Problem 1:
Two events
"occur" at the same place in the laboratory frame of reference and are
separated in time by 3 seconds.
(a) What is the spatial
distance between these events in a rocket frame moving with respect to the
laboratory frame, in which the events are separated in time by 5 seconds?
(b) What is the relative speed of the moving frame and the laboratory
frame?
Solution:
 | Concepts, principles, relations that apply to the
problem: Relativistic kinematics, the Lorentz transformation |
| Why do they apply? We are given the space-time coordinates of two events in one reference frame. We are asked to transform to another reference frame. |
| How do they apply? (a) Let K be the laboratory frame and K' be the rocket frame. K K' c2Dt2 - Dx2 = c2Dt'2 - Dx'2. c2(3s)2 - 0 = c2(5s)'2 - Dx'2. Dx'2 = c2(16s2) Dx' = c(4s) = 12´108m (b) In the laboratory frame the two events "occur" at the same place. In the rocket frame the laboratory place has moved 12´108m in 5s. The relative speed of the two frames is therefore (12´108m)/5s = (4/5)c. Alternatively: dt = dt/g. 3s = 5s/g = 5s(1 - v2/c2)1/2. v = (4/5)c. |
| Details of the calculation: None |
Problem 2:
A pyramid in the Egyptian desert (see the figure) has sides of length l
inclined at an angle q according to Egyptian
references. An archeologist at the base of the pyramid begins moving up the side
of the pyramid and reaches the top of the pyramid in time T according to
observers on the ground. A spaceship, moving in the positive x-direction,
is approaching the pyramid at a relativistic velocity v.
(a) According to observers in the spaceship, how far did the archeologist
move and how long did it take him to reach to top of the pyramid?
(b) Give two calculations of the invariant space-time interval
Ds2 for the archaeologists ascent:
one, using the inertial frame of the spaceship, and the other, using the
inertial frame of the ground.
For part (c), assume that the archeologist moved up the pyramid at a constant
velocity v0 = l/T. Do not neglect v0 compared
to c.
(c) Calculate the proper time required for the archeologist's ascent.
Solution:
| Concepts, principles, relations that apply to the
problem: Relativistic kinematics, the Lorentz transformation, |
| Why do they apply? The spaceship moves with uniform relativistic velocity vi with respect to the ground. The ground and the spaceship are two inertial reference frames, moving with respect to each other. We are given the space-time coordinates of two events in one reference frame. We are asked to transform to another reference frame. |
| How do they apply? (a) Let K' be the frame of the spaceship and K the frame of the observer on the ground.  where (x0,x1,x2,x3) = (ct,x,y,z), b = v/c, g = (1 - b2)-1/2. Assume that at t = t' = 0 the origins of the two coordinate systems coincide at r = r' = 0. In K the archeologist's position 4-vector when he reaches the top is (cT, lcosq, lsinq, 0). In K' it is (cT', x', y', 0), with cT' = gcT - gblcosq, x' = -gbcT + glcosq, y' = lsinq. T' = g(T - (v/c2)lcosq) is the time it takes the archeologist to get to the top in K'. He moves a distance d' = [g2(lcosq -vT)2 + (lsinq)2]1/2. |
| Details of the calculation: (b) In K: ds2 = c2dt2 - |dr|2 = c2T2 - (lcosq)2 - (lsinq)2 = c2T2 - l2. In K': ds2 = c2dt'2 - |dr'|2 = c2g2(T - (v/c2)lcosq)2 - g2(lcosq -vT)2 - (lsinq)2 = g2[c2T2 -2bcTlcosq + b2(lcosq)2 - (lcosq)2- (vT)2 +2lvcosqT] - (lsinq)2 = g2[c2T2(1 - v2/c2) - (lcosq)2(1 - b2)] - (lsinq)2 = c2T2 - l2. (c) The proper time is calculated in the archeologist's rest frame. He moves with constant velocity v0 = l/T with respect to an observer at rest at the foot of the pyramid. dt = dt/garch = T(1-v02/c2)1/2 = (T2 - l2/c2)1/2. |
Problem 3:
A person on Earth observes two rocket ships moving
directly toward each other and colliding. At time t = 0 in the
Earth frame, the Earth observer determines that rocket 1, traveling to the right
at v1 = 0.8c, is at point a, and rocket 2 is at
point b, traveling to the left at v2 = 0.6c. They are
separated by a distance 4.2´108m.
(a) In the Earth frame, how much time will pass before the rockets
collide?
(b) How fast is rocket 2 approaching in rocket 1’s frame? How fast
is rocket 1 approaching in rocket 2’s frame?
