% %% � � � , , ,, � �� �� e ee @ @@ l l l Q QQ HHPPPXXX hhhh (((( ��� IFT Instituto de F́ısica Teórica Universidade Estadual Paulista DISSERTAÇÃO DE MESTRADO IFT–D.003/15 Application of effective field theory and Grassmann variables to model compact binaries with spin Gilberto Mautner Orientador Riccardo Sturani Coorientador Nathan Berkovits Julho de 2015 ii Agradecimentos Agradeço às instituições que me formaram, inspiraram e capacitaram: ao Colégio Rio Branco da Fundação de Rotarianos de São Paulo, que me ofereceu formação de excelência em condições extremamente favoráveis; ao Colégio I. L. Peretz, que me apresentou pela primeira vez ao mundo fascinante da F́ısica, além de me formar como indiv́ıduo; ao Instituto Tecnológico de Aeronáutica (ITA) e ao Instituto de F́ısica Teórica (IFT-UNESP) por terem me fornecido, gratuitamente, ensino de excelência em graduação e pós-graduação. Espero poder devolver ao Brasil e à sociedade tudo que recebi destas instituições. Agradeço a todos os que me apoiaram nesta inusitada empreitada de concluir um mestrado em F́ısica Teórica após os 40 anos de idade, voltando a estudar, por pura paixão, após quase 20 anos na indústria: à toda minha famı́lia, pela compreensão e pelo suporte; aos meus colegas da Locaweb, por cuidarem com tanto carinho e competência da empresa que ajudei a criar; e à minha amiga Ana Maria Zampieri, por me ter me provado que este projeto era fact́ıvel. Eis o resultado! Agradeço aos professores do IFT e ICTP-SAIFR. Em especial, ao Nathan Berkovits, que me deu todo apoio para ser aceito no programa do Mestrado apesar de condições tão fora do padrão; e ao meu orientador Riccardo Sturani, por ter me transmitido seu conhecimento vasto de uma forma muito dinâmica e didática. Foi um privilégio poder interagir e aprender com vocês. À minha esposa, Lulu: por me dar uma famı́lia linda, pelo seu apoio, pelo seu amor, por você existir, por tudo! iii iv Resumo Começamos por revisar um modo de calcular a energia potencial entre dois corpos sem spin até segunda ordem nas velocidades, através da aplicação de uma Teoria de Campos Efetiva que modela a interação entre eles por meio do campo gravitacional. Estendendo este método para que leve em conta também os graus de liberdade de spin, e descrevendo tais graus de liberdade através de variáveis de Grassmann, calculamos termos adicionais no potencial advindos das interações spin-spin e spin-orbital. Finalmente, calculamos a energia como função da frequência de rotação do sistema binário, incluindo o efeito destes termos relacionados a spin. Tais resultados devem tornar-se cada vez mais relevantes para posśıveis medições em observatórios de ondas gravitacionais tais como LIGO [1] e VIRGO [2]. Palavras-chave: Teoria de Campos Efetiva, Supersimetria, Variáveis de Grassmann, Sistemas Binários v vi Abstract We review how an approximation of the potential between two spin-less bod- ies is calculated, up to second order in the velocities, by employing effective field theory methods to model their interaction though the gravitational field. By extending this methodology to take into account their spin degrees of free- dom and by describing such spin degrees of freedom in terms of Grassmannian variables, we calculate the additional terms in the potential due to spin-spin and spin-orbital interactions. Finally, we calculate the energy as a function of rotational frequency of a binary system, including the effect of these spin related terms. Such results are becoming increasingly relevant in possible measurements by gravitational observatories like LIGO [1] or VIRGO [2]. Keywords: Effective Field Theory, Supersymmetry, Grassmann Variables, Binary Systems vii viii Contents 1 Introduction 1 2 Summary of Effective Field Theory 3 2.1 Conceptual Basis . . . . . . . . . . . . . . . . . . . . . . . . . 3 2.2 A toy model . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3 EFT for two gravitationally interacting spin-less particles 9 3.1 Quadratic field terms in the action . . . . . . . . . . . . . . . 9 3.1.1 φ Green’s function . . . . . . . . . . . . . . . . . . . . 11 3.1.2 ~A Green’s function . . . . . . . . . . . . . . . . . . . . 12 3.1.3 σ Green’s function . . . . . . . . . . . . . . . . . . . . 12 3.2 Source term . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.3 Calculation of potential terms . . . . . . . . . . . . . . . . . . 14 4 The Case of Spinning Particles 19 4.1 Usage of anti-commuting Grassmann variables for spin degrees of freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 4.2 Reparametrization invariant action and new pairings . . . . . 23 5 Energy as a function of frequency 29 5.1 Calculation of EIH corrections . . . . . . . . . . . . . . . . . . 29 5.2 Calculation of corrections related to Spin . . . . . . . . . . . . 33 A Mathematica notebook calculations 37 A.1 φ field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 A.1.1 Einstein Hilbert term of order φ2 . . . . . . . . . . . . 38 A.1.2 Gauge fixing term of order φ2 . . . . . . . . . . . . . . 38 A.2 A field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 A.2.1 Einstein Hilbert term of order A2 . . . . . . . . . . . . 38 A.2.2 Gauge fixing term of order A2 . . . . . . . . . . . . . . 39 A.3 σ field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 ix A.3.1 Einstein Hilbert term of order σ2 . . . . . . . . . . . . 39 A.3.2 Gauge fixing term of order σ2 . . . . . . . . . . . . . . 40 x Chapter 1 Introduction Our goal in this work is to calculate classical quantities associated with a compact binary system, such as potential and total energy. We’ll apply two methods that have been separately developed and put them together to tackle this problem: effective field theories and the application of Grassmann variables to model spin degrees of freedom. We begin by briefly reviewing the effective field theory (EFT) method in chapter 2 and applying it to a toy model of two point particles interacting through a scalar field which will serve to illustrate the basic concept of gener- ating an effective action by “integrating out” the field and revealing potential terms that depend only on the particles’ trajectories. This will also serve as a prototype to the problem involving the real gravitational field. Then, in chapter 3 we model a system of two point particles, still spinless, interacting through a real gravitational field. We employ EFT to integrate out the mediating gravitational field and give corrections to the Newtonian potential energy, which will be calculated here up to order 1PN. The obtained result is, effectively, the Einstein Infeld Hoffmann post-Newtonian potential as a function of particles’ positions and velocities. In chapter 4, section 4.1, we review a technique by which spin degrees of freedom of point particles can be modeled by anticommuting Grassmann variables in the Lagrangian. We show that they seem to describe spin by making rotation transformations and seeing them appear in a sum with or- bital angular momentum that is preserved along rotations. From this result, in section 4.2, we construct a reparametrization invariant action and then employ EFT to a system of two spinning particles whose interaction is me- diated by the gravitational field. With that in hand we are able to generate new pairings which will give rise to new potential energy terms in the ef- fective action related to spin-spin and spin-orbit couplings between the two bodies. 