Linearized Gravity February 9, 2026
Reference: Bardeen (1980), PRD 22, 1882
1. Perturbation Expansion ¶ In cosmological perturbation theory, we decompose the metric and the stress‑energy tensor into a homogeneous, isotropic background plus small perturbations:
g μ ν = g ˉ μ ν + δ g μ ν , T μ ν = T ˉ μ ν + δ T μ ν , g_{\mu \nu} = \bar{g}_{\mu \nu} + \delta g_{\mu \nu}, \qquad
T_{\mu \nu} = \bar{T}_{\mu \nu} + \delta T_{\mu \nu}, g μν = g ˉ μν + δ g μν , T μν = T ˉ μν + δ T μν , where:
g ˉ μ ν \bar{g}_{\mu \nu} g ˉ μν
T ˉ μ ν \bar{T}_{\mu \nu} T ˉ μν
δ g μ ν \delta g_{\mu \nu} δ g μν δ T μ ν \delta T_{\mu \nu} δ T μν
2. The Gauge Problem ¶ The split into “background” and “perturbation” is not unique.spatial average on each constant‑time hypersurface.slicing ) is coordinate‑dependent, so the correspondence between the perturbed quantity δ g μ ν \delta g_{\mu \nu} δ g μν g ˉ μ ν \bar{g}_{\mu \nu} g ˉ μν
Consider an infinitesimal coordinate change:
x μ ⟶ x ~ μ = x μ + ϵ μ ( x ) , ∣ ϵ μ ∣ ≪ 1. x^\mu \;\longrightarrow\; \tilde{x}^\mu = x^\mu + \epsilon^\mu(x),
\qquad |\epsilon^\mu| \ll 1 . x μ ⟶ x ~ μ = x μ + ϵ μ ( x ) , ∣ ϵ μ ∣ ≪ 1. Under such a transformation, the metric perturbation (see the details in
δ g ~ μ ν = δ g μ ν − ∇ ˉ μ ϵ ν − ∇ ˉ ν ϵ μ , \delta \tilde{g}_{\mu \nu}
= \delta g_{\mu \nu} - \bar{\nabla}_{\!\mu}\epsilon_{\nu}
- \bar{\nabla}_{\!\nu}\epsilon_{\mu}, δ g ~ μν = δ g μν − ∇ ˉ μ ϵ ν − ∇ ˉ ν ϵ μ , where ∇ ˉ \bar{\nabla} ∇ ˉ g ˉ μ ν \bar{g}_{\mu \nu} g ˉ μν ϵ μ = g ˉ μ ν ϵ ν \epsilon_\mu = \bar{g}_{\mu\nu}\epsilon^\nu ϵ μ = g ˉ μν ϵ ν δ g μ ν \delta g_{\mu \nu} δ g μν gauge‑invariant ; different choices of ϵ μ \epsilon^\mu ϵ μ
The same logic applies to any matter or geometric quantity. For a generic tensor field T T T
δ T ~ = δ T − L ϵ T ˉ . \delta \tilde{T} = \delta T - \mathcal{L}_\epsilon \bar{T}. δ T ~ = δ T − L ϵ T ˉ . Explicitly, for the cases relevant to cosmology:
To clarify the concept, let us denote by U U U real perturbed spacetime and by U ˉ \bar{U} U ˉ homogeneous background spacetime .
A gauge is a mapping between U U U U ˉ \bar{U} U ˉ U U U x μ x^\mu x μ
x μ : U ⟶ R 4 . x^\mu : U \longrightarrow \mathbb{R}^4 . x μ : U ⟶ R 4 . A change of gauge corresponds to changing the identification between R 4 \mathbb{R}^4 R 4 U U U
The perturbation written in the coordinate system x μ x^\mu x μ same coordinate values x μ x^\mu x μ different physical point M ∈ U M \in U M ∈ U R 4 \mathbb{R}^4 R 4 U U U
In practice, a pure gauge transformation is simply an infinitesimal coordinate transformation that does not affect the background – it only changes the way we split a given physical spacetime into background plus perturbation. Such transformations are the source of the gauge ambiguity.
2.4 Constructing Gauge‑Invariant Variables ¶ To obtain physically meaningful quantities, one must combine components of δ g μ ν \delta g_{\mu \nu} δ g μν δ T μ ν \delta T_{\mu \nu} δ T μν gauge‑invariant variables .
The following simple 2‑dimensional example illustrates how a coordinate transformation can “absorb” a perturbation, making a perturbed universe appear perfectly homogeneous. It also shows why gauge‑invariant combinations are necessary.
A 2D Toy Example ¶ Consider a two‑dimensional spacetime with coordinates ( t , x ) (t,x) ( t , x )
d s 2 = − d t 2 + [ 1 + h ( x ) ] d x 2 , ds^2 = -dt^2 + \bigl[1 + h(x)\bigr] dx^2, d s 2 = − d t 2 + [ 1 + h ( x ) ] d x 2 , where h ( x ) h(x) h ( x )
{ t ′ = t + ϵ sin x , x ′ = x , ϵ ≪ 1. \begin{cases}
t' = t + \epsilon \sin x ,\\[4pt]
x' = x ,
\end{cases}
\qquad \epsilon \ll 1 . { t ′ = t + ϵ sin x , x ′ = x , ϵ ≪ 1. Task :ϵ \epsilon ϵ h ( x ) h(x) h ( x ) d s 2 = − d t ′ 2 + d x ′ 2 ds^2 = -dt'^2 + dx'^2 d s 2 = − d t ′2 + d x ′2
First, compute the differentials:
d t = d t ′ − ϵ cos x ′ d x ′ , d x = d x ′ . \begin{aligned}
dt &= dt' - \epsilon \cos x' \, dx', \\
dx &= dx'.
\end{aligned} d t d x = d t ′ − ϵ cos x ′ d x ′ , = d x ′ . 3. Metric perturbations ¶ Most general metric perturbations read:
d s 2 = a 2 ( τ ) [ ( 1 + 2 ϕ ) d τ 2 − 2 B i d x i d τ − ( ( 1 − 2 ψ ) δ i j + H i j ) d x i d x j ] \boxed{
ds^2 = a^2(\tau) \left[ (1+2\phi) d\tau^2 - 2 B_i dx^i d\tau - ((1 - 2\psi)\delta_{ij} + H_{ij}) dx^i dx^j \right]
} d s 2 = a 2 ( τ ) [ ( 1 + 2 ϕ ) d τ 2 − 2 B i d x i d τ − (( 1 − 2 ψ ) δ ij + H ij ) d x i d x j ] 3.1 Scalar, Vectors and Tensors ¶ SVT decomposition separates perturbations into scalar, vector, and tensor parts.
At linear order they are independent, so we can study each one separately.
In practice we care mostly about scalar and tensor modes.
Vector modes are not produced in standard inflation and quickly decay as the Universe expands.
They would create a preferred direction, which contradicts the observed homogeneity and isotropy of the early Universe.
For any vector field B \mathbf{B} B R 3 \mathbb{R}^3 R 3
B = ∇ B + B ~ \mathbf{B} = \nabla B + \mathbf{\tilde{B}} B = ∇ B + B ~ where:
∇ B \nabla B ∇ B gradient (curl-free, “longitudinal”)
B ~ \mathbf{\tilde{B}} B ~ divergence-free (∇ ⋅ B ~ = 0 \nabla \cdot \mathbf{\tilde{B}} = 0 ∇ ⋅ B ~ = 0
In Fourier space: B i ( k ) = i k i B ( k ) + B ~ i ( k ) B_i(\mathbf{k}) = i k_i B(\mathbf{k}) + \tilde{B}_i(\mathbf{k}) B i ( k ) = i k i B ( k ) + B ~ i ( k )
i k i B ( k ) i k_i B(\mathbf{k}) i k i B ( k ) k \mathbf{k} k
B ~ i ( k ) \tilde{B}_i(\mathbf{k}) B ~ i ( k ) k ⋅ B ~ ( k ) = 0 \mathbf{k} \cdot \mathbf{\tilde{B}}(\mathbf{k}) = 0 k ⋅ B ~ ( k ) = 0
For a symmetric tensor h i j h_{ij} h ij
The Trace
h a a = 3 C ⇒ C = 1 3 h a a h^a_a = 3C \quad \Rightarrow \quad C = \frac{1}{3}h^a_a h a a = 3 C ⇒ C = 3 1 h a a This is a scalar , the simplest piece.
