ENGI 5432
1.5
1.5 Conversions between Coordinate Systems
Page 1-32
Conversions between Coordinate Systems
In general, the conversion of a vector F  Fx ˆi  Fy ˆj  Fz kˆ from Cartesian coordinates
(x, y, z) to another orthonormal coordinate system (u, v, w) in 3 (where “orthonormal”
means that the new basis vectors aˆ u , aˆ v , aˆ w are mutually orthogonal and of unit length) is
given by F  F ˆi  F ˆj  F kˆ  F aˆ  F aˆ  F aˆ .
x
y
z

u u
v v

w w
 
 


However, Fu  F aˆ u  Fx ˆi  Fy ˆj  Fz kˆ aˆ u  ˆi aˆ u Fx  ˆj aˆ u Fy  kˆ aˆ u Fz .
Fv and Fw are defined similarly in terms of the Cartesian components Fx , Fy , Fz .
In matrix form
 ˆi aˆ u ˆj aˆ u kˆ aˆ u   F 
 Fu 
  x
 F    ˆi aˆ
ˆ
ˆ
 v j aˆ v k aˆ v   Fy  .
 v
ˆ
ˆ ˆ
 Fw 
ˆ ˆ  F 
ˆ
 i a w j a w k a w   z 
The matrices on the right hand side of the equation will contain a mixture of expressions
in the new (u, v, w ) and old (x, y, z ) coordinates. This needs to be converted into a
set of expressions in (u, v, w ) only.
Example 1.5.1
Express the vector F = y i  x j + z k in cylindrical polar coordinates.
x y plane: coordinates

x y plane: basis vectors
x   cos 
ˆi ˆ  ˆi
y   sin 
ˆj ˆ  11 cos     sin 
2
z  z
kˆ ˆ  11 cos 2  0
ˆ cos   cos 


ENGI 5432
1.5 Conversions between Coordinate Systems
Page 1-33
Example 1.5.1 (continued)


ˆi ˆ  11 cos      sin  **
2
ˆi kˆ  0
ˆj ˆ  11 cos   cos 
ˆj kˆ  0
kˆ ˆ  11 cos 2  0
kˆ kˆ  1
The coefficient conversion matrix from Cartesian to cylindrical polar is therefore
 cos  sin  0 
  sin  cos  0 


 0
0
1 
Letting c  cos  , s  sin  : F  y ˆi  x ˆj  z kˆ   s ˆi   c ˆj  z kˆ
Fpolar
  c s  s c  0 
 c s 0   s 
 0 






   s c 0     c      s  s  c  c  0       

 0 0 1   z 
 0  0  z    z 
Therefore
F  y ˆi  x ˆj  z kˆ    ˆ  z kˆ
** This result can be obtained from the trigonometric identity
cos  A  B   cos A cos B  sin A sin B
Setting A  2 and B   ,


cos 2    cos 2 cos   sin 2 sin   0  1 sin 
ENGI 5432
1.5 Conversions between Coordinate Systems
Page 1-34
We can also generate the coordinate transformation matrix from Cartesian coordinates
(x, y, z ) to spherical polar coordinates (r, ,  ).
[ is the declination (angle down from the north pole, 0     ) and
 is the azimuth (angle around the equator 0    2 ).]
[Vertical] Plane containing z-axis and radial vector r :
z  r cos 
The projection of the radial vector r  r rˆ
onto the plane z = r cos  has length
r sin 
The angle between r̂ and k̂ is 
 kˆ rˆ  cos 
The angle between ˆ and k̂ is 2  
 kˆ ˆ  cos 2     sin 


[Horizontal] Plane z = r cos  :
The projection of r rˆ onto the x axis ( î )
is x   r sin   cos    r rˆ  ˆi
 ˆi rˆ  sin  cos 
r =
x, y, z

r sin  cos , r sin  sin  , r cos .
The projection of r rˆ onto the y axis ( ĵ )
is y   r sin   sin    r rˆ  ˆj
 ˆj rˆ  sin  sin 
 ˆj ˆ  cos 
The angle between ˆ and ĵ is 
The angle between ˆ and î is 2  
 ˆi ˆ  cos 2     sin 
The remaining three entries in the coordinate conversion matrix can be found in a similar
way.


ENGI 5432
1.5 Conversions between Coordinate Systems
Page 1-35
The conversion matrix from Cartesian to spherical polar coordinates is then
 ˆi  rˆ ˆj  rˆ kˆ  rˆ 
 sin  cos 



ˆi  ˆ ˆj  ˆ kˆ  ˆ    cos  cos 
ˆ ˆ ˆ ˆ ˆ ˆ 
  sin 
 i   j   k   
sin  sin 
cos  sin 
cos 
cos  
 sin  
0 
Example 1.5.2
Convert F  y ˆi  x ˆj to spherical polar coordinates.
Let c1  cos  , s1  sin  , c2  cos  , s2  sin 
 y  rs1s2 ,  x   rs1c2
 Fr   s1c2
 
F   F    c1c2
 F    s2
 
s1s2
c1s2
c2
c1   r s1s2
 s1   r s1c2
0   0




 r s 2 c s  s c    0 
1
2 2
2 2


   r s1c1  c2 s2  s2 c2     0 

  r s 
2
2
1 
 r s1   s2  c2   
Therefore
F   r sin  ˆ
Expressions for the gradient, divergence, curl and Laplacian operators in any orthonormal
coordinate system will follow in section 1.7.
ENGI 5432
1.5 Conversions between Coordinate Systems
Page 1-36
Summary for Coordinate Conversion:
To convert a vector expressed in Cartesian components vx ˆi  v y ˆj  vz kˆ into the
equivalent vector expressed in cylindrical polar coordinates v ˆ  v ˆ  v kˆ , express


the Cartesian components vx, vy, vz in terms of   ,  , z  using
x =  cos  ,
y =  sin  ,
z
z = z ; then evaluate
v   cos  sin  0  vx 
 v     sin  cos  0 v 
  
  y
 vz   0
0
1   vz 
Use the inverse matrix to transform back to Cartesian coordinates:
 vx  cos   sin  0 v 
v    sin  cos  0  v 
 y 
 
 vz   0
0
1   vz 
To convert a vector expressed in Cartesian components vx ˆi  v y ˆj  vz kˆ into the
equivalent vector expressed in spherical polar coordinates v rˆ  v ˆ  v ˆ , express the
r
Cartesian components vx, vy, vz in terms of  r, ,   using
x  r sin  cos  , y  r sin  sin  , z  r cos  ; then evaluate
 vr   sin  cos 
  
v   cos  cos 
 v    sin 
 
sin  sin 
cos  sin 
cos 
cos 
 sin 
0


  vx 
 v 
  y
  vz 
Use the inverse matrix to transform back to Cartesian coordinates:
 vx  sin  cos  cos  cos   sin    vr 
v    sin  sin  cos  sin  cos   v 
 y 
 
 vz   cos 
 sin 
0   v 
Note that, in both cases, the transformation matrix A is orthogonal, so that A1 = AT.
This is generally true for transformations between orthonormal coordinate systems.
ENGI 5432
1.6
1.6 Basis Vectors in Other Coordinate Systems
Page 1-37
Basis Vectors in Other Coordinate Systems
In the Cartesian coordinate system, all three basis vectors are absolute constants:
d ˆ
d ˆ
d ˆ
i 
j 
k  0
dt
dt
dt
The derivative of a vector is then straightforward to calculate:
d
df
df
df
f1 ˆi  f 2 ˆj  f 3 kˆ  ˆi 1  ˆj 2  kˆ 3
dt
dt
dt
dt


But many non-Cartesian basis vectors are not constant.
Cylindrical Polar:
ˆ  cos  ˆi  sin  ˆj
ˆ   sin  ˆi  cos  ˆj
kˆ  kˆ
Let
v 
dv
dt
then
ˆ    sin    ˆi   cos    ˆj   ˆ
ˆ    cos    ˆi    sin    ˆj    ˆ
kˆ  0
Therefore if a vector F is described in cylindrical polar coordinates
F  F ˆ  F ˆ  Fz kˆ , then

