 − v 2 ) 2} .We could solve Eq. (4.59) for the new field χ in the case where ξ < 0. However, it is moreinsightful to solve the differential equation in two different regimes. First consider theregime where the field φ is small, i.e.φ ≪ M P√−ξ(small field approximation). (4.60)Take a typical value ξ = −10 4 , then we can safely say that the small field approximation isvalid when the Higgs field φ < 10 −2 M P . So this approximation works well up to an energyscale of ∼ 10 16 GeV. In the small field approximation, Ω 2 ≃ 1 anddχ≃ 1,dφχ ≃ φ. (4.61)