modern physic forsci and eng 2nd e

www.elsolucionario.net Chapter For a particle Newton’s second law says F = ma = m d2 x/dt2 i + d2 y/dt2 j + d2 z/dt2 k Using Equation (2.1) and taking the necessary derivatives: d2 x d2 x = dt2 dt2 d2 y d2 y = dt2 dt2 d2 z d2 z = dt2 dt2 Therefore F = ma = m d2 x d2 x d2 x i + j + k =F dt2 dt2 dt2 From Equation (2.1) p=m dx dy dz i+ j+ k dt dt dt In a Galilean transformation dx dx = −v dt dt dy dy = dt dt dz dz = dt dt Therefore p=m dy dz dx j+ k=p +v i+ dt dt dt However, because p =m dy dz dx i+ j+ k dt dt dt the same form is clearly retained, given the velocity transformation dx /dt = dx/dt − v Using the vector triangle shown, the net speed√of the light coming toward mirror D is c2 − v and the same on the return trip Therefore the total time is t2 = v c2 – v2 θ 2l2 distance =√ speed c − v2 Notice that sin θ = v/c, so θ = sin−1 (v/c) ∼ = v/c As in Problem 3, sin θ = v1 /v2 , so θ = sin−1 (v1 /v2 ) and v = www.elsolucionario.net v22 − v12 c CHAPTER When the apparatus is rotated by 90◦ , the situation is equivalent, except that we have effectively interchanged l1 and l2 Interchanging l1 and l2 in Equation (2.3) leads to Equation (2.4) * Let n = the number of fringes shifted From the analysis in the text n= c (∆t − ∆t) v (l1 + l2 ) = λ c2 λ Solving for v and noting that l1 + l2 = 22 m, v=c nλ = 3.00 × 108 m/s l1 + l2 (0.005) (589 × 10−9 m) = 3.47 km/s 22 m * Letting l1 → l1 − β (where β = v/c) the text equation for t1 becomes t1 = l1 − β 2 l1 = c (1 − β) c − β2 which is identical to t2 when l1 = l2 , so ∆t = as required Because the Lorentz transformations depend on c (and the fact that c is the same constant for all inertial frames), different values of c would necessarily lead two observers to different conclusions about (say) the order or positions of two spacetime events, in violation of Postulate Let an observer in S send a light signal along the +x-axis with speed c According to the Galilean transformations, an observer in S would measure the speed of the signal to be dx dx = −v =c−v dt dt Therefore the speed of light cannot be constant under the Galilean transformations www.elsolucionario.net CHAPTER 10 From the Principle of Relativity, we know that the correct transformation must be of the form (assuming y = y and z = z ) x = ax − bt x = ax + bt The spherical wavefront equations give us ct = (ac − b) t ct = (ac + b) t Substitute the second wavefront equation into the first: ct = (ac + b)(ac − b) t/c c2 = (ac + b)(ac − b) = a2 c2 − b2 Now v is the speed of the origin of the x -axis with respect to the x-axis We can find that speed by setting x = which gives = ax − bt, or v = x/t = b/a, or equivalently b = av Substituting this into the equation above for c2 yields c = a c − a2 v = a c − v Solving for a: a= 1− v2 c2 =γ This expression, along with b = av, can be substituted into the original expressions for x and x to obtain x = γ (x − vt) x = γ (x + vt ) which