MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Depatment Physics 8.033 Decembe 5, 003 Poblem Set 10 Solutions Poblem 1 M s y x test paticle The figue above depicts the geomety of the poblem. The position of the test paticle elative to the cental mass (labeled M ) is give by the vecto + s. The exact elative acceleation a, also called the tidal acceleation, between the feely falling test paticle and the oigin of the x y coodinate system is whee To fist ode in s/, we have a = ( + s) +, + s 3 3 + s 3 = ( + s + s) 3/. ( s ) + s 3 3 1 3. Keeping only those tems that ae fist ode in s, we obtain ( s ) a + s 3. 3 = [s 3(ˆ s) ] ˆ, 3 whee ˆ is a unit vecto paallel to. In the x y coodinate system, we see that ˆ y ˆ and s = xˆ x + y y, ˆ and thus a (x x ˆ y y) ˆ. 3 1
The same fist ode esult may be obtained in a simple fashion by diffeentiating the acceleation / 3 about the point (x, y) = (0, 0): d d a 3 3 5, whee d( 3 ) = 3 d/ 5 has been used. Identifying d s, we see that the ealie esult has been ecoveed. Note that a test paticle on the y axis appeas to be epelled fom the point (0,0), while a test paticle on the x axis appeas to be attacted to the oigin of the x y coodinate systems. Theefoe, the tidal acceleation about the point (0,0) will defom a cicula ing of feely falling paticles, initially at est elative to (0,0), into an elliptical shape, as shown in the figue below. y x The figue below shows the initial tajectoy of fou points located on a cicle centeed at the oigin. (0,y) (0, x) (0,x) (0, y)
Poblem The fist equation (G 00 = 0) gives and thus d [(1 e Λ )] = 0, d (1 e Λ ) = A, whee the constant of integation, which takes the value A = /c. Theefoe, e Λ = 1 A/. Afte a bit of manipulation, the second equation (G 11 = 0) becomes Substituting fo e Λ, we find dφ Λ = e 1. d ( dφ A ) 1 = 1 1 d = A 1, which has the simple solution A Φ = ln 1, o A e Φ = 1, whee the second constant of integation has been set to 0. Poblem 3 The figue below shows how the integated shell adius, s 0, vaies with the coodinate adius,. 3
Poblem 4 Conside a photon that is emitted fom adius 1 in the Schwazschild geomety, and eceived at adius. If the enegy of the emitted photon is E 1, then the enegy, E, measued upon eception of the photon is given by 1/ 1 S / 1 E = E 1, 1 S / whee S = /c is the Schwazschild adius. (a) An eceive at infinity has =, so that 1/ S E = 1 E NS, NS whee NS = 10 km is the neuton sta adius and E NS is 6 kev. Plugging in the numbes, we find E 4.6 kev. (b) We have (c) Fom pats (a) and (b), we see that 1/ 1 S / NS E = E NS. 1 S / 1/ 1/ S S E = 1 E = 1 E NS, just as in pat (a). The enegies of the X ay photons ae unchanged as they pass the obseve at adius, and so it is as if the photons tavel unpetubed fom the suface of the neuton sta to the obseve at infinity. NS Poblem 5 The fist thee queies involve appoximations to the elation 1/ 1/ M M = 1 1, Eath satellite Eath satellite whee both M/ satellite and M/ Eath ae small quantities. (Q1) The task hee is to deive an appoximate elation to second ode in small quantities. Howeve, one must be caeful in doing this, and the esult in the book is missing two tems. The coect second ode esult is satellite M M M M 1 M 3 M 1 + +. Eath satellite Eath satellite Eath satellite Eath 4
