Page 87 - Intro to Tensor Calculus

82

Figure 1.3-12. Angles between curvilinear coordinates.

2
2
1
1
3
3
Solution: Let y = x, y = y, y = z and x = u, x = v, x = w denote the Cartesian and curvilinear
coordinates respectively. With reference to the figure 1.3-12 we can interpret the intersection of the surfaces
v = c 2 and w = c 3 as the curve ~r = ~r(u, c 2 ,c 3 ) which is a function of the parameter u. By moving only along
r
∂~
r
this curvewehave d~ = du and consequently
∂u
r
r
∂~ ∂~ 1 2
2
ds = d~r · d~r = · dudu = g 11 (dx ) ,
∂u ∂u
or
1 2
d~ d~r dx
r
1= · = g 11 .
ds ds ds
1 1
dx
This equation shows that the vector = √ is a unit vector along this curve. This tangent vector can
ds g 11
r
be represented by t r = √ 1 δ .
(1) g 11 1
The curve which is defined by the intersection of the surfaces u = c 1 and w = c 3 has the unit tangent
r
vector t r = √ 1 δ . Similarly, the curve which is defined as the intersection of the surfaces u = c 1 and
(2) g 22 2
r
v = c 2 has the unit tangent vector t r = √ 1 δ . The cosine of the angle θ 12 , which is the angle between the
(3) g 33 3
unit vectors t r and t r , is obtained from the result of equation (1.3.25). We find
(1) (2)
1 1
p q p q g 12
cos θ 12 = g pq t t = g pq √ δ √ δ = √ √ .
(1) (2) 1 2
g 11 g 22 g 11 g 22
For θ 13 the angle between the directions t i and t i we find
(1) (3)
g 13
cos θ 13 = √ √ .
g 11 g 33
Finally, for θ 23 the angle between the directions t i and t i we find
(2) (3)
g 23
cos θ 23 = √ √ .
g 22 g 33
When θ 13 = θ 12 = θ 23 =90 , we have g 12 = g 13 = g 23 = 0 and the coordinate curves which make up the
◦
curvilinear coordinate system are orthogonal to one another.
In an orthogonal coordinate system we adopt the notation
2
2
g 11 =(h 1 ) , g 22 =(h 2 ) , g 33 =(h 3 ) 2 and g ij =0,i 6= j.