1 Finally, in chapter 5, we use the previously computed results to calculate the energy associated with system of two spinning bodies with all corrections associated both with 1PN corrections and terms coming from the spin-spin and spin-orbit coupling. Then, in the particular case of circular orbits, we show that we can express this energy in the more useful format of a function of the rotational frequency without dependence on coordinates such as the distance between the two bodies. 2 Chapter 2 Summary of Effective Field Theory 2.1 Conceptual Basis Consider the partition function: Z[J ] = ∫ Dφe( i~ ) ∫ d4xL(φ,∂µφ,J) Suppose that the action S = ∫ d4xL(φ, ∂µφ, J) varies by large amounts of ~ over small displacements in φ. Then, the partition function will be dominated by the trajectories near the classical one, which minimizes the action. Further from these trajectories, the exponent oscillates very rapidly and contributions cancel out. So, in the classical limit: lim ~→0 Z[J ] = N e( i~ )S(φcl,J) Now, making ~ = 1 and defining iSeff (J) ≡ ln(Z[J ]) (2.1) We can see that iSeff (J) = iS(φcl, J) + quantum corrections + const. In this way, we have achieved a result which, apart from quantum corrections, gives us the classical action in terms of the sources as the logarithm of the partition function. If the ~ dependent terms are very small, this effectively enables us to use all the techniques developed for path integrals, such as perturbation theory and Feynman diagrams, to extract terms from the action such as the kinetic energy, potential energy etc. 3 2.2 A toy model In this section we’ll apply the concepts just described above to the case of sources whose interaction is mediated by a scalar field with no mass and no self-interaction. That is, S = Sscalar + Sint + Ssource Where Sscalar = ∫ d4x(−1 2 ∂µφ∂µφ) And Sint = ∫ d4xφ(x)J(x) In this case, the partition function can be calculated exactly by Gaussian integration. We have: Z[J ] = eiSsource ∫ Dφ e i ∫ d4x{+ 1 2 φ (∂µ∂ µ)︸ ︷︷ ︸ ≡∆ φ+ iε 2 φ2+φJ} = N eiSsourcee− i 2 J ·∆−1·J where J ·∆−1 · J ≡ ∫∫ d4x1d 4x2J(x1)∆(x1, x2)J(x2) (2.2) And N is a normalization constant. The term + iε 2 φ2 has been added to make the integral converge and will result in the iε term in the denominator below, which implies the time symmetric solution for ∆−1: ∆−1(x1, x2) = ∫ d4k (2π)4 e−ik·(x1−x2) −k2 + iε (2.3) ⇒ ∆∆−1(x1, x2) = ∫ d4k 2π e−ik·(x1−x2) = δ4(x1 − x2) Where k2 ≡ kµkµ, therefore −k2 = (k0)2 − k2. Now we want to model the presence of two point particles. Therefore we want: J(x, t) ∝ m1δ 3(x− x1(t)) +m2δ 3(x− x2(t)) From the scalar term we see that [φ] = L−1 is required to make the action term dimensionless. So we need that [J ] = L−3 to make the interaction term dimensionless as well. Since δ3(x) is already of dimension L−3 we need to 4 offset the dimension of the mass factor. Let’s do that by dividing m by an arbitrary constant Λ of dimension L−1. We choose: Λ ≡ (32πGN)−1/2 (2.4) Where GN is Newton’s gravitational constant. Note that Λ is arbitrary and that’s OK as long as we aren’t trying to create an experimentally proven theory; we’re just playing with a toy model to demonstrate the methodology of effective field theory. So now we define: J(x, t) = m1 Λ δ3(x− x1(t)) + m2 Λ δ3(x− x2(t)) And the source term is: Ssource = −m1 ∫ dτ1 −m2 ∫ dτ2 From (2.1), (2.2) and (2.3) we conclude that: Seff = −i ln(Z[J ]) = Ssource− − i(− i 2 ) m1m2 Λ2 ∫∫ d4x d4x′δ3(x− x1(t))δ3(x′ − x2(t′)) ∫ d4k (2π)4 e−ik·(x−x ′) −k2 + iε − − i(− i 2 ) m2 1 Λ2 ∫∫ d4x d4x′δ3(x− x1(t))δ3(x′ − x1(t′)) ∫ d4k (2π)4 e−ik·(x−x ′) −k2 + iε + + 1↔ 2 It can be seen that the terms proportional to m2 1 ( m2 2 ) are constants with no dependence on the particles’ trajectories, so they can be discarded because they won’t influence the equations of motion. So, apart from these innocuous terms we have: Seff = Ssource − m1m2 Λ2 ∫ dt dt′ ∫ dk0 2π e+i k0(t−t′) ∫ k e−ik·(x1(t)−x2(t′)) (k0)2 − k2 (2.5) Now we make use of the expansion: 1 k2 − (k0)2 = 1 k2 ( 1 + (k0)2 k2 + (k0)4 k4 + . . . ) 5 We shall see below that this expansion can be done as each successive term will generate higher orders in v2. So, in zeroth order in k0: Seff ⊃ ∫ dt m1m2 Λ2 ∫ k e−ik·(x1(t)−x2(t)) k2 = ∫ dt m1m2 Λ2 1 4πr = ∫ dt 8GNm1m2 r Where ∫ k ≡ ∫ d3k (2π)3 And r ≡ |r| ≡ |x1 − x2| This gives the following potential term: V ⊃ −8GNm1m2 r Which, apart from a factor of 8, resembles the Newtonian gravitational po- tential. We don’t need to worry about this extra factor because as said before we’re not trying to deduce that gravity is a scalar field, which we know is not. We are performing this calculation simply to exercise the method of calcu- lating potential terms starting from partition functions in a simpler scenario before we go to the more complicated (and real) gravitational case. We can also calculate the higher order terms in (k0)2. For example, we can calculate the term of first order in (k0)2. To do that, first notice that 1 k2 − (k0)2 = 1 k2 ( 1 + (k0)2 k2 + (k0)4 k4 + . . . ) can be written inside the integral in (2.5) as 1 k2 ( 1 + ∂t∂t′ k2 + (∂t∂t′) 2 k4 + . . . ) Where the time derivatives are applied to e+i k0(t−t′). Then, to first order in (k0)2/k2 we’ll have the term: Seff ⊃ m1m2 Λ2 ∫ dt dt′ ∫ dk0 2π (∂t∂t′e +i k0(t−t′)) ∫ k e−ik·(x1(t)−x2(t′)) k4 6 After integrating by parts both with respect to t and t′: Seff ⊃ m1m2 Λ2 ∫ dt dt′δ(t− t′) ∫ k kiv1i(t)kjv2j(t ′) e−ik·(x1(t)−x2(t′)) k4 = m1m2 Λ2 ∫ dt v1i(t)v2j(t)(−∂ri∂rj) ∫ k e−ik·r(t) k4 (2.6) Here we can see, as stated previously, that each consecutive term in the expansion in (k0)2/k2 generates a higher order term in velocities. We use that ∫ k e−ik·r k2α = r2α−3π−3/22−2αΓ(3/2− α) Γ(α) Which, for α = 2, gives: ∫ k e−ik·r k4 = − r 8π Also, since ∂ri∂rjr = δij r − rirj r3 Then (2.6) becomes: Seff ⊃ 32πGNm1m2 ∫ dt v1iv2j 1 8π ( δij r − rirj r3 ) From which we get the following additional potential terms: V ⊃ 4GNm1m2 ( − v1 · v2 r + rirj r3 vivj ) Although seemingly this potential appears to imply action at a distance (since velocities and positions of the particles are measured at the same time), this actually comes from trading knowledge about the particles’ positions at all times for knowing all higher derivatives in the Taylor expansion and, then, since we’re assuming velocities much smaller than 1, we’re allowing ourselves to treat the problem perturbatively and consider only the first leading orders. 