Unformatted text preview: istered in a reference frame
attached to the particle. In this reference frame the two events occur at the same location.
The proper lifetime is equal to the contracted distance L, which is measured in this reference
frame, divided by the speed v of the particle, or ∆t0 = L / v .
d. The particle’s contracted lifetime ∆t is related to its proper lifetime ∆t0 by the
timedilation formula, ∆t = ∆t0 / 1 − v 2 / c 2 (Equation 28.1). SOLUTION
a. The proper distance is L0 = 1.05 × 10−3 m .
b. The distance measured by a hypothetical person traveling with the particle is Chapter 28 Problems L = L0 1 − v2
c 2 = (1.05 × 10−3 m ) 1 − ( 0.990c )2
c 2 = 1.48 × 10−4 m 1485 (28.2) c. The proper lifetime ∆t0 is equal to the contracted distance L divided by the speed v of the
particle:
L
1.48 × 10−4 m
∆t0 = =
= 4.98 ×10−13 s
v ( 0.990 ) ( 3.00 ×108 m/s )
1444 24444
4
3
0.990c d. The dilated lifetime is ∆t = ∆t0 ( 4.98 × 10–13 s ) =
= 3.53 × 10 –12 s (28.1)
v2
( 0.990 c )2
1– 2
1–
c
c2
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15. REASONING AND SOLUTION The side x will have its length contracted as viewed from
the rocket. Taking the proper length to be L0 = x, we can find the dilated length L = x′ from
Equation 28.2:
2 2 v 0.730 c x′ = x 1 – = x 1 – = 0.683 x
c
c
The new angle, can be found using tan θ′ = y/x′ = y/(0.683 x) = (tan 30.0°)/(0.683) = 0.845. θ ′ = tan −1 0.845 = 40.2°
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16. REASONING To the observer at rest relative to the cube, its length, width, and height are
all equal to the proper length L0 = 0.11 m of one of the cube’s sides. Suppose that the
moving observer is moving parallel to the width of the cube and, therefore, measures a
contracted length L for the width. Note, however, that this observer still measures the proper
length L0 for the other two dimensions of the cube, since they are perpendicular to the v2
c2
(Equation 28.2), where v is the speed of the observer relative to the cube. Accordingly, the
volume V = L0 × L × L0 = L2 L of the cube is smaller for the moving observer than it is for
0
the observer at rest relative to the cube. Given the mass m of the cube, then, the moving
direction of motion. The shortened width of the cube is given by L = L0 1 − 1486 SPECIAL RELATIVITY observer calculates a density of ρ = m
m
= 2 (Equation 11.1) for the cube that is greater
V L0 L than the density of glass.
SOLUTION Substituting L = L0 1 − v2
m
(Equation 28.2) into ρ = 2 (Equation 11.1), we
2
c
L0 L obtain ρ= m
=
L2 L
0 m ( L20 ) L0 v2
1− 2
c m =
L3
0 (1) v2
1− 2
c Rearranging Equation (1) and solving for the quantity v2/c2, we find that
m
v2
= 1− 2
c
ρ L3
0 2 or m
v2 3 = 1− 2 ρL c 0 or m
v2
= 1− 3 c2 ρ L0 2 (2) Taking the square root of both sides of Equation (2) and solving for v yields
2 2 m
3.2 kg = 0.951c
or
v = c 1− 3 = c 1− ρL 7800 kg/m3 ( 0.11 m )3 0 ______________________________________________________________________________ m
v
= 1− 3 ρL c 0 2 ( ) 17. REASONING The first twin, traveling at the higher speed v1 = 0.900c, arrives at the distant
planet first. Thereafter, the first twin is at rest relative to the earth, and ages at the same rate
as people back on the earth. As measured by an observer on the earth, the dilated time
intervals ∆t1, ∆t2 for the journeys of the two twins are related to the proper time intervals
∆t0
∆t01, ∆t02 for their journeys by ∆t =
(Equation 28.1), so we have that
2
v
1− 2
c
∆t01 = ∆t1 1 − 2
v1 c2 and ∆t02 = ∆t2 1 − 2
v2 c2 (1) The proper time intervals ∆t01 and ∆t02 in Equations (1) give the aging experienced by each
twin during the journey. The additional aging that the first twin undergoes between the time
of his arrival at the distant planet and the arrival of the second twin is the difference between Chapter 28 Problems 1487 the time of the second twin’s journey and the time of the first twin’s journey, both as
measured by an observer on the earth: Additional aging = ∆t2 − ∆t1 (2) d
(Equation 2.1) to determine the time intervals for each twin’s
∆t
journey as measured from the earth, where d = 12.0 lightyears is the distance between the
earth and the distant planet. The distance d and the time intervals ∆t1, ∆t2 are all measured
by an observer on the earth, so we do not need to apply the special theory of relativity.
Therefore, we have that
Lastly, we will use v = ∆t1 = d
v1 and ∆t2 = d
v2 (3) SOLUTION
a. When the second twin arrives at the distant planet, the final age of the first twin is given
by A1 = A + ∆t01 + ∆t2 − ∆t1, where A = 19.0 years is the initial age of both twins and we
have used Equation (2) to take into account the additional aging. The second twin’s final age
is the initial age A of the twin plus the time ∆t02 that elapses during the journey, as measured
by that twin: A2 = A + ∆t02. Therefore, the net difference between their ages will be A2 − A1 = A + ∆t02 − ( A + ∆t01 + ∆t2 − ∆t1 ) = ∆t02 − ∆t01 − ∆t2 + ...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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