ty of Simultaneity
lightning bolts shown in Fig. 37.5a are
ver on the train. Show that they are not
ver on the ground. Which lightning strike
measure to come first?
ty of Time Intervals
on 1m+2, an unstable particle, lives on
measured in its own frame of reference)
uch a particle is moving, with respect to
eed of 0.900c, what average lifetime is
ry? (b) What average distance, measured
e particle move before decaying?
ocket travel relative to the earth so that time
” to half its rate as measured by earth-based
jet planes approach such speeds?
past Mars with a speed of 0.985c relative
net. When the spaceship is directly overe Martian surface blinks on and then off.
easures that the signal light was on for
server on Mars or the pilot on the spaceime? (b) What is the duration of the light
ot of the spaceship?
n 1p-2 is an unstable particle with an
* 10 -8 s (measured in the rest frame of
is made to travel at very high speed relaverage lifetime is measured in the labora7
s. Calculate the speed of the pion
of c. (b) What distance, measured in the
travel during its average lifetime?
r space utility vehicle at a constant speed
pilot flies past you in her spaceracer at a
Coruscant at a speed of 0.600c. A scientist on Coruscant measures
the length of the moving spacecraft to be 74.0 m. The spacecraft
later lands on Coruscant, and the same scientist measures the
length of the now stationary spacecraft. What value does she get?
37.10 . A meter stick moves past you at great speed. Its motion
relative to you is parallel to its long axis. If you measure the length
of the moving meter stick to be 1.00 ft (1 ft = 0.3048 m2 —for
example, by comparing it to a 1-foot ruler that is at rest relative to
you—at what speed is the meter stick moving relative to you?
37.11 .. Why Are We Bombarded by Muons? Muons are
unstable subatomic particles that decay to electrons with a mean
lifetime of 2.2 ms. They are produced when cosmic rays bombard
the upper atmosphere about 10 km above the earth’s surface, and
they travel very close to the speed of light. The problem we want
to address is why we see any of them at the earth’s surface.
(a) What is the greatest distance a muon could travel during its
2.2-ms lifetime? (b) According to your answer in part (a), it would
seem that muons could never make it to the ground. But the 2.2-ms
lifetime is measured in the frame of the muon, and muons are moving very fast. At a speed of 0.999c, what is the mean lifetime of a
muon as measured by an observer at rest on the earth? How far
would the muon travel in this time? Does this result explain why
we find muons in cosmic rays? (c) From the point of view of the
muon, it still lives for only 2.2 ms, so how does it make it to the
ground? What is the thickness of the 10 km of atmosphere through
which the muon must travel, as measured by the muon? Is it now
clear how the muon is able to reach the ground?
37.12 . An unstable particle is created in the upper atmosphere
from a cosmic ray and travels straight down toward the surface of
the earth with a speed of 0.99540c relative to the earth. A scientist
at rest on the earth’s surface measures that the particle is created at
an altitude of 45.0 km. (a) As measured by the scientist, how much
time does it take the particle to travel the 45.0 km to the surface of
the earth? (b) Use the length-contraction formula to calculate the
distance from where the particle is created to the surface of the
earth as measured in the particle’s frame. (c) In the particle’s
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 1 − u 2 /c 2 gives u = c 1 − (l/l0 ) 2 = c 1 − (0.3048/1.00)2 = 0.9524c = 2.86 × 108 m/s.
37.11.
IDENTIFY and SET UP: The 2.2 µs lifetime is ∆t0 and the observer on earth measures ∆t. 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.
(b) ∆t =
∆t0
2
1 − u /c
2
=
2.2 × 10−6 s
1 − (0.999)
2
= 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 1 − u 2 /c 2 = (10 km) 1 − (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, ∆t0 .
EXECUTE: (a) t =
l0
45.0 × 103 m
=
= 1.51 × 10−4 s
v (0.99540)(3.00 × 108 m/s)
(b) l = l0 1 − u 2 /c 2 = (45.0 km) 1 − (0.99540) 2 = 4.31 km
(c) time dilation formula: ∆t0 = ∆t 1 − u 2 /c 2 = (1.51 × 10−4 s) 1 − (0.99540) 2 = 1.44 × 10−5 s
l
4.31 × 103 m
=
= 1.44 × 10−5 s
8
v (0.99540)(3.00 × 10 m/s)
The two results agree.
from ∆l : t =
(b) v¿ = 0.900c; (c) v¿ = 0.990c?
37.16 . Space pilot Mavis zips past Stanley at a constant speed
relative to him of 0.800c. Mavis and Stanley start timers at zero
when the front of Mavis’s ship is directly above Stanley. When
Mavis reads 5.00 s on her timer, she turns on a bright light under
the front of her spaceship. (a) Use the Lorentz coordinate transformation derived in Example 37.6 to calculate x and t as measured
by Stanley for the event of turning on the light. (b) Use the time
dilation formula, Eq. (37.6), to calculate the time interval between
the two events (the front of the spaceship passing overhead and
turning on the light) as measured by Stanley. Compare to the value
of t you calculated in part (a). (c) Multiply the time interval by
Mavis’s speed, both as measured by Stanley, to calculate the distance she has traveled as measured by him when the light turns on.
Compare to the value of x you calculated in part (a).
37.17 .. A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an
observer on Tatooine, the cruiser is traveling away from the planet
with a speed of 0.600c. The pursuit ship is traveling at a speed of
0.800c relative to Tatooine, in the same direction as the cruiser.
(a) For the pursuit ship to catch the cruiser, should the velocity of
the cruiser relative to the pursuit ship be directed toward or away
from the pursuit ship? (b) What is the speed of the cruiser relative
to the pursuit ship?
37.18 . An extraterrestrial spaceship is moving away from the
earth after an unpleasant encounter with its inhabitants. As it
departs, the spaceship fires a missile toward the earth. An observer
on earth measures that the spaceship is moving away with a speed
of 0.600c. An observer in the spaceship measures that the missile
is moving away from him at a speed of 0.800c. As measured by an
observer on earth, how fast is the missile approaching the earth?
37.19 .. Two particles are created in a high-energy accelerator
and move off in opposite directions. The speed of one particle, as
measured in the laboratory, is 0.650c, and the speed of each particle relative to the other is 0.950c. What is the speed of the second
particle, as measured in the laboratory?
37.20 .. Two particles in a high-energy accelerator experiment
are approaching each other head-on, each with a speed of 0.9520c
as measured in the laboratory. What is the magnitude of the veloc-
that the rocket is approaching with a speed of 0.3
speed of the spaceship relative to Arrakis? Is the s
toward or away from Arrakis?
Section 37.6 The Doppler Effect
for Electromagnetic Waves
37.24 . Electromagnetic radiation from a star is
earth-based telescope. The star is moving away fro
speed of 0.600c. If the radiation has a frequency o
in the rest frame of the star, what is the frequency
observer on earth?