 
 
F  F ˆ  F ˆ  F ˆ  F ˆ  Fz kˆ  0




 F  F ˆ  F  F ˆ   Fz  kˆ

ENGI 5432
1.6 Basis Vectors in Other Coordinate Systems
Page 1-38
In particular, the displacement vector is r  t     t  ˆ  0 ˆ  z  t  kˆ , so that the
velocity vector is
dr
d
d ˆ
dz ˆ
v 

ˆ  
 
k
dt
dt
dt
dt
Example 1.6.1
Find the velocity and acceleration in cylindrical polar coordinates for a particle travelling
along the helix x = 3 cos 2t , y = 3 sin 2t , z = t .
Cylindrical polar coordinates: x   cos  ,
y
  2  x 2  y 2 , tan  
x
 2  9cos2 2t  9sin 2 2t  9
tan  
3sin 2t
 tan 2t
3cos 2t
  3
   2t
z t
 v 
y   sin  ,
zz
  0
 2

z 1
dr
 0 ˆ  3  2 ˆ  1kˆ  6ˆ  kˆ
dt
[The velocity has no radial component – the helix remains the same distance from the
z axis at all times.]
a 
dv
 6ˆ  kˆ   6 ˆ  0   12 ˆ
dt
[The acceleration vector points directly at the z axis at all times.]
ENGI 5432
1.6 Basis Vectors in Other Coordinate Systems
Page 1-39
Spherical Polar Coordinates
Vertical plane containing z-axis and radial vector r :
rˆ  sin  cos  ˆi  sin  sin  ˆj  cos  kˆ
ˆ  cos  cos  ˆi  cos  sin  ˆj  sin  kˆ
Equatorial plane (   0 ):
[ sin  ˆ is the projection of r̂ onto the equatorial
plane.]
ˆ   sin  ˆi  cos  ˆj
[This reproduces the three rows of the coordinate conversion matrix in section 1.5.]
d rˆ 
d
d  ˆ
  cos 
cos   sin  sin 
i
dt
dt
dt 

d
d  ˆ 
d  ˆ

  cos 
sin   sin  cos 
 j    sin 
k
dt
dt 
dt 


d rˆ
d ˆ
d ˆ


  sin 

dt
dt
dt

d ˆ 
d
   sin 
cos   cos  sin 
dt
dt

d

   sin 
sin   cos  cos 
dt

dˆ
d
d ˆ

 
rˆ  cos 

dt
dt
dt
d  ˆ
i
dt 
d  ˆ 
d  ˆ
 j    cos 
k
dt 
dt 

ENGI 5432
1.6 Basis Vectors in Other Coordinate Systems
Page 1-40
Spherical Polar Coordinates (continued)
d ˆ 
d  ˆ 
d  ˆ
   cos 
 i    sin 
j
dt
dt 
dt 


But sin  rˆ  cos  ˆ  sin 2  cos  ˆi  sin 2  sin  ˆj  sin  cos 
 cos 2  cos  ˆi  cos 2  sin  ˆj  sin  cos 
 cos  ˆi  sin  ˆj

kˆ
kˆ
d ˆ
d
  sin  rˆ  cos  ˆ
dt
dt


In particular, the displacement vector is r  r rˆ , so that the velocity vector is
dr
dr
d rˆ
dr
d ˆ 
 d
v 

rˆ  r

rˆ  r  ˆ  sin 

dt
dt
dt
dt
dt 
 dt
dr
d ˆ
d ˆ
 v 
rˆ  r
  r sin 

dt
dt
dt
It can be shown that the acceleration vector in the spherical polar coordinate system is
 d 2r
  d  2  d   2 2  
dv
a 
  2  r 

