37 RELATIVITY 37.1 IDENTIFY and SET UP: Consider the distance A to O′ and B to O′ as observed by an observer on the ground (Figure 37.1) Figure 37.1 EXECUTE: Simultaneous to observer on train means light pulses from A′ and B′ arrive at O′ at the same time To observer at O light from A′ has a longer distance to travel than light from B′ so O will conclude that the pulse from A( A′) started before the pulse at B ( B′) To observer at O bolt A appeared to strike 37.2 first EVALUATE: Section 37.2 shows that if they are simultaneous to the observer on the ground then an observer on the train measures that the bolt at B′ struck first IDENTIFY: Apply Eq (37.8) SET UP: The lifetime measured in the muon frame is the proper time Δt0 u = 0.900c is the speed of the muon frame relative to the laboratory frame The distance the particle travels in the lab frame is its speed in that frame times its lifetime in that frame EXECUTE: (a) γ = = 2.29 Δt = γ Δt0 = (2.29) (2.20 × 10−6 s) = 5.05 × 10−6 s − (0.9) (b) d = vΔt = (0.900)(3.00 × 108 m/s)(5.05 × 10−6 s) = 1.36 × 103 m = 1.36 km 37.3 EVALUATE: The lifetime measured in the lab frame is larger than the lifetime measured in the muon frame IDENTIFY and SET UP: The problem asks for u such that Δt0 /Δt = EXECUTE: Δt = Δt0 − u /c 2 ⎛1⎞ gives u = c − (Δt0 /Δt )2 = (3.00 × 108 m/s) − ⎜ ⎟ = 2.60 × 108 m/s; ⎝2⎠ u = 0.867 c Jet planes fly at less than ten times the speed of sound, less than about 3000 m/s Jet planes fly at much 37.4 lower speeds than we calculated for u IDENTIFY: Time dilation occurs because the rocket is moving relative to Mars SET UP: The time dilation equation is Δt = γΔt0 , where t0 is the proper time EXECUTE: (a) The two time measurements are made at the same place on Mars by an observer at rest there, so the observer on Mars measures the proper time © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 37-1 37-2 Chapter 37 (b) Δt = γΔt0 = 37.5 (75.0 μ s) = 435 μs − (0.985) EVALUATE: The pulse lasts for a shorter time relative to the rocket than it does relative to the Mars observer (a) IDENTIFY and SET UP: Δt0 = 2.60 × 10−8 s; Δt = 4.20 × 10−7 s In the lab frame the pion is created and decays at different points, so this time is not the proper time EXECUTE: Δt = Δt0 − u /c says − u2 ⎛ Δt ⎞ =⎜ ⎟ c ⎝ Δt ⎠ 2 ⎛ 2.60 × 10−8 s ⎞ u ⎛ Δt ⎞ = 1− ⎜ ⎟ = 1− ⎜ = 0.998; u = 0.998c −7 ⎟ ⎜ ⎟ c ⎝ Δt ⎠ ⎝ 4.20 × 10 s ⎠ EVALUATE: u < c, as it must be, but u/c is close to unity and the time dilation effects are large (b) IDENTIFY and SET UP: The speed in the laboratory frame is u = 0.998c; the time measured in this frame is Δt , so the distance as measured in this frame is d = uΔt EXECUTE: d = (0.998)(2.998 × 108 m/s)(4.20 × 10−7 s) = 126 m 37.6 EVALUATE: The distance measured in the pion’s frame will be different because the time measured in the pion’s frame is different (shorter) IDENTIFY: Apply Eq (37.8) SET UP: For part (a) the proper time is measured by the race pilot γ = 1.667 EXECUTE: (a) Δt = 1.20 × 108 m (0.800)(3.00 × 108 m/s) = 0.500 s Δt0 = Δt γ = 0.500 s = 0.300 s 1.667 (b) (0.300 s)(0.800c) = 7.20 × 107 m 37.7 37.8 37.9 1.20 × 108 m = 0.500 s (0.800)(3 × 108 m/s) EVALUATE: The two events are the spaceracer passing you and the spaceracer reaching a point 1.20 × 108 m from you The timer traveling with the spaceracer measures the proper time between these two events IDENTIFY and SET UP: A clock moving with respect to an observer appears to run more slowly than a clock at rest in the observer’s frame The clock in the spacecraft measurers the proper time Δt0 Δt = 365 days = 8760 hours EXECUTE: The clock on the moving spacecraft runs slow and shows the smaller elapsed time (c) You read Δt0 = Δt − u /c = (8760 h) − (4.80 × 106 /3.00 × 108 )2 = 8758.88 h The difference in elapsed times is 8760 h − 8758.88 h = 1.12 h IDENTIFY and SET UP: The proper time is measured in the frame where the two events occur at the same point EXECUTE: (a) The time of 12.0 ms measured by the first officer on the craft is the proper time Δt0 (b) Δt = gives u = c − (Δt0 /Δt )2 = c − (12.0 × 10−3 /0.190)2 = 0.998c 2 − u /c EVALUATE: The observer at rest with respect to the searchlight measures a much shorter duration for the event IDENTIFY and SET UP: l = l0 − u /c The length measured when the spacecraft is moving is l = 74.0 m; l0 is the length measured in a frame at rest relative to the spacecraft EXECUTE: l0 = l − u /c = 74.0 m − (0.600c/c)2 = 92.5 m © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Relativity 37-3 EVALUATE: l0 > l The moving spacecraft appears to an observer on the planet to be shortened along the 37.10 direction of motion IDENTIFY and SET UP: When the meterstick is at rest with respect to you, you measure its length to be 1.000 m, and that is its proper length, l0 l = 0.3048 m EXECUTE: l = l0 − u /c gives u = c − (l/l0 ) = c − (0.3048/1.00)2 = 0.9524c = 2.86 × 108 m/s 37.11 IDENTIFY and SET UP: The 2.2 μs lifetime is Δt0 and the observer on earth measures Δt The atmosphere is moving relative to the muon so in its frame the height of the atmosphere is l and l0 is 10 km EXECUTE: (a) The greatest speed the muon can have is c, so the greatest distance it can travel in 2.2 × 10−6 s is d = vt = (3.00 × 108 m/s)(2.2 × 10−6 s) = 660 m = 0.66 km Δt0 (b) Δt = − u /c = 2.2 × 10−6 s − (0.999) = 4.9 × 10−5 s d = vt = (0.999)(3.00 × 108 m/s)(4.9 × 10−5 s) = 15 km In the frame of the earth the muon can travel 15 km in the atmosphere during its lifetime (c) l = l0 − u /c = (10 km) − (0.999)2 = 0.45 km 37.12 In the frame of the muon the height of the atmosphere is less than the distance it moves during its lifetime IDENTIFY and SET UP: The scientist at rest on the earth’s surface measures the proper length of the separation between the point where the particle is created and the surface of the earth, so l0 = 45.0 km The transit time measured in the particle’s frame is the proper time, Δt0 EXECUTE: (a) t = l0 45.0 × 103 m = = 1.51 × 10−4 s v (0.99540)(3.00 × 108 m/s) (b) l = l0 − u /c = (45.0 km) − (0.99540)2 = 4.31 km (c) time dilation formula: Δt0 = Δt − u /c = (1.51 × 10−4 s) − (0.99540) = 1.44 × 10−5 s l 4.31 × 103 m = = 1.44 × 10−5 s v (0.99540)(3.00 × 108 m/s) The two results agree IDENTIFY: Apply Eq (37.16) SET UP: The proper length l0 of the runway is its length measured in the earth’s frame The proper time from Δl : t = 37.13 Δt0 for the time interval for the spacecraft to travel from one end of the runway to the other is the time interval measured in the frame of the spacecraft EXECUTE: (a) l0 = 3600 m l = l0 − (b) Δt = u2 c2 (4.00 × 107 m/s) (3.00 × 108 m/s) = (3600 m)(0.991) = 3568 m l0 3600 m = = 9.00 × 10−5 s u 4.00 × 107 m/s (c) Δt0 = 3568 m l = = 8.92 × 10−5 s u 4.00 × 107 m/s 8.92 × 10−5 s = 9.00 × 10−5 s The result from length 0.991 γ contraction is consistent with the result from time dilation IDENTIFY: The astronaut lies along the motion of the rocket, so his height will be