Today in Astronomy 142: general relativity and the Universe

Transcription

1 Tody in Astronomy 14: generl reltivity nd the Universe The Robertson- Wlker metric nd its use. The Friedmnn eqution nd its solutions. The ges nd ftes of flt universes The cosmologicl constnt. Glxy cluster formtion in ΛCDM model (Jürg Diemnd, U. Zürich). 18 April 013 Astronomy 14, Spring 013 1

2 Enhnced interrogtion of the Universe Now we know tht the Universe is expnding, isotropic nd homogeneous, nd ruled by grvity nd kinetic energy of glxies nd clusters. So this rises severl questions: How cn we use the observtions of glxy motions nd distribution, long with our theories of grvity, to determine the density of mss nd energy on lrge scles? Wht is the ge of the Universe? Wht is the fte of the Universe? How will the expnsion chnge with time? 18 April 013 Astronomy 14, Spring 013

3 A Newtonin Universe? In n infinite, homogeneous nd isotropic universe, the grvittionl ccelertion g should be zero everywhere (verged over lrge enough scles), since mss would be eqully ttrcted to ll directions in such universe. The only wy to do tht if spce is Eucliden nd grvity is Newtonin is if spce is completely empty. Reson: Through ny point one cn drw lrge sphere. The ccelertion g t this point due to mtter outside this sphere is zero (tht mtter is uniform, infinite, ). The ccelertion due to mtter within this sphere is ( ) g = GM r r Thus g = 0 only if the sphere is empty (M(r) = 0), nd if ll other such spheres re empty (by homogeneity!). 18 April 013 Astronomy 14, Spring

4 Generl-reltivistic universes The Universe is not empty, so it cnnot be Newtonin. Fortuntely the generl theory of reltivity offers good description, nd indeed the lrge-scle structure of the Universe ws the first problem to which Einstein pplied GR. GR itself, nd the solution to the Einstein field eqution (EFE) which is GR s fundmentl expression, is something you will begin to lern in AST 31. (Not here.) The EFE itself is nonliner second-order differentil eqution involving symmetric second-rnk tensors. It cn be solved uniquely, like ny other differentil eqution, if given enough boundry conditions. The solution to the EFE is second-rnk tensor clled the metric, which describes the mesures nd curvtures of spcetime under the given boundry conditions. 18 April 013 Astronomy 14, Spring 013 4

5 Generl-reltivistic universes (continued) In turn, to metric tensor corresponds n bsolute intervl between events. For instnce, in flt spcetime with no msses present in which limit GR converges to specil reltivity the (infinitesiml) bsolute intervl is ds = c dt dx dy dz where dx, dy, dz, dt re infinitesiml intervls between two events, mesured in one reference frme. The combintion is independent of reference frme (i.e. is bsolute), s you will show in Homework #9. This is sometimes clled the Minkowski intervl. 18 April 013 Astronomy 14, Spring 013 5

6 Generl-reltivistic universes (continued) Similrly, the Schwrzschild metric for spcetime outside the bounds of sphericl mss M gives rise to the following bsolute intervl, expressed this time in sphericl coordintes: ( 1 ) ds = GM rc c dt dr r dθ r sin θdφ 1 GM rc where r > GM rc. Most of the expressions for spcetime structure outside blck hole, introduced on 14 Februry 013, were derived from this bsolute intervl. 18 April 013 Astronomy 14, Spring 013 6

7 Generl-reltivistic universes (continued) The solution to the Einstein field equtions for n isotropic nd homogeneous Universe is clled the Robertson-Wlker metric. To this metric corresponds n bsolute intervl given by dr = ( ) + θ sin θ φ + 1 Kr ds c dt R t r d r d scle fctor: rdius of curvture of the Universe. = ±1 or 0, depending upon curvture. r, θ, φ: sphericl comoving coordintes (dimensionless) 18 April 013 Astronomy 14, Spring 013 7

8 Generl-reltivistic universes (continued) The scle fctor R which ppers in the R-W intervl is in turn the solution to the (modified) Friedmnn eqution, one component of the EFE for homogeneity/isotropy: R 8π G Λ c ρm = K R 3 3 R R R = t Mss density Other terms s before. 18 April 013 Astronomy 14, Spring Λ: cosmologicl constnt. Not originlly prt of generl reltivity; plced in the EFE d hoc by Einstein to permit the EFE to hve sttic (timeindependent) solutions. R = 0 for Λ= 3Kc R 8 πgρ. M

9 Generl-reltivistic universes (continued) Comprison of mesurements of glxy motions nd distributions ( R, Rr ), with solutions of these equtions, cn be used to determine ρm, Λ nd K. K is the sign of spce (not spcetime) curvture. Exmples in -D: K = 1 for sphericl surfce, 0 for plne, nd -1 for hyperboloidl surfce. From Nick Strobel s Astronomy Notes 18 April 013 Astronomy 14, Spring 013 9

