Transformation Properties of
데 시터르 공간의 선형화된 유도 방정식의 해는 tsamis와 woodard가 찾은 해다. 이들은 데 시터르 공간에서 파워 법적 양자중력학의 해를 구하는 데에 성공하였지만, 게이지로 사용되지 않은 배경에서 얻어진 솔루션이 게이지 조건을 지키지 않는다는 문제점이 있었다.
본 논문에서는 데 시터르 군의 효과를 계산하기 위해 compensate 게이지 변환을 사용하여 데 시터르 공간에서 선형화된 유도 방정식의 해가 데 시터르 군에 대한 변환 하에 닫혀있다는 것을 보여준다. 또한, 데 시터르 군의 효과를 계산하는 데 사용되는 compensate 게이지 변환의 효과를 분석한다.
데 시터르 공간에서 선형화된 유도 방정식의 해를 데 시터르 군의 효과로 변환시키기 위해서는 세 가지 단계가 필요하다. 첫 번째 단계는 데 시터르 군의 효과를 계산하는 데 사용되는 compensate 게이지 변환을 계산한다. 두 번째 단계는 게이지 조건을 지키지 않는 솔루션에 대해 compensate 게이지 변환을 적용하여 새로운 솔루션을 얻는다. 세 번째 단계는 새로운 솔루션에 대한 게이지 조건을 다시 적용하여 원래의 솔루션과 동일한 결과를 얻는다.
이 논문에서는 데 시터르 공간에서 선형화된 유도 방정식의 해가 데 시터르 군의 효과 하에 닫혀있다는 것을 보여주고, 데 시터르 군의 효과로 변환시키기 위한 compensate 게이지 변환을 계산한다.
한글 요약 끝
영어 요약 시작:
We show that the set of solutions to the linearized gauge fixed equations of motion is closed under de Sitter transformations. The effect of a de Sitter transformation on an arbitrary solution is calculated explicitly, and it is shown that the solution is transformed back into itself after applying three steps: first, we calculate the effect of the transformation on the solution using a compensating gauge transformation; second, we apply this transformation to the solution and obtain a new solution; third, we reapply the gauge conditions to the new solution and obtain the original solution.
We use a conformal coordinate system to study quantum effects on de Sitter space. This system is only valid for half of the full de Sitter space manifold, so it is not obvious that solutions to the linearized equations will transform into themselves under the de Sitter group.
We calculate the effect of a de Sitter transformation on an arbitrary solution using a compensating gauge transformation. The result shows that the solution is transformed back into itself after applying three steps: first, we calculate the effect of the transformation on the solution; second, we apply this transformation to the solution and obtain a new solution; third, we reapply the gauge conditions to the new solution and obtain the original solution.
We also show that the set of solutions to the linearized gauge fixed equations is closed under de Sitter transformations. This means that any solution can be transformed into another solution using a de Sitter transformation.
The effect of a de Sitter transformation on an arbitrary solution is calculated explicitly, and it is shown that the solution is transformed back into itself after applying three steps: first, we calculate the effect of the transformation on the solution; second, we apply this transformation to the solution and obtain a new solution; third, we reapply the gauge conditions to the new solution and obtain the original solution.
This result will be useful for classifying solutions into irreducible representations of the de Sitter group. The starting point for any such analysis is the transformation law for the solution, which is given by equation (28).
We thank R. P. Woodard for suggesting this problem and for helpful discussions. This work has been supported by the department of energy under contract DOE-AS05-80ER-10713.
영어 요약 끝
Transformation Properties of
arXiv:hep-th/9209121v1 29 Sep 1992VPI-IHEP-92/12Transformation Properties ofLinearized de Sitter Gravity SolutionsGary KleppeDepartment of PhysicsVirginia Polytechnic Instituteand State UniversityBlacksburg VA 24060KLEPPE @ VTINTE . BITNETKLEPPE%VTINTE.DNET@VTSERF.CC.VT.EDUABSTRACT: The effect of de Sitter transformations on Tsamis and Woodard’s solutionsto the linearized gauge fixed equations of motion of quantum gravity in a de Sitter spacebackground is worked out explicitly.
It is shown that these solutions are closed under thetransformations of the de Sitter group. To do this it is necessary to use a compensatinggauge transformation to return the transformed solution to the original gauge.
De Sitter space is the unique maximally symmetric solution to Einstein’s gravitationalequations with a positive cosmological constant. Therefore, it is interesting to study quantumeffects on de Sitter space, as they may provide a solution [1] to the cosmological constantproblem [2].
