Mathematical Physics Vol 1

Chapter 4. Field theory

130

it follows that ∇ · ( A + B )=

∂ ∂ z k · [( A 1 + B 1 ) i +( A 2 + B 2 ) j +( A 3 + B 3 ) k ]=

∂ ∂ x

∂ ∂ y

i +

j +

∂ ( A 1 + B 1 ) ∂ x

∂ ( A 2 + B 2 ) ∂ y

∂ ( A 3 + B 3 ) ∂ z

=

+

+

=

∂ A 1 ∂ x

∂ A 2 ∂ y

∂ A 3 ∂ z

∂ B 1 ∂ x

∂ B 2 ∂ y

∂ B 3 ∂ z

=

+

+

+

+

+

=

= ∇ · A + ∇ · B .

b)

∇ · ( φ A )= ∇ · ( φ A 1 i + φ A 2 j + φ A 3 k )= = ∂ ( φ A 1 ) ∂ x + ∂ ( φ A 2 ) ∂ y + ∂ ( φ A 3 ) ∂ z =

∂φ ∂ x

∂φ ∂ y ∂φ ∂ y

∂φ ∂ z

∂ A 1 ∂ x

∂ A 2 ∂ y

∂ A 3 ∂ z

φ +

φ +

φ =

A 1 +

A 2 +

A 3 +

=

=

k · ( A 1 i + A 2 j + A 3 k )+

∂φ ∂ x

∂φ ∂ z

i +

j +

+ φ ·

∂ z

∂ A 1 ∂ x

∂ A 2 ∂ y

∂ A 3

+

+

=

=( ∇ φ ) · A + φ ( ∇ · A ) .

Problem52 Show that ∇ 2 1

r

= △

1 r

= 0, where r = p x 2 + y 2 + z 2 .

Solution

∇ 2

1 r

= =

∂ 2 ∂ z 2 p

x 2 + y 2 + z 2

∂ 2 ∂ x 2 ∂ 2 ∂ x 2

∂ 2 ∂ y 2 ∂ 2 ∂ y 2

− 1

+

+

=

∂ 2 ∂ z 2

( x 2 + y 2 + z 2 ) − 1 2 .

+

+

Given that

∂ ∂ x

x 2 + y 2 + z 2 − 1

2 = − x x 2 + y 2 + z 2 − 3 2 ,

it follows that ∂ 2 ∂ x 2

2 x 2 − y 2 − z 2 ( x 2 + y 2 + z 2 )

∂ ∂ x

( x 2 + y 2 + z 2 ) − 1

( − x ( x 2 + y 2 + z 2 ) − 3

(4.153)

2 =

2 )=

.

5 2

Analogously we obtain

∂ 2 ( x 2 + y 2 + z 2 ) − 1 2 ∂ y 2

2 y 2 − x 2 − z 2 ( x 2 + y 2 + z 2 )

(4.154)

=

,

5 2