Solution manual for modern physics 6th edition by tipler

The interior is dark until t = 2s at which time the spherical wave that reflected from the inner surface at t = 1s returns to the center showing the entire inner surface of the sphere

2 2.74 102

1.83 103.00 10 /

(b) From Equation 1-6 the correction

2 2

2L v t

1-4 (a) This is an exact analog of Example 1-1 with L = 12.5 m, c = 130 mph, and v = 20

mph Calling the plane flying perpendicular to the wind plane #1 and the one flying

parallel to the wind plane #2, plane #1 win will by Δt where

3 3

1-5 (a) In this case, the situation is analogous to Example 1-1 withL 3 108m,

first reflected light ring Then the interior is dark again

(b) In the frame of the seated observer, the spherical wave expands outward at c in all directions The interior is dark until t = 2s at which time the spherical wave (that reflected from the inner surface at t = 1s) returns to the center showing the entire inner

surface of the sphere in reflected light, following which the interior is dark again

1-6 Yes, you will see your image and it will look as it does now The reason is the second

postulate: All observers have the same light speed In particular, you and the mirror are

in the same frame Light reflects from you to the mirror at speed c relative to you and the mirror and reflects from the mirror back to you also at speed c, independent of your

motion

1-7

2 2

1-8 (a) No Results depends on the relative motion of the frames

(b) No Results will depend on the speed of the proton relative to the frames (This answer anticipates a discussion in Chapter 2 If by “mass”, the “rest mass” is implied, then the answer is “yes”, because that is a fundamental property of protons.)

(Problem 1-8 continued)

(c) Yes This is guaranteed by the 2nd postulate

(d) No The result depends on the relative motion of the frames

(e) No The result depends on the speeds involved

(f) Yes Result is independent of motion

(g) Yes The charge is an intrinsic property of the electron, a fundamental constant

1-9 The wave from the front travels 500 m at speed c + (150/3.6) m/s and the wave from the

rear travels at c – (150/3.6) m/s As seen in Figure 1-14, the travel time is longer for the

wave from the rear

8

3.00 10 / 150 / 3.6 / 3.00 10 / 150 / 3.6 /

3 10 150 / 3.6 3 10 150 / 3.6500

A B C have moved to the * marks, according to the observer in S

(a) According to an Sobserver, the wavefronts arrive simultaneously at AandB

(b) According to an S observer, the wavefronts do not arrive at AandCsimultaneously

(c) The wavefront arrives at Afirst, according to the S observer, an amount Δt before

  

0.2 1.0206 0.4 1.0911 0.6 1.2500 0.8 1.6667 0.85 1.8983 0.90 2.2942 0.925 2.6318 0.950 3.2026 0.975 4.5004 0.985 5.7953 0.990 7.0888 0.995 10.0125

4.0/ 1.898 2.0 10 0.85 75 / 3.756 10

Therefore, t2 – t1 = 0 and no fringe shift is expected

1-15 (a) Let frame S be the rest frame of Earth and frame Sbe the spaceship moving at speed

v to the right relative to Earth The other spaceship moving to the left relative to Earth

at speed u is the “particle” Then v = 0.9c and ux = −0.9c

(b) Calculating as above with 4

184.0/ 1.898 3.756 10 0.85 9.537 10 / 2.0 10

1 0.9 0.9 /

x x

2 8

3.0 10 / 3.0 10 / 6.0 10 /

6.0 10 /

1 103.0 10 / 3.0 10 /

2 2

3 2

1-17 (a) As seen from the diagram, when the observer in the rocket (S) system sees 1 c∙s

tick by on the rocket’s clock, only 0.6 c∙s have ticked by on the laboratory clock

(b) When 10 seconds have passed on the rocket’s clock, only 6 seconds have passed on the laboratory clock

x y

u u

x

y y

1-19 By analogy with Equation 1-23,

10 /

2.5 10 2.5 10 3.15 10

3600 /1.46 10 40.5

the x axis is L The

length OB on the ct axis

(b) In the laboratory frame the length is contracted to LL p/ and the round trip time is

(c) Yes The time ts measured in the spaceship is the proper time interval τ From time dilation (Equation 1-26) the time interval in the laboratory t L  ; therefore,

2 2

21

p L

L t

1-27 Using Equation 1-28, with and

1

0.5000.500where tan 43.9

2 2

In S: Both and have components in the direction

sin 25 2 sin 25 0.84 and cos 25 4 cos 25 3.63

2.76 1.69 3.24 (in z direction) is unchanged, so 2 (between c and xy-plane) tan 1.69 /

1 590 / 650

1 590 / 650Similarly, for green 525 0.210

and for blue 460 0.333

c v

and t Hans t Heidi1 in years.

