Length Contraction (“Moving Rods Contract”) Problem 1 7, page 45 •At what speed does a meter stick move if its length is observed to shrink to 0 5 m? Solution
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Solved Problems in Special Relativity
Charles Asman, Adam Monahan and Malcolm McMillan
Department of Physics and Astronomy
University of British Columbia,
Vancouver, British Columbia, Canada
Fall 1999; revised 2011 by Malcolm McMillan
Given here are solutions to 24 problems in Special Relativity.The solutions were used as a learning-tool for students in the introductory undergraduate course Physics
200Relativity and Quantagiven by Malcolm McMillan at UBC during the 1998 and 1999 Winter Sessions.
The solutions were prepared in collaboration with Charles Asman and Adam Monaham who were graduate students in the Department of Physics at that time. The problems are from Chapter 1Relativityof the course textModern Physicsby Raymond A. Serway, Clement J. Moses and Curt A. Moyer, Saunders College Publishing, 2nd ed., (1997).Standard Inertial FramesWe use the standard inertial framesSandS0which are set up such that thexandx0axes coincide and
theyandy0axes andzandz0axes are parallel. Seen fromS,S0moves in the positivex-direction with speedvand, seen fromS0,Smoves in the negativex0-direction with speedv. Furthermore, it is imagined that in
each inertial frame there is an innite set of recording clocks at rest in the frame and synchronized with each
other. Clocks in both frames are set to zero when the originsOandO0coincide. Time Dilation (\Moving Clocks Run Slow")Problem 1.6, page 45 At what speed does a clock move if it runs at a rate which is one-half the rate of a clock at rest?SolutionWe assume that the clock is at rest inS0. As observed by stationary observers inS, the clock moves in the
positivex-direction with speedv. Text Eq. (1.30): t= t0+vx0c
2 (1) relates the timetmeasured inSwith the timet0measured inS0where =1p12(2) and =v=c(3) Let t0be a time interval measured by an observer at rest inS0. (t0is a proper time. t0is measured when x0= 0.) 1 Let tbe the time interval measured by observers at rest inS. Then t= t0(4) and therefore =s1t0t 2 (5)It follows from Eq. (5) that= 0:866 when t= 2t0.
Eq. (4) indicates that the time interval tmeasured by observers at rest inSis larger than the time interval
t0measured by an observer at rest with respect to the clock. That is, \moving clocks run slow".It is important to note that Eq. (4) relates clock readings on a single clock inS0with clock readings on two
separate clocks inS."Moving clocks run slow" is illustrated by the LIght Pulse Clock. For this clock, a light pulse is directed
along the positivey0axis and re ected back to its starting point. The traversal time is recorded as t0. InS,the clock moves along the positivexaxis with speedv. A stationary observer records the time the light pulse
starts and a second stationary observer, farther to the right along thexaxis, records the time when the light
pulse returns to its starting point. This traversal time is recorded as t. Because of the sideways motion, the
light pulse travels farther inSthan inS0. Since the speed of light is the same in both frames, it follows that
the tis larger than t0. The moving clock runs slow. Length Contraction (\Moving Rods Contract")Problem 1.7, page 45 At what speed does a meter stick move if its length is observed to shrink to 0.5 m?SolutionWe assume that the meter stick is at rest inS0. As observed by stationary observers inS, the meter stick
moves in the positivex-direction with speedv. Text Eq. (1.25): x 0= (xvt) (6) relates the positionx0measured inS0with the positionxmeasured inS. Let x0be the length of the meter stick measured by an observer at rest inS0. (x0is the proper length of the meter stick.) The meter stick is moving with speedvalong thexaxis inS. To determine its length inS, the positionsof the front and back of the meter stick are observed by two stationary observers inSat the same time. The
length of the meter stick as measured inSis the distance xbetween the two stationary observers at t= 0.
Then x0= x(7) where is given by Eq. (2). It follows that =s1xx0 2 (8)It follows from Eq. (8) that= 0:866 when x= x0=2.