(Q) The mass and adius of the Eath ae 6 10 4 kg and 6400 km, espectively, and the adius of the satellite s obit is 7000 km. Using these numbes, we find that M/ Eath 7 10 10 and M/ satellite 10 10. Theefoe, the second ode tems that appea in Q1 ae a facto of 10 10 smalle than the fist ode tems, and so the equation can be safely appoximated to only fist ode: satellite M M +. Eath 1 satellite Eath (Q3) Substituting the appopiate numbes into the fist ode esult, we obtain satellite 1 + 5 10 10. Eath A spatial esolution of 1 m coesponds to a light tavel time of 3 ns. The factional timing diffeence due to geneal elativity between clocks on the Eath and on the satellite is 0.5 ns, and so the GPS system should take this effect into account. (Q4) Fo this poblem we must use somewhat moe accuate values fo the mass of the Eath (5.974 10 4 kg) and the adius of the Eath (6370 km), as well as fo the adius of the 1 hou satellite obit (6600 km). The accumulated time diffeence between the two clocks is given by M M Δt satellite Δt Eath = Δt Eath 45700 ns, Eath satellite whee Δt Eath = 86400 s. (Q5) Given the otation peiod, P, the otation speed of a point on the Eath s equato is v Eath = πr/p 463 m s 1 1.5 10 6 c (Q6) The esulting expession is 1/3 πm v satellite =, T whee both M and T ae in units of metes. In these units M = 0.44 cm and T = 1.3 10 13 m (Q7) Using the fomula above, we find v satellite 1.3 10 5. Also, we have satellite = cv satellite T /π 7000 km. (Q8) Note that vsatellite is of the same ode of magnitude as the quantities M/ satellite and M/ Eath. Theefoe, we should cetainly keep tems popotional to v satellite when computing the elative timing diffeences between the clock on the satellite and the clock on the Eath. Fo completeness, we should also include tems popotional to v Eath, even though it is quite a bit smalle. The fist ode esult fo satellite / Eath is clealy satellite M M 1 1. + vsatellite + v Eath Eath 1 satellite Eath (Q9) Substituting the elevant numbes, we find that afte one day (as measued on the Eath) Δt satellite Δt Eath 38600 ns. 5
Poblem 6 (a) The constant density is given by ρ = M/V, so that 4π M = V ρ = 3 ρ. 3 In geometic units, whee M has units of metes, 4πG M = 3 ρ. 3c (b) Setting = M, we find 4πGρ M = 8M 3, 3c so that 1/ 3c M =. 8πGρ Fo ρ = 1 kg m 3, we have M 1.3 10 13 m. (c) The mass of the black hole in metes is about twice the adius of the sola system. In conventional units, the mass of the black hole is 10 40 kg, o 5 10 9 M. Poblem 7 (a) Fo a black hole of mass 1 M, we find that R S 3 km, and fo a black hole of mass 10 6 M, R S 3 10 6 km. (b) We have Fo = 3R S, we obtain 1/ 1/ 3 3 P = π = π. c R S R s Fo a 10 6 M black hole, the peiod is P 460 s. P = π 54 c M 4.6 10 4 s. (1) M Poblem 8 Fo a timelike geodesic, the pope time, dτ, is elated to the infinitesimal coodinate diffeences, and d, by dτ = f () f 1 ()d, whee f () = 1 M/ < 1. Theefoe, the coodinate (bookkeepe) adial velocity, d/, of the feely falling clock is given by d = f ()[1 f ()E ], 6
whee E f ()/dτ is the conseved enegy like quantity along the geodesic. If the clock is falling inwad, then d/ < 0, so that d = f () 1 f ()E. Now, the pope adial length, d shell, measued by a shell obseve is elated to the bookkeepe s adial length, d, by d shell = f 1/ ()d. Likewise, the pope time, shell, measued by a shell obseve is elated to the bookkeepe s time,, by shell = f 1/. Theefoe, the adial velocity, d shell / shell, measued by a shell obseve is elated to the bookkeepe s velocity, d/, by so that fo the adially falling clock, d shell d = f 1 (), shell d shell = 1 f ()E. shell (a) When the clock is fist dopped fom adius 0, we see that /dτ = f 1/ ( 0 ), so that E = f 1/ ( 0 ). It follows that [ ] 1/ d f () = f () 1 f (0 ), fom which d shell / shell follows immediately. (b) The shell velocity at = 5 km, of a wench that is dopped fom est at 0 = 40 km is given by [ ] 1/ d shell f (5 km) = 1 shell f (40 km) 1/ 1 = 0.45. 5 Clealy, if the wench is dopped fom infinity, it should have a adial speed at = 5 km that is lage than if the wench is dopped fom = 40 km. This is, in fact, the case, since d shell shell [ ] 1/ f (5 km) = 1 f ( ) 1/ 1/ 10 km = = 0.63. 5 km 5 (c) The bookkeepe velocity is d d shell = f (5 km) shell 1/ 10 1 = 1 0.7. () 5 5 7
