Solutions to Special Relativity P758: 2, 3, 6, 9, 11, 20, 27, 35, 43, 78 2. A spaceship passes you at a speed of 0.750c. You measure its length to be 28.2 m. How long would it be when at rest? Solution: The length at rest is its proper length, and Lp   L  1
1  0.752  28.2 m  42.6 m 3. A certain type elementary particle travels at a speed of 2.70×108 m/s. At this speed, the average lifetime measured to be 4.76×10‐6 s. What is the particle’s lifetime at rest? Solution: The lifetime at rest is the proper time, and Tp  T   4.76  106 s  1  0.92  2.07  106 s . 6. Suppose you decide to travel to a star 75 light‐years away. How fast would you have to travel so the distance would be only 25 light‐years? Solution: According to the problem, the Lorentz gamma factor is   75 25  3 . So the speed can be obtained as v  c 1   2  0.943c  2.83  108 m/s . 9. A certain star is 95.0 light‐years away. How long would it take a spacecraft traveling 0.96c to reach that star from Earth, as measured by observer: (a) on Earth, (b) on the spacecraft? (c) What is the distance traveled according to observers on the spacecraft? (d) What will the spacecraft occupants compute their speed to be from the results of (b) and (c)? Solution: (a) In this case, the time required is t1  L v  95 0.96  99.0 year . (b) In this case, the distance traveled is L  Lp  , and   10.962 
0.5
 3.57 . So the time recorded reads t2  Lp  v  t1   27.7 year . (c) According to (b) The distance is Lp   95.0 3.57 lightyears  26.6 lightyears . (d) The speed computed by the spacecraft occupants is v  L t2  0.96 c . 10. A friend of yours travels by you in her fast sports car at a speed of 0.660c. It is measured in your frame to be 4.80 m long and 1.25 m high. (a) What will be its length and height at rest? (b) How many seconds would you say elapsed on your friend’s watch when 20.0 s passed on yours? (c) How fast did you appear to be traveling according to your friend? (d) How many seconds would she say elapsed on your watch when she saw 20.0 s pass on hers? 11. How fast must an average pion be moving to travel 15 m before it decays? The average lifetime, at rest, is 2. 6×10‐8 s. Solution: Assuming the speed of the pion is v, then we get vTp  Lp  ,or  v  Lp Tp . From this equation, one can figure out that v 
Lp Tp
L T c
2
p
2
p
2
c  0.887c  2.66 108 m/s . 20. If a particle moves in the xy plane of system S with speed u in a direction that makes an angle θ with x axis, show that it makes an angle θ’ in S’ given by tan  '   sin   1  v 2 c 2
 cos  v u  .
Proof: The Lorentz gamma factor is   1  u 2 c 2 , and the x and y component of the
u y 1  v 2 c2
ux  v
and u y ' 
velocity are ux ' 
, therefore one gets
1  ux v c 2
1  ux v c 2
tan  '  u y ' ux '  u y
 ux  v   sin 
1  v 2 c2
 cos  v / u  , QED.
27. (a) A particle travels at v = 0.10c. By what percentage will a calculation of its momentum be wrong if you use classical formula? (b) Repeat for v = 0.50 c. Solution: The classical momentum of the particle is p  m0v , and the Lorentz gamma factor reads   1  v 2 c 2 
1/2
. (a) Lorentz gamma factor in this case is   1.005 , so the percentage is 1.005  1 1.005  0.5 %. (b) For this case,   1.1547 , and the percentage is 1.1547  1 1.1547  13.4% . 35. Show that when the kinetic energy of a particle equals its rest energy, the speed of the particle is about 0.866c. Proof: According to the problem, the Lorentz gamma factor is   1  1    2 . So, the speed of the particle is v  c 1   2 
1/2
 0.866c . QED. 43. Two identical particles of rest mass m approach each other at equal and opposite speeds, v. The collision is completely inelastic and results in a single particle at rest. What is the rest mass of the new particle? How much energy was lost in the collision? How much kinetic energy is lost in this collision? Solution: The kinetic energy loss in this collision is K  2   1 mc 2 , and the rest energy of the resultant particle is E M ,R  2 mc 2  M  2 m , so there are no energy loss in the collision. 78. A slab of glass moves to the right with speed v. A flash of light is emitted at point A (Fig. 32‐17) and passes through the glass arriving at point B a distance L away. The glass has thickness d in the reference frame where it is at rest, and the speed of light in the glass is c/n. How long does it take the light to go from point A to point B according to an observer at rest with respect to points A and B? Check your answer for the case v = 0, n = 1, and for v = c. Solution: The speed of light in the medium measured by an observer at rest in the medium is c n and the speed of light in the medium measured by an observer at rest with respect to the points A reads v 
to B is t 
c nv
c  1  vn c 
 
. So the time required for the light go from A 2
1   c n  vc
n  1  v cn 
Ld 
 d  v  . For v = 0, n = 1, one has t  L c ; and for v = c, t  L c . c