## Geometry and Coordinate Systems

n ifforni n Ysü>lsxPucn lYa Suvarai

(The geometrical identity has great meaning)

This part of the appendix contains some additional material on coordinate systems and geometry.

### C.1 Rotation of Coordinate Systems

Consider two right-handed Cartesian coordinate systems with the same origin. Let the second system with coordinates (x', y', z') be generated by a counterclockwise rotation of the first one with coordinates (x, y, z) around its z-axis by an angle a. This situation is demonstrated in Fig. C.1. Then a point (x, y, z) in the first system has the coordinates (x', y', z') in the second system, and they can be computed according to cos a sin a 0 - sin a cos a 0 0 0 1

with

The explicit formula reads

The matrix Rz(a) is an orthogonal matrix, i.e.,

Its inverse is the matrix

which may be used to compute (x, y, z) as a function of (xy', z').

Some caution is needed when coordinate systems are connected by several rotations around different axes. As commutativity is violated in most cases, i.e.,

the sequence of rotations has to be considered very carefully.

C.2 Volume and Surface Elements in Spherical Coordinates

Let us start with the derivation of the volume element. The derivation of the volume element in spherical coordinates is based on the «-dimensional substitution rule in calculus:

Jg(I) Jl where g(v) describes the coordinate transformation from coordinates v to u, I is the domain of integration in the coordinates v, and g'(v) is the Jacobian matrix associated with g. In our case, the vectors u and v represent the Cartesian and spherical polar coordinates 0 and ^ introduced in Sect. 3.1.1, i.e.,

y I = g(r,0,f), g(r,0,f) = r I sinf sin0 |, (C.2.2)

I = Ir x I0 X I< = [0, TO) X [0, 180°] X [0, 360°]. (C.2.3) The Jacobian matrix is d (v) / cos < sin 0 r cos < cos 0 — r sin < sin 0

g(v):=-= I sin < sin 0 r sin < cos 0 r cos < sin 0 |, (C.2.4)

and the determinant is det g'(v) = r2 sin 0. (C.2.5)

Note that due to 0 < 0 < 180° the expression r2 sin 0 is always nonnegative. Therefore, the volume element in spherical polar coordinates follows as dV = I det g'(v)| dv = r2 sin 0 dr d0d<. (C.2.6)

The derivation of the surface element in the form (3.1.5) requires some formulas from vector calculus and differential geometry. At first we review that for three-dimensional vectors a, b, and c the scalar product a ■ (b x c) can be represented by a determinant according to

a3 b3 c3

where b x c denotes the cross product or outer product defined as b x c = (b2c3 - b3c2, b3c1 - b1 