Lecture X: External fields and generation of gravitational waves

Transcription

1 Lecture X: External fields and generation of gravitational waves Christopher M. Hirata Caltech M/C , Pasadena CA 91125, USA (Dated: November 12, 2012) I. OVEVIEW Having examined weak field gravity and the associated experimental tests, we now turn our attention to the external field produced by a system (e.g. a star or the solar system) in linearized gravity. This will include both the gravitational analogues of electric and magnetic multipole moments, as well as gravitational waves. The discussion in the beginning part of this lecture makes sense only for weak perturbations around Minkowski spacetime. Later we will generalize the concepts to make sense in asymptotically flat spacetime i.e. spacetime that looks like Minkowski far from the system, but may have strongly curved regions inside of it (e.g. a black hole). The recommended reading for this lecture is: MTW II. GEEN S FUNCTIONS IN LOENTZ GAUGE We found that in Lorentz gauge the trace-reversed metric perturbation is given via the relation h µν = 16πT µν. (1) We would like to formally solve this equation using a Green s function approach. That is, we wish to construct the Green s function G(x α ) such that G(x α ) = δ (4) (x α ), (2) and then by the principle of superposition we may write the metric perturbation as an integral over the Green s function: h µν (x α ) = 16π G(x α y α )T µν (y α )d 4 y. (3) How are we to find G? There is unfortunately no unique answer! After all, we could add any function f with f = 0 to G and it would still satisfy Eq. (2). However, in most situations there is a physical choice: we want the arded Green s function G, which is zero for x 0 = t < 0. This corresponds to the solution in which there is no incoming gravitational radiation. The use of the arded Green s function is however not a necessity but a particular solution to Einstein s equations. [Note: The notion of a arded Green s function is slightly tricky for theories with a conserved source, since in any physical system there was nonzero T µν somewhere back into the infinite past. However it is clear that if we start with a stationary mass distribution and no incident gravitational waves, and then allow some complex dynamics to occur, then a arded Green s function will respect causality and not radiate gravitational waves into the past.] The easiest method to obtain the arded Green s function is to expand Eq. (2) in spherical coordinates (t, r, θ, φ). We note that for a function G with spherical symmetry, Since the 4-dimensional δ function can be written as G = 2 t G + 2 G = 2 t G + 1 r 2 r(rg) = 1 r ( 2 t + 2 r)(rg). (4) δ (4) (x α 1 ) = δ(t)δ(r ǫ) 4πǫ 2, (5) Electronic address: chirata@tapir.caltech.edu

2 with the last factor introduced to give proper normalization for small but positive ǫ, we have ( 2 t + 2 r)(rg) = r G = δ(t)δ(r ǫ) 1 4πǫ. (6) This applies at r > 0; at r = 0 we have the boundary condition rg = 0. We may extend the range of validity to r < 0 by enforcing the boundary condition with a mirror source and ensuring the odd symmetry of rg: ( t 2 + r)(rg) 2 δ(r ǫ) δ(r + ǫ) = δ(t) = 1 4πǫ 2π δ(t)δ (r) = 1 2π r[δ(t)δ(r)]. (7) This equation can be solved via the definition Ψ(t, r) = r rgdr, so that rg = rψ. Then: or switching to a characteristic coordinate system, u = r t and v = t + r, This has the solution ( t r )Ψ = 1 δ(t)δ(r), (8) 2π 4 2 Ψ u v = 1 δ(u)δ(v). (9) π Ψ = 1 4π Θ( u)θ(v) = 1 Θ(t r)θ(t + r), (10) 4π which is obviously arded. Taking the derivative to get back rg yields the arded Green s function G (x α ) = 1 δ(t r)θ(t). (11) 4πr The arded solution for the metric is then 1 h µν(x sp, t) = 4 r T µν(y sp, t r)d 3 y sp, (12) where r x sp y sp. emember that the arded solution is only one possible solution for the metric for a given matter source. But the difference between the true solution and the arded solution can be found by noting that if both are soutions of h µν = 16πT µν for the same source, then Therefore, we may write ( h µν h µν = h µν ) = 0. (13) homo h µν + h µν, (14) homo where h µν is a homogeneous or gravitational wave solution (solution for zero matter source). We will see that Eq. (12) contains gravitational waves, but they are only outgoing (as one can see from the t r time argument). homo Therefore setting h µν to zero is equivalent to saying that no external gravitational waves are incident on a system. Usually this is a good approximation! [Warning: The outgoing gravitational waves in the above formulation are not necessarily in transverse-traceless gauge.] 2 III. MULTIPOLE EXPANSION Far from a source, it is common to do a multipole expansion: essentially a power-series expansion of Eq. (12) in powers of y sp. This can be done for both relativistic and nonrelativistic sources, but we will focus on nonrelativistic sources (e.g. binary stars) here. We will also examine only the lowest-order multipoles since these (i) correspond to conserved quantities and (ii) the next-lowest terms carry the dominant source of gravitational radiation from a nonrelativistic obect. We will use (t, x i ) for the position of the point at which the metric is measured, y i as the position of a point in the source, and then define = x sp, n i = x i /, and r = x sp y sp. The source is enclosed in a region of characteristic size L and evolves on a dynamical timescale t dyn.