Lorentz-contracted SET UP:The doctor in the rocket measures his proper length l0 EVALUATE: 37.14 = (3600 m) − = 0.991, so Eq (37.8) gives Δt = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 37-4 Chapter 37 EXECUTE: (a) l0 = 2.00 m l = l0 − u /c = (2.00 m) − (0.850) = 1.05 m The person on earth would measure his height to be 1.05 m l 2.00 m (b) l = 2.00 m l0 = = = 3.80 m This is not a reasonable height for a human 2 − u /c − (0.850) 37.15 (c) There is no length contraction in a direction perpendicular to the motion and both observers measure the same height, 2.00 m EVALUATE: The length of an object moving with respect to the observer is shortened in the direction of the motion, so in (a) and (b) the observer on earth measures a shorter height IDENTIFY: Apply Eq (37.23) G G SET UP: The velocities v′ and v are both in the + x-direction, so v′x = v′ and vx = v v′ + u 0.400c + 0.600c = = 0.806c + (0.400)(0.600) + uv′/c v′ + u 0.900c + 0.600c = = 0.974c (b) v = + (0.900)(0.600) + uv′/c v′ + u 0.990c + 0.600c (c) v = = = 0.997c + (0.990)(0.600) + uv′/c EVALUATE: Speed v is always less than c, even when v′ + u is greater than c IDENTIFY: Apply Eq (37.6) and the equations for x and t that are developed in Example 37.6 SET UP: S is Stanley’s frame and S ′ is Mavis’s frame The proper time for the two events is the time interval measured in Mavis’s frame γ = 1.667 (γ = 5/3 if u = (4/5)c) EXECUTE: (a) In Mavis’s frame the event “light on” has space-time coordinates x′ = and t ′ = 5.00 s, so from the result of Example 37.6, x = γ ( x′ + ut ′) and EXECUTE: (a) v = 37.16 ux′ ⎞ ⎛ t = γ ⎜ t ′ + ⎟ ⇒ x = γ ut ′ = 2.00 × 109 m, t = γ t ′ = 8.33 s c ⎠ ⎝ (b) The 5.00-s interval in Mavis’s frame is the proper time Δt0 in Eq (37.6), so Δt = γΔt0 = 8.33 s, the same as in part (a) (c) (8.33 s)(0.800c) = 2.00 × 109 m, which is the distance x found in part (a) EVALUATE: Mavis would measure that she would be a distance (5.00 s)(0.800c) = 1.20 × 109 m from Stanley when she turns on her light In Eq (37.16), l0 = 2.00 × 109 m and l = 1.20 × 109 m 37.17 IDENTIFY: The relativistic velocity addition formulas apply since the speeds are close to that of light v −u SET UP: The relativistic velocity addition formula is v′x = x uv − 2x c EXECUTE: (a) For the pursuit ship to catch the cruiser, the distance between them must be decreasing, so the velocity of the cruiser relative to the pursuit ship must be directed toward the pursuit ship (b) Let the unprimed frame be Tatooine and let the primed frame be the pursuit ship We want the velocity v′ of the cruiser knowing the velocity of the primed frame u and the velocity of the cruiser v in the unprimed frame (Tatooine) vx − u 0.600c − 0.800c = = −0.385c uvx − (0.600)(0.800) 1− c The result implies that the cruiser is moving toward the pursuit ship at 0.385c EVALUATE: The nonrelativistic formula would have given −0.200c, which is considerably different from the correct result v′x = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Relativity 37.18 37-5 IDENTIFY: The observer on the spaceship measures the speed of the missile relative to the ship, and the earth observer measures the speed of the rocketship relative to earth v′ + u SET UP: u = 0.600c v′x = −0.800c vx = ? vx = x + uv′x /c v′x + u −0.800c + 0.600c −0.200c = = = −0.385c The speed of the missile in the 0.520 + uv′x /c + (0.600)(−0.800) earth frame is 0.385c EVALUATE: The observers on earth and in the spaceship measure different speeds for the missile because they are moving relative to each other IDENTIFY and SET UP: Reference frames S and S ′ are shown in Figure 37.19 EXECUTE: vx = 37.19 Frame S is at rest in the laboratory Frame S ′ is attached to particle Figure 37.19 u is the speed of S ′ relative to S; this is the speed of particle as measured in the laboratory Thus u = +0.650c The speed of particle in S ′ is 0.950c Also, since the two particles move in opposite directions, moves in the − x′-direction and v′x = −0.950c We want to calculate vx , the speed of particle in frame S; use Eq (37.23) v′ + u −0.950c + 0.650c −0.300c EXECUTE: vx = x = = = −0.784c The speed of the second 2 − 0.6175 + uv′x /c + (0.950c)(−0.650c)/c 37.20 particle, as measured in the laboratory, is 0.784c EVALUATE: The incorrect Galilean expression for the relative velocity gives that the speed of the second particle in the lab frame is 0.300c The correct relativistic calculation gives a result more than twice this IDENTIFY and SET UP: Let S be the laboratory frame and let S ′ be the frame of one of the particles, as shown in Figure 37.20 Let the positive x-direction for both frames be from particle to particle In the lab frame particle is moving in the + x -direction and particle is moving in the − x -direction Then u = 0.9520c and vx = −0.9520c v′x is the velocity of particle relative to particle vx − u −0.9520c − 0.9520c = = −0.9988c The speed of particle relative to − uvx /c − (0.9520c)(−0.9520c)/c particle is 0.9988c v′x < shows particle is moving toward particle EXECUTE: v′x = Figure 37.20 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 37-6 Chapter 37 37.21 IDENTIFY: The relativistic velocity addition formulas apply since the speeds are close to that of light v −u SET UP: The relativistic velocity addition formula is v′x = x uv − 2x c EXECUTE: In the relativistic velocity addition formula for this case, v′x is the relative speed of particle 37.22 37.23 with respect to particle 2, v is the speed of particle measured in the laboratory, and u is the speed of particle measured in the laboratory, u = −v v′ v − ( −v ) 2v v′x = = x v − 2v + v′x = and (0.890c )v − 2c 2v + (0.890c3 ) = − ( −v)v/c + v /c c This is a quadratic equation with solution v = 0.611c (v must be less than c ) EVALUATE: The nonrelativistic result would be 0.445c, which is considerably different from this result IDENTIFY and SET UP: Let the starfighter’s frame be S and let the enemy spaceship’s frame be S ′ Let the positive x-direction for both frames be from the enemy spaceship toward the starfighter Then u = +0.400c v′ = +0.700c v is the velocity of the missile relative to you v′ + u 0.700c + 0.400c = = 0.859c EXECUTE: (a) v = + (0.400)(0.700) + uv′/c (b) Use the distance it moves as measured in your frame and the speed it has in your frame to calculate the 8.00 × 109 m = 31.0 s time it takes in your frame t = (0.859)(3.00 × 108 m/s) IDENTIFY and SET UP: The reference frames are shown in Figure 37.23 S = Arrakis frame S ′ = spaceship frame The object is the rocket Figure 37.23 u is the velocity of the spaceship relative to Arrakis vx = +0.360c; v′x = +0.920c (In each frame the rocket is moving in the positive coordinate direction.) v −u Use the Lorentz velocity transformation equation, Eq (37.22): v′x = x − uvx /c EXECUTE: v′x = u= ⎛ v v′ ⎞ ⎛ v v′ ⎞ so v′x − u ⎜ x x ⎟ = vx − u and u ⎜ − x x ⎟ = vx − v′x c ⎠ − uvx /c ⎝ c ⎠ ⎝ vx − u vx − v′x 0.360c − 0.920c 0.560c = =− = −0.837c 0.6688 − vx v′x /c − (0.360c )(0.920c )/c The speed of the spacecraft relative to Arrakis is 0.837c = 2.51 × 108 m/s The minus sign in our result for 37.24 u means that the spacecraft is moving in the –x-direction, so it is moving away from Arrakis EVALUATE: The incorrect Galilean expression also says that the spacecraft is moving away from Arrakis, but with speed 0.920c − 0.360c = 0.560c IDENTIFY: There is a Doppler effect in the frequency of the radiation due to the motion of the star c −u f0 SET UP: The star is moving away from the earth, so f = c+u EXECUTE: f = c − 0.600c f = 0.500 f = (0.500)(8.64 × 1014 Hz) = 4.32 × 1014 Hz c + 0.600c © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Relativity 37.25 37-7 EVALUATE: The earth observer measures a lower frequency than the star emits because the star is moving away from the earth c+u IDENTIFY and SET UP: Source and observer are approaching, so use Eq (37.25): f = f Solve c −u for u, the speed of the light source relative to the observer ⎛c+u⎞ (a) EXECUTE: f = ⎜ ⎟ f0 ⎝ c −u ⎠ (c − u ) f = (c + u ) f 02 and u = c( f − f 02 ) f + f 02 ⎛ ( f /f ) − ⎞ = c⎜ ⎜ ( f/f ) + ⎟⎟ ⎝ ⎠ λ0 = 675 nm, λ = 575 nm ⎛ (675 nm/575 nm) − ⎞ u =⎜ c = 0.159c = (0.159)(2.998 × 108 m/s) = 4.77 × 107 m/s; definitely speeding ⎜ (675 nm/575 nm) + ⎟⎟ ⎝ ⎠ (b) 4.77 × 107 m/s = (4.77 × 107 m/s)(1 km/1000 m)(3600 s/1 h) = 1.72 × 108 km/h Your fine would be 37.26 37.27 $1.72 × 108 (172 million dollars) EVALUATE: The source and observer are approaching, so f > f and λ < λ0 Our result gives u < c, as it must IDENTIFY: There is a Doppler effect in the frequency of the radiation due to the motion of the source c+u SET UP: f > f so the source is moving toward you f = f0 c−u c+u EXECUTE: ( f/f ) = c ( f / f ) − ( f/ f ) u = c + u c −u c[( f/f ) − 1] ⎡ (1.25) − ⎤ u= =⎢ ⎥ c = 0.220c, toward you ( f/f )2 + ⎣⎢ (1.25) + ⎦⎥ EVALUATE: The difference in frequency is rather large (1.25 times), so the motion of the source must be a substantial fraction of the speed of light (around 20% in this case) IDENTIFY: The speed of the proton is a substantial fraction of the speed of light, so we must use the relativistic formula for momentum p γv SET UP: p = γ mv p0 = γ mv0 = v/v0 = 2.00 p0 γ 0v0 EXECUTE: γ = 37.28 1 − v02 /c = 1 − (0.400) = 1.0911 γ = 1 − (0.800)2 = 1.667 ⎛ 1.667 ⎞ p = p0 (2) ⎜ ⎟ = 3.06 p0 ⎝ 1.091 ⎠ EVALUATE: The speed doubles but the momentum more than triples IDENTIFY and SET UP: γ = If γ is 1.0% greater than then γ = 1.010, if γ is 10% greater − v /c than then γ = 1.10 and if γ is 100% greater than then γ = 2.00 EXECUTE: v = c − 1/γ (a) v = c − 1/(1.010)2 = 0.140c (b) v = c − 1/(1.10) = 0.417c (c) v = c − 1/(2.00) = 0.866c © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 37-8 Chapter 37 37.29 IDENTIFY: Apply Eqs (37.27) and (37.32) SET UP: For a particle at rest (or with v c ), a = F/m mv EXECUTE: (a) p = = 2mv − v /c v2 3 = − ⇒ v2 = c2 ⇒ v = c = 0.866c 4 c v (b) F = γ 3ma = 2ma ⇒ γ = ⇒ γ = (2)1/3 so = 22/3 ⇒ = − 2−2/3 = 0.608 2 c − v /c EVALUATE: The momentum of a particle and the force required to give it a given acceleration both increase without bound as the speed of the particle approaches c IDENTIFY: The speed of the proton is a substantial fraction of the speed of light, so we must use the relativistic form of Newton’s second law G G ma SET UP: F and v are along the same line, so F = (1 − v /c )3/2 ⇒ = − v /c ⇒ 37.30 ma EXECUTE: (a) F = (1 − v /c )3/2 = (1.67 × 10−27 kg)(2.30 × 108 m/s ) [1 − (2.30 × 108 / 3.00 × 108 ) ]3/2 = 1.45 × 10−18 N; − x-direction F 1.45 × 10−18 N = = 8.69 × 108 m/s m 1.67 × 10−27 kg EVALUATE: The acceleration in part (b) is much greater than the acceleration given in the problem because the proton starting at rest is not relativistic IDENTIFY: When the speed of the electron is close to the speed of light, we must use the relativistic form of Newton’s second law ma SET UP: When the force and velocity are parallel, as in part (b), F = In part (a), v c (1 − v /c )3/2 so F = ma (b) a = 37.31 EXECUTE: (a) a = (b) γ = (1 − v /c )1/2 F = (1 − [2.50 × 108 /3.00 × 108 ]2 )1/2 = 1.81 5.49 × 1015 m/s = 9.26 × 1014 m/s (1.81)3 EVALUATE: The acceleration for low speeds is over times greater than it is near the speed of light as in part (b) IDENTIFY and SET UP: The force is found from Eq (37.32) or Eq (37.33) EXECUTE: (a) Indistinguishable from F = ma = 0.145 N a= 37.32 F 5.00 × 10−15 N = = 5.49 × 1015 m/s m 9.11× 10−31 kg mγ = (b) γ 3ma = 1.75 N (c) γ 3ma = 51.7 N (d) γ ma = 0.145 N, 0.333 N, 1.03 N 37.33 EVALUATE: When v is large, much more force is required to produce a given magnitude of acceleration when the force is parallel to the velocity than when the force is perpendicular to the velocity IDENTIFY: Apply Eq (37.36) SET UP: The rest energy is mc EXECUTE: (a) K = mc 2 − v /c ⇒ − mc = mc 1 − v /c =2⇒ v2 c = 0.866c =1− ⇒ v = 4 c © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Relativity (b) K = 5mc ⇒ 37.34 =6⇒ 37-9 35 v2 =1− ⇒ v = c = 0.986c 36 36 c − v /c EVALUATE: If v c, then K is much less than the rest energy of the particle IDENTIFY: At such a high speed, we must use the relativistic formulas for momentum and kinetic energy SET UP: mμ = 207 me = 1.89 × 10−28 kg v is very close to c and we must use relativistic expressions p= mv − v /c EXECUTE: Using K = mc , K= p= mc − v /c mv − v /c − v /c 2 = − mc (1.89 × 10−28 kg)(0.999)(3.00 × 108 m/s) − (0.999) = 1.27 × 10−18 kg ⋅ m/s − mc gives ⎛ ⎞ K = (1.89 × 10−28 kg)(3.00 × 108 m/s)2 ⎜ − 1⎟ = 3.63 × 10−10 J ⎜ − (0.999) ⎟ ⎝ ⎠ EVALUATE: The nonrelativistic values are pnr = mv = 5.66 × 10−20 kg ⋅ m/s and K nr = 12 mv = 8.49 × 10−12 J Each relativistic result is much larger 37.35 IDENTIFY and SET UP: Use Eqs (37.38) and (37.39) EXECUTE: (a) E = mc + K , so E = 4.00mc means K = 3.00mc = 4.50 × 10−10 J (b) E = ( mc ) + ( pc) ; E = 4.00mc , so 15.0( mc ) = ( pc )2 p = 15mc = 1.94 × 10−18 kg ⋅ m/s (c) E = mc / − v /c E = 4.00mc gives − v /c = 1/16 and v = 15/16c = 0.968c 37.36 EVALUATE: The speed is close to c since the kinetic energy is greater than the rest energy Nonrelativistic expressions relating E, K, p and v will be very inaccurate IDENTIFY: Apply the work energy theorem in the form W = ΔK SET UP: K is given by Eq (37.36) When v = 0, γ = EXECUTE: (a) W = ΔK = (γ f − 1)mc = (4.07 × 10−3 ) mc (b) (γ f − γ i ) mc = 4.79mc (c) The result of part (b) is far larger than that of part (a) EVALUATE: The amount of work required to produce a given increase in speed (in this case an increase of 0.090c) increases as the initial speed increases 37.37 IDENTIFY: Use E = mc to relate the mass increase to the energy increase (a) SET UP: Your total energy E increases because your gravitational potential energy mgy increases EXECUTE: ΔE = mg Δy ΔE = (Δm)c so Δm = ΔE/c = mg ( Δy )/c Δm/m = ( g Δy )/c = (9.80 m/s )(30 m)/(2.998 × 108 m/s) = 3.3 × 10−13% This increase is much, much too small to be noticed (b) SET UP: The energy increases because