(c) How much time will elapse in 1’s frame from the
time rocket 1 passes point a until collision? How much time will elapse
in 2’s frame from the time rocket 2 passes point b until collision?
Solution:
| Concepts, principles, relations that apply to the
problem: Relativistic Kinematics, velocity addition, proper time |
| Why do they apply? The spaceships move with relativistic speeds. We are given the relative speed of the spaceships with respect to Earth and are asked to add their velocities to find their speed with respect to each other. We are asked to find the time interval separating two events in the earth frame and then transform to another frame where those two events have the same space coordinate. |
| How do they apply? (a) 0.6ct + 0.8ct = 4.2´108m. t = (4.2´108m)/1.4c = 1s is the time that will pass on Earth before the ships collide. (b) Earth is moving with velocity -v1i with respect to rocket 1. In the Earth frame rocket 2 is moving with velocity -v2i. The velocity addition formula yields v21 = -(v1+v2)/(1 + v1v2/c2)i = -0.95ci for the velocity of rocket 2 in the frame of rocket 1. Similarly, v12 = 0.95ci. |
| Details of the calculation: (c) The time interval between passing point a and the collision is the proper time interval in the frame of rocket 1. tac = t/g1 = t (1 - 0.82)1/2 = 1s ´ 0.6 = 0.6s. The time interval between passing point b and the collision is the proper time interval in the frame of rocket 2. tbc = t/g2 = t (1 - 0.62)1/2 = 1s ´ 0.8 = 0.8s. |
Problem 4:
The half-life of a p+ meson at rest is 2.5 10-8s. A beam of p+ mesons is generated at a point 15 m from a detector. Only 1/2 of the p+ mesons live to reach the detector. What is the speed of the p+ mesons?
Solution:
| Concepts, principles, relations that apply to the
problem: Relativistic kinematics, the proper time interval |
| Why do they apply? Given the half-life Dt of the p+ meson in its rest frame (the proper time) we are asked to find the relative speed of a frame in which the half-life is Dt = 15m/v. |
| How do they apply? gDt = 15m/v, gv = 15m/Dt. Define v0 = 15m/Dt. Then v2/(1 - v2/c2) = v02, v2 = (v02 - v02v2/c2), v2(1 + v02/c2) = v02. v2 = v02/(1 + v02/c2), v = 2.68*108m/s = 0.894c. Details of the calculation: None |
Problem 5:
In the reference frame of an outside observer two particles move towards each other, both with relativistic speed v . The angle between them is 2q as shown in the figure below. What is the speed of one of the particles as viewed by the other?
Solution:
| Concepts, principles, relations that apply to the
problem: Relativistic Kinematics, velocity addition A particle moves in K with velocity u = dr/dt. K' moves with respect to K with velocity v. The particle's velocity in K’, u' = dr'/dt', is given by u'|| = (u|| - v)/(1 - v×u/c2), u'^ = u'^/(g(1 - v×u/c2)), where parallel and perpendicular refer to the direction of the relative velocity v. |
| Why do they apply? The particles move with relativistic speeds. We are given the speed of the particles with respect to an inertial frame and are asked to add their velocities to find their speed with respect to each other. |
| How do they apply? Consider a frame moving with velocity vcos(q) along the x-axis. In that frame u'|| = 0 for each of the particles, u'^ = ±vsinq/(g(1-v2cos2q/c2)) = ±gvsinq, the particles approach each other while moving perpendicular to the x'-axis. |
| Details of the calculation: To find the speed of particle 1 in the rest frame of the particle 2 we denote this rest frame by K''. K'' moves with respect to K' with speed v' = -gvsinq, and the particle 1 moves in K' with speed u' = gvsinq. In K'' particle 1 moves with speed u'' = ( gvsinq + gvsinq)/(1 + (gvsinq)2/c2) = 2gvsinq/(1 + ( gvsinq)2/c2), with g = (1-v2cos2q/c2)-1/2. |
Problem 6:
A space ship is moving to the east at a speed of 0.9c relative to the earth. A second spaceship is moving to the west at a speed of 0.8c relative to the earth. What is the speed of one spaceship relative to the other?