Eq. (7) indicates that the length xof an object measured by observers at rest inSis smaller than thelength x0measured by an observer at rest with respect to the meter stick. That is, \moving rods contract".
It is important to note that Eq. (7) compares an actual length measurement inS0(a proper length) with a
length measurement determined at equal times on two separate clocks inS. 2 Decay of aMeson: Time Dilation (\Moving Clocks Run Slow")Problem 1.11, page 45The average lifetime of ameson in its own frame of reference is 26.0 ns. (This is its proper lifetime.)
If themeson moves with speed 0:95cwith respect to the Earth, what is its lifetime as measured by an observer at rest on Earth? What is the average distance it travels before decaying as measured by an observer at rest on Earth?SolutionWe take theSframe to be attached to the Earth and theS0frame to be the rest frame of themeson.
It follows from Eq. (4) that t= 83:3 ns when t0= 26:0 ns andv= 0:95c.The average distance travelled before decaying as measured by an observer at rest on Earth isvt= 24:0m.
Time Dilation for a Slow Moving Object (\Moving Clocks Run Slow")Problem 1.12, page 46An atomic clock is placed in a jet airplane. The clock measures a time interval of 3600 s when the jet moves
with speed 400 m/s.How much larger a time interval does an identical clock held by an observer at rest on the ground measure?
SolutionWe take theSframe to be attached to the Earth and theS0frame to be the rest frame of the atomic clock.
It follows from Eq. (2) that
'1 +2=2 (9) and from Eq. (4) that t= tt0'2t0=2 (10) It follows thatt= 3:2 ns whenv= 400 m=s and t0= 3600 s. Muon Decay: Time Dilation (\Moving Clocks Run Slow")Problem 1.14, page 46The muon is an unstable particle that spontaneously decays into an electron and two neutrinos. If the
number of muons att= 0 isN0, the numberNat timetisN=N0et=(11)
where= 2:20s is the mean lifetime of the muon. Suppose the muons move at speed 0:95c.What is the observed lifetime of the muons?
How many muons remain after traveling a distance of 3.0 km?Solution3
We take theSframe to be attached to the Earth and theS0frame to be the rest frame of the muon. It follows from Eq. (4) that t= 7:046s when t0= 2:2s and= 0:95. A muon at this speed travels 3:0 km in 10:53s. After travelling this distance,Nmuons remain from an initial population ofN0muons whereN=N0et==N0e10:53=7:046= 0:225N0(12)
Length Contraction and RotationProblem 1.15, page 46 A rod of lengthL0moves with speedvalong the horizontal direction. The rod makes an angle0with respect to thex0axis. Determine the length of the rod as measured by a stationary observer.Determine the anglethe rod makes with thexaxis.
SolutionWe take theS0frame to be the rest frame of the rod. A rod of lengthL0inS0makes an angle0with thex0axis. Its projected lengths x0and y0are x0=L0cos0(13) y0=L0sin0(14) In a frameSin which the rod moves at speedvalong thexaxis, the projected lengths xand yare given by Eq. (7) and y0= y(15) which equation follows from text Eq. (1.26): y0=y(16)
The lengthLof the rod as measured by a stationary observer inSisL=p(x)2+ (y)2=L012cos20
1=2:(17)
The rod makes an anglewith thexaxis inSwhere
tan= y=x= tan0:(18) The rod inSappears contracted and rotated. See also text Fig. (1.14) on page 19 of the text.Relativistic Doppler ShiftProblem 1.18, page 46
How fast and in what direction must galaxyAbe moving if an absorption line found at wavelength 550nm (green) for a stationary galaxy is shifted to 450 nm (blue) (a "blue-shift") for galaxyA? How fast and in what direction is galaxyBmoving if it shows the same line shifted to 700 nm (red) (a "red shift")?Solution4
GalaxyAis approaching since an absorption line with wavelength 550nm for a stationary galaxy is shifted
to 450 nm. To nd the speedvat whichAis approaching, we use text Eq. (1.13): f obs=s1 +1fsource(19) and=c=fto write obs=s11 +source(20) from which =2source2obs2source+2obs(21)
It follows that= 0:198 whensource= 550 nm andobs= 450 nm. GalaxyBis receding since the same absorption line is shifted to 700 nm. Proceeding as above, =2obs2source2obs+2source(22)
It follows that= 0:237 whensource= 550 nm andobs= 700 nm. Lorentz Velocity TransformationProblem 1.20, page 46Two spaceships approach each other, each moving with the same speed as measured by a stationary observer
on the Earth. Their relative speed is 0:70c, Determine the velocities of each spaceship as measured by the stationary observer on Earth. SolutionText Eq. (1.32) gives the Lorentz velocity transformation: u0x=uxv1uxv=c2(23)
whereuxis the velocity of an object measured in theSframe,u0xis the velocity of the object measured in the
S0frame andvis the velocity of theS0frame along thexaxis ofS.
