Chapter 35: Relativity

Bauer/Westfall: University Physics, 1E

Chapter 35: Relativity

In-Class Exercises

35.1. c, d, e 35.2. a, c, d, e 35.3. e 35.4. a) True b) False c) True 35.5. a 35.6. b, c, d 35.7. a 35.8. d

Multiple Choice

35.1. a 35.2. d 35.3. c 35.4. d 35.5. a 35.6. d 35.7. c 35.8. c

Questions

35.9. A direct corollary of Einstein’s special theory of relativity postulates that no entity or interaction in the universe can propagate with a speed greater than the speed of light in vacuum. Therefore, instantaneous effects of events originating at one point in space on another point in space are impossible. The translational motion of a perfectly rigid object would imply that, by moving one end of the object, the other end of the object would also move instantaneously, without any time delay. This contradicts Einstein’s theory. 35.10. The y-axis is the time given in μs and the x-axis is the ‘distance’ x / c given in units of μs also, since the speed of light can be written c 3.00 1082 m/s  3.00 10 m/μs  0.300 km/μs. To hit the target, the world line from t 13 μs of the person (Eddie and/or Martin) must lie inside the past light cone of the target at x  0 and t  0. As seen in the diagram, Eddie’s world line is inside the past light cone of the target from t 13 μs to t 2 km / 0.3 km/μs  20 / 3 μs and so Eddie could hit the target. However, Martin’s world line lies outside of the light cone for all time after t  13 μs and so he could not have hit the target. Eddie and Martin find out the target has been hit at the point where their individual world lines intersect the light cone from the target at the origin at some time after the target is hit at t = 0. As shown in the diagram, Eddie finds out the target has been hit at t  20 / 3 μs and Martin finds out it has been hit at t 5km/0.3km/μs  50/3 μs

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35.11. If the lens was situated perfectly, there would be indeed be a halo, since the alignment is typically not exact, we see arcs instead. Likewise, the curvature is a result of the mass, so if the object does not have a uniform mass distribution, different rays would be affected non-uniformly. 35.12. In the relativistic limit, velocities must be added relativistically (using the Lorentz transformation), not classically (using the Galilean transformation), as your friend is suggesting. Let F be the frame of the rocket and F be the frame of the Earth. The torpedo has a speed of uc'2/3 with respect to the rocket (frame F ) and the rocket travels at a speed of vc 2/3 with respect to Earth (frame F ). According to the Lorentz transformation the velocity, u, of the torpedo in the Earth’s frame is uv'12 2/3cc   2/3 uc . 1'/ vu c2 14/9/ cc2213 This is less than the speed of light, so no violation of the theory of relativity occurs. 35.13. Yes, the observer still sees the positive charge attracted to the wire. If the positive charge is moving, with  velocity v in the lab frame, parallel to the current, then it is actually moving anti-parallel to electrons,  which have velocity u in the lab frame. Since the positive charge sees only a magnetic field, this must mean that the wire is electrically neutral, i.e. there are equal positive charges (ion cores) per unit length as there are negative charges per unit length. When the wire is seen in the reference frame of the positive charge, the positive charge is stationary while the ion cores are moving away from the positive charge with  velocity v. The electrons are also moving away from the positive charge with a velocity uv uv'.  1/ vu c2 Both the electrons and ion cores have their separation contracted due to their velocities. Since the electrons are; however, moving faster than the ion cores, their separation is smaller than the separation of

1235 Bauer/Westfall: University Physics, 1E

the ion cores, meaning the positive charge now sees a net electric charge in any given length of wire and is therefore, attracted to the wire via an electric force instead of the magnetic force in the lab frame. 35.14. The pilot of the rocket sees the garage length contracted. At the speed of the rocket the value of  is: 1/2 1/2 2 v2 0.866c   11  2. cc22  The rocket pilot therefore thinks that the garage has a length that is reduced by the  factor of 2; that is, LL/2 /  /4.

35.15. Since the rod makes an angle with the x-axis, it has a projected length on both the x and y axes. Since the velocity is in the x-direction, only the projection of the length on the x-axis will be contracted, meaning the y-projection length remains unchanged. Since the angle is given by   tan1  yx / , as x decreases, the angle increases as viewed by an observer on the ground. 35.16. The primary reason that this presents no contradiction is that the two observations are made in reference frames that are not equivalent. As such, the measurements cannot be directly compared simply by making comparison of observed dimensions. The Earth’s shape is distorted from the usual spherical shape due to the fact that length contraction that occurs in the direction of the observers motion only – perpendicular to the axis of rotation for the first astronaut and along the axis of rotation for the second astronaut. If the two observers really want to compare what they’ve seen, they must exchange information that includes their own relative speed and direction with respect to the Earth. 35.17. The Lorentz transformation for the positions relating the coordinates in the moving frame (primed coordinates) to our reference frame (unprimed coordinates) takes the form x ', xvt with  and z-coordinates unchanged, and  given by 1/2 v2  1.2 c Hence, the moving clock at x ' 0 has coordinate x  vt and the clock at x '  l has coordinate x vt l /. The time readings are then related by the Lorentz transformation, vx tt'.  c2 For the clock at x ' 0 the reading is vvt  vtt2 1  tt'1.  ttt  ttt   0 22 2 2  cc  For the clock at x'  l the reading is vvtl /  vvl2 1 vltlv tt'1.  tt  t t  1 222222 ccccc  These results display two important effects. First, time dilation is apparent, as the advance of the t ' values is slowed compared to the advance of t by factor 1/ . Second, relativity of simultaneity is also manifest, as the readings on the moving clocks – which are synchronized in their own reference frame – differ by lv/ c2 at fixed time t in our reference frame. The clock behind in position is “ahead” in time reading. That is, “the same time” at different positions is a reference-frame-dependent notion. This effect is often overlooked, but most purported relativistic kinematics are resolved unambiguously once it is take into account.

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35.18. Velocities are added using the relativistic velocity transformation. Assume that the velocities are along the x-axis. Then the transformation equation is uv x  y uc , 1/ uv c2 1 xy where x and y represent the fractions of the speed of light of the two sub-light velocities being added. Now, since x 1, it follows that x2 1. Multiply both sides of this inequality by 1 y2 (which is positive since y 1 ), to obtain x2211.yy  2Expand, and add the negative terms to the opposite sides to get xy221. x 22y Subtract 2xy from both sides, to yield: xxyy22212. xyxy 22Factoring both 22 sides as squares gives the inequality: x yxy1 . Divide both sides by the right-hand side (which 2 xy   is positive since xy 1 ) which results in the inequality 2 1. Taking square roots of both sides 1 xy xy preserves the inequality (with absolute values), so 1. It follows that the velocity added 1 xy xy xy relativistically is still less than c, since uccc  . 11xy xy

35.19. Classically, conservation of kinetic energy in an elastic collision for identical particles of mass m means that 111 mv2220, mv  mv 2221i 1f 2f

Where v1i is the velocity before the collision and v2i and v2f are the velocities after the collision. If the particles have the same mass this reduces to vvv222, which can only be true if the velocities are 1i 1f 2f   perpendicular (since conservation of momentum requires also that vvv1i 1f 2f ). Let the energy and momentum of the originally moving particle be E and p . Let the two particles have total energies after

the collision of E1 and E2 , and momenta after the collision of p1 and p2 , respectively. Energy- momentum conservation implies the relationships: EmcE2 E 12 pp12 p. The term Epc222 is a scalar invariant so it is the same before and after the collision, implying: 2 22222 2 2 Emc pc E12 E c p 12 p c 2 2 24 22 2 2 22 22 2 E222 Emc mc pc E112212 EE E pc pc ppc 12 222 2 24222222 2 E pc222 Emc  mc  E11 pc E 22 pc EE 1212 ppc Using the term Epcmc22224, this reduces to 22mc24 Emc 2 222 mc 24 EE ppc 2 12 12 22 Emc E12 E p 12 p c

Hence, the dot product of the momenta p1 and p2 is given by 22 ppc12 EE 12 Emc 22 EEmcE11 Emc.

