Einstein equations is not easy to solve. But we can study a set of approximate solutions using linear perturbation theory around the Minkowski metric.
A remarkable consequence of relativistic gravitation is that distortions of spacetime can be generated by the motions of matter, then propagate through spacetime.
Linearized Einstein Equation ¶ Metric perturbation ¶ Consider a region of spacetime, where the geometry is only slightly distorted compared to Minkowski geometry
g μ ν = η μ ν + h μ ν , ∣ h μ ν ∣ ≪ 1 , ∣ ∂ ρ h μ ν ∣ ≪ 1 g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}, \qquad |h_{\mu \nu}| \ll 1, \qquad |\partial^\rho h_{\mu \nu}| \ll 1 g μν = η μν + h μν , ∣ h μν ∣ ≪ 1 , ∣ ∂ ρ h μν ∣ ≪ 1 We describe the curvature of spacetime by retaining the first order approximation (linear terms).
Curvature of spacetime ¶ First order of Christoffel coefficients reads
Γ μ ν λ = 1 2 η λ σ ( ∂ μ h σ ν + ∂ ν h σ μ − ∂ σ h μ ν ) \Gamma^\lambda_{\mu \nu} = \dfrac{1}{2} \eta^{\lambda \sigma}(\partial_\mu h_{\sigma \nu} + \partial_\nu h_{\sigma \mu} - \partial_\sigma h_{\mu \nu}) Γ μν λ = 2 1 η λσ ( ∂ μ h σ ν + ∂ ν h σ μ − ∂ σ h μν ) Because Γ Γ = O ( 2 ) \Gamma \Gamma = \mathcal{O}(2) ΓΓ = O ( 2 )
R μ ρ ν λ = ∂ ρ Γ μ ν λ − ∂ μ Γ ν ρ λ R^\lambda_{\mu \rho \nu} = \partial_\rho \Gamma^\lambda_{\mu \nu} - \partial_\mu \Gamma^\lambda_{\nu \rho} R μ ρ ν λ = ∂ ρ Γ μν λ − ∂ μ Γ ν ρ λ The Ricci tensor to the first order
R μ ν = ∂ ρ Γ μ ν ρ − ∂ μ Γ ν ρ ρ = ∂ σ ∂ ( μ h ν ) σ − 1 2 □ h μ ν − 1 2 ∂ μ ∂ ν h \begin{align*}
R_{\mu \nu} &= \partial_\rho \Gamma^\rho_{\mu \nu} - \partial_\mu \Gamma^\rho_{\nu \rho} \\
&= \partial^\sigma \partial_{(\mu}h_{\nu)\sigma} - \dfrac{1}{2} \square h_{\mu \nu} - \dfrac{1}{2} \partial_\mu \partial_\nu h
\end{align*} R μν = ∂ ρ Γ μν ρ − ∂ μ Γ ν ρ ρ = ∂ σ ∂ ( μ h ν ) σ − 2 1 □ h μν − 2 1 ∂ μ ∂ ν h where
∂ σ ∂ ( μ h ν ) σ = 1 2 ( ∂ σ ∂ μ h ν σ + ∂ σ ∂ ν h μ σ ) \partial^\sigma \partial_{(\mu}h_{\nu)\sigma} = \dfrac{1}{2} \left( \partial^\sigma \partial_{\mu}h_{\nu \sigma} + \partial^\sigma \partial_{\nu}h_{\mu \sigma} \right) ∂ σ ∂ ( μ h ν ) σ = 2 1 ( ∂ σ ∂ μ h ν σ + ∂ σ ∂ ν h μ σ )
□ = ∂ ρ ∂ ρ \square = \partial^\rho \partial_\rho □ = ∂ ρ ∂ ρ
h = η μ ν h μ ν h = \eta^{\mu \nu} h_{\mu \nu} h = η μν h μν
The Ricci scalar to the first order
R = η μ ν R μ ν = ∂ μ ∂ ν h μ ν − □ h R = \eta^{\mu \nu} R_{\mu \nu} = \partial^\mu \partial^\nu h_{\mu \nu} - \square h R = η μν R μν = ∂ μ ∂ ν h μν − □ h Introduce trace-reversed perturbation
h ˉ μ ν = h μ ν − 1 2 η μ ν h \boxed{
\bar h_{\mu \nu} = h_{\mu \nu} - \dfrac{1}{2} \eta_{\mu \nu} h } h ˉ μν = h μν − 2 1 η μν h Proof 1 (Trace-reversed perturbation)
h ˉ = η μ ν h ˉ μ ν = η μ ν h μ ν − 1 2 η μ ν η μ ν h = h − 1 2 4 h = − h \bar h = \eta^{\mu \nu} \bar h_{\mu \nu} = \eta^{\mu \nu} h_{\mu \nu} - \dfrac{1}{2} \eta^{\mu \nu} \eta_{\mu \nu} h = h - \dfrac{1}{2} 4 h = -h h ˉ = η μν h ˉ μν = η μν h μν − 2 1 η μν η μν h = h − 2 1 4 h = − h The trace is indeed reversed !