10 Concordnce with the book I prefer slightly different symbols nd sign conventions thn re used in the textbook. It doesn t mtter which you use, but here s how to trnslte one to the other: R & P Dn d ( ) Rt ( ) c,0 ds dx + dy + dz d tr κ K R & P: Scle fctor (t) is dimensionless, comoving rdius r hs dimensions of length. Dn: Scle fctor R(t) hs dimensions of length, comoving rdil coordinte r is dimensionless. 18 April 013 Astronomy 14, Spring

11 How to use the Robertson-Wlker intervl, nd be convinced tht we re on the right trck, using GR Exmple 1 Proper distnce: clculte the distnce between two glxies t some time t (i.e. for dt = 0), choosing both to lie long the x xis (so θ = φ = dθ = dφ = 0): dr ds = c dt d = d = R ( t ) 1 Kr r r Rtrcsin r if K 1 = + dr d Rt = = ( ) = Rtr ( ) if K= Kr Rt ( ) rcsinh r if K= 1 So the dimensionless rdil coordinte r is relted to distnce in n intuitive wy if K = 0 (i.e. if the Universe is flt). 18 April 013 Astronomy 14, Spring ( )

12 How to use the Robertson-Wlker intervl (continued) Exmple Clculte the expnsion speed, if r 1 (tht is, if one views nerby glxy): rcsin r = r + r + r + r, nd rcsinh r == r r + r r, so 6 40 = Rtr ( ) (ll curvtures); d dr Rt ( ) vr = v= = r Rtr ( ) = Ht ( ) dt dt R( t) The lst result is of course just Hubble s Lw. Our usul vlue, -1-1 of the Hubble constnt, H 0 = 74. km sec Mpc, is the present vlue of Rt Rt ( ) ( ). 18 April 013 Astronomy 14, Spring 013 1

13 Exmple 3 How to use the Robertson-Wlker intervl (continued) Smll distnces, r 1, s functions of time: 1 Rt ( 1 ) = Rtr ( ) = 0 Rt ( 0) But if this works for smll interglctic distnces, it should work even better for wvelengths, which re usully quite smll distnces. Suppose light is emitted t time t 1 nd detected t time t 0 ; then its wvelengths t those two epochs re relted by λ0 λ 1 = Rt ( 0) Rt ( 1). But tht rtio of wvelengths is relted to redshift by ( λ0 λ1) λ1 = z, so Rt ( 0 ) once gin recovering 1 + z =. Rt fmilir result ( ) 1 18 April 013 Astronomy 14, Spring

14 How to use the Friedmnn eqution: criticl density Combine the source terms nd use the RW result bove: R 8π G Λ c ρm K R 3 + = 8π G R 8π G H u= K 3c c R Exmple 4. Suppose universe were exctly flt (K = 0), described by Eucliden geometry; tht would correspond to specil vlue of the energy density t ech time: 18 April 013 Astronomy 14, Spring π G 3c H H u 0. c = uc = 3c 8π G,

15 Criticl density (continued) This criticl density hs current vlue in our Universe of u c0 0 3c H 9 3 = = erg cm. 8πG As it turns out tht our Universe is very nerly flt, it is customry to define normlized energy densities in terms of the criticl density: ρmc 8πGρM Λc Λ M Λ 8 c c 3H π c 3H u Ω= Ω = = Ω = = u u Gu So Ω=Ω +Ω = 1 M Λ in flt universe. 18 April 013 Astronomy 14, Spring

16 How to use the Friedmnn eqution: the constnts We cn cst the Friedmnn eqution in n esier-to-use form, in terms of the present-dy normlized densities, by noting few things bout the constnts it contins. Cosmologicl constnt, Λ. At the present time (subscript zero) we hve, ccording to the previous pge, Λ 3= 3H Ω Λ 3 = H 0 ΩΛ 0. Mss density, ρ M. Since universe stys homogeneous nd isotropic s it expnds or contrcts, the mss contined within sphere of rdius R is constnt: 3 3 ρ R = ρ R, so M M R0 3H0ΩM0 R0 R0 M = ρm0 = = H 3 3 0ΩM0 3 8πG 8πG 8πG ρ 3 3 R 3 8π G R R 18 April 013 Astronomy 14, Spring

17 The constnts (continued) Curvture, K. K is constnt, so we cn evlute it from the FE written for the present time: 3 R0 c H0 H0ΩM0 H 3 0Ω Λ0 = K, or R R K H R = ( Ω M0 +ΩΛ0 1. ) c H0R0 c is positive definite, so the sign of K is determined by the Ωs: the universe is positively curved if Ω M0 +Ω Λ0 > 1, nd negtively-curved if < 1. Put ll these into the FE, multiply through by R, nd we get: 18 April 013 Astronomy 14, Spring