2 Energy E1 can take values from mc to E (as can E2 ). Therefore, the function on the right-hand side of this equation increases monotonically from zero to the value

1237 Bauer/Westfall: University Physics, 1E

2 1 2 221 Emc for mc E1  E  mc , 4 2

1 2 and decreases monotonically back to zero for EmcE 1 E. It is never negative over the allowed 2 2 range of E1. This implies pp12 0, with equality only for Emc1  or EE1  , i.e., only if one of the particles remains at rest after the collision. Otherwise the dot product is positive, meaning the two particles emerge from the collision on trajectories forming an acute angle. Therefore, it is not necessary for the velocities of the two particles to be perpendicular. 35.20. The spaceship is accelerating, and since special relativity deals only with objects moving with constant velocity, one might think that general relativity is required to solve this problem. However, the fact that the spaceship is accelerating is irrelevant since at any point in the trajectory, its velocity is constant. Since the direction of the speed is constantly changing, the length will also appear to be warped along the curvature of the orbit. The observed length of the spaceship is

L 22 LLvcL0 1  /  1  0.800  0.600 L .  00 0 So, the length would look to be 60.0% of the original length.

Problems

35.21. The speed of light converted from SI to ft/ns is:

881 s 3.2808 ft c 2.9979 10 m/s  2.9979 10 m/s9  0.984 ft/ns. 10 ns 1 m You can see that our result is quite close to 1 foot per nanosecond, which makes this a great way to visualize the speed of light: light moves about a foot in a time interval of a billionth of a second!

35.22. Convert the acceleration due to gravity from SI units into units of ly/year2 .

365.25 days 24 hours 3600 s 7 1 year 1 year 3.1556 10 s 1 year 1 day 1 hour 2 7 2 1 ly 3.1556 10 s 2 g 9.81 m/s15  1.03 ly/year 9.461 10 m 1 year Just like in problem 35.21, the numerical coefficient comes out to be very close to 1. However, unlike the answer in 35.21, the answer to the present problem is more of a curiosity than a useful number for any practical purposes. 35.23. The boat has a velocity of v with respect to the water. The velocity of the water is u downstream. So in order for the boat to directly cross the river, the boat must be headed upstream at an angle such that the velocity of the boat with respect to the ground is vu22 . The cross-stream time across the river of width D with this velocity is 2D tcs  . vu22 Going upstream, the boat has velocity vu , and going downstream it is vu . Over a distance D, the upstream-downstream time is: DDDv  u Dv  u 2 Dv t   . ud vu vu v uvu vu22

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t 2/Dvu22 vu 22 The ratio of times is then: cs . 22 tvud 2/Dv v u

1 1 1 1 5 35.24. For v  0.8c  4 c,       1.6667. 5 2 2 1v / c 14/5 116 / 25 9/25 3

35.25. (a) Another astronaut on the ship sees the meter stick in the same (rest) frame as the astronaut holding the stick and so its length remains unchanged at one meter. (b) For a ship moving at vc 0.50 , the length of the meter stick as measured by an observer on Earth is

L 22 LLvc0 1  /  1.00 m 1  0.50 cc /  0.87 m.  0

35.26. (a) According to a clock on Earth the trip takes 8 L 3.84 10 m t 0  2.6 s. v 0.50 3.00 108 m/s (b) According to a clock on the spaceship the trip takes,

t 22 ttvc 1 /  2.56 s 1  0.50 cc /   2.2 s. 0  (c) On the ship, the distance to the Moon is contracted to L :

L 22 LDvc0 1  /  3.84  1088 m 1  0.50 cc /  3.3 10 m.  

35.27. The time that passes in the rest frame of the Earth is t  30. yr. The time that passes in the mother’s

frame is t0 10. yr. Therefore, 2 tt001 22 t  t0    1/  vc    1/ vc tt  1/2 1/2 2 2 t0 10.  vccc110.94.     t 30.

35.28. The muon’s lifetime t when it is moving at vc 0.90 will be longer than t  2.2 μs when it is at rest in the laboratory frame due to time dilation: 2.2 106 s t0   6 tt 0   5.0  10 s. 22 1/vc 10.90 cc/

35.29. The fire truck of length L0 10.0 m is traveling fast enough so a stationary observer sees its length contracted to L  8.00 m. Therefore, 1/2 1/2 22 2 L 22   0 LL 8.00 m L L0 1/  vc    1/ vc  v 1  c  1 c  0.600. c  LL10.0 m 00   (a) The time taken from the garage’s point of view is

L 8.00 m 8 tg   4.44  10 s. v 0.600 3.00 108 m/s (b) From the fire truck’s perspective the length of the garage will be contracted to

1239 Bauer/Westfall: University Physics, 1E

L0 22 LLvc0 1  /  8.00 m 1  0.600 cc /  6.40 m.  Therefore, the truck will not fit inside the garage from the fire truck’s point of view since the length of the truck from its rest frame is 10.0 m.

35.30. The rest frame time taken by Phileas Fogg is t0  80 days, while time dilation makes the time seem like t 81 days. Therefore 2 1/2 tt1 22 t t  00   1/  vc   1/. vc 0    tt   Therefore, 1/2 1/2 2 2 t0 80 days vc11  cc  0.16. t 81 days  

35.31. THINK: The planet is L0  35 ly away, but the astronauts cannot travel as fast as c and hence will take

longer than 35 years in the NASA (Earth) reference frame while it will take only t0 25 years in the astronauts’ reference frame. The astronauts will see the distance as being contracted. SKETCH:

RESEARCH: The time it takes to reach the planet as observed from Earth is tLv 0 /. The relationship 1/2 2 between t and t is tt  , where  1/vc . 0 0   SIMPLIFY: 2 LLt  2 000    (a) tt0  2 vv1/ vc 222 1/2 LLL2    2 vvcvvvtcLc2220001/ 1 /       00 ttct000    L (b) L  0 .  CALCULATE: 1/2 2 25 years (a) Since Lc/35 years, vcc1  0.81373 . 0 35 years 

