Assignment 7, solutions

Problem 1:

The radius of the galaxy is 3*1020 m, measured in its own rest frame.
(a)  If the time it takes a spaceship to cross the entire galaxy is 300 years measured in the spaceship’s rest frame, what is the relative speed of spaceship and galaxy?
(b)  How much time elapses on Earth during this trip?

Solution:

Concepts, principles, relations that apply to the problem:
 Relativistic kinematics, the proper time interval, Lorentz contraction
Why do they apply?
We are given the proper time interval between two events and the space coordinate of these two events in another frame.
How do they apply?
(a)  t = 300 years.  This is the time interval between when opposite edges of the galaxy pass the spaceship.
L = L0/g is the diameter of the galaxy in the frame of the spaceship.
L0 = 6*1020 m = 6.34*104 ly.
v = L/t. 
v2 = (1 - v2/c2)L02/t2,  v2/c2 = L02/(c2t2 + L02).
v/c = 0.9999888.
(b)  t = L0/v = 6.34*104 years.
Details of the calculation:
 None

Problem 2:

A solar prominence (a large bright feature extending outwards from the sun's surface) was observed 10 minutes following a volcanic eruption on Earth.  Can those two events be related?  Why or why not?  The Earth-Sun distance is approximately 1.5 ·108 km.

Solution:

Concepts, principles, relations that apply to the problem:
The space-time interval
Why do they apply?
If the space-time interval c2Dt2 – Dx2 between two events is negative (space-like), then no causal relationship can exist.
How do they apply?
The space-time interval between these two events is space-like.
Event 1: x = - 1. 5 ·1011 m, t = (600 s - 1. 5 ·1011 m/3 ·108 m/s} = 100 s
Event 2: x = 0, t = 0.
c2Dt2 – Dx2 = [(3 ·108 *100)2 – 2.25·1022] m2 = -2.16·1022 < 0.
Not enough time passes for there to be a cause-effect relationship between these two events.  The eruption occurs before the flare in the rest frame of the solar system, but cannot have caused the flare.
Details of the calculation:
One can also argue that since the light from the flare arrives 10 minutes after the volcano erupts, and it takes light ~8 minute to travel from the sun to the earth, the flare occurred 2 minutes after the volcanic eruption.  But a signal from the eruption could not have arrived at the sun earlier than ~ 8 minutes after the eruption, so there is no way that the eruption in any way triggered the flare.

Problem 3:

In the year 2310 construction of the large space colony “Heavy Metal“ is finally completed.  The distance between the “East” and “West” edges of the colony is 400 km in the rest frame of the colony.  To celebrate the end of construction, two flares are lighted at midnight during a New Year’s celebration.  The light from the flares reaches an observer at the center of the colony at the same time.  A fast spacecraft flying with a constant speed of v = 0.95c “East” to “West” is directly over the “East” edge of “Heavy Metal”.  Let Event 1 be flare is lighted on “East” edge and Event 2 be flare is lighted on “West” edge.  In the reference frame of the spacecraft, does Event 1 occur before, after, or at the same time as Event 2?  Explain your reasoning!

Solution:

Concepts, principles, relations that apply to the problem:
The Lorentz transformation, the simultaneity problem
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?
Let the x axis point from “East” to “West”.
Put the origins of the coordinate systems of the colony and the spacecraft at the “East” edge and synchronizes the clocks at t = 0 when the flare is lighted on “East” edge.  Event 1 has coordinates x = 0, t = 0 in both frames.  Event 2 has coordinates x = 400 km, t = 0 in the rest frame of the colony.
To find the time t’ the flare is lighted on the “West” edge in the frame of the galaxy we use ct’ = gct - gbx = - gb(400 km).
t’ is negative, Event 2 occurred before event 1 in the frame of the spaceship.
Details of the calculation:
 None

Problem 4:

An atomic clock is taken to the North Pole, while another stays at the Equator.  How far will they be out of synchronization after a year has elapsed?

Solution:

Concepts, principles, relations that apply to the problem:
 Relativistic kinematics, the proper time interval
Why do they apply?
We are given the proper time interval between two events in frame 1 and are asked to find the time interval in another frame moving with speed v with respect to frame 1.
How do they apply?
The clock on the North Pole is in an inertial frame (frame 1), the clock on the equator is in an accelerating frame (frame 2).
The accelerating frame is not an inertial frame.  It accelerates with respect to an imagined second frame, which is at any instant moving alongside the accelerating frame with identical instantaneous velocity v, but is not accelerating.  The imagined frame provides a momentary inertial frame of reference, relative to which the acceleration is v2/r.
After a time interval Dt measured in the imagined ship, the rocket ship will have velocity (v2/r)Dt in a direction perpendicular to v with respect to the imagined frame.
The speed with respect to frame 1 will be v' = (v2 + (v2/r)2Dt2)1/2 = v as Dt ŕ 0.
The time interval measured in frame 1 is dt = gdt.  Since g is constant, t = gt.
t = 365*24*3600s,  v = 2p*6.37*106m/86400s, 1/g  = (1 - v2/c2)1/2 @ 1 - v2/(2c2).
t - t  = t - t(1 - v2/(2c2)) = t v2/(2c2) = 3.8*10-5 s.
Details of the calculation:
None

