Mathematical Physics Vol 1

Chapter 4. Field theory

204

i . Given that

where f i , i = 1 , 2 , 3 are unambiguous functions of u

∂ r ∂ u 2

∂ r ∂ u 3

∂ r ∂ u 1

d u 1 +

d u 2 +

d u 3 =

d r =

1 + h

2 + h

3 ,

= h 1 e 1 d u

2 e 2 d u

3 e 3 d u

it follows (see p. 85, (4.26))

1 + h

2 + h

3 .

d φ = ∇ φ · d r = h 1 f 1 d u

2 f 2 d u

3 f 3 d u

(4.224)

On the other hand, the total differential of the scalar function φ ( u i ) is

∂φ ∂ u 1

∂φ ∂ u 2

∂φ ∂ u 3

d u 1 +

d u 2 +

d u 3 .

d φ =

(4.225)

From (4.224) and (4.225) it follows that

∂φ ∂ u 1

∂φ ∂ u 2

∂φ ∂ u 3

1 h 1

1 h 2

1 h 3

f 1 =

f 2 =

f 3 =

(4.226)

,

,

.

Substituting the values (4.226) into (4.223) we obtain

∂φ ∂ u 1

∂φ ∂ u 2

∂φ ∂ u 3

e 2 h 2

e 3 h 3

e 1 h 1

∇ φ =

+

+

=

=

∂ ∂ u 3

∂ ∂ u 1

∂ ∂ u 2

e 1 h 1

e 2 h 2

e 3 h 3

φ .

+

+

Thus, the ∇ operator, in orthogonal coordinates, is

∂ ∂ u 1

∂ ∂ u 2

∂ ∂ u 3

e 1 h 1

e 2 h 2

e 3 h 3

∇ ≡

+

+

.

Exercise 158 Let u 1 , u 2 and u 3 be orthogonal coordinates. Prove that a) | ∇ u p | = h − 1 p , b) e p = E p , p = 1 , 2 , 3.

Solution a) Introducing the substitution φ = u 1 in relations on page 203, we obtain ∇ u 1 =

e 1 h 1

.

The magnitude of this vector is | ∇ u 1 | = | e 1 | h 1 this procedure for φ = u 2 and φ = u 3 yields | ∇ u 2 | = | e 2 | h 2

= h − 1

1 , because | e 1 | = 1. Repeating

= h − 1 2

e 3 | h 3

| ∇ u 3 | = |

= h − 1

3 ,