Linearised Theory and Gravitational Waves

Question 1

Weak Gravity

Answer 1

-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

Question 2

hμν

Answer 2

-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

Question 3

Connection Coefficients

Answer 3

• sub in gμν = ημν + hμν

- eliminate any h² terms

Question 4

Riemann Tensor

Answer 4

-any Γ x Γ terms are O(h²)~0

Question 5

Newtonian Limit

Answer 5

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

Question 6

Geodesic Equation for Massive Particles

Overview

Answer 6

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

Question 7

Geodesic Equation for Massive Particles

μ=0

Answer 7

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

Question 8

Geodesic Equation for Massive Particles

μ=i

Answer 8

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

Question 9

Acceleration

Newton

Answer 9

ai = xi’’ = -∇i φ

Question 10

Acceleration

GR

Answer 10

ai = xi’’ = 1/2 h00,i c²
=>
-∇i φ = 1/2 h00,i c²
h00 = -2φ/c²

Question 11

Stress Tensor

Answer 11

-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

Question 12

Field Equation

Not in Terms of R

Answer 12

Rαβ = K [Tαβ - 1/2 gαβ T]

Question 13

Gravitational Waves

Answer 13

-use only h < < η and look at Riemann

Question 14

Transformation of η

Answer 14

• η is invariant under any Lorentz transformation

- in weak fields, all transforms are ‘near’ Lorentz

Question 15

Transformation of h

Answer 15

hμν’ = Λμ^α Λν^β hαβ

Question 16

Gauge Transformations

Answer 16

-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)

Question 17

Choices of Gauge Transformation

Answer 17

• 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

Question 18

Gauge Transformations

Field Equation

Answer 18

-to simplify the field equations, only need Lorentz gauge

Question 19

Gauge Transformations

Gμγ

Answer 19

-use Lorentz gauge to cancel terms first
-then define:
hαβ_ = hαβ - 1/2 ηαβ h
-where hαβ_ is the trace-reverse of h

Question 20

Tμγ in a Vacuum

Answer 20

Tμγ = 0

Question 21

Gravity Waves in a Vacuum

Answer 21

-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

Question 22

Gravity Waves

Solution

Answer 22

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)

Question 23

Gravity Waves

Conditions

Answer 23

-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

Question 24

Gravity Waves

A

Answer 24

-4x4 matrix
-outer ring of zeros
-inner four components top left to bottom right:
Axx, Axy, Axy, -Axx

Question 25

Gravity Waves

ΔL

Answer 25

-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

Question 26

Gravity Waves

Acceleration

Answer 26

-no acceleration in the T-T gauge