Introduction to General Relativity and Gravitational Waves

Transcription

1 Introduction to General Relativity and Gravitational Waves Patrick J. Sutton Cardiff University International School of Physics Enrico Fermi Varenna, 2017/07/03-04

2 Suggested reading James B. Hartle, Gravity: An Introduction to Einstein s General Relativity, Eanna E. Flanagan and Scott A. Hughes, The basics of gravitational wave theory, New J.Phys. 7 (2005) 204 Living Reviews in Relativity an open-access online journal of invited reviews

3 GW150914

4 GR: Gravity as Geometry LIGO-G v1 Sutton GR and GWs

5 Geometry = measuring distances Pythagoras s formula: the line element of flat space. ds 2 = dx 2 + dy 2 + dz = (dx dy dz) Flat spacetime. dx dy dz (1) ds 2 = (c dt) 2 + dx 2 + dy 2 + dz = (cdt dx dy dz) cdt dx dy dz (2) Note: Everyone uses units in which c = 1 (and G = 1).

6 Minkowski spacetime Types of spacetime interval: ds 2 > 0 spacelike (A C) = 0 null (light cones) < 0 timelike (A B) Key physical concept: proper time along a worldline. τ = dτ ds 2 (3) The proper time is the time elapsed as measured by an observer moving on that worldline.

7 Curved spacetime Switch from Cartesian to general coordinates x α = (x 0, x 1, x 2, x 3 ) with line element ds 2 = (dx 0 dx 1 dx 2 dx 3) g 00 g 01 g 02 g 03 g 10 g 11 g 12 g 13 g 20 g 21 g 22 g 23 g 30 g 31 g 32 g 33 dx 0 dx 1 dx 2 dx 3 (4) The matrix g µν (x α ) is called the metric. Properties: symmetric: : g µν = g νµ a function of position in spacetime: g µν = g µν (x α ). All information on the geometry of the spacetime is contained in the metric. The inverse matrix is denoted with raised indices: g µν (g µν ) 1.

8 Coordinate transformations There are no preferred coordinates in General Relativity. Spacetime intervals ds 2 are invariant under coordinate transformations. Exercise: Use the invariance of ds 2 to show that under the coordinate transformation x α x α the metric transforms as g αβ = g µν x µ x α x ν x β (5) Exercise: Use eqn (5) to show that the line element of flat spacetime in spherical coordinates is ds 2 = dt 2 + dr 2 + r 2 (dθ 2 + sin 2 θdφ 2 ) (6)

9 Examples Schwarzschild spacetime: a non-rotating, uncharged black hole of mass M ( ds 2 = 1 2M r ) ( dt M r ) 1 dr 2 +r 2 (dθ 2 +sin 2 θdφ 2 ) (7) Friedmann-Lemaitre-Robertson-Walker spacetime: a homogeneous isotropic universe with scale factor a(t) [ ] dr ds 2 = dt 2 + a 2 2 (t) 1 kr 2 + r 2 (dθ 2 + sin 2 θdφ 2 ) (8)

10 Physical Consequences of Curved Spacetime Hypothesis: freely falling test masses move along worldlines of extremal proper time. τ = ds 2 = gµν dx µ dx ν (9) Exercise: Show that the Euler-Lagrange equations become this geodesic equation: d 2 x α dτ 2 where the Christoffel symbols are + dx β dx γ Γα βγ dτ dτ = 0 (10) Γ α βγ = 1 2 gαδ ( β g γδ + γ g δβ δ g βγ ) (11) and α is short-hand for / x α.

11 Einstein Field Equations The metric components g µν (x α ) are determined by the Einstein field equations, where: R µν 1 2 g µνr + Λg µν = 8πT µν, (12) T µν is the stress-energy-momentum tensor that describes all of the matter and fields in the spacetime; Λ is the cosmological constant; R µν and R are the Ricci tensor and Ricci scalar, R = g µν R µν, R µν = R α µαν ; (13) R µ ναβ is the Riemann tensor: R µ ναβ = x α Γµ νβ x β Γµ να+γ µ λαγ λ νβ Γ µ λβγ λ να. (14)

12 Exercise: Show that in vacuum (T µν = 0) the Einstein equations reduce to R µν = 0. (15) The Einstein equations are a set of 10 coupled, non-linear, second-order, hyperbolic-elliptic partial differential equations for the metric components g αβ. There is no systematic way to solve such systems. Very few analytic solutions exist. These correspond to situations with a high degree of symmetry. E.g.: The exact solution for the two-body problem is not known. Analytic solutions exist for the Einstein equation linearised around flat spacetime; e.g., for spacetimes describing weak static gravitational fields or weak gravitational waves.

