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where we use sin
2
θ
+ cos
2
= 1 to eliminate
. Now the righthand side can be written as
v
vc
c
2
2
2
2
1
1
1
1
−
=−
−
−
L
N
M
O
Q
P
(/)
,
so
1
1
11
1
1
2
22
2
−
=
F
H
G
I
K
J
−
F
H
G
I
K
J
+
F
H
G
I
K
J
L
N
M
M
O
Q
P
P
+
'
cos
'
sin
.
h
mc
λλ
λ
φφ
Now we rewrite Eq. 388 as
h
mc
1
1
1
2
−
F
H
G
I
K
J
+=
−
'
.
If we square this, then it can be directly compared with the previous equation we obtained
for [1 – (
v
/
c
)
2
]
–1
. This yields
2
2
1
1c
o
s
s
i
n
1
.
hh
mc
mc
⎡⎤
⎛⎞
⎛
⎞
⎛
⎞
⎛
⎞
−+=
−
+
+
⎢⎥
⎜⎟
⎜
⎟
⎜
⎟
⎜
⎟
′′
′
λ
⎝⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎣⎦
We have so far eliminated
and
v
. Working out the squares on both sides and noting that
sin
2
φ
+ cos
2
φ
= 1, we get
= =
'(
c
o
s
)
.
−−
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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