sin    rˆ
  dt   dt 

 dt
dt



2
 1 d  2 d 

r  d 
 
r
 
sin 2  ˆ



 r dt  dt 

2  dt 


 1 d  2 d 2   ˆ
 
sin    
r

 r sin  dt  dt
d 2x ˆ d 2 y ˆ d 2z ˆ
i 
j 2k !
Compare this to the Cartesian equivalent a 
dt 2
dt 2
dt
ENGI 5432
1.6 Basis Vectors in Other Coordinate Systems
Page 1-41
Example 1.6.2
Find the velocity vector v for a particle whose displacement vector r, in spherical polar
coordinates, is given by r = 4,  = t ,  = 2t , (0 < t < ) .
r  4 ,   t ,   2t 
dr
0,
dt
d
1,
dt
d
2
dt
dr
d ˆ
d ˆ
rˆ  r
  r sin 

dt
dt
dt
 0 rˆ  4 1 ˆ  4  sin t   2 ˆ
v 

v  t   4 ˆ  8sin t ˆ
[This describes a path spiralling around a sphere of radius 4, from pole to pole.]
Summary:
Cylindrical Polar:
d
d ˆ
ˆ 

dt
dt
d ˆ
d
  
ˆ
dt
dt
d ˆ
k  0
dt
r   ˆ  z kˆ
 v   ˆ   ˆ  z eˆ z
Spherical Polar:
d rˆ
d ˆ
d ˆ

  sin 

dt
dt
dt
d ˆ
d
d ˆ
 
rˆ  cos 

dt
dt
dt
d ˆ
d
  sin  rˆ  cos  ˆ
dt
dt


r  r rˆ
 v  r rˆ  r ˆ  r sin   ˆ
ENGI 5432
1.7
1.7 Gradient Operator in Other Coordinate Systems
Page 1-42
Gradient Operator in Other Coordinate Systems
For any orthogonal curvilinear coordinate system (u1, u2, u3) in 3,
1 r
the unit tangent vectors along the curvilinear axes are eˆ i  Ti 
,
hi  ui
r
.
ui
where the scale factors hi 


The displacement vector r can then be written as r  u1eˆ 1  u 2eˆ 2  u3eˆ 3 , where
the unit vectors ê i form an orthonormal basis for 3.
0
eˆ i  eˆ j   ij  
1
i  j 
i  j 
[ ij is the “Kronecker delta”.]
The differential displacement vector dr is (by the Chain Rule)
r
r
r
dr 
du1 
du2 
du3  h1 du1 eˆ1  h2 du2 eˆ 2  h3 du3 eˆ 3
u1
u2
u3
and the differential arc length ds is
2
2
2
ds 2  dr dr   h1 du1    h2 du2    h3 du3 
The element of volume dV is
 x, y, z 
du1du 2 du3
u1 , u 2 , u3 
dV  h1h2 h3 du1du 2 du3 

Gradient operator
Gradient
Divergence
x
u1
x
u 2
x
u 3
y
u1
y
u 2
y
u 3
z
u1
z
du1du 2 du 3
u 2
z
u 3
eˆ 
eˆ1 
eˆ 
 2
 3
h1 u1
h2 u2
h3 u3
eˆ V
eˆ V
eˆ V
V  1
 2
 3
h1 u1
h2 u2
h3 u3
 
 F 
1    h2 h3 f1    h3 h1 f 2    h1 h2 f3  




h1 h2 h3 
u1
u2
u3

ENGI 5432
1.7 Gradient Operator in Other Coordinate Systems
F 
Curl
Laplacian
 2V 
1
h1 h2 h3
1
h1 h2 h3
h1 eˆ1

u1
h2 eˆ 2

u2
h3 eˆ 3

u3
h1 f1
h2 f 2
h3 f3
Page 1-43
   h2 h3 V 
  h3 h1 V 
  h1 h2 V  


 




 u1  h1 u1  u2  h2 u2  u3  h3 u3  
Cartesian:
hx = hy = hz = 1 .
Cylindrical polar:
h  hz  1 , h   .
Spherical polar:
hr  1, h  r , h  r sin  .
The familiar expressions then follow for the Cartesian coordinate system.
In cylindrical polar coordinates, naming the three basis vectors as ˆ , ˆ, kˆ , we have:
r   ˆ  0ˆ  z kˆ   , 0, z
The relationship to the Cartesian coordinate system is
x   cos  ,
y   sin  , z  z
  2  x 2  y 2 , tan  
One scale factor is
r
h 


