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I. EINSTEIN FIELD EQUATION A. Vacuum Equation

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I. EINSTEIN FIELD EQUATION A. Vacuum Equation I. EINSTEIN FIELD EQUATION In order to construct the equation for describing the relation between matter/energy and spacetime curvature, one has impose two requirements such that • The equation must be reduced to Newtonian theory. • The curvature part must contain the metric tensor gµν and its derivatives such as ρ ρ and while the matter/enegy should be proportional to the EMT . Γµν;R µσν;Rµν R Tµν From the rst requirement, the important equation in Newtonian theory in the Poisson equation r2Φ = 4πGρ Which can be derived from the Gauss's law and leave for the student in Exercise. As we discussed before, the component of the metric tensor is proportional to the gravitational potential g00 / Φ. Therefore, in order to reduce the master equation into the Poisson equation, the curvature part must be proportional to the second derivative of the metric. These quantities are ρ and . R µσν;Rµν R A. Vacuum equation For the vacuum equation, the matter/energy part vanishes and then the curvature part will be vanished ρ (1) f(R µσν;Rµν;R) = 0: One may rstly guess for this equation such that ρ (2) R µσν = 0 (?): However, one found that this may not be possible since this equation provides the at spacetime nearby the massive source. So that we can make a further guess by Rµν = 0 (?): (3) This is a good choice since the Ricci tensor has ten dof. like the metric tensor, hoping that ten dof. of the metric transfer to ten dof. of Ricci tensor through their second derivatives making from source nearby. 1 B. Equation with source Now let us consider the equation with source. By adding the EMT, the equation may be written as Rµν = kTµν (?); (4) where k is the proportional constant. This still be the good choice since the index symmetry also satisfy. However, as we discussed before, the EMT obeys the conservation equation µν rµT = 0 while the Ricci tensor does not satisfy in general. As a result, one may nd other quantities to satisfy this condition as well as maintain the mentioned requirements. Fortunately, from the Bianchi indentity, r[λRρσ]µν = 0, it serve us the conservation quantity as follows 1 rµG = 0;G = R − Rg ; (5) µν µν µν 2 µν µ where Gµν is called Einstein tensor. Note that the derivation of r Gµν = 0 from the Bianchi identity is leave in the Exercise. As a result, the equation can be constructed as Gµν = kTµν: (6) Next task for this construction is that we have to nd the proportional constant k as well as check whether this equation satises the vacuum equation or not. To perform this evaluation, let us take the trace of the above equation as follows 1 R − (4)R = kT; 2 R = −kT (7) Substituting R from this equation into Eq. (6), one obtains 1 R + kT g = kT ; µν 2 µν µν 1 R = k T − T g : (8) µν µν 2 µν From this equation, one can see that the vacuum equation still be satised where the source is eliminated, Tµν = T = 0. 2 Now, we will nd the proportional constant by taking the Newtonian limit into Eq. (8). As a result, the EMT can be written as 0 1 0 1 1 v1 v2 v3 1 0 0 0 B C B C B v1 v1v1 v1v2 v1v3 C B 0 0 0 0 C µν B C B C (9) T = ρ B C ∼ ρ B C : B v2 v2v1 v2v2 v2v3 C B 0 0 0 0 C @ A @ A v3 v3v1 v3v2 v3v3 0 0 0 0 Then we have 0 1 1 0 0 0 B C 1 1 B 0 1 0 0 C 1 µν µν B C µν T − T g = ρ B C = ρδ ; 2 2 B 0 0 1 0 C 2 @ A 0 0 0 1 1 1 T − T g = ρδ : (10) µν 2 µν 2 µν Now let us consider the left hand side of Eq. (8), ρ ρ λ ρ λ ρ (11) Rµν = @ρΓ µν − @νΓ µρ + Γ µνΓ λρ − Γ µρΓ λν: By using the weak eld limit gµν = ηµν +hµν and then keeping only rst order perturbations, ρ does not contain the zeroth order of the perturbation metric as Γµν hµν 1 Γρ = gρσ (@ h + @ h − @ h ) : (12) µν 2 µ νσ ν µσ σ µν Thus the last 2 terms on the left of (11) is neglected in our consideration. The the rst order of the Ricci tensor is (1) ρ ρ Rµν = @ρΓµν − @νΓρµ; 1 1 = @ ηρσ (@ h + @ h − @ h ) − @ ηρσ@ h : (13) ρ 2 µ νσ ν µσ σ µν ν 2 µ ρσ (1) Since the h00 is much more than other components, the dominant components of Rµν is 1 1 R(1) = @ ηρσ (@ h + @ h − @ h ) − @ ηρσ@ h ; 00 ρ 2 0 0σ 0 0σ σ 00 0 2 0 ρσ 1 ' − ηρσ@ @ h ; 2 ρ σ 00 1 = − @ @ih ; (h is time-independent.) 