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Linearized Einstein Field Equations General Relativity Fall 2019 Lecture 15: Linearized Einstein field equations Yacine Ali-Ha¨ımoud October 17th 2019 SUMMARY FROM PREVIOUS LECTURE We are considering nearly flat spacetimes with nearly globally Minkowski coordinates: gµν = ηµν + hµν , with jhµν j 1. Such coordinates are not unique. First, we can make Lorentz transformations and keep a µ ν globally-Minkowski coordinate system, with hµ0ν0 = Λ µ0 Λ ν0 hµν , so that hµν can be seen as a Lorentz tensor µ µ µ ν field on flat spacetime. Second, if we make small changes of coordinates, x ! x − ξ , with j@µξ j 1, the metric perturbation remains small and changes as hµν ! hµν + 2ξ(µ,ν). By analogy with electromagnetism, we can see these small coordinate changes as gauge transformations, leaving the Riemann tensor unchanged at linear order. Since we will linearize the relevant equations, we may work in Fourier space: each Fourier mode satisfies an independent equation. We denote by ~k the wavenumber and by k^ its direction and k its norm. We have decomposed the 10 independent components of the metric perturbation according to their transformation properties under spatial rotations: there are 4 independent \scalar" components, which can be taken, for instance, ^i ^i^j to be h00; k h0i; hii, and k k hij { or any 4 linearly independent combinations thereof. There are 2 independent ilm^ ilm^ ^j transverse \vector" components, each with 2 independent components: klh0m and klhmjk { these are proportional to the curl of h0i and to the curl of the divergence of hij, and are divergenceless (transverse to the ~ TT Fourier wavenumber k). Finally, there is a \tensor" mode hij , which is the transverse-trace-free part of hij, ~ and is obtained by double-projecting hij transverse to k, and subtracting the trace. This piece has 2 independent components. Explicitly, we have TT TT hij = Pijmnhmn; (1) where the TT-projection operator is given by 1 PTT ≡ P P − P P ;P ≡ δ − k^ k^ : (2) ijmn im jn 2 ij mn ij ij i j Let us write explicitly the gauge transformation equations in Fourier space: replace @j by ikj: h00 ! h00 + 2@0ξ0; h0j ! h0j + @0ξj + ikjξ0; hjl ! hjl + 2ik(jξl): (3) GAUGE-INVARIANT METRIC PERTURBATIONS While a gauge transformation in electromagnetism amounts to providing one scalar function, a gauge transforma- tion in linearized GR amounts to providing 4 functions ξ0; ξ~. These 4 functions can be decomposed in 2 scalars 0 ^ i ^j k ξ ; kiξ and 1 transverse vector ijkk ξ . Therefore, we expect that out of the 4 scalars components, only 2 linear combinations are gauge-invariant. Similarly, out of the 2 vector modes, only 1 linear combination is gauge-invariant. Finally, since there is no way to make a TT mode out of scalars and vectors, we expect, and will show explicitly, that the \tensor" mode is gauge-invariant. The two scalar gauge-invariant variables are not unique (any linear combination is also gauge-invariant). We'll see that these two make expressions particularly simple (for reference these are related to the gauge-invariant Bardeen potentials in cosmological perturbation theory): " # 1 1 2ik^ 3 1 Ψ ≡ k^ k^ h − h ; Φ ≡ − h + j @ h − @2 k^ k^ h − h : (4) 4 j l jl jj 2 00 k 0 0j 2k2 0 j l jl 3 jj I encourage you to explicitly check gauge-invariance { and please report likely typos! 2 There is only one gauge-invariant transverse vector mode, defined up to a normalization constant: i vi ≡ ilmk^ h + @ h k^ (5) l 0m k 0 mj j Indeed, under a gauge transformation, the change in the parenthesis is a pure gradient, so its curl is zero. Finally, since a gauge transformation cannot add a TT part to the metric: the TT part of the metric pertur- bation is gauge-invariant, much like the transverse-vector part of the vector potential is in electromagnetism. I encourage you to show this explicitly using the TT projection operator applied to a gauge transformation. So, to summarize, gauge freedom implies that there are only 6 physical degrees of freedom in the metric perturbation (that we could tell right away just from counting the number of coordinates). For linearized GR, we can moreover explicilty identify these degrees of freedom and classify them as 2 scalar modes, 1 transverse vector mode, and 1 transverse-traceless \tensor" mode. Note: it is always possible to set the metric to be Minkowski and with vanishing first derivatives at any given point by using a LICS. So it is no surprise that the gauge-invariant variables are defined with at least two derivatives of