The Einstein tensor
G μ ν = R μ ν − 1 2 η μ ν R = ∂ σ ∂ ( μ h ν ) σ − 1 2 □ h μ ν − 1 2 ∂ μ ∂ ν h − 1 2 η μ ν ∂ σ ∂ ρ h ρ σ + 1 2 η μ ν □ h = ∂ σ ∂ ( μ h ˉ ν ) σ − 1 2 □ h ˉ μ ν − 1 2 η μ ν ∂ σ ∂ ρ h ˉ ρ σ \begin{align*}
G_{\mu \nu} &= R_{\mu \nu} - \dfrac{1}{2} \eta_{\mu \nu} R \\
&= \partial^\sigma \partial_{(\mu}h_{\nu)\sigma} - \dfrac{1}{2} \square h_{\mu \nu} - \dfrac{1}{2} \partial_{\mu} \partial_\nu h - \dfrac{1}{2} \eta_{\mu \nu} \partial^\sigma \partial^\rho h_{\rho \sigma} + \dfrac{1}{2} \eta_{\mu \nu} \square h \\
&= \partial^\sigma \partial_{(\mu} \bar h_{\nu)\sigma} - \dfrac{1}{2} \square \bar h_{\mu \nu} - \dfrac{1}{2} \eta_{\mu \nu} \partial^\sigma \partial^\rho \bar h_{\rho \sigma}
\end{align*} G μν = R μν − 2 1 η μν R = ∂ σ ∂ ( μ h ν ) σ − 2 1 □ h μν − 2 1 ∂ μ ∂ ν h − 2 1 η μν ∂ σ ∂ ρ h ρ σ + 2 1 η μν □ h = ∂ σ ∂ ( μ h ˉ ν ) σ − 2 1 □ h ˉ μν − 2 1 η μν ∂ σ ∂ ρ h ˉ ρ σ The curvature of spacetime depends only on the second derivative of perturbation h μ ν h_{\mu \nu} h μν
Gauge Choice ¶ By a change of coordinates, the metric g μ ν g_{\mu \nu} g μν h μ ν h_{\mu \nu} h μν ∣ h μ ν ′ ∣ ≪ 1 |h'_{\mu \nu}| \ll 1 ∣ h μν ′ ∣ ≪ 1 infinitestimal coordinate transformation :
x μ → x ′ μ = x μ + ξ μ ( x ) x^\mu \to x'^\mu = x^\mu + \xi^\mu (x) x μ → x ′ μ = x μ + ξ μ ( x ) We will find
h μ ν ( x ) = h μ ν ′ ( x ′ ) + ∂ μ ξ ν + ∂ ν ξ μ h_{\mu \nu}(x) = h'_{\mu \nu}(x') + \partial_\mu \xi_\nu + \partial_\nu \xi_\mu h μν ( x ) = h μν ′ ( x ′ ) + ∂ μ ξ ν + ∂ ν ξ μ Proof 2 (transformation of
h μ ν h_{\mu \nu} h μν )
Under the transformation law of metric
g μ ν = ∂ x ′ ρ ∂ x μ ∂ x ′ σ ∂ x ν g ρ σ ′ ( x ′ ) = ( δ μ ρ + ∂ μ ξ ρ ) ( δ ν σ + ∂ ν ξ σ ) ( η ρ σ + h ρ σ ′ ( x ′ ) ) = η μ ν + h μ ν ′ ( x ′ ) + ∂ μ ξ ν + ∂ ν ξ μ \begin{align*}
g_{\mu \nu} &= \dfrac{\partial x'^\rho}{\partial x^\mu} \dfrac{\partial x'^\sigma}{\partial x^\nu} g'_{\rho \sigma}(x') \\
&= (\delta^\rho_\mu + \partial_\mu \xi^\rho) (\delta^\sigma_\nu + \partial_\nu \xi^\sigma)(\eta_{\rho \sigma} + h'_{\rho \sigma}(x')) \\
&= \eta_{\mu \nu} + h'_{\mu \nu}(x') + \partial_\mu \xi_\nu + \partial_\nu \xi_\mu
\end{align*} g μν = ∂ x μ ∂ x ′ ρ ∂ x ν ∂ x ′ σ g ρ σ ′ ( x ′ ) = ( δ μ ρ + ∂ μ ξ ρ ) ( δ ν σ + ∂ ν ξ σ ) ( η ρ σ + h ρ σ ′ ( x ′ )) = η μν + h μν ′ ( x ′ ) + ∂ μ ξ ν + ∂ ν ξ μ Also,
g μ ν = η μ ν + h μ ν ( x ) g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}(x) g μν = η μν + h μν ( x ) And