2 i 00 00 Φ = r2 : (14) c2 3 Substituting results into Eq. (8), one obtains 1 r2Φ = kρ. (15) 2 By comparing this equation to the Poisson equation r2Φ = 4πGρ, the constant k can be written as k = 8πG: (16) Finally, the Einstein equation is completely constructed as 1 G = R − gµνR = 8πGT : (17) µν µν 2 µν II. LAGRANGIAN FORMULATION IN GR Motivaton - Most of physical theories can be expressed in terms of Lagrangian formulation. - It is convenient to identify conserved quantities. - It is very useful to generalize the theory. A. Review of Classical Field Theory • Classical Mechanics: The action of a particle is expressed as Z S = dt L(q(t); q_(t)): (18) where q and q_ are the position and velocity of the particle. • Variational Principle: in nature, the physical system will evolve along the path such that the variation of action is zero, δS = 0. Z δS = dt δL; Z @L @L = dt δq + δq_ : (19) @q @q_ Since the variation of path is independent of time, we can write dq d δq_ = δ = δq: (20) dt dt 4 Then integrating by part for the second term in parenthesis in (19), Z @L d @L d @L δS = dt − δq + δq : (21) @q dt @q_ dt @q_ We are interested in the last term in the square bracket. Since @L is continuous, we can @q_ δq integrate out as Z d @L @L @L dt δq = δq − δq : (22) dt @q_ @q_ t2 @q_ t1 where t1 and t2 are the initial and nal time of consideration. From the fact that we consider the variation of the path between any two points. The position at the ends of the path (q(t1) and q(t2)) must be xed as illustrated below Figure 1. Variation of paths We thus have the conditions (23) δqjt1 = δqjt2 = 0: Eventually, the variational principle gives us the equation of motion of the particle, @L d @L δS = 0; ! − = 0; (24) @q dt @q_ which is well-known as the Euler-Lagrange equation. • Classical Field theory: The action for the eld Φa(x) can be written as Z 4 a a a S = d x L(Φ (x);@µΦ (x);@µ@νΦ (x);:::): (25) 5 Then @L @L @L (26) δS = 0; ! a − @µ a + @µ@ν a − ::: = 0: @Φ @(@µΦ ) @(@µ@νΦ ) It is very important to note that the condition of xing the end points in classical mechanics is equivalent to take the boundary surface to innity and then the eld becomes zero in classical eld theory. • Classical Field theory in curved spacetime: moving to consider in curved spacetime, we have to generalize some quantities as follows 4 p 4 @µ ! rµ; ηµν ! gµν; d x ! −gd x: (27) The action becomes Z 4 p a a S = d x −gL(Φ (x); rµΦ (x);:::); Z p = d4x L ; L = −gL: (28) The equation is @L @L (29) δS = 0; ! a − rµ a + ::: = 0: @Φ @(rµΦ ) Example: A massive canonical scalar eld Z 4 p µ 2 2 S = d x −g −∇µφr φ − m φ : (30) Then the variation with respect to φ is (gµν is xed under this consideration) Z 4 p µν 2 δφS = d x −g −g (rµδφrνφ + rµφrνδφ) − 2m φδφ ; Z 4 p µ 2 = d x −g (−2) r φrµδφ + m φδφ ; Z 4 p µ µ 2 = d x −g (−2) rµ (r φδφ) − rµr φ − m φ δφ : (31) R 4 p µ Considering the term d x −g rµ (r φδφ). The integral over proper volume, Σ of the 4-divergence can be written as the integral over its boundary, @Σ via the Gauss's theorem, Z Z p 4 p µ 3 µ d x −g rµ (r φδφ) = d x h (r φδφ) ; (32) Σ @Σ where h is the determinant of metric hµν describing the geometry on the boundary @Σ. Since R 4 p µ the variation of the eld vanishes at the boundary, Then, d x −g rµ (r φδφ) = 0. 6 R 4 p µ Moreover, it is also be considered the boundary term, d x −g rµ (r φδφ) in other form. By the relation between 4-divergence of the vector and partial derivative, 1 p r V µ = p @ −g V µ ; (33) µ −g µ we have Z Z 4 p µ 4 p µ p µ d x −g rµ (r φδφ) = d x @µ −gr φδφ = −gr φδφ = 0: (34) boundary It is more familiar consideration as in classical eld theory.
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