Of course perturbative quantum gravity is nonrenormalizable, but it can stillbe used as a low-energy effective theory to study infrared effects.Recently the propagator for quantum gravity in a de Sitter space background has beenfound [3], and the linearized gauge-fixed equations of motion solved [4].This has beenaccomplished by using a conformal coordinate system which only covers half of the full deSitter space manifold. However, the de Sitter group, the fundamental set of symmetriesof the de Sitter background, contains transformations which take us out of this restrictedsubmanifold.
For this reason, and also because the gauge which was used is not de Sitterinvariant, it is not obvious that the solutions of the equations of motion will transform intothemselves under the de Sitter group.In this paper we will show that these gauge-fixed solutions still possess de Sitter invari-ance. We will show that since the gauge used is not invariant, it is necessary to combine deSitter transformations with gauge transformations which return the de Sitter transformedsolutions to the chosen gauge.
We work out in detail the de Sitter transformation of anarbitrary solution, and show explicitly that the set of solutions is closed under the de Sittertransformations.D dimensional de Sitter space may be defined as the surface of constant length 1/H fromthe origin of D + 1 dimensional Minkowski space. The D coordinates x0, xi necessary todescribe de Sitter space may then be constructed as functions of the Minkowski embeddingspace coordinates X0, Xi, XD.
We will use the conformal coordinate system of Tsamisand Woodard [3]:u ≡x0 =√X · XH(X0 + XD)(1a)2
xi =XiH(X0 + XD)(1b)The inverse of this mapping isX0 = 1 + H2(xixi −u2)2H2u(2a)Xi = xiHu(2b)XD = 1 −H2(xixi −u2)2H2u(2c)The induced de Sitter space metric is thends2 = −du2 + dxidxiH2u2,(3)hence the name “conformal”.The de Sitter group is the Lorentz group of the D + 1 dimensional embedding space. Itwill be convenient to parametrize it as follows:δX0 = 12(di −ai)Xi + bXD(4a)δXi = 12di(X0 + XD) −12ai(X0 −XD) + cijXj(4b)δXD = bX0 −12(ai + di)Xi(4c)Using (1), it is straightforward to calculate the effect of these transformations on the deSitter coordinates:δu = −bu + Haiuxi(5a)δxi =12H di + 12Hai(u2 −xjxj) + Hajxixj + cijxj −bxi(5b)Note that these rules may be written in a pseudo-covariant formδxµ =12H dµ + cµν −bxµ −12Haµxνxν + Haνxµxν(6)if we take all zero components of a, c, and d to be zero.3
Now we allow the metric to vary about the de Sitter space background:gµν = ˆgµν + κhµν(7)whereˆgµν =1H2u2ηµν(8a)andhµν =1Hu3−D/2χµν(8b)It is not hard to show that the background metric ˆg is invariant under the de Sitter trans-formations (5). The perturbation h will transform as a tensor, so the rescaled quantity χwill transform asδχµν = −χµν,ρ δxρ −(D2 −3) 1uχµν δx0 −χρν δxρ,µ −χρµ δxρ,ν(9)We now wish to investigate the effect of these transformations on the gauge fixed solutionsto the equation of motion for χ.
Following Tsamis and Woodard [4], these solutions are foundby seperating χ into three parts, asχij = ǫija +1D−3δij(−ǫkka + ǫc)(10a)χ0i = ǫib(10b)χ00 = ǫc(10c)The gauge is fixed by setting Fµ = 0, whereFµ = χρµ,ρ −12χρρ,µ −D−22uχ0µ −D−24uδ0µ χρρ(11)This condition may be enforced by settingǫc = ǫib = ǫiia = ǫij,ja= 0. (12)4
This set of conditions is stronger than Fµ = 0, but for solutions to the linearized fieldequations, there is always enough residual gauge invariance to impose (12) after imposingFµ = 0. Then the equation of motion for ǫa isDaǫa ≡∂2 + D(D−2)4u2ǫa = 0(13)This equation is solved byǫija =ZdD−1k(2π)D−1hAij(⃗k) χa(u, ⃗x;⃗k) + c.c.i(14)where the basis functions χa have the formχa(u, ⃗x,⃗k) ≡q12πkuH(1)D−12(ku) exp(i⃗k · ⃗x −i kH + iD4 π)(15)Since (12) is clearly not de Sitter invariant, δχ as given by (9) will not obey the samegauge conditions as the original χ.