7 8 Chapter 3 EFT for two gravitationally interacting spin-less particles In the previous section we illustrated the usage of EFT to calculate the potential in a simple case involving a scalar field that mediates the interaction between two spin-less particles. Despite some resemblance to the Newtonian potential obtained in the end, we know that gravity is a rank 2 tensor field, not a scalar. In this chapter we will take this into account and calculate the potential terms given by all degrees of freedom of the gravitational field. In order to tackle this more complicated problem we will break it into three parts: First obtain the quadratic field terms of the action in terms of components of gravity taken from a useful ansatz (section 3.1). Second, we will calculate the source terms in the action, which, contrary to the last chapter, now play a role in the potential because of the non-flat metric (section 3.2). Third, calculate the potential terms up to 1PN order by contracting the fields in the composite (field + source) action (section 3.3) 3.1 Quadratic field terms in the action Here we’ll use the following variable redefinition for the metric as prescribed by [4] and is inspired by Kaluza-Klein, although it is used here for a different purpose than adding an extra dimension to unify gravity and electromag- netism, but to create a hierarchy among the gravity component fields in 4 dimensions: gµν = e2φ/Λ ( −1 Aj/Λ Ai/Λ e−cdφ/Λγij − AiAj/Λ2 ) (3.1) 9 with γij ≡ δij + σij/Λ and cd ≡ 2d−1 d−2 . So far we are simply rewriting gµν in terms of new components φ, Aj and σij. The total action is given by: S = SEH + SGF + Spp = Squad + Spp Where SEH refers to the standard Einstein Hilbert action, expanded to quadratic order in the fields, SGF is the harmonic gauge fixing term which we can find, for example, in [6] and Spp refers to the two point particles’ world-lines which will be detailed in the next section. In this section we’ll calculate the sum Squad = SEH + SGF up to quadratic order where SEH = 1 16πGN ∫ dt ddx √ −gR (3.2) SGF = −1 32πGN ∫ dt ddxgµνΓ µΓν (3.3) With Γµ ≡ Γµνρg νρ. The calculations were performed with the help of the Mathematica notebook as explained in Appendix A. We got: Squad = ∫ dt ddx { 4 [ − (~∇φ)2 + φ̇2 ] + [ F 2 ij 2 + (~∇ · ~A)2 − ~̇A 2 ] + [ − 1 4 σ̇2 + 1 4 (~∇σ)2 + 1 2 σ̇2 ij − 1 2 (~∇σij)2 ]} (3.4) Now we calculate the Green’s functions associated with each field in the above action. These Green’s functions arise when fields in two different space-time coordinates are “paired up” such as, for example, in the following expression: GF ( x1(t1); t1,x2(t2); t2 ) ≡ 1 Z[φ, J = 0] ∫ Dφ eiS[φ]φ ( x1(t1), t1 ) φ ( x2(t2), t2 ) = 1 Z[φ, J = 0] { δ iδJ ( x1(t1), t1 ) δ iδJ ( x2(t2), t2 )∣∣∣∣ J=0 Z[φ, J ] } Where Z[φ, J ] ≡ ∫ Dφ ei ( S[φ]+ ∫ dt ddxφ(x,t)J(x,t) ) In the last chapter, we have calculated the partition function for a scalar field. Given: Sscalar = ∫ d4x(−1 2 ∂µφ∂µφ) 10 the partition function can be calculated exactly giving: Z[J ] = ∫ Dφ e i ∫ d4x{+ 1 2 φ (∂µ∂ µ)︸ ︷︷ ︸ ≡∆ φ+φJ} = N e− i 2 J ·∆−1·J where J ·∆−1 · J ≡ ∫∫ d4x1d 4x2J(x1)∆(x1, x2)J(x2) (3.5) And ∆−1(x1, x2) = ∫ d4k (2π)4 e−ik·(x1−x2) −k2 + iε In this case the Green’s function will be: GF ( x1(t1); t1,x2(t2); t2 ) = 1 Z[φ, J = 0] { δ iδJ ( x1(t1), t1 ) δ iδJ ( x2(t2), t2 )∣∣∣∣ J=0 Z[φ, J ] } = 1 N 1 i2 N ( − i∆−1 ( x1(t1); t1,x2(t2); t2 )) =− i ∫ dk0 2π ∫ k e−ik 0(t2−t1)+ik· ( x2(t2)−x1(t1) ) k2 − (k0)2 − iε (3.6) Observe that GF as defined obeys GF = i∆−1 so that it obeys GF∆(x1, x2) = iδ4(x1 − x2) as expected for Green’s functions in Field Theory. Note, also, that we’re simply reusing the results obtained in the last chapter in (2.2) and (2.3) but this time Ssource is not factored out. It will be expanded in powers of the fields and, when summed to the action, will cause the appearance of 2-point and 4-point functions etc. modifying the potential as will become clear in section 3.3. So we follow ahead with the plan and calculate the Green’s functions associated with each field appearing in the quadratic action (3.4). 3.1.1 φ Green’s function The only difference between the quadratic term in φ and the scalar field example of last section is the factor of 4 instead of 1/2. Since the Green’s function is the inverse of the operator, it should be one eighth of the one obtained in (3.6). Then G (φ) F (x, t) = − i 8 ∫ dk0 2π ∫ k e−ik 0t+ik·x k2 − (k0)2 − iε (3.7) 11 3.1.2 ~A Green’s function Take a look at the ~A term in the integrand in the quadratic field action (3.4). It is: [ F 2 ij 2 + (~∇ · ~A)2 − ~̇A 2 ] First note that: F 2 ij 2 = 1 2 (∂iAj − ∂jAi)(∂iAj − ∂jAi) =∂iAj∂iAj − ∂jAi∂jAi =− Ai(δij∇2 − ∂i∂j)Aj Where we integrated by parts in the last step. We also have: (~∇ · ~A)2 = ∂iAi∂jAj = −Ai∂i∂jAj And finally: − ~̇A 2 = + ~A · ∂ 2 ∂t2 ~A = Ai ∂2 ∂t2 δijAj Putting it all together we find that:[ F 2 ij 2 + (~∇ · ~A)2 − ~̇A 2 ] = 1 2 (−2)Ai∂ µ∂µδijAj Here the operator 1 2 ∂µ∂µ is multiplied by a factor of −2 whose inverse is −1/2 and there’s also an outer product with δij whose inverse is δij itself. So we conclude that: G (A) F (x, t)ij = i 2 δij ∫ dk0 2π ∫ k e−ik 0t+ik·x k2 − (k0)2 − iε (3.8) 3.1.3 σ Green’s function After integrating by parts, the quadratic terms in σ present in the action (3.4) can be written in the following form:[ σii ( − 1 4 ∂µ∂µ ) σkk + σij ( 1 2 ∂µ∂µ ) σij ] =σij ( − 1 2 δijδkl )( 1 2 ∂µ∂µ ) σkl + σij ( 1 2 δikδjl + 1 2 δilδjk )( 1 2 ∂µ∂µ ) σkl =σijMijkl ( 1 2 ∂µ∂µ ) σjk 12 With Mijkl ≡ 1 2 δikδjl + 1 2 δilδjk − 1 2 δijδkl To find the inverse operator, we have to multiply the already found inverse of (1/2 ∂µ∂µ) by the inverse of M . Such inverse should by a matrix M−1 ijkl such that M−1 ijklMklmn = 1 2 (δimδjn + δinδjn) The right hand side is the identity matrix with double and symmetric indices ij and mn. It can be checked that the solution is given by: M−1 ijkl = 1 2 ( δikδjl + δilδjk − 2 d− 2 δijδkl ) Where d is the number of spatial dimensions. Therefore the Green’s function for d = 3 is: G (σ) F (x, t)ijkl = −i 2 (δikδjl + δilδjk − 2δijδkl) ∫ dk0 2π ∫ k e−ik 0t+ik·x k2 − (k0)2 − iε (3.9) 3.2 Source term We will now write the source term for each particle which will be summed with the quadratic action term calculated in last section. The expression for a particle’s action is Spp = −m ∫ dτ = −m ∫ √ −gµνdxµdxν = −m ∫ dt √ −g00 − 2g0ivi − gijvivj Using the definition of the metric components given in (3.1) this action be- comes: Spp = −m ∫ dt eφ/Λ √ 1− 2 Ai Λ vi − e−cdφ/Λ(v2 + σij Λ vivj − AiAj Λ2 vivj) What we’re going to do next is expand the integrand in powers of the fields φ, A and σ. In this way, as will become apparent in the next sec- tion, expressions of type ∫ Dφ eiSquadeiSpp can be expanded into ones of type∫ Dφ eiSquadφ(x1)φ(x2) which in turn are equal to the Green’s function cal- culated in the section above. 13 Up to second order we have: ex ≈ 1 + x+ x2 2 √ 1 + x ≈ 1 + x 2 − x2 8 We then get, up to order v2: Spp ≈ −m ∫ dt ( 1 + φ Λ + φ2 2Λ2 ) × × ( 1− Ai Λ vi + 3AiAj 4Λ2 vivj − v2 2 − σij 2Λ vivj + cdφ 2Λ v2 + cdφσij 2Λ2 vivj − c2 dφ 2 4Λ2 v2 ) + . . . = −m ∫ dt [√ 1− v2 + φ Λ ( 1 + 3 2 v2 ) + φ2 2Λ2 ( 1 +O(v2) ) − Ai Λ vi ( 1− 3Aj 4Λ vj ) − σij 2Λ vivj ( 1− cdφ Λ )] (3.10) Where we used cd = 4. Only terms up to order v2 are displayed. We also omitted terms with products of 3 or more fields, such as, for example, φAiAj or φ2σij. This omission is fine because, as will be detailed in the following section, these terms give rise to vanishing or quantum contributions, which can be neglected. 