2
2
 x    y   z 

 
 

        
2
2
2

 
  
  

  cos        sin       z  

 
  
  


cos 2   sin 2   0  1
In a similar way, we can confirm that h  
2
and hz  1 .
y
x
ENGI 5432
1.7 Gradient Operator in Other Coordinate Systems
Page 1-44
In cylindrical polar coordinates,
dV  h h hz d  d dz   d  d dz

ds 2   h d    h d
2
  ˆ
 F 
 f

F 
2V 
2
  hz dz    d      d    dz 
2
2
2
2

ˆ 


 kˆ

 
z
V  ˆ


V
ˆ V
V

 kˆ

 
z


    1 f    11 f
 1  f z  





1   1 


z


1

f


 fz
1  f

 
z
ˆ
 ˆ
kˆ

1   1 
f


 f

z
fz
1
1     1 V

    1 
 2V
1 V


2

 
   11 V    1  V  



 
      z  1 z  
1  2V
 2V
 2 2 
 
z 2
All of the above are undefined on the z-axis ( = 0), where there is a coordinate
singularity. However, by taking the limit as   0 , we may obtain well-defined values
for some or all of the above expressions.
ENGI 5432
1.7 Gradient Operator in Other Coordinate Systems
Page 1-45
Example 1.7.1
Given that the gradient operator in a general curvilinear coordinate system is
 eˆ 
eˆ 
eˆ  
   1
 2
 3
 , why isn’t the divergence of
h2 u2
h3 u3 
 h1 u1
 1 F1
1 F2
1 F3 
F  F1 eˆ 1  F2 eˆ 2  F3 eˆ 3 equal, in general, to 


?
h2 u2
h3 u3 
 h1 u1
The quick answer is that the differential operators operate not just on the components
F1 , F2 , F3 , but also on the basis vectors eˆ1 , eˆ 2 , eˆ 3 . In most orthonormal coordinate
systems, these basis vectors are not constant.
The divergence therefore contains
additional terms.
 e1 
e 
e  
 2
 3

   F1e1  F2e 2  F3e3  
h

u
h

u
h

u
2
2
3
3 
 1 1
 e1  e1 F1 F1
 e   e  e F
F
 e   e  e F
F
e 
 e1  1    2 1 1  1 e2  1    3 1 1  1 e3  1  

u1   h2 u2 h2
u2   h3 u3 h3
u3 
 h1 u1 h1
 e1  e2 F2 F2
e   e e
 e1  2    2 2

 h1 u1
h1
u1   h2

 e1  e3 F3 F3
e   e e
 e1  3    2 3

 h1 u1
h1
u1   h2

F2 F2
 e   e  e F
F
e 
 e 2  2    3 2 2  2 e3  2  
u2 h2
u2   h3 u3
h3
u3 
F3 F3
 e   e  e F F
e 
 e 2  3    3 3 3  3 e3  3  =
u2 h2
u2   h3 u3 h3
u3 
 1 F1 F1
e   F
e   F
e 

 e1  1    1 e2  1    1 e3  1  
u1   h2
u2   h3
u3 
 h1 u1 h1
 F2
 e   1 F2 F2
e   F
e 
 e1  2   
 e2  2    2 e3  2  
u1   h2 u2 h2
u2   h3
u3 
 h1
 F3
e   F
 e   1 F3 F3
e 
 e1  3    3 e2  3   
 e3  3 
u1   h2
u2   h3 u3 h3
u3 
 h1
For Cartesian coordinates, all derivatives of any basis vector are zero, which leaves the
familiar Cartesian expression for the divergence. But for most non-Cartesian coordinate
systems, at least some of these partial derivatives are not zero.
More complicated
expressions for the divergence therefore arise.
ENGI 5432
1.7 Gradient Operator in Other Coordinate Systems
Page 1-46
Example 1.7.1 (continued)
For cylindrical polar coordinates, we have
 1 F
F
 ˆ   F ˆ  ˆ   F ˆ  ˆ 
  ˆ 