It will be necessary to restore (12) by gauge transforma-tions. Thus the full effect of the de Sitter transformation on a gauge fixed soultion will bemade up of three parts: δ1, given by (9); a gauge transformation δ2 to restore Fµ = 0; andanother gauge transformation δ3 to restore (12).We now calculate explicitly the effect of (9) on the ǫ’s.
We findδ1ǫija = −dk2H χij,k + ckl(xkχij,l + δikχjl + δjkχil)+ b(xkχij,k + uχij,0 + (3 −D2 )χij) + Hakh−12u2χij,k + 12xlxlχij,k(16a)−xkxlχij,l −uxkχij,0 + (1 −D2 )xkχij + xiχjk + xjχik −xlδikχjl −xlδjkχiliδ1ǫib = −Hajuχij(16b)δ1ǫc = 0(16c)Consequently, the change in the gauge fixing quantity Fµ isδ1F0 = 0(17a)5
δ1Fi = 2Hakχik(18b)We wish to find a gauge transformation δ2 which cancels these results.Under a gaugetransformation characterized by a parameter eµ(u, ⃗x), the variation of χ isδ2 χµν = −eµ,ν −eν,µ −D−22u (eµδ0ν + eνδ0µ) + 2uηµνe0(19)From (19), we see that if we choose e0 = 0, then χ00 = 0 will be preserved. Then we haveδ2 ǫc = 0(20a)δ2 ǫib = −ei,0 −D−22u ei(20b)δ2 ǫija = −ei,j −ej,i + δijek,k(20c)soδ2F0 = 0(21a)δ2Fi = −Daei(21b)Therefore we need to look for an ei such thatDaei = 2Hakχik(22)A solution isei = HuαjZdD−1k(2π)D−1Aijk2∂∂u + D−22uχa(23)We can add any e′i such that Dae′i = 0.
Let us use this residual gauge freedom to take ǫb tozero. Calculating what we have so far,(δ1 + δ2)ǫib = −(D −1)HαjZdD−1k(2π)D−1Aijk2 χa(24)This can be easily cancelled by a gauge transformation δ3 parametrized bye′i ≡−(D −1)HαjZdD−1k(2π)D−1Aijk2 χa(25)6
We have now completely restored the original gauge conditions. What remains is thestraightforward but tedious calculation of δǫa.
We will not give the details of this calculation,except to note that there will be many factors of ⃗x, which one must get rid of by writingxk exp(i⃗k · ⃗x) = −i ∂∂kk exp(i⃗k · ⃗x)(26)and integrating by parts on kk. This of course generates derivatives of the Hankel functionH(1)(ku), but in the end these are all eliminated and one is left withδǫija =ZdD−1k(2π)D−1 δAij χa(u, ⃗x;⃗k)(27)withδAij = −i2H kkdkAij + cklkk ∂Aij∂kl + cikAjk + cjkAik + bikH + 2 −32D −k · ∂∂kAij+Hak−iD(D−2)kk8k2+ kkHk −ikk2H2 +iD2 + kH ∂∂kk −ikk2∂2∂kl∂kl + ikl∂2∂kk∂klAij(28)+HakiDki2k2 + kiHk + i ∂∂kiAjk −iHai∂Ajl∂kl + HakiDkj2k2 + kjHk + i ∂∂kjAik −iHaj ∂Ail∂klIt is straightforward to verify that this result satisfies the gauge conditionsδAii = kjδAij = 0(29)In conclusion, we have shown that the set of solutions to the linearized gauge fixedequations of motion are closed under de Sitter transformations, and have given the explicittransformation laws for these solutions.
It would be interesting to classify these solutionsinto irreducible representations of the de Sitter group. Equation (28) will be the startingpoint for any such analysis.I thank R. P. Woodard for suggesting this problem, and for helpful discussions.
This workhas been supported by the department of energy under contract DOE-AS05-80ER-10713.7
References1)N. C. Tsamis and R. P. Woodard, “Relaxing the Cosmological Constant”, Floridapreprint UFIFT-92-23.2)S. Weinberg, Rev. Mod.
Phys. 61 (1989) 1.3)N. C. Tsamis and R. P. Woodard, “The Structure of Perturbative Quantum Gravity ona De Sitter Background”, Florida preprint UFIFT-92-14, to appear in Comm.
Math.Phys.4)N. C. Tsamis and R. P. Woodard, “Mode Analysis and Ward Identities for PerturbativeQuantum Gravity in de Sitter Space”, Florida preprint UFIFT-92-20, to appear inPhys. Lett.
B292, Oct. 15 1992 issue.8
출처: arXiv:9209.121 • 원문 보기