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3714
Chapter 37
Solve this equation for
v
:
2
2
22
1/
vF
t
vc
m
⎛⎞
=
⎜⎟
−
⎝⎠
and
2
2
(1
/
)
Ft
vv
c
m
=−
2
1
Ft
Ft
v
mc
m
⎛
⎞
+=
⎜
⎟
⎝
⎠
so
2
2
2
(/)
1( / )
Ft m
Ft
Ft mc
m c
F t
==
++
As
2 2
,
1,
Ft
Ft
t
mc
Ft
→∞
→
→
+
so
.
→
EVALUATE:
Note that
Ft
+
is always less than 1, so
<
always and
v
approaches
c
only when
.
t
37.64.
Setting
0
x
=
in Eq.(37.21), the first equation becomes
xu
t
γ
′=−
and the last, upon multiplication by
,
c
becomes
.
ct
ct
′ =
Squaring and subtracting gives
2
2
γ
()
,
ct
x
ut
′′
−=
−
=
or
8
4.53 10 m.
xc
t t
=
×
37.65.
(a) IDENTIFY
and
SET UP:
Use the Lorentz coordinate transformation (Eq.37.21) for
11
(,)
xt
and
:
1
,
t
x
uc
−
′ =
−
2
t
x
−
′ =
−
2
1
/
,
tu
t
−
′ =
−
2
2
/
t
−
′ =
−
Same point in
S
′
implies
12
.
xx
=
What then is
21
?
ttt
′′′
Δ= −
EXECUTE:
=
implies
112 2
t xu
t
−=−
2 1
ut t
x
x
and
xx x
u
tt
t
−Δ
From the time transformation equations,
2
1
(/
)
x
c
Δ= − =
Δ−Δ
−
Using the result that
x
u
t
Δ
=
Δ
gives
2
1
((
)
/
(
(
)
)
)
1( )
/
(
( ) )
x
t
c
c
′
Δ=
Δ
Δ
2
)
/
(
(
)
)
)
() ()
/
t
x
t
c
tx
c
Δ
′
Δ
Δ−
Δ
2
2
/
() ( /
)
,
/
c
x
c
c
Δ
′
= Δ
Δ
as was to be shown.
This equation doesn’t have a physical solution (because of a negative square root) if
) ()
t
Δ>
Δ
or
.
t
Δ≥Δ
(b) IDENTIFY
and
SET UP:
Now require that
=
(the two events are simultaneous in
S
′
) and use the Lorentz
coordinate transformation equations.
EXECUTE:
=
implies
/
/
xc tu
−
2
u
c
−
so
2
x
c
Δ
and
2
u
x
Δ
=
Δ
From the Lorentz transformation equations,
1
.
xxx
t
−
Using the result that
2
/
uct x
=ΔΔ
gives
2
1
)
/
)
1(
)
/
(
)
c
t
x
ct x
′
Δ
Δ
−Δ Δ
22 2
)
/
)
x
c
t
x
t
Δ
′
Δ
Δ
t
c
t
t
′
−
Δ
(c) IDENTIFY
and
SET UP:
The result from part (b) is
c
t
′
Δ= Δ
−
Δ
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3715
Solve for
22
2
2
: (
)
(
)
( )
tx
x
c
t
′
ΔΔ=
Δ−Δ
EXECUTE:
2
2
8
8
(
)
(
)
(5.00 m)
(2.50 m)
1.44 10 s
2.998 10 m/s
xx
t
c
−
′
Δ−
Δ
−
Δ=
=
=
×
×
EVALUATE:
This provides another illustration of the concept of simultaneity (Section 37.2): events observed to
be simultaneous in one frame are not simultaneous in another frame that is moving relative to the first.
37.66.
(a)
80.0 m s is nonrelativistic, and
2
1
186 J.
2
Km
v
==
(b)
21
5
(1
)
1
.
3
1
1
0
J
.
mc
γ
−=
×
(c)
In Eq. (37.23), c)
88
7
2.20 10 m s,
1.80 10 m s,and so
7.14 10 m s.
vu
v
′
=×
=
−
×
(d)
20.0 m
13.6 m.
=
(e)
8
8
20.0 m
9.09 10 s.
2.20 10 m s
−
×
(f)
8
13.6 m
6.18 10 s, or
6.18 10 s.
2.20 10 m s
t
tt
−−
′′
×
=
=
×
×
37.67.
IDENTIFY
and
SET UP:
An increase in wavelength corresponds to a decrease in frequency
(/
)
,
fc
λ
=
so the
atoms are moving away from the earth. Receding, so use Eq.(37.26):
0
cu
ff
−
=
+
EXECUTE:
Solve for
u
:
2
0
(/ )
(
)
ff cu cu
+=−
and
2
0
2
0
1(/ )
uc
⎛⎞
−
=
⎜⎟
+
⎝⎠
00
/ ,
/
fc f c
λλ
so
/
/
=
8
0
0
1 (
/ )
1 (656.3/953.4)
0.357
1.07 10 m/s
1 (
/ )
1 (656.3/953.4)
c
c
=
=
×
++
EVALUATE:
The relative speed is large, 36% of
c
. The cosmological implication of such observations will be
discussed in Section 44.6.
37.68.
The baseball had better be moving nonrelativistically, so the Doppler shift formula (Eq.(37.25)) becomes
0
(1 (
)).
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 Spring '06
 Buchler
 Physics

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