3.3 Calculation of potential terms Now we will find the effective action terms which will be present in the potential. We start from (2.1) substituting the results from (3.10): iSeff = ln ∫ DφDADσ eiSquad { 1 + · · ·+ i2 2 × [ − m1 Λ ∫ dt1 ( φ ( 1 + 3 2 v2 1 ) + φ2 2Λ − Aivi1 ) − m2 Λ ∫ dt2 ( φ ( 1 + 3 2 v2 2 ) + φ2 2Λ − Aivi2 ) + . . . ]2 + i3 6 [ − m1 Λ ∫ dt1(φ(1 + . . . ))− m2 Λ ∫ dt2 ( · · ·+ φ2 2Λ + . . . )]3 + . . . } (3.11) 14 We see that Green’s functions will start to appear when we combine (“pair up”) fields of the same type at different space-time points. For example, taking the term linear in φ in the second line and multiplying by the analog term in the third line will give rise to the following term: eiSeff ⊃ ∫ DφDADσ eiSquad [ 1− m1m2 Λ2 ∫∫ dt1dt2φ ( x1(t1), t1 ) φ ( x2(t2), t2 ) × ( 1 + 3 2 v2 1 )( 1 + 3 2 v2 2 )] = N +N [ − m1m2 Λ2 ∫∫ dt1dt2G (φ) F ( x1(t1)− x2(t2), t1 − t2 ) × ( 1 + 3 2 v2 1 )( 1 + 3 2 v2 2 )] Using what we found for φ Green’s function in (3.7) and approximating ln(1 + x) ≈ 1 + x, we get, apart from an additive constant: Seff ⊃ m1m2 8Λ2 ∫∫ dt1dt2 ∫ dk0 2π ∫ k e−ik 0(t1−t2)+ik· ( x1(t1)−x2(t2) ) k2 − (k0)2 ( 1 + 3 2 v2 1 )( 1 + 3 2 v2 2 ) We now use the same prescription of section 2.2 with 1 k2 − (k0)2 = 1 k2 ( 1 + (k0)2 k2 + (k0)4 k4 + . . . ) = 1 k2 ( 1 + ∂t1∂t2 k2 + (∂t1∂t2)2 k4 + . . . ) Up to order v2 will we get: Seff ⊃ ∫ dt m1m2 8Λ2 [ 1 4πr ( 1 + 3 2 v2 1 + 3 2 v2 2 ) − vi1vi2∂ri∂rj ( − r 8π )] = ∫ dt GNm1m2 r [ 1 + 3 2 v2 1 + 3 2 v2 2 + 1 2 ( v1 · v2 − (v1 · r̂)(v2 · r̂) )] (3.12) Where we have used that vi1v i 2∂ri∂rjr = vi1v j 2 ( δij r − rirj r3 ) And the definition of Λ given in equation (2.4). Notice that in (3.11) we’re hiding a lot of terms. That’s because their pairing won’t generate relevant contributions. These are the criteria accord- ing to which we include pairings: 15 1. We only want terms which contain pairs of the same field. Terms with an odd power of any field will give zero because ∫ d[φ(x)]eiSquad(φ(x))n vanishes for n odd. 2. We only want contributions up to order v2. Because of this, for ex- ample, terms proportional to σij from (3.10) aren’t considered because their pairing is of order v4. 3. We aren’t including quantum contributions. The power of ~ is pro- portional to the number of vertices minus the number of internal lines. Then, for example, pairing φ2 from two different terms will give a power of ~ superior to the other terms. Notice, however, that the cubic term which includes one φ2 and two different φ’s will give a non-quantum contribution because the extra vertex will be compensated by an addi- tional internal line, as we will see below. Leaving aside the contributions excluded by the above considerations, we have the term calculated in (3.12) plus others. The next one will come from the pairing of the A · v terms: eiSeff ⊃ ∫ DφDADσ eiSquad [ 1 + i2 2 × 2 m1m2 Λ2 ∫∫ dt1dt2AiAjv i 1(t1)vj2(t2) ] = N +N { − m1m2 Λ2 ∫∫ dt1dt2G (A) F ( x1(t1)− x2(t2), t1 − t2 ) ij vi1(t1)vj2(t2) } From (3.8) we substitute the Green’s function for the A field. We see that only the order zero expansion in (k0)2 is needed here since the expression is already quadratic in v. Then: iSeff ⊃ − ∫ dt m1m2 Λ2 ∫ dt1dt2 i 2 δij δ(t1 − t2) 1 4πr vi1(t1)vj2(t2) = − ∫ i 2 m1m2 Λ2 1 4πr v1 · v2 From which we obtain: Seff ⊃ − ∫ dt 4GNm1m2 r v1 · v2 (3.13) 16 There is yet another way to combine terms in (3.11) which is to contract two φ’s from the fourth line with the φ2 from the same line: iSeff ⊃ ln ∫ DφDADσeiSquad [ 1 + i3 6 × −3m2 1m2 2Λ4 ∫∫∫ dtA dtB dtC φ(x1, tA)φ(x1, tB)φ2(x2, tC) ] (3.14) Where the factor of 3 in the second line is due to the three different ways to generate the product in the integrand from the cubic term in (3.11). To save space let’s use the following notations: A↔ (x1, tA) B ↔ (x1, tB) C ↔ (x2, tC) φAφBφ 2 C ≡ φ(x1, tA)φ(x1, tB)φ2(x2, tC) δJA ≡ δ δJ(x1, tA) δJB ≡ δ δJ(x1, tB) δJC ≡ δ δJ(x2, tC) J ·GF · J ≡ ∫∫ dt1 dt2 d 3x1 d 3x2 J(x1, t1)GF ( x1(t1)− x2(t2), t1 − t2 ) J(x2, t2) GF · J(A) ≡ ∫ dt ddxGF ( x1(tA)− x(t), tA − t ) J(x, t) GF (A,C) ≡ GF ( x1(tA)− x2(tC), tA − tC ) Then Z[J ] = ∫ DφDADσeiSquad+ ∫ dt ddxφJ = N e− 1 2 J ·GF ·J And∫ DφDADσeiSquadφAφBφ2 C = 1 i4 δJAδJBδ 2 JC Z[J ] ∣∣∣∣ J=0 =N δJAδJBδ 2 JC e− 1 2 J ·GF ·J = N [ δJAδJBδJC ( −GF · J(C) e− 1 2 J ·GF ·J )] =N δJAδJB {[( GF · J(C) )2 −GF (C,C) ] e− 1 2 J ·GF ·J } =N δJA {[ 2 ( GF · J(C) ) GF (B,C) +GF (C,C) ( GF · J(B) ) +O(J2) ] e− 1 2 J ·GF ·J }∣∣∣∣∣ J=0 =N [ 2GF (A,C)GF (B,C) +GF (C,C)GF (A,B) ] 17 Now substituting the result above, together with φ Green’s function ob- tained in (3.7) into (3.14) we have, to order v0: iSeff ⊃ ln { N −N im2 1m2 128Λ4 ∫∫∫ dtA dtB dtC ∫ dk0 2π e−ik 0(tA−tC) ∫ dk0′ 2π e−ik 0′(tB−tC) ∫ k eik· ( x1(tA)−x2(tC) ) k2 ∫ k′ eik ′· ( x1(tB)−x2(tC) ) k′2 } iSeff ⊃ −i m2 1m2 128Λ4 ∫ dt [ ∫ k eik·(x1−x2) k2 ]2 = − i 128 m2 1m2(32πGN)2 × 1 (4πr)2 = −iG 2 Nm 2 1m2 2r2 Including the 1↔ 2 combination we get: Seff ⊃ − ∫ dt G2 Nm1m2(m1 +m2) 2r2 (3.15) Putting together equations (3.12), (3.13) and (3.15) we find the Einstein Infeld Hoffman corrections to the Newtonian potential: VEIH = −GNm1m2 2r [ 3(v2 1+v2 2)−7v1·v2−(v1·r̂)(v2·r̂) ] + G2 Nm1m2(m1 +m2) 2r2 (3.16) Essentially, what we’ve done so far is calculating the O(v2) (1PN) correc- tions to the Newtonian potential using field theoretic methods. We will now extend this technique to the case of spinning bodies. Note, also, that the last term has scale proportional to (GM/r)2 and is, therefore, of the same PN order since each additional factor of (GM/r) is of the same order of v2. 18 Chapter 4 The Case of Spinning Particles 4.1 Usage of anti-commuting Grassmann vari- ables for spin degrees of freedom The problem of calculating additional terms in the above potential due to spinning degrees of freedom has already been solved in EFT as seen, for example, in [7]. However, here we will devise a different approach leading to the same results, which is based on the modeling of the spin degrees of freedom by anti-commuting Grassmann variables as depicted in [8]. In this section we will make a brief summary of this method, and in the following section we will apply it to our problem of two bodies interacting through the gravitational field. The point particle action written in a reparametrization invariant way is given by: Spp = ∫ dτ √ −ẋµẋµ As explained, for example, in [9], the action can be put in another useful form by introducing a one-dimensional world-line metric γττ (τ) and its square root η(τ) ≡ √ −γττ (τ). In 4 dimensions we would call η a tetrad or vierbein; since it depends only on the world-line parameter τ we call it the einbein. The action: S ′pp = 1 2 ∫ dτ ( η−1ẋµẋµ − η m2 ) also has reparametrization invariance. By varying η we find the equation of motion: η2 = −ẋµẋµ/m2 (4.1) When substituting this back into S ′pp we get back the first form Spp. 19 If τ is the proper time then from the