We take theSframe to be attached to the Earth and theS0frame to be attached to the spaceship moving to the right with velocityv. The other spaceship has velocityux=vinSand velocityu0x=0:70cinS0.It follows from Eq. (23) that
0:70 =21 +2(24)
solving which yields= 0:41. As measured by the stationary observer on Earth, the spaceships are moving
with velocities0:41c. Lorentz Velocity TransformationProblem 1.22, page 46 A stationary observer on Earth observes spaceships A and B moving in the same direction toward the Earth. Spaceship A has speed 0:5cand spaceship B has speed 0:80c. Determine the velocity of spaceship A as measured by an observer at rest in spaceship B. 5Solution
We take theSframe to be attached to the Earth and theS0frame to be attached to spaceship B moving with velocityv=0:8calong thexaxis. Spaceship A has velocityux=0:50cinS.It follows from Eq. (23) that spaceship A has velocityu0x= 0:50cinS0. Spaceship A moves with velocity
0:5cas measured by an observer at rest in spaceship B.
Speed of Light in a Moving MediumProblem 1.23, page 46The motion of a medium such as water in
uences the speed of light. This eect was rst observed by Fizeau in 1851. Consider a light beam passing through a horizontal column of water moving with velocityv.Determine the speeduof the light measured in the lab frame when the beam travels in the same direction
as the ow of the water. Determine an approximation to this expression valid whenvis small.SolutionWe assume that the light beam and the tube carrying water are oriented along the positivex-direction of
the lab frame. The speed of lightuin the lab frame is related to the speed of lightu0in a frame moving with
the water by text Eq. (1.34): u=u0+v1 +u0v=c2(25) wherevis the speed of the water in the lab frame. Nowu0=c=n, wherenis the index of refraction of water, so u=cn1 +n1 +=n
(26) Using (1 +=n)1'(1=n) (27) it follows that u'cn +v 11n 2 :(28)This equation agrees with Fizeau's experimental result. The equation shows that the Lorentz velocity trans-
formation and not the Galilean velocity is correct for light. Lorentz Velocity Transformations for Two ComponentsProblem 1.25, page 47 As seen from Earth, two spaceships A and B are approaching along perpendicular directions. If A is observed by a stationary Earth observer to have velocityuy=0:90cand B to have velocity u x= +0:90c, determine the speed of ship A as measured by the pilot of ship B.Solution6
We take theSframe to be attached to the Earth and theS0frame to be attached to spaceship B moving with= 0:90 along thexaxis. Spaceship A has velocity componentsux= 0;uy=0:90cinS.Eq. (23) and text Eq. (1.33):
u 0y=uy (1uxv=c2)(29) give the velocity components of spaceship A inS0, from which u0x=v=0:90c(30)
u0y=uy=
=0:39c(31) so u0=q(ux0)2+ (uy0)2= 0:98c:(32)
Relativistic Form of Newton's Second LawProblem 1.28, page 47 Consider the relativistic form of Newton's Second Law.Show that whenFis parallel tov
F=m 1v2c 23=2dvdt
(33) wheremis the mass of the object andvis its speed.SolutionThe forceFon a particle with rest massmis the rate of change its momentumpas given by text Eq. (1.36):
F=dpdt
(34) where as given by text Eq. (1.35): p= mv(35) where =1p1v2=c2(36)wherevis the velocity of the particle. Eq. (34) withpgiven by Eq. (35) is the relativistic generalization of
Newton's Second Law.