Problem 5:

(a) Observer A is stationary in a laboratory on Earth’s surface and observer B is in a laboratory moving at high velocity parallel to Earth’s surface (neglect the curvature of the earth in the subsequent considerations) and parallel to the x axis.  Assume that the observers each witness two separate lightning strikes on the surface of the earth lying along the x axis.
(i) Describe concisely a practical scheme whereby observer A can determine whether the two lighting strikes occurred simultaneously in her reference frame.
(ii) Show that if observer A concludes from her observations that the strikes were simultaneous, observer B generally must conclude from his observations of the same events that the strikes are not simultaneous, thus demonstrating that simultaneity is a relative and not absolute concept.
(b) Einstein was strongly inspired by Maxwell’s theory of electromagnetism and viewed Galilean invariance for transformations between reference frames as being seriously flawed because it was inconsistent with Maxwell’s theory.  The special theory of relativity removed this inconsistency.  Explain the meaning of the preceding two statements.

Solution:

Concepts, principles, relations that apply to the problem:
The Lorentz transformation, the simultaneity problem
Why do they apply?
We are asked to demonstrate that simultaneity is a relative and not absolute concept.
How do they apply?
(a) A lightning strike at time t produces light, which travels through the atmosphere which speed c/n (n ~1) and sound, which travels through the atmosphere with speed vs.
Let tl be the time at which the light signal arrives at A and ts the time at which the sound signal arrives at A.
d = vs (ts - t),  d = (c/n)(tl - t).
Eliminate d and solve for t.
t = (vsts – (c/n)tl)/(vs – c/n).
If t is the same for both lightning strikes, then they occurred simultaneously in the frame of observer A.
In the fame of observer A the coordinates of event 1 are (ct1, x1, y1, z1) =  (0, d1, 0, 0).  (We set t = 0 at the time of the strike.) 
The coordinates of event 2 are (ct2, x2, y2, z2) =  (0, d2, 0, 0). 
In the frame of observer B we have from the Lorentz transformation,

,

that the coordinates for event 1 are (ct1’, x1’, y1’, z1’) =  (-gbd1, gd1, 0, 0), and the coordinates for event 2 are (ct2’, x2’, y2’, z21’) =  (-gbd2, gd2, 0, 0).
Since d1 is not equal to d2, we have t1’ not equal to t2’.  The two lightning strikes do not occur simultaneously in the frame of observer B.
(b)  Maxwell’s equations predict that electromagnetic fields satisfy a wave equation in free space.  The equations do not contain any reference to a medium, and the propagation speed of any electromagnetic wave through the vacuum is c, regardless of the relative motion of the source and the observer.  The Galilean transformation predicts that the speed of electromagnetic waves depends on the relative motion of the source and the observer.
The special theory of relativity removes this inconsistency.

The postulates of special relativity are:
I. The laws of nature are the same in all inertial reference frames.
II. In vacuum, light propagates with respect to any inertial frame and in all directions with the universal speed c.  This speed is a constant of nature.

The second postulate of the special theory of relativity, however, goes beyond just Maxwell’s equations.  Important parts of physics, such as the physics of elementary particles, cannot possibly be explained in terms of electrodynamics.

Details of the calculation:
 None

Problem 6:

For an observer on Earth, the star Sirius is 8.8 ly away.  Assume Sirius is stationary with respect to Earth.  At t = 0, a spaceship moving with velocity v = 0.95ci towards Sirius passes Earth.  Earth and the ship synchronize clocks. 
An observer on Earth tells the following story at a much later time:
At t = 0, just as the spaceship passed Earth, a flare erupted on Sirius.  The light from the flare reached Earth at t = 8.8 y.  At t = 0.9 years a comet hit Earth.  At t = 9.26 years the spaceship passed Sirius and sent us a light signal.  The signal reached us at 18.06 years.
How does an observer on the ship tell the same story?

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 5 events in one reference frame.  We are asked to transform to another reference frame.
How do they apply?
 