in turn can be solved for t and t to complete the transformations *11 When v > mc2 , we have √ Ecm ∼ = 2Kmc2 = (0.938 GeV) (7000 GeV) = 114.5 GeV b) For colliding beams the available energy is the sum of the two beam energies, or 14 TeV This is an improvement over the fixed-target result by a factor of 14000 GeV = 122 114.5 GeV 173 www.elsolucionario.net Chapter 15 * From Newton’s second law we have for a pendulum of length L F = mG g sin θ = mI a = mI L d2 θ dt2 mG g d2 θ mG g sin θ ∼ θ = = dt mI L mI L where we have made the small angle approximation sin θ ∼ = θ This is a simple harmonic oscillator equation with solution θ = θ0 cos(ωt) where θ0 is the amplitude and the angular frequency is mG g ω= mI L The period of oscillation is T = 2π mI L = 2π ω mG g Therefore two masses with different ratios mI /mG will have different small-amplitude periods ∆ν = gHν (9.80 m/s2 ) (4 × 105 m) (108 s−1 ) = = 4.36 × 10−3 Hz 2 c (2.998 × 10 m/s) ∆ν GM =− ν c 1 − r1 r2 =− GM GM (r2 − r1 ) = (r1 − r2 ) r r2 c r r2 c Use r1 − r2 = H and let r1 ∼ = r2 = r From classical mechanics g = GM/r2 , so ∆ν = gHν c2 GM ∆T =− T c 1 − r1 r2 We use r2 = 6378 km and r1 = 6378 km +10 km = 6388 km ∆T T (6.673 × 10−11 m3 · kg−1 · s−2 ) (5.98 × 1024 kg) 1 − 6388 × 10 m 6378 × 103 m (2.998 × 10 m/s) = 1.09 × 10−12 = − which is the same as in the example, to three significant digits 174 www.elsolucionario.net CHAPTER 15 The distance d is the sum of the radii of the earth’s orbit and Venus’s orbit (assuming circular orbits) d = 149.6 × 109 m + 108.2 × 109 m = 258.8 × 109 m The round-trip time is t= 2d (258.8 × 109 m) = = 1726.5 s c 2.998 × 108 m/s The percent change is therefore 200 × 10−6 s (100 %) = 1.16 × 10−5 % 1726.5 s * If the photon “falls” at a rate g then during the time taken to travel a distance x = 40, 000 km it has fallen a distance d : 1 d = gt2 = 9.80 m/s2 2 × 107 m 2.998 × 108 m/s = 8.72 cm Using the mass and radius of the sun ∆ν GM (6.673 × 10−11 m3 · kg−1 · s−2 ) (1.99 × 1030 kg) = = = 2.123 × 10−6 ν rc2 (6.96 × 108 m) (2.998 × 108 m/s)2 The wavelength is affected by the same factor, so the redshift of the given wavelength is ∆λ = 2.123 × 10−6 (550 nm) = 1.17 × 10−3 nm As in the previous problem ∆ν (6.673 × 10−11 m3 · kg−1 · s−2 ) (5 × 1030 kg) GM = = 0.371 = ν rc2 (104 m) (2.998 × 108 m/s)2 ∆λ = (0.371) (550 nm) = 204 nm Let us assume that g is constant over this short distance Using E = hν we find ∆ν = gHν (9.80 m/s2 ) (22.5 m) (14.4 × 103 eV) gHE = = = 8541 Hz c2 c2 h (2.998 × 108 m/s)2 (4.136 × 10−15 eV · s) The percentage change is (14.4 × 103 8541 Hz (100 %) = 2.45 × 10−13 % eV) / (4.136 × 10−15 eV · s) 175 www.elsolucionario.net CHAPTER 15 10 rs = 2GM (6.673 × 10−11 m3 · kg−1 · s−2 ) (7.35 × 1022 kg) = = 1.09 × 10−4 m c2 (2.998 × 108 m/s)2 *11 rs = 2GM (6.673 × 10−11 m3 · kg−1 · s−2 ) (1.90 × 1027 kg) = = 2.82 m c2 (2.998 × 108 m/s)2 *12 hc3 (6.626 × 10−34 J · s) (2.998 × 108 m/s) T = = 8πkGM 8π (1.381 × 