Download Modern Physics Solution Textbook and more Exercises Physics in PDF only on Docsity!
Chapter 1. Problem Solutions
- If the speed of light were smaller than it is, would relativistic phenomena be more or less conspicuous than they are now?
- An athlete has learned enough physics to know that if he measures from the earth a time interval on a moving spacecraft, what he finds will be greater than what somebody on the spacecraft would measure. He therefore proposes to set a world record for the 100-m dash by having his time taken by an observer on a moving spacecraft. Is this a good idea?
Sol All else being the same, including the rates of the chemical reactions that govern our brains and bodies, relativisitic phenomena would be more conspicuous if the speed of light were smaller. If we could attain the absolute speeds obtainable to us in the universe as it is, but with the speed of light being smaller, we would be able to move at speeds that would correspond to larger fractions of the speed of light, and in such instances relativistic effects would be more conspicuous.
Sol Even if the judges would allow it, the observers in the moving spaceship would measure a longer time, since they would see the runners being timed by clocks that appear to run slowly compared to the ship's clocks. Actually, when the effects of length contraction are included (discussed in Section 1.4 and Appendix 1), the runner's speed may be greater than, less than, or the same as that measured by an observer on the ground.
- Two observers, A on earth and B in a spacecraft whose speed is 2.00 x 10^8 m/s, both set their watches to the same time when the ship is abreast of the earth. (a) How much time must elapse by A 's reckoning before the watches differ by 1.00 s? (b) To A , B 's watch seems to run slow. To B , does A 's watch seem to run fast, run slow, or keep the same time as his own watch?
Sol Note that the nonrelativistic approximation is not valid, as v / c = 2/3. (a) See Example 1.1. In Equation (1.3), with t representing both the time measured by A and the time as measured in A 's frame for the clock in B 's frame to advance by to, we need
from which t = 3.93 s. (b) A moving clock always seems to run slower. In this problem, the time t is the time that observer A measures as the time that B's clock takes to record a time change of t o.
0255 1 00 s 3
2 2
2 (^0) = ×.^ =.
− = − − t t c
v t t t
- A galaxy in the constellation Ursa Major is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the corresponding wavelength measured by astronomers on the earth? Sol See Example 1.3; for the intermediate calculations, note that
v c
c c v c
o
o
o +
where the sign convention for v is that of Equation (1.8), which v positive for an approaching source and v negative for a receding source. For this problem,
. , .
30 10 m/s
150 10 km/s 8
7 = − ×
×
c
v
so that
578 nm
550 nm
v c
v c
λ λ o
- A spacecraft receding from the earth emits radio waves at a constant frequency of 109 Hz. If the receiver on earth can measure frequencies to the nearest hertz, at what spacecraft speed can the difference between the relativistic and classical Doppler effects be detected? For the classical effect, assume the earth is stationary. Sol This problem may be done in several ways, all of which need to use the fact that when the frequencies due to the classical and relativistic effects are found, those frequencies, while differing by 1 Hz, will both be sufficiently close to v o = 109 Hz so that v o could be used for an approximation to either. In Equation (1.4), we have v = 0 and V = - u , where u is the speed of the spacecraft, moving away from the earth ( V < 0). In Equation (1.6), we have v = u (or v = - u in Equation (1.8)). The classical and relativistic frequencies, v c and v r respectively, are
( / ) u c
u c u c
u c c (^) u c r o o +
2 ν ν^0 ν ν ν
The last expression for v o, is motivated by the derivation of Equation (1.6), which essentially incorporates the classical result (counting the number of ticks), and allows expression of the ratio
. 1 ( / )^2
r u c
c −
ν
ν
- If the angle between the direction of motion of a light source of frequency vo and the direction from it to an observer is 0, the frequency v the observer finds is given by
where v is the relative speed of the source. Show that this formula includes Eqs. (1.5) to (1.7) as special cases. Sol The transverse Doppler effect corresponds to a direction of motion of the light source that is perpendicular to the direction from it to the observer; the angle θ = ±π/2 (or ± 90 o), so cos θ = 0, and which is Equation (1.5). For a receding source, θ = π (or 180o), and cos θ = 1. The given expression becomes
ν = νo 1 − v^2 / c^2 ,
v c
v c v c
v c o o +
ν ν ν
which is Equation (1.8). For an approaching source, θ = 0, cos θ = 1, and the given expression becomes
v c
v c v c
v c o o −
ν ν ν
which is Equation (1.8).