  
  k

1
   
   1
z 
 1 
 F
 ˆ   1 F F ˆ  ˆ   F ˆ  ˆ 
 
 ˆ
 
  k

    

   1
z 
 1
 Fz
 kˆ   Fz ˆ  kˆ   1 Fz Fz ˆ  kˆ 


k
 ˆ 
   
 
   
   1 z
1
z 
 1
But none of the basis vectors varies with  or z * and the basis vector k̂ is absolutely
constant. Therefore the divergence becomes
 1 F
 F
 ˆ 
 0     ˆ 

   0 
 
 1 
  
 1 F
F
 ˆ 
 1 Fz

   0   0   0  
 0
  ˆ
 0   
 

 
 1 z



F
F
 F ˆ ˆ 
 ˆ
 ˆ 
But
 ˆ    ˆ 
      

 

 
 

and
 ˆ
  ˆ

F
 F

 ˆ 
   ˆ 
   ˆ    ˆ    0
 


 
So we recover the cylindrical polar form for the divergence,
div F 
*
F


F


1 F Fz

 
z
As shown here, the basis vectors ̂ and ˆ
clearly vary with  but do not change with  .
k is an absolute constant.
ENGI 5432
1.7 Gradient Operator in Other Coordinate Systems
Page 1-47
In spherical polar coordinates, naming the three basis vectors as rˆ , ˆ, ˆ, we have:
r  r rˆ  0ˆ  0 ˆ  r , 0, 0
The relationship to the Cartesian coordinate system is
x  r sin  cos  , y  r sin  sin  , z  r cos  .
One of the scale factors is
r
h 

2

2
 x    y   z 
     
        
2
2
2

 
  
  

  r sin  cos       r sin  sin       r cos   
 
  
  


 r cos cos  
2
  r cos  sin     r sin  
2
2
2
 r cos2   cos2   sin 2    sin 2   r cos2   sin 2   r
In a similar way, we can confirm that hr  1 and h  r sin  .
dV  1 r  r sin   dr d d  r 2 sin  dr d d
ds 2  1 dr    r d    r sin  d   dr 2  r 2 d 2  r 2 sin 2  d 2
2
 
2
2
rˆ 
ˆ 
ˆ



1 r
r 
r sin  
V  rˆ
V
ˆ V
ˆ V


r
r 
r sin  

   r  r sin  f    r sin  1 f   1 r f
1
r




1 r  r sin  
r




  r 2 fr 
 f 
  sin  f 
1
 sin 

 2
r
r
r sin  
r

 


 fr
2
1  f
cot 
1  f

 fr 

f 
r
r
r 
r
r sin  
 F 
 


ENGI 5432
F 
1.7 Gradient Operator in Other Coordinate Systems
1
r sin 
2
1
V  2
r sin 
2
rˆ

r
fr
r ˆ


r f
r sin  ˆ


r sin  f
   r 2 sin  V    r sin  V    r V  
 





 r    r     r sin    
 r  1
1 
  2 V
 sin 
r
2
r sin  
r  r
 2V
2 V
1  2V



r 2
r r
r 2  2

Page 1-48
  
V 
1   V  

 sin 



  sin      
  

cot  V
1
 2V

r 2 
r 2 sin 2   2
All of the above are undefined on the z-axis (sin  = 0), where there is a coordinate
singularity. However, by taking the limit as sin   0 , we may obtain well-defined
values for some or all of the above expressions.
ENGI 5432
1.7 Gradient Operator in Other Coordinate Systems
Page 1-49
Example 1.7.2
A vector field has the equation, in cylindrical polar coordinates (, , z),
k
k
F  n eˆ   n ˆ


Find the divergence of F and the value of n for which the divergence vanishes for all
 0.
In cylindrical polar coordinates,
div F 
F


F


Fz
1 F

 
z
F  k  n , F  Fz  0
 div F   kn   n1 
k  n
and clearly div F  0 when
F 
k


 0  0  k 1  n    n1
n 1.
ˆ is therefore a source-free field everywhere except on the z axis.
ENGI 5432
1.7 Gradient Operator in Other Coordinate Systems
Example 1.7.3
In spherical polar coordinates,
F  r , ,    f   cot  rˆ  2 f   ˆ  g  r ,   ˆ ,
where f () is any differentiable function of  only
and g (r, ) is any differentiable function of r and only.
Find the divergence of F.
Fr  f   cot  , F   2 f   , F  g  r,  
For spherical polar coordinates,

  r 2 Fr 
 F
  sin  F 
1
 sin 
∇F  2
r
r
r sin  
r








  r 2 f   cot  
  2sin  f   
 g  r ,  
1
 sin 

∇F  2
r
r
r sin  
r

 




2
f   cot    r 
2 f     sin 

2
r
r
r sin 


 0
2 f   cot 
2 f   cot 

 0
r
r
everywhere (except possibly on the z axis, where r sin   0 ).
Page 1-50
ENGI 5432
1.7 Gradient Operator in Other Coordinate Systems
Page 1-51
Example 1.7.4



Find curl sin ˆ  ˆ , where ,  are the two angular coordinates in the standard
spherical polar coordinate system.
F 

1
r sin 
2
1
r sin 
2
 F 
rˆ

r
fr
r ˆ


r f
r sin  ˆ


r sin  f
 2r sin  cos  0 rˆ


1
r sin 
2
rˆ

r
0
r ˆ


r sin 
r sin  ˆ


r sin 2 
 r  0  sin 2  ˆ  r sin   sin   0  ˆ
1
2 cos  rˆ  sin  ˆ  sin  ˆ
r


ENGI 5432
1.7 Gradient Operator in Other Coordinate Systems
Page 1-52
Central Force Law
If a potential function V(x, y, z), (due solely to a point source at the origin) depends only
on the distance r from the origin, then the functional form of the potential can be
deduced. Using spherical polar coordinates:
V (r, , ) = f (r)
1 d  df 
d2 f
2 df
 2V  2  r 2  

2
r dr  dr 
dr
r dr
But, in any regions not containing any sources of the vector field, the divergence of the
vector field F  V (and therefore the Laplacian of the associated potential function V)
must be zero. Therefore, for all r  0,
d2 f
2 df

 0
2
dr
r dr
Solve this ODE by reduction of order:
Let
y
df
dr
then



df
 y  Br 2
dr

f 
dy
2
 y  0
dr
r


dy
dr
 2
y
r
 ln y   2 ln r  C  ln Br 2

Br 1
 A
1
 V  r , ,    A 
B
r
OR (a much faster solution!)
 2V  0


dV B

dr r 2
1 d  2 dV
r
r 2 dr  dr

0

 V  A 
B
r
 r2
dV
B
dr
 r  0
ENGI 5432
1.7 Gradient Operator in Other Coordinate Systems
Page 1-53
Gravity is an example of a central force law, for which the potential function must be of
B
the form V  r ,  ,    A  . The zero point for the potential is usually set at infinity:
r
B

lim V  lim  A    A  0
r 
r 
r

The force per unit mass due to gravity from a point mass M at the origin is
F   V  
GM
rˆ
r2
But, in spherical polar coordinates,
V  rˆ


V
ˆ V
ˆ V
dV
B


 rˆ
 rˆ 2
r
r 
r sin  
dr
r
GM
B
  2
2
r
r
 B  GM
Therefore the gravitational potential function is
V r   
GM
r
The electrostatic potential function is similar, with a different constant of proportionality.
END OF CHAPTER 1