potential energy is stored in the compressed spring EXECUTE: ΔE = ΔU = 12 kx = 12 (2.00 × 104 N/m)(0.060 m) = 36.0 J Δm = (ΔE )/c = 4.0 × 10−16 kg Energy increases so mass increases The mass increase is much, much too small to be noticed © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 37-10 37.38 Chapter 37 EVALUATE: In both cases the energy increase corresponds to a mass increase But since c is a very large number the mass increase is very small IDENTIFY: Apply Eq (37.38) SET UP: When the person is at rest her total energy is E0 = mc EXECUTE: (a) E = 2mc , so − v /c = v2 v2 = − ⇒ = ⇒ v = c 3/4 = 0.866c = 2.60 × 108 m/s 4 c c (b) E = 10mc 2, so 37.39 = 10 − v2 = v 99 99 ⇒ 2= = 0.995c = 2.98 × 108 m/s .v = c 100 100 100 c c − v /c EVALUATE: Unless v approaches c, the total energy of an object is not much greater than its rest energy IDENTIFY and SET UP: The energy equivalent of mass is E = mc ρ = 7.86 g/cm3 = 7.86 × 103 kg/m3 For a cube, V = L3 EXECUTE: (a) m = E c2 = 1.0 × 1020 J (3.00 × 108 m/s)2 = 1.11 × 103 kg m m 1.11 × 103 kg so V = = = 0.141 m3 L = V 1/ = 0.521 m = 52.1 cm V ρ 7.86 × 103 kg/m3 EVALUATE: Particle/antiparticle annihilation has been observed in the laboratory, but only with small quantities of antimatter IDENTIFY: With such a large potential difference, the electrons will be accelerated to relativistic speeds, so we must use the relativistic formula for kinetic energy ⎛ ⎞ SET UP: K = ⎜ − 1⎟ mc The classical expression for kinetic energy is K = 12 mv ⎜ ⎟ 2 ⎝ − v /c ⎠ (b) ρ = 37.40 EXECUTE: For an electron mc = (9.11× 10−31 kg)(3.00 × 108 m/s) = 8.20 × 10−14 J K = 7.50 × 105 eV = 1.20 × 10−13 J (a) K mc +1 = 1 − v /c 1 − v /c = 1.20 × 10−13 J 8.20 × 10−14 J + = 2.46 v = c − (1/ 2.46)2 = 0.914c = 2.74 × 108 m/s (b) K = 12 mv gives v = 37.41 2K 2(1.20 × 10−13 J) = = 5.13 × 108 m/s m 9.11 × 10−31 kg EVALUATE: At a given speed the relativistic value of the kinetic energy is larger than the nonrelativistic value Therefore, for a given kinetic energy the relativistic expression for kinetic energy gives a smaller speed than the nonrelativistic expression IDENTIFY and SET UP: The total energy is given in terms of the momentum by Eq (37.39) In terms of the total energy E, the kinetic energy K is K = E − mc (from Eq 37.38) The rest energy is mc EXECUTE: (a) E = (mc ) + ( pc) = [(6.64 × 10−27 )(2.998 × 108 ) ]2 + [(2.10 × 10−18 )(2.998 × 108 )]2 J E = 8.67 × 10−10 J (b) mc = (6.64 × 10−27 kg)(2.998 × 108 m/s) = 5.97 × 10−10 J K = E − mc = 8.67 × 10−10 J − 5.97 × 10−10 J = 2.70 × 10−10 J (c) K mc = 2.70 × 10−10 J 5.97 ì 1010 J = 0.452 â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Relativity 37-11 EVALUATE: The incorrect nonrelativistic expressions for K and p give K = p /2m = 3.3 × 10−10 J; 37.42 the correct relativistic value is less than this IDENTIFY: Since the final speed is close to the speed of light, there will be a considerable difference between the relativistic and nonrelativistic results 1 SET UP: The nonrelativistic work-energy theorem is F Δx = mv − mv02 , and the relativistic formula 2 for a constant force is F Δx = (γ − 1)mc EXECUTE: (a) Using the classical work-energy theorem and solving for Δx, we obtain m(v − v02 ) (0.100 × 10−9 kg)[(0.900)(3.00 × 108 m/s)]2 = = 3.65 m 2F 2(1.00 × 106 N) (b) Using the relativistic work-energy theorem for a constant force, we obtain Δx = Δx = For the given speed, γ = Δx = 37.43 (γ − 1)mc F = 2.29, thus 1−0.9002 (2.29 − 1)(0.100 × 10−9 kg)(3.00 × 108 m/s) = 11.6 m (1.00 × 106 N) EVALUATE: (c) The distance obtained from the relativistic treatment is greater As we have seen, more energy is required to accelerate an object to speeds close to c, so that force must act over a greater distance IDENTIFY and SET UP: The nonrelativistic expression is K nonrel = 12 mv and the relativistic expression is K rel = (γ − 1)mc EXECUTE: (a) v = × 107 m/s ⇒ γ = K rel = (γ − 1)mc = 5.65 × 10−12 J 1 = 1.0376 For m = mp , K nonrel = mv = 5.34 × 10−12 J − v /c 2 K rel = 1.06 K nonrel (b) v = 2.85 × 108 m/s; γ = 3.203 K nonrel = mv = 6.78 × 10−11 J; K rel = (γ − 1)mc = 3.31 × 10−10 J; K rel /K nonrel = 4.88 EVALUATE: K rel /K nonrel increases without bound as v approaches c 37.44 IDENTIFY: Since the speeds involved are close to that of light, we must use the relativistic formula for kinetic energy ⎛ ⎞ − 1⎟ mc SET UP: The relativistic kinetic energy is K = (γ − 1)mc = ⎜ ⎜ ⎟ 2 ⎝ − v /c ⎠ EXECUTE: (a) ⎛ ⎞ ⎛ ⎞ 1 K = (γ − 1)mc = ⎜ − 1⎟ mc = (1.67 × 10−27 kg)(3.00 × 108 m/s) ⎜ − 1⎟ ⎜ ⎟ 2 ⎜ − (0.100c/c) ⎟ ⎝ − v /c ⎠ ⎝ ⎠ ⎛ ⎞ − 1⎟ = 7.56 × 10−13 J = 4.73 MeV K = (1.50 × 10 −10 J) ⎜ ⎝ − 0.0100 ⎠ ⎛ ⎞ − 1⎟ = 2.32 × 10−11 J = 145 MeV (b) K = (1.50 × 10−10J) ⎜ ⎜ − (0.500)2 ⎟ ⎝ ⎠ ⎛ ⎞ − 1⎟ = 1.94 × 10−10 J = 1210 MeV (c) K = (1.50 × 10−10 J) ⎜ ⎜ − (0.900)2 ⎟ ⎝ ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 37-12 Chapter 37 (d) ΔE = 2.32 × 10−11 J − 7.56 × 10−13 J = 2.24 × 10−11 J = 140 MeV (e) ΔE = 1.94 × 10−10 J − 2.32 × 10−11 J = 1.71 × 10−10 J = 1070 MeV (f) Without relativity, K = mv The work done in accelerating a proton from 0.100c to 0.500c in the 1 nonrelativistic limit is ΔE = m(0.500c)2 − m(0.100c) = 1.81 × 10−11 J = 113 MeV 2 The work done in accelerating a proton from 0.500c to 0.900c in the nonrelativistic limit is 37.45 1 ΔE = m(0.900c) − m(0.500c) = 4.21 × 10−11 J = 263 MeV 2 EVALUATE: We see in the first case the nonrelativistic result is within 20% of the relativistic result In the second case, the nonrelativistic result is very different from the relativistic result since the velocities are closer to c IDENTIFY and SET UP: Use Eq (23.12) and conservation of energy to relate the potential difference to the kinetic energy gained by the electron Use Eq (37.36) to calculate the kinetic energy from the speed EXECUTE: (a) K = qΔV = eΔV ⎛ ⎞ − 1⎟ = 4.025mc = 3.295 × 10−13 J = 2.06 MeV K = mc ⎜ ⎜ ⎟ 2 ⎝ − v /c ⎠ ΔV = K/e = 2.06 × 106 V (b) From part (a), K = 3.30 × 10−13 J = 2.06 MeV 37.46 EVALUATE: The speed is close to c and the kinetic energy is four times the rest mass IDENTIFY: The total energy is conserved in the collision SET UP: Use Eq (37.38) for the total energy Since all three particles are at rest after the collision, the final total energy is Mc + mc The initial total energy of the two protons is γ Mc m 9.75 EXECUTE: (a) Mc + mc = γ Mc ⇒ γ = + =1+ = 1.292 2M 2(16.7) Note that since γ = v 1 = 0.6331 , we have that = − = − 2 c γ (1.292) − v /c (b) According to Eq (37.36), the kinetic energy of each proton is ⎛ 1.00 MeV ⎞ K = (γ − 1) Mc = (1.292 − 1)(1.67 × 10−27 kg)(3.00 × 108 m/s)2 ⎜⎜ −13 ⎟ ⎟ = 274 MeV ⎝ 1.60 × 10 J ⎠ ⎛ 1.00 MeV ⎞ (c) The rest energy of η is mc = (9.75 × 10−28 kg)(3.00 × 108 m/s) ⎜⎜ −13 ⎟ ⎟ = 548 MeV ⎝ 1.60 × 10 J ⎠ EVALUATE: (d) The kinetic energy lost by the protons is the energy that produces the η 0, 548 MeV = 2(274 MeV) 37.47 IDENTIFY: Use E = mc to relate the mass decrease to the energy produced SET UP: kg is equivalent to 2.2 