θ
ν ν ( / )cos
v c
v c o (^) −
- An astronaut whose height on the earth is exactly 6 ft is lying parallel to the axis of a spacecraft moving at 0.90 c relative to the earth. What is his height as measured by an observer in the same spacecraft? By an observer on the earth?
- How much time does a meter stick moving at 0.100 c relative to an observer take to pass the observer? The meter stick is parallel to its direction of motion.
Sol The astronaut’s proper length (height) is 6 ft, and this is what any observer in the spacecraft will measure. From Equation (1.9), an observer on the earth would measure
Sol The time will be the length as measured by the observer divided by the speed, or
L = Lo 1 − v^2 / c^2 =( 6 ft) 1 −( 0. 90 )^2 = 2. 6 ft
332 10 s
0100 30 10 m/s
1 1 00 m 101008
8
(^2 22) −
= ×
×
v
L v c
v
L
t o
- A woman leaves the earth in a spacecraft that makes a round trip to the nearest star, 4 light- years distant, at a speed of 0.9 c. Sol The age difference will be the difference in the times that each measures the round trip to take, or
( ) ( 1 1 09 ) 5 yr.
4 yr ∆ = 2 1 − 1 −^2 2 = 2 − −.^2 = .
v / c v
L
t o
- All definitions are arbitrary, but some are more useful than others. What is the objection to defining linear momentum as p = m v instead of the more complicated p = γ m v? Sol It is convenient to maintain the relationship from Newtonian mechanics, in that a force on an object changes the object's momentum; symbolically, F = d p / dt should still be valid. In the absence of forces, momentum should be conserved in any inertial frame, and the conserved quantity is p = - γ m v , not m v
- Dynamite liberates about 5.4 x 10^6 J/kg when it explodes. What fraction of its total energy content is this? Sol For a given mass M , the ratio of the mass liberated to the mass energy is
8 2
6
30 10 m/s
5 4 10 J/kg −
= ×
× ×
× ×
M
M
29. At what speed does the kinetic energy of a particle equal its rest energy? Sol If the kinetic energy K = E o = mc^2 , then E = 2 mc^2 and Equation (1.23) reduces to 2 1
− v^2 / c^2 = (γ = 2 in the notation of Section 1.7). Solving for v ,
260 10 m/s 2
v = 3 c =. ×^8
- An electron has a kinetic energy of 0.100 MeV. Find its speed according to classical and relativistic mechanics. Sol Classically,
188 10 m/s. 911 10 kg
2 2 0200 MeV 160 10 J/eV 8 31
19 = × ×
× × ×
− . .
m e
K
v
Relativistically, solving Equation (1.23) for v as a function of K ,
2
2
2
2
2 2 2
1
K m c
c m c K
m c c E
m c v c e e
e e
- How much work (in MeV) must be done to increase the speed of an electron from 1.2 x 10^8 m/s to 2.4 X 10^8 m/s? Sol The difference in energies will be, from Equation (1.23),
0294 MeV 1 1 2 30
0 511 MeV^1
2 2
2 2 1
2 2 2
2
− v c v c
mec
37. Prove that ½γ mv^2 , does not equal the kinetic energy of a particle moving at relativistic speeds. Sol Using the expression in Equation (1.20) for the kinetic energy, the ratio of the two quantities is
2 2
(^22) 2
1
1 1
c v c
v c
v K
mv γ
γ γ
An alternative derivation of the mass-energy formula E O = mc^2 , also given by Einstein, is based on the principle that the location of the center of mass (CM) of an isolated system cannot be changed by any process that occurs inside the system. Figure 1.27 shows a rigid box of length L that rests on a frictionless surface; the mass M of the box is equally divided between its two ends. A burst of electromagnetic radiation of energy E o is emitted by one end of the box. According to classical physics, the radiation has the momentum p = E o/ c , and when it is emitted, the box recoils with the speed v ≈ E 0 1 Mc so that the total momentum of the system remains zero. After a time t ≈ L / c the radiation reaches the other end of the box and is absorbed there, which brings the box to a stop after having moved the distance S. If the CM of the box is to remain in its original place, the radiation must have transferred mass from one end to the other. Show that this amount of mass is m = E O 1 c^2_._ Sol Measured from the original center of the box, so that the original position of the center of mass is 0, the final position of the center of mass is
2 2 2 2
−^ +
M − m L S M m L S
Expanding the products and canceling similar terms [( M /2)( L /2), mS ], the result MS = mL is obtained. The distance 5 is the product vt , where, as shown in the problem statement, v ≈ E / Mc (approximate in the nonrelativistic limit M >> Elc^2 ) and t ≈ L / c. Then,
c
E
c
L
Mc
E
L
M
L
m = MS = =
- Find the momentum of an electron whose kinetic energy equals its rest energy of 511 keV Sol When the kinetic energy of an electron is equal to its rest energy, the total energy is twice the rest energy, and Equation (1.24) becomes 4 m (^) e^4 c^4 = me^4 c^4 + p^2 c^2 , or p = 3 ( mec^2 )/ c = 3 ( 511 keV/ c )= 1. 94 GeV/ c
The result of Problem 1-29 could be used directly; γ = 2, v = ( /2) c , and Equation (1.17) gives p = m e c , as above.
