This preview shows page 1. Sign up to view the full content.
46. We rewrite Eq. 399 as
h
mλ
−
h
mλ
0
cos
φ
=
v
p
1
−
(
v/c
)
2
cos
θ,
and Eq. 3910 as
h
mλ
0
sin
φ
=
v
p
1
−
(
v/c
)
2
sin
θ.
We square both equations and add up the two sides:
µ
h
m
¶
2
"
µ
1
λ
−
1
λ
0
cos
φ
¶
2
+
µ
1
λ
0
sin
φ
¶
2
#
=
v
2
1
−
(
v/c
)
2
,
where we use sin
2
θ
+cos
2
θ
= 1 to eliminate
θ
. Now the righthand side can be written as
v
2
1
−
(
v/c
)
2
=
−
c
2
·
1
−
1
1
−
(
v/c
)
2
¸
,
so
1
1
−
(
v/c
)
2
=
µ
h
mc
¶
2
"
µ
1
λ
−
1
λ
0
cos
φ
¶
2
+
µ
1
λ
0
sin
φ
¶
2
#
+1
.
Now we rewrite Eq. 398 as
h
mc
µ
1
λ
−
1
λ
0
¶
+1=
1
p
1
−
(
v/c
)
2
.
If we square this, then it can be directly compared with the previous equation we obtained for [1
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

Click to edit the document details