Solution:
| Concepts, principles, relations that apply to the
problem: Relativistic Kinematics, velocity addition |
| Why do they apply? The spaceships move with relativistic speeds. We are given the relative speed of the spaceships with respect to Earth and are asked to add their velocities to find their speed with respect to each other. |
| How do they apply? Earth is moving with velocity -v1i = -0.9c i with respect to rocket 1. In the Earth frame rocket 2 is moving with velocity -v2i = -0.8c i The velocity addition formula yields v21 = -(v1+v2)/(1 + v1v2/c2)i = -0.988ci for the velocity of rocket 2 in the frame of rocket 1. The speed of one spaceship relative to the other is 0.988c. |
| Details of the calculation: None |
Problem 7:
Consider three reference frames. A meter stick is at rest in frame K2.
It is positioned on the x-axis, from x = 0 to x = 1m.
Frame K2 moves with velocity v = v2j
in the positive y-direction with respect to frame K1.
Frame K3 moves with velocity v = v3i
in the positive x-direction with respect to frame K1.
(a) Find the velocity of the stick in K3.
(b) Find the length of the stick in K3.
(c) Find the angle a the
stick makes with the x-axis in K3.
Solution:
| Concepts, principles, relations that apply to the
problem: Relativistic kinematics | |||||||||||||||
| Why do they apply? We are asked to relate observations in an inertial frame K1 to observations in an inertial frame K3. | |||||||||||||||
| How do they apply? (a) The stick is at rest in K2. K2 moves with respect to K1 with speed v2 in the +y-direction. So the stick moves with velocity v2j in K1. It is oriented parallel to the x-axis, so its length in K1 is 1m. K3 moves with velocity v3i with respect to K1. The velocity of the stick in K3 can be found from the velocity addition formulas. Assume that K’ is moving with velocity vi with respect to K. A particle moves in K with velocity u = dr/dt. The particle's velocity in K’, u' = dr'/dt', is given by u'|| = (u|| - v)/(1 - v×u/c2) where parallel and perpendicular refer to the direction of the relative velocity v.
Here v = v3i, u|| = 0, u^
= v2j.
We can also proceed by transforming the coordinates of events. Let us look
at the positions of the left and the right end of the stick at t = 0 and at t =
t1 in K1 and K3.
From events 1 and 3 we find vx' = Dx'/Dt'
= -g3b3ct/g3t
= -g3b3c
= -v3. | |||||||||||||||
| Details of the calculation: (b) In K3 at t' = 0 the left edge of the stick is at x' = 0, y' = 0. At t' = -g3b3L/c the right edge of the stick is at x' = g3L, y' = 0. The stick travels with velocity v' = -v3i + (v2/g3)j. Therefore at t' = 0 the right edge of the stick is at x' = g3L - v3g3b3L/c, y' = (v2/g3)g3b3L/c, or x' = g3L(1 - b32) = L/g3, y' = b2b3L. The length of the stick in K3 is (x'2 + y'2)1/2 = L(1 - b32 + (b2b3)2)1/2. (c) tana' = y'/x' = b2b3g3. |
Problem 8:
An observer moves horizontally away from a flashlight with a
speed 0.6c.
(a) The flashlight is pointed in the direction of the observer and emits
light. Prove that the speed of the light determined by the observer is exactly
c.
(b) The flashlight is now turned perpendicular to the direction of motion for
the observer and light is emitted. Demonstrate that the observer again will
deduce that the speed of the light as measured in her frame is c.
(c) What angle will the light velocity vector make with the horizontal axis in
the observer’s frame for case b?
Solution:
| Concepts, principles, relations that apply to the
problem: Velocity addition: |
| Why do they apply? A particle moves in K with velocity u = dr/dt. K' moves with respect to K with velocity v. The particle's velocity in K’, u' = dr'/dt', is given by u'|| = (u|| - v)/(1 - v×u/c2), u'^ = u^/(g(1 - v×u/c2)), where parallel and perpendicular refer to the direction of the relative velocity v. |
| How do they apply? (a) Let the observer move towards the right, in the positive x-direction. K is the frame of the flashlight, K’ the frame of the observer, u and v are parallel to each other. u = ci, v = vi, u’i = (c – v)/(1- cv/c2) = ci. (b) Now v×u = 0. u'|| = -v, u'^ = c/g, u’2 = v2 + c2(1 - v2/c2) = c2, u’ = c. (c) |tanq| = |u'^ / u'||| = c/(vg) = 1.33. q = (180 - 53)º. |
| Details of the calculation: None |
Problem 9:
A space-time diagram is a plot of position versus time. It is always drawn in an inertial frame. Consider the space-time diagram shown below.
(a) Assume a particle moves along the path OPB from O to B.
Describe the motion in the inertial frame in which the space-time diagram is
drawn. Find the proper time interval (the time interval in the rest frame
of the particle) between events O and B.
(b) Assume a particle moves along the path ORB from O to B.