h μ ν ′ ( x ′ ) = h μ ν ′ ( x + ξ ) ≃ h μ ν ′ ( x ) + ξ ρ ∂ ρ h μ ν ′ ( x ) ⏟ O ( 2 ) ≃ h μ ν ′ ( x ) h'_{\mu \nu}(x') = h'_{\mu \nu}(x + \xi) \simeq h'_{\mu \nu}(x) + \underbrace{\xi^\rho \partial_\rho h'_{\mu \nu}(x)}_{\mathcal{O}(2)} \simeq h'_{\mu \nu}(x) h μν ′ ( x ′ ) = h μν ′ ( x + ξ ) ≃ h μν ′ ( x ) + O ( 2 ) ξ ρ ∂ ρ h μν ′ ( x ) ≃ h μν ′ ( x ) Therefore,
h μ ν ( x ) = h μ ν ′ ( x ′ ) + ∂ μ ξ ν + ∂ ν ξ μ h_{\mu \nu}(x) = h'_{\mu \nu}(x') + \partial_\mu \xi_\nu + \partial_\nu \xi_\mu h μν ( x ) = h μν ′ ( x ′ ) + ∂ μ ξ ν + ∂ ν ξ μ We have
h μ ν → h μ ν ′ = h μ ν − ∂ μ ξ ν − ∂ ν ξ μ h ˉ μ ν → h ˉ μ ν ′ = h ˉ μ ν − ∂ μ ξ ν − ∂ ν ξ μ + ∂ λ ξ λ η μ ν \begin{align*}
h_{\mu \nu} \to h'_{\mu \nu} &= h_{\mu \nu} - \partial_\mu \xi_\nu - \partial_\nu \xi_\mu \\
\bar h_{\mu \nu} \to \bar h'_{\mu \nu} &= \bar h_{\mu \nu} - \partial_\mu \xi_\nu - \partial_\nu \xi_\mu + \partial_\lambda \xi^\lambda \eta_{\mu \nu}
\end{align*} h μν → h μν ′ h ˉ μν → h ˉ μν ′ = h μν − ∂ μ ξ ν − ∂ ν ξ μ = h ˉ μν − ∂ μ ξ ν − ∂ ν ξ μ + ∂ λ ξ λ η μν By consequent,
∂ μ h ˉ μ ν → ∂ μ h ˉ μ ν ′ = ∂ μ h ˉ μ ν − □ ξ ν \partial^\mu \bar h_{\mu \nu} \to \partial^\mu \bar h'_{\mu \nu} = \partial^\mu \bar h_{\mu \nu} - \square \xi_\nu ∂ μ h ˉ μν → ∂ μ h ˉ μν ′ = ∂ μ h ˉ μν − □ ξ ν We can use the freedom of gauge choice to impose a particular condition on h μ ν h_{\mu \nu} h μν Lorenz gauge :
∂ μ h ˉ μ ν = 0 \boxed{
\partial^\mu \bar h_{\mu \nu} = 0
} ∂ μ h ˉ μν = 0 If ∂ μ h ˉ μ ν ≠ 0 \partial^\mu \bar h_{\mu \nu} \neq 0 ∂ μ h ˉ μν = 0 x ′ μ x'^\mu x ′ μ ∂ μ h ˉ μ ν = 0 \partial^\mu \bar h_{\mu \nu} = 0 ∂ μ h ˉ μν = 0 □ ξ ν = ∂ μ h ˉ μ ν \square \xi_\nu = \partial^\mu \bar h_{\mu \nu} □ ξ ν = ∂ μ h ˉ μν ξ ν \xi_\nu ξ ν
The Einstein tensor is significantly simplified
G μ ν = − 1 2 □ h ˉ μ ν G_{\mu \nu} = -\dfrac{1}{2} \square \bar h_{\mu \nu} G μν = − 2 1 □ h ˉ μν Propagation of gravitational waves ¶ In the absence of matter, the momentum energy tensor is null T μ ν = 0 T_{\mu \nu} = 0 T μν = 0
□ h ˉ μ ν = 0 \square \bar h_{\mu \nu} = 0 □ h ˉ μν = 0 This equation mirrors Maxwell’s equations in the Lorenz gauge, featuring the same d’Alembertian operator and therefore implying propagation at the speed of light . The key distinction is that gravitational waves are described by a rank-2 tensor field h μ ν h_{\mu\nu} h μν rank-1 vector field A μ A_\mu A μ