lbs and ton = 2000 lbs W = J/s EXECUTE: (a) E = mc , m = E/c = (3.8 × 1026 J)/(2.998 × 108 m/s) = 4.2 × 109 kg = 4.6 × 106 tons (b) The current mass of the sun is 1.99 × 1030 kg, so it would take it (1.99 × 1030 kg)/(4.2 × 109 kg/s) = 4.7 × 1020 s = 1.5 × 1013 years to use up all its mass 37.48 EVALUATE: The power output of the sun is very large, but only a small fraction of the sun’s mass is converted to energy each second IDENTIFY and SET UP: The astronaut in the spaceship measures the proper time, since the end of a swing Δt0 occurs at the same location in his frame Δt = − u /c © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Relativity EXECUTE: (a) Δt0 = 1.50 s Δt = Δt0 − u /c = 1.50 s − (0.75c/c) 37-13 = 2.27 s (b) Δt = 1.50 s Δt0 = Δt − u /c = (1.50 s) − (0.75c/c) = 0.992 s 37.49 EVALUATE: The motion of the spaceship makes a considerable difference in the measured values for the period of the pendulum! (a) IDENTIFY and SET UP: Δt0 = 2.60 × 10−8 s is the proper time, measured in the pion’s frame The time measured in the lab must satisfy d = cΔt , where u ≈ c Calculate Δt and then use Eq (37.6) to calculate u EXECUTE: Δt = Δt0 d 1.90 × 103 m Δt so (1 − u /c )1/2 = and = = 6.3376 × 10−6 s Δt = 2 Δt c 2.998 × 108 m/s − u /c ⎛ Δt ⎞ (1 − u /c ) = ⎜ ⎟ Write u = (1 − Δ )c so that (u/c )2 = (1 − Δ) = − 2Δ + Δ ≈ − 2Δ since Δ is small ⎝ Δt ⎠ 2 ⎛ Δt ⎞ ⎛ 2.60 × 10−8 s ⎞ ⎛ Δt ⎞ Using this in the above gives − (1 − 2Δ) = ⎜ ⎟ Δ = ⎜ ⎟ = ⎜ = 8.42 × 10−6 −6 ⎟ ⎜ ⎟ t Δ Δ t × 3376 10 s ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ EVALUATE: An alternative calculation is to say that the length of the tube must contract relative to the moving pion so that the pion travels that length before decaying The contracted length must be ⎛l ⎞ l = cΔt0 = (2.998 × 108 m/s)(2.60 × 10−8 s) = 7.7948 m l = l0 − u /c so − u /c = ⎜ ⎟ Then ⎝ l0 ⎠ 2 1⎛ l ⎞ ⎛ 7.7948 m ⎞ −6 u = (1 − Δ )c gives Δ = ⎜ ⎟ = ⎜ ⎟ = 8.42 × 10 , which checks ⎝ l0 ⎠ ⎜⎝ 1.90 × 103 m ⎟⎠ (b) IDENTIFY and SET UP: E = γ mc Eq (37.38) 1 EXECUTE: γ = = = = 244 2 2Δ − u /c 2(8.42 × 10−6 ) E = (244)(139.6 MeV) = 3.40 × 104 MeV = 34.0 GeV 37.50 EVALUATE: The total energy is 244 times the rest energy IDENTIFY and SET UP: The proper length of a side is l0 = a The side along the direction of motion is shortened to l = l0 − v /c The sides in the two directions perpendicular to the motion are unaffected by the motion and still have a length a 37.51 EXECUTE: V = a 2l = a − v /c IDENTIFY and SET UP: There must be a length contraction such that the length a becomes the same as b; l0 = a, l = b l0 is the distance measured by an observer at rest relative to the spacecraft Use Eq (37.16) and solve for u l b EXECUTE: = − u /c so = − u /c ; a l0 a = 1.40b gives b/ 1.40b = − u /c and thus − u /c = 1/(1.40)2 37.52 u = − 1/(1.40)2 c = 0.700c = 2.10 × 108 m/s EVALUATE: A length on the spacecraft in the direction of the motion is shortened A length perpendicular to the motion is unchanged IDENTIFY and SET UP: The proper time Δt0 is the time that elapses in the frame of the space probe Δt is the time that elapses in the frame of the earth The distance traveled is 42.2 light years, as measured in the earth frame c ⎛ ⎞ EXECUTE: Light travels 42.2 light years in 42.2 y, so Δt = ⎜ ⎟ (42.2 y) = 42.5 y ⎝ 0.9930c ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 37-14 Chapter 37 Δt0 = Δt − u /c = (42.5 y) − (0.9930) = 5.0 y She measures her biological age to be 19 y + 5.0 y = 24.0 y EVALUATE: Her age measured by someone on earth is 19 y + 42.5 y = 61.5 y 37.53 IDENTIFY and SET UP: The total energy E is related to the rest mass mc by E = γ mc EXECUTE: (a) E = γ mc , so γ = 10 = 1 − (v/c) ⇒ v γ −1 v 99 = ⇒ = = 0.995 100 c c γ (b) ( pc) = m 2v 2γ 2c , E = m 2c 4γ ⇒ E − ( pc) E = − (v /c) = 0.01 = 1% EVALUATE: When E mc 2, E → pc 37.54 IDENTIFY and SET UP: The clock on the plane measures the proper time Δt0 Δt = 4.00 h = 4.00 h(3600 s/1 h) = 1.44 × 104 s Δt0 Δt = − u /c EXECUTE: and Δt0 = Δt − u /c u small so c − u /c = (1 − u /c )1/ ≈ − ⎛ u2 ⎞ u2 ; thus t t Δ = Δ ⎜⎜1 − 2⎟ ⎟ c2 ⎝ 2c ⎠ The difference in the clock readings is Δt − Δt0 = ⎞ u2 1⎛ 250 m/s −9 Δt = ⎜⎜ ⎟ (1.44 × 10 s) = 5.01 × 10 s The clock on the plane has the c2 ⎝ 2.998 × 108 m/s ⎟⎠ shorter elapsed time EVALUATE: Δt0 is always less than Δt ; our results agree with this The speed of the plane is much less 37.55 than the speed of light, so the difference in the reading of the two clocks is very small IDENTIFY: Since the speed is very close to the speed of light, we must use the relativistic formula for kinetic energy ⎛ ⎞ − 1⎟ and the relativistic mass SET UP: The relativistic formula for kinetic energy is K = mc ⎜ ⎜ ⎟ 2 ⎝ − v /c ⎠ m is mrel = − v /c EXECUTE: (a) K = × 1012 eV = 1.12 × 10−6 J Using this value in the relativistic kinetic energy formula ⎛ ⎞ − 1⎟ which gives and substituting the mass of the proton for m, we get K = mc ⎜ ⎜ ⎟ 2 ⎝ − v /c ⎠ − v /c = 7.45 × 103 and − v2 c = (7.45 × 10 ) Solving for v gives − since c + v ≈ 2c Substituting v = (1 − Δ)c, we have − v 2 = v2 c = (c + v)(c − v) c = 2(c − v) , c 2(c − v) 2[c − (1 − Δ)c] = = 2Δ Solving for Δ c c c 1 − v /c (7.45 × 103 )2 = = × 10−9 , to one significant digit gives Δ = 2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Relativity (b) Using the relativistic mass formula and the result that 37.56 − v /c 37-15 = 7.45 × 103 , we have ⎛ ⎞ m mrel = = m⎜ ⎟ = (7 × 103 ) m, to one significant digit ⎜ 2 2 ⎟ − v /c ⎝ − v /c ⎠ EVALUATE: At such high speeds, the proton’s mass is over 7000 times as great as its rest mass E ⎛ ⎞ IDENTIFY and SET UP: The energy released is E = ( Δm)c Δm = ⎜ ⎟ (12.0 kg) Pav = t ⎝ 10 ⎠ The change in gravitational potential energy is mg Δy ⎛ ⎞ EXECUTE: (a) E = ( Δm)c = ⎜ ⎟ (12.0 kg)(3.00 × 108 m/s) = 1.08 × 1014 J ⎝ 10 ⎠ (b) Pav = E 1.08 × 1014 J = = 2.70 × 1019 W t 4.00 × 10−6 s E 1.08 × 1014 J = = 1.10 × 1010 kg g Δy (9.80 m/s )(1.00 × 103 m) EVALUATE: The mass decrease is only 1.2 grams, but the energy released is very large c IDENTIFY and SET UP: In crown glass the speed of light is v = Calculate the kinetic energy of an n electron that has this speed 2.998 × 108 m/s = 1.972 × 108 m/s EXECUTE: v = 1.52 (c) E = ΔU = mg Δy m = 37.57 K = mc (γ − 1) mc = (9.109 × 10−31 kg)(2.998 × 108 m/s)2 = 8.187 × 10−14 J(1 eV/1.602 × 10−19 J) = 0.5111 MeV γ= − v /c = − ((1.972 × 10 m/s)/(2.998 × 108 m/s)) = 1.328 K = mc (γ − 1) = (0.5111 MeV)(1.328 − 1) = 0.168 MeV 37.58 EVALUATE: No object can travel faster than the speed of light in vacuum but there is nothing that prohibits an object from traveling faster than the speed of light in some material IDENTIFY: Apply conservation of momentum to the process of emitting a photon SET UP: A photon has zero rest mass and for it E = pc EXECUTE: (a) v = 37.59 p ( E/c) E = = , where the atom and the photon have the same magnitude of m m mc momentum, E/c E c, so E mc (b) v = mc EVALUATE: The rest energy of a hydrogen atom is