37.25 . Tell It to the Judge. (a) How fa
approaching a red traffic light 1l = 675 nm2 for
low 1l = 575 nm2? Express your answer in term
light. (b) If you used this as a reason not to get a
a red light, how much of a fine would you g
Assume that the fine is $1.00 for each kilometer p
speed exceeds the posted limit of 90 km>h.
37.26 . A source of electromagnetic radiation
radial direction relative to you. The frequency yo
times the frequency measured in the rest frame of
is the speed of the source relative to you? Is th
toward you or away from you?
Section 37.7 Relativistic Momentum
37.27 . A proton has momentum with magnit
speed is 0.400c. In terms of p0, what is the magn
ton’s momentum when its speed is doubled to 0.8
37.28 . When Should You Use Relativity? A
relativistic calculations usually involve the quan
appreciably greater than 1, we must use relativistic
of Newtonian ones. For what speed v (in terms of
g (a) 1.0% greater than 1; (b) 10% greater than 1;
than 1?
37.29 . (a) At what speed is the momentum of a
great as the result obtained from the nonrelativisti
Express your answer in terms of the speed of lig
applied to a particle along its direction of motion.
the magnitude of force required to produce a g
twice as great as the force required to produce the
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′ = 0 and t ′ = 5.00 s,
so from the result of Example 37.6, x = γ ( x′ + ut ′) and
ux′ ⎞
⎛
t = γ ⎜ t ′ + 2 ⎟ ⇒ 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 ∆t0 in Eq. (37.6), so ∆t = γ∆t0 = 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
.
uvx
1− 2
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 1 − (0.600)(0.800)
1− 2
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.
to the pursuit ship?
37.18 . An extraterrestrial spaceship is moving away from the
earth after an unpleasant encounter with its inhabitants. As it
departs, the spaceship fires a missile toward the earth. An observer
on earth measures that the spaceship is moving away with a speed
of 0.600c. An observer in the spaceship measures that the missile
is moving away from him at a speed of 0.800c. As measured by an
observer on earth, how fast is the missile approaching the earth?
37.19 .. Two particles are created in a high-energy accelerator
and move off in opposite directions. The speed of one particle, as
measured in the laboratory, is 0.650c, and the speed of each particle relative to the other is 0.950c. What is the speed of the second
particle, as measured in the laboratory?
37.20 .. Two particles in a high-energy accelerator experiment
are approaching each other head-on, each with a speed of 0.9520c
as measured in the laboratory. What is the magnitude of the velocity of one particle relative to the other?
37.21 .. Two particles in a high-energy accelerator experiment
approach each other head-on with a relative speed of 0.890c. Both
particles travel at the same speed as measured in the laboratory.
What is the speed of each particle, as measured in the laboratory?
37.22 .. An enemy spaceship is moving toward your starfighter
with a speed, as measured in your frame, of 0.400c. The enemy
ship fires a missile toward you at a speed of 0.700c relative to the
Section 37.7 Relativist
37.27 . A proton has mo
speed is 0.400c. In terms o
ton’s momentum when its s
37.28 . When Should You
relativistic calculations usu
appreciably greater than 1, w
of Newtonian ones. For wha
g (a) 1.0% greater than 1; (b
than 1?
37.29 . (a) At what speed
great as the result obtained f
Express your answer in ter
applied to a particle along i
the magnitude of force req
twice as great as the force re
when the particle is at rest
speed of light.
37.30 . As measured in an
in the !x-direction at a spe
(magnitude and direction) i
the - x -direction that has m
magnitude of acceleration d
to a proton that is initially a
u is the speed of S ′ relative to S; this is the speed of particle 1 as measured in the laboratory. Thus
u = +0.650c. The speed of particle 2 in S ′ is 0.950c. Also, since the two particles move in opposite
directions, 2 moves in the − x′-direction and v′x = −0.950c. We want to calculate vx , the speed of particle 2
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
2
1 − 0.6175
1 + uv′x /c
1 + (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 1 to particle 2. In the
lab frame particle 1 is moving in the + x -direction and particle 2 is moving in the − x-direction. Then
u = 0.9520c and vx = −0.9520c. v′x is the velocity of particle 2 relative to particle 1.
EXECUTE: v′x =
vx − u
=
−0.9520c − 0.9520c
= −0.9988c. The speed of particle 2 relative to
1 − uvx /c 2 1 − (0.9520c)(−0.9520c )/c 2
particle 1 is 0.9988c. v′x < 0 shows particle 2 is moving toward particle 1.
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.
matter reactor, how much mass of matter and antimatter fuel
would be consumed yearly? (b) If this fuel had the density of iron
(7.86 g>cm32 and were stacked in bricks to form a cubical pile,
how high would it be? (Before you get your hopes up, antimatter
reactors are a long way in the future—if they ever will be feasible.)
37.40 .. Electrons are accelerated through a potential difference
of 750 kV, so that their kinetic energy is 7.50 * 10 5 eV. (a) What
is the ratio of the speed v of an electron having this energy to the
speed of light, c? (b) What would the speed be if it were computed
from the principles of classical mechanics?
37.41 . A particle has rest mass 6.64 * 10 -27 kg and momentum
2.10 * 10 -18 kg # m>s. (a) What is the total energy (kinetic plus
rest energy) of the particle? (b) What is the kinetic energy of the
particle? (c) What is the ratio of the kinetic energy to the rest
energy of the particle?
37.42 .. A 0.100-mg speck of dust is accelerated from rest to a
speed of 0.900c by a constant 1.00 * 10 6 N force. (a) If the nonrelativistic mechanics is used, how far does the object travel to
reach its final speed? (b) Using the correct relativistic treatment of
Section 37.8, how far does the object travel to reach its final
speed? (c) Which distance is greater? Why?
ment? (b) If each swing takes 1.50 s as
mission control on earth, how long will i
astronaut in the spaceship?
37.49 . After being produced in a colli
particles, a positive pion 1p+2 must trav
tube to reach an experimental area. A p
lifetime (measured in its rest frame) of 2
are considering has this lifetime. (a) How
it is not to decay before it reaches the end
be very close to c, write u = 11 - ¢2c
terms of ¢ rather than u.) (b) The p
139.6 MeV. What is the total energy of t
lated in part (a)?
37.50 .. A cube of metal with sides of
frame S with one edge parallel to the x-axi
has volume a 3. Frame S¿ moves along the
measured by an observer in frame S¿, w
metal cube?
37.51 ... The starships of the Solar Fede
symbol of the federation, a circle, while
Empire are marked with the empire’s s
Relativity
37-11
EVALUATE: The incorrect nonrelativistic expressions for K and p give K = p 2 /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
1
SET UP: The nonrelativistic work-energy theorem is F ∆x = mv 2 − mv02 , and the relativistic formula
2
2
for a constant force is F ∆x = (γ − 1) mc 2 .
