Wong, Lemuel – Homework 13 – Due: Apr 30 2007, 11:00 pm – Inst: Matt Mackie
1
This
printout
should
have
13
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
NOTE the due time is 11pm central, which
is MIDNIGHT our time.
Also, you may
need to use
E
=
mc
2
,
and that electon
volts are converted to joules using 1eV =
1
.
602
×
10

19
J.
001
(part 1 oF 1) 10 points
Use the relativistic coordinate transFormation
(
x,y,z,t
)
←
(
x
0
,y
0
,z
0
,t
0
) shown and given
below where the latter Frame
S
0
, (
x
0
,y
0
,z
0
,t
0
),
has a velocity
v
in the positive direction rela
tive to the Frame
S
, (
x,y,z,t
).
y
x
z
S
y'
x'
z'
S'
v
The relativistic coordinate transFormation
is
x
=
x
0
x
0
=
x
y
=
y
0
y
0
=
y
z
=
γ
(
z
0
+
v t
0
)
z
0
=
γ
(
z

v t
)
t
=
γ
µ
t
0
+
β
c
z
0
¶
t
0
=
γ
µ
t

β
c
z
¶
.
Given
T
0
= 39 s,
c
= 2
.
99792
×
10
8
m
/
s,
v
= 2
.
69813
×
10
8
m
/
s,
β
≡
v
c
, and
γ
≡
r
1
1

β
2
, what is the observed time interval
T
=
t
2

t
1
in the stationary reFerence Frame
given
T
0
=
t
0
2

t
0
1
in the moving reFerence
Frame?
Correct answer: 89
.
4721 s.
Explanation:
Given :
β
=
v
c
=
2
.
69813
×
10
8
m
/
s
2
.
99792
×
10
8
m
/
s
= 0
.
9
γ
=
r
1
1

β
2
=
s
1
1

(0
.
9)
2
= 2
.
29416
In the moving reFerence Frame
S
0
suppose
there is an oscillator with a period oF
T
0
=
t
0
2

t
0
1
. In the stationary reFerence Frame
S
an
observer measures the period to be
T
=
t
2

t
1
.
The oscillator is resting (
z
0
constant) in
the moving reFerence Frame
S
0
.
ThereFore,
we can exclude the
z
0
coordinate, by trans
Forming From the moving reFerence Frame;
i.e.
,
z
0
2
=
z
0
1
=
z
0
, into the stationary reFerence
Frame. Applying the equation
t
=
γ
µ
t
0

β
c
z
0
¶
,
we have
T
=
t
2

t
1
=
γ
µ
t
0
2

β
c
z
0
¶

γ
µ
t
0
1

β
c
z
0
¶
=
γ
(
t
0
2

t
0
1
)
=
γ T
0
=
1
p
1

β
2
T
0
.
ThereFore
T
=
γ T
0
= (2
.
29416) (39 s)
= 89
.
4721 s
.
Thus an observer in Frame
S
measures a
longer period than is measured in
S
0
. IF the
period is that oF a clock, the observer in
S
would say it is running too slowly, For while
he has measured a time oF one minute, the
clock in
S
0
has not yet reached one minute.
This time dilation is with the length con
traction: iF a rod oF length
L
is at rest in
S
,
its length as measured From
S
0
is
L
0
=
1
γ
L
;
iF the oscillator is at rest in
S
, its period as
measured From
S
0
is
T
0
=
γ T
.
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