equation of motion η = 1/m. In this case we can write S ′pp = 1 2 ∫ dτ(mẋµẋµ −m) Where we no longer have reparametrization invariance. In order to be consis- tent with the notation in [8] we will use another parameter s = τ/2m which is also proportional to the proper time. We can then rewrite the action as S ′pp = ∫ ds ( 1 4 ẋµẋµ −m2 ) (4.2) So that L = 1 4 ẋµẋµ We don’t need to include the m2 term in the Lagrangian since, in the partic- ular case when s is proportional do the proper time, it is constant and thus doesn’t influence the equations of motion. But we should keep this in mind as we’ll put it back later when we will need to express the action in another parameter different from the particle’s proper time. Ravndal proposes in [8] that we modify the Lagrangian in the following way to describe the additional spinning degrees of freedom, first in the case of a free particle (gµν = ηµν): L = 1 4 (ẋµẋµ − iξµξ̇µ) Where he introduces the four Grassmann variables ξµ = ξµ(s) which anti- commute: {ξµ, ξν} = 0 The hint that ξµ expresses the spin degrees of freedom comes from observing that the Lagrangian is invariant under Lorentz transformations: δxµ = εµνx ν δξµ = εµνξ ν Where εµν = −ενµ are infinitesimal parameters. By Noether’s theorem the conserved quantity associated with this sym- metry is given by K = δxµpµ + δξµζµ Where pµ and ζµ are the conjugate momenta: pµ ≡ ∂L ∂xµ ζµ ≡ ∂L ∂ξµ 20 From which we get then the conserved quantity K as: K = −1 2 εµν [ (xµpν − xνpµ) + i 4 (ξµξν − ξνξµ) ] We see then that Jµν = Lµν + Sµν is conserved where Lµν = xµpν − xνpµ is the orbital angular momentum and Sµν = i 4 [ξµ, ξν ] (4.3) is the spin angular momentum. This gauge fixed action is also verified to be SUSY invariant by expressing it in the form of super-variables: Xµ(s, θ) = xµ(s) + iθξµ(s) Where the SUSY transformations are induced by: θ → θ + ε s→ s− iεθ So that xµ + iθξµ → xµ − iεθẋµ + i(θ + ε)(ξµ − iεθξ̇µ) = xµ + iεξµ︸︷︷︸ δxµ +iθ(ξµ + εẋµ︸︷︷︸ δξµ ) Which makes evident the component transformations: δxµ = iεξµ δξµ = εẋµ (4.4) We define the fermionic operator Q such that δQX µ = δxµ + iθδξµ = iεξµ − iεθẋµ ≡ −iεQXµ Implying Q = θ ∂ ∂s + i ∂ ∂θ We see that Q satisfies the supersymmetry algebra: {Q,Q} = 2i∂s 21 And it is also convenient to define the operator: D ≡ θ∂s − i∂θ satisfying the anti-commutation relation: {Q,D} = 0 Finally, we can show that our Lagrangian can be written as: L = 1 4 (ẋµẋµ − iξµξ̇µ) = 1 4 ∫ dθẊµDXµ which can be checked by substituting D, Xµ and using the standard rules for integration of Grassmann variables:∫ dθ = 0 ∫ dθ θ = 1 Following the rationale in [8], once we have the Lagrangian for the free particle using super-variables, under the presence a gravitational field we must simply exchange ηµν → gµν . That is, L = 1 4 ∫ dθ gµν(X)ẊµDXν obtaining the correct equations in motion. When expanding the Lagrangian in component form, remember that gµν(X) is expressed with a super-variable in the argument, therefore it has to be expanded like: gµν(x+ iθξ) = gµν(x) + iθξλ∂λgµν(x) Then the Lagrangian in component form is expressed as L = 1 4 gµν(ẋ µẋν − iξµξ̇ν)− i 4 gµν,λẋ µξνξλ (4.5) With the conjugate momenta being given by pµ = ∂L ∂xµ = 1 2 ẋµ − 1 2 SλνΓλνµ ζµ = ∂L ∂ξµ = i 4 ξµ 22 One important fact which will be used later on is that there is a conserved quantity associated with symmetries (4.4). Under these transformations, the Lagrangian (4.5) changes by the total derivative: δL = i 4 ε d ds (gµνξ µẋν) Implying the conserved quantity: K = δxµpµ + δξµζµ − i 4 ε d ds (gµνξ µẋν) = 1 2 iεξµẋµ That is, ξµẋµ is conserved. 4.2 Reparametrization invariant action and new pairings Now we take a detour from [8]. Here we have a different objective. Instead of deriving the equations of motion we want to write the action from the reference frame of an observer far away from the two particles. Therefore we cannot use the proper time so we have to go back to the reparametrization invariant form of the action. First remember we had a m2 term in (4.2) coming from the reparametriza- tion invariant expression, so we put it back in the Lagrangian (4.5): S = 1 4 ∫ ds [ gµν(ẋ µẋν − iξµξ̇ν)− igµν,λẋµξνξλ − 4m2 ] Then change the parameter to τ = 2ms obtaining: S = 1 2 ∫ dτ [ gµν(mẋµẋν − i 2 ξµξ̇ν)− i 2 gµν,λẋ µξνξλ −m ] Here we are still using the proper time as the parameter for the world-line of the particle and the m term wouldn’t change the equations of motion. To go back to the reparametrization invariant way, first notice that η(τ) = 1/m and since, under any change of parameter, η(τ)dτ = η′(t)dt, we have: dτ = mη′(t)dt⇒ d dτ = 1 mη′(t) d dt Therefore, making the change of variable from τ to t (and dropping the prime from η): S = 1 2 ∫ dt [ gµν ( η−1ẋµẋν − i 2 ξµξ̇ν ) − i 2 gµν,λẋ µξνξλ − η m2 ] 23 Now we can see that the m2 term influences the equations of motion since it couples with η. Also, comparing to (4.1) we see that the equation of motion of η is unchanged since the new terms containing ξ don’t couple with η. So we get the same result which we repeat here: η2 = −gµν ẋµẋν m2 Substituting η back into the action we get: S = ∫ dt [ −m √ −gµν ẋµẋν − i 4 gµνξ µξ̇ν − i 4 gµν,λẋ µξνξλ ] The two new terms in the particle’s action, compared to the spin-less case, will generate new pairings in the partition function and therefore modify the potential. In order to calculate this effect we have to expand them into the gravitational component fields as below: − i 4 gµνξ µξ̇ν = − i 4 e2φ/Λ [ − ξ0ξ̇0 + Ai Λ (ξ0ξ̇i + ξiξ̇0) + e−cdφ/Λ ( ξaξ̇a + σij Λ ξiξ̇j ) − AiAj Λ2 ξiξ̇j ] (4.6a) − i 4 gµν,λẋ µξνξλ = − i 4 [ g00,iξ 0ξi + g0i,0ξ iξ0 + g0i,jξ iξj + gi0,jv iξ0ξj + gij,0v iξjξ0 + gij,kv iξjξk ] (4.6b) Let us first focus on the O(v0) terms. Let’s consider that the conserved quantity ξµẋµ is zero, which is allowed given that its square will always be zero. Then, we should have ξ0 = ξivi. Since, for now, we’re interested only in O(v0) corrections, the terms containing ξ0 don’t matter. Continuing further to the second part (4.6b) we see that after disconsid- ering terms containing ξ0 or of O(v) we’re left only with: − i 4 [ g0i,jξ iξj ] = − i 4 [ 2 Λ ∂jφ e 2φ/λAi Λ + e2φ/Λ∂jAi Λ ] ξiξj We see above that the only term whose pairing generates non quantum con- tributions will be the one in zeroth order in φ and first order in A, that is: − i 4 ∂jAi Λ ξiξj (4.7) 24 This pairing can be calculated as follows:∫ DφDADσeiSquad ∂jAi(x) Λ ∂j′Ai′(x ′) Λ = 1 Λ2 ∂j∂j′ ∫ DφDADσeiSquadAi(x)Ai′(x ′) = 1 Λ2 ∂j∂j′ ( − 1 2 δii′ )[ − i ∫ k dk0 2π e−ik 0(t−t′)+ik· ( x(t)−x′(t′) ) k2 − (k0)2 ] = i 2Λ2 δii′ ∫ k dk0 2π e−ik 0(t−t′)+ik· ( x(t)−x′(t′) ) k2 − (k0)2 kjkj′ (4.8) Where we used that ∂j∂j′ = ikj(−ikj′) = kjkj′ We’ll also use (4.3) to define the spatial components of spin: Sk = 1 2 εijkSij ⇒ 1 2 i ξiξj = Sij = εijkSk ⇒ ξiξj = −2i εijkSk Now we are in conditions to start evaluating pairings. The first one we can look at is the pairing between two terms containing ∂jAi. Substituting (4.7) and the above definition for the spatial components of spin: iSeff = ln ∫ DφDADσ eiSquad { 1 + · · ·+ i2 2 × [ − i 