EXECUTE: (a) Using the classical work-energy theorem and solving for ∆x, we obtain
m(v 2 − v02 ) (0.100 × 10−9 kg)[(0.900)(3.00 × 108 m/s)]2
∆x =
=
= 3.65 m.
2F
2(1.00 × 106 N)
(b) Using the relativistic work-energy theorem for a constant force, we obtain
∆x =
For the given speed, γ =
∆x =
37.43.
(γ − 1)mc 2
.
F
1
= 2.29, thus
1−0.9002
(2.29 − 1)(0.100 × 10−9 kg)(3.00 × 108 m/s) 2
= 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 2 and the relativistic expression is
K rel = (γ − 1)mc 2 .
EXECUTE: (a) v = 8 × 107 m/s ⇒ γ =
K rel = (γ − 1)mc 2 = 5.65 × 10−12 J.
(b) v = 2.85 × 108 m/s; γ = 3.203.
1
1
= 1.0376. For m = mp , K nonrel = mv 2 = 5.34 × 10−12 J.
2
1 − v 2 /c 2
K rel
= 1.06.
K nonrel
kinetic energy is much greater than its rest energy. (a) What is the
speed of a particle (expressed as a fraction of c) such that the total
energy is ten times the rest energy? (b) What is the percentage difference between the left and right sides of Eq. (37.39) if 1mc222 is
neglected for a particle with the speed calculated in part (a)?
37.54 .. Everyday Time Dilation. Two atomic clocks are
carefully synchronized. One remains in New York, and the other is
loaded on an airliner that travels at an average speed of 250 m>s
and then returns to New York. When the plane returns, the elapsed
time on the clock that stayed behind is 4.00 h. By how much will
the readings of the two clocks differ, and which clock will show
the shorter elapsed time? (Hint: Since u V c, you can simplify
21 - u 2 >c2 by a binomial expansion.)
37.55 . The Large Hadron Collider (LHC). Physicists and
engineers from around the world have come together to build the
largest accelerator in the world, the Large Hadron Collider (LHC)
at the CERN Laboratory in Geneva, Switzerland. The machine will
accelerate protons to kinetic energies of 7 TeV in an underground
ring 27 km in circumference. (For the latest news and more information on the LHC, visit www.cern.ch.) (a) What speed v will protons
reach in the LHC? (Since v is very close to c, write v = 11 - ¢2c
and give your answer in terms of ¢.) (b) Find the relativistic mass,
m rel, of the accelerated protons in terms of their rest mass.
37.56 . CP A nuclear bomb containing 12.0 kg of plutonium
explodes. The sum of the rest masses of the products of the explosion is less than the original rest mass by one part in 10 4. (a) How
much energy is released in the explosion? (b) If the explosion
takes place in 4.00 ms, what is the average power developed by the
bomb? (c) What mass of water could the released energy lift to a
height of 1.00 km?
37.57 . CP C̆erenkov Radiation. The Russian physicist P. A.
C̆erenkov discovered that a charged particle traveling in a solid
with a speed exceeding the speed of light in that material radiates
electromagnetic radiation. (This is analogous to the sonic boom
produced by an aircraft moving faster than the speed of sound in
air; see Section 16.9. C̆erenkov shared the 1958 Nobel Prize for this
given by x¿ 2 = c2t¿ 2. Use the Lorentz coo
transform this equation to an equation in
result is x 2 = c2t 2; that is, the motion ap
frame of reference S as it does in S¿; the
be spherical in both frames.
37.62 . In certain radioactive beta decay
cle (an electron) leaves the atomic nucleu
the speed of light relative to the decaying
moving at 75.00% the speed of light in
frame, find the speed of the emitted electr
tory reference frame if the electron is emi
tion that the nucleus is moving and (b)
from the nucleus’s velocity. (c) In each cas
the kinetic energy of the electron as meas
frame and (ii) the reference frame of the d
37.63 .. CALC A particle with mass m a
constant force F will, according to Newto
to accelerate without bound; that is, as t S
according to relativistic mechanics, the pa
c as t S q . [Note: A useful integral
x> 21 - x 2.]
37.64 .. Two events are observed in a
occur at the same space point, the second
first. In a second frame S¿ moving relative
observed to occur 2.35 s after the first
between the positions of the two events a
37.65 ... Two events observed in a fra
positions and times given by 1x 1, t 12 a
(a) Frame S¿ moves along the x-axis just
events occur at the same position in S¿. S
interval ¢t¿ between the two events is giv
¢t¿ =
B
1¢t22 - a
¢
where ¢x = x 2 - x 1 and ¢t = t 2 ¢x 7 c ¢t, there is no frame S¿ in whic
The difference in the clock readings is
2
⎞
1 u2
1⎛
250 m/s
4
−9
∆t − ∆t0 =
∆t = ⎜⎜
⎟ (1.44 × 10 s) = 5.01 × 10 s. The clock on the plane has the
2
8
2c
2 ⎝ 2.998 × 10 m/s ⎟⎠
shorter elapsed time.
EVALUATE: ∆t0 is always less than ∆t ; 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
− 1⎟ and the relativistic mass
SET UP: The relativistic formula for kinetic energy is K = mc 2 ⎜
⎜
⎟
2 2
⎝ 1 − v /c
⎠
m
is mrel =
.
1 − v 2 /c 2
EXECUTE: (a) K = 7 × 1012 eV = 1.12 × 10−6 J. Using this value in the relativistic kinetic energy formula
⎛
⎞
1
− 1⎟ which gives
and substituting the mass of the proton for m, we get K = mc 2 ⎜
⎜
⎟
2 2
⎝ 1 − v /c
⎠
1
1 − v 2 /c 2
= 7.45 × 103 and 1 −
v2
c
2
=
1
3
(7.45 × 10 )
. Solving for v gives 1 −
2
since c + v ≈ 2c. Substituting v = (1 − ∆)c, we have 1 −
v2
2
=
v2
c
2
=
(c + v)(c − v)
c
2
=
2(c − v)
,
c
2(c − v) 2[c − (1 − ∆)c]
=
= 2∆. Solving for ∆
c
c
c
1
Relativity
1 − v 2 /c 2 (7.45 × 103 )2
=
= 9 × 10−9 , to one significant digit.
gives ∆ =
2
2
1
= 7.45 × 103 , we have
(b) Using the relativistic mass formula and the result that
2 2
1 − v /c
37-15
⎛
⎞
1
= m⎜
⎟ = (7 × 103 ) m, to one significant digit.
⎜
2 2
2 2 ⎟
1 − v /c
⎝ 1 − v /c ⎠
EVALUATE: At such high speeds, the proton’s mass is over 7000 times as great as its rest mass.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
E
⎛ 1 ⎞
No portion of
this material
beU
reproduced,
in any form
or by any
without
from the
m = ⎜ 4in⎟ writing
(12.0 kg).
37.56.
IDENTIFY
andmay
SET
P: The energy
released
is Emeans,
= ( ∆m
)c 2 . ∆permission
Pav publisher.