3 3 We begin by expanding 1/r as a power series in y: 1 r = [(x i y i )(x i y i )] 1/2 = [x i x i 2x i y i + y i y i ] 1/2 A. The trace-reversed perturbation = (x i x i ) 1/2 1 2 (x ix i ) 3/2 ( 2x i y i + y i y i ) (x ix i ) 5/2 (2x i y i ) = 1 n iy i 2 + n in (3y i y y k y k δ ) , (15) with a fractional error of order (L/) 3, where L is the typical size scale of the source. We may also Taylor-expand T µν : T µν (y sp, t r) = T µν (y sp, t ) + n i y i T µν,0 (y sp, t ) n in y i y T µν,00 (y sp, t )..., (16) with a fractional error of order L 3 /t 3 dyn = V 3 or L 2 /t dyn = V L/, where V is the typical velocity scale of the source. It then follows that the arded solution for the metric perturbation is [ 1 h µν (x sp, t) = 4 n iy i 2 + n ] in (3y i y y k y k δ ) [ T µν (y sp, t ) + n i y i T µν (y sp, t ) + 1 ] 2 n in y i y T µν,00 (y sp, t ) d 3 y sp. (17) The lowest-order fractional errors are V 2, V L/, and 1/ 3. It is useful to consider the behavior of each metric component at large distances from the source. We first see that 1 00 (x sp, t) = 4 h 1 ρ d 3 y sp n i 2 y i ρ d 3 y sp + n i y i ρ d 3 y sp + 3n in 2 3 (y i y 1 3 y ky k δ )ρ d 3 y sp n in y i y ρ d 3 y sp +..., (18) where the on the right-hand side indicates evaluation at the arded time t. Now the first integral is simply the (conserved) mass M of the system. (Technically this is the energy, but the difference only arises at the next order in velocity, which we have dropped.) The second integral is the mass dipole moment, MY i = y i ρ d 3 y sp, (19) where Y i is the center of mass. Finally, defining the momentum density F = T 0 and noting that F, = ρ the third integral is MẎi = y i ρd 3 y sp = y i F, d 3 y sp = y i, F d 3 y sp = F i d 3 y sp = P i, (20) where P i is the total momentum. Under normal circumstances, we will work in the center of mass frame, in which Y i and P i are zero. Finally, we define the mass quadrupole moment of the system to be Q = (y i y 1 3 y ky k δ )ρ d 3 y sp = I 1 3 I kkδ. (21) (Here I is the moment of inertia tensor.) Then Eq. (17) reduces to h 00 (x) = 4 M + 6Q n i n n in Ï.... (22) This is simply the standard quadrupolar formula familiar from Newtonian physics, plus a new term that contains Ï. But note that in fully nonlinear general relativity, the leading correction to the M/ formula is not the quadrupole