Remove the Trace
Define the traceless part:
S i j = h i j − C δ i j S_{ij} = h_{ij} - C\delta_{ij} S ij = h ij − C δ ij Now S a a = 0 S^a_a = 0 S a a = 0
Decompose S i j S_{ij} S ij
“What’s the most general traceless symmetric tensor I can build from derivatives of scalar/vector functions?”
Consider all possible derivatives:
From a scalar E E E :
∂ i ∂ j E but this has trace! \partial_i\partial_j E \quad \text{but this has trace!} ∂ i ∂ j E but this has trace! ∇ 2 E = ∂ a ∂ a E \nabla^2 E = \partial^a\partial_a E ∇ 2 E = ∂ a ∂ a E traceless :
∂ i ∂ j E − 1 3 δ i j ∇ 2 E \partial_i\partial_j E - \frac{1}{3}\delta_{ij}\nabla^2 E ∂ i ∂ j E − 3 1 δ ij ∇ 2 E This is pure scalar origin but traceless.
From a vector F i F_i F i :
The symmetric derivative ∂ ( i F j ) = 1 2 ( ∂ i F j + ∂ j F i ) \partial_{(i}F_{j)} = \frac{1}{2}(\partial_i F_j + \partial_j F_i) ∂ ( i F j ) = 2 1 ( ∂ i F j + ∂ j F i )
This also has a trace ∂ a F a \partial^a F_a ∂ a F a
So we must impose ∂ a F a = 0 \partial^a F_a = 0 ∂ a F a = 0
Tensor h ~ i j \tilde{h}_{ij} h ~ ij :
What remains is h ~ i j \tilde{h}_{ij} h ~ ij
Vector decomposition
B i = ∂ i b ⏟ scalar + b i ^ ⏟ vector \boxed{
B_i = \underbrace{\partial_i b}_{\text{scalar}} + \underbrace{\hat{b_i}}_{\text{vector}}
} B i = scalar ∂ i b + vector b i ^
with ∂ i B i ^ = 0 \partial^i \hat{B_i} = 0 ∂ i B i ^ = 0
Rank-2 tensor decomposition
H i j = 2 ∂ ⟨ i ∂ j ⟩ μ ⏟ scalar + 2 ∂ ( i A ^ j ) ⏟ vector + 2 H ^ i j ⏟ tensor \boxed{
H_{ij} = \underbrace{2 \partial_{\langle i} \partial_{j \rangle} \mu}_{\text{scalar}} + \underbrace{2 \partial_{( i} \hat{A}_{j )}}_{\text{vector}} + \underbrace{2 \hat{H}_{ij}}_{\text{tensor}}
} H ij = scalar 2 ∂ ⟨ i ∂ j ⟩ μ + vector 2 ∂ ( i A ^ j ) + tensor 2 H ^ ij
where
∂ ⟨ i ∂ j ⟩ μ ≡ ( ∂ i ∂ j − 1 3 δ i j ∇ 2 ) μ ∂ ( i A ^ j ) ≡ 1 2 ( ∂ i A ^ j + ∂ j A ^ i ) \begin{align*}
\partial_{\langle i} \partial_{j\rangle} \mu & \equiv\left(\partial_i \partial_j-\frac{1}{3} \delta_{i j} \nabla^2\right) \mu \\
\partial_{(i} \hat{A}_{j)} & \equiv \frac{1}{2}\left(\partial_i \hat{A}_j+\partial_j \hat{A}_i\right)
\end{align*} ∂ ⟨ i ∂ j ⟩ μ ∂ ( i A ^ j ) ≡ ( ∂ i ∂ j − 3 1 δ ij ∇ 2 ) μ ≡ 2 1 ( ∂ i A ^ j + ∂ j A ^ i )
with ∂ i A i ^ = 0 \partial^i \hat{A_i} = 0 ∂ i A i ^ = 0 ∂ i H i j ^ = 0 \partial^i \hat{H_{ij}} = 0 ∂ i H ij ^ = 0 E i i ^ = 0 \hat{E^i_i} = 0 E i i ^ = 0
The 10 degrees of freedom of the metric is decomposed into
scalars : ϕ , ψ , b , μ → 4 dof vectors : b i ^ , A i ^ → 2 dof tensors : H i j ^ → 2 dof \begin{align*}
&\text{scalars}: \phi, \psi, b, \mu &\to \quad &\text{4 dof} \\
&\text{vectors}: \hat{b_i}, \hat{A_i} &\to \quad &\text{2 dof} \\
&\text{tensors}: \hat{H_{ij}} &\to \quad &\text{2 dof}
\end{align*} scalars : ϕ , ψ , b , μ vectors : b i ^ , A i ^ tensors : H ij ^ → → → 4 dof 2 dof 2 dof
Ψ ≡ A + H ( B − E ′ ) + ( B − E ′ ) ′ , Φ ^ i ≡ E ^ i ′ − B ^ i , E ^ i j , Φ ≡ − C − H ( B − E ′ ) + 1 3 ∇ 2 E . \begin{align*}
\Psi & \equiv A+\mathcal{H}\left(B-E^{\prime}\right)+\left(B-E^{\prime}\right)^{\prime}, \quad \hat{\Phi}_i \equiv \hat{E}_i^{\prime}-\hat{B}_i, \quad \hat{E}_{i j}, \\
\Phi & \equiv-C-\mathcal{H}\left(B-E^{\prime}\right)+\frac{1}{3} \nabla^2 E .
\end{align*} Ψ Φ ≡ A + H ( B − E ′ ) + ( B − E ′ ) ′ , Φ ^ i ≡ E ^ i ′ − B ^ i , E ^ ij , ≡ − C − H ( B − E ′ ) + 3 1 ∇ 2 E . The impact of gauge transformations can be computed from previous formulas for tensors
of Lorentz:
δ g μ ν ′ = δ g μ ν − ∂ μ ϵ ρ g ˉ ρ ν − ∂ ν ϵ ρ g ˉ μ ρ − ϵ ρ ∂ ρ g ˉ μ ν \delta g'_{\mu \nu} = \delta g_{\mu \nu} - \partial_\mu \epsilon^\rho \bar g_{\rho \nu} - \partial_\nu \epsilon^\rho \bar g_{\mu \rho} - \epsilon^\rho \partial_\rho \bar g_{\mu \nu} δ g μν ′ = δ g μν − ∂ μ ϵ ρ g ˉ ρ ν − ∂ ν ϵ ρ g ˉ μ ρ − ϵ ρ ∂ ρ g ˉ μν Let
x μ → x μ + ϵ μ x^\mu \to x^\mu + \epsilon^\mu x μ → x μ + ϵ μ where
{ ϵ 0 = α ϵ i = ∂ i β + β ^ i \begin{cases}
\epsilon^0 &= \alpha \\
\epsilon^i &= \partial^i \beta + \hat \beta^i
\end{cases} { ϵ 0 ϵ i = α = ∂ i β + β ^ i We have
{ ϕ ′ = ϕ − α ˙ − H α b ′ = b + β ˙ − α ψ ′ = ψ + 1 3 ∇ 2 β + H α μ ′ = μ − β \begin{cases}
\phi' &= \phi - \dot \alpha - \mathcal{H} \alpha \\
b' &= b + \dot \beta - \alpha \\
\psi' &= \psi + \dfrac{1}{3} \nabla^2 \beta + \mathcal{H} \alpha \\
\mu' &= \mu - \beta
\end{cases} ⎩ ⎨ ⎧ ϕ ′ b ′ ψ ′ μ ′ = ϕ − α ˙ − H α = b + β ˙ − α = ψ + 3 1 ∇ 2 β + H α = μ − β where . = ∂ ∂ η . = \dfrac{\partial}{\partial \eta} . = ∂ η ∂
Proof 1 (Scalar transformation laws)
Given g ˉ μ ν = ( a 2 0 0 − a 2 δ i j ) \bar g_{\mu \nu} = \begin{pmatrix}
a^2 & 0 \\ 0 & -a^2 \delta_{ij}