4 ∫ dt ∂jAi Λ (−2i)εijkSk − i 4 ∫ dt′ ∂j′Ai′ Λ (−2i)εi′j′k′Sk′ + . . . ]2 + . . . } (Note that we’re using unprimed and primed indices to denote the two particles to make notation cleaner). Using (4.8) we get: iSeff ⊃ − i 8Λ2 δii′ ∫ dt dt′δ(t− t′) ∫ k ( 1 + ∂t∂t′ k2 + . . . ) eik· ( x(t)−x′(t′) ) k2 kjkj′ × εijkSk ( x(t), t ) εi′j′k′S ′ k′ ( x′(t′), t′ ) 25 Using that: δii′ εijk εi′j′k′ = εijk εij′k′ = δjj′ δkk′ − δjk′ δj′k We then get the following terms with zeroth order in velocities: iSeff ⊃ − i 8Λ2 S · S′ ∫ dt ∫ k eik·(x−x ′) + i 8Λ2 SjS ′ k ∫ dt ∫ k eik·(x−x ′) k2 kkkj = i 8Λ2 S · S′ ∫ dt δ3(x− x′)− i 8Λ2 SjS ′ k ∫ dt ∂rj∂rk ( 1 4πr ) Where we have used that ∂rj∂rke ik·r = −kjkk eik·r. The first term can be ignored since we’re dealing with bodies interacting over a long distance compared to their own radii. To calculate the second term we note that: ∂rj∂rk ( 1 r ) = −δjk r3 + 3 rjrk r5 So we obtain: iSeff ⊃ ∫ dt i 32πΛ2 [ S · S′ r3 − 3 r3 (S · r̂)(S′ · r̂) ] Substituting Λ = (32πGN)−1/2 we finally obtain the spin-spin contribu- tion to the potential (changing notation for particles to indices 1 and 2): VSS = GN r3 [ − S1 · S2 + 3(S1 · r̂)(S2 · r̂) ] (4.9) Matching the result previously obtained for the spin-spin potential by [7]. Now we look at contributions of O(v). The first possibility arises from combining the A·v term from (3.11) with the O(v0) term due to the spinning degrees of freedom (4.7): iSeff = ln ∫ DφDADσ eiSquad { 1 + · · ·+ i2 2 × [ − i 4 ∫ dt1 ∂jAi Λ (−2i)εijkS (1) k + m2 Λ ∫ dt2 ( Alv (2) l ) + . . . ]2 + . . . } + 1↔ 2 Using the Green’s function forA calculated in (3.8) and that ∂/∂x (1) j eik·(x (1)−x(2)) = −∂/∂x(2) j eik·(x (1)−x(2)) = ∂/∂rj e ik·(x(1)−x(2)) with r ≡ x(1) − x(2): 26 iSeff ⊃ − 1 2 · 2 ( − i 4 ) m2 Λ2 ∫∫ dt1 dt2 [ (−2i) εijkS (1) k v (2) l ∂rj i 2 δil ( 1 + ∂t1∂t2 k2 + . . . ) × δ ( t1 − t2 ) ∫ k eik· ( x1(t1)−x2(t2) ) k2 ] − 1↔ 2 ⇒ iSeff ⊃ ∫ dt [ i 4Λ2 m2v (2) i εijkS (1) k ∂rj 1 4πr ] − 1↔ 2 = i ∫ dt 2GN [ −1 r3 εijkS (1) k m2rjv (2) i ] − 1↔ 2 (4.10) There is yet another possible contribution of O(v). It comes from pair- ing O(v0) terms in (3.11) with O(v) terms in (4.6b). We see that the non quantum contributions will come from the pairing of φ. Expanding the O(v) terms in (4.6b) to first order in φ we obtain: − i 4 [ g00,iξ 0ξi + g0i,0ξ iξ0 + gij,kv iξjξk ] =− i 4 { ∂i(−e2φ/Λ)ξ0ξi + ∂t ( e2φ/ΛAi Λ ) ξiξ0 + ∂k [ e(2−cd)φ/Λ ( δij + σij Λ ) − e2φ/ΛAiAj Λ2 ] viξjξk } =− i 4 [ − 2 Λ (∂iφ)ξ0ξi + 2− cd Λ (∂kφ)δijv iξjξk ] =− i 4 [ − 2 Λ (∂iφ)(−2i)Sjivj − 2 Λ (∂kφ)vi(−2i)Sik ] = 2 Λ (∂iφ)εjikSkvj In the first line, we neglected the terms of orders different than O(v). In the fourth line, we expanded only to first order in φ and no orders in the other fields (non quantum contributions). In the fifth line, we used that ξµξν = (−2i)Sµν and cd = 4. In the last line, we used that Sji = εjikSk. The pairing with −m2 Λ ∫ dt φ will produce: 27 iSeff ⊃ ∫ DφDADσeiSquad { i2 2 · 2 · 2m2 Λ2 ∂ (1) i ∫∫ dt1 dt2 φ(x1)εjikS (1) k v (1) j φ(x2) } + 1↔ 2 Where x1 and x2 are the four-vector coordinates of particles 1 and 2. Making use of the Green’s function for φ calculated in (3.7), and noting that we don’t want to go beyond O(v), so that we can approximate 1/(k2 − (k0)2) ≈ 1/k2 we obtain: iSeff ⊃ − im2 4Λ2 ∂ri ∫ dt 1 4πr εjikS (1) k v (1) j − 1↔ 2 = i ∫ dt 2GN [ 1 r3 εjikS (1) k m2riv (1) j ] − 1↔ 2 (4.11) Where we used that ∂ (1) i = −∂(2) i = ∂ri Summing the two contributions (4.10) and (4.11) we obtain: iSeff ⊃ i ∫ dt 2GN [ 1 r3 εijkS (1) i m2rj ( v (2) k − v (1) k )] − 1↔ 2 = −i ∫ dt 2GNm2 r3 S(1) · L(2,1) + 1↔ 2 (4.12) Where we define L(2,1) ≡ m2(r2 − r1)× (v2 − v1) to be the orbital angular momentum of particle 2 relative to particle 1 and vice-versa for L(1,2). This is the spin-orbit contribution to the potential: VSO = 2GN r3 (S(1) · L(2,1) + S(2) · L(1,2)) (4.13) We’ll see later on that this formula is incomplete. Another contribution will come out from the zeroth order term proportional to ξ0ξ̇0 in the action. We’ll see how it changes the spin-orbit potential in (5.11). 28 Chapter 5 Energy as a function of frequency Having achieved the results for the Einstein Infeld Hoffmann corrections to the Newtonian potential in Chapter 3 (3.16); and spin-spin and spin-orbit corrections in Chapter 4 (4.9) and (4.13) it is straightforward to compute the energy by simply adding these corrections to the Newtonian expression. However, this is not very useful since we would be expressing the energy in terms of the radius r, which is not an observable. If we restrict ourselves to circular orbits, we can express the energy as a function of the frequency of rotation. This can be done in the following way: 1. We compute the equations of motion from the Lagrangian with correc- tions 2. The radial component of the acceleration is, by definition, −ω2(r,v) where ω is the angular frequency 3. We use v = ωr to eliminate v and thus we can express r as a function of ω and therefore we obtain E(ω). We will apply this method first to the EIH corrections and then to the spin corrections. 5.1 Calculation of EIH corrections As outlined above we start by computing the equations of motion. Other than the potential terms. From the point particle action (3.10) we see that a term −m ∫ dt √ 1− v2 goes unaltered to the effective action Seff . We see 29 from (3.16) that VEIH is of order 1PN. To match that we expand the kinetic term to O(v4) so that: S = ∫ dt { 1 2 (m1v 2 1 +m2v 2 2) + 1 8 (m1v 4 1 +m2v 4 2) + GNm1m2 r + GNm1m2 2r [ 3(v2 1 + v2 2)− 7v1 · v2 − (v1 · r̂)(v2 · r̂) ] + G2 Nm1m2(m1 +m2) 2r2 } The equation of motion for particle 1 is, applying Euler Lagrange: d dt ( ∂L ∂vi1 ) = ∂L ∂xi1 ⇒ m1a i 1 +m1a j 1v j 1v i 1 + 1 2 m1v 2 1a i 1 − 3GN m1m2 r3 rj(v j 1 − v j 2)vi1 + 3GN m1m2 r ai+ 7 2 GN m1m2 r3 rj(v j 1 − v j 2)vi2 − 7 2 GN m1m2 r ai2 + d dt ( − 1 2 GN m1m2 r3 rivj2r j ) =∓GN m1m2 r3 ri ∓ 1 2 GN m1m2 r3 ri [ 3(v2 1 + v2 2)− 7v1 · v2 − (v1 · r̂)(v2 · r̂) ] ∓ 1 2 GN m1m2 r [ − (v1 · r̂)(v2 · r̂) + vi1 r (v2 · r̂) ] ±G2 N m1m2(m1 +m2) r4 ri1 Where the upper sign applies to particle 1, and the lower sign applies to particle 2 (also exchanging indices 1 and 2 for particle 2). These equations of motion seem complicated but by recurring to the hy- pothesis of pure circular motion, by which ri(v i 1−vi2) = 0, many terms cancel out. Another simplification can be done by using center of mass (CM) coordi- nates. They are defined by: rCM ≡ m1r1 +m2r2 m1 +m2 M ≡ m1 +m2 µ ≡ m1m2 M Then v1 = vCM + m2 M v v2 = vCM − m1 M v By symmetry under boosts we can assume that vCM = 0. After subtracting EOMs for particles 1 and 2, canceling multiples of r ·v and converting to CM coordinates we reach the following equation: 2µai + µaj(1− 2η)vjvi + 1 2 µai(1− 2η)v2 + 13 2 µ GM r ai =− µri [ 2 GM r3 + 3 GM r3 v2 + Gµ r3 v2 − 2 (GM)2 r4 ] 30 Where η ≡ µ/M . Note that we’re at level 1PN so, in the second and third terms, we can substitute the Newtonian solution for ai and aj. With that substitution, the second term cancels out because it is proportional to rjvj. Replacing, in the third