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(1
)(1
)
)
AC
AB
BC
AB
BC
AC
β
ββ
βββ
−++
=
−−+
and expand:
1 –
β
AC
+
β
AB
+
β
BC
–
β
AC
β
AB
–
β
AC
β
BC
+
β
AB
β
BC
–
β
AB
β
BC
β
AC
=
1 +
β
AC
–
β
AB
–
β
BC
–
β
AC
β
AB
–
β
AC
β
BC
+
β
AB
β
BC
+
β
AB
β
BC
β
AC
We note that several terms are identical on both sides of the equals sign, and thus cancel,
which leaves us with
–
β
AC
+
β
AB
+
β
BC
–
β
AB
β
BC
β
AC
=
β
AC
–
β
AB
–
β
BC
+
β
AB
β
BC
β
AC
which can be rearranged to produce
22
AB
BC
AC
AB
BC
AC
+
=+
.
The lefthand side can be written as
2
β
AC
(1 +
β
AB
β
BC
)
in which case it becomes clear
how to obtain the result from part (a) [just divide both sides by 2
(1
+
β
AB
β
BC
)
].
67. Interpreting
v
AB
as the
x
component of the velocity of
A
relative to
B
, and defining the
corresponding speed parameter
AB
=
v
AB
/
c
, then the result of part (a) is a straightforward
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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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