4 term ( 3 ) but rather the nonlinearity of the theory ( M 2 / 2 this is the term that is necessary to obtain the perihelion precession of Mercury). In some cases, such as the orbit of Mercury around the Sun, the relativistic M 2 / 2 term dominates over the quadrupole effect, but for a satellite orbiting the Earth (with its much more flattened shape) the opposite is true. We may further write down the linearized formula for h 0i, including terms through second order in V : h 0i (x) = 4 F i (y sp )d 3 y sp 4 n 2 y F i (y sp )d 3 y sp + 4 n 4 y F i d 3 y sp +... ; (23) the first integral is P i and hence vanishes. The second integral may be simplified using the rule that the angular momentum is S k = ǫ k y F i (y sp )d 3 y sp (24) so that ǫ klm S k = ǫ klm ǫ k y F i (y sp )d 3 y sp = (δ li δ m δ l δ mi ) y F i (y sp )d 3 y sp = 2 y [m F l] (y sp )d 3 y sp. (25) Furthermore, I = y i y ρd 3 y sp = y i y F k,k d 3 y sp = (y i,k y F k + y,k y i F k )d 3 y sp = 2 y ( F i) d 3 y sp. (26) Therefore, decomposing the integral in h 0i into its symmetric and antisymmetric parts: 4 n 2 y F i (y sp )d 3 y sp = 4 n [ ] 2 y [ F i] (y sp )d 3 y sp + y ( F i) (y sp )d 3 n y sp = 2ǫ k S k I n 2. (27) Substituting in this, and its time derivative, gives n h 0i (x) = 2ǫ k S k I n 2 2ǫ n kṡk n 2Ï (28) Using conservation of angular momentum, we drop the Ṡk term, leaving n h 0i (x) = 2ǫ k S k I n n 2Ï (29) Finally, in the case of the space-space components we will write only the lowest term, h (x) = 4 T d 3 y sp. (30) This can be simplified by noting that from Eq. (26): Ï = 2 y ( F i) d 3 y sp = 2 y ( T i)k,k d 3 y sp = 2 T (i y ),k d 3 y sp = 2 T d 3 y sp. (31) Then: h (x) = Ï. (32) The far-field metric perturbation from a source according to the above equations can be found by reversing the trace of h ; we need h = 4 M Ïkk n in Ï + 6 Q n i n (33)

5 Including this with trace reversal gives h 00 = 2 M Ïkk n in Ï + 3 Q n i n n h 0i = 2ǫ k S k 2 + n 2İ 2 2n Ï +... and h = 2 M δ Ï Ïkkδ + 1 n kn l Ï kl δ 3 Q n i n (34) This looks at first glance to have several pieces: there is the familiar Newtonian potential; there is a piece associated with the angular momentum in the time-space part (gravitomagnetism); and there is a set of outward-propagating waves (gravitational waves!) associated with time variation of the quadrupole moment. 5 B. The radiation terms You will note that the gravitational waves ( I terms) in the far field do not look like the ones we studied: in particular they have time-space and time-time components and are not orthogonal to n i (the outward direction). At large distances from the source, the leading-order (1/) wave terms are 00 = 1 Ïkk + 1 n in Ï, 0i = 2 n Ï, and = 2 Ï 1 Ïkkδ + 1 n kn l Ï kl δ (35) (all arded). Fortunately, a gauge transformation can be used to eliminate most of these terms. ecall that under a change of coordinates ξ µ, the metric tensor changed by h µν = ξ µ,ν ξ ν,µ. We first perform a change of the time coordinate t, ξ 0 = 1 2 I kk n in I, (36) leaving the spatial coordinates fixed (ξ i = 0). Then the time-time component changes by h 00 = 2 ξ 0 = 1 Ïkk 1 n in Ï. (37) To find the correction h 0i = ξ 0,i, we recall that since I is evaluated at the arded time, its spatial derivative contains a time derivative: Thus: After applying this transformation, the outgoing wave is i I kl = I kl i (t ) = I kl i = I kl n i. (38) 00 = 0, h 0i = 1 2 n iïkk n in n k Ï k. (39) 0i = 1 2 n iïkk n in n k Ï k 2 n Ï, and = 2 Ï 1 Ïkkδ + 1 n kn l Ï kl δ (40) (again arded). Finally, we may introduce a gauge transformation in the spatial coordinates to eliminate 0i. We want h 0i = ξ i to cancel 0i, so we choose ξ i to be the time-integral of 0i : ξ i = 1 2 n ii kk n in n kik 2 n I. (41)