\end{pmatrix} g ˉ μν = ( a 2 0 0 − a 2 δ ij ) { ϵ 0 = α ϵ i = ∂ i β + ϵ j k i ∂ j β k \begin{cases}
\epsilon^0 &= \alpha \\
\epsilon^i &= \partial^i \beta + \epsilon^i_{jk}\partial^j \beta^k
\end{cases} { ϵ 0 ϵ i = α = ∂ i β + ϵ jk i ∂ j β k
00-component
δ g 00 ′ = δ g 00 − ϵ 0 α g ˉ 00 − ∂ 0 ϵ 0 g ˉ 00 − ϵ 0 ∂ 0 g ˉ 00 2 a 2 ϕ ′ = 2 a 2 ϕ − 2 α ˙ a 2 − 2 a ˙ a a 2 α ϕ ′ = ϕ − α ˙ − H α \begin{align*}
\delta g_{00}' &= \delta g_{00} - \epsilon^0 \alpha \bar g_{00} - \partial_0 \epsilon^0 \bar g_{00} - \epsilon^0 \partial_0 \bar g_{00} \\
2a^2 \phi ' &= 2 a^2 \phi - 2 \dot \alpha a^2 - 2 \dfrac{\dot a}{a} a^2 \alpha \\
&\boxed{
\phi' = \phi - \dot \alpha - \mathcal{H} \alpha
}
\end{align*} δ g 00 ′ 2 a 2 ϕ ′ = δ g 00 − ϵ 0 α g ˉ 00 − ∂ 0 ϵ 0 g ˉ 00 − ϵ 0 ∂ 0 g ˉ 00 = 2 a 2 ϕ − 2 α ˙ a 2 − 2 a a ˙ a 2 α ϕ ′ = ϕ − α ˙ − H α 0 i 0i 0 i
δ g 0 i ′ = δ g 0 i − ∂ 0 ϵ j g ˉ j i − ∂ i ϵ 0 g ˉ 00 a 2 ∂ i b ′ = a 2 ∂ i b + ∂ 0 ( ∂ j β ) a 2 δ i j − ∂ i ϵ 0 a 2 ∂ i b ′ = ∂ i b + ∂ 0 ∂ i β − ∂ i α b ′ = b + β ˙ − α \begin{align*}
\delta g_{0i}' &= \delta g_{0i} - \partial_0 \epsilon^j \bar g_{ji} - \partial_i \epsilon^0 \bar g_{00} \\
a^2 \partial_i b' &= a^2 \partial_i b + \partial_0(\partial^j \beta) a^2 \delta_{ij} - \partial_i \epsilon^0 a^2 \\
\partial_i b' &= \partial_i b + \partial_0 \partial_i \beta - \partial_i \alpha \\
&\boxed{b' = b + \dot \beta - \alpha}
\end{align*} δ g 0 i ′ a 2 ∂ i b ′ ∂ i b ′ = δ g 0 i − ∂ 0 ϵ j g ˉ ji − ∂ i ϵ 0 g ˉ 00 = a 2 ∂ i b + ∂ 0 ( ∂ j β ) a 2 δ ij − ∂ i ϵ 0 a 2 = ∂ i b + ∂ 0 ∂ i β − ∂ i α b ′ = b + β ˙ − α i j ij ij
δ g i j ′ = δ g i j − ∂ i ϵ k g ˉ j k − ∂ j ϵ k g ˉ i k − ϵ 0 ∂ 0 g ˉ i j a 2 [ 2 ψ ′ − H i j ′ ] = a 2 [ 2 ψ − H i j ] + ∂ i ∂ k β a 2 δ j k + ∂ j ∂ k β a 2 δ i k + α 2 a ˙ a a 2 δ i j 2 ψ ′ δ i j − H i j ′ = 2 ψ δ i j − H i j + 2 ∂ i ∂ j β + 2 H α δ i j \begin{align*}
\delta g_{ij}' &= \delta g_{ij} - \partial_i \epsilon^k \bar g_{jk} - \partial_j \epsilon^k \bar g_{ik} - \epsilon^0 \partial_0 \bar g_{ij} \\
a^2[2\psi' - H'_{ij}] &= a^2[2\psi - H_{ij}] + \partial_i \partial^k \beta a^2 \delta_{jk} + \partial_j \partial^k \beta a^2 \delta_{ik} + \alpha 2 \dfrac{\dot a}{a} a^2 \delta_{ij} \\
2 \psi' \delta_{ij} - H'_{ij} &= 2\psi \delta_{ij} - H_{ij} + 2\partial_i \partial_j \beta + 2 \mathcal{H} \alpha \delta_{ij}
\end{align*} δ g ij ′ a 2 [ 2 ψ ′ − H ij ′ ] 2 ψ ′ δ ij − H ij ′ = δ g ij − ∂ i ϵ k g ˉ jk − ∂ j ϵ k g ˉ ik − ϵ 0 ∂ 0 g ˉ ij = a 2 [ 2 ψ − H ij ] + ∂ i ∂ k β a 2 δ jk + ∂ j ∂ k β a 2 δ ik + α 2 a a ˙ a 2 δ ij = 2 ψ δ ij − H ij + 2 ∂ i ∂ j β + 2 H α δ ij Contracting with δ i j \delta^{ij} δ ij H i j H_{ij} H ij H i j ′ H'_{ij} H ij ′
6 ψ ′ = 6 ψ + 2 ∇ 2 β + 6 H α ψ ′ = ψ + 1 3 ∇ 2 β + H α \begin{align*}
&6 \psi' = 6 \psi + 2\nabla^2 \beta + 6 \mathcal{H} \alpha \\
&\boxed{\psi' = \psi + \dfrac{1}{3} \nabla^2 \beta + \mathcal{H} \alpha}
\end{align*} 6 ψ ′ = 6 ψ + 2 ∇ 2 β + 6 H α ψ ′ = ψ + 3 1 ∇ 2 β + H α i j ij ij
From the previous part, we have
2 ( ψ ′ − ψ ) δ i j − H i j ′ = − H i j + 2 ∂ i ∂ j β + 2 H α δ i j 2 ( 1 3 ∇ 2 β + H α ) δ i j − H i j ′ = − H i j + 2 ∂ i ∂ j β + 2 H α δ i j H i j ′ = H i j − 2 ∂ i ∂ j β + 2 3 ∇ 2 β δ i j 2 ( ∂ i ∂ j − 1 3 ∇ 2 ) μ ′ = 2 ( ∂ i ∂ j − 1 3 ∇ 2 ) μ − 2 ∂ i ∂ j β + 2 3 ∇ 2 β δ i j 2 ( ∂ i ∂ j − 1 3 ∇ 2 ) μ ′ = 2 ( ∂ i ∂ j − 1 3 ∇ 2 ) ( μ − β ) \begin{align*}
2(\psi' - \psi) \delta_{ij} - H'_{ij} &= - H_{ij} + 2 \partial_i \partial_j \beta + 2\mathcal{H} \alpha \delta_{ij} \\
2 \left( \dfrac{1}{3} \nabla^2 \beta + \mathcal{H} \alpha \right) \delta_{ij} - H'_{ij} &= - H_{ij} + 2 \partial_i \partial_j \beta + 2\mathcal{H} \alpha \delta_{ij} \\
H'_{ij} &= H_{ij} - 2\partial_i \partial_j \beta + \dfrac{2}{3} \nabla^2 \beta \delta_{ij} \\
2 \left( \partial_i \partial_j - \dfrac{1}{3} \nabla^2 \right) \mu' &= 2 \left( \partial_i \partial_j - \dfrac{1}{3} \nabla^2 \right) \mu - 2\partial_i \partial_j \beta + \dfrac{2}{3} \nabla^2 \beta \delta_{ij} \\
2 \left( \partial_i \partial_j - \dfrac{1}{3} \nabla^2 \right) \mu' &= 2 \left( \partial_i \partial_j - \dfrac{1}{3} \nabla^2 \right) (\mu - \beta) \\
\end{align*} 2 ( ψ ′ − ψ ) δ ij − H ij ′ 2 ( 3 1 ∇ 2 β + H α ) δ ij − H ij ′ H ij ′ 2 ( ∂ i ∂ j − 3 1 ∇ 2 ) μ ′ 2 ( ∂ i ∂ j − 3 1 ∇ 2 ) μ ′ = − H ij + 2 ∂ i ∂ j β + 2 H α δ ij = − H ij + 2 ∂ i ∂ j β + 2 H α δ ij = H ij − 2 ∂ i ∂ j β + 3 2 ∇ 2 β δ ij = 2 ( ∂ i ∂ j − 3 1 ∇ 2 ) μ − 2 ∂ i ∂ j β + 3 2 ∇ 2 β δ ij = 2 ( ∂ i ∂ j − 3 1 ∇ 2 ) ( μ − β ) So
μ ′ = μ − β \boxed{\mu' = \mu - \beta} μ ′ = μ − β 3.3 Gauge fixing ¶ To solve the gauge problem, we fix the gauge and keep track of all perturbations (metric and matter).