term, the Newtonian acceleration ai = −GMri/r3 and canceling factors of µ we have: ai [ 2 + 13 2 GM r ] =− ri [ 2 GM r3 + 3 GM r3 v2 + η GM r3 v2 − 2 (GM)2 r4 + ( η − 1 2 ) GM r3 v2 ] (5.1) Let us define two auxiliary dimensionless variables: α ≡ GM r x ≡ (GMω)2/3 Our task is, then, to find α as a function of x. Note that, at Newtonian level: ai = −ω2ri = −GM r3 ri ⇒ (GMω)2 = ( GM r )3 ⇒ x = α So we’re looking for perturbative corrections for α(x), that is, α(x) = x+O(x2) + . . . Since we want only the next to leading order correction we want to find d2α/dx2|x=0 in the expansion: α(x) ≈ dα dx ∣∣∣∣ x=0 x+ 1 2 d2α dx2 ∣∣∣∣ x=0 x2 (5.2) Where we already know that dα/dx|x=0 = 1. Replacing α and x in (5.1) we get: 2α4 − 2α3 + (4− 2η)x3α + 2x3 = 0 If we derive this equation twice, apply (5.2) and set x = 0 we find: d2α dx2 ∣∣∣∣ x=0 = 2− 2 3 η 31 So (5.2) now reads as: α = x [ 1 + ( 1− 1 3 η ) x ] (5.3) Another useful formula is: v2 = ω2r2 = x3 α2 = x[ 1 + ( 1− 1 3 η ) x ]2 = x [ 1− 2 ( 1− η 3 ) x ] (5.4) Now we have everything at hand to compute the energy as a function of frequency with corrections up to 1PN using (3.16). First, it’s important to take into account the terms up to O(v4) in the expansion of −m √ 1− v2 because we’re working at level 1PN. This will gen- erate the following terms in the Lagrangian: 1 2 (m1v 2 1 +m2v 2 2) + 1 8 (m1v 4 1 +m2v 4 2) = 1 2 µ m2 +m1 M v2 + 1 8 µ m3 2 +m3 1 M3 v4 =− 1 2 µ [ − v2 + ( 3 4 η − 1 4 ) v4 ] Where we have used center of mass coordinates with vCM = 0. Summing with the potential at (3.16) we get the Lagrangian: L = −1 2 µ [ − 2 GM r − GM r (3 + η)v2 + ( GM r )2 − v2 + ( 3 4 η − 1 4 ) v4 ] Where we’re assuming that v · r = 0 (circular orbit). Now, since energy is given by E = ∂L ∂v v − L And substituting (5.3) and (5.4) we obtain, up to O(x2), E = −1 2 µx [ 1− ( 3 4 + η 12 ) x ] = −1 2 µ(GMω)2/3 [ 1− ( 3 4 + µ 12M ) (GMω)2/3 ] (5.5) Which is the Newtonian energy plus the 1PN correction. 32 5.2 Calculation of corrections related to Spin Let’s start with the spin-spin contribution (4.9). Stating the Euler Lagrange equation in terms of center of mass coordinates we obtain: ∂L ∂vi =µai = ∂ ∂ri { G r3 [ S1 · S2 − 3(S1 · r̂)(S2 · r̂) ]} + . . . =− 3Gri r5 [ S1 · S2 − 3(S1 · r̂)(S2 · r̂) ] − 3G r3 [ − 2(S1 · r̂)(S2 · r̂) ri r2 + Si1 r (S2 · r̂) + Si2 r (S1 · r̂) ] + . . . So we conclude that aSS = − 3G µr4 { r̂ [ S1 · S2 − 5(S1 · r̂)(S2 · r̂) ] + S1(S2 · r̂) + S2(S1 · r̂) } (5.6) One could ask why we hadn’t included terms with higher orders of v in the left hand side, that is, why we considered only the Newtonian kinetic term 1 2 µv2 in ∂L ∂vi . The answer is that the right hand side is entirely (at least) of order 1PN. If the objects are small, that is R1,2 is of order m1,2, and, let’s say, their rotational speeds are closer to unity, then their spins are of order m1,2v1,2R1,2 = m2 1,2. Then: G r3 S1S2 ∼ µ G r3 Mm1m2 ∼ µ ( GM r )3 Which is already of order 2PN. If the rotational speeds are small, then the correction is even less significant. If the problem involved objects with sizes comparable to r, which isn’t the case here, one could show that the correction would still be at least of order 1PN. Coming back to (5.6), the next step is to look at the radial component which is what interests us here, since we want to find ω though the identity −ω2r = a · r̂. Projecting (5.6) along the radial direction see that: aSS · r̂ = − 3G µr4 [ S1 · S2 − 3(S1 · r̂)(S2 · r̂) ] (5.7) In order to express the spin projections along the axis perpendicular to the orbit (i.e., along the orbital angular momentum of the system), we first see that: S1 · S2 = (S1 · r̂)(S2 · r̂) + (S1 · v̂)(S2 · v̂) + (S1 · l̂)(S2 · l̂) 33 But, taking the angular average, the first two terms are equal. Then we conclude that (S1 · r̂)(S2 · r̂) = 1 2 [ S1 · S2 − (S1 · l̂)(S2 · l̂) ] In this way (5.7) becomes: aSS · r̂ = + 3G 2µr4 [ S1 · S2 − 3(S1 · l̂)(S2 · l̂) ] (5.8) Just for abbreviation let’s define d ≡ S1 · S2 − 3(S1 · l̂)(S2 · l̂) Now we can calculate the change in radius due to the spin-spin potential, relative to the Newtonian case, by adding the new radial acceleration term: − ω2r = (aNEWTON + aSS) · r̂ = −GM r2 + 3G 2µr4 d ⇒ ω2 − GM r3 + 3G 2µr5 d = 0 We can solve this equation perturbatively since we’re looking for the first order correction. We can then state the equation above as: ω2 − GM r3 + 3G 2µr3 1 r2 NEWTON d = 0 (5.9) Where we use the Newtonian level solution (GMω)2 = ( GM rNEWTON )3 The solution to (5.9) is given by: r = (GM)1/3 ω2/3 ( 1− 3 2 ω4/3 µG2/3M5/3 d )1/3 = (GM)1/3 ω2/3 ( 1− 1 2 ω4/3 µG2/3M5/3 d ) (5.10) Where in the last line we expanded (1 + x)1/3 ≈ 1 + x/3 since we want only the first order term in d. 34 Now we turn to the spin-orbit contribution. We could just start from (4.13) but there is something else to consider. When looking at the terms added to the action in (4.6a), consider what happens to the term: i 4 ξ0ξ̇0 Where it is taken at leading order. It gets unaltered to the effective action and it does have a contribution. From the constraint ξ0 = ξivi we find (e.g. for particle 1): ξ0 1 ξ̇ 0 1 = ξi1 v i 1 ξ̇ j 1 v j 1 + ξi1 v i 1 ξ j 1 a j 1 From the EOM of ξ one can see that the first term has contributes in higher order than O(v2) but the second term does have a contribution to be con- sidered when taking the Newtonian order for acceleration. Taking the center of mass coordinate where vi1 = m2 M vi we get: i 4 ξ0 1 ξ̇ 0 1 = i 4 ξi1 v i 1 ξ j 1 a j 1 = 1 2 εijkS (1) k m2 2 M2 vi ( − GM r3 rj ) Now it becomes evident that the spin-orbit potential deducted from (4.12) is incomplete. We have to add this new term. After this correction the formula for VSO becomes: VSO = 2G r3 µ(r×v) · [( 1+ 3m1 4m2 ) S2+ ( 1+ 3m2 4m1 ) S1 ] ≡ G r3 µ(r×v) ·~χ (5.11) Where, following [7], we define: ~χ ≡ ( 2 + 3m1 2m2 ) S2 + ( 2 + 3m2 2m1 ) S1 Now we can follow the same path we did for the spin-spin contribution. We find the equation of motion: aSO = G r3 [ 3r̂(r̂× v) · ~χ− 2v × ~χ+ 3r̂ · v(r̂× ~χ) ] (5.12) Taking the radial component of this acceleration plus the Newtonian one and comparing with −ωr we find: ω2 − GM r3 + Gω r3 l̂ · ~χ = 0 Giving the first order correction to the radius: r = (GM)1/3 ω2/3 ( 1− 1 3 ω M l̂ · ~χ ) (5.13) 35 Now we can start calculating E(ω). First let’s do it taking into account the spin-spin and spin-orbit contributions which have higher orders. Noticing from (5.12) that aSO · v = 0 the conserved energy is: E = 1 2 µv2 − µGM r + VSS For any power of r we just need to approximate [rNEWTON(1 + δ)]n ≈ rNEWTON n(1 + nδ) so it becomes straightforward to get all contributions from the spin-spin and spin-orbit to the energy: E(ω) = −1 2 µ(GMω)2/3 { 1+ 4 3 ω M l̂ · ~χ+ ω4/3 µG2/3M5/3 [ S1 ·S2−3(S1 · l̂)(S2 · l̂) ]} Now summing the contribution previously calculated in (5.5) we can state the final result: E(ω) = −1 2 µ(GMω)2/3 { 1 + 4 3 ω M l̂ · ~χ+ ω4/3 µG2/3M5/3 [ S1 · S2 − 3(S1 · l̂)(S2 · l̂) ] − ( 3 4 + µ 12M ) (GMω)2/3 } (5.14) And that concludes our work. 36 Appendix A Mathematica notebook calculations In this section we detail how