13 Linearised gravity A weak gravitational field in GR is a spacetime for which there exist global coordinates x α such that g αβ = η αβ + h αβ, h αβ 1 (16) where η αβ = diag( 1, 1, 1, 1) are the components of the Minkowski metric. Thus, a weak gravitational field differs only slightly from flat spacetime. The quantities h αβ are perturbations or deviations of the metric away from flat spacetime.

14 A word about coordinate transformations: It is always possible to find coordinates for which the above decomposition is not valid e.g., flat spacetime in spherical polar coordinates does not satisfy (16), even though the gravitational field is identically zero! The set of coordinates x α in which (16) holds is not unique. It is possible to make an infinitesimal coordinate transformation x α x α for which the decomposition with respect to the new set of coordinates still holds. We ll often refer to these infinitesimal coordinate transformations as gauge transformations.

15 Note that for weak gravitational fields, one typically raises and lowers indices with the background Minkowski metric η αβ and η αβ, and not with g αβ and g αβ. For example, h α β η αµ h µβ, h αβ η αµ η βν h µν. (17) The only exception is g αβ, which still denotes the inverse of g αβ, not η αµ η βν g µν. To first order, g αβ = η αβ h αβ. (18)

16 Exercise: Show that to first order in h αβ the Riemann tensor has components R αβµν = 1 2 ( µ β h αν µ α h βν + ν α h βµ ν β h αµ ) Exercise: Show that to first order in h αβ the Ricci tensor has components: where (19) R αβ = 1 2 ( h αβ + α V β + β V α ) (20) := η αβ α β = 2 t (21) is the D Alembertian (or wave operator) and V α := β h β α 1 2 αh β β (22)

17 Simplification: It is always possible to find a set of coordinates for which V α := β h β α 1 2 αh β β = 0 (23) This is sometimes called the Loren(t)z condition (in analogy with the gauge condition in electromagnetism). Importance: If V α = 0 then the vacuum Einstein equation for a weak gravitational field in this gauge is simply h αβ = 0 (24) Thus, the metric perturbations satisfy the flat space wave equation. The solutions can therefore be interpreted as gravitational waves.

18 Existence Proof for the Lorenz Gauge Consider an infinitesimal coordinate transformation x α := x α + ξ α (x) (25) where ξ α are slowly varying functions: α ξ β 1. To first-order, the transformation matrix from x α to x µ is x µ x α = δµ α ξµ x α (26) Thus, to first order, the metric components transform as g αβ = x µ x α x ν x β g µν = g αβ α ξ β β ξ α (27) h αβ = h αβ α ξ β β ξ α (28) Since α ξ β 1, it follows that h αβ 1, so the new coordinates x α are also valid coordinates for a weak gravitational field.

19 Exercise: Show that under this infinitesimal coordinate transformation V α = V α ξ α (29) so that V α = 0 ξ α = V α (30) Since is just the wave operator in flat spacetime, one can always find a solution of ξ α = V α. Thus, if V α 0 in the original coordinates x α, we can always find new coordinates x α for which the Lorenz condition V α = 0 is satisfied.

20 Exercise: Show that under an infinitesimal coordinate transformation the components of the Riemann tensor R µανβ given by eqn. (19) are unchanged to first-order. This shows that the curvature of a weak-field spacetime, and so any physical predictions such as geodesic deviation, are unchanged to first-order by an infinitesimal coordinate transformation.

21 Solving the Wave Equation The most general solution to h αβ = 0 is a linear combination of sinusoidal plane wave solutions: h αβ = a αβ exp(ik x) (31) where a αβ and k α are constants satisfying: η αβ k α k β = 0, (32) k β a β α 1 2 k αa β β = 0. (33) The first condition, from the wave equation, says that a gravitational wave propagates along a null direction (i.e., with the speed of light); the second condition is just the Lorenz condition (23) expressed in terms of a αβ and k α.