2 (b) L 35 ly 1 0.81373  = 20.343 ly ROUND: The answers should be given to two significant figures. (a) vc 0.81 (b) L  20. ly DOUBLE-CHECK: The velocity found for the astronauts is less than the speed of light and the distance of the planet from the perspective of the astronauts does contract; so these values are reasonable. Also, the astronauts believe that 25 years pass during their trip. Their length contracted distance to the planet is

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20.343 ly. This means their speed in terms of c during the trip is 20.343 ly / 25 yr  0.81c which agrees with the value found. 35.32. THINK: Since the velocity of frame F is in the x-direction, the projection of the length of the rod on the x- axis will experience a contraction, while the projection on the y-axis will remain unchanged. The angle

that the meter stick makes with the x-axis changes from 0  37 to 1  45 in frame F. Trigonometry can give equations relating the angles to the speed and length. SKETCH:

RESEARCH: In both frames, LLyy Lsin01  L sin . In frame F, LLx  cos0 , and in frame F,

LLx cos1 . The x-axis contraction is given by LLxx  / . SIMPLIFY:

sin0 (a) LLsin01 sin LL . In frame F, the x-axis projection is sin1 2 2 LLLLsin0000x sin cos1  tan LLxcos1   tan11 tan    tan  1 1/2 22 2 tan  tan 1/vc00  v 1  c . tan tan 11  sin (b) The length of the rod in frame F is LL  0 . sin1 CALCULATE: 1/2 2 tan 37 (a) vcc10.6574  tan 45  sin 37 (b) L 1.00 m 0.8511m sin 45 ROUND: The answers should be rounded to two significant figures. (a) vc 0.66 (b) L  0.85 m DOUBLE-CHECK: The velocity does not exceed the speed of light and the length does contract; therefore, the answers are reasonable. 35.33. THINK: The tip of the triangle is the direction of the speed, vc 0.400 , so that only the length, L  50.0 m, will be contracted and the width, w  20.0 m, is not affected. The length of the ship L is not the same as the length of a side of the ship l . Relate the observed angle   to the speed of the ship.

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SKETCH:

RESEARCH: The lengths are related to the angles, in both frames, by lwcos  / 2, lwcos  / 2, Ll sin , Ll sin  , and tan  2Lw / . The length of the ship contracts by LL  / . SIMPLIFY: Determine l in terms of l : w cos llcos cos l  l . 2cos  The contracted length is then

Llsin tan 2 L 2 Llsin  l cos tan   tan   1 vc / .  w

1 2L 2 Therefore,  ()vvc tan 1 / . w The plot of the angle between the base and side of the ship as a function of the speed of the ship as measured by a stationary observer is shown below.

250. m 2 CALCULATE:   vc0.40 tan1  1  0.40 cc /  77.69  20. m ROUND: To three significant figures,  vc 0.400  77.7 . DOUBLE-CHECK: As v approaches c, the expression under the square root approaches zero and hence the angle will also approach zero. This agrees with the graph where the angle is smaller at higher velocities. When vc , the side of the ship would effectively contract to zero, thus making an angle of zero with the width.

1242 Chapter 35: Relativity

35.34. Since the light whose rest wavelength, 0  480 nm, appears as   660 nm, it is red-shifted, so you must be travelling away from the light. cv cv  22   2cv  2  cv 00cv cv 0 22 22 660 nm 480 nm vc0  cc 0.31 22  2 2 0 660 nm 480 nm

35.35. The light with wavelength 0  650 nm is blue-shifted and appears as   520 nm, as expected since the driver is travelling towards the light. Therefore, cv cv  22   2cv  2  cv 00cv cv 0 22 22 650 nm 520 nm vcv0  cc 0.22 . 22  2 2 0 650 nm 520 nm You would have been traveling 0.22c, or 22% of the speed of light. This explanation would likely result in a speeding ticket!

35.36. Since the light has a rest wavelength of 0  532 nm and must appear to have   560 nm, it must be red- shifted so it must travel away from the meteor. cv cv  22   2cv  2  cv 00cv cv 0 22 22 560 nm 532 nm vc0  cc 0.051 22  2 2 0 560 nm 532 nm

1000 m 1 h 35.37. Since the car, moving with a speed v 32.0 km/h 8.889 m/s, is moving away from 1 km 3600 s

the radar of frequency f0 10.6 GHz, the shift in frequency is,

8 cv 3.00 10 m/s 8.889 m/s fff   f 1  10.6  109 Hz  1  314.078 Hz. 00 8 cv 3.00 10 m/s 8.889 m/s   Therefore, the frequency is red-shifted by 314 Hz. 35.38. THINK: Since the spaceship is moving towards the station, the wavelength will be blue-shifted, resulting

in the original wavelength of 0  632.8 nm being reduced to   514.5 nm. Using the relativistic formula for wavelength shift the speed of the ship can be deduced. SKETCH:

RESEARCH: Since the ship is moving towards the station, the relevant formula for wavelength shift is

cv   0  0 . The shift parameter is by definition: z  . cv 0

1243 Bauer/Westfall: University Physics, 1E

cv cv 22 SIMPLIFY:  22   2cv  2 cv v0 c 00 0  22 cv cv 0  22 632.8 nm 514.5 nm 514.5 nm   632.8 nm CALCULATE: vcc0.20405 , z 0.186946 22 632.8 nm 632.8 nm 514.5 nm ROUND: To four significant figures, vc 0.2041 and z   0.1869. DOUBLE-CHECK: The velocity is less than the speed of light and the shift parameter is negative, which is what it should be for blue shift, so it makes sense.

35.39. In Sam’s reference frame, each event occurs at the following points: xA  0 m, t A  0 s, x B  500. m and

tB  0 s. To find the timing of the events in Tim’s reference frame, use the Lorentz transformation  2  ttvxc  /. Therefore, tA  0s and

vx 0.999 500. m 5 tB  3.73 10 s. 2 2 c 2.9979 108 m/s 1 0.999 (a) Therefore, Tim experiences event B before event A. (a) For Tim, event A occurs 3.73 105 s after event B. 35.40. Let an inertial reference frame F be at rest and let another inertial reference frame F move at a constant speed v along a common x-axis with respect to reference frame F. According to the relativistic velocity addition formula, uv u  1/ vu c2 cv cv cc v uc u     c, vc v cv 11 c2 c as required. Thus, the result is independent of the specific value of v. 35.41. Let all speeds be in a common x-direction. Let frame F be the ground and frame F be the frame of your car. The speed of your car with respect to the ground is v  50.0 m/s and the speed of the oncoming car is u 50.0 m/s in frame F. Using the relativistic velocity transformation, the relative speed of the oncoming car is uv 50.0 m/s  50.0 m/s u   99.99999999999862 m/s  100. m/s. 2 2 1/ uv c 8 1 50.0 m/s 50.0 m/s / 2.9979  10 m/s The relative velocity is about the same as a Galilean velocity transformation uuvu 2100m/s,  since the speed of the cars is so small compared to the speed of light. In order to detect a difference, fourteen significant figures would need to be kept. This shows how close the values are. 35.42. Assuming all speeds are measured along the same direction, let vc 0.90 be the speed of the ship (frame F ) relative to Earth (frame F) and let uc  0.50 be the speed of the missile relative to the ship. The speed of the missile as seen from the Earth is given by uv  0.50cc   0.90  uc 0.97 . 1vu / c22 1 0.90 c 0.50 c / c