3.4 Newton Gauge ¶ Choose
b ^ i = 0 constant-time hypersurfaces orthogonal to worldlines of observers at rest μ = 0 induced geometry of the constant-time is isotropic \begin{align*}
\hat b_i &= 0 \quad \text{\small constant-time hypersurfaces orthogonal to worldlines of observers at rest} \\
\mu &= 0 \quad \text{\small induced geometry of the constant-time is isotropic}
\end{align*} b ^ i μ = 0 constant-time hypersurfaces orthogonal to worldlines of observers at rest = 0 induced geometry of the constant-time is isotropic The metric
d s 2 = a 2 ( η ) [ ( 1 + 2 ϕ ) d η 2 − ( 1 − 2 ψ ) δ i j d x i d x j ] \boxed{
ds^2 = a^2(\eta) [(1 + 2\phi) d\eta^2 - (1-2\psi) \delta_{ij} dx^i dx^j]
} d s 2 = a 2 ( η ) [( 1 + 2 ϕ ) d η 2 − ( 1 − 2 ψ ) δ ij d x i d x j ] 4. Linearized Einstein Equation ¶ We will work with Newton gauge.
Connection Coefficients ¶ The connection coefficients associated with the Newton-gauge metric are
Γ 00 0 = H + ϕ ˙ Γ i 0 0 = ∂ i ϕ Γ 00 i = δ i j ∂ j ϕ Γ i j 0 = H δ i j − [ ψ ˙ + 2 H ( ψ + ϕ ) ] δ i j Γ j 0 i = [ H − ψ ˙ ] δ j i Γ j k i = − 2 δ ( j i ∂ k ) ψ + δ j k δ i l ∂ l ψ \begin{align*}
\Gamma^0_{00} &= \mathcal{H} + \dot \phi \\
\Gamma^0_{i0} &= \partial_i \phi \\
\Gamma^i_{00} &= \delta^{ij} \partial_j \phi \\
\Gamma^0_{ij} &= \mathcal{H}\delta_{ij} - [\dot \psi + 2 \mathcal{H}(\psi + \phi)]\delta_{ij} \\
\Gamma^i_{j0} &= [\mathcal{H} - \dot \psi] \delta^i_j \\
\Gamma^i_{jk} &= -2 \delta^i_{(j} \partial_{k)} \psi + \delta_{jk} \delta^{il} \partial_l \psi \\
\end{align*} Γ 00 0 Γ i 0 0 Γ 00 i Γ ij 0 Γ j 0 i Γ jk i = H + ϕ ˙ = ∂ i ϕ = δ ij ∂ j ϕ = H δ ij − [ ψ ˙ + 2 H ( ψ + ϕ )] δ ij = [ H − ψ ˙ ] δ j i = − 2 δ ( j i ∂ k ) ψ + δ jk δ i l ∂ l ψ Proof 2 (Connection coefficients)
The equation for connection coefficients
Γ μ ν α = 1 2 g α β ( ∂ μ g β ν + ∂ ν g μ β − ∂ β g μ ν ) \begin{align*}
\Gamma^\alpha_{\mu \nu} &= \dfrac{1}{2} g^{\alpha \beta} (\partial_\mu g_{\beta \nu} + \partial_\nu g_{\mu \beta} - \partial_\beta g_{\mu \nu})
\end{align*} Γ μν α = 2 1 g α β ( ∂ μ g β ν + ∂ ν g μ β − ∂ β g μν ) We will only consider everything to the first order.
Γ 00 0 = 1 2 g 00 ∂ 0 g 00 = 1 2 a − 2 ( 1 − 2 ϕ ) 2 a 2 [ H ( 1 + 2 ϕ ) + ϕ ˙ ] = H + ϕ ˙ Γ i 0 0 = 1 2 g 00 ∂ i g 00 = 1 2 a − 2 ( 1 − 2 ϕ ) 2 a 2 ∂ i ϕ = ∂ i ϕ Γ 00 i = − 1 2 g i j ∂ j g 00 = 1 2 a − 2 ( 1 + 2 ψ ) δ i j 2 a 2 ∂ j ϕ = δ i j ∂ j ϕ Γ i j 0 = − 1 2 g 00 ∂ 0 g i j = 1 2 a − 2 ( 1 − 2 ϕ ) 2 a 2 [ H ( 1 − 2 ψ ) − ψ ˙ ] δ i j = H δ i j − [ ψ ˙ + 2 H ( ϕ + ψ ) ] δ i j Γ j 0 i = 1 2 g i k ∂ 0 g j k = 1 2 a − 2 ( 1 + 2 ψ ) δ i k 2 a 2 [ H ( 1 − 2 ψ ) − ψ ˙ ] δ j k = [ H − ψ ˙ ] δ j i Γ j k i = 1 2 g i l [ ∂ j g l k + ∂ k g j l − ∂ l g j k ] = 1 2 a − 2 ( 1 + 2 ψ ) δ i l 2 a 2 [ − ∂ j ψ δ l k − ∂ k ψ δ j l + ∂ l ψ δ j k ] = − 2 δ ( j i ∂ k ) ψ + δ j k δ i l ∂ l ψ \begin{align*}
\Gamma^0_{00} &= \dfrac{1}{2} g^{00} \partial_0 g_{00} \\
&= \dfrac{1}{2} a^{-2} (1 - 2\phi) 2a^2[ \mathcal{H} (1 + 2\phi) + \dot \phi ] \\
&= \boxed{\mathcal{H} + \dot \phi} \\
\Gamma^0_{i0} &= \dfrac{1}{2} g^{00} \partial_i g_{00} \\
&= \dfrac{1}{2} a^{-2} (1 - 2\phi) 2 a^2 \partial_i \phi \\
&= \boxed{\partial_i \phi} \\
\Gamma^i_{00} &= - \dfrac{1}{2} g^{ij} \partial_j g_{00} \\
&= \dfrac{1}{2} a^{-2} (1 + 2 \psi) \delta^{ij} 2 a^2 \partial_j \phi \\
&= \boxed{\delta^{ij} \partial_j \phi} \\
\Gamma^0_{ij} &= -\dfrac{1}{2} g^{00} \partial_0 g_{ij} \\
&= \dfrac{1}{2} a^{-2} (1 - 2\phi) 2a^2[\mathcal{H}(1-2\psi) - \dot \psi] \delta_{ij} \\
&= \boxed{\mathcal{H}\delta_{ij} - [\dot \psi + 2 \mathcal{H}(\phi + \psi)]\delta_{ij}} \\
\Gamma^i_{j0} &= \dfrac{1}{2} g^{ik} \partial_0 g_{jk} \\
&= \dfrac{1}{2} a^{-2} (1+2\psi) \delta^{ik} 2a^2[\mathcal{H}(1-2\psi) - \dot \psi] \delta_{jk} \\
&= \boxed{[\mathcal{H} - \dot \psi] \delta^i_j} \\
\Gamma^i_{jk} &= \dfrac{1}{2} g^{il}[\partial_j g_{lk} + \partial_k g_{jl} - \partial_l g_{jk}] \\
&= \dfrac{1}{2} a^{-2} (1+2\psi) \delta^{il} 2 a^2[-\partial_j \psi \delta_{lk} - \partial_k \psi \delta_{jl} + \partial_l \psi \delta_{jk}] \\
&= \boxed{-2 \delta^i_{(j} \partial_{k)} \psi + \delta_{jk} \delta^{il} \partial_l \psi}