22 Transverse traceless gauge The Lorenz gauge does not completely fix the coordinates. A further infinitesimal coordinate transformation with x α x α = x α + η α (x) (34) preserves the Lorenz gauge condition. η α = 0 (35) We can exploit this additional coordinate freedom to set h ti = 0, η αβ h αβ = 0 (36) in these coordinates. Such a choice of coordinates is called the transverse traceless gauge (or TT gauge for short).

23 Exercise: Consider the infinitesimal coordinate transformation defined by η α = B α exp(ik x) (37) where k is the same null vector as in (31). Show that under this coordinate transformation with h αβ h αβ = a αβ exp(ik x) (38) a αβ = a αβ ik α B β ik β B α (39) Exercise: Explicitly find B α satisfying the TT gauge conditions a ti = 0, η αβ a αβ = 0 (40) [Hint: Contract a ti = 0 with k i and solve for B i k i in terms of B t ; then substitute this expression for B i k i into η αβ a αβ = 0 to solve for B t ; finally, substitute the solution for B t back into a ti = 0 to find B i.]

24 In the TT gauge, the Lorenz condition eqn. (23) reduces to β h β α = 0. Thus, in the TT gauge there are 8 conditions on the 10 independent components of h αβ : h ti = 0, η αβ h αβ = 0, β h β α = 0 (41) This leaves only 2 independent components of h αβ. In terms of a αβ and k α, we have a ti = 0, η αβ a αβ = 0, k β a β α = 0 (42) The remaining two independent components of a αβ correspond to the two independent polarisation states of a gravitational wave, typically denoted h + and h.

25 For example, take k α = (ω, 0, 0, ω), corresponding to a plane monochromatic gravitational wave with angular frequency ω propagating in the +z-direction. Then eqns (42) become a ti = 0, a zi = 0, a tt = 0, a xx + a yy = 0. (43) These show that the perturbations are transverse to the direction of propagation. The metric perturbations h αβ in the TT gauge are thus h αβ = h + h 0 0 h h (44)

26 The corresponding line element is ds 2 = dt 2 + (1 + h + ) dx 2 + (1 h + ) dy 2 + 2h dx dy + dz 2 (45) The most general solution of the linearised field equation is a superposition of solutions of the form (44) having different propagation directions, frequencies, and amplitudes for h +, h.

27 6 Interpretation: The Effect of GWs Emitted by a massive object, or group of objects, whose shape or orientation changes rapidly with time Waves travel away from the source at the speed of light Waves deform space itself, stretching it first in one direction, then in the Gravitational perpendicular waves direction are deformations of space itself, stretching it first in one direction, then in the perpendicular direction. Plus polarization Time Cross polarization

28 Exercise: Consider two particles at rest at (x, y, z) = (0, 0, 0) and (L, 0, 0). A plus polarized gravitational wave of frequency f and amplitude h 0 1 passes by, propagating in the z direction: h ab (t, x, y, z) = h 0 sin(2πf [t z c ]) (46) Show that the distance d measured along the x-axis between the two particles as the wave passes is given by [ d = ] 2 h 0 sin(2πft) L. (47)

29 binaries (neutron-star -hole) - chirps Expected sources of gravitational waves stochastic (random) ation Inspiraling background - binary systems: Systems that spiral toward one another and eventually coalesce due to AP 2003 the dataenergy lost in GWs. SN 1987A GRB / accreting BH image: K. Thorne The component objects need to be compact (e.g., neutron stars or black holes) and the inspiral needs to be in its final stages (last few minutes) in order for the GWs to be detectable by Earth-based interferometers. 11

30 Ws s Bursts f sinen-star dom) nd - - Unmodeled burst sources: GWs produced by supernovae, gamma ray bursters, or other sources for which we do not know the gravitational waveform. The waveform may be too difficult to calculate due to complicated (or unknown) initial conditions, or numerical relativity has not yet been able to solve the Einstein field equations for the strong-field case of interest. SN SN 1987A 1987A GRB GRB / accreting / accreting BH BH

31 Sources of GWs Deformed pulsars - steady sine- Periodic sources: Continuous sources of GWs (GRBs) - short bursts of such as pulsars with a non-trivial quadrupole radiation moment (e.g., there is a mountain on the surface of a neutron star that is wave not signal aligned with the axis of rotation). Stochastic (random) GWs: Remnant gravitational waves from the big bang, or the superposition of GWs produced by many unresolved astrophysical sources (e.g. distant supernovae or inspiral events). Supernovae, Gamma-ray Bursts Inspiraling binaries (neutron-sta or black-hole) - chirps Big Bang - stochastic (random) GW radiation background - like CMB WMAP 2003 data

32 Stress-Energy-Momentum Tensor Schematically: T αβ = energy density momentum density energy flux stress tensor (48) The stress energy tensor is symmetric: T αβ = T βα. Momentum density is equivalent to energy flux. Conservation law: α T αβ = 0.