35.43. (a) The total distance travelled, as measured by Alice is

1244 Chapter 35: Relativity

L 2 Lcc0 2 3.25 ly 1  0.65 /  4.940 ly  4.9 ly.  L 4.940 ly (b) The total time duration for the trip as measured by Alice is t  7.6 years. vc0.65

35.44. THINK: The spaceship that Alice boards travels at a speed of uc 0.650 to a station L0  3.25 ly away. The question asks for the speed v Alice must travel so that she measures a relative speed of uc 0.650 on the return journey. In Alice’s frame, the distance of the return flight will be length contracted. The relativistic velocity transformation and length contraction formulae can be used to solve the problem. SKETCH:

RESEARCH: (a) The relativistic velocity transforms as uv u  . 1/ vu c2

(b) The time of the return flight as measured by Alice is tLv /, where LL 0 /  is the length contracted distance in her frame. SIMPLIFY: (a) The speed of the spaceship is given by uv uvu uvu uu u u uv uu  v v . 1/uv c22 c c 2 1 uu / c 2

LL 2 (b) The time for Alice’s return flight is tvc001/.  vv CALCULATE: (a) To Alice, the Earth is moving toward her with a speed of uc  0.650 , so 0.650cc  0.650  vc0.91388 . 0.650cc 0.650 1 c2 (b) The time duration of the flight as measured by Alice is

3.25 ly 2 tcc1 0.91388 /  1.4438 years. 0.91388c ROUND: The answers should be given to three significant figures. (a) As required, the velocity of the ship relative to the Earth is vc 0.914 . (b) The duration of Alice’s return flight as measured by her is t 1.44 years. DOUBLE-CHECK: The speed vc 0.914 gives 0.650cc   0.914  uc 0.650 . 1  0.650ccc 0.914 / 2

1245 Bauer/Westfall: University Physics, 1E

35.45. THINK: The arrow has a velocity of uc  0.300 in Robert’s reference frame. The railroad car has a

length of L0 100. m and travels at a speed of vc 0.750 . The velocity transformation equations and the equation for length contraction can be used to determine the values observed by Jenny. SKETCH:

RESEARCH: As observed by Jenny, L (a) the railroad car is length contracted: L  0 ,  uv  (b) the velocity of the arrow is given by the inverse relativistic velocity transformation: u  , 1/ vu c2 vx  (c) the time of the arrow’s flight is given by the inverse Lorentz transformation: tt 2 , and c (d) the distance traveled by the arrow is given by the inverse Lorentz transformation: x  xvt.

SIMPLIFY: Here x  L0 is the length of the railroad car and tLu  0 /  is the time of the arrow`s flight in Robert`s frame of reference. As observed by Jenny,

2 (a) LL0 1/, vc

L0 1 v (c) the time taken by the arrow to cover the length of the car is t , and 2  2 1/ vc u c

L0 v (d) the distance covered by the arrow is x 1. 2  1/ vc u CALCULATE:

2 (a) Lcc100. m 1 0.750 /  66.14 m 0.300cc   0.750  (b) uc0.85714 1  0.750ccc 0.300 / 2 100. m 1 (c) t 0.750 2.059 106 s 2  2.9979 108 m/s 1 0.750cc / 0.300 100. m 0.750c (d) x 1529.2m 2  10.750/ cc 0.300c ROUND: The answers should be given to three significant figures. As observed by Jenny, (a) the railroad car is L  66.1 m long, (b) the velocity of the arrow is uc 0.857 , (c) the time it takes the arrow to cover the length of the railroad car is t  2.06 μs, and (d) the arrow covers a distance of x  529 m.

1246 Chapter 35: Relativity

DOUBLE-CHECK: The railroad car length is contracted from Jenny’s viewpoint, as expected. Multiplying the answer to part (b) by the answer to part (c): x 0.8571 2.9979 1086 m/s 2.059 10 s 529 m, as found in part (d). So, the answers are consistent. 35.46. THINK: The speed of an object can be described by the relation vc tanh where  is known as the rapidity. The question asks to prove that two velocities adding via the Lorentzian rule, corresponds to adding the rapidity of the two velocities. The question also asks for the Lorentz transformation of two coordinate systems using the rapidity. The Lorentz transformation equations can be used to solve this problem. SKETCH:

RESEARCH: uu  12 (a) The Lorentzian rule for adding two velocities is v 2 . The Lorentz transformation between 1/ uu12 c two frames with relative velocity v in the x direction is given by the equations x xct, yy  , zz  , and ttxc  /. Velocities that add according to the Lorentzian rule correspond to adding the rapidity of each:

uu12 ctanh 1 c tanh 2 vctanh12 2 1/ uu12 c 1 tanh12 tanh (b) For the derivation it is useful to know that the hyperbolic tangent is related to exponentials by eex  x tanhx  . eex  x The following relations are also useful: sinh 1sech22 tanh and tanh  . cosh SIMPLIFY: (a) According to the Lorentzian rule,  ee1122 ee  tanh tanh 1122 cctanh12 tanh 12ee ee vcc2 1122 1 cc tanh12 tanh / c1 tanh12 tanh ee ee 1  ee1122 ee    1122 1122   ee ee ee ee 22ee12 12  c cctanh  ,      12 ee1122 ee e1122eee 22ee12 12 as required. (b) If vc tanh then,

11 11 22  1/vc 22  1tanh/ c c   1tanhsechcosh2222   and vctanh   tanh . cc The Lorentz transformation becomes xxctx    cosh  cosh tanh ctxct  cosh  sinh , yy  ,

1247 Bauer/Westfall: University Physics, 1E

zz  , and xx ttxct  / cosh cosh tanh  t cosh sinh cc CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: Note that the transformation is similar to a transformation from one coordinate system to another where they differ by the angle  :

x xycos sin and yx   sin y cos .

35.47. The relativistic momentum is pmv  . If the momentum is equal to pmc then mv mc or 222 11ccccc    11 2v . 2 2    1/ vc vvvv1/ vc    2 This can be left in exact form, or written as vc 0.707 .

35.48. (a) The energy of the electron is Emc  2 . For the energy to be 10 times greater than its rest energy of 2 Emc0  ,

11992  =101/vc   v c  0.995 c 2 1/ vc 100 100

(b) The momentum is pmv 10 0.511 MeV/ c2  99 /100 c  5.08 MeV/ c .

35.49. The kinetic energy of the colliding beams in the center-of-mass reference frame is related to the fixed- target equivalent, or lab reference frame by 2 2 2 K cm 2 197 100. GeV lab cm     6 KK42  4 197 100. GeV    4.02 10 GeV. mcp 197 1.00 GeV  This is an incredibly large energy. 35.50. The work done on the proton is equal to the change in kinetic energy of the proton.  22 21 2 W K  mcpp  mc  11 mc p   mc p 2  1/vc  2 1 27 8 1 1.672 10 kg 2.9979 10 m/s 2   10.997/cc 11.14477 GeV 11.14 GeV.

2 1 35.51. The energy of the proton is Emc p 938 MeV  1200 MeV. 2 10.61/ cc

1248 Chapter 35: Relativity

35.52. THINK: Two protons in an accelerator are on a head-on collision course. In the lab reference frame (frame F) the protons reach a speed of vc 0.9972 . The relativistic velocity transformation and the relativistic formula for kinetic energy can be used to solve the problem. SKETCH:

RESEARCH: The speed of the proton in the other proton’s rest frame (frame F ) is given by uv u  . The kinetic energy of a relativistic particle is Kmc 1. 2 The mass of a proton is 1/ vu c2 2 mcp  938.27 MeV/ . SIMPLIFY: Let u denote the speed of the proton in the lab frame. In the proton reference frame, the speed of the other proton is vv  2v  u 22. 1/vv c 1/ vc The kinetic energy K of the protons in the lab reference frame is the sum of the kinetic energy of each proton:  22 21 2 KK12 K112121. mc p   mc p     mc p   mc p 2  1/vc  221 The kinetic energy K in the proton reference frame is Kmc  11.pp   mc 2  1/uc CALCULATE: 20.9972 c  (a) uc2 0.999996 10.9972/ cc  1 22 (b) Kcc2 1 938.27 MeV/  23217.35 MeV 2   10.9972/cc  1 22 (c) Kcc 1 938.27 MeV/ 333689.6 MeV 2   1 0.999996cc / ROUND: (a) To six significant figures, the speed of one proton with respect to another is uc  0.999996 . (b) To four significant figures, in the lab reference frame, the particles have a kinetic energy of K  23220 MeV. (c) To four significant figures, in the proton’s reference frame, the other proton has a kinetic energy of 333700 MeV. DOUBLE-CHECK: These are typical speeds and energies for protons to have in proton accelerators. 35.53. THINK: Electrons acquire kinetic energy as they accelerate through the potential difference. The speed acquired by the electron after moving through this potential can be found and then the appropriate

1249 Bauer/Westfall: University Physics, 1E

classical and relativistic formulae can be used to find the total energy and momentum. Many of the answers only make sense if they are given to three significant figures, so rounding will be nonstandard. SKETCH:

RESEARCH: (a) The kinetic energy gained by the electron in moving through the potential difference V is equal to the work done by the potential difference: WKqV  .

(b) The kinetic energy of a relativistic particle is KE 1. 0

(c) The relativistic values for the total energy and momentum are EER0  and pmvR   . Classically, 1 these values are given by EK mv2 and pmv . CC2 CC

The rest mass energy of an electron is E0  511 keV. SIMPLIFY: (a) KeV

(b) The speed of the particle is found using the relativistic formula KE 1: 0

2 2 2 2 1 KE  E KE00 E KKE 2 0  00vc1. cc  2 2 EKEKE00 0 1/ vc  KE0 (c) The relativistic values for the total energy and momentum are

EEKER0   0, and 2  KE 00 E KKE2 0 2 pmvR  2 cKKEc 2/.0 EKE00c  Classically, the total energy and momentum are

EKC  , and

pCC mv m2/ K m  2 Km  2 KE 0 /. c CALCULATE: (a) Ke5.00 kV 5.00 keV

2 5.00 keV 2 5.00 keV 511 keV (b) vcc0.1389 5.00 keV 511 keV

(c) ER 5.00 keV  511 keV 516 keV

2 pccR 5.00 keV 2 5.00 keV 511 keV /  71.659 keV/

EC  5.00 keV

pccC 2 5.00 keV 511 keV / 71.484 keV/ ROUND: (a) The kinetic energy that the electron acquires is K  5.00 keV.

1250 Chapter 35: Relativity

(b) The electron has a speed of vc 0.139 , thus the electron will have only a small difference between its classical and relativistic values, but this can still be considered a relativistic speed.

(c) The relativistic and classical energies are ER  516 keV and 5.00 keV, respectively. (The difference is due to the fact that the relativistic energy includes the rest energy). The relativistic and classical momenta

are pcR  71.7 keV/ and pcC  71.5 keV/ , respectively. DOUBLE-CHECK: The classical and relativistic momenta are similar, as expected for such a low speed. 35.54. The momentum before the collision must equal the momentum after the collision.

pp pp  1212 111mv 111 mv 222 mv . The ratio is 0.700cc  0.500   22 mvv  1 0.700cc / 1 0.500 cc / 21111 7.63. mv122  0.200 c 2 1  0.200cc /

35.55. THINK: Two particles collide inelastically. One particle has a mass of mm1  and momentum pmc1  .

The second particle has a mass of mm2  22. Conservation of energy and momentum can be used with the relativistic energy equation to determine the speed and mass of the new particle. SKETCH:

RESEARCH: The relativistic momentum is p   mv. The energy of the particles is Epcmc22224 after the collision. SIMPLIFY:

(a) The speed of the projectile with momentum pmc1  before the collision is given by

mc22 222 c v111111 c1/ vc  vc /1/   vc  2 v  c v . m 2 (b) The total energy is conserved before and after the collision. Therefore,

EEfi 22 24 22 24 22 24 pc Mc pc11 mc pc 22 mc

2 2 pc22 Mc 24 mc c 2  mc 24022 m c 4  p22 c M 24 c222 mc 2  mc 2 pc22 Mc 2432 mc 2 pMcmc2222218

1251 Bauer/Westfall: University Physics, 1E

From conservation of momentum, pp 1  mc. Therefore, the above equation becomes: 2 mc M22 c18 m 22 c Mc2217 mc 22 Mm 17 Note that there is more mass than there was before the collision. Some kinetic energy has become mass energy. (c) Using the conservation of momentum

2 22 22 22 p p1  Mv  mc  Mv  mc1/  v c  M v  m c  m v mc mc c v   22 2 2 18 Mm 17mm CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: It is reasonable that the speed of the new particle is smaller than the speed of the projectile. The momentum of the new particle is given by

Mv  17mc / 18  17 17 pMv   mc  mcmc . 1(/) vc2218 1  1/18 18 1 1/18/ cc This is the initial momentum of the projectile, as expected by conservation of momentum. 35.56. THINK: To derive the Lorentz transformation for momentum, follow Derivation 35.3. In this case, the momentum is similar to the position coordinates and the energy is analogous to the time. SKETCH:

RESEARCH: The energy is given by Emc  2 and the momentum is given by p  mv . In order to use vE the energy as a momentum, it must be of the form pE . c2 c SIMPLIFY: In frame F, the vectors are   p  Ev OA p , OA  x and OO  . x  c2   Using the equation OA O A OO gives p Ev Ev x  pppxxx    . (1)  cc22 For frame F, the vectors are  p   Ev OA  x , OA  p , and OO  .  x c2   Using the equation OA O A OO gives