10−23 J/K) (6.673 × 10−11 m3 · kg−1 · s−2 ) (1.99 × 1030 kg) = 3.87 × 10−7 K 13 Rearranging the formula given in the previous problem, hc3 (6.626 × 10−34 J · s) (2.998 × 108 m/s) M= = = 2.63×1021 kg −23 −11 −1 −2 8πkGT 8π (1.381 × 10 J/K) (6.673 × 10 m · kg · s ) (293 K) which is about 2.63 × 1021 kg = 32 × 10−9 1.99 × 1030 kg solar masses 2GM (6.673 × 10−11 m3 · kg−1 · s−2 ) (2.63 × 1021 kg) rs = = = 3.91 × 10−6 m 2 c (2.998 × 10 m/s) *14 Set the change in the photon’s energy equal to the change in gravitational potential energy: ∆E = h ∆ν = − GM m GM m − − = −GM m − r1 r2 r1 r where M is the mass of the earth and m is the equivalent mass of the photon Now m = E/c2 = hν/c2 , so GM hν 1 h ∆ν = − − c2 r r2 GM ∆ν =− − ν c r1 r2 176 www.elsolucionario.net CHAPTER 15 15 Using the formula from the previous problem and recalling that the earth’s radius is 6378 km, ∆ν GM 1 = − − ν c r1 r2 (6.673 × 10−11 m3 · kg−1 · s−2 ) (5.98 × 1024 kg) 1 = − − 6.678 × 106 m 6.378 × 106 m (2.998 × 108 m/s) = 3.127 × 10−11 Therefore ∆ν = 3.127 × 10−11 294 × 106 Hz = 9.19 × 10−3 Hz 16 g = GM/r2 which can be differentiated to give dg = − 2GM dr r3 For a small change let dg ∼ = ∆r = m Also notice that the distance from the = |dg| and dr ∼ center of the earth is 6.378 × 10 m +3 × 105 m = 6.678 × 106 m Then (6.673 × 10−11 m3 · kg−1 · s−2 ) (5.98 × 1024 kg) ∆g ∼ (3 m) = 8.04 × 10−6 m/s2 = (6.678 × 10 m) This is about 10−7 g, so it is a very small effect 17 If we use λ = h/mc for a relativistic particle of mass m , we have λ= h 2πGm = πrs = mc c2 Solving for m we have m= hc = 2πG (6.626 × 10−34 J · s) (2.998 × 108 m/s) = 2.18 × 10−8 kg −11 −1 −2 2π (6.673 × 10 m · kg · s ) The Planck energy is EPl = mc2 = 2.18 × 10−8 kg 2.998 × 108 m/s = 1.96 × 109 J = 1.22 × 1028 eV 18 a) The combination of G, h, and c that has the right units is λPl = Gh = c3 (6.673 × 10−11 m3 · kg−1 · s−2 ) (6.626 × 10−34 J · s) = 4.05 × 10−35 m (2.998 × 10 m/s) b) hc h 1240 × 10−9 eV · m = = 1.02 × 10−34 m λ= = 28 mc mc 1.22 × 10 eV which is the same order of magnitude as (a) 177 www.elsolucionario.net CHAPTER 15 19 The combination of constants that gives time is tPl = Gh = c5 (6.673 × 10−11 m3 · kg−1 · s−2 ) (6.626 × 10−34 J · s) = 1.35 × 10−43 s (2.998 × 10 m/s) The time for light to travel the Planck length is t= λPl = c Gh = 1.35 × 10−43 s c5 as we found in this problem 20 As in Problem 15 ∆ν GM 1 = − − ν c r1 r2 (6.673 × 10−11 m3 · kg−1 · s−2 ) (5.98 × 1024 kg) = − (2.998 × 108 m/s)2 1 × − 3.587 × 107 m + 6.378 × 106 m 6.378 × 106 m = 5.91 × 10−10 Therefore ∆ν = 5.91 × 10−10 × 109 Hz = 1.18 Hz 178 www.elsolucionario.net Chapter 16 * 1 pc = au tan 1.496 × 1011 m/au = 3.086 × 1016 m ly = 2.9979 × 108 m/s (365.25 d/y) (86400 s/d) = 9.461 × 1015 m pc = 3.086 × 1016 m = 3.26 ly 9.461 × 1015 m/ly As in Example 16.3 we have = exp (∆mc2 /kT ) so ln = ∆mc2 /kT Then T = ∆mc2 939.56563 MeV − 938.27231 MeV = 7.71 × 109 K = k ln (8.617 × 10−11 MeV/K) (ln 7) Using mc2 = 135 MeV = kT T = 135 MeV = 1.57 × 1012 K 8.617 × 10−11 MeV/K From Figure 16.8 the time associated with this temperature is about 10−1 s Similarly, number of d 10 MeV − MeV = exp number of u (8.617 × 10−11 MeV/K) (1014 K) = 1.00 1300 MeV − 10 MeV number of u = exp number of c (8.617 × 10−11 MeV/K) (1014 K) = 1.16 number of u 115 MeV − 10 MeV = exp number of s (8.617 × 10−11 MeV/K) (1014 K) = 1.01 115 MeV − MeV number of d = exp number of s (8.617 × 10−11 MeV/K) (1014 K) = 1.01 number of u or d 1300 MeV − MeV = exp number of c (8.617 × 10−11 MeV/K) (1014 K) 174000 MeV − MeV number of u or d = exp number of t (8.617 × 10−11 MeV/K) (1014 K) = 5.9 × 108 number of u or d 4250 MeV − MeV = exp number of b (8.617 × 10−11 MeV/K) (1014 K) 179 www.elsolucionario.net = 1.16 = 1.64 CHAPTER 16 number of s 1300 MeV − 115 MeV = exp number of c (8.617 × 10−11 MeV/K) (1014 K) number of s 174000 MeV − 115 MeV = exp number of t (8.617 × 10−11 MeV/K) (1014 K) = 1.15 = 5.8 × 108 number of s 4250 MeV − 115 MeV = exp number of b (8.617 × 10−11 MeV/K) (1014 K) number of c 174000 MeV − 1300 MeV = exp number of t (8.617 × 10−11 MeV/K) (1014 K) = 5.1 × 108 number of c 4250 MeV − 1300 MeV = exp number of b (8.617 × 10−11 MeV/K) (1014 K) number of b 174000 MeV − 4250 MeV = exp number of t (8.617 × 10−11 MeV/K) (1014 K) = 1.62 = 1.41 = 3.6 × 108 mc2 = kT so we have electron: T = muon: T = 0.5110 MeV mc2 = = 5.93 × 109 K k 8.617 × 10−11 MeV/K mc2 105.66 MeV = = 1.23 × 1012 K k 8.617 × 10−11 MeV/K As in the previous problem we have for the eV/c2 mass: T = for the 10−4 eV/c2 mass: eV mc2 = = 5.80 × 104 K k 8.617 × 10−5 eV/K T = 10−4 eV mc2 = = 1.16 K k 8.617 × 10−5 eV/K Note that the answer for mc2 = 10−4 eV is lower than the present temperature! * The π + (E0 = 140 MeV) is more massive than the π (E0 = 135 MeV), so the π + was formed first With ∆mc2 = k∆T we have ∆T = ∆mc2 MeV = = 5.80 × 1010 K −11 k 8.617 × 10 MeV/K * Set the deuteron binding energy 2.22 MeV equal to kT : T = 2.22 MeV 2.22 MeV = = 2.58 × 1010 K k 8.617 × 10−11 MeV/K 180 www.elsolucionario.net CHAPTER 16 Set the hydrogen binding energy 13.6 eV equal to kT : T = 13.6 eV 13.6 eV = = 1.58 × 105 K k 8.617 × 10−5 eV/K 10 0.3GN m2 3.9¯ h2 N 5/3 = V 4/3 2mV 5/3 V 1/3 6.5¯ h2 3.9¯ h2 = 1/3 = 0.6m3 N 1/3 G N mG 11 ρ= mass 2M 3M = = volume 2πR3 πR where M = 1.99 × 1030 kg (mass of sun) and R = 11 km 3M (1.99 × 1030 kg) 17 = kg/m3 = 14 × 10 2πR3 2π (11 × 10 m) ρ= For the nucleon (or nucleus, assuming uniform nuclear density) we have mp (1.67 × 10−27 kg) 17 ρ= = kg/m3 = 2.31 × 10 −15 πr0 4π (1.2 × 10 m) and the neutron star is about three times as dense 12 We see from Equation (16.8) that V 1/3 is proportional to N −1/3 However, V 1/3 is proportional to R, so R is proportional to N −1/3 The reason this makes sense is that the gravitational pressure increases not in proportion to N but rather in proportion to N 5/3 [see Equation (16.7)] 13 a) (N m)2 P = 0.3G = 0.3G V 4/3 