the metric. LINEARIZED EINSTEIN TENSOR The Riemann tensor takes the form Riemann ∼ @Γ + ΓΓ, where Γ are the Christoffel symbols. To linear order in hµν , we only need to keep the first term. Furthermore, we only need to compute the Christoffel symbol at linear order in hµν . This implies 1 R = (@ @ h + @ @ h − @ @ h − @ @ h ) ; (6) αµβν 2 µ β αν ν α βµ µ ν αβ α β µν which is the same expression as that in a LICS, derived in lecture 11. From this we obtain the Ricci tensor, by contracting the first and third indices: 1 R = Rα = (@ @αh + @ @αh − @ @ h − h ) ; (7) µν µαν 2 µ αν ν αµ µ ν µν µ µν µ µν where h ≡ h µ ≡ η hµν is the trace of hµν (obtained using the Minkowski metric) and ≡ @ @µ = η @µ@ν is the D'Alembertian operator. The Ricci scalar is obtained by taking the trace: µ α β R = R µ = @ @ hαβ − h: (8) From this we obtain the Einstein tensor at linear order in metric perturbations: 1 G = @ @αh + @ @αh − @ @ h − h − η (@α@βh − h) : (9) µν 2 µ αν ν αµ µ ν µν µν αβ EINSTEIN FIELD EQUATIONS (EFES) We first note that we can also decompose Gµν in scalars, vectors and tensors, and that each kind can only include the components of hµν of the same kind: at linear order one cannot make a transverse vector mode out of a scalar, and vice-versa. Let us now explicitly write the EFEs. The 00 equation can be rewritten in terms of our gauge-invariant variable Ψ, 2 r Ψ = 4πT00: (10) This is just Poisson's equation. As in the case of electromagnetism, the (linearized) Bianchi identity G0i;i = G00;0, ,ν consistent with the conservation of the stress-energy tensor Tµν = 0 (at linear order), implies that the divergence of the 0i equation is nothing but the time-derivative of the 00 equation, and therefore does not carry any additional information. The curl of the 0i equation can be written in terms of the gauge-invariant vector field as 2 i ilm^ r v = 16π klT0m (11) 3 Out of the 6 purely spatial equations, 3 are redundant with the G0i equations, again, from the contracted Bianchi 1 identity. We are left with three independent equations. The first one can be taken to be Gij − 3 δijGkk, which, upon taking the double gradient, gives us [please report typos!] 1 r4(Φ − Ψ) = −12π@ @ T − δ T : (12) i j ij 3 ij kk The right-hand-side is proportional to the anisotropic stress, i.e. the stress tensor Tij minus its isotropic (i.e. diag- onal) part, proportional to the pressure. This equation shows that Φ = Ψ if there are no anisotropic stresses. Finally, the transverse-trace-free part of Gij gives us the following equation for the TT part of the metric pertur- bation: TT TT TT TT hij = −16πTij ;Tij ≡ PijmnTmn: (13) As expected, the EFEs only provide information about the 6 gauge-invariant, physical degrees of freedom. To fix all 10 components of hµν one needs to impose 4 additional and freely specifiable gauge (or coordinate) conditions. Constraints and dynamics Now, let us consider the character of these equations. The first three are constraint equations: they do not involve any time derivatives. It is easier to see this explicitly in the transverse gauge, which is the generalization of the Coulomb gauge in electromagnetism: this gauge is defined by the 4 conditions 1 @ h = 0 = @ h − δ h : (14) i 0i i ij 3 ij kk First, one needs to show explicitly that such a gauge choice is indeed possible, and it is: starting from a coordinate 2 i 1 system in which the gauge condition is not satisfied, make a gauge transformations such that r ξ + 3 @jξl;l = 1 2 0 i −2@j(hij − 3 δijhkk) and r ξ = @ih0i + @0ξ;i. The metric perturbation in these new coordinates will satisfy the transverse gauge conditions. 1 1 i ilm^ In the transverse gauge, we have Ψ = − 6 hkk and Φ = − 2 h00, and v = klh0m, i.e. no time derivatives appear, and the equations for Φ; Ψ and vi are clearly purely spatial, constraint equations. TT Finally, the TT part is a truly dynamical equation for the gauge-invariant tensor mode hij . Therefore, just like electromagnetism, GR has two dynamical degrees of freedom, the TT part of the metric. These are the famous gravitational waves, which propagate at the speed of light (they satisfy the wave equation in vaccum), and can exist even in vacuum, while all other components can be set to zero in vacuum by appropriate gauge choices. While we have shown all this in linearized gravity, let's now mention how this carries over to non-linear GR. First, it remains true at the non-linear level that there are 6 physical degrees of freedom, due to the 4 coordinate degrees of freedom.
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