
Note that there is now a certain
ambiguity: You describe the same vector for
an ¥ set of values for Q and f, because you
always can add n·2p (n = 1,2,3...) to
any of the two angles and obtain the same result. 


This has a first consequence if you
do an integration. Lets look at the ubiquitous case of normalizing a wave
function y (x,y,z) by demanding that 


¥
ó
õ
–¥ 
¥
ó
õ
–¥ 
¥
ó
õ
–¥ 
y (x,y,z) · dxdydz = 1 



In spherical coordinates, we
have 


¥
ó
õ
0 
2p
ó
õ
0 
p
ó
õ
0 
y (r,j,Q) · dr dj dQ = 1 




You no longer integrate from
¥ to ¥ with respect to the angles, but from 0
to 2p for j
and from 0 to p for Q because this covers all of space. Notice the different upper bounds! 

Lets try this by computing the volume
V_{R} of a sphere with radius R. This is always done by
summing over all the differential volume elements dV inside the body
defined by some equation 


In Cartesian coordinates we have for
the volume element dV = dxdydz, and for the integral:



¥
ó
õ
–¥ 
¥
ó
õ
–¥ 
¥
ó
õ
–¥ 
??? dxdydz 




Well, if you can just formulate the integral, let alone solving it, you
are already doing well! 

In spherical coordinates we first
have to define the volume element. This is relatively easily done by looking at
a drawing of it: 






An incremental increase in the three
coordinates by dr, dj, and dQ produces the volume element dV which is close
enough to a rectangular body to render its volume as the product of the length
of the three sides. 

Looking at the basic geometry, the length of the
three sides are identified as
dr, r · dQ, and r ·
sinQ · dj, which
gives the volume element 







dV 
= 
r^{2} · sinQ · dr ·
dQ · dj 








The volume of our sphere thus results
from the integral 





V_{r} = 
¥
ó
õ
0 
2p
ó
õ
0 
p
ó
õ
0 
r^{2} · sinQ · dr dj dQ = 2p · 
¥
ó
õ
0 
p
ó
õ
0 
r^{2} · sinQ · dr dQ = 2p 
· [–cos p + cos 0] · 
¥
ó
õ
0 
r^{2} · dr 
V_{r} = 
2p · 
[2] · 
1/3R^{3} 
= (4/3) · p ·
R^{3} 
q.e.d. 







Not extremely easy, but no problem
either. 

Next, consider
differential
operators, like div, rot, or more general,
and
^{2}
(= D). 


Lets just look at D to see what happens. We have (for some function
U) 





Cartesian
coordinates 
D = 
¶^{2}U

+ 
¶^{2}U

+ 
¶^{2}U

¶x^{2} 
¶y^{2} 
¶z^{2} 







Spherical
Coordinates 
D = 
¶^{2}U
¶r^{2} 
+ 
2
r 
· 
¶U
¶r 
+ 
1
r^{2} · sin^{2} · Q 
· 
¶^{2}U
¶j^{2} 
+ 
1
r^{2} 
· 
¶^{2}U
¶Q^{2} 
+ 
cotg Q
r^{2} 
· 
¶U
¶Q 







Looks messy, OK, but it is
still a lot easier to work with this D
operator than with its Cartesian counterpart for problems with spherical
symmetry; witness the
solution of
Schrödingers equation for the Hydrogen atom. 