It is clea that if the wench is dopped fom infinity, the bookkeepe s adial speed will be geate than if the wench is dopped fom = 40 km, since the same facto, f (5 km), multiplies the shell speed in the two cases, while the shell speed is lage if the wench is dopped fom infinity. The actual speed in bookkeepe coodinates fo a wench dopped fom infinity is 0.6(/5) 1/ 0.38. (d) The speed in bookkeepe s coodinates appoaches 0 as the wench neas the hoizon, while a shell obseve nea the hoizon measues a speed nea 1 (speed of light). (e) Fom pat (a), we know that v shell = [1 f ()/f( 0 )] 1/, so that [ ] 1/ f ( 0 ) E shell =, f () whee the mass m of the falling object has been divided out. We also know fom pat (a) that E = f 1/ ( 0 ), so that E shell = f 1/ ()E. The same esult is obtained in Poblem 1, but unde moe geneal cicumstances. Poblem 9 In conventional units, the gavitational acceleation is d = g conv =. Measuing time in metes, we find g =. c We identify /c with the mass M, given in geometic units, so that g M/. Clealy, since both M and ae given in length units, g is an invese length. In these units, the gavitational acceleation at the suface of the eath is g E 10 16 m. (b) In geneal elativity, we know that the adial velocity measued by a shell obseve is d shell / shell = [1 f ()/f( 0 )] 1/. Diffeentiating both sides with espect to t shell, we find [ ] 1/ d shell M f () 1 d () = 1. shell f ( 0 ) f ( 0 ) shell Substituting d = f 1/ ()d shell and the function fo d shell / shell, and evaluating the esult at = 0, we obtain d shell ( g shell = M f 1/ 0 ) ( 0 ). 0 shell As 0 M, the acceleation blows up. Fo lage 0, the appoximate esult is M M g shell 1 +. 0 0 (c) Substituting 0 = 4M, we see that g shell = /16M, whee M 10 km fo a black hole that is thee times the mass of the Sun. This acceleation is 1.4 10 11 g E. The Newtonian esult is 1 10 11 g E. 8
(d) The mass and adius of a typical NS ae 1.4 M (M.1 km) and 10 km, espectively, which places the suface of the NS at 4.8M. The gavitational acceleation at the suface is. 10 11 g E. (e) It is easy to see that f 1/ (M + d) (M/d) 1/, so that g shell (8M d) 1/. Setting g shell = 7g E 7 10 13 km 1, we find M.6 10 3 km fo d = 1 km, which coesponds to a black hole of mass 1.7 10 3 M, which exceeds the mass in the obsevable univese. Poblem 10 This poblem is a diect extension of Poblem 8. Fa fom the black hole, the conseved quantity E = f ()/dτ is equivalent to the (special) elativistic enegy pe unit mass, so that E = γ, whee γ epesent the speed of the paticle, via γ = (1 v ) 1/. Theefoe, fom Poblem 8, we see that and d 1/ = f ()[1 f ()γ ], d shell = [1 f ()γ ] 1/. shell Poblem 11 Recall the fist ode elation between the elapsed time of a satellite and the elapsed time of a clock on Eath (see Poblem 5 of Poblem Set 10): satellite M M 1 1 = 1 + vsatellite + v Eath, Eath satellite Eath whee M is the mass of the Eath in geometic units. Fo a satellite in a cicula obit, v satellite = (M/ satellite ) 1/. Also, let the otation speed at the suface of the Eath be v Eath = ɛ(m/ Eath ) 1/, whee ɛ = 0.06. Now, satellite M M 1 M ɛ M = 1 + + Eath satellite Eath ) ( satellite Eath 3 M ɛ M = 1 + 1 +. satellite Eath Since we ae demanding that the two clocks un at the same ate, we have satellite / Eath = 1, and thus s 3 ɛ 1 3 = 1 +. E We see that the otation of the Eath does not play a significant ole in this calculation. The adius of the satelite is given as: s = 9567 km. 9