about 940 MeV and typical energies of photons emitted by atoms are a few eV, so E mc is typical If this is the case, then treating the motion of the atom nonrelativistically is an accurate approximation IDENTIFY and SET UP: Let S be the lab frame and S ′ be the frame of the proton that is moving in the + x-direction, so u = +c/2 The reference frames and moving particles are shown in Figure 37.59 The other proton moves in the − x-direction in the lab frame, so v = −c/ A proton has rest mass mp = 1.67 × 10−27 kg and rest energy mpc = 938 MeV EXECUTE: (a) v′ = v−u − uv/c = −c/2 − c/2 − (c/2)(−c/2)/c The speed of each proton relative to the other is =− 4c c © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 37-16 Chapter 37 (b) In nonrelativistic mechanics the speeds just add and the speed of each relative to the other is c mc (c) K = − mc 2 − v /c (i) Relative to the lab frame each proton has speed v = c/2 The total kinetic energy of each proton is 938 MeV K= − (938 MeV) = 145 MeV ⎛1⎞ 1− ⎜ ⎟ ⎝ 2⎠ (ii) In its rest frame one proton has zero speed and zero kinetic energy and the other has speed c In this 938 MeV frame the kinetic energy of the moving proton is K = − (938 MeV) = 625 MeV ⎛4⎞ 1− ⎜ ⎟ ⎝5⎠ (d) (i) Each proton has speed v = c/2 and kinetic energy mc 938 MeV ⎛1 ⎞ K = mv = ⎜ m ⎟ (c/2) = = = 117 MeV 8 ⎝2 ⎠ (ii) One proton has speed v = and the other has speed c The kinetic energy of the moving proton is 938 MeV K = mc = = 469 MeV 2 EVALUATE: The relativistic expression for K gives a larger value than the nonrelativistic expression The kinetic energy of the system is different in different frames Figure 37.59 37.60 IDENTIFY: The protons are moving at speeds that are comparable to the speed of light, so we must use the relativistic velocity addition formula SET UP: S is lab frame and S ′ is frame of proton moving in + x-direction vx = −0.600c In lab frame each proton has speed α c u = +α c vx = −α c vx = −0.600c + α c v′x + u = = −α c − 0.600α ′ + uvx /c EXECUTE: (1 − 0.600α )(−α ) = −0.600 + α 0.600α − 2α + 0.600 = Quadratic formula gives α = 3.00 or α = 0.333 Can’t have v > c so α = 0.333 Each proton has speed 0.333c in the earth frame EVALUATE: To the earth observer, the protons are separating at 2(0.333c ) = 0.666c, but to the protons they are separating at 0.600c © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Relativity 37.61 37.62 37-17 IDENTIFY and SET UP: Follow the procedure specified in the problem EXECUTE: x′2 = c 2t ′2 ⇒ ( x − ut ) γ = c 2γ (t − ux/c ) ⎛ u⎞ ⇒ x − ut = c(t − ux/c ) ⇒ x ⎜1 + ⎟ = x(u + c) = t (u + c) ⇒ x = ct ⇒ x = c 2t ⎝ c⎠ c EVALUATE: The light pulse has the same speed c in both frames IDENTIFY and SET UP: Let S be the lab frame and let S ′ be the frame of the nucleus Let the + x-direction be the direction the nucleus is moving u = 0.7500c v′ + u 0.9995c + 0.7500c = = 0.999929c EXECUTE: (a) v′ = +0.9995c v = + uv′/c + (0.7500)(0.9995) −0.9995c + 0.7500c = −0.9965c + (0.7500)(−0.9995) (c) emitted in same direction: ⎛ ⎞ ⎛ ⎞ 1 (i) K = ⎜ − 1⎟ mc = (0.511 MeV) ⎜ − 1⎟ = 42.4 MeV ⎜ ⎟ 2 ⎜ − (0.999929) ⎟ ⎝ − v /c ⎠ ⎝ ⎠ (b) v′ = −0.9995c v = ⎛ ⎞ ⎛ ⎞ 1 − 1⎟ mc = (0.511 MeV) ⎜ − 1⎟ = 15.7 MeV (ii) K ′ = ⎜ ⎜ ⎟ 2 ⎜ − (0.9995)2 ⎟ ⎝ − v /c ⎠ ⎝ ⎠ (d) emitted in opposite direction: ⎛ ⎞ ⎛ ⎞ 1 (i) K = ⎜ − 1⎟ mc = (0.511 MeV) ⎜ − 1⎟ = 5.60 MeV ⎜ ⎟ 2 ⎜ ⎟ ⎝ − v /c ⎠ ⎝ − (0.9965) ⎠ 37.63 ⎛ ⎞ ⎛ ⎞ 1 (ii) K ′ = ⎜ − 1⎟ mc = (0.511 MeV) ⎜ − 1⎟ = 15.7 MeV ⎜ ⎟ 2 ⎜ − (0.9995)2 ⎟ ⎝ − v /c ⎠ ⎝ ⎠ IDENTIFY and SET UP: Use Eq (37.30), with a = dv/dt , to obtain an expression for dv/dt Separate the variables v and t and integrate to obtain an expression for v(t ) In this expression, let t → ∞ EXECUTE: a = dv F = (1 − v /c )3/ (One-dimensional motion is assumed, and all the F, v and a refer to dt m x-components.) dv ⎛F⎞ = ⎜ ⎟ dt 2 3/ (1 − v /c ) ⎝m⎠ Integrate from t = 0, when v = 0, to time t, when the velocity is v t⎛ dv v F⎞ ∫ (1 − v 2/c2 )3/ =∫0 ⎜⎝ m ⎟⎠ dt Since F is constant, t⎛ F⎞ Ft ∫ ⎜⎝ m ⎟⎠ dt = m In the velocity integral make the change of variable y = v/c; then dy = dv/c dv v ∫ (1 − v 2/c )3/ v Thus 2 = c∫ = v/c dy v/c (1 − y )3/2 ⎡ ⎤ y v = c⎢ = 1/2 ⎥ − v /c ⎣ (1 − y ) ⎦ Ft m − v /c Solve this equation for v: v2 ⎛ Ft ⎞ ⎛ Ft ⎞ = ⎜ ⎟ and v = ⎜ ⎟ (1 − v /c ) m − v /c ⎝ ⎠ ⎝m⎠ 2 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 37-18 37.64 Chapter 37 ⎛ ⎛ Ft ⎞2 ⎞ ⎛ Ft ⎞ ( Ft/m) Ft ⎟= v ⎜1 + ⎜ so v = =c ⎜ ⎝ mc ⎟⎠ ⎟ ⎜⎝ m ⎟⎠ 2 + ( Ft/mc) m c + F 2t ⎝ ⎠ Ft Ft As t → ∞, → → 1, so v → c 2 2 m c +F t F 2t Ft EVALUATE: Note that is always less than 1, so v < c always and v approaches c only 2 m c + F 2t when t → ∞ IDENTIFY: Apply the Lorentz coordinate transformation SET UP: Let t and t ′ be time intervals between the events as measured in the two frames and let x and x′ be the difference in the positions of the two events as measured in the two frames EXECUTE: Setting x = in Eq (37.21), the first equation becomes x′ = −γ ut and the last, upon multiplication by c, becomes ct ′ = γ ct Squaring and subtracting gives c 2t ′2 − x′2 = γ 2t (c − u ) But γ = c /(c − v ), so γ 2t (c − v ) = c 2t Therefore, c 2t ′2 − x′2 = c 2t and x′ = c t′2 − t = 4.53 × 108 m 37.65 EVALUATE: We did not have to calculate the speed u of frame S ′ relative to frame S (a) IDENTIFY and SET UP: Use the Lorentz coordinate transformation (Eq 37.21) for ( x1, t1 ) and ( x2 , t2 ): x1 − ut1 x1′ = t1′ = − u /c t1 − ux1/c x2 − ut2 , x2′ = , t2′ = − u /c t2 − ux2 /c − u /c − u /c Same point in S ′ implies x1′ = x2′ What then is Δt ′ = t2′ − t1′ ? x1′ = x′2 implies x1 − ut1 = x2 − ut2 EXECUTE: x2 − x1 Δx = t2 − t1 Δt From the time transformation equations, ( Δt − uΔx/c ) Δt ′ = t2′ − t1′ = − u /c Δx gives Using the result that u = Δt Δt ′ = (Δt − (Δx ) /((Δt )c )) − (Δx) /((Δt ) c ) u (t2 − t1) = x2 − x1 and u = Δt ′ = Δt ′ = Δt (Δt ) − ( Δx)2 /c ( Δt )2 − (Δx)2 /c 2 (Δt ) − (Δx) /c ( Δt − ( Δx)2 / ((Δt )c )) = ( Δt )2 − (Δx/c) , as was to be shown This equation doesn’t have a physical solution (because of a negative square root) if (Δx / c) > (Δt )2 or Δx ≥ cΔt (b) IDENTIFY and SET UP: Now require that t2′ = t1′ (the two events are simultaneous in S ′ ) and use the Lorentz coordinate transformation equations EXECUTE: t2′ = t1′ implies t1 − ux1/c = t2 − ux2 /c c Δt ⎛x −x ⎞ ⎛ Δx ⎞ t2 − t1 = ⎜ 2 ⎟ u so Δt = ⎜ ⎟ u and u = Δx ⎝ c ⎠ ⎝c ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Relativity 37-19 From the Lorentz transformation equations, ⎛ ⎞ Δx′ = x2′ − x1′ = ⎜ ⎟ ( Δx − u Δt ) ⎜ 2 ⎟ ⎝ − u /c ⎠ Using the result that u = c 2Δt/Δx gives Δx′ = (Δx − c (Δt ) /Δx ) 2 − c ( Δt ) /( Δx) Δx ′ = Δx ′ = Δx 2 (Δx) − c (Δt ) (Δx) − c (Δt ) ( Δx)2 − c ( Δt )2 (Δx − c (Δt ) /Δx) = ( Δx) − c (Δt ) (c) IDENTIFY and SET UP: The result from part (b) is Δx′ = (Δx) − c ( Δt )2 Solve for Δt: (Δx′) = (Δx) − c (Δt ) (5.00 m) − (2.50 m) ( Δx) − (Δx′) = = 1.44 × 10−8 s c 2.998 × 108 m/s EVALUATE: This provides another illustration of the concept of simultaneity (Section 