= .
t
⎝ 10 ⎠
The change in gravitational potential energy is mg ∆y.
mrel =
m
⎛ 1 ⎞
EXECUTE: (a) E = ( ∆m)c 2 = ⎜ 4 ⎟ (12.0 kg)(3.00 × 108 m/s) 2 = 1.08 × 1014 J.
⎝ 10 ⎠
(b) Pav =
E 1.08 × 1014 J
=
= 2.70 × 1019 W.
t 4.00 × 10−6 s
(c) E = ∆U = mg ∆y. m =
E
1.08 × 1014 J
=
= 1.10 × 1010 kg.
g ∆y (9.80 m/s 2 )(1.00 × 103 m)
o kinetic energies of 7 TeV in an underground
mference. (For the latest news and more informasit www.cern.ch.) (a) What speed v will protons
Since v is very close to c, write v = 11 - ¢2c
er in terms of ¢.) (b) Find the relativistic mass,
ted protons in terms of their rest mass.
lear bomb containing 12.0 kg of plutonium
of the rest masses of the products of the explooriginal rest mass by one part in 10 4. (a) How
leased in the explosion? (b) If the explosion
ms, what is the average power developed by the
ass of water could the released energy lift to a
kov Radiation. The Russian physicist P. A.
ed that a charged particle traveling in a solid
ding the speed of light in that material radiates
diation. (This is analogous to the sonic boom
craft moving faster than the speed of sound in
9. C̆erenkov shared the 1958 Nobel Prize for this
the minimum kinetic energy (in electron volts)
ust have while traveling inside a slab of crown
order to create this C̆erenkov radiation?
n with energy E is emitted by an atom with
ls in the opposite direction. (a) Assuming that the
can be treated nonrelativistically, compute the
atom. (b) From the result of part (a), show that
much less than c whenever E is much less than
of the atom.
periment, two protons are shot directly toward
oving at half the speed of light relative to the
t speed does one proton measure for the other
to accelerate without bound; that is, as t S q , v S q . Show that
according to relativistic mechanics, the particle’s speed approaches
c as t S q . [Note: A useful integral is 1 11 - x 22-3>2 dx =
x> 21 - x 2.]
37.64 .. Two events are observed in a frame of reference S to
occur at the same space point, the second occurring 1.80 s after the
first. In a second frame S¿ moving relative to S, the second event is
observed to occur 2.35 s after the first. What is the difference
between the positions of the two events as measured in S¿?
37.65 ... Two events observed in a frame of reference S have
positions and times given by 1x 1, t 12 and 1x 2, t 22, respectively.
(a) Frame S¿ moves along the x-axis just fast enough that the two
events occur at the same position in S¿. Show that in S¿, the time
interval ¢t¿ between the two events is given by
¢t¿ =
B
1¢t22 - a
¢x 2
b
c
where ¢x = x 2 - x 1 and ¢t = t 2 - t 1. Hence show that if
¢x 7 c ¢t, there is no frame S¿ in which the two events occur at
the same point. The interval ¢t¿ is sometimes called the proper
time interval for the events. Is this term appropriate? (b) Show that
if ¢x 7 c ¢t, there is a different frame of reference S¿ in which
the two events occur simultaneously. Find the distance between the
two events in S¿; express your answer in terms of ¢x, ¢t, and c.
This distance is sometimes called a proper length. Is this term
appropriate? (c) Two events are observed in a frame of reference
S¿ to occur simultaneously at points separated by a distance of
2.50 m. In a second frame S moving relative to S¿ along the line
joining the two points in S¿, the two events appear to be separated
by 5.00 m. What is the time interval between the events as measured in S? [Hint: Apply the result obtained in part (b).]
37.64.
m c +F t
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 = 0 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 2 (c 2 − u 2 ). But
γ 2 = c 2 /(c 2 − v 2 ), so γ 2t 2 (c 2 − v 2 ) = c 2t 2 . Therefore, c 2t ′2 − x′2 = c 2t 2 and x′ = c t′2 − t 2 = 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′ =
1 − u 2 /c 2
t1 − ux1/c 2
1 − u 2 /c 2
t2 − ux2 /c 2
, t2′ =
1 − u 2 /c 2
1 − u 2 /c 2
Same point in S ′ implies x1′ = x2′ . What then is ∆t ′ = t2′ − t1′ ?
x1′ = x′2 implies x1 − ut1 = x2 − ut2
EXECUTE:
x2 − x1 ∆x
=
t2 − t1 ∆t
From the time transformation equations,
1
( ∆t − u∆x/c 2 )
∆t ′ = t2′ − t1′ =
1 − u 2 /c 2
∆x
gives
Using the result that u =
∆t
1
∆t ′ =
(∆t − (∆x ) 2 /((∆t )c 2 ))
2
1 − (∆x) /((∆t ) 2 c 2 )
u (t2 − t1) = x2 − x1 and u =
∆t ′ =
∆t ′ =
∆t
(∆t ) 2 − ( ∆x)2 /c 2
( ∆t )2 − (∆x)2 /c 2
2
2
(∆t ) − (∆x) /c
Relativity
x2 − ut2
, x2′ =
2
( ∆t − ( ∆x)2 / ((∆t )c 2 ))
= ( ∆t )2 − (∆x/c) 2 , as was to be shown.
2
or
This equation doesn’t have a physical solution (because of a negative square root) if (∆x / c) 2 > (∆t )37.66.
∆x ≥ c∆t.
(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 2 = t2 − ux2 /c 2
c 2 ∆t
⎛x −x ⎞
⎛ ∆x ⎞
t2 − t1 = ⎜ 2 2 1 ⎟ u so ∆t = ⎜ 2 ⎟ u and u =
∆x
⎝ c
⎠
⎝c ⎠
From the Lorentz transformation equations,
⎛
⎞
1
∆x′ = x2′ − x1′ = ⎜
⎟ ( ∆x − u∆t ).
⎜
2 2 ⎟
⎝ 1 − u /c ⎠
Using the result that u = c 2∆t/∆x gives
1
∆x′ =
(∆x − c 2 (∆t ) 2 /∆x )
1 − c 2 ( ∆t ) 2 /( ∆x) 2
∆x ′ =
∆x ′ =
∆x
2
2
(∆x) − c (∆t )
2
(∆x) 2 − c 2 (∆t ) 2
2
2
( ∆x) − c ( ∆t )
2
(∆x − c 2 (∆t ) 2 /∆x)
= ( ∆x) 2 − c 2 (∆t ) 2
(c) IDENTIFY and SET UP: The result from part (b) is ∆x′ = (∆x) 2 − c 2 ( ∆t )2 .
Solve for ∆t: (∆x′) 2 = (∆x) 2 − c 2 (∆t ) 2
(5.00 m) 2 − (2.50 m) 2
( ∆x) 2 − (∆x′) 2
=
= 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: ∆t =
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.
