Mathematical Physics Vol 1

4.6 Examples

203

or

∂ r ∂ u 1 ·

∂ r ∂ u 3

∂ r ∂ u 2 ×

∇ u 1 · ∇ u 2 × ∇ u 3 = 1 ,

which was to be proved.

Exercise 156 Prove that the square of the arc element, in generalized coordinates, can be expressed as follows d s 2 = 3 ∑ p = 1 3 ∑ q = 1 g pq d u p d u q .

Solution We have

∂ r ∂ u 1

∂ r ∂ u 2

∂ r ∂ u 3

d u 1 +

d u 2 +

d u 3 = α

1 + α

2 + α

3 ,

d r =

1 d u

2 d u

3 d u

so that

d s 2 = d r · d r = α

1 ) 2 + α

1 d u 2 + α

1 d u 3 2 d u 3

1 · α 1 ( d u

1 · α 2 d u 2 · α 2 ( d u 3 · α 2 d u

1 · α 3 d u 2 · α 3 d u 3 · α 3 ( d u

2 d u 1 + α 3 d u 1 + α

2 ) 2 + α

+ α 2 · α 1 d u + α 3 · α 1 d u

3 d u 2 + α

3 ) 2

3 ∑ p = 1

3 ∑ q = 1

= pq = α p · α q . This expression is called the fundamental quadratic form or metric form . The values g pq are called metric coefficients and are symmetrical ( g pq = g qp ). If g pq = 0 for p̸ = q then the coordinate system is orthogonal. Then g 11 = h 2 1 , g 22 = h 2 2 , g 33 = h 2 3 . g pq d u p d u q gde je g

4.6.9 Gradient, divergence and rotor in generalized orthogonal coordinates

Exercise 157 Determine ∇ φ in generalized orthogonal coordinates.

Solution Let

∇ φ = f 1 e 1 + f 2 e 2 + f 3 e 3 ,

(4.223)