1252 Chapter 35: Relativity

p Ev x p . (2)  x c2 p Ev E v  x  Substituting from equation (1) for px into equation (2) gives  px . Solving for E :  cc22 cc22p c 2111  x EpEpEEvpxxx    221.  vv v  1 11  From   , it is easy to show that 2211.  Therefore, EEvp  x . Of course, for 1  2 

motion in one dimension (the x-direction), pyy p and ppzz  . Thus the Lorentz transformation for momentum and energy is established. CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: This result matches with the required expressions. 35.57. THINK: The Lorentz transformations for energy and momentum in the frame F can be used to write the quantity Epc222 in terms of the values in the unprimed frame F. SKETCH: Not required. RESEARCH: The Lorentz transformations are   2   EEvp  x , ppvEcxx  /, pyy p and ppzz . SIMPLIFY: Apply the transformations: 2 2 22  2 22  22  22 22 2 2 2 22 22 E pc E pcxyz pc pc E vp x  p x  vEc/  c  pc yz pc 22 2 222 222 2 222 2 22 22 EEvpvppcpvEvEcpcpc22/xxxx          yz 2 22222222221/vc E  v  c p  pc pc   xy z 22 2222222221/vcE  1/ vcpcpcpc Epc222 .     xyz CALCULATE: Not required. ROUND: Not required. DOUBLE-CHECK: The statement in the problem has been proved using only the Lorentz transformation equations. One could check the result for special and limiting cases. For example, if v = 0 then  = 1 and

the Lorentz transformations reduce to EE  and ppx  x , so the result holds. When p = 0 in the frame F, 2 2 22  2 22  22  22 2 2 2 2 2 E pc E pcxyz pc pc E vEc/  c

2 22EvEc 222/1/. 2  2  vcEE 2  2

35.58. The gravitational potential at the surface of the Earth – taking the potential to be zero at infinity – is the

same as would be produced by a point mass m at the center of the Earth. Hence, the desired ratio is: 6.674 1011 m 3 / kg s 2 5.9736 10 24 kg  Gm     10 22  2 6.962  10 , ccr 86  2.998 10 m/s 6.371 10 m a dimensionless quantity. The deviation from flat space-time geometry produced by the Earth’s gravitation is rather small. 35.59. (a) Using the formula for the Schwarzschild radius, the Schwarzschild radius corresponding to the mass of the Sun is

1253 Bauer/Westfall: University Physics, 1E

11 3 2 30 2GM 2 6.674 10 m / kg s 1.989 10 kg S  rS 222.954 km, c 2.998 108 m/s a characteristic size scale for stellar-mass black holes. (b) The Schwarzschild radius corresponding to a proton mass is

11 3 2 27 2Gm 2 6.674 10 m / kg s 1.673 10 kg P 54 rS 222.485 10 m. c 2.998 108 m/s This is much smaller than the femtometer size scale usually associated with protons: it is orders of magnitude smaller than the Planck scale (see Chapter 39), generally considered the smallest scale on which our basic notions of length make sense. Hence, it is unlikely that a proton could usefully be described via a classical black-hole geometry.

1 2 35.60. The time dilation between the Earth and the satellite is tt00 1.   t The difference is 2 2 3 114.00 10 m/s  2118.90 10 s/ Earth second 89.0 ps/ Earth second . 222.9979 108 m/s  This corresponds to a difference of 7.69 106 s/day 8 μs/day.

2GM 35.61. The Schwarzschild radius of a black hole is R  . The black hole at the center of the Milky Way in S c2 Example 12.4 was found to be 3.72 106 solar masses. The mass of the Sun is 1.989 1030 kg. The Schwarzschild radius of this black hole is 2 6.674 1011 N m 2 / kg 2 3.72 10 6 1.989 10 30 kg     6 AU RS 2910.99 10 m 2.9979 108 m/s 149.60 10 m  0.0735 AU.

Additional Problems

35.62. In the garage’s reference frame, the limousine is length contracted. The speed required for it to fit into the garage is

L 222 L0 L1/  vc  LL /  1/  vc  00

2 2 vLLc1 /0  1 35.0 ft / 50.0 ft cc  0.71 . In the limousine’s reference frame, the length of the garage is length contracted by a factor of  = 50.0/35.0 = 1.43.

2 35.63. The relativistic momentum of an electron is given by pmvmvvcR  ee/1  / where me is the mass

of the electron. The classical momentum is pmvC  . Therefore, the percentage difference between the classical and relativistic momenta is pp mvmv p RC100 % ee 100 %   1 100 % pmvC e For an electron moving at vc2.00 108 m/s  2 / 3 ,

1254 Chapter 35: Relativity

22  1/ 1vc /  1/ 1 2 / 3  1.342. Its relativistic momentum is 31 8 22 pmvR  ee1.342 mc (2 / 3)  0.8944 9.109  10 kg 3.00  10 m/s  2.44 10 kg m/s, which differs from its classical value by p  1100%   34%. For an electron moving at vc2.00 1035 m/s  2.00  10 / 3.00 ,

2 2  1/ 1vc /  1/ 1 2.00 105 /3.00  1.000 Its relativistic momentum is 5531827 pmvR  ee1.000 m 2.00  10 / 3 c  9.109  10 2.00  10 / 3 3.00  10 m/s  1.82 10 kg m/s. This does not differ appreciably from its classical value since  1100%   0 to many decimal places. For small velocities, the classical momentum of the electron is a good approximation. 35.64. Let the Earth be frame F and rocket A be the moving frame F . The speed of rocket B in frame F is then uc 0.95 . The speed of frame F with respect to frame F is vc 0.75 . The speed of rocket B relative to rocket A is then uv 0.95cc   0.75  uc  0.70 . 1uv / c22 1 0.95 c 0.75 c / c

2 2 35.65. The Newtonian and relativistic kinetic energies of a particle are KmvN  1/2 and KmcR  1, respectively. In Newtonian mechanics, the difference in their kinetic energy is

11122 22 1 2 22 KmvmvmvvN12 120.9999  0.9900 mc 222 2 1 0.999922 0.9900 0.511 MeV/cc 22  5.03 keV. 2 The difference using special relativity is  22 211 2 KmcmcmcR111   2    12    mc 22 1/vc 1/ vc 12 11 0.511 MeV  32.5 MeV. 22 1 0.9999 1 0.9900 35.66. (a) The clock of the friend waiting in B will show a longer time interval due to time dilation. The person traveling experiences time “slowing down” relative to a stationary observer.