Now v=vut(37)whereutis a unit vector along the tangent of the particle's trajectory, and the accelerationaof the particle is
a=dvdt =dvdt ut+v2 un(38)whereunis a unit vector orthogonal toutdirected towards the centre of curvature of the trajectory andis
the radius of curvature of the trajectory, so dpdt m 2dvdt ut+v2 un (39) WhenF=Fut(40)
7 that is, whenFis parallel tov, if follows that =1(41) That is, the particle moves in a straight line, and a= ~a= 3(42) where a=dvdt (43) and ~a=Fm (44) Relativistic Form of Newton's Second Law: Particle in an Electric FieldProblem 1.29, page 47 A charged particle moves along a straight line in a uniform electric eldEwith speedv.If the motion and the electric eld are both in thexdirection, show that the magnitude of the acceleration
of the chargeqis given by a=dvdt =qEm 1v2c 2 3=2 (45) Discuss the signicance of the dependence of the acceleration on the speed.If the particle starts from restx= 0 att= 0, nd the speed of the particle and its position after a time
thas elapsed.Comment of the limiting values ofvandxast! 1.
SolutionWhen a particle of chargeqmoves in an electric eldE, the forceFon the particle isF= qE. If the
particle moves in the direction ofE, thenFandvare parallel. Accordingly, Eq. (42) holds with ~a=qE=m.
It follows from Eq. (42) thata!0 asv!cand also thatF'mawhenvc(46)
which is the nonrelativistic result.WhenFis constant and the particle starts from rest att= 0, its speedv(t) is found by integrating Eq. (42):
Z v(t) 0dv0(1v02=c2)3=2= ~at(47)
to be v(t) =~atp1 + (~at=c)2:(48)It follows thatv(t)!cast! 1and also that
v(t)'~atwhen ~atc(49) which is the non-relativistic result. The positionx(t) of the particle is found by integratingv=dx=dt: x(t) =Z t 0 v(t0)dt0=p1 + (~at=c)21 c2=~a:(50) 8It follows thatx!ctast! 1and also that
x(t)'12 ~at2when ~atc(51) which is the nonrelativistic result. The positionx(v) is found by integratinga(v) =dv=dt=vdv=dx: x(v)x(0) =Z v x(0)v0dv0a(v0)= (
1)c2=~a:(52)
It follows thatx(v)x(0)! 1asv!cand also that
v2'2~a[x(v)x(0)] whenvc(53)
which is the nonrelativistic result. Relativistic Form of Newton's Second Law: Particle in a Magnetic FieldProblem 1.30, page 47 The forceFon a particle with rest massmand chargeqmoving with velocityvin a magnetic eldBisF=qvB:(54)
If the particle moves in a circular orbit with a xed speedvin the presence of a constant magnetic eld,
use Newton's Second Law to show that the frequency of its orbital motion is f=qB2m 1v2c 2 1=2 (55) SolutionWhen a particle moves with constant speed, that is, when dvdt = 0 (56) it follows from Eqs. (34) and (39) that v(57) where ~!=qBsinm (58) whereis the angle thatvmakes withB. For a proton moving perpendicular to a 1:00 T magnetic eld, ~!= 95:8 MHz.The right side of Eq. (57) is constant. Accordingly, the radius of curvatureof the particle's trajectory
changes to accommodate changes in the magnetic eldB.WhenBis constant (that is, time-independent and homogeneous), it follows from Eq. (57) that the particle
moves in a circle with radius r= v=~!(59) and speed v= ~!r= =~!rp1 + (~!r=c)2:(60) The angular frequency!=v=rof the circular motion is =~!p1 + (~!r=c)2(61) 9 as above.It follows thatr! 1asv!cand and also that
v'~!rwhenvc(62) which is the nonrelativistic result.It follows also that
!'~!when ~!rc(63) For a proton moving perpendicular to a 1:00 T magnetic eld, this requires thatr3:13 m.The above results limit the range of speeds attainable in a conventional particle-accelerating cyclotron