Event:
    		coordinates in K (Earth):
(x0,x1,x2,x3) = (ct,x,y,z)   		coordinates in K' (ship):
(x'0,x'1,x'2,x'3) = (ct',x',y', z')
#1:  Ship passes Earth (0, 0, 0, 0) (0, 0, 0, 0)
#2:  Flare erupts on Sirius (0, l2, 0, 0) (ct2', l2', 0, 0)
#3:  Comet his Earth (ct3, 0, 0, 0) (ct3', l3', 0, 0)
#4:  Ship passes Sirius (ct4, l4, 0, 0) (ct3', 0, 0, 0)
#5:  Signal reaches Earth (ct5, 0, 0, 0) (ct5', l5', 0, 0)

Event 2:
l2 = 8.8 ly, g = (1 - b2)-1/2 = 3.2, 
ct2' = gct2 - gbx2 = (-3.2)(0.95)(8.8 y),  t2' = - 26.8 years.
l2' = gl2' = 28.2 ly.
Event 3:
t3 = 0.9 y.
ct3' = gct3 - gbx3 = gct3, t3' = (3.2)(0.9 y) = 2.9 years.
l3' = -gbct3 = -(3.2)(0.95)(0.9)ly = -2.74 ly. (Note: c = 1ly/y)
Event 4:
t4 = 9.26 y, l4 = 8.8 ly, ct4' = gct4 - gbx4, t4' = 3.2[9.26 y - (0.95)(8.8 y)] = 2.9 y.
The distance between the Earth and the ship as measured from the ship is d = (8.8 ly)/g = 2.7 ly.
Event 5:
t5 = 18.06 y.  t5' = gt5 = (3.2)(18.06 y) = 57.8 y.
l5' = -gbct5 = -(3.2)(0.95)(18.06 ly) = -54.9 ly.

Details of the calculation:
The story told by an observer on the ship:
A flare erupted on Sirius when the ship was 28.2 ly from Sirius, 26.8 years before we reached Earth.  At t = 0 we passed Earth.
A comet hit earth at 2.88 years after we had passed Earth, when it lay 2.74 ly behind us.  At the same time we passed Sirius and sent a light signal back to Earth.  That signal reached Earth at t = 57.8 y, when it lay 54.9 ly behind us.

A space-time diagram depicting events 1-5 in K and K' is shown below.  The events are dots labeled 1-5.  For a reference on how to construct space-time diagrams click here.

Problem 7:

Imagine a train going through a tunnel.  The train and the tunnel both have length l in their own frame. The train moves through the tunnel with speed v.  There is a bomb at the front of the train which is designed to explode when the front of the train passes out the far end of the tunnel.  However, there is a disarming sensor located on the back of the train which will disarm the bomb just as the back of the train enters the near end of the tunnel.  Will the bomb explode?

Solution:

Concepts, principles, relations that apply to the problem:
Relativistic kinematics, the space-time interval
Why do they apply?
The space-time interval between two events is ds.  ds2 = c2dt2 - |dr|2.
ds2 is invariant under a Lorentz transformation.
If the space time interval between two events is positive (time-like) in one frame it is positive in all frames.  Event 1 can influence event 2.  There exists a frame in which the two events "happen" at the same place.
If the space time interval between two events is negative (space-like) in one frame it is negative in all frames.  Event 1 cannot influence event 2.  There exists a frame in which the two events "happen" at the same time.
How do they apply?
In the rest frame of the train, the tunnel moves with speed v.  It has length l/g < l.  Let us work in the rest frame of the train.
Let event 1 describe the front of of the train emerging from the tunnel (or the rear of the tunnel passing the front of the train).  Let the coordinates of the event be x = 0, t = 0.
Let event 2 describe the rear of the train entering the tunnel (or the front of the tunnel passing the rear of the train).  The coordinates for this event are x = l, t = (l - l/g)/v = (l/v)(1 - 1/g) > 0.
Since in that frame event 2 "happens" after event 1, there is no way event 2 can influence event 1 in this frame, and therefore in any frame.  The bomb will explode.
Details of the calculation:
One might argue that in the frame of the tunnel, the train appears contracted by a factor 1/g, so the rear of the train will enter the tunnel before the front passes out, and the bomb will be disarmed.  However, this line of reasoning ignores the finite time that any disarming signal must take to move from the rear of the train to the bomb at the front.  The bomb can be disarmed only if a signal traveling at c emitted from the rear of the tunnel at the instant the rear of the train passes, reaches the far end of the tunnel before the train does.  Working in the frame of the tunnel, the signal takes a time l/c, and the train takes a time [(l - l/g)/v].  For the bomb not to explode we need: l/c < [(l - l/g)/v] , which simplifies to Ö{1 + v/c} < Ö{1 - v/c} , which is false.  The bomb explodes.