Describe the motion in the inertial frame in which the space-time diagram is
drawn. Find the proper time interval (the time interval in the rest frame
of the particle) between events O and B.
(c) Comment on the implication for space travel that you can deduce from your
analysis of this simple example.
Solution:
| Concepts, principles, relations that apply to the
problem: The proper time interval |
| Why do they apply? We are asked to find the proper time interval for different paths in a space-time diagram. |
| How do they apply? (a) The particle is at rest in the inertial frame in which the space-time diagram is drawn. The proper time interval between events O and B is (Dct)2 = (Dct)2 - (Dx)2 = (Dct)2, since Dx = 0. Dct = 10/3. (b) The particle moves with uniform velocity in the positive x-direction from O to R. It accelerates briefly to change direction, and then moves with uniform velocity in the negative x-direction from R to B. The proper time interval between events O and R is (Dct)2 = (Dct)2 - (Dx)2. Dct = 5/3, Dx = 4/3, Dct = 1. Similarly, the proper time interval between events R and B is Dct = 1. The proper time interval between events O and B is Dct = 2. (c) In the Lorentz geometry, the proper time interval between two specified events along a curved world line is always less than or equal to the proper time interval along a straight world line. A curved world line implies acceleration. An observer who has moved away from a place and has accelerated to come back has aged less than an observer who has remained stationary in an inertial frame. |
| Details of the calculation: None |
Problem 10:
Assume a rocket ship leaves the earth in the year 2020. One of a set of
twins born in 2000 remains on earth, the other rides in the rocket. The
rocket ship is so constructed that it has an acceleration g in its own rest
frame to make the occupants feel at home. It accelerates in a straight
line path for 5 years as measured by its own clock, decelerates at the
same rate for 5 more years years, turns around, accelerates for 5 years,
decelerates for 5 years, and lands on earth. The twin in the rocket is 40
years old.
(a) What year is it on earth?
(b) How far away from earth did the rocket ship travel?
Solution:
| Concepts, principles, relations that apply to the
problem: The proper time interval |
| Why do they apply? We are asked to compare proper time intervals in an inertial and an accelerating frame. |
| How do they apply? The accelerating rocket ship is not an inertial frame. It accelerates with respect to an imagined second ship, which is at any instant happening to move alongside with identical instantaneous velocity v, but is not accelerating. The imagined ship provides a momentary inertial frame of reference, relative to which the acceleration is g. After a time interval Dt measured in the imagined ship, the rocket ship will have velocity gDt with respect to the imagined ship. The velocity with respect to earth will be v' = (v + gDt)/(1 + vgDt/c2). v'/c = (v/c + (g/c)Dt)/(1 + vgDt/c2). To express this equation in terms of the boost parameter we write v'/c = tanh(B), v/c = tanh(B1), (g/c)Dt = tanh(B2). Then tanh(B) = (tanh(B1) + tanh(B2))/(1 + tanh(B1)tan(B2)) = tanh(B1 + B2) If after each time interval Dt we imagine a new second ship, not accelerating, but moving with identical instantaneous velocity next to the rocket ship, we have after n time intervals nDt = t that tanh(B) = tanh(nB2). As Dt --> dt we have tanh(B2) --> B2 , or B2 = (g/c)dt. Therefore tanh(B) = tanh((ng/c)dt), B = (g/c)t. g(t) = (1 - (v'/c)2)-1/2 = (1 - tanh2(B))-1/2 = cosh(B). Dt = òt1t2 g(t)dt, since dt = dt/g(t). Dt = òt1t2 cosh(gt/c)dt = (c/g)sinh(gt/c)|t1t2, since ò cosh(x)dx = sinh(x). If t1 = 0 and t2 = 5years, then Dt = (c/g)sinh(gt2/c), Dt = [(3*108m/s)/(9.8m/s2)]sinh[(1.75*108s)(9.8m/s2)/(3*108m/s)] = 2.6*109s = 84 years. |
| Details of the calculation: (a) 2020 + 4*84 = 2356 is the year on earth. (b) dx = v' dt = c tanh(B) dt = c tanh(B)cosh(B)dt. Dx = còt1t2 tanh(gt/c)cosh(gt/c)dt = còt1t2 sinh(gt/c)dt. If t1 = 0 and t2 = 5years, then Dx = (c2/g)(cosh(gt2/c) - 1). Dx = [(3*108m/s)2/(9.8m/s2)](cosh(5.12) - 1) = 7.6*1017m = 80 ly. The ship has traveled 2*80 ly = 160 ly away from earth. |