33 Trace-Reversed Amplitude When solving the linearised equations in vacuum, it was useful to introduce the Lorenz condition V α := β h β α 1 2 αh β β = 0. The equations simplify if we introduce the trace-reversed amplitude h αβ h αβ 1 2 η αβh γ γ. (49) Then, the Lorenz condition simplifies to β hβ α = 0. (50)

34 Generation of Gravitational Waves With a source term, the linearised Einstein equation is h αβ = 16πT αβ (51) Using the Green s function for the d Alembertian gives T h αβ (t, x) αβ (t x x, x ) = 4 x x d 3 x (52) 4 T αβ (t r, x )d 3 x (53) r where r = x. Exercise: Using the conservation law for the stress tensor, β T αβ = 0, show that the spatial components are h ij (t, x) 2 d 2 r dt 2 ρ(t r, x ) x i x j d 3 x, (54) where ρ = T 00 is the mass-energy density of the source.

35 Example: Binary Systems Order-of-magnitude estimate of GW amplitude:! = 2"/P I 2MR 2 (55) Ï 2MR 2 Ω 2 (56) #"!"!" #" Kepler s third law for a circular binary: M 1 + M 2 = Ω 2 (R 1 + R 2 ) 3 (57) h M2 rr M5/3 r ( ) 4π 2/3 P (58) It can be shown that the dominant frequency of the GWs is twice the orbital frequency, f GW = 2f orbit = 2/P.

36 Exercise: For a neutron-star binary (M 1.4M ) at 5 kpc with P = 1 hr show that h Exercise: For the same system with P = 0.02 s (giving f GW = 2f orbit = 100 Hz, in the sensitive band of LIGO) show that h at a distance of 15 Mpc approximately the distance of the Virgo cluster of galaxies. Exercise: Show the orbital separation R 100 km when P = 0.02 s. Thus, we can only hope to detect inspirals of compact binary systems (e.g., NS-NS, NS-BH, or BH-BH) with Earth-based interferometers like LIGO.

37 Example: Distorted Pulsar Consider a spinning neutron star of radius R with a non-spherical deformation ( mountain ) of mass δm on the equator. If the angular velocity is Ω, then " = 2#/P I δm R 2 (59) Ï δm R 2 Ω 2 (60) The GW amplitude is approximately!"!#" h 2δMR2 Ω 2 r (61) Exercise: For a star at 1 kpc with δm = 10 6 M, a spin frequency of 50 Hz, and a stellar radius of 10 km, show that the GW amplitude at Earth is h

38 Energy in GWs The energy flux (power/area) or energy density in GWs can be estimated using the formula F c 3 32π 2 G ḣ 2 c3 8G h2 f 2 (62) where f = ω/2π is the frequency of the GW (assumed monochromatic) and h is the RMS amplitude. Recall that energy flux or energy density in electromagnetism is E 2 + B 2. In GR, the metric components play the role of gravitational potential, so their derivatives play the role of the field; hence F ḣ 2. Exercise: Show that c 5 G = Watts (63) This equals 1 in geometric units (c = 1 = G).

39 Exercise: GW had a peak amplitude of h at f 200 Hz. Show that the corresponding energy flux is F 10 3 W m 2 (64) This is approximately the energy flux in electromagnetic waves that we receive from the full moon despite GW being at an estimated distance of 400 Mpc!

40 A final word: Astrophysics GWs vs. with EM waves GWs vs. EM Electromagnetic waves Accelerating charge Wavelength small compared to sources! images Absorbed, scattered, dispersed by matter 10 MHz and up Gravitational Waves Accelerating aspherical mass Wavelength large compared to sources! no spatial resolution Very small interaction; matter is transparent 10 khz and down Very different information, mostly mutually exclusive. Difficult to predict GW sources based on EM observations. 12