Mathematical Physics Vol 1

Chapter 4. Field theory

202

and by a scalar product with ∇ u 1 , we obtain ∇ u 1 · d r = d u 1 = ∇ u 1 · ∂ r ∂ u 1

d u 1 + ∇ u 1 ·

∂ r ∂ u 2

d u 2 + ∇ u 1 ·

∂ r ∂ u 3

d u 3 ,

or

∂ r ∂ u 1

∂ r ∂ u 2

∂ r ∂ u 3

∇ u 1 ·

∇ u 1 ·

∇ u 1 ·

= 1 ,

= 0 ,

= 0 .

Similarly, it can be shown that

∂ r ∂ u p

∂ r ∂ u p

∇ u 2 ·

3 ·

= δ 2 p and ∇ u

= δ 3 p .

Exercise 155

Prove that

∂ r ∂ u 1 ·

∂ r ∂ u 3

∂ r ∂ u 2 ×

∇ u 1 · ∇ u 2 × ∇ u 3 = 1 .

Solution In the previous example we have shown that ∂ r ∂ u 1 , ∂ r ∂ u 2 , ∂ r ∂ u 3

and ∇ u 1 , ∇ u 2 , ∇ u 3

are reciprocal vectors. Let us denote the corresponding Jacobians by

J = ∂ ( u 1 , u 2 , u 3 ) ∂ ( x , y , z ) . The Jacobians are equal to corresponding mixed products (see Example 153, p. 199) J = ∂ r ∂ u 1 · ∂ r ∂ u 2 × ∂ r ∂ u 3 , j = | ∇ u 1 · ∇ u 2 × ∇ u 3 | . Further, according to the theorem that states that a mixed product of three vectors yields a volume, and the mixed product of the reciprocal vectors yields the reciprocal volume (see Example 20c, on p. 66), we obtain ∂ ( x , y , z ) ∂ ( u 1 , u 2 , u 3 ) and j =

∂ u 1 ∂ x ∂ u 2 ∂ x ∂ u 3 ∂ x

∂ u 1 ∂ y ∂ u 2 ∂ y ∂ u 3 ∂ y

∂ u 1 ∂ z ∂ u 3 ∂ z ∂ u 3 ∂ z

∇ u 1 · ∇ u 2 × ∇ u 3 =

= j

and

1 d V

d V = j · J = 1 ,