1
EXECUTE: (a) 80.0 m/s is nonrelativistic, and K = mv 2 = 186 J.
2
(b) K = (γ − 1)mc 2 = 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 =
© 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.
37-19
(e)
l0
γ
=
20.0 m
γ
20.0 m
2.20 × 108 m/s
(f) ∆t0 =
∆t
γ
= 13.6 m.
= 9.09 × 10−8 s.
= 6.18 × 10−8 s
13.6 m
he wavelengths of light emitted by hydrogen
al laboratory conditions is l = 656.3 nm, in the
electromagnetic spectrum. In the light emitted
axy this same spectral line is observed to be
l = 953.4 nm, in the infrared portion of the
st are the emitting atoms moving relative to the
proaching the earth or receding from it?
ng Speed by Radar. A baseball coach uses a
asure the speed of an approaching pitched baseends out electromagnetic waves with frequency
ures the shift in frequency ¢ƒ of the waves
moving baseball. If the fractional frequency
a baseball is ¢ƒ>ƒ0 = 2.86 * 10 -7, what is the
km>h? (Hint: Are the waves Doppler-shifted a
reflected off the ball?)
avel? Travel to the stars requires hundreds or
, even at the speed of light. Some people have
can get around this difficulty by accelerating the
onauts) to very high speeds so that they will age
lation. The fly in this ointment is that it takes a
gy to do this. Suppose you want to go to the
t Betelgeuse, which is about 500 light-years
r is the distance that light travels in a year.) You
onstant speed in a 1000-kg rocket ship (a little
, in reality, is far too small for this purpose. In
ows, calculate the time for the trip, as measured
and by astronauts in the rocket ship, the energy
and the energy needed as a percentage of U.S.
is 1.0 * 10 20 J). For comparison, arrange your
howing vrocket, t earth, t rocket, E (in J), and E (as %
rocket ship’s speed is (a) 0.50c; (b) 0.99c;
e basis of your results, does it seem likely that
will invest in such high-speed space travel any
ship moving at constant speed u relative to us
signal at constant frequency ƒ0. As the spaces, we receive a higher frequency ƒ; after it has
a lower frequency. (a) As the spaceship passes
aneously moving neither toward nor away from
v =
c
+ kV
n
where n = 1.333 is the index of refraction of water. Fizeau called
k the dragging coefficient and obtained an experimental value of
k = 0.44. What value of k do you calculate from relativistic transformations?
CHALLENGE PROBLEMS
37.73 ... CALC Lorentz Transformation
for Acceleration.
Using a method analogous to the one in the text to find the Lorentz
transformation formula for velocity, we can find the Lorentz transformation for acceleration. Let frame S¿ have a constant x-component
of velocity u relative to frame S. An object moves relative to frame
S along the x-axis with instantaneous velocity vx and instantaneous acceleration ax. (a) Show that its instantaneous acceleration
in frame S¿ is
a¿x = ax a1 -
u2
c
b
2
3>2
a1 -
uvx
c2
b
-3
[Hint: Express the acceleration in S¿ as a¿x = dv¿x >dt¿. Then use
Eq. (37.21) to express dt¿ in terms of dt and dx, and use Eq. (37.22)
to express dv¿x in terms of u and dvx. The velocity of the object in S
is vx = dx>dt.] (b) Show that the acceleration in frame S can be
expressed as
ax = a¿x a1 -
u2
c
b
2
3>2
a1 +
uv¿x
c2
b
-3
where v¿x = dx¿>dt¿ is the velocity of the object in frame S¿.
37.74 ... CALC A Realistic Version of the Twin Paradox. A
rocket ship leaves the earth on January 1, 2100. Stella, one of a
pair of twins born in the year 2075, pilots the rocket (reference
frame S¿ ); the other twin, Terra, stays on the earth (reference frame
S ). The rocket ship has an acceleration of constant magnitude g in
its own reference frame (this makes the pilot feel at home, since it
simulates the earth’s gravity). The path of the rocket ship is a
straight line in the +x-direction in frame S. (a) Using the results of
Challenge Problem 37.73, show that in Terra’s earth frame S, the
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
1 + (V/nc)
1+ 2
nc
(1 + V/nc) −1 ≈ (1 − V/nc ). This gives
v ≈ ((cn) + V )(1 − (V/nc)) = ( nc/n) + V − (V/n2 ) − (V 2 /nc) ≈
37.73.
c ⎛
1 ⎞
1 ⎞
⎛
+ ⎜ 1 − 2 ⎟V , so k = ⎜1 − 2 ⎟ . For water,
n ⎝ n ⎠
⎝ 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 2 ). dv′ =
dv
+
2
2
2
dt ′
(1 − uv/c ) (1 − uv/c ) c 2
⎛
⎛ 1 − u 2 /c 2 ⎞
dv′
1
v−u
1
(v − u )u/c 2 ⎞
⎛u ⎞
′
+
=
+
⎟ = dv ⎜⎜
⎜ ⎟ . dv = dv ⎜⎜
2
2 2⎟
⎟
dv 1 − uv/c 2 (1 − uv / c 2 )2 ⎝ c 2 ⎠
(1 − uv/c 2 ) 2 ⎟⎠
⎝ 1 − uv/c
⎝ (1 − uv/c ) ⎠
(1 − u 2 /c 2 )
dv
2 2
1
(1 − uv/c 2 ) 2 dv (1 − u /c )
a′ =
=
= a (1 − u 2 /c 2 )3/2 (1 − uv/c 2 ) −3.
2
2
2
2
dt
γ dt − uγ dx/c
(1 − uv/c ) γ (1 − uv/c )
(b) Changing frames from S ′ → S just involves changing
2
2 3/ 2
a → a ′, v → − v′ ⇒ a = a ′ (1 − u /c )
−3
⎛ uv′ ⎞
⎜⎝1 + 2 ⎟⎠ .
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′ = 0. The acceleration as seen in the rocket is given to be a′ = g , and so the acceleration as measured on
⎛ u2 ⎞
du
the earth is a =
= g ⎜1 − 2 ⎟
⎜ c ⎟
dt
⎝
⎠
(b) With v1 = 0 when t = 0,
dt =
du
1
.
g (1 − u 2 /c 2 )3/2
t1
3/ 2
.
1
v1
∫0 dt = g ∫0
du
(1 − u 2 /c 2 )
. t1 =
3/2
v1
g
1 − v12 /c 2
.
500 light-years
in a year.) You
ket ship (a little
this purpose. In
ip, as measured
ship, the energy
centage of U.S.
n, arrange your
J), and E (as %
50c; (b) 0.99c;
eem likely that
pace travel any
u relative to us
. As the spacey ƒ; after it has
paceship passes
nor away from
, and derive an
e frequency we
ase, successive
ver and so they
[Hint: Express the acceleration in S¿ as a¿x = dv¿x >dt¿. Then use
Eq. (37.21) to express dt¿ in terms of dt and dx, and use Eq. (37.22)
to express dv¿x in terms of u and dvx. The velocity of the object in S
is vx = dx>dt.] (b) Show that the acceleration in frame S can be
expressed as
ax = a¿x a 1 -
u2
c
b
2
3>2
a1 +
uv¿x
c2
b
-3
where v¿x = dx¿>dt¿ is the velocity of the object in frame S¿.
37.74 ... CALC A Realistic Version of the Twin Paradox. A
rocket ship leaves the earth on January 1, 2100. Stella, one of a
pair of twins born in the year 2075, pilots the rocket (reference
frame S¿ ); the other twin, Terra, stays on the earth (reference frame
S ). The rocket ship has an acceleration of constant magnitude g in
its own reference frame (this makes the pilot feel at home, since it
simulates the earth’s gravity). The path of the rocket ship is a
straight line in the + x-direction in frame S. (a) Using the results of
Challenge Problem 37.73, show that in Terra’s earth frame S, the
rocket’s acceleration is
du
u 2 3>2
= ga1 - 2 b
dt
c
where u is the rocket’s instantaneous velocity in frame S. (b) Write
the result of part (a) in the form dt = ƒ1u) du, where ƒ1u) is a
function of u, and integrate both sides. (Hint: Use the integral
given in Problem 37.63.) Show that in Terra’s frame, the time
when Stella attains a velocity v1x is
t1 =
v1x
g21 - v1x2>c2
(c) Use the time dilation formula to relate dt and dt¿ (infinitesimal
time intervals measured in frames S and S¿, respectively). Combine this result with the result of part (a) and integrate as in part
(b) to show the following: When Stella attains a velocity v1x relative to Terra, the time t¿1 that has elapsed in frame S¿ is
t¿1 =
v1x
c
arctanha
b
g
c
Here arctanh is the inverse hyperbolic tangent. (Hint: Use the integral given in Challenge Problem 5.124.) (d) Combine the results of
parts (b) and (c) to find t 1 in terms of t¿1, g, and c alone. (e) Stella
accelerates in a straight-line path for five years (by her clock),
slows down at the same rate for five years, turns around, accelerates for five years, slows down for five years, and lands back on
the earth. According to Stella’s clock, the date is January 1, 2120.
What is the date according to Terra’s clock?
37.75 . . . CP Determining
the Masses of Stars. Many Figure P37.75
of the stars in the sky are actum
ally binary stars, in which two
stars orbit about their common
center of mass. If the orbital
R
speeds of the stars are high
enough, the motion of the stars
can be detected by the Doppler
To the
a binary system are u
circular, and the plan
the line of sight from
37.76 ... CP CALC
Consider the Galile
x¿ = x - vt and t¿ =
magnetic waves in a v
0 2E
where E represents t
using the Galilean tra
found to be
2
v2 0 E1x¿,
a1 - 2 b
c
0x¿ 2
This has a different f
Galilean transformati
all physical laws ha
frames. (Hint: Expre
0>0x¿ and 0>0t¿ by us
part (a), but use
Eqs. (37.21), and sho
same form as in fram
0 2E1
0
Explain why this sho
both frames S and S¿.
37.77 ... CP Kaon
particles can be create
ticles with stationary
1
(1 − uv/c 2 ) 2 dv (1 − u /c )
=
= a (1 − u 2 /c 2 )3/2 (1 − uv/c 2 ) −3.
2
2
2
2
dt (1 − uv/c ) γ (1 − uv/c )
γ dt − uγ dx/c
(b) Changing frames from S ′ → S just involves changing
a′ =
−3
⎛ uv′ ⎞
a → a ′, v → − v′ ⇒ a = a ′ (1 − u 2 /c 2 )3/ 2 ⎜1 + 2 ⎟ .
⎝
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′ = 0. The acceleration as seen in the rocket is given to be a′ = g , and so the acceleration as measured on
⎛ u2 ⎞
du
the earth is a =
= g ⎜1 − 2 ⎟
⎜ c ⎟
dt
⎝
⎠
(b) With v1 = 0 when t = 0,
37-22
1
37du
dtChapter
.
=
g (1 − u 2 /c 2 )3/2
t1
3/ 2
.
1
v1
∫0 dt = g ∫0
du
2
2 3/2
(1 − u /c )
. t1 =
v1
g 1 − v12 /c 2
.
(c) dt ′ = γ dt = dt / 1 − u 2 /c 2 , so the relation in part (b) between dt and du, expressed in terms of dt ′ and
1
du
1
du
.
du, is dt ′ = γ dt =
=
2
2
3/2
© Copyright 2012 Pearson Education, Inc.
under all copyright laws as they currently exist.
g material
(1 − u reserved.
/c ) This
(1 − u 2 /cis2 protected
)2
1 − u 2 /c 2Allg rights
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
c
⎛v ⎞
arctan h ⎜ 1 ⎟ . For those who
g
⎝c⎠
wish to avoid inverse hyperbolic functions, the above integral may be done by the method of partial
1 ⎡ du
c ⎛ c + v1 ⎞
du
du ⎤
′
fractions; gdt ′ =
which
integrates
to
= ⎢
+
t
ln ⎜
,
=
⎟.
1
(1 + u/c)(1 − u/c) 2 ⎣1 + u/c 1 − uc ⎥⎦
2 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
1 − (v1/c) 2 = 1/ cosh ( gt1′/c), using the appropriate indentities for hyperbolic functions. Using this in the
expression found in part (b), t1 =
c tanh ( gt1′/c)
c
= sinh ( gt1′/c), which may be rearranged slightly as
g 1/ cosh ( gt1′/c) g
gt1
v1 e gt1′/c − e − gt1′/c
⎛ gt ′ ⎞
= sinh ⎜ ⎟ . If hyperbolic functions are not used, v1 in terms of t1′ is found to be
=
c e gt1′/c + e − gt1′/c
c
⎝ c ⎠
which is the same as tanh( gt1′/c). Inserting this expression into the result of part (b) gives, after much
algebra, t1 =
c gt1′/c − gt1′/c
(e
−e
), which is equivalent to the expression found using hyperbolic functions.
2g
(c) dt ′ = γ dt = dt / 1 − u 2 /c 2 , so the relation in part (b) between dt and du, expressed in terms of dt ′ and
1
du
1
du
.
du, is dt ′ = γ dt =
=
2
2
3/2
g (1 − u 2 /c 2 ) 2
1 − u 2 /c 2 g (1 − u /c )
c
⎛v ⎞
arctan h ⎜ 1 ⎟ . For those who
g
⎝c⎠
wish to avoid inverse hyperbolic functions, the above integral may be done by the method of partial
1 ⎡ du
c ⎛ c + v1 ⎞
du
du ⎤
fractions; gdt ′ =
= ⎢
+
ln ⎜
, which integrates to t1′ =
⎟.
⎥
(1 + u/c)(1 − u/c) 2 ⎣1 + u/c 1 − uc ⎦
2 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
1 − (v1/c) 2 = 1/ cosh ( gt1′/c), using the appropriate indentities for hyperbolic functions. Using this in the
expression found in part (b), t1 =
c tanh ( gt1′/c)
c
= sinh ( gt1′/c), which may be rearranged slightly as
g 1/ cosh ( gt1′/c) g
gt1
v1 e gt1′/c − e − gt1′/c
⎛ gt ′ ⎞
′
= sinh ⎜ ⎟ . If hyperbolic functions are not used, v1 in terms of t1 is found to be
=
c e gt1′/c + e − gt1′/c
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 5 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 7 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 0 = 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))
⎬⇒ 2
c + (u − v) ⎪
f − (c − (u − v)) = f 02 (c + (u − v))
f− =
f0 ⎪
c − (u − v) ⎭
f+ =
2
2
2
This has a simplest
different case
formofthan
the wave equation
the
S. Hence
a spectroscopic
binaryinstar:
two identical
stars,
gral given in Challenge Problem 5.124.) (d) Combine the results of
Galilean transformation
violates
the
first
relativity
postulate
that
each with mass m, orbiting their center of mass in a circle of radius R.
parts (b) and (c) to find t 1 in terms of t¿1, g, and c alone. (e) Stella
all physical
have
the stars’
sameorbits
form isinedge-on
all inertial
Thelaws
plane
of the
to thereference
line of sight of an
accelerates in a straight-line path for five years (by her clock),
frames. (Hint:
Express
the
derivatives
and
in
terms
of hydrogen
0>0x
0>0t
observer on the earth. (a) The light produced by heated
slows down at the same rate for five years, turns around, accelerby
use
of
the
chain
rule.)
(b)
Repeat
the
analysis
of
0>0x¿ and 0>0t¿
gas in a laboratory on the earth has a frequency of 4.568110 *
ates for five years, slows down for five years, and lands back on
part (a), but
use the Lorentz coordinate transformations,
10 14 Hz. In the light received from the stars by a telescope on the
the earth. According to Stella’s clock, the date is January 1, 2120.
Eqs. (37.21), and show that in frame S¿ the wave equation has the
earth, hydrogen light is observed to vary in frequency between
What is the date according to Terra’s clock?
same form as in frame S: 14
.
.
.
4.567710 * 10 Hz and 4.568910 * 10 14 Hz. Determine whether
37.75
CP Determining
the binary
start¿2
system1 as
a whole
the Masses of Stars. Many Figure P37.75
0 2E1x¿,
t¿2 is moving toward or away from
0 2E1x¿,
- 2 this motion,
= and
0 the orbital speeds of the
the earth, the
of the stars in the sky are actu2 speed of
2
m
0x
c
0t¿
stars. (Hint: The speeds involved are much less than c, so you may
ally binary stars, in which two
use the approximate result
given in Section 37.6.)
= in
u>cvacuum
stars orbit about their common
Explain why this shows that the speed¢ƒ>ƒ
of light
is c in
(b)
The
light
from
each
star
in
the
binary
system
varies from its
center of mass. If the orbital
both frames S and S¿.
R
to itsInminimum
frequency
andnew
back again in
speeds of the stars are high
37.77 ... maximum
CP Kaon frequency
Production.
high-energy
physics,
11.0
Determine
the orbital
radius Rprojectile
and the mass
enough, the motion of the stars
particles can
be days.
created
by collisions
of fast-moving
par- m of each
Give your
answer
for mofinthe
kilograms
and as of
a multiple
of the
can be detected by the Doppler
To the
ticles with star.
stationary
particles.
Some
kinetic energy
the
30
mass of
valueAof R to the
kg.ofCompare
1.99 *the10mass
shifts of the light they emit.
earth
incident particle
is the
usedsun,
to create
the new the
particle.
11
distance
from
the
earth
to
the
sun,
1.50
*
10
(This technique
m.
Stars for which this is the case
proton–proton collision can result in the creation
of a Problems
negative
m
Challenge
1259
actually
usedkaon
in astronomy
to determine the masses of stars.
In
are called spectroscopic binary
kaon 1K -2isand
a positive
1K +2
practice, the problem is more complicated because the two stars in
stars. Figure P37.75 shows the
- identical,
simplest case
of au is
spectroscopic
binary star: two
identical
stars,S. (b) Write a binaryp system
where
the rocket’s instantaneous
velocity
in frame
the orbits are usually not
+ p Sare
p +usually
p + Knot
+ K+
each with mass
their(a)center
of mass
circle
radius
m, orbiting
R. ƒ1u) is a circular, and the plane of the orbits is usually tilted with respect to
the result
of part
in the
form indta =
where
ƒ1u)ofdu,
Calculate
kinetic
The plane of
the
stars’
orbits
is
edge-on
to
the
line
of
sight
of
an
function of u, and integrate both sides. (Hint: Use the (a)
integral
thethe
lineminimum
of sight from
the energy
earth.) of the incident proton
thattime
will allow
to occur
if the second
(target)
...reaction
observer ongiven
the earth.
(a) The37.63.)
light produced
by heated
hydrogen
in Problem
Show that
in Terra’s
frame, the
37.76this
CP CALC
Relativity
and the
Waveproton
Equation. (a)
493.7
MeV,
is
initially
at
rest.
The
rest
energy
of
each
kaon
is
gas in a laboratory
on attains
the earth
has a frequency
of 4.568110 *
when Stella
a velocity
Consider the Galilean transformation along the x-direction:
v1x is
14
and
the
rest
It is use-for electro10 Hz. In the light received from the stars by
and proton
In938.3
frameMeV.
wave equation
x¿energy
= x -ofvteach
t¿ = t. is
S the (Hint:
v1xa telescope on the
work in waves
the frame
in whichis the total momentum is
= vary in frequency between ful here tomagnetic
earth, hydrogen light is observedt 1 to
in a vacuum
14
14 - v1x2>c2
g
21
transformation
must be used to
4.567710 * 10 Hz and 4.568910 * 10 Hz. Determine whether zero. But note that the Lorentz
2
0to
E1x,
t2 in the zero0 2E1x, t2 frame
1
relate
the
velocities
in
the
laboratory
those
the binary star
system
as a whole
moving to
toward
from
(c) Use
the time
dilationis formula
relateordtaway
and dt¿
(infinitesimal
= 0
2does this
2 calculated
2
total-momentum frame.) (b) How
minimum
the earth, the
speed
of
this
motion,
and
the
orbital
speeds
of
the
0x
c
0t
time intervals measured in frames S and S¿, respectively). Comstars. (Hint:bine
Thethis
speeds
involved
areresult
muchof
less
than
so you
may askinetic
result
with the
part
(a)c,and
integrate
in partenergy compare with the total rest mass energy of the crethe electric
field
in theare
wave.
that by
E represents
kaons?where
(c) Suppose
that instead
the two
protons
bothShow
in
use the approximate
given
in attains
Sectiona velocity
37.6.) vatedrela¢ƒ>ƒ = u>c
(b) to showresult
the following:
When
Stella
1x
using
the
Galilean
transformation
the
wave
equation
in
frame
S¿ is
(b) The light
eachthe
startime
in the
binary
system varies
from
tivefrom
to Terra,
has elapsed
in frame
t¿1 that
S¿ isits motion with velocities of equal magnitude and opposite direcfound
to be combined kinetic energy of the two prominimum
maximum frequency to its minimum frequency and back again in tion. Find the
reaction to occur.
