# Spherical Coordinates

| For many mathematical problems, it is far easier to use spherical coordinates instead of Cartesian ones.
|  | In essence, a vector r (we drop the underlining here) with the Cartesian coordinates (x,y,z) is expressed in spherical coordinates by giving its distance from the origin (assumed to be identical for both systems) |r|, and the two angles j and Q between the direction of r and the x- and z-axis of the Cartesian system.
|  | This sounds more complicated than it actually is: j and Q are nothing but the geographic longitude and latitude. The picture below illustrates this.
| This is simple enough, for the translation from one system to the other one we have the equations
x = | r · sinQ · cosj |  | r = | (x2 + y2 + z2)½
--- | --- | --- | --- | ---
y = | r · sinQ · sinj |  | j = | arctg (y/x)
z = | r · cos Q |  | Q = | arctg | (x2 + y2 + z2 )½ z
| Not particularly difficult, but not so easy either.

| 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 djdQ = 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 VR 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 | = | r2 · sinQ· dr · dQ· dj
--- | --- | ---
| The volume of our sphere thus results from the integral
Vr = | ¥ ó õ 0 | 2p ó õ 0 | p ó õ 0 | r2 · sinQ · dr djdQ= 2p · | ¥ ó õ 0 | p ó õ 0 | r2 · sinQ · dr dQ = 2p | · [–cos p + cos 0] · | ¥ ó õ 0 | r2 · dr
Vr = | 2p · | [2] · | 1/3R3 | = (4/3) · p · R3 | 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= | ¶2U | + | ¶2U | + | ¶2U
--- | --- | --- | --- | --- | ---
¶x2 | ¶y2 | ¶z2
Spherical Coordinates
D= | ¶2U ¶r2 | + | 2 r | · | ¶U ¶r | + | 1 r2 · sin2· Q | · | ¶2U ¶j2 | + | 1 r2 | · | ¶2U ¶Q2 | + | cotg Q r2 | · | ¶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.

| Looking back now on our treatment of the orientation polarization, we find yet another way of expressing spherical coordinates for problems with particular symmetry:
|  | We use a solid angle W and its increment dW.
|  | A solid angle W is defined as the ratio of the area on a unit sphere that is cut out by a cone with the solid angle W to the total surface of a unit sphere (=4pR2==4p for R=1).
|  | A solid angle of 4p therefore is the same as the total sphere, and a solid angle of p is a cone with a (plane) opening angle of 120o (figure that out our yourself).
| An incremental change of a solid angle creates a kind of ribbon around the opening of the cone defined by W. This is shown below
| Relations with spherical symmetry where the value of Q does not matter - i.e. it does not appear in the relevant equations - are more elegantly expressed with the solid angle W.
|  | That is the reason why practically all text books introduce Q in the treatment of the polarization orientation. And in order to be compatible with most text books, that was what we did in the main part of the Hyperscript.
|  | Of course, eventually, we have to replace Q and dQ by the basic variables that describe the problem, and that is only the angle d in our problem (same thing as the angle j here).
| Expressing dQ in terms of d is easy (compare the picture in the main text)
|  | The radius ot the circle bounded by the dQ ribbon is r·sind=sind because we have the unit sphere, and its width is simply dd.
|  | Its incremental area is thus the relation that we used in the main part.
dQ = 2p · sind · dd

With frame

3.2.4 Orientation Polarization

© H. Föll (Advanced Materials B, part 1 - script)