\end{align*} Γ 00 0 Γ i 0 0 Γ 00 i Γ ij 0 Γ j 0 i Γ jk i = 2 1 g 00 ∂ 0 g 00 = 2 1 a − 2 ( 1 − 2 ϕ ) 2 a 2 [ H ( 1 + 2 ϕ ) + ϕ ˙ ] = H + ϕ ˙ = 2 1 g 00 ∂ i g 00 = 2 1 a − 2 ( 1 − 2 ϕ ) 2 a 2 ∂ i ϕ = ∂ i ϕ = − 2 1 g ij ∂ j g 00 = 2 1 a − 2 ( 1 + 2 ψ ) δ ij 2 a 2 ∂ j ϕ = δ ij ∂ j ϕ = − 2 1 g 00 ∂ 0 g ij = 2 1 a − 2 ( 1 − 2 ϕ ) 2 a 2 [ H ( 1 − 2 ψ ) − ψ ˙ ] δ ij = H δ ij − [ ψ ˙ + 2 H ( ϕ + ψ )] δ ij = 2 1 g ik ∂ 0 g jk = 2 1 a − 2 ( 1 + 2 ψ ) δ ik 2 a 2 [ H ( 1 − 2 ψ ) − ψ ˙ ] δ jk = [ H − ψ ˙ ] δ j i = 2 1 g i l [ ∂ j g l k + ∂ k g j l − ∂ l g jk ] = 2 1 a − 2 ( 1 + 2 ψ ) δ i l 2 a 2 [ − ∂ j ψ δ l k − ∂ k ψ δ j l + ∂ l ψ δ jk ] = − 2 δ ( j i ∂ k ) ψ + δ jk δ i l ∂ l ψ Ricci tensor ¶ Ricci tensor can be expressed in terms of the connection as
R μ ν = ∂ λ Γ μ ν λ − ∂ ν Γ μ λ λ + Γ λ ρ λ Γ μ ν ρ − Γ μ λ ρ Γ ν ρ λ R_{\mu \nu} = \partial_\lambda \Gamma^\lambda_{\mu \nu} - \partial_\nu \Gamma^\lambda_{\mu \lambda} + \Gamma^\lambda_{\lambda \rho} \Gamma^\rho_{\mu \nu} - \Gamma^\rho_{\mu \lambda} \Gamma^\lambda_{\nu \rho} R μν = ∂ λ Γ μν λ − ∂ ν Γ μ λ λ + Γ λ ρ λ Γ μν ρ − Γ μ λ ρ Γ ν ρ λ Substituting the perturbed connection coefficients, we find
R 00 = − 3 H ˙ + ∇ 2 ϕ + 3 H ( ϕ ˙ + ψ ˙ ) + 3 ψ ¨ R 0 i = 2 ∂ i ( ψ ˙ + H ϕ ) R i j = [ H ˙ + 2 H 2 − ψ ¨ + ∇ 2 ψ − 2 ( H ˙ + 2 H 2 ) ( ϕ + ψ ) − H ϕ ˙ − 5 H ψ ˙ ] δ i j + ∂ i ∂ j ( ψ − ϕ ) \begin{align*}
R_{00} &= -3 \dot{\mathcal{H}} + \nabla^2 \phi + 3 \mathcal{H}(\dot \phi + \dot \psi) + 3 \ddot{\psi}\\
R_{0i} &= 2 \partial_i(\dot \psi + \mathcal{H} \phi)\\
R_{ij} &= [\mathcal{\dot H} + 2\mathcal{H}^2 - \ddot{\psi} + \nabla^2 \psi - 2(\mathcal{\dot H} + 2\mathcal{H}^2)(\phi + \psi) - \mathcal{H}\dot \phi - 5 \mathcal{H} \dot \psi]\delta_{ij} + \partial_i \partial_j(\psi - \phi)\\
\end{align*} R 00 R 0 i R ij = − 3 H ˙ + ∇ 2 ϕ + 3 H ( ϕ ˙ + ψ ˙ ) + 3 ψ ¨ = 2 ∂ i ( ψ ˙ + H ϕ ) = [ H ˙ + 2 H 2 − ψ ¨ + ∇ 2 ψ − 2 ( H ˙ + 2 H 2 ) ( ϕ + ψ ) − H ϕ ˙ − 5 H ψ ˙ ] δ ij + ∂ i ∂ j ( ψ − ϕ ) Proof 3 (perturbed Ricci tensors)
R 00 = ∂ ρ Γ 00 ρ − ∂ 0 Γ 0 ρ ρ + Γ 00 α Γ α ρ ρ − Γ 0 ρ α Γ 0 α ρ R_{00} = \partial_\rho \Gamma^\rho_{00} - \partial_0 \Gamma^\rho_{0\rho} + \Gamma^\alpha_{00} \Gamma^\rho_{\alpha \rho} - \Gamma^\alpha_{0\rho} \Gamma^\rho_{0 \alpha} R 00 = ∂ ρ Γ 00 ρ − ∂ 0 Γ 0 ρ ρ + Γ 00 α Γ α ρ ρ − Γ 0 ρ α Γ 0 α ρ The term with ρ = 0 \rho=0 ρ = 0 ρ = i \rho = i ρ = i
R 00 = ∂ i Γ 00 i − ∂ 0 Γ 0 i ρ + Γ 00 α Γ α i i − Γ 0 i α Γ 0 α i = ∂ i Γ 00 i − ∂ 0 Γ 0 i i + Γ 00 0 Γ 0 i i + Γ 00 j Γ j i i ⏟ O ( 2 ) − Γ 0 i 0 Γ 00 i ⏟ O ( 2 ) − Γ 0 i j Γ 0 j i = ∇ 2 ϕ − 3 ∂ 0 ( H − ψ ˙ ) + 3 ( H + ϕ ˙ ) ( H − ψ ˙ ) − ( H − ψ ˙ ) 2 δ i j δ j i = − 3 H ˙ + ∇ 2 ϕ + 3 H ( ϕ ˙ + ψ ˙ ) + 3 ψ ¨ R 0 i = ∂ ρ Γ 0 i ρ − ∂ i Γ 0 ρ ρ + Γ ρ α ρ Γ 0 i α − Γ 0 α ρ Γ i ρ α = [ ∂ i ( ϕ ˙ − ψ ˙ ) ] − [ ∂ i ϕ ˙ − 3 ∂ i ψ ˙ ] + [ 5 H ∂ i ϕ − 3 H ∂ i ψ ] − [ 3 H ∂ i ( ϕ − ψ ) ] = 2 ∂ i ( ψ ˙ + H ϕ ) R i j = ∂ ρ Γ i j ρ − ∂ j Γ i ρ ρ + Γ ρ α ρ Γ i j α − Γ i α ρ Γ j ρ α = [ H ˙ − ψ ¨ − 2 H ˙ ( ϕ + ψ ) − 2 H ( ϕ ˙ + ψ ˙ ) + ∇ 2 ψ ] δ i j − 2 ∂ i ∂ j ψ + ∂ i ∂ j ϕ − 3 ∂ i ∂ j ψ + 4 H 2 δ i j + [ H ϕ ˙ − 7 H ψ ˙ − 8 H 2 ( ϕ + ψ ) ] δ i j + 2 H 2 δ i j − 4 H [ ψ ˙ + H ( ϕ + ψ ) ] δ i j = [ H ˙ + 2 H 2 − ψ ¨ + ∇ 2 ψ − 2 ( H ˙ + 2 H 2 ) ( ϕ + ψ ) − H ϕ ˙ − 5 H ψ ˙ ] δ i j + ∂ i ∂ j ( ψ − ϕ ) \begin{align*}
R_{00} &= \partial_i \Gamma^i_{00} - \partial_0 \Gamma^\rho_{0i} + \Gamma^\alpha_{00} \Gamma^i_{\alpha i} - \Gamma^\alpha_{0i} \Gamma^i_{0 \alpha} \\
&= \partial_i \Gamma^i_{00} - \partial_0 \Gamma^i_{0i} + \Gamma^0_{00} \Gamma^i_{0 i} + \underbrace{\Gamma^j_{00} \Gamma^i_{ji}}_{\mathcal{O(2)}} - \underbrace{\Gamma^0_{0i} \Gamma^i_{00}}_{\mathcal{O(2)}} - \Gamma^j_{0i} \Gamma^i_{0 j} \\
&= \nabla^2 \phi - 3 \partial_0 (\mathcal{H} - \dot \psi) + 3(\mathcal{H} + \dot \phi)(\mathcal{H} - \dot \psi) - (\mathcal{H} - \dot \psi)^2 \delta^j_i \delta^i_j \\
&= \boxed{-3 \dot{\mathcal{H}} + \nabla^2 \phi + 3 \mathcal{H}(\dot \phi + \dot \psi) + 3 \ddot{\psi}} \\
R_{0i} &= \partial_\rho \Gamma^\rho_{0i} - \partial_i \Gamma^\rho_{0\rho} + \Gamma^\rho_{\rho \alpha} \Gamma^\alpha_{0i} - \Gamma^\rho_{0\alpha} \Gamma^\alpha_{i\rho} \\