Poblem 1 Stat with the geneal expession fo dτ / in the Schwazschild metic: [ ( dτ = 1 M ) ( 1 M ) 1 ] 1/ d dφ, Now, ecall fom poblem 8: d = f d s s = 1 M Futhemoe, poblem 8 tells us the pope time, s, measued by a shell obseve is elated to the bookkeepe s time,, by s = f 1/. This implies that: 1 dφ 1 dφ M dφ = f = 1 s s Now, plug into the oiginal equation the esults fo [ ] 1/ dτ M M d M dφ = 1 1 1, s s M Pull out the common facto of 1 to obtain: { [ ]} 1/ dτ M d s dφ = 1 1. s s Now, we note that d s s and dφ d [ ] 1/ d s dφ + s s. This gives: is the speed of an obiting paticle measued by a shell obseve at bookkeepe adius. Fom Poblem 8, we aleady know that d s / s is the adial component of the velocity measued by a shell obseve. With the same easoning as in Poblem 8, we see fom the metic that dφ is the pope displacement of the paticle in the φ diection tavesed in time s, as measued by the shell obseve. Theefoe, dφ/ s is the φ component of the velocity measued by the shell obseve. It follows that the tem in squae backets that appeas on the second line in the poblem set is indeed γ, in the sense that the shell obseve measues a total s elativistic enegy of γ s m 0, whee m 0 is the est mass of the obiting paticle. Hence, 1/ Remembe, dτ M 1 = 1. γ shell 1 E M M = 1 = 1 γ shell, m 0 dτ 10
So Finally, 1 1 M M E = 1 m 0 γ shell = 1 E shell, Poblem 13 Equations (5 14) and (5 15) in the Taylo & Wheele book ae the equations of motion fo a photon moving in the Schwazschild geomety, given an impact paamete, b: [ ] 1/ d b = ±f () 1 f () ; dφ b = ±f (). (a) Combining the above esults, we immediately find that [ ] 1/ [ d b ] 1/ [ ] 1/ dφ M b + = f () 1 + (1 f ()) = f () 1 +. 3 (b) Fo puely adial motion, b = 0, and the adial velocity in bookkeepe coodinates is simply d/ = ±f (), less than the speed of light in the absence of the mass. (c) The integal that we must evaluate is d Δt =. f () In the weak field limit, f 1 () 1 + M/, so that Δt Δ + M ln, 1 1 whee Δ = 1. Fo 1 = 10 8 m, = 10 10 m, and M = 1 M, we find that Δt Δ + 5 10 5 s and Δt 33 s. In the absence of the mass, the tavel time is just Δt = Δ, which is 5 10 5 s shote than the pevious esult. 11
Poblem 14 (a) Fo a satellite in a cicula obit, d = 0, and the pope time measued on the satellite is given by dτ = f () dφ. Recall that L = dφ/dτ, and thus [ ] 1 dτ L = f () 1 +. Fo a paticle in a cicula obit, it can be shown that (see Taylo & Wheele) L = M 3, so that dτ 3M = 1. (b) Fo the case of the minimum stable cicula obit, dτ / = 1/ 0.71. (c) Afte 10 yeas have gone by at a lage distance fom the black hole, 7 yeas have gone by fo the spaceship that is obiting at = 6M. Theefoe, the black hole does not make fo a vey inteesting time machine unless the spaceship waits fo a vey long time befoe emeging. Inode to get the same effect in flat space fo a ocket, we need γ = 10 which 7 implies β 0.71. Poblem 15 In the poblem set, the minus sign in font of L /m should be changed to a plus sign. Let E and L be the enegy and angula momentum pe unit mass, espectively, so that Setting ṙ = 0, we obtain the quadatic, 1 L E = ṙ +. which has the solutions L + = 0, E E [ ] 1/ L E ± = 1 ± 1 +. E G M Now, define a = /E, so that [ ] 1/ L ± = a 1 ± 1. a 1
Given that ± = a(1 ± e), we see that the tem in paenthesis is e, and theefoe L = a(1 e ). The following optional poblem is fo discussion in ecitation sections, and not to be handed in: Optional Poblem A Solution to be povided. Optional Poblem B (a) Fist, it is clea fom the figue that β = θ α. Also, fom the definitions of the angles in tems of the elevant distances, we see that α = δd LS /D S, so that D LS β = θ δ DS 4 D LS = θ c b = θ θe. (b) Multiply the above equation by θ to obtain a quadatic: θ θ θβ θ E, which has the solutions 1 θ ± = [β ± (β + 4θE) 1/ ] The figue below illustates the angula positions of the two images. D S θ + S β L O θ 13