M2 πR3 4/3 (1.99 × 1030 kg) = 0.3 6.67 × 10−11 m3 · kg−1 · s−2 π 3 4/3 (6.96 × 108 m) = 5.00 × 1013 N/m2 b) P = 0.3G M2 πR3 33 = 3.21 × 10 −11 4/3 = 0.3 6.67 × 10 −1 m · kg N/m2 181 www.elsolucionario.net ·s −2 (1.99 × 1030 kg) π (1.1 × 104 m)3 4/3 CHAPTER 16 *14 v = HR = 65 km/s × 109 ly Mpc *15 R= Mpc 3.26 × 106 ly = 7.98 × 104 km/s ∼ = 0.27c v 15000 km/s = = 231 Mpc or about 752 Mly H 65 km/s/Mpc 16 For a redshift of 3.8 we have λ/λ0 = 4.8, so 4.8 = 1+β 1−β which can be solved to find β = 0.917 Then R= v 0.917 (299790 km/s) = = 4230 Mpc or about 13.8 Gly H 65 km/s/Mpc 17 a) × 10−28 kg/m3 nucleon 1.67 × 10−27 kg = 0.18 nucleons/m3 b) R = (16.6 ly) 9.46 × 1015 m/ly = 1.57 × 1017 m 60 1.99 × 1030 kg nucleon = 7.15 × 1058 nucleons 1.67 × 10−27 kg The nucleon density is the number of nucleons divided by the volume, or 7.15 × 1058 = 4.41 × 106 nucleons/m3 17 m)3 π (1.57 × 10 The nucleon density in our neighborhood is much larger than it is for the universe as a whole 18 Using (16.2) we see that H = a da dt Substituting (da/dt)2 from Equation (16.14) H2 = a2 8πGρ kc2 8π Gρa2 − kc2 = − 3 a 182 www.elsolucionario.net CHAPTER 16 19 H = (21 km/s/Mly) 106 Mly = 2.22 × 10−18 s−1 15 (9.46 × 10 m) ρc = (2.22 × 10−18 s−1 ) 3H = = 8.82 × 10−27 kg/m3 = 8.82 × 10−30 g/cm3 −11 −1 −2 8πG 8π (6.67 × 10 m · kg · s ) 20 We use H = 65 km/s/Mpc or H = (65000 m/s/Mpc) Mpc = 1063 × 10−18 s−1 3.086 × 1022 m a) For critical density ρ = × 10−27 kg/m3 : q= 4πρG 4π (9 × 10−27 kg/m3 ) (6.67 × 10−11 m3 · kg−1 · s−2 ) = = 0.57 3H (2 1063 × 10−18 s−1 )2 b) With the observed mass density × 10−28 kg/m3 we have q= 4π (3 × 10−28 kg/m3 ) (6.67 × 10−11 m3 · kg−1 · s−2 ) 4πρG = = 0.019 3H (2 1063 × 10−18 s−1 )2 Note that at critical density we should get q = 0.5, so the value of H used in part (a) brings us close to critical density 21 a) a = Ctn d2 a/dt2 = n(n − 1)Ctn−2 da/dt = nCtn−1 q = −a d2 a/dt2 da dt =− an(n − 1)Ctn−2 an(n − 1) n(n − 1) =− n =− 2 2n−2 nC t n Ct n2 where in the last step we used the definition a = Ctn From this we see that there is deceleration (q > 0) only if < n < b) nCtn−1 da n = = n a dt Ct t with < n < There is an inverse dependence on time H= 22 From Wien’s law λmax T = 2.898 × 10−3 m·K λmax = 2.898 × 10−3 m · K = 1.07 mm 2.7 K 183 www.elsolucionario.net CHAPTER 16 *23 For H = 55 km/s/Mpc we have Mpc = 1.78 × 10−18 s−1 3.086 × 1022 m H = (55000 m/s/Mpc) ρc = (1.78 × 10−18 s−1 ) 3H = = 5.67 × 10−27 kg/m3 8πG 8π (6.67 × 10−11 m3 · kg−1 · s−2 ) For H = 85 km/s/Mpc we have Mpc = 2.75 × 10−18 s−1 3.086 × 1022 m H = (85000 m/s/Mpc) ρc = (2.75 × 10−18 s−1 ) 3H = = 1.35 × 10−26 kg/m3 8πG 8π (6.67 × 10−11 m3 · kg−1 · s−2 ) 24 a) the only combination of the constants that gives time is m = 12 , n = 12 , and l = 25 b) = (6.67 × 10−11 m3 · kg−1 · s−2 ) (6.626 × 10−34 J · s) = 1.35 × 10−43 s (2.998 × 10 m/s) Gh = c5 25 A complete derivation can be found in Section 2.10 26 Let dm be the mass of a spherical shell of thickness