which relies, as with Eq. (63), on a constant-frequency accelerating potential to increase particle speeds and a
time-independent homogeneous magnetic eld to make particles move in circles. This limitation is overcome at the TRIUMF cyclotron on the UBC campus which accelerates protons to520 MeV (0:75c), and has a diameter of 17:1 m. This is accomplished by increasing the magnetic eld with
radius to accommodate the Lorentz factor . For more information on TRIUMF, see http://www.triumf.ca. Relativistic Form of Newton's Second Law: Particle in a Magnetic FieldProblem 1.31, page 47 Show that the momentum of a particle having chargeemoving in a circle of radiusRis given byp= 300BR wherepis in MeV/c,Bis in teslas andRis in meters.SolutionIt follows from Eqs. (35) and (59) that the momentumpof a particle of chargeemoving perpendicular to
a constant magnetic eldBis p=eBR(64) whereRis the radius of the circular orbit. Usinge= 1:6021019C, MeV = 1:6021013J andc=3:00108m=s, it follows that
p= 300BR(65) wherepis in MeV/c,Bis in teslas andRis in meters. Relativistic Kinematics: Energy-Momentum RelationshipProblem 1.32, page 47 Show that the energy-momentum relationshipE2=p2c2+m2c4follows fromE= mc2andp= mv. SolutionSee alsoNotes on a Few Topics in Special Relativityby Malcolm McMillan.It follows fromp=
mvand Eq. (36) that v=cp1 + (mc=p)2(66) and =p1 + (p=mc)2(67) which withE= mc2yields E=pp2c2+m2c4(68)
It follows from Eq. (66) that a particle with rest massm= 0 travels at the speed of lightc. 10 It follows from Eq. (68) that the energyEand momentumpof a particle with rest massm= 0 are related byE=pc(69)
Relativistic Kinematics for an ElectronProblem 1.40, page 47 Electrons in projection television sets are accelerated through a potential dierence of 50 kV.Calculate the speed of the electrons using the relativistic form of kinetic energy assuming the electrons
start from rest. Calculate the speed of the electrons using the classical form of kinetic energy. Is the dierence in speed signicant in the design of this set? SolutionA particle with rest massmmoving with speedvhas kinetic energyKgiven by text Eq. (1.42): K= (1)mc2(70)
where is given by Eq. (36) from which it follows that v=cp1[1 + (K=mc2)2]:(71)It follows from Eq. (70) that
K'12 mv2(72) whenvc, which is the nonrelativistic result.An electron (rest energy 511keV) moving through a potential dierenceV= 50kV acquires a kinetic energy
K= 50 keV.
It follows from Eq. (71) thatv= 0:413c. The nonrelativistic expression Eq. (72) yieldsv= 0:442cwhich is
6% greater than the correct relativistic result.
Does it make a dierence which number is used when building a TV set? If the distance between thelament o which electrons are boiled and the phosphor screen of a TV set is 50 cm, then the dierence in
travel times calculated relativistically and nonrelativistically is 26ns. This time dierence is too small to make
a signicant dierence in the operation of a TV set. Relativistic Kinematics: Lorentz InvariantProblem 1.41, page 47 The quantityE2p2c2is an invariant quantity in Special Relativity. This means thatE2p2c2has the same value in all inertial frames even thoughEandphave dierent values in dierent frames.Show this explicitly by considering the following case: A particle of massmis moving in the +xdirection
with speeduand has momentumpand energyEin the frameS. IfS0is moving at speedvin the standard way, determine the momentump0and energyE0observed inS0, and show thatE02p02c2=E2p2c2.