&= [\partial_i(\dot \phi - \dot \psi) ]-[ \partial_i \dot \phi - 3 \partial_i \dot \psi ] + [5 \mathcal{H} \partial_i \phi - 3 \mathcal{H} \partial_i \psi] - [3 \mathcal{H} \partial_i(\phi - \psi)] \\
&= \boxed{2 \partial_i (\dot \psi + H \phi)} \\
R_{ij} &= \partial_\rho \Gamma^\rho_{ij} - \partial_j \Gamma^\rho_{i\rho} + \Gamma^\rho_{\rho \alpha} \Gamma^\alpha_{ij} - \Gamma^\rho_{i\alpha} \Gamma^\alpha_{j\rho} \\
&= \bigl[\dot{\mathcal{H}} - \ddot\psi - 2\dot{\mathcal{H}}(\phi+\psi) - 2\mathcal{H}(\dot\phi+\dot\psi) + \nabla^2\psi\bigr]\delta_{ij} - 2\partial_i\partial_j\psi \\
& \quad + \partial_i\partial_j\phi - 3\partial_i\partial_j\psi \\
& \quad + 4\mathcal{H}^2\delta_{ij} + \bigl[\mathcal{H}\dot\phi - 7\mathcal{H}\dot\psi - 8\mathcal{H}^2(\phi+\psi)\bigr]\delta_{ij} \\
& \quad + 2\mathcal{H}^2\delta_{ij} - 4\mathcal{H}\bigl[\dot\psi + \mathcal{H}(\phi+\psi)\bigr]\delta_{ij} \\
&= \boxed{[\mathcal{\dot H} + 2\mathcal{H}^2 - \ddot{\psi} + \nabla^2 \psi - 2(\mathcal{\dot H} + 2\mathcal{H}^2)(\phi + \psi) - \mathcal{H}\dot \phi - 5 \mathcal{H} \dot \psi]\delta_{ij} + \partial_i \partial_j(\psi - \phi)}
\end{align*} R 00 R 0 i R ij = ∂ i Γ 00 i − ∂ 0 Γ 0 i ρ + Γ 00 α Γ α i i − Γ 0 i α Γ 0 α i = ∂ i Γ 00 i − ∂ 0 Γ 0 i i + Γ 00 0 Γ 0 i i + O ( 2 ) Γ 00 j Γ ji i − O ( 2 ) Γ 0 i 0 Γ 00 i − Γ 0 i j Γ 0 j i = ∇ 2 ϕ − 3 ∂ 0 ( H − ψ ˙ ) + 3 ( H + ϕ ˙ ) ( H − ψ ˙ ) − ( H − ψ ˙ ) 2 δ i j δ j i = − 3 H ˙ + ∇ 2 ϕ + 3 H ( ϕ ˙ + ψ ˙ ) + 3 ψ ¨ = ∂ ρ Γ 0 i ρ − ∂ i Γ 0 ρ ρ + Γ ρ α ρ Γ 0 i α − Γ 0 α ρ Γ i ρ α = [ ∂ i ( ϕ ˙ − ψ ˙ )] − [ ∂ i ϕ ˙ − 3 ∂ i ψ ˙ ] + [ 5 H ∂ i ϕ − 3 H ∂ i ψ ] − [ 3 H ∂ i ( ϕ − ψ )] = 2 ∂ i ( ψ ˙ + H ϕ ) = ∂ ρ Γ ij ρ − ∂ j Γ i ρ ρ + Γ ρ α ρ Γ ij α − Γ i α ρ Γ j ρ α = [ H ˙ − ψ ¨ − 2 H ˙ ( ϕ + ψ ) − 2 H ( ϕ ˙ + ψ ˙ ) + ∇ 2 ψ ] δ ij − 2 ∂ i ∂ j ψ + ∂ i ∂ j ϕ − 3 ∂ i ∂ j ψ + 4 H 2 δ ij + [ H ϕ ˙ − 7 H ψ ˙ − 8 H 2 ( ϕ + ψ ) ] δ ij + 2 H 2 δ ij − 4 H [ ψ ˙ + H ( ϕ + ψ ) ] δ ij = [ H ˙ + 2 H 2 − ψ ¨ + ∇ 2 ψ − 2 ( H ˙ + 2 H 2 ) ( ϕ + ψ ) − H ϕ ˙ − 5 H ψ ˙ ] δ ij + ∂ i ∂ j ( ψ − ϕ ) Ricci scalar ¶ Ricci scalar can be computed as
R = 1 a 2 [ − 6 ( H ˙ + H 2 ) + 2 ∇ 2 ϕ − 4 ∇ 2 ψ + 12 ( H ˙ + H 2 ) ϕ + 6 ψ ¨ + 6 H ( ϕ ˙ + 3 ψ ˙ ) ] \boxed{
R = \dfrac{1}{a^2} \left[ -6(\mathcal{\dot H} + \mathcal{H}^2) + 2\nabla^2 \phi - 4\nabla^2 \psi + 12(\mathcal{\dot H} + \mathcal{H}^2)\phi + 6 \ddot{\psi} + 6 \mathcal{H}(\dot \phi + 3 \dot \psi) \right]} R = a 2 1 [ − 6 ( H ˙ + H 2 ) + 2 ∇ 2 ϕ − 4 ∇ 2 ψ + 12 ( H ˙ + H 2 ) ϕ + 6 ψ ¨ + 6 H ( ϕ ˙ + 3 ψ ˙ ) ] Proof 4 (Ricci scalar)
R = g μ ν R μ ν = g 00 R 00 + 2 g 0 i R 0 i ⏟ O ( 2 ) + g i j R i j \begin{align*}
R &= g^{\mu \nu} R_{\mu \nu} \\
&= g^{00} R_{00} + 2 \underbrace{g^{0i} R_{0i}}_{\mathcal{O(2)}} + g^{ij} R_{ij}
\end{align*} R = g μν R μν = g 00 R 00 + 2 O ( 2 ) g 0 i R 0 i + g ij R ij We have
a 2 R = ( 1 − 2 ϕ ) R 00 − ( 1 + 2 ψ ) δ i j R i j = ( 1 − 2 ϕ ) [ − 3 H ˙ + ∇ 2 ϕ + 3 H ( ϕ ˙ + ψ ˙ ) + 3 ψ ¨ ] − 3 ( 1 + 2 ψ ) [ H ˙ + 2 H 2 − ψ ¨ + ∇ 2 ψ − H ϕ ˙ − 5 H ψ ˙ − 2 ( H ˙ + 2 H 2 ) ( ϕ + ψ ) ] − ( 1 + 2 ψ ) ∇ 2 ( ψ − ϕ ) \begin{align*}
a^2R &= (1-2\phi)R_{00} - (1+2\psi)\delta^{ij}R_{ij} \\
&= (1-2\phi)\bigl[-3\dot{\mathcal{H}} + \nabla^2\phi + 3\mathcal{H}(\dot\phi + \dot\psi) + 3\ddot\psi\bigr] \\
& \quad - 3(1+2\psi) \left[\dot{\mathcal{H}} + 2\mathcal{H}^2 - \ddot\psi + \nabla^2\psi - \mathcal{H}\dot\phi - 5\mathcal{H}\dot\psi - 2(\dot{\mathcal{H}}+2\mathcal{H}^2)(\phi+\psi)\right] \\
& \quad - (1 + 2\psi) \nabla^2 (\psi - \phi)
\end{align*} a 2 R = ( 1 − 2 ϕ ) R 00 − ( 1 + 2 ψ ) δ ij R ij = ( 1 − 2 ϕ ) [ − 3 H ˙ + ∇ 2 ϕ + 3 H ( ϕ ˙ + ψ ˙ ) + 3 ψ ¨ ] − 3 ( 1 + 2 ψ ) [ H ˙ + 2 H 2 − ψ ¨ + ∇ 2 ψ − H ϕ ˙ − 5 H ψ ˙ − 2 ( H ˙ + 2 H 2 ) ( ϕ + ψ ) ] − ( 1 + 2 ψ ) ∇ 2 ( ψ − ϕ ) Cancel the non-linear term
a 2 R = − 6 ( H ˙ + H 2 ) + 2 ∇ 2 ϕ − 4 ∇ 2 ψ + 12 ( H ˙ + H 2 ) ϕ + 6 ψ ¨ + 6 H ( ϕ ˙ + 3 ψ ˙ ) a^2 R = -6(\mathcal{\dot H} + \mathcal{H}^2) + 2 \nabla^2 \phi - 4 \nabla^2 \psi + 12(\mathcal{\dot H} + \mathcal{H}^2)\phi + 6 \ddot{\psi} + 6 \mathcal{H} (\dot \phi + 3 \dot \psi) \\ a 2 R = − 6 ( H ˙ + H 2 ) + 2 ∇ 2 ϕ − 4 ∇ 2 ψ + 12 ( H ˙ + H 2 ) ϕ + 6 ψ ¨ + 6 H ( ϕ ˙ + 3 ψ ˙ ) Einstein tensors ¶ From the equation