(b) The time dilation is given by tt 0 . Since the velocity of the airplane is small compared to the 1 speed of light,  can be approximated as  1. 2 The difference in time between the two clocks is 2 2 112402  m/s tt000 t t  t 03.00 h 3600 s/h  3.5 ns. 223.00 108 m/s

35.67. The mass can be found from the energy:

15.0 4.00 1012 J 24E     Emc m 22 6.78 10 kg  0.678 g. c 3.00 108 m/s

1255 Bauer/Westfall: University Physics, 1E

35.68. The speed can be found using the equation for length contraction:

2 L0 22 2 290.0 cm L L0001/  vc   vc/  1/  LL   v 1/  LL  c  1 c  0.436. c  100. cm

35.69. Using the relativistic velocity transformation, the speed of object A relative to object B as measured by an observer on object B is vv 0.600cc  0.600   AB uc22 0.882 . 1vvAB / c 1 0.600 c 0.600 c / c

35.70. The length contraction factor is one-third so   3. Therefore, the relative velocity is

111822 31/vc   1 vc /  v c 2 1/ vc 99 3

35.71. The average speed on the trip, which took 40.0 hours to travel 2200.0 miles, was 55.0 mph. Since the 1 velocity of the vehicle is small compared to the speed of light,  can be approximated as  1. 2 2 Therefore, the difference in time between your watch and your professor’s watch (your watch runs slow) is 2 1 h 1609.3 m 55.0 mph 112 3600 s 1 mi tt000 t t  t 0 40.0 h 3600. s/h  0.484 ns 223.00 108 m/s   This amount of time is very tiny and could not be a reason for being late. 35.72. Because of the second postulate of relativity, both observers measure the speed of light to be the same. (a) The speed of light measured on the spaceship is c. (b) The speed of light measured on the asteroid is also c. 35.73. The distance of 100. ly was measured by someone on one of the space stations. Someone on the spaceship will measure a different distance, one that is shorter according to the formula for length contraction,

LL 0 /. The time it takes to travel from one space station to the next as measured by someone on the spaceship is

L LL 22100. ly tvccc00 1  /  1  0.950 /  32.8684 years  32.9 years. 1 vvv 0.950 c As seen by someone on the space station, the time will be L 100. ly t  105 years. 2 vc0.950

35.74. The electron gains kinetic energy from the potential: KmcqV 1. 2  Solving for the velocity v : 2 qV11 qV qV 1mc2 qV 1  1  1 22222  mc1/ vc mc1/ vc  mc

2 2 1 qV  1/vc 22  v  11   c . qV mc 1 2 mc

1256 Chapter 35: Relativity

The rest mass energy of the electron is mc2  0.511 MeV and the potential energy is qV e1.0 106 V  1.0 MeV. Thus the electron attains a speed of

2 1.0 MeV vcc11  0.94.  0.511MeV

L 35.75. As seen by those on the ship, the round trip distance is length contracted to L  0 , where 

Lc0 / 4000.0 yr. If the speed of the ship is v and the journey must take only t  40.000 yr then the required speed is

LL00 22222 c vvcctLvcvv1/   /0  tt 2 1/ ct L0

c vc0.99995 . 2 1  40.000 yr / 4000.0 yr

35.76. THINK: The particle is moving at a speed of vc 0.800 . The mass of the particle is unknown, but the momentum of the particle is p 1.00 1020 N s. This is all that is required to find the energy of the particle. SKETCH:

RESEARCH: The energy and momentum of a relativistic particle are Emc  2 and pmv  respectively. ppc2 SIMPLIFY: Emcmc 22  mv v 20 8 pc2 1.00 10 N s 2.9979  10 m/sc CALCULATE: E  3.747  1012 J  23.392 MeV vc0.800 ROUND: To three significant figures, the energy of the particle is 3.75 1012 J or 23.4 MeV. DOUBLE-CHECK: This is a typical energy for a high energy particle. For vc 0.800 , the value of  is found to be 5/3. Hence the mass of the particle is mpv/0.2510 28 kg, which is a reasonable mass for an atomic particle. Using this mass, the energy of the particle is Emc 2288212(5/ 3)(0.25  10 kg)(2.9979  10 m/s)  3.747 10 J, which agrees with the calculated value. 35.77. THINK: The running back is travelling at 55.0% the speed of light relative to the field. He throws the ball to a receiver running at 65.0% the speed of light relative to the field in the same direction. The speed of the ball relative to the running back is 80.0% the speed of light. The relativistic velocity transformation can be used to find the speed that the receiver perceives the ball to be travelling at. Recall that the speed of light is the same in all reference frames.

1257 Bauer/Westfall: University Physics, 1E

SKETCH:

RESEARCH: The velocity of the ball with respect to the running back is ucx  0.800 . The velocity of the

running back with respect to the field is vcrb  0.550 . The inverse Lorentz transformation can be used to

find the velocity ux of the ball in the field frame: uv   x,rb rb ux 2 . 1/ uvx ,rb rb c Using a Lorentz transform gives the speed of the ball relative to the receiver: uv   x rec ux,rec 2 , 1/ uvx rec c

where vcrec  0.650 is the velocity of the receiver relative to the field. SIMPLIFY: Not required. CALCULATE: 0.800cc   0.550  0.9375cc   0.650  (a) ucuc0.9375 0.7360 xx1  0.800ccc 0.550 / 2 ,rec 1  0.9375ccc 0.650 / 2 (b) Photons travel at the speed of light and the speed of light is the same in any reference frame; therefore, the photons would appear to be travelling at the speed of light to the receiver. ROUND: (a) To three significant figures, the speed of the ball perceived by the receiver is 8 ucx,re c 0.736 2.21 10 m/s. DOUBLE-CHECK: The calculated value of the football’s relative speed was less than the speed of light as it must be, since no massive object can travel at the speed of light.

14 35.78. THINK: The C electrons have kinetic energy KE 0.3050 , where E0 is the rest energy. The baseline between the detectors is x 2.0 m. Find the necessary timing accuracy needed by the detectors to show that the expression for the relativistic momentum, and not the expression for the non-relativistic momentum, is correct. SKETCH: Not required. 2 RESEARCH: The rest energy is Emc0  . The non-relativistic momentum is pmvnr e nr , and the non- 2 relativistic kinetic energy is Kmvnr 1/2 e nr , where me is the electron’s mass. The non-relativistic

velocity vnr can be determined from these equations. The relativistic momentum is pmvrer  , where

2  1/ 1vcr / . The relativistic kinetic energy is KE 1. 0 The relativistic velocity vr can be determined from these equations. Finally, the time needed to travel a distance x is txv /. SIMPLIFY: Non-relativistic case: 1 KmvEv2 0.305  2 0.305 Em / nr2 e nr 0 nr 0 e 2 Substituting Emc0e into the equation gives

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2 x x vmcmctnr2 0.305 e / e 0.610 nr vnr 0.610c Relativistic case:

KEE1 00  0.305  1.305 2 1 22c 22 2  1.305 1.305 1.305 cvr c 2 22  cv r 1/ vcr

22 22 1.305 1xx 1.305  1.305vcvcct22 1.305 1 0.6425 rr  22 r 1.305vcr  1.305 1

2 2.0 m 2.0 m 1.305  CALCULATE: tnr 8.5358 ns, tr 2 10.376 ns 0.610 3.00 108 m/s 3.00 108 m/s 1.305 1

ROUND: To two significant figures, tnr  8.5 ns and tr 10. ns. By comparison of the calculated values

tnr and tr , the necessary timing accuracy is on the order of 1 ns.

DOUBLE-CHECK: The calculated values for tnr and tr had the correct units.

35.79. THINK: The spacecraft travels a distance of d 1.00 103 ly in a time of t  20.0 hrs as measured by

an observer stationed on Earth. The length of the journey, t0 , as measured by the captain of the spacecraft will be shorter due to time dilation. SKETCH: Not required. RESEARCH: The speed of the spacecraft is given by vd / t . The expression for time dilation is given

by tt 0 .

