Baixe Resolução Marion [Mec.Clássica] - Capítulo 14 e outras Manuais, Projetos, Pesquisas em PDF para Mecânica, somente na Docsity! CHAPTER 14 The Special Theory of Relativity 14-1. Substitute Eq. (14.12) into Eqs. (14.9) and (14.10): 1 1 v x x x c γ = −′  1   (1) (1 1 1 v x x x c  = + = )′ ′   ′γ γ γ (2) From (1) 1 1 1 x v x c γ′  = −   From (2) 1 1 1 1 x vx c γ ′ =  +   So 1 1 1 v vc c γ γ  − =     +   or 2 2 1 1 v c γ = − 461 462 CHAPTER 14 14-2. We introduce cosh , sinhy y v cα α≅ ≅ and substitute these expressions into Eqs. (14.14); then 1 1 1 2 2 3 3 cosh sinh cosh sinh ; x x ct x t t a c x x x x α α α = −′    = −′    = =′ ′  (1) Now, if we use cosh α = cos (iα) and i sinh α = sin (iα), we can rewrite (1) as ( ) ( ) ( ) ( ) 1 1 1 cos sin sin cos x x i ict i ict x i ict i α α α α = +′   = − +′  (2) Comparing these equations with the relation between the rotated system and the original system in ordinary three-dimensional space, 1 1 2 2 1 2 3 3 cos sin sin cos x x x x x x x x θ θ θ θ = +′   = − +′  =′  (3) x2x2′ x1 x1′ θ We can see that (2) corresponds to a rotation of the 1x ict− plane through the angle iα. 14-3. If the equation ( ) ( ) 2 2 2 2 ,1 , x ict x ict c t ψ ψ ∂ ∇ − ∂ 0= (1) is Lorentz invariant, then in the transformed system we must have ( ) ( ) 2 2 2 2 ,1 , x ict x ict c t ψ ψ ∂ ′ ′ ∇ −′ ′ ′ ∂ 0= ′ (2) where 2 2 2 2 2x y z 2 2 ∂ ∂ ∂ ∇ = + +′ ∂ ∂ ∂′ ′ ′ (3) We can rewrite (2) as ( )24 2 1 , 0 x ict xµ µ ψ = ∂ ′ ′ = ∂ ′∑ (4) THE SPECIAL THEORY OF RELATIVITY 465 ( ) (1 2 1 2 1 222 2 1 1 v t t x x cv c )t t  − = − − −′ ′    − (2) We also have 1 2x x− = (3) ( )1 2v t t− = (4) ( )1 2v t t− =′ ′ ′ (5) Multiplying (2) by v and using (3), (4), and (5), we obtain the FitzGerald-Lorentz contraction: 2 21 v c = −′ (6) 14-5. The “apparent shape” of the cube is that shape which would be recorded at a certain instant by the eye or by a camera (with an infinitesimally short shutter speed!). That is, we must find the positions that the various points of the cube occupy such that light emitted from these points arrives simultaneously at the eye of the observer. Those parts of the cube that are farther from the observer must then emit light earlier than those parts that are closer to the observer. An observer, looking directly at a cube at rest, would see just the front face, i.e., a square. When in motion, the edges of the cube are distorted, as indicated in the figures below, where the observer is assumed to be on the line passing through the center of the cube. We also note that the face of the cube in (a) is actually bowed toward the observer (i.e., the face appears convex), and conversely in (b). (a) Cube moving toward the observer. (a) Cube moving away from the observer. 