We need to define the number of physical degrees of freedom that actually propagate. The Lorenz condition does not completely fix the gauge because any non-trivial solution ξ ν \xi_\nu ξ ν □ ξ ν = 0 \square \xi_\nu= 0 □ ξ ν = 0 ξ ν \xi_\nu ξ ν h μ ν h_{\mu \nu} h μν
h = 0 , h 0 i = 0 h = 0, \qquad h_{0i} = 0 h = 0 , h 0 i = 0 The first condition lead to
h ˉ μ ν = h μ ν , \bar h_{\mu \nu} = h_{\mu \nu}, h ˉ μν = h μν , we can summarize the restrictions imposed on the perturbation of the metric in the form
h T T = 0 , ∂ μ h μ ν T T = 0 , h 0 i T T = 0 h^{TT} = 0, \qquad \partial^\mu h^{TT}_{\mu \nu} = 0, \qquad h^{TT}_{0i} = 0 h TT = 0 , ∂ μ h μν TT = 0 , h 0 i TT = 0 which define tranverse, traceless gauge .
To solve the equation of motion
□ h μ ν T T = 0 \square h^{TT}_{\mu \nu} = 0 □ h μν TT = 0 We move to Fourier space and write the solutions as the superposition of plane waves of the form
h μ ν T T = H μ ν e i k σ x σ h^{TT}_{\mu \nu} = H_{\mu \nu} e^{i k_\sigma x^\sigma} h μν TT = H μν e i k σ x σ Equation of motion (23) light-like geodesics of gravitational waves:
η μ ν k μ k ν = 0 \eta_{\mu \nu} k^\mu k^\nu = 0 η μν k μ k ν = 0 Choose z z z
k μ = ( ω , 0 , 0 , ω ) k^\mu = (\omega, \, 0, \, 0, \, \omega) k μ = ( ω , 0 , 0 , ω ) From the gauge conditions (22)
H = 0 , k μ H μ ν = ω ( H 0 ν + H 3 ν ) , H 0 i = 0 H = 0, \qquad k^\mu H_{\mu \nu} = \omega(H_{0\nu} + H_{3\nu}), \qquad H_{0i} = 0 H = 0 , k μ H μν = ω ( H 0 ν + H 3 ν ) , H 0 i = 0 The matrix H μ ν H_{\mu \nu} H μν
H μ ν = ( 0 0 0 0 0 H + H × 0 0 H × − H + 0 0 0 0 0 ) H_{\mu \nu} = \begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & H_+ & H_\times & 0 \\
0 & H_\times & - H_+ & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix} H μν = ⎝ ⎛ 0 0 0 0 0 H + H × 0 0 H × − H + 0 0 0 0 0 ⎠ ⎞ characterized by two degrees of freedom, analogous to two polarizations of EM waves.
Proof 3 (matrix
H μ ν H_{\mu \nu} H μν )
We have
H 0 ν = − H 3 ν \begin{align*}
H_{0\nu} &= -H_{3\nu}
\end{align*} H 0 ν = − H 3 ν and H 0 i = 0 H_{0i} = 0 H 0 i = 0
{ H 00 = − H 30 = 0 H 0 i = − H 3 i = 0 \begin{cases}
H_{00} &= -H_{30} = 0 \\
H_{0i} &= -H_{3i} = 0
\end{cases} { H 00 H 0 i = − H 30 = 0 = − H 3 i = 0 We have
H = 0 H 00 = H 11 + H 22 + H 33 H 11 = − H 22 = H + \begin{align*}
H &= 0 \\
H_{00} &= H_{11} + H_{22} + H_{33} \\
H_{11} &= - H_{22} = H_{+}
\end{align*} H H 00 H 11 = 0 = H 11 + H 22 + H 33 = − H 22 = H + What remains are
H 12 = H 21 = H × H_{12} = H_{21} = H_\times H 12 = H 21 = H × Detection of gravitational waves ¶