G μ ν = R μ ν − 1 2 R g μ ν G_{\mu \nu} = R_{\mu \nu} - \dfrac{1}{2} R g_{\mu \nu} G μν = R μν − 2 1 R g μν This is just a basic substitution problem:
δ G 0 0 = 2 a − 2 [ ∇ 2 ψ − 3 H ( ψ ˙ + H ϕ ) ] δ G 0 i = − 2 a − 2 ∂ i ( ψ ˙ + H ϕ ) \begin{align*}
\delta G^0_0 &= 2 a^{-2} \, [ \nabla^2 \psi - 3 \mathcal{H} (\dot \psi + \mathcal{H}\phi) ] \\
\delta G^i_0 &= -2 a^{-2} \, \partial^i (\dot \psi + \mathcal{H} \phi) \\
\end{align*} δ G 0 0 δ G 0 i = 2 a − 2 [ ∇ 2 ψ − 3 H ( ψ ˙ + H ϕ )] = − 2 a − 2 ∂ i ( ψ ˙ + H ϕ ) For δ G j i \delta G^i_j δ G j i P i j ≡ k j ^ k i ^ − 1 3 δ i j P^j_i \equiv \hat{k^j} \hat{k_i} - \dfrac{1}{3} \delta^j_i P i j ≡ k j ^ k i ^ − 3 1 δ i j
P i j δ G j i = 2 3 a − 2 k 2 ( ψ − ϕ ) P^j_i \delta G^i_j = \dfrac{2}{3} a^{-2} k^2 (\psi - \phi) P i j δ G j i = 3 2 a − 2 k 2 ( ψ − ϕ ) Proof 5 (Einstein tensors)
G 0 0 = g 00 ( R 00 − 1 2 g 00 δ R ) = 1 a 2 ( 1 − 2 ϕ ) R 00 − 1 2 R ⇒ δ G 0 0 = 2 a − 2 [ ∇ 2 ψ − 3 H ( ψ ˙ + H ϕ ) ] G 0 i = g i j [ R j 0 − 1 2 g j 0 R ] = g i j R j 0 ⇒ δ G 0 i = − 2 a − 2 ∂ i ( ψ ˙ + H ϕ ) \begin{align*}
G^0_0 &= g^{00} \left(R_{00} - \dfrac{1}{2} g_{00} \delta R\right) \\
&= \dfrac{1}{a^2} (1-2\phi) R_{00} - \dfrac{1}{2} R \\
\Rightarrow \delta G^0_0 &= \boxed{2 a^{-2} \, [ \nabla^2 \psi - 3 \mathcal{H} (\dot \psi + \mathcal{H}\phi) ]} \\
G^i_0 &= g^{ij} [R_{j0} - \dfrac{1}{2} g_{j0} R] \\
&= g^{ij} R_{j0} \\
\Rightarrow \delta G^i_0 &= \boxed{-2 a^{-2} \partial^i (\dot \psi + \mathcal{H} \phi)}
\end{align*} G 0 0 ⇒ δ G 0 0 G 0 i ⇒ δ G 0 i = g 00 ( R 00 − 2 1 g 00 δ R ) = a 2 1 ( 1 − 2 ϕ ) R 00 − 2 1 R = 2 a − 2 [ ∇ 2 ψ − 3 H ( ψ ˙ + H ϕ )] = g ij [ R j 0 − 2 1 g j 0 R ] = g ij R j 0 = − 2 a − 2 ∂ i ( ψ ˙ + H ϕ ) Matter perturbation ¶ Consider perturbations of energy momentum tensors
T 0 0 = ρ ˉ + δ ρ T 0 i = ( ρ ˉ + P ˉ ) v i T j i = − ( P ˉ + δ P ) δ j i − Π j i \begin{align*}
T^0_0 &= \bar \rho + \delta \rho \\
T^i_0 &= (\bar \rho + \bar P) v^i \\
T^i_j &= -(\bar P + \delta P) \delta^i_j - \Pi^i_j
\end{align*} T 0 0 T 0 i T j i = ρ ˉ + δ ρ = ( ρ ˉ + P ˉ ) v i = − ( P ˉ + δ P ) δ j i − Π j i where
v i v^i v i bulk velocity .
Π j i \Pi^i_j Π j i anisotropic stress (a transverse, traceless tensor)
Define momentum density q i q^i q i
q i = ( ρ ˉ + P ˉ ) v i q^i = (\bar \rho + \bar P) v^i q i = ( ρ ˉ + P ˉ ) v i We can apply SVT decomposition
v i = ∂ i v + v ^ i , ∂ i v ^ i = 0 q i = ∂ i q + q ^ i , ∂ i q ^ i = 0 Π i j = ∂ ⟨ i ∂ j ⟩ + ∂ ( i Π ^ j ) + Π ^ i j \begin{align*}
v_i &= \partial_i v + \hat v_i, \qquad \partial_i \hat v^i = 0 \\
q_i &= \partial_i q + \hat q_i, \qquad \partial_i \hat q^i = 0 \\
\Pi_{ij} &= \partial_{\langle i} \partial_{j \rangle} + \partial_{(i} \hat \Pi_{j)} + \hat \Pi_{ij}
\end{align*} v i q i Π ij = ∂ i v + v ^ i , ∂ i v ^ i = 0 = ∂ i q + q ^ i , ∂ i q ^ i = 0 = ∂ ⟨ i ∂ j ⟩ + ∂ ( i Π ^ j ) + Π ^ ij For scalar perturbations, the Fourier components of bulk velocity and anisotropic stress can be rewritten as
v i ( η , k ) = i k i ^ v ( η , k ) Π i j ( η , k ) = − ( ρ ˉ + P ˉ ) k ^ ⟨ i k ^ j ⟩ σ ( η , k ) \begin{align*}
v^i (\eta, \mathbf{k}) &= i \hat{k^i} v(\eta, \mathbf{k}) \\
\Pi^{ij}(\eta, \mathbf{k}) &= - (\bar \rho + \bar P) \hat{k}^{\langle i} \hat{k}^{j \rangle} \sigma(\eta, \mathbf{k})
\end{align*} v i ( η , k ) Π ij ( η , k ) = i k i ^ v ( η , k ) = − ( ρ ˉ + P ˉ ) k ^ ⟨ i k ^ j ⟩ σ ( η , k ) where
k ^ = k ∣ k ∣ \mathbf{\hat k} = \dfrac{\mathbf{k}}{|\mathbf{k}|} k ^ = ∣ k ∣ k
k ^ ⟨ i k ^ j ⟩ ≡ k ^ ⟨ i k ^ j ⟩ − 1 3 δ i j \hat{k}^{\langle i} \hat{k}^{j \rangle} \equiv \hat k^{\langle i} \hat k^{j \rangle} - \dfrac{1}{3} \delta^{ij} k ^ ⟨ i k ^ j ⟩ ≡ k ^ ⟨ i k ^ j ⟩ − 3 1 δ ij
Einstein equation for scalar perturbations ¶ δ G ν μ = 8 π G δ T ν μ \delta G^\mu_{\nu} = 8\pi G \delta T^\mu_{\nu} δ G ν μ = 8 π G δ T ν μ 00-component ¶ The 00-component of Einstein equation reads
∇ 2 ψ ⏟ Newtonian − 3 H ( ψ ˙ + H ϕ ) ⏟ GR correction = 4 π G a 2 δ ρ \boxed{
\underbrace{\nabla^2 \psi}_{\text{Newtonian}} - \underbrace{3 \mathcal{H} (\dot \psi + \mathcal{H} \phi)}_{\text{GR correction}} = 4 \pi G a^2 \delta \rho
} Newtonian ∇ 2 ψ − GR correction 3 H ( ψ ˙ + H ϕ ) = 4 π G a 2 δ ρ This is generalized relativistic Poission equation.