466 CHAPTER 14 14-6. K′ v x1 x2 K We transform the time t at the points and in the K system into the K′ system. Then, 1x 2x 1 1 2 2 2 2 vx t t c vx t t c  = −′        = −′      γ γ (1) From these equations, we have ( )1 2 1 2 2 1x x t t t v v x c c − − = − = − ∆′ ′ ′ γ γ 2∆ = (2) 14-7. K K′ v x Suppose the origin of the K′ system is at a distance x from the origin of the K system after a time t measured in the K system. When the observer sees the clock in the K′ system at that time, he actually sees the clock as it was located at an earlier time because it takes a certain time for a light signal to travel to 0. Suppose we see the clock when it is a distance from the origin of the K system and the time is t in K and 1 1t′ in K′. Then we have ( ) 1 1 2 1 1 v t t c c t t tv x t v γ  = −′       − =   =  =  (1) We eliminate , t , and x from these equations and we find 1 THE SPECIAL THEORY OF RELATIVITY 467 1 1 v t c γ t = −′    (2) This is the time the observer reads by means of a telescope. 14-8. The velocity of a point on the surface of the Earth at the equator is ( )8 4 4 2 6.38 10 cm2 8.64 10 sec 4.65 10 cm/sec eRv ππ τ × × = = × = × (1) which gives 4 6 10 4.65 10 cm/sec 1.55 10 3 10 cm/sec v c −×= = = × × β (2) According to Eq. (14.20), the relationship between the polar and equatorial time intervals is 2 2 1 1 21 t t t β β ∆ ∆ = ≅ ∆ +′ −   (3) so that the accumulated time difference is 2 1 2 t t β t∆ = ∆ − ∆ = ∆′ (4) Supplying the values, we find ( ) ( ) ( )6 71 1.55 10 3.156 10 sec/yr 10 yr 2 −× × × × 2∆ = × (5) Thus, 0.0038 sec∆ = (6) 14-9. w dm′ m + dm v + dv The unsurprising part of the solution to the problem of the relativistic rocket requires that we apply conservation of momentum, as was done for the nonrelativistic case. The surprising, and key, part of the solution is that we not assume the mass of the ejected fuel is the same as the mass lost from the rocket. Hence ( ) ( ) ( ) wp mv d m dm v dv dm wγ γ γ γ+ + + += = ′ (1) where –dm is the mass lost from the rocket, dm′ is the mass of the ejected fuel, ( ) ( )21w v V vV c≡ − − is the velocity of the exhaust with respect to the inertial frame, and 2 21 1w w cγ ≡ − . One can easily calculate 3d dγ γ β β= , ad after some algebra one obtains 470 CHAPTER 14 ( ) ( ) ( ) 12 8 2 2 2.5 10 m/s 100 m .4 sec 1 2.5 3 .22 sec v t t x c c γ µ µ  = −′     × −    = − = The speed observed by the racer is 82.5 10 m/sv t t ′ = = = × ′ 14-13. ( ) 1 22 1.5 s 1 0.999 22.4 t t t γ µ γ − ∆ = ∆′ ∆ = = − Therefore 34 st µ∆ ′ . 14-14. K K′ receiver v source In K, the energy and momentum of each photon emitted are 00 and h E h p c ν ν= = Using Eq. (14.92) to transform to K′: ( ) 01 1 0 0 ; h E h E vp p c v h h c ν ν γ γ ν ν  = = − = −′     = +   THE SPECIAL THEORY OF RELATIVITY 471 So 0 0 02 1 1 1 11 v c  = +   + + = = −− ν ν γ β β ν ν ββ which agrees with Eq. (14.31). 14-15. From Eq. (14.33) 0 1 1 β ν ν β − = + Since cλ ν= 0 1 1 β λ λ β − = + or 0 1 1 β λ λ β − = + With 0 656.3 nmλ = and 4 8 4 10 3 10 β × = × , λ = 656.4 nm. S o the shift is 0.1 nm toward the red (longer wavelength). 