6 6 Now we find that the change to the purely spatial metric is h = ξ i, ξ,i = n ξi + n i ξ The overall amplitude of the outgoing gravitational wave is then = 1 n in Ï kk + 1 n in n k n l Ï kl 2 n n k Ï ik 2 n in k Ï k. (42) 00 = 0, = 0, and = 1 ) (2Ï Ïkkδ + n k n l Ï kl δ + n i n Ï kk + n i n n k n l Ï kl 2n n k Ï ik 2n i n k Ï k 0i (43) (all arded, as usual). This is the usual form for the amplitude of emitted gravitational waves. It is straightforward to check that Eq. (43) corresponds to a transverse-traceless tensor. To see that it is transverse, note that n i = 1 ) (2Ïn i Ïkkn + n i n k n l Ï kl + n Ï kk + n n k n l Ï kl 2n n k n i Ï ik 2n k Ï k = 0. (44) To see that it is traceless, take ii = 1 ) (2Ïii 3Ïkk + 3n k n l Ï kl + Ïkk + n k n l Ï kl 2n i n k Ï ik 2n i n k Ï ik = 0. (45) The equation for may look messy (it has 7 terms) but in fact the first one is the key: the others are simply proections that guarantee the transverse-traceless nature. C. The non-radiation terms In addition to gravitational radiation, the metric can contain the non-radiation terms, the leading ones of which are associated with the mass M and angular momentum (3) S: ( ds 2 = 1 2 M ) ( dt 2 S k + 4ǫ k n 2dt dxi M ) [(dx 1 ) 2 + (dx 2 ) 2 + (dx 3 ) 2 ] +... (46) The mass terms M we have already discussed extensively: indeed, the mass M of a system is measurable using ordinary Newtonian dynamics, e.g. using Kepler s 3rd law. The angular momentum terms (3) S are a bit trickier: they lead to gravitomagnetism. A contribution h 0i to the time-space part of the metric in linear theory leads to a contribution to the Christoffel symbol: Γ i 0 = 1 2 h 0,i h 0i, = h 0[i,], (47) so that there is an additional contribution to the second derivative of a coordinate: d2 x i ( ) 2 dt dτ 2 = 2h 0[i,]ẋ 2h 0[i,]ẋ. (48) dτ Therefore if we define the 3-vector (3) A (the gravitomagnetic 3-vector potential) with components h 0i, we see that an obect has an apparent acceleration given by (3) a gravitomagnetic = (3) v ( (3) A). (49) Thus the h 0i term, generated by the total angular momentum of an obect, is exactly analogous in its low-velocity limit to the magnetic field interaction in electrodynamics. There is a difference though it is common in both nature and in the laboratory for magnetic forces to dominate over electric forces. This is because charges of both signs exist; a current-carrying wire has equal numbers of protons and electrons, but there is a drift of the electrons relative to the protons (or nuclei). Thus the wire s electric field may be small and the magnetic forces become dominant. For gravitation, no such circumstance occurs: ρv in most systems is small compared with ρ.