Linearised Theory and Gravitational Waves Flashcards
Weak Gravity
-weak gravity is nearly special relativity
-this can be written as a small perturbation of the Minkowski metric:
gμν = ημν + hμν
-the inverse is:
g^μν = η^μν - h^μν
hμν
-small perturbation
-if h is small then its determinant is also small:
det(hμν) < < 1
-anything of order h² ~ 0
-any time derivative of hμν is zero
Connection Coefficients
- sub in gμν = ημν + hμν
- eliminate any h² terms
Riemann Tensor
-any Γ x Γ terms are O(h²)~0
Newtonian Limit
1) weak fields: gμν = ημν + hμν 2) static gravitational fields so no time derivatives of h: d/dτ (h ...) = d/dt (h ...) = 0 3) low velocity test particles: v < < c (i.e. dx^i/dτ < < dt/dτ) => -can ignore dx^i/dτ in the geodesic equation and just keep dt/dτ
Geodesic Equation for Massive Particles
Overview
d²x^μ/dτ² + Γ^μ_αβ dx^α/dτ dx^β/dτ = 0 -since v < < c, can set: dx^α/dτ = U^α ~ (c, 0, 0, 0) -so only need to consider the Γ^μ_00 term: d²x^μ/dτ² + Γ^μ_00 dx^0/dτ dx^0/dτ = 0
Geodesic Equation for Massive Particles
μ=0
d²x^μ/dτ² + Γ^μ_00 dx^0/dτ dx^0/dτ = 0 -but Γ^0_00 = 0 => d²x^0/dτ² = 0 => d²(ct)/dτ² = 0
Geodesic Equation for Massive Particles
μ=i
d²x^i/dτ² + Γ^i_00 dx^0/dτ dx^0/dτ = 0
- sub in x^0 = ct and dx^i/dτ = dx^i/dt dt/dτ
- cancel terms
- sub in Γ^i_00 = -1/2 η^ij h_oo,j
Acceleration
Newton
ai = xi’’ = -∇i φ
Acceleration
GR
ai = xi’’ = 1/2 h00,i c²
=>
-∇i φ = 1/2 h00,i c²
h00 = -2φ/c²
Stress Tensor
-if v < < c then p ~ 0 => Tμν = (ρ + p/c²) Uμ Uν + p gμν ~ ρ Uμ Uν => T = g^μν Tc = ρ g^μν Uμ Uν = - ρ c²
Field Equation
Not in Terms of R
Rαβ = K [Tαβ - 1/2 gαβ T]
Gravitational Waves
-use only h < < η and look at Riemann
Transformation of η
- η is invariant under any Lorentz transformation
- in weak fields, all transforms are ‘near’ Lorentz
Transformation of h
hμν’ = Λμ^α Λν^β hαβ
Gauge Transformations
-problem: too many degress of freedom in the equations
-solution: pick gauge to simplify this:
x^μ -> x^μ’ + ε^μ(x)
-with |ε^α_,β| < < O(h)
Choices of Gauge Transformation
- a gauge transformation is not unique
- any combination satisfying the gauge condition is acceptable
- so can choose a gauge to impose a choice of coordinate of coordinate system and reduce the degrees of freedom
- there are three typical choices:
1) Synchronous
2) Lorentz
3) Transverse-traceless
Gauge Transformations
Field Equation
-to simplify the field equations, only need Lorentz gauge
Gauge Transformations
Gμγ
-use Lorentz gauge to cancel terms first
-then define:
hαβ_ = hαβ - 1/2 ηαβ h
-where hαβ_ is the trace-reverse of h
Tμγ in a Vacuum
Tμγ = 0
Gravity Waves in a Vacuum
-for gravitational waves the standard wave equation for a wave propagating at the speed of light is recovered
=>
-gravity waves propagate in a vacuum like light waves and at the speed of light
Gravity Waves
Solution
hμγ_ = A^μγ exp(i kα x^α)
-where A is a symmetric rank (2,0) tensor independent of x^α with 10 non-trivial components in principle (4 diagonal and 6 off-diagonal)
Gravity Waves
Conditions
-unless h^μγ_ = 0 (i.e. no waves) we must have:
kα k^α = 0
-where k^α = (ω/c, k1, k2, k3)
-h^μγ_ obeys the wave equation with ω=c|k|
-we also require that the solution satisfies the Lorentz gauge
=> h^αβ_,β_ = 0
-so A^αβ kβ = 0 and A and k are ‘orthogonal
-also impose the transverse-traceless gauge
=>
A^α_α = η^αβ Aαβ = ηαβ A^αβ = 0 (traceless)
AND
A^α0 = 0 (transverse)
AND (trivially)
h_ = 0
Gravity Waves
A
-4x4 matrix
-outer ring of zeros
-inner four components top left to bottom right:
Axx, Axy, Axy, -Axx
Gravity Waves
ΔL
-two particles at rest at x=0 and x=ε, y=z=0
-there separation is ΔL
ΔL ~ [1 + 1/2 hxx(x=0)] ε
-appears to suggest that ΔL is changing as waves pass through
-this is because of the gauge condition
-the points themselves (x=0 & x=ε) have not changed but the shortest path between them has
-lengths stretch
Gravity Waves
Acceleration
-no acceleration in the T-T gauge