14-16. θ θ′ Earth star v K′ K Consider a photon sent from the star to the Earth. From Eq. (14.92) ( )1E E vpγ= −′ also ( )1E E vpγ= +′ ′ 472 CHAPTER 14 Now 00 1 1, , cos , cos hh E h p p c c E h νν ν ν θ= = = − = −′ ′ θ ′ Substituting yields ( )0 1 cosν ν γ β θ= + and ( )0 1 cosν γ ν β θ= − ′ Thus ( ) ( ) 21 cos 1 cosβ θ β θ γ −+ − ′ = 1 2 2cos cos cos cos 1β θ β θ β θ θ β+ − − = −′ ′ cos cos cos cosθ θ β θ θ β− − =′ ′ − Solving for cos θ yields cos cos 1 cos θ β θ β θ −′ = − ′ where angle in earth’s frame angle in star’s frame v cβ θ θ = = =′ 14-17. From Eq. (14.33) 0 1 1 β ν ν β − = + Since cν λ= , 0 1 1 β λ λ β + = − We have 01.5λ λ= . This gives 5 13 β = or 81.2 10 m/secv = × THE SPECIAL THEORY OF RELATIVITY 475 2 80 80 1 0.3 0.95 years 3 3  − =    So Those on Earth age 26.7 years. The astronaut ages 25.4 years. 14-21. ( ) ( ) ( ) 0 0 3 22 2 2 0 3 22 2 1 2 2 1 1 1 1 1 md F m dt m β β β β β ββ β β   − −      = = +    − −   −        = +  − −  v v v v v (1) If we take (this does not mean v v1 1v=v e 2 3 0= = ), we have ( ) ( ) 1 1 1 01 1 0 13 2 3 22 2 21 1 1 v v v mv c c v m v β β β     = + = =  − − −    1F m (2) 02 221 t m 2F v m v β = = − (3) 03 321 t m 3F v m v β = = − (4) 14-22. The total energy output of the sun is ( )3 21.4 10 W m 4dE R dt −= × ⋅ × 2π (1) where is the mean radius of the Earth’s orbit around the sun. Therefore, 111.50 10 mR = × 263.96 10 W dE dt × (2) The corresponding rate of mass decrease is 92 1 4.4 10 kg s dm dE dt c dt 1−= × ⋅ (3) The mass of the sun is approximately 1. , so this rate of mass decrease can continue for a time 3099 10 kg× 476 CHAPTER 14 30 13 9 1 1.99 10 yr 1.4 10 yr 4.4 10 kg s T − × = × ⋅ × (4) Actually, the lifetime of the sun is limited by other factors and the sun is expected to expire about years from now. 94.5 10× 14-23. From Eq. (14.67) ( ) 2 2 2 2 0 2 2 0 0 2 0 2 2 2 2 2 2 p c E E E T E E T T p c T mc T = − = + − = + = + 14-24. The minimum energy will occur when the four particles are all at rest in the center of the mass system after the collision. Conservation of energy gives (in the CM system) 22 4p pE m c= or 2,CM 02 2p pE m c= = E which implies γ = 2 or 3 2β = To find the energy required in the lab system (one proton at rest initially), we transform back to the lab ( )1E E vpγ= +′ ′ (1) The velocity of K′(CM) with respect to K(lab) is just the velocity of the proton in the K′ system. So u = v. Then ( ) ( ) 2 21 CMv p v mu mv mc 2vp γ γ γ= = = =′ β Since γ = 2, 3 2β = , 1 0 3 2 vp E=′ Substituting into (1) THE SPECIAL THEORY OF RELATIVITY 477 lab 0 0 0 0 3 7 2 2 2 2 E E E Eγ    = + = =       7E 2 2 The minimum proton energy in the lab system 7 , of which 6 is kinetic energy.p pm cis m c 14-25. Let B z 0B= x yv v= +v i j Then 0 0 0 0 0 0 x y y x q q v v B q v B v B × =  = −  i j k v B i j ( ) ( )d dq m dt dt γ= × = =F v B p v gives ( )0 y xqBd v vdt m= −v iγ j Define 0q B mω γ≡ Thus and x y y xv v v vω ω= = − or 2x y xv vω ω= = − v and 2y x yv vω ω= − = − v t t So cos sin cos sin x y v A t B v C t D ω ω ω ω = + = + Take ( )0xv v= , ( )0yv = 0 . Then A = v, C = 0. Then ( ) ( )0 0x y 0v vω= = ( ) ( )0 0 0, y xv v B D v = − = − → = = − vω ω 480 CHAPTER 14 0.115v c≤ 7 The classical kinetic energy will be within 1% of the correct for 0 3.5 10 m/sec, independent of mass.v ×value ≤ ≤ 14-29. 0E Eγ= For ( ) 9 6 0 4 1 22 2 10 2 30 10 eV 0.51 10 eV, 5.88 10 1 or 1 1 1 1 1 1.4 10 2 E E γ γ β β β γ − − = × × × = = − − = − × γ− ( )101 1.4 10 c 0.99999999986 c v −= − × = 14-30. A neutron at rest has an energy of 939.6 MeV. Subtracting the rest energies of the proton (938.3 MeV) and the electron (0.5 MeV) leaves 0.8 MeV. Ot her than rest energies 0.8 MeV is available. 14-31. 0.98c θ θ Conservation of energy gives 2 pE Eπ = where E energy of each photon (Cons. of p = yp implies that the photons have the same energy). THE SPECIAL THEORY OF RELATIVITY 481 Thus 0 2 pE Eγ = 0 2 135 MeV 339 MeV 2 2 1 0.98 p E E γ = = = − Th e energy of each photon is 339 MeV. Conservation of gives xp mv 2 cos where momentum of each photonp pp pγ θ= = ( ) ( ) ( ) 2 2 135 Mev/c 0.98 c s 0.98 2 1 0.98 339 MeV/c = = − θco 1cos 0.98 11.3θ −= = ° 14-32. From Eq. (14.67) we have 2 2 20E E p c− = 2 2 With , this reduces to 0E E T= + 2 202E T T p c+ = Using the quadratic formula (taking the + root since T ≥ 0) gives 2 2 20 0T E p c E= + − Substituting pc = 1000 MeV ( )0 electron 0.5 MeVE = ( )0 proton 938 MeVE = gives electron proton 999.5 MeV 433 MeV T T = = 482 CHAPTER 14 14-33. 120˚ 120˚ afterbefore p e n ν Conservation of yp gives sin 60 sin 60 or e ep p p pν ν° = ° = Conservation of gives xp cos 60 cos 60p e ep p pν p= ° + ° = So e pp p pν p= = ≡ Conservation of energy gives 0n e pE E E Eν= + + 2 2 2 2 2 20 0 0n e pE E p c E p c= + + + + pc (1) Substituting 0 939.6 MeVnE = 0 938.3 MeVpE = 0 0.5 MeVeE = and solving for pc gives p = 0.554 MeV/c 0.554 MeV/cp ep p pν= = = Substituting into 0 2 2 2 0 0 T E E E p c E = − = + − gives ( )0 0E ν = 4 0.554 MeV 2 10 MeV, or 200 eV 0.25 MeV p e T T T ν − = = × = THE SPECIAL THEORY OF RELATIVITY 485 ( ) 2 3 22 0 1 mc d kx dx β β β + = − (7) This is easily integrated to give 2 2 2 1 21 mc kx E β + = − (8) where E is the constant of integration. The value of E is evaluated for some particular point in phase space, the easiest being x = a; β = 0: 2 1 2 E mc ka= + 2 (9) From (8) and (9), 2 2 2 2 1 1 2 21 mc kx mc ka β + = + − 2 (10) Eliminating 2β from (10), we have ( ) ( ) ( ) ( ) 2 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 4 2 m c mc k a x k mc a x k a x k mc a x = −  + −    + −  = −  + −   β (11) and, therefore, ( ) ( ) ( ) 2 2 2 2 2 2 2 2 41 2 k a x mc k a xdx c dt mc k a x − + − + − β = = (12) The period will then be four times the integral of dt = dt(x) from x = 0 to x = a: ( ) ( ) 2 2 2 2 2 2 20 2 1 24 1 4 a k a x m mc dx k k a x a x mc  + −  − + − ∫τ = (13)  Since x varies between 0 and a, the variable x a takes on values in the interval 0 to 1, and therefore, we can define sin x a φ = (14) from which 486 CHAPTER 14 2 2 cos a x a φ − = (15) and 2 2dx a x dφ= − (16) We also define the dimensionless parameter, 22 a k mc ≡κ (17) Using (14) – (17), (13) transforms into ( )2 22 2 2 0 1 2 cos2 1 cos a d c + = + ∫ π κ φ τ φ κ κ φ (18) Since 2 2 1ka mc for the weakly relativistic case, we can expand the integrand of (18) in a series of powers of κ : ( ) ( ) ( ) 2 2 2 2 2 2 1 22 2 2 2 2 2 1 2 cos 1 2 cos 1 cos 21 cos 1 1 2 cos 2 3 1 cos 2 +   ≅ + −  +  ≅ + −   = + κ φ κκ φ φ κ φ κ φ κ φ (19) Substitution of (19) into (18) yields 2 2 2 0 2 0 2 3 1 cos 2 3 1 sin 2 2 2 a d c a a c c  ≅ +    = + +   ∫ π π τ κ φ κ π κ φ φ κ φ (20) Evaluating (20) and substituting the expression for κ from (17), we obtain 2 2 3 2 8 m a k c π τ π= + k m (21) or, 2 0 2 3 1 16 ka mc τ τ   = +    (22) THE SPECIAL THEORY OF RELATIVITY 487 14-38. ( ) ( ) (for constant) dp d F mu dt dt d m u m dt γ γ = = = = 2 21 d u m dt u c    =  −  ( ) ( ) ( ) 1 2 1 22 2 2 2 2 2 2 1 1 1 u u c u u c ducm dtu c −  − − − −     −     = ( ) 3 22 21 dum u c dt − = − Thus ( ) 3 22 21duF m u c dt − = − 14-39. The kinetic energy is 2 2 2 4 20 0T p c m c m c= + − (1) For a momentum of 100 MeV/c, ( )24proton 10 931 931 936 931 5 MeV= + − ≅ − =T (2) ( )24electron 10 0.51 0.51 100 0.5 99.5 MeV= + − ≅ − =T (3) In order to obtain γ and β, we use the relation 2 2 2 0 0 21 m c E mc m cγ β = = = − (4) so that 2 0 E m c γ = (5) and 2 1 1β γ = − (6) electron 100 200 0.51 γ = ≅ (7) 490 CHAPTER 14 ( ) ( ) ( ) ( ) 2 2 2 2 2 1 1 2 2 2 2 1 1 2 2 1 1 2 1 1 2 cos 1 sin 2 cos 1 cos sin sin cos cos cos 1 cos cos 1 1 cos + + = + − + = + + − + = + − + = + − − ψ γ ψ ψ γ ψ ψ γ ψ ψ γ ψ γ γ ψ ψ γ γ ψ ψ (11) provides us with the desired result: ( ) ( ) 2 1 2 0 1 1 2 cos 1 1 cos T T ψ γ γ ψ = + − − (12) Notice that the shape of the curve changes when T m , i.e., when 21 0c> 1 2γ > . 0.2 0 0˚ 30˚ 60˚ 90˚ 0.4 0.6 0.8 1.0 T T 1 0 T1 = 0.1 GeV T1 = 1 GeV T1 = 10 GeV ψ 14-42. φ θ hν hν′ γm e c2 y x From conservation of energy, we have 2 2e eh m c m c hν γ ν+ = + ′ (1) Momentum conservation along the x axis gives cos cose h h m v c c ν ν θ γ′= + φ (2) Momentum conservation along the y axis gives sin sine h m v c ν γ φ ′= θ (3) THE SPECIAL THEORY OF RELATIVITY 491 In order to eliminate φ, we use (2) and (3) to obtain 1 cos cos sin sin e e h h m v c c h m v ν ν φ θ γ ν φ θ γ ′  = −      ′ =  (4) Then, 2 2 2 2 2 2 2 1 s sin 1 2 cos e h h h h m v c c c c ν ν ν ν co φ φ θ γ  ′ ′       + = = + − +                 (5) Since 2 2 1 1 v c γ = − and 2 1 c v γ γ = − we have ( )2 2 2 2 1v cγ γ= − (6) Substituting γ from (1) into (6), we have ( ) ( 2 22 2 2 2 2 e e h h v m m c = − + − )′ ′γ ν ν ν ν (7) From (5) and (7), we can find the equation for ν′: ( ) ( 2 2 2 2 22 cos 2 e h h h h h hm c c c c c ν ν ν ν )θ ν ν ν ν′ ′     + − = − + −  ′ ′              (8) or, ( ) 2 22 2 1 cose m c m c h h 2 eν θ ν ν   + − =′    (9) Then, ( )2 1 1 1 cos e h m c ν νν θ      =′  + −    (10) or, ( ) 1 21 1 cos e E E E m c θ −   = + −′     (11) The kinetic energy of the electron is 492 CHAPTER 14 ( ) 2 2 2 1 1 1 1 cos e e e c m c h h E E m c γ ν νT m θ  = − = − = −′   + −    ( ) 2 2 2 1 cos 1 1 cose e E T Em c m c θ θ − =   + −    (12)