- Find the speed and momentum (in GeV/ c ) of a proton whose total energy is 3.500 GeV
Sol Solving Equation (1.23) for the speed v in terms of the rest energy E O and the total energy E ,
v = c 1 −( Eo / E )= c 1 −( 0. 938 / 3. 500 )^2 = 0. 963 c
numerically 2.888 x 10^8 m/s. (The result of Problem 1-32 does not give an answer accurate to three significant figures.) The value of the speed may be substituted into Equation (1.16) (or the result of Problem 1-46), or Equation (1.24) may be solved for the magnitude of the momentum,
p = ( E / c )^2 −( Eo / c )^2 = ( 3. 500 GeV/ c )^2 −( 0. 938 GeV/ c )^2 = 3. 37 GeV/ c
- A particle has a kinetic energy of 62 MeV and a momentum of 335 MeV/ c. Find its mass (in MeV/ c^2 ) and speed (as a fraction of c ). Sol From E = mc^2 + K and Equation (1.24),
Expanding the binomial, cancelling the m^2 c^4 term, and solving for m ,
( ) 2 4 2 2 2 2 mc + K = m c + p c
2
2 2 2
2 2 874 MeV 2 62 MeV
335 MeV 62 MeV 2
c c K c
pc K m =
The particle's speed may be found any number of ways; a very convenient result is that of Problem 1-46, giving
c. c. mc K
pc c E
p v c 036 874 MeV 62 MeV
335 MeV 2
- A spacecraft moving in the + x direction receives a light signal from a source in the xy plane. In the reference frame of the fixed stars, the speed of the spacecraft is v and the signal arrives at an angle θ to the axis of the spacecraft. (a) With the help of the Lorentz transformation find the angle θ ' at which the signal arrives in the reference frame of the spacecraft. (b) What would you conclude from this result about the view of the stars from a porthole on the side of the spacecraft?
Sol (a) A convenient choice for the origins of both the unprimed and primed coordinate systems is the point, in both space and time, where the ship receives the signal. Then, in the unprimed frame (given here as the frame of the fixed stars, one of which may be the source), the signal was sent at a time t = - r / c , where r is the distance from the source to the place where the ship receives the signal, and the minus sign merely indicates that the signal was sent before it was received. Take the direction of the ship's motion (assumed parallel to its axis) to be the positive x -direction, so that in the frame of the fixed stars (the unprimed frame), the signal arrives at an angle 0 with respect to the positive x -direction. In the unprimed frame, x = r cos θ and y = r sin θ. From Equation (1.41),
cos ( / )
cos ( / )
1 2 / 2 1 2 2 1 v^2 c^2
v c
r
v c
r r c
v c
x vt
x
and y’ = y = r sin θ. Then,
- A man on the moon sees two spacecraft, A and B , coming toward him from opposite directions at the respective speeds of 0.800 c and 0.900 c. (a) What does a man on A measure for the speed with which he is approaching the moon? For the speed with which he is approaching B? (b) What does a man on B measure for the speed with which he is approaching the moon? For the speed with which he is approaching A? Sol (a) If the man on the moon sees A approaching with speed v = 0.800 c, then the observer on A will see the man in the moon approaching with speed v = 0.800 c. The relative velocities will have opposite directions, but the relative speeds will be the same. The speed with which B is seen to approach A , to an observer in A , is then
cos ( / )
(cos ( / ))/ /
sin
tan
v c
v c
x v c v c
y
2 2 2 2
sin 1
and arctan
(b) From the form of the result of part (a), it can be seen that the numerator of the term in square brackets is less than sin θ , and the denominator is greater than cos θ , and so tan θ and θ’ < θ when v ≠ 0. Looking out of a porthole, the sources, including the stars, will appear to be in the directions close to the direction of the ship’s motion than they would for a ship with v = 0. As v ‡ c , θ’ ‡0, and all stars appear to be almost on the ship’s axis(farther forward in the field of view).
c c vV c
V v V x
x x (^) 1 0 800 0900 0988