Therefore, 0.95c0.45c t Heidi 0.95 and c t Heidi 1.90 ; y t Hans 0.90y

(a) In her reference frame S, Heidi has aged t when she and Hans meet

2 6

1

where 1.021

1.02 11

3 2 2

2 1 1

(Problem1-34 continued)

In his reference frame S, Hans has aged t when he and Heidi meet

The difference in their ages will be 1.697 0.290 1.407  y1.4y

(b) Heidi will be the older

Information could only be transmitted by modulating the beam’s frequency or intensity,

but the modulation could move along the beam only at speed c, thus arriving at the moon

only at that rate

1-36 (a) Using Equation 1-28 and Problem 1-20(b)

Time lost by satellite clock = / 3.15 10 2.06 10 / 2 0.00668 6.68

1-37

The train is A from you when the headlight disappears, where

1-38 (a)   t  t0 For the time difference to be 1 ,s    t t0 1s

(b)

Where v is the relative speed of the planes flying opposite directions The speed of

each plane was 1230 /m s/ 2615m s/ 2210km h/ 1380mph

coscos (Equation 1-41)

1 coswhere half-angle of the beam in 30

0.75

3.0tan14.1

2 8 2

9

273 10 Using the same substitution as in (a)

1 1/ 273 10 and the circumference of Earth 40, 000 , so4.0 10 or 4.0 10 / , and

2 273 104.0 10 / 2 / 273 10 , or

Since (c v ) (c v )1 for v > 1, the distance the information must travel to reach

the front of the rod is L p; therefore, the rod has extended beyond its proper length

1 /

p p

tan

x y

y x

(c) As v   c, , the maximum length of the rod  also

1-41 (a) Alpha Centauri is 4 c∙y away, so the traveler went 2 

diff diff diff

t  m c   s So, for N 0 = 50,000 pions initially, at the end of 50m

(Equation 1-33)1.5 10 3 10 / 2 102.25 10 3.6 10 1.89 101370

1-45     22

11/ 1 1/ (See Problem 1-20)

(Problem 1-47 continued)

(e)

1-48 In Doppler radar, the frequency received at the (approaching) aircraft is shifted by

approximately f / f0 v c/ Another frequency shift in the same direction occurs at the receiver, so the total shift    7

32 / 3.2 10 /period 115

3.2 103.2 10

Simultaneous images of star#1 in opposition

will appear at Earth when L is at least as large

as:

c L

tanuy/ux0.745 / 0.286 111 with respect to the  x axis.

1-54 This is easier to do in the xy and x y planes Let the center of the meterstick, which is

parallel to the x-axis and moves upward with speed v y in S, at

0.5

y y

1-55

1-56 The solution to this problem is essentially the same as Problem 1-53, with the manhole

taking the place of the meterstick and with the addition of the meterstick moving to the

right along the x-axis Following from Problem 1-53, the manhole is titled up on the right

and so the meterstick passes through it; there is no collision

1-57 (a)

(b) For simultaneity in S , t2 t1, or T vD c/ 2  v c/ cT D/

(c) If D < cT, then  2  2  

positive because v/c < 1 Thus,  2

t t  T vD c is always positive

(d) Assume T D c/ with  cc Then

This changes sign at v c/ c c/  which is still smaller than 1 For any larger v

still smaller than c)  2

75min 0.6 75min 45 min.

v  c  c The signal travels for 45 min to reach the S observer and arrives at 75 min + 45 min = 120 min on the S clock

(b) The observer in S sends his first signal at 60 min and its subsequent wavefront is

found at xc t 60 min  The Sobserver is at x vt 0.6ctand receives the

wavefront when these x positions coincide, i.e., when

The confirmation signal sent by the Sobserver is sent at that time and place, taking

90 min to reach the observer in S It arrives at 150 min + 90 min = 240 min

(c) Observer in S:

The Sobserver makes identical observations

1-59 Clock at r moves with speed ur, so time dilation at that clock’s location is:

60 min 0.60.4 60 min

60 min / 0.4 150 min0.6 0 min 90 min

2And,

2

r r

1-61

1-62 (a) Apparent time A B T/ 2 t A t B and apparent time B A T/ 2 t A t B where

t A = light travel time from point A to Earth and t B = light travel time from point B to

Earth

(b) Star will appear at A and B simultaneously when t BT/ 2t A or when the period is:

1-63 The angle of uwith the xaxis is: