College of Engineering and Computer Science
Mechanical Engineering Department
ME 692 – Computational Fluid Dynamics
Spring 2002 Ticket: 57541 Instructor: Larry Caretto
Solutions to Sixth Homework Assignment
1. Review the grid transformation notes leading to equation [20] on page 5.
Show that the remaining, unique, off-diagonal terms, g13 and g23, which are
not shown in equation [20] are zero.
The metric coefficients, gij are given by equation [18] from the grid-transformation
notes, where yi are the Cartesian coordinates and xi represent the new
coordinate system:
g ij 
yk ym
y y y y y y
 km  1 1  2 2  3 3
xi x j
xi x j xi x j xi x j
[18]
The necessary partial derivatives for the cylindrical coordinates are shown in
equation [19] from the notes.
y1
 cos( x2 )
x1
y 2
 sin( x2 )
x1
y3
0
x1
y1
  x1 sin( x2 )
x2
y 2
 x1 cos( x2 )
x2
y3
0
x2
y1
0
x3
y2
0
x3
y3
1
x2
[19]
Using equation [18] and the appropriate derivatives from [19] gives the following
results, which shows that the remaining off-diagonal metric coefficients are zero.
g13 
y1 y1 y 2 y 2 y3 y3


 cos( x2 )[0]  sin( x 2 )[0]  [0][1]  0
x1 x3 x1 x3 x1 x3
g 23 
y1 y1 y 2 y 2 y3 y3


  x1 sin( x 2 )[0]  x1 cos( x 2 )[0]  [0][1]  0
x 2 x3 x 2 x3 x 2 x3
2. For the spherical polar system the three coordinates are x1, the distance
from the origin to a point on a sphere, x2, the counterclockwise angle on
the x-y plane from the x axis to the projection of the r coordinate on the x-y
plane, and x3 = the angle from the vertical axis to the line from the origin to
the point. (These coordinates are more conventionally called r, θ, and φ.)
The transformation equations from Cartesian coordinates (y1, y2, and y3) to
spherical polar coordinates are given by the following equations: x1 =
y12 + y22 + y32, x2 = tan-1(y2/y1), and x3 = tan-1[ y12 + y22/y3]. The inverse
Engineering Building Room 2303
E-mail: lcaretto@csun.edu
Mail Code
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Phone: 818-677-6448
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Page 2
ME 692, L. S. Caretto, Spring 2002
Homework Six Solutions
transformation to obtain Cartesian coordinates from spherical polar
coordinates is: y1 = x1 cos(x2)sin(x3), y2 = x1 sin(x2)sin(x3), and y3 =
x1cos(x3). Find all components of the metric tensor for this transformation.
Verify that this is an orthogonal coordinate system. What are the three
possible differential areas for this system? What is the volume element for
this system?
The solution to this equation requires the application of equation [18] from the
notes, shown in problem one, to the coordinate transformation spherical
coordinates. The partial derivatives required for this computation are the same
as those shown in equation [19], but applied to the coordinate transformation
given in the problem statement. Those partial derivatives are shown below.
Note that these derivatives use the transformation equations for transforming the
spherical coordinate system back into Cartesian coordinates.
y1
y1
y1
 cos( x2 ) sin( x3 )
  x1 sin( x2 ) sin( x3 )
 x1 cos( x2 ) cos( x3 )
x1
x2
x3
y 2
y 2
y 2
 sin( x2 ) sin( x3 )
 x1 cos( x2 ) sin( x3 )
 x1 sin( x2 ) cos( x3 )
x1
x2
x3
y3
y3
y3
 cos( x3 )
0
  x1 sin( x3 )
x1
x2
x2
Using these derivatives in equation [18], gives the following metric coefficients.
g11 
y1 y1 y 2 y 2 y3 y3


 cos 2 ( x2 ) sin 2 ( x3 )  sin 2 ( x2 ) sin 2 ( x3 ) 
x1 x1 x1 x1 x1 x1


cos 2 ( x3 )  sin 2 ( x3 ) cos 2 ( x2 )  sin 2 ( x2 )  cos 2 ( x3 )  sin 2 ( x3 )  cos 2 ( x3 )  1
g12 
y1 y1 y 2 y 2 y3 y3


  x1 sin( x2 ) cos( x2 ) sin 2 ( x3 )
x1 x2 x1 x2 x1 x2
 x1 sin( x2 ) cos( x2 ) sin 2 ( x3 )  cos( x3 )[0]  0
g13 
y1 y1 y 2 y 2 y3 y3


 x1 cos 2 ( x2 ) cos( x3 ) sin( x3 ) 
x1 x3 x1 x3 x1 x3
x1 sin 2 ( x2 ) cos( x3 ) sin( x3 )  x1 cos( x3 ) sin( x3 ) 


x1 cos( x3 ) sin( x3 ) cos 2 ( x2 )  sin 2 ( x2 )  1  0
g 22 

y1 y1 y 2 y 2 y3 y3
2
2


  x1 sin( x2 ) sin( x3 )  x1 cos( x2 ) sin( x3 )  [0]2
x2 x2 x2 x2 x2 x2

 x12 sin 2 ( x2 )  cos 2 ( x2 ) sin 2 ( x3 )  x12 sin 2 ( x3 )
Homework Six Solutions
g 23 
ME 692, L. S. Caretto, Spring 2002
Page 3
y1 y1 y 2 y 2 y3 y3


  x12 sin( x2 ) sin( x3 ) cos( x2 ) cos( x3 ) 
x2 x3 x2 x3 x2 x3
x12 sin( x2 ) sin( x3 ) cos( x2 ) cos( x3 )  x1 sin( x3 )[0]  0
g 33 
y1 y1 y 2 y 2 y3 y3
2
2


 x1 cos( x2 ) cos( x3 )  x1 sin( x2 ) cos( x3 )
x3 x3 x3 x3 x3 x3


  x1 sin( x3 )  x12 cos ( x3 ) x1 cos 2 ( x2 )  sin 2 ( x2 )  x12 sin 2 ( x3 )
2
 x12 cos 2 ( x3 )  x12 sin 2 ( x3 )  x12
We have the usual symmetry relationship for the off-diagonal coefficients so that
g21 = g12 = 0, g31 = g13 = 0, and g32 = g23 = 0.
The differential areas are found from equation [26] in the notes, which is copied
below.
(dS ) i  g jj g kk  g 2jk dx j dxk
(no summation)
i, j , k cyclic
i  1, 2, 3
[26]
Using the metric coefficients just found, we obtain the following differential areas.
2
(dS )1  g 22 g 33  g 23
dx2 dx3  x12 sin 2 ( x3 ) x12  0dx2 dx3  x12 sin( x3 )dx2 dx3
(dS ) 2  g 33 g11  g132 dx2 dx3  x12 (1)  0dx3 dx1  x1dx3 dx1
(dS ) 3  g11 g 22  g122 dx1dx2  (1) x12 sin 2 ( x3 )  0dx1dx2  x1 sin( x3 )dx1dx2
In these equations, (dS)i represents the differential area facing the ith coordinate
direction. For example, x1 corresponds to the r coordinate direction. So the
equation that we have just derived tells us that the differential area in this
direction is r2sin(φ)dφdθ, where φ is the polar angle and θ is the angle in the x-y
plane. This differential area represents the differential area on a spherical
surface, a distance r from the origin, when the change in the polar angle creates
a differential length of rdφ on the surface of the sphere and the angle in the x-y
plane creates a differential path length rsin(φ)dθ.
The differential volume element for this coordinate system is found from equation
[30] in the notes, which is copied below.
dV 
g dx1dx2 dx3  Det ( g ij ) dx1dx2 dx3
[30]
For an orthogonal coordinate system, such as the one we have here, where all
the off-diagonal gij are zero the determinant is simply the product of the diagonal
terms. This gives the differential volume element by the following equation.
dV  g11 g12 g13 dx1dx2 dx3  (1) x12 sin 2 ( x3 ) x12 dx1dx2 dx3  x12 sin( x3 )dx1dx2 dx3