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Adjacent dots on the screen may light up as if the dot had moved from one position to the next at a speed
greater than the speed of light, but in fact, no dot moved.
Each lit dot is produced by a different electron.
The image of a moving object created by the successive impacts of the electrons on the screen is not an object.
The information carried by the electrons moves from the electron gun to the screen at a speed below that of
light.
It is possible for the electron beam in a television picture tube to move across the screen at a speed faster than
the speed of light. Why does this not contradict special relativity?
Problem 1.2
Physics 107
When G(v) is "near" 1, relativistic effects are not obvious.
When G(v) deviates significantly from 1,
relativistic effects are more obvious.
Notice the rapid dropoff in GG(v)
Notice how G(v) falls rather slowly.
0
1
.
10
8
2
.
10
8
3
.
10
8
0
0.5
1
GG v
( )
v
0
1
.
10
8
2
.
10
8
3
.
10
8
0
0.5
1
G v
( )
v
Plot:
Plot:
GG v
( )
1
v
2
cc
2

=
G v
( )
1
v
2
c
2

=
v
0 c
⋅
0.01 c
⋅
,
c
..
=
cc
3 10
7
⋅
=
c
3 10
8
⋅
=
Set the range of v:
Set the range of v:
This column is for c=3x10
7
.
This column is for c=3x10
8
.
If c were smaller, then G(v) would differ from unity at much lower velocities v thus making relativistic effects
more conspicuous than they are now.
Let's plot G(v) as a function of v/c to see the effect.
The square box at the end of this equation indicates it is just text; i.e., not "live."
G v
( )
1
v
2
c
2

≡
All of the phenomena of special relativity depend upon the factor G(v):
If the speed of light were smaller than it is, would relativistic phenomena be more or less conspicuous than
they are now?
Problem 1.1
Physics 107
1
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Something to think about:
what time would an observer who remained stationary on earth measure?
t
0
28.566
=
t
0
t
1
v
c
2

⋅
=
t
40
=
v
0.7 c
⋅
=
c
1
=
Assign values and solve:
t
0
t
1
v
c
2

⋅
=
t
t0
1
v
c
2

=
Solution:
we apply the time dilation equation.
An observer on the spacecraft measures a dilated time t = 40
min (why is it the dilated time?).
We need to calculate the proper time, as measured by the driver of the car.
An observer on a spacecraft moving at 0.7c relative to the earth finds that a car takes 40 minutes to make a
trip. How long does the trip take to the driver of the car?
Problem 1.4
Physics 107
This answer ignores the complication of length contraction, which we will get into later in this chapter.
Attempting to account for length contraction would introduce the problem of simultaneity (see the polebarn
paradox in the class notes).
The time measured by the spacecraft observer would not be useable by an earth
observer.
If the observer in the spacecraft times the run by watching a clock in the spacecraft, it appears as if the clock
on the earth ran slow, so that, in fact, MORE time elapsed during the run.
The spacecraft observer would
actually measure a longer time.
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 Fall '09
 Abra
 pH

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