For subhorizon modes (k ≫ H k \gg \mathcal{H} k ≫ H
∣ ∇ 2 ψ ∣ ≫ ∣ 3 H ( ψ ˙ + H ϕ ) ∣ (in Fourier modes ∣ ∇ 2 ∣ ≃ k 2 ) | \nabla^2 \psi | \gg |3 \mathcal{H} (\dot \psi + \mathcal{H} \phi)| \quad \text{(in Fourier modes } |\nabla^2| \simeq k^2) ∣ ∇ 2 ψ ∣ ≫ ∣3 H ( ψ ˙ + H ϕ ) ∣ (in Fourier modes ∣ ∇ 2 ∣ ≃ k 2 )
So we reduce to Poisson equation
∇ 2 ψ ≈ 4 π G a 2 δ ρ \nabla^2 \psi \approx 4 \pi G a^2 \delta \rho ∇ 2 ψ ≈ 4 π G a 2 δ ρ The GR correction term is important in superhorizon scales.
0i-component ¶ The scalar component
δ T 0 i = ∂ i q = ( ρ ˉ + P ˉ ) ∂ i v \delta T^i_0 = \partial^i q = (\bar \rho + \bar P) \partial^i v δ T 0 i = ∂ i q = ( ρ ˉ + P ˉ ) ∂ i v Einstein equation reads
ψ ˙ + H ϕ = − 4 π G a 2 q \boxed{
\dot \psi + \mathcal{H} \phi = - 4 \pi G a^2 q
} ψ ˙ + H ϕ = − 4 π G a 2 q ij-component ¶ The tracefree part of the stress tensor for scalar perturbation
P i j T j i = − P i j Π j i = ( ρ ˉ + P ˉ ) ( k ^ j k ^ i − 1 3 δ i j ) ( k ^ i k ^ j − 1 3 δ j i ) σ = 2 3 ( ρ ˉ + P ˉ ) σ \begin{align*}
P^j_i T^i_j = -P^j_i \Pi^i_j &= (\bar \rho + \bar P) (\hat k^j \hat k_i - \dfrac{1}{3} \delta^j_i) (\hat k^i \hat k_j - \dfrac{1}{3} \delta^i_j) \sigma \\
&= \dfrac{2}{3} (\bar \rho + \bar P) \sigma
\end{align*} P i j T j i = − P i j Π j i = ( ρ ˉ + P ˉ ) ( k ^ j k ^ i − 3 1 δ i j ) ( k ^ i k ^ j − 3 1 δ j i ) σ = 3 2 ( ρ ˉ + P ˉ ) σ The i j ij ij
where σ \sigma σ
Perfect fluids have zero anisotropic stress.
Dark matter and baryons can be described as perfect fluids.
Photons started as a perfect fluid, but developed an anisotropic stress component during the matter-dominated era when their energy density is subdominant.
The only relevant source for anisotropic stress in the early universe is neutrino (But the level is small so we will ignore them for this course ).
Evolution equation for metric potential ¶ The equation governs the evolution of the metric potential ψ \psi ψ
ϕ ¨ + 3 H ϕ ˙ + ( 2 H ˙ + H 2 ) ϕ = 4 π G a 2 δ P \boxed{
\ddot \phi + 3\mathcal{H} \dot \phi + (2\mathcal{\dot H} + \mathcal{H}^2) \phi = 4\pi G a^2 \delta P
} ϕ ¨ + 3 H ϕ ˙ + ( 2 H ˙ + H 2 ) ϕ = 4 π G a 2 δ P Proof 6 (Evolution equation for metric potential)
Given the traceless Π i j \Pi^{ij} Π ij
δ T i i = − δ P δ i i − Π i i = − 3 δ P \delta T^i_i = - \delta P \delta^i_i - \Pi^i_i = - 3 \delta P δ T i i = − δ P δ i i − Π i i = − 3 δ P We have
δ G μ μ = − δ R ⇒ G i i = − δ R − δ G 0 0 \begin{align*}
\delta G^\mu_\mu &= - \delta R \\
\Rightarrow G^i_i &= - \delta R - \delta G^0_0
\end{align*} δ G μ μ ⇒ G i i = − δ R = − δ R − δ G 0 0 with
δ R = 2 a − 2 [ ∇ 2 ϕ − 2 ∇ 2 ψ + 6 ( H ˙ + H 2 ) ϕ + 3 ψ ¨ + 3 H ( ϕ ˙ + 3 ψ ˙ ) ] \delta R = 2 a^{-2} \left[ \nabla^2 \phi - 2 \nabla^2 \psi + 6(\mathcal{\dot H} + \mathcal{H^2}) \phi + 3 \ddot{\psi} + 3 \mathcal{H}(\dot{\phi} + 3 \dot{\psi}) \right] δ R = 2 a − 2 [ ∇ 2 ϕ − 2 ∇ 2 ψ + 6 ( H ˙ + H 2 ) ϕ + 3 ψ ¨ + 3 H ( ϕ ˙ + 3 ψ ˙ ) ]
δ G 0 0 = 2 a − 2 [ ∇ 2 ψ − 3 H ( ψ ˙ + H ϕ ) ] \delta G^0_0 = 2 a^{-2} \left[ \nabla^2 \psi - 3 \mathcal{H}(\dot{\psi} + \mathcal{H} \phi) \right] δ G 0 0 = 2 a − 2 [ ∇ 2 ψ − 3 H ( ψ ˙ + H ϕ ) ]
In the absence of anisotropic stress σ = 0 \sigma = 0 σ = 0 ψ = ϕ \psi = \phi ψ = ϕ (66)
Combining everything, we have
δ G i i = − 6 a − 2 [ ψ ¨ + 3 H ψ ˙ + ( 2 H ˙ + H 2 ) ψ ] \delta G^i_i = -6 a^{-2} \left[ \ddot{\psi} + 3 \mathcal{H} \dot{\psi} + (2\mathcal{\dot{H}} + \mathcal{H^2}) \psi \right] δ G i i = − 6 a − 2 [ ψ ¨ + 3 H ψ ˙ + ( 2 H ˙ + H 2 ) ψ ] We have
δ G i i = 8 π G δ T i i \delta G^i_i = 8 \pi G \delta T^i_i δ G i i = 8 π G δ T i i which gives
ϕ ¨ + 3 H ϕ ˙ + ( 2 H ˙ + H 2 ) ϕ = 4 π G δ P \boxed{\ddot{\phi} + 3 \mathcal{H} \dot{\phi} + (2\mathcal{\dot{H}} + \mathcal{H^2}) \phi = 4 \pi G \delta P} ϕ ¨ + 3 H ϕ ˙ + ( 2 H ˙ + H 2 ) ϕ = 4 π G δ P The right-hand side is sourced by the total pressure perturbation δ P \delta P δ P
If anisotropic stress is present, an additional source term appears, and ϕ ≠ ψ \phi \neq \psi ϕ = ψ