46. We rewrite Eq. 39-9 as

h

mλ

−

h

mλ

cos φ =

v

1

− (v/c)

2

cos θ ,

and Eq. 39-10 as

h

mλ

sin φ =

v

1

− (v/c)

2

sin θ .

We square both equations and add up the two sides:

h

m

2

1

λ

−

1

λ

cos φ

2

+

1

λ

sin φ

2

=

v

2

1

− (v/c)

2

,

where we use sin

2

θ + cos

2

θ = 1 to eliminate θ. Now the right-hand 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

=

h

mc

2

1

λ

−

1

λ

cos φ

2

+

1

λ

sin φ

2

+ 1 .

Now we rewrite Eq. 39-8 as

h

mc

1

λ

−

1

λ

+ 1 =

1

1

− (v/c)

2

.

If we square this, then it can be directly compared with the previous equation we obtained for [1

−

(v/c)

2

]

−1

. This yields

h

mc

1

λ

−

1

λ

+ 1

2

=

h

mc

2

1

λ

−

1

λ

cos φ

2

+

1

λ

sin φ

2

+ 1 .

We have so far eliminated θ and v. Working out the squares on both sides and noting that sin

2

φ+cos

2

φ =

1, we get

λ

− λ = ∆λ =

h

mc

(1

− cos φ) .