we got to (3.4) using a Mathematica notebook whose original file can be downloaded from http://tiny.cc/MautnersMasterThesis. The notebook makes use of the “EinS” package ([5]) which can be obtained at http://rcswww.urz.tu-dresden.de/~klioner/eins.html. Observe that the object definitions of the fields A, φ and σ are made such that the Λ denominator is incorporated into them. Therefore, the obtained expressions in quadratic order have to be multiplied by (1/Λ2). For example, √ −gR→ 1 Λ2 √ −gR = 1 32πGN √ −gR Therefore, the expression obtained for √ −gR has to be multiplied by 2 to give the Einstein Hilbert action integrand since according to (3.2): SEH = ∫ dt ddx 1 16πGN √ −gR Accordingly, the expression − √ −g(gµνΓ µΓν) in the Mathematica file gives, after adjusting by the factor 1/Λ2, the correct factor of 1/(32πGN) in (3.3). So, in the Mathematica file, the gauge fixing term is divided by 2 so that it has the same overall factor of (1/2) as the Einstein Hilbert term. To summarize: the results of the Mathematica notebook are half the correct ones so we have to multiply everything by 2. We now explain how we got the results for each of the component fields. 37 http://tiny.cc/MautnersMasterThesis http://rcswww.urz.tu-dresden.de/~klioner/eins.html A.1 φ field A.1.1 Einstein Hilbert term of order φ2 This is located in the “RicciPAS0” entry in the file, which shows the expan- sion of √ −gR up to second order in the A field and zeroth order in σ. The terms quadratic in the φ field can be extracted as the ones proportional to epsP 2. Note that E4 epsP φ when expanded gives additional factors of epsP . We find the terms: −2(~∇φ)2 + 18φ̇2 − 6φ̈(−4φ) = −2(~∇φ)2 − 6φ̇2 Where we integrated the third term by parts. A.1.2 Gauge fixing term of order φ2 This is located in the “GaugeFixPAS0” entry and the only term quadratic in epsP 2 is: 8φ̇2 Therefore, after summing the EH and GF terms quadratic in φ and mul- tiplying by 2 as explained at the beginning of this section we obtain: Squad ⊃ ∫ dt ddx 4[−(~∇φ)2 + φ̇2] A.2 A field A.2.1 Einstein Hilbert term of order A2 We look again at the “RicciPAS0” entry in the file, which shows the expansion of √ −gR up to second order in the A field and zeroth order in σ. The terms quadratic in the A field can be extracted as the ones proportional to epsA2 and they are: −2 ~̇A 2 + 1 2 Aa,bAa,b − 2AaÄa − 1 2 Aa,bAb,a 38 The first and third terms cancel each other after integrating by parts. The second and fourth terms can be simplified in the following way: 1 2 ( Aa,bAa,b − Aa,bAb,a ) = 1 4 ( Aa,bAa,b + Ab,aAb,a − Aa,bAb,a − Ab,aAa,b ) = 1 4 (Aa,b − Ab,a)2 = 1 4 F 2 ab A.2.2 Gauge fixing term of order A2 This is located in the “GaugeFixPAS0” entry and the terms quadratic in epsA2 are: −1 2 ~̇A 2 + 1 2 (~∇ · ~A)2 Summing the EH and GF terms and multiplying by 2 we obtain: Squad ⊃ ∫ dt ddx [ F 2 ij 2 + (~∇ · ~A)2 − ~̇A 2 ] A.3 σ field A.3.1 Einstein Hilbert term of order σ2 The calculations involving σ are more complicated, so here we calculate sep- arately √ −g and R and then multiply them manually. The calculation for √ −g can be found at entry “SDetgS2” which expands it only up to second order in σ without any expansion in A and φ. It gives: 1 + σ 2 − 1 4 σ2 ab + 1 8 σ2 Where it is understood that σ2 ab is equal to the contraction ∑ a,b σabσab and σ ≡ Tr σ = ∑ a σaa. The calculation of R up to second order in σ and no expansions in A and φ is located at entry “RiccisclS2”. It has terms of order σ, which get multiplied by the order σ term in √ −g, and terms of order σ2, which get 39 multiplied by the 1 from √ −g. The desired terms quadratic in σ in √ −gR, after simplifying, are: √ −gR in order σ2 =− 1 4 σ̇2 + 1 4 (~∇σ)2 + 1 4 σ̇2 ij − 1 4 (~∇σij)2 + 1 2 σσbc,bc + (σab,a) 2 − 1 2 (σab,cσac,b) A.3.2 Gauge fixing term of order σ2 First we calculate Γα ≡ Γαµνg µν . This is performed at cell “GamGFS2”. We can see from the result that only terms of order σ and beyond are obtained. Since we’re interested in the contraction gµνΓ µΓν , only the O(σ) term from Γα matters. Also, only the zeroth order of gµν (that is, ηµν) can be used if we truncate the result to orders less than 2 in σ. The same applies to √ −g which is therefore considered to be 1 for this calculation. Then, in order σ2 we’ll have: 1 2 (−1) √ −ggµνΓµΓν =− 1 2 ΓαΓα =− 1 2 ( −∂ασ 2 + ∂aσαa )( −∂ασ 2 + ∂aσαa ) = σ̇2 8 − (~∇σ)2 8 + 1 2 ∂bσ∂aσab − 1 2 (∂aσab) 2 Where α is a 4-vector index and a is a 3D vector index. Also, indices are lowered and uppered with the Lortentz metric. Remember, also, that the (1/2) factor accounts for the fact that the Einstein Hilbert term obtained in the notebook is half the correct one, so we’ll divide the GF term by 2, sum it with the EH term and after multiply the final result by 2. Now we can sum the EH and GF terms. After integrating by parts, the fifth term in the EH expression cancel with the third term in the GF expression. The last two terms of EH plus the last term of GF give (1/2)[(σab,a) 2 − σab,cσac,b] =(1/2)[σba,aσbc,c − σba,cσbc,a] =(1/2)[−σba,caσbc + σba,caσbc] =0 Where we integrated by parts in the third line. 40 Therefore, summing the remaining terms we’re left with (after multiplying by 2 as explained before): Squad ⊃ ∫ dt ddx [ − 1 4 σ̇2 + 1 4 (~∇σ)2 + 1 2 σ̇2 ij − 1 2 (~∇σij)2 ] 41 42 Bibliography [1] A. Abramovici et al., Science 256, 325 (1992). [2] C. Bradaschia et al., Nucl. Instrum. Meth. A 289, 518 (1990). [3] S. Foffa and R. Sturani, “Effective field theory methods to model compact binaries,” Class. Quantum Grav. 31 043001 (2014), [arXiv:1309.3474 [gr-qc]]. [4] B. Kol and M. Smolkin, “Classical Effective Field Theory and Caged black holes,” Phys. Rev. D 77 064033 (2008), [arXiv:0712.2822 [hep- th]]. [5] S. A. Klioner, “EinS: a Mathematica package for computations with indexed objects,” [arXiv:gr-qc/0011012] [6] B. Kol and M. Smolkin, “Einstein’s action and the harmonic gauge in terms of Newtonian fields,” Phys. Rev. D 85 044029 (2012), [arXiv:1009.1876 [hep-th]] [7] R.A. Porto, “Post-Newtonian corrections to the motion of spinning bod- ies in NRGR,” Phys. Rev. D 73 104031 (2006) [arXiv:gr-qc/0511061] [8] F. Ravndal, “Supersymmetric Dirac particles in external fields,” Phys. Rev. D 21, 2823 (1980) [9] J. Polchinski, String Theory, Cambridge University Press, Vol.1, pp.10- 11, 1998. 43 Introduction Summary of Effective Field Theory Conceptual Basis A toy model EFT for two gravitationally interacting spin-less particles Quadratic field terms in the action Green's function Green's function Green's function Source term Calculation of potential terms The Case of Spinning Particles Usage of anti-commuting Grassmann variables for spin degrees of freedom Reparametrization invariant action and new pairings Energy as a function of frequency Calculation of EIH corrections Calculation of corrections related to Spin Mathematica notebook calculations field Einstein Hilbert term of order 2 Gauge fixing term of order 2 A field Einstein Hilbert term of order A2 Gauge fixing term of order A2 field Einstein Hilbert term of order 2 Gauge fixing term of order 2