dr, so dm = ρ d(vol) = 4πρr2 dr dV = − GM dm r where M = 43 ρπr3 is the mass inside radius r Thus dV = −Gρ2 R V = 16π r dr dV = − 16π ρ2 R5 G 15 Now ρ = 3M/4πR , so V =− 16π ρ2 R5 G 15 9M 16π R6 =− 27 From both figures it appears that t0 ∼ = τ /2 184 www.elsolucionario.net 3GM 5R CHAPTER 16 28 In general as in Example 16.7 we have ∆t = (d/2c) (mc2 /E) d (distance) (50 kpc) 3.086 × 1019 m = = 2.57 × 1012 s 2c (3.00 × 108 m/s) kpc Note that 2.57 × 1012 s = (2.57 s) (10 MeV/10 eV)2 , so mc2 10 eV distance ∆t = (2.57 s) 50 kpc 10 MeV E 29 We know from Section 9.7 that the energy flux rate Φ = σT is related to the energy density ρe by a factor of c/4, such that Φ = (c/4)ρe But using E = mc2 we have ρe = ρrad c2 , so c c σT = ρe = ρrad c2 4 or, rearranging ρrad = 4σT c3 30 ρrad 4σT 4 (5.67 × 10−8 W · m−2 · K−4 ) (3 K)4 = = = 6.80 × 10−31 kg/m3 c (3.00 × 108 m/s)3 This is substantially less than the matter density of about × 10−28 kg/m3 , so the universe is matter dominated 31 At 700,000 years the temperature is about 108 K Then ρmatter = ρrad = 4σT 4 (5.67 × 10−8 W · m−2 · K−4 ) (108 K) = = 0.84 kg/m3 c3 (3.00 × 10 m/s) This is slightly less than the density of air at STP, so the universe was quite dense! ρ rad (kg/m3) 1070 1030 10–10 10–50 10–40 10–30 10–20 10–10 100 Time (s) 185 www.elsolucionario.net 1010 1020 CHAPTER 16 32 The 300 day mark is almost exactly 200 days after the peak, so use t = 200 d in the exponential decay formula exp −(ln 2)t/t1/2 56 Ni: exp −(ln 2)t/t1/2 = exp (−(ln 2) (200 d) / (6.1 d)) = 1.35 × 10−10 56 Co: exp −(ln 2)t/t1/2 = exp (−(ln 2) (200 d) / (77.1 d)) = 0.166 The cobalt must be primarily responsible, with some contribution from the nickel *33 Redshift = ∆λ 582.5 nm − 121.6 nm = 3.79 = λ0 121.6 nm 3.79 = 1+β −1 1−β so β = 0.92 and v = 0.92c *34 From the Doppler effect we know λ = λ0 ∆λ 1+β λ0 + ∆λ =1+ =1+z = 1−β λ0 λ0 35 Using the binomial expansion on the result of the previous problem β 1+z ∼ = + + 1+ β + ∼ =1+β so we see that to first order z ∼ = β 36 5.34 = 1+β −1 1−β so β = 0.951 and v = 0.951c R= v 0.951 (299790 km/s) = = 4390 Mpc H 65 km/s/Mpc 37 Q = (2mp − md − me ) u · c2 = 0.43 MeV There are three particles in the final state, but it is possible for the deuteron and positron to have negligible energy, in which case the neutrino energy is 0.43 MeV 186 www.elsolucionario.net ... at the energy difference between levels in hydrogen we see that E2 − E1 = 10.2 eV, E3 − E1 = 12.1 eV, and E4 − E1 = 12.8 eV There is enough energy to excite only to the second or third level In... frame, and E0 is the rest energy of the electron From conservation of energy and momentum: hν momentum: = pe = c Ee2 − E0 2 or c energy: hν + Ee = 3E0 or hν = Ee2 − E0 2 hν = 3E0 − Ee Squaring and. .. eV) and because of the energy difference E= it comes from n = with E5 = −0.54 eV 25 In general the ground state energy is Z E0 a) E = 12 E0 = 13.6 eV The reduced mass does not change this result