18 or, with The constnts (continued) R 8π G Λ c ρm = K R 3 3 R 3 R0 0 M0 0 Λ0 0 0 M0 Λ0 ( 1) R H Ω H Ω R = H R Ω +Ω R R R0 R = H 0 1+ΩM0 1 +ΩΛ0 1 R R 0 R 0 RR 0 (normlized scle fctor; = 1 t present), 1 ( ) = H0 1+ΩM0 1 +ΩΛ April 013 Astronomy 14, Spring

19 Use of the Friedmnn eqution: flt universe The Friedmnn eqution is seprble nd directly integrble. Exmple 5. Suppose universe were flt: 0 Ω 1, Ω =Ω, Ω = 1 Ω. M0 Λ0 Solve for reltion between time nd the normlized scle fctor. d 1 ( )( ) H dt = +Ω + Ω Ω = H0 1+ Ω+ 1 Ω +Ω Ω H0 Ω 1 Ω 3 0 = H + Ω = 1 +. Ω 18 April 013 Astronomy 14, Spring

20 Substitute: Multiply through by A flt universe (continued) dx Ω 3 1 Ω 1 x = = Ω d Ω [ ] ( ) nd use the chin rule: Now we re redy to tke the squre root, seprte, nd integrte. Convenient to obtin t(x) rther thn x(t): 18 April 013 Astronomy 14, Spring x = 0 1 Ω Ω s = 0. 1 Ω 3 H0 Ω 3 dx d = + d dt Ω dx dt ( dx d) x = H0 ( 1 Ω) x ( 1 x )

21 A flt universe (continued) This integrl cn be done nlyticlly with couple of substitutions of vribles (see below); the solution is Let s plot ginst t(): t ([ 1 Ω] Ω) 13 1 x t = dt = dx H 1 Ω 1 + x Ω 3 1 Ω 3 t( ) = ln H0 1 Ω Ω Ω 3 18 April 013 Astronomy 14, Spring 013 1

22 A flt universe (continued) Age since the Big Bng is the time from the present t which the = 0: Ω M Age (Gyr) Scle fctor (1 = present dy) Flt universes Λ0 1 1 H0 = 74. km sec Mpc 0 Ω 1 Ω M0 =Ω Ω = 1 Ω The first of these universes hs n ge not fr from tht of n empty universe, t = 1 H0 = 13. Gyr (see tomorrow s recittion) Empty 0.5 Ω= Time from present (Gyr) 18 April 013 Astronomy 14, Spring 013

23 A flt universe (continued) Fte: no flt universes with Ω Λ0 0 collpse. All expnd exponentilly, forever, except the puremtter flt Universe (Ω Λ0 = 0, Ω M0 = 1), for which the expnsion continues forever but not so fst 3 ( t ). Scle fctor (1 = present dy) H 0 Flt universes 1 1 = 74. km sec Mpc 0 Ω 1 Ω M0 =Ω Ω = 1 Ω Λ Time from present (Gyr) Empty 0.5 Ω= April 013 Astronomy 14, Spring 013 3

24 Notes on the cosmologicl constnt Despite the ppernce tht it functions just s nother energy density, the cosmologicl constnt is different from the density term(s). The totl mss of physicl universe is constnt energy is conserved, fter ll so in n expnding universe the mss density decreses monotoniclly with time. But Λ, being constnt, does not. As the volume of n expnding universe increses, the totl energy it represents, which we cll drk energy, increses. The drk energy in universe is not necessrily conserved. This increse is responsible for the exponentil, ccelerting expnsion in flt universes with nonzero Ω Λ. 18 April 013 Astronomy 14, Spring 013 4

25 Tht integrl It s in Wolfrm Alph, but mzingly not in Grdshteyn nd Rhyzik. Recll tht secs θ = 1 + tn θ : 13 ([ 1 Ω] Ω) t = = = 0 ([ 1 Ω] Ω) 0 x dx 1 + x u 3 du [ ] ( ) ( ) ([ ] ) [ ] θ θ ( ) rctn 1 Ω Ω rctn 1 Ω Ω sec d = 3 3 Substitute: u= x, du = u = tn θ 0 18 April 013 Astronomy 14, Spring Ω Ω 1 3 Substitute: u = tn θ, du = sec θ = 0 rctn [ 1 Ω] Ω 3 xdx 1 3 θ dθ secθ dθ

26 Tht integrl (continued) The next step requires clssic dirty trick: d ( tnθ + secθ tn sec sec tn sec ) secθ = sec θ = =!! tn sec tn sec tn sec t = rctn 1 θ + θ θ θ + θ dθ θ + θ θ + θ θ + θ 1 ([ Ω] Ω) 3 0 ( θ θ) rctn 1 0 ( tnθ + secθ) 3 tnθ + secθ = ln tn + sec 3 d ([ Ω] Ω) 1 3 Agin, use secθ = 1 + tn θ : ( ) 1 Ω 1 Ω = ln Ω Ω April 013 Astronomy 14, Spring 013 6