How does2 this
11.0 days. Determine the orbital radiuscR and the vmass
m of each tons that will allow2 the
1x
0 2E1x¿, t¿2
0 2E1x¿, t¿2
0 E1x¿, t¿2
1total
2v
v
b
t¿1 = arctanh a
energy+ compare with the
rest
star. Give your answer for m in kilograms
b
= 0
a 1 - 2kinetic
g and as acmultiple of the calculated minimum
2
2
2
0x¿0t¿
30
c
0x¿
c
c
0t¿ 2
mass of the sun, 1.99 * 10 kg. Compare the value of R to the mass energy of the created kaons? (This example shows that
when
colliding beams of particles are used instead of a stationary
distance from
thearctanh
earth toisthe
1.50hyperbolic
* 10 11 m.tangent.
(This technique
Here
the sun,
inverse
(Hint: Use the
inteThis has a different form than the wave equation in S. Hence the
target,
is actually gral
usedgiven
in astronomy
to determine
the masses
of stars. In
in Challenge
Problem 5.124.)
(d) Combine
the results ofthe energy requirements for producing new particles are
(a) Calcul
that will a
is initially
and the re
ful here to
zero. But
relate the
total-mom
kinetic en
ated kaon
motion w
tion. Find
tons that
calculated
mass ener
when coll
target, the
reduced su
(e) After the first acceleration period (of 5 years by Stella’s clock), the elapsed time on earth is
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 7 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 0 = 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))
⎬⇒ 2
c + (u − v) ⎪
f − (c − (u − v)) = f 02 (c + (u − v))
f− =
f0 ⎪
c − (u − v) ⎭
f+ =
(u + v) =
( f + /f 0 ) 2 − 1
2
( f + /f 0 ) + 1
c and (u − v) =
( f −2 /f 02 ) − 1
( f −2 /f 02 ) + 1
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 2 ) mv 2
4 Rv 2 4(5.96 × 109 m)(3.94 × 104 m/s) 2
Relativity
37-23
=
⇒
=
=
= 5.55 × 1029 kg = 0.279 msun .
m
−
2
11
2
2
R
G
(2 R )
6.672 × 10 N ⋅ m /kg
EVALUATE: u and v are both much less than c, so we could have used the approximate expression
∆f = ± f 0vrev /c, where vrev is the speed of the source relative to the observer.
©
CopyrightIDENTIFY
2012 Pearson
All rights
reserved. Thisspecified
material isinprotected
under all copyright laws as they currently exist.
37.76.
andEducation,
SET UP:Inc.Apply
the procedures
the problem.
No portion ofEthis
material
may
be
reproduced,
in
any
form
or
by
any
means,
without
permission
in writing
′
′
′
′) = the
F ( x′, tfrom
f ( xpublisher.
( x′, t ′), t ( x′, t ′))
XECUTE: For any function f = f ( x, t ) and x = x ( x , t ), t = t ( x , t ′), let
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
actum
h two
mmon
rbital
R
high
stars
ppler
To the
emit.
earth
case
m
inary
s the
scopic binary star: two identical stars,
heir center of mass in a circle of radius R.
its is edge-on to the line of sight of an
The light produced by heated hydrogen
earth has a frequency of 4.568110 *
ved from the stars by a telescope on the
bserved to vary in frequency between
.568910 * 10 14 Hz. Determine whether
whole is moving toward or away from
s motion, and the orbital speeds of the
volved are much less than c, so you may
t ¢ƒ>ƒ = u>c given in Section 37.6.)
ar in the binary system varies from its
minimum frequency and back again in
orbital radius R and the mass m of each
m in kilograms and as a multiple of the
10 30 kg. Compare the value of R to the
he sun, 1.50 * 10 11 m. (This technique
my to determine the masses of stars. In
ore complicated because the two stars in
0 E1x¿, t¿2
0x 2
-
1 0 E1x¿, t¿2
c2
0t¿ 2
= 0
Explain why this shows that the speed of light in vacuum is c in
both frames S and S¿.
37.77 ... CP Kaon Production. In high-energy physics, new
particles can be created by collisions of fast-moving projectile particles with stationary particles. Some of the kinetic energy of the
incident particle is used to create the mass of the new particle. A
proton–proton collision can result in the creation of a negative
kaon 1K -2 and a positive kaon 1K +2
p + p S p + p + K- + K+
(a) Calculate the minimum kinetic energy of the incident proton
that will allow this reaction to occur if the second (target) proton
is initially at rest. The rest energy of each kaon is 493.7 MeV,
and the rest energy of each proton is 938.3 MeV. (Hint: It is useful here to work in the frame in which the total momentum is
zero. But note that the Lorentz transformation must be used to
relate the velocities in the laboratory frame to those in the zerototal-momentum frame.) (b) How does this calculated minimum
kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in
motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur. How does this
calculated minimum kinetic energy compare with the total rest
mass energy of the created kaons? (This example shows that
when colliding beams of particles are used instead of a stationary
target, the energy requirements for producing new particles are
reduced substantially.)
∂2E
∂x 2
37.77.
37-24
−
1 ∂2E
c 2 ∂t 2
⎛ v2 ⎞ ∂2E
⎛ v2 1 ⎞ ∂2E ∂2E 1 ∂2E
= γ 2 ⎜1 − 2 ⎟ 2 + γ 2 ⎜ 4 − 2 ⎟ 2 = 2 − 2
= 0.
⎜ 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 ∂ 2 /∂t 2
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
Chapter
37
kaons as well. In this situation the kinetic energy of the protons must equal the total rest energy of the two
kaons.
m
EXECUTE: (a) 2(γ cm − 1)mpc 2 = 2mk c 2 ⇒ γ cm = 1 + k = 1.526. The velocity of a proton in the center of
mp
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
2
No portion of this material may be reproduced, in any γform
cm −or1by any means, without permission in writing from the publisher.
momentum frame is then vcm = c
2
γ cm
= 0.7554c.
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
1
v′ + u
= 3.658 ⇒ K lab = (γ lab − 1) mpc 2 = 2494 MeV.
=
= 0.9619c ⇒ γ lab =
vlab =
2
uv′
2
v
1 + 2 1 + vcm
1 − 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.