2 t0 2 SIMPLIFY: tttvcttdct  00  1/    1/   2  1/ vc CALCULATE: Since dc/1.00103 yr,

2 1.00 1033 yr 8.766 10 hr/yr  t0 20.0 hr 1   17.977 hr. 20.0 hr 

ROUND: To three significant figures, t0 18.0 hr. DOUBLE-CHECK: The time measured by the captain is shorter than the time measured by the observer on the Earth. This makes sense because the captain is traveling at the same speed as the spacecraft (e.g. the captain is at rest with respect to the spacecraft). According to the time dilation theory, a moving clock runs slower than a clock at rest.

35.80. THINK: A hypothetical particle with rest mass mc1.000 GeV/ 2 and kinetic energy K  1.000 GeV collides with an identical particle at rest. The two particles fuse to form a single new particle. Total energy and momentum are both conserved in the collision. Find (a) the momentum p and speed v of the first

particle and (b) the rest mass mnew and speed vnew of the new particle. SKETCH:

1259 Bauer/Westfall: University Physics, 1E

2 2 RESEARCH: The total energy is EmcEK  0 , where Emc0 1.000 GeV and KE 1. 0 The relationship between energy and momentum is given by Epcmc22224. SIMPLIFY: (a) The momentum of the first particle is given by 22 22224222 2EE 0 EpcmcpcE 0 p c2 2 22 pEKEcEKKc000/2  /. The speed of the first particle is given by  EE2 1 00 KEKEvc100 1 . 222 KE 0 1/vc 1/ vc 

(b) The rest mass mnew of the new particle can be found by using the relationship between energy and momentum: Epc222 Epcmcm222242  new new  m Epcc 2222/. new new new newc4 new new new By energy and momentum conservation, the newly formed particle has the same total energy and momentum as the two original particles did prior to the collision, so

EEKnew 23.000 0  GeV,

and ppnew  , which was found in part (a). The speed of the new particle is given by: 22 222 mvnew new2 mv new new mcv new new pmvnew new new  p new 2 222 1/ vc cv new 1/ vcnew new

22 2 2 222 pcnew pcnew pv new new mcv new new  v new . 22 2 mcnew p new CALCULATE:

2 (a) pcc2 1.000 GeV 1.000 GeV 1.000 GeV / 1.73205 GeV/

2 1.000 GeV vcc1  0.86603 2  1.000 GeV 1.000 GeV

2 2 22 2 (b) mccccnew 3.000 GeV 1.73205 GeV/ /  2.44949 GeV/ 1.73205 GeV/cc vcnew 0.57735 2 2 2.44949 GeV/cc22  1.73205 GeV/ c ROUND: To four significant figures, (a) pc1.732 GeV/ , vc 0.8660 2 (b) mcnew  2.449 GeV/ , vcnew  0.5774 DOUBLE-CHECK: The mass of the new particle is on the same order as the mass of a proton, 2 mcp  0.938 GeV/ , so it is reasonable. The calculated speeds are large, but are realistic for small masses.

35.81. THINK: In considering accelerating bodies with special relativity, the acceleration experienced by the moving body is constant; that is, in each increment of the body’s own proper time, d , the body acquires

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velocity increment dv gd as measured in the body’s frame (the inertial frame in which the body is momentarily at rest). Given this interpretation, (a) Write a differential equation for the velocity v of the body, moving in one spatial dimension, as measured in the inertial frame in which the body was initially at rest (the “ground frame”). (b) Solve this equation for vt , where both v and t are measured in the ground frame. (c) Verify that the solution behaves appropriately for small and large values of t. (d) Calculate the position of the body xt , as measured in the ground frame. (e) Identify the trajectory of the body on a Minkowski diagram with coordinates x and ct, as measured in the ground frame. (f) For g  9.81 m/s2 , calculate how much time t it takes the body to accelerate from rest to 70.7% of c, as measured in the ground frame, and how much ground-frame distance, x, the body covers in this time. SKETCH: Not required. RESEARCH: In moving from the ground frame to the next frame, the body’s velocity was incremented by dv. Since we are interested in a differential equation for the velocity as measured in the ground frame, an inverse Lorentz transformation from the next frame to the ground frame is necessary: uv vdv next uvdvground 22 . 1/uvcnext 1/ vdvc The increment of the body’s proper time d is related to the increment of ground-frame time dt by time 1/2 2 dilation, dvcdt 1/ . The trajectory of the body in a space-time diagram will be determined by examining the position as a function of time, which is determined in part (d). SIMPLIFY: (a) Ignoring squares and higher powers of differentials,

vgd  vgd 2 2 vdv  vgd1   ...  vg 1 vc / d  ..., or dv g1/ v c d . 1/ vgd c22 c     But the increment of proper time d is related to the increment of ground-frame time dt by time dilation so the differential equation, in terms of ground frame quantities, becomes 1/2 222 dvgvcdgvcvcdt1/   1/ 1/  3/2 dv 2 gvc1/ dt   (b) The above differential equation separates, yielding

tvt dv vt   gdt3/2 gt 1/2 . 00 2 2 1/ vc  1/ vt c gt This is readily solved, giving vt 1/2 for the ground-frame velocity of the accelerating body 2 1/ gt c  as a function of ground-frame time. (c) For gt c, i.e., the Newtonian limit, the above result takes the form vt   gt, exactly as expected. The relativistic limit, as time approaches infinity is: gt gt limvt lim c . tt  22 1/ gt c gt / c

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That is, the velocity of the accelerating body asymptotically approaches c, as expected. (d) The position follows from the velocity through integration: t 1/2 1/2 22 22222 ccttgt dt ccc  gt    gt        xt vt dt 1/2 11   gg002 ggcgc 1/ gt c      0 2 (e) The above result implies the simple relation x242ct c/. g The right-hand side is constant. Hence, the trajectory is a branch of a hyperbola on a Minkowski diagram. (f) Consider the ground-frame speed as a function of ground-frame time from part (b), gt vt . 2 1/ gt c The time t required for the body to accelerate from rest to vc 0.707 is given by:

gt 22 v vvgtcgtt2 1/ . 2   2 1/gt c g 1/ v c The ground-frame distance travelled in this time is xxtx  0. As stated in the problem, the ground-frame position at ground-frame time t  0 is x0/.  cg2 Then

1/2 2 2222  cccgt vc/ x 111.    2  gc gg1/ vc   CALCULATE: 0.707 2.998 108 m/s (f) t 3.055 107 s 353.6 days, 2 9.81m/s2 1  0.707cc /

2 2 8  2.998 10 m/s 0.707cc / x 1   1  3.793  1015 m  0.4009 ly 2 2 9.81 m/s 10.707/ cc  ROUND: The answers should be quoted to three significant figures: (f) t  354 days, and x 0.401 ly. DOUBLE-CHECK: The motion of an object with constant proper acceleration in special relativity should be described by a hyperbola, as found in parts (d) and (e). The values found in part (f) are reasonable considering the relatively slow acceleration of 9.81m/s2 .

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