37.2): events observed to be simultaneous in one frame are not simultaneous in another frame that is moving relative to the first IDENTIFY: Apply the relativistic expressions for kinetic energy, velocity transformation, length contraction and time dilation SET UP: In part (c) let S ′ be the earth frame and let S be the frame of the ball Let the direction from EXECUTE: Δt = 37.66 Einstein to Lorentz be positive, so u = −1.80 × 108 m/s In part (d) the proper length is l0 = 20.0 m and in part (f) the proper time is measured by the rabbit EXECUTE: (a) 80.0 m/s is nonrelativistic, and K = mv = 186 J (b) K = (γ − 1)mc = 1.31 × 1015 J (c) In Eq (37.23), v′ = 2.20 × 108 m/s, u = −1.80 × 108 m/s, and so v = 7.14 × 107 m/s (d) l = (e) l0 γ = 20.0 m γ 20.0 m 2.20 × 108 m/s (f) Δt0 = Δt γ = 13.6 m = 9.09 × 10−8 s = 6.18 × 10−8 s 37.67 13.6 m = 6.18 × 10 −8 s 2.20 × 108 m/s IDENTIFY and SET UP: An increase in wavelength corresponds to a decrease in frequency ( f = c/λ ), so EVALUATE: In part (f) we could also calculate Δt0 as Δt0 = the atoms are moving away from the earth Receding, so use Eq (37.26): f = c −u f0 c+u ⎛ − ( f/f )2 ⎞ EXECUTE: Solve for u: ( f/f ) (c + u ) = c − u and u = c ⎜ ⎜ + ( f/f )2 ⎟⎟ ⎠ ⎝ f = c/λ , f = c/λ0 so f/f = λ0 /λ ⎛ − (λ0 /λ )2 ⎞ ⎛ − (656.3/953.4) ⎞ = = 0.357c = 1.07 × 108 m/s u = c⎜ c ⎟ ⎜⎜ 2⎟ ⎜ + (λ /λ ) ⎟ ⎟ ⎝ ⎠ ⎝ + (656.3/953.4) ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 37-20 37.68 Chapter 37 EVALUATE: The relative speed is large, 36% of c The cosmological implication of such observations will be discussed in Chapter 44 IDENTIFY: The baseball is moving toward the radar gun, so apply the Doppler effect as expressed in Eq (37.25) SET UP: The baseball had better be moving nonrelativistically, so the Doppler shift formula (Eq (37.25)) becomes f ≅ f (1 − (u/c)) In the baseball’s frame, this is the frequency with which the radar waves strike the baseball, and the baseball reradiates at f But in the coach’s frame, the reflected waves are Doppler shifted again, so the detected frequency is f (1 − (u/c)) = f (1 − (u / c)) ≈ f (1 − 2(u / c)) EXECUTE: Δf = f (u/c) and the fractional frequency shift is u= Δf = 2(u/c) f0 Δf (2.86 × 10−7 ) c= (3.00 × 108 m) = 42.9 m/s = 154 km/h = 92.5 mi/h f0 EVALUATE: u c, so using the approximate expression in place of Eq (37.25) is very accurate 37.69 IDENTIFY and SET UP: 500 light years = 4.73 × 1018 m The proper distance l0 to the star is 500 light years The energy needed is the kinetic energy of the rocket at its final speed d 4.73 × 1018 m = 3.2 × 1010 s = 1000 y EXECUTE: (a) u = 0.50c Δt = = u (0.50)(3.00 × 108 m/s) The proper time is measured by the astronauts Δt0 = Δt − u /c = 866 y ⎛ ⎞ − mc = (1000 kg)(3.00 × 108 m/s)2 ⎜ − 1⎟ = 1.4 × 1019 J ⎜ − (0.500) ⎟ − v /c ⎝ ⎠ This is 14% of the U.S yearly use of energy d 4.73 × 1018 m = 1.6 × 1010 s = 505 yr, Δt0 = 71 y (b) u = 0.99c Δt = = u (0.99)(3.00 × 108 m/s) K= mc ⎛ ⎞ K = (9.00 × 1019 J) ⎜ − 1⎟ = 5.5 × 1020 J ⎜ − (0.99)2 ⎟ ⎝ ⎠ This is 5.5 times (550%) the U.S yearly use d 4.73 × 108 m = 1.58 × 1010 s = 501 y, Δt0 = 7.1 y (c) u = 0.9999c Δt = = u (0.9999)(3.00 × 108 m/s) 37.70 ⎛ ⎞ K = (9.00 × 1019 J) ⎜ − 1⎟ = 6.3 × 1021 J ⎜ − (0.9999) ⎟ ⎝ ⎠ This is 63 times (6300%) the U.S yearly use EVALUATE: The energy cost of accelerating a rocket to these speeds is immense IDENTIFY and SET UP: For part (a) follow the procedure specified in the hint For part (b) apply Eqs (37.25) and (37.26) EXECUTE: (a) As in the hint, both the sender and the receiver measure the same distance However, in our frame, the ship has moved between emission of successive wavefronts, and we can use the time T = 1/f as the proper time, with the result that f = γ f > f 1/2 (b) Toward: f1 = f c+u ⎛ + 0.758 ⎞ = 345 MHz ⎜ ⎟ c−u ⎝ − 0.758 ⎠ = 930 MHz and f1 − f0 = 930 MHz − 345 MHz = 585 MHz 1/2 Away: f = f c −u ⎛ − 0.758 ⎞ = 345 MHz ⎜ ⎟ c+u ⎝ + 0.758 ⎠ = 128 MHz and f − f = −217 MHz (c) f3 = γ f = 1.53 f = 528 MHz, f3 − f = 183 MHz © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Relativity 37-21 EVALUATE: The frequency in part (c) is the average of the two frequencies in part (b) A little algebra shows that f3 is precisely equal to ( f1 + f )/2 37.71 IDENTIFY: We need to use the relativistic form of Newton’s second law because the speed of the proton is close to the speed of light G G 1 v2 SET UP: F and v are perpendicular, so F = γ ma = γ m γ = = = 1.512 R − v /c − (0.750)2 [(0.750)(3.00 × 108 m/s]2 = 2.04 × 10−13 N 628 m EVALUATE: If we ignored relativity, the force would be 2.04 × 10−13 N = 1.35 × 10−13 N, which is substantially less than the relativistic force Frel /γ = 1.512 IDENTIFY: Apply the Lorentz velocity transformation SET UP: Let the tank and the light both be traveling in the + x -direction Let S be the lab frame and let S ′ be the frame of the tank of water (c/n) + V (c/n) + V = For V c, EXECUTE: In Eq (37.23), u = V , v′ = (c/n) v = cV + (V/nc) 1+ nc EXECUTE: F = (1.512)(1.67 × 10−27 kg) 37.72 (1 + V/nc) −1 ≈ (1 − V/nc ) This gives v ≈ ((cn) + V )(1 − (V/nc)) = ( nc/n) + V − (V/n2 ) − (V /nc) ≈ 37.73 c ⎛ ⎞ ⎞ ⎛ + ⎜ − ⎟V , so k = ⎜1 − ⎟ For water, n ⎝ n2 ⎠ ⎝ n ⎠ n = 1.333 and k = 0.437 EVALUATE: The Lorentz transformation predicts a value of k in excellent agreement with the value that is measured experimentally IDENTIFY and SET UP: Follow the procedure specified in the hint dv v −u u dv EXECUTE: (a) a′ = dt ′ = γ (dt − udx/c ) dv′ = dv + dt ′ (1 − uv/c ) (1 − uv/c ) c ⎛ ⎛ − u /c ⎞ dv′ v−u (v − u )u/c ⎞ ⎛u ⎞ + = + ⎟⎟ = dv ⎜⎜ ⎜ ⎟ dv′ = dv ⎜⎜ 2 2 2⎟ 2 ⎟ dv − uv/c (1 − uv/c ) ⎠ (1 − uv / c ) ⎝ c ⎠ ⎝ − uv/c ⎝ (1 − uv/c ) ⎠ (1 − u /c ) 2 (1 − uv/c ) dv (1 − u /c ) a′ = = = a (1 − u /c )3/2 (1 − uv/c ) −3 2 dt γ dt − uγ dx/c (1 − uv/c ) γ (1 − uv/c ) (b) Changing frames from S ′ → S just involves changing dv −3 ⎛ uv′ ⎞ a → a ′, v → − v′ ⇒ a = a ′ (1 − u /c )3/ ⎜1 + ⎟ ⎝ c ⎠ EVALUATE: a′x depends not only on a x and u, but also on vx , the component of the velocity of the 37.74 object in frame S IDENTIFY and SET UP: Follow the procedures specified in the problem EXECUTE: (a) The speed v′ is measured relative to the rocket, and so for the rocket and its occupant, v′ = The acceleration as seen in the rocket is given to be a′ = g , and so the acceleration as measured on ⎛ u2 ⎞ du = g ⎜1 − ⎟ ⎜ c ⎟ dt ⎝ ⎠ (b) With v1 = when t = 0, 3/ the earth is a = dt = du g (1 − u /c )3/2 t1 v1 ∫0 dt = g ∫0 du (1 − u /c )3/2 t1 = v1 g − v12 /c © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 37-22 Chapter 37 (c) dt ′ = γ dt = dt / − u /c , so the relation in part (b) between dt and du, expressed in terms of dt ′ and du du du, is dt ′ = γ dt = = 2 g (1 − u /c )3/2 g (1 − u /c ) − u /c c ⎛v ⎞ arctan h ⎜ ⎟ For those who g ⎝c⎠ wish to avoid inverse hyperbolic functions, the above integral may be done by the method of partial ⎡ du c ⎛ c + v1 ⎞ du du ⎤ fractions; gdt ′ = = ⎢ + ln ⎜ , which integrates to t1′ = ⎟ ⎥ (1 + u/c)(1 − u/c) ⎣1 + u/c − uc ⎦ g ⎝ c − v1 ⎠ Integrating as above (perhaps using the substitution z = u/c ) gives t1′ = (d) Solving the expression from part (c) for v1 in terms of t1, (v1/c) = tanh( gt1′/c), so that − (v1/c) = 1/ cosh ( gt1′/c), using the appropriate indentities for hyperbolic functions Using this in the expression found in part (b), t1 = c ( gt1′/c) c = sinh ( gt1′/c), which may be rearranged slightly as g 1/ cosh ( gt1′/c) g gt1 v e gt1′/c − e − gt1′/c ⎛ gt ′ ⎞ = sinh ⎜ ⎟ If hyperbolic functions are not used, v1 in terms of t1′ is found to be = gt ′/c − gt ′/c c e +e c ⎝ c ⎠ which is the same as tanh( gt1′/c) Inserting this expression into the result of part (b) gives, after much c gt1′/c − gt1′/c (e −e ), which is equivalent to the expression found using hyperbolic functions 2g (e) After the first acceleration period (of years by Stella’s clock), the elapsed time on earth is algebra, t1 = c sinh ( gt1′/c) = 2.65 × 109 s = 84.0 yr g The elapsed time will be the same for each of the four parts of the voyage, so when Stella has returned, Terra has aged 336 yr and the year is 2436 (Keeping more precision than is given in the problem gives February of that year.) EVALUATE: Stella has aged only 20 yrs, much less than Terra IDENTIFY: Apply the Doppler effect equation SET UP: At the two positions shown in the figure given in the problem, the velocities of the star relative to the earth are u + v and u − v, where u is the velocity of the center of mass and v is the orbital velocity t1′ = 37.75 EXECUTE: (a) f = 4.568110 × 1014 Hz; f + = 4.568910 × 1014 Hz; f − = 4.567710 × 1014 Hz c + (u + v) ⎫ f0 ⎪ c − (u + v) ⎪ f +2 (c − (u + v)) = f 02 (c + (u + v)) ⎬⇒ c + (u − v) ⎪ f − (c − (u − v)) = f 02 (c + (u − v)) f− = f0 ⎪ c − (u − v) ⎭ f+ = (u + v) = ( f + /f ) − ( f + /f ) + c and (u − v) = ( f −2 /f 02 ) − ( f −2 /f 02 ) + c u + v = 5.25 × 104 m/s and u − v = −2.63 × 104 m/s This gives u = +1.31 × 104 m/s (moving toward at 13.1 km/s) and v = 3.94 × 104 m/s (b) v = 3.94 × 104 m/s; T = 11.0 days 2π R = vt ⇒ (3.94 × 104 m/s)(11.0 days)(24 hrs/day)(3600 sec/hr) = 5.96 × 109 m This is about 2π 0.040 times the earth-sun distance R= Also the gravitational force between them (a distance of 2R) must equal the centripetal force from the center of mass: (Gm ) mv Rv 4(5.96 × 109 m)(3.94 × 104 m/s) = ⇒ = = = 5.55 × 1029 kg = 0.279 msun m R G (2 R ) 6.672 × 10−11 N ⋅ m /kg © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Relativity 37-23 EVALUATE: u and v are both much less than c, so we could have used the approximate expression Δf = ± f 0vrev /c, where vrev is the speed of the source relative to the observer 37.76 IDENTIFY and SET UP: Apply the procedures specified in the problem EXECUTE: For any function f = f ( x, t ) and x = x ( x′, t ′), t = t ( x′, t ′), let F ( x′, t ′) = f ( x ( x′, t ′), t ( x′, t ′)) and use the standard (but mathematically improper) notation F ( x′, t ′) = f ( x′, t ′) The chain rule is then ∂f ( x′, t′) ∂f ( x, t ) ∂x′ ∂f ( x′, t ′) ∂t ′ , = + ∂x ∂x′ ∂x ∂t ′ ∂x ∂f ( x′, t ′) ∂f ( x, t ) ∂x′ ∂f ( x′, t ′) ∂t ′ = + ∂t ∂x′ ∂t ∂x′ ∂t In this solution, the explicit dependence of the functions on the sets of dependent variables is suppressed, ∂f ∂f ∂x′ ∂f ∂t ′ ∂f ∂f ∂x′ ∂f ∂t ′ = + = + , and the above relations are then ∂x ∂x′ ∂x ∂t ′ ∂x ∂t ∂x′ ∂t ∂t ′ ∂t ∂x′ ∂x′ ∂t ′ ∂t ′ ∂E ∂E ∂2E ∂2E = 1, = −v, = and = Then, = , and = For the time derivative, (a) ∂x ∂t ∂x ∂t ∂x ∂x′ ∂x ∂x′ ∂E ∂E ∂E To find the second time derivative, the chain rule must be applied to both terms; that is, = −v + ∂t ∂x′ ∂t ′ ∂ ∂E ∂2E ∂2E , = −v + ∂t ∂x′ ∂t ′∂x′ ∂x′ ∂ ∂E ∂2E ∂2E = −v + ∂t ∂t ′ ∂x′∂t ′ ∂t ′2 Using these in ∂2E ∂2E ∂t , collecting terms and equating the mixed partial derivatives gives ∂2E ∂2E ∂2E ∂2E , and using this and the above expression for gives the result v − + ∂x′∂t ′ ∂t ′2 ∂x′2 ∂t ∂x′2 ∂x′ ∂x′ ∂t ′ ∂t ′ (b) For the Lorentz transformation, =γ, = γ v, = γ v/c and =γ ∂x ∂t ∂x ∂t The first partials are then = v2 ∂E ∂E v ∂E ∂E ∂E ∂E =γ −γ = −γ v +γ , ∂x ∂x′ ∂x′ ∂t ′ c ∂t ′ ∂t and the second partials are (again equating the mixed partials) ∂2E ∂x =γ2 ∂2E ∂2E v2 ∂2E v ∂2E +γ2 − 2γ 2 2 ∂x′ c ∂t ′ c ∂x′∂t ′ 2 ∂2E 2∂ E ∂ E + − γ γ v ∂x′∂t ′ ∂t ∂x′2 ∂t ′2 Substituting into the wave equation and combining terms (note that the mixed partials cancel), ∂2E ∂x 37.77 − ∂2E c ∂t = γ 2v ⎛ v2 ⎞ ∂2E ⎛ v2 ⎞ ∂2E ∂2E ∂2E = γ ⎜1 − ⎟ + γ ⎜ − ⎟ = − = ⎜ c ⎟ ∂x′ ⎜c ∂x′ c ⎟⎠ ∂t ′ c ∂t ′2 ⎝ ⎠ ⎝ EVALUATE: The general form of the wave equation is given by Eq (32.1) The coefficient of the ∂ /∂t term is the inverse of the square of the wave speed This coefficient is the same in both frames, so the wave speed is the same in both frames IDENTIFY: Apply conservation of total energy, in the frame in which the total momentum is zero (the center of momentum frame) SET UP: In the center of momentum frame, the two protons approach each other with equal velocities (since the protons have the same mass) After the collision, the two protons are at rest─but now there are kaons as well In this situation the kinetic energy of the protons must equal the total rest energy of the two kaons © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 37-24 Chapter 37 EXECUTE: (a) 2(γ cm − 1)mpc = 2mk c ⇒ γ cm = + momentum frame is then vcm = c mk = 1.526 The velocity of a proton in the center of mp γ cm −1 = 0.7554c γ cm To get the velocity of this proton in the lab frame, we must use the Lorentz velocity transformations This is the same as “hopping” into the proton that will be our target and asking what the velocity of the projectile proton is Taking the lab frame to be the unprimed frame moving to the left, u = vcm and v′ = vcm (the velocity of the projectile proton in the center of momentum frame) 2vcm v′ + u = 3.658 ⇒ K lab = (γ lab − 1) mpc = 2494 MeV = = 0.9619c ⇒ γ lab = vlab = uv′ v + + vcm − lab c c2 c2 2494 MeV K (b) lab = = 2.526 2mk 2(493.7 MeV) (c) The center of momentum case considered in part (a) is the same as this situation Thus, the kinetic energy required is just twice the rest mass energy of the kaons K cm = 2(493.7 MeV) = 987.4 MeV EVALUATE: The colliding beam situation of part (c) offers a substantial advantage over the fixed target experiment in part (b) It takes less energy to create two kaons in the proton center of momentum frame © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... × 10 13 J (a) K mc +1 = 1 − v /c 1 − v /c = 1.20 × 10 13 J 8.20 × 10−14 J + = 2.46 v = c − (1/ 2.46)2 = 0.914c = 2.74 × 108 m/s (b) K = 12 mv gives v = 37.41 2K 2(1.20 × 10 13 J) = = 5 .13 ×... − (0.750)2 [(0.750)(3.00 × 108 m/s]2 = 2.04 × 10 13 N 628 m EVALUATE: If we ignored relativity, the force would be 2.04 × 10 13 N = 1.35 × 10 13 N, which is substantially less than the relativistic... 1)(1.67 × 10−27 kg)(3.00 × 108 m/s)2 ⎜⎜ 13 ⎟ ⎟ = 274 MeV ⎝ 1.60 × 10 J ⎠ ⎛ 1.00 MeV ⎞ (c) The rest energy of η is mc = (9.75 × 10−28 kg)(3.00 × 108 m/s) ⎜⎜ 13 ⎟ ⎟ = 548 MeV ⎝ 1.60 × 10 J ⎠ EVALUATE: