Mathematical Physics Vol 1

4.6 Examples

211

( u 1 , u 2 , u 3 ) system, are given by ( x = x 1 ( u 1 , u 2 , u 3 ) , x = x 2 ( u 1 , u 2 , u 3 ) ,

1 , u 2 , u 3 ) , 1 , u 2 , u 3 ) ,

1 , u 2 , u 3 ) 1 , u 2 , u 3 ) .

y = y

1 ( u

z = z

1 ( u

(4.227)

y = y

2 ( u

z = z

2 ( u

Then there exists a direct transformation from system ( u 1 , u 2 , u 3 ) into system ( u 1 , u 2 , u 3 ) definedby u 1 = u 1 ( u 1 , u 2 , u 3 ) , u 2 = u 2 ( u 1 , u 2 , u 3 ) , u 3 = u 3 ( u 1 , u 2 , u 3 ) , (4.228) and vice versa. Based on the first set of equations, we obtain

∂ r ∂ u 2 ∂ r ∂ u 2

∂ r ∂ u 3 ∂ r ∂ u 3

∂ r ∂ u 1 ∂ r ∂ u 1

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

or

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

By equalizing the right sides we obtain α 1 d u 1 + α 2 d u 2 + α 3 d u

3 = α

1 + α

2 + α

3 .

1 d u

2 d u

3 d u

(4.229)

From (4.228)it follows that

∂ u 1 ∂ u 2 ∂ u 2 ∂ u 2 ∂ u 3 ∂ u 2

∂ u 1 ∂ u 3 ∂ u 2 ∂ u 3 ∂ u 3 ∂ u 3

∂ u 1 ∂ u 1 ∂ u 2 ∂ u 1 ∂ u 3 ∂ u 1

d u 1 +

d u 2 +

d u 3

d u 1 =

d u 2 =

d u 1 +

d u 2 +

d u 3

d u 3 =

d u 1 +

d u 2 +

d u 3 .

By substituting into (4.229) and equalizing the coefficients next to d u 1 , d u 2 and d u 3 , we obtain     α 1 = α 1 ∂ u 1 ∂ u 1 + α 2 ∂ u 2 ∂ u 1 + α 3 ∂ u 3 ∂ u 1 , α 2 = α 1 ∂ u 1 ∂ u 2 + α 2 ∂ u 2 ∂ u 2 + α 3 ∂ u 3 ∂ u 2 , α 3 = α 1 ∂ u 1 ∂ u 3 + α 2 ∂ u 2 ∂ u 3 + α 3 ∂ u 3 ∂ u 3 . (4.230) Now A can be expressed in the two systems A = C 1 α 1 + C 2 α 2 + C 3 α 3 = C 1 α 1 + C 2 α 2 + C 3 α 3 , (4.231) ) are contravariant components of vector A in the two systems. Substituting the equation (4.230) into equation (4.231) we obtain C 1 α 1 + C 2 α 2 + C 3 α 3 = C 1 α 1 + C 2 α 2 + C 3 α 3 = = C 1 ∂ u 1 ∂ u 1 + C 2 ∂ u 1 ∂ u 2 + C 3 ∂ u 1 ∂ u 3 α 1 + C 1 ∂ u 2 ∂ u 1 + C 2 ∂ u 2 ∂ u 2 + C 3 ∂ u 2 ∂ u 3 α 2 + + C 1 ∂ u 3 ∂ u 1 + C 2 ∂ u 3 ∂ u 2 + C 3 ∂ u 3 ∂ u 3 α 3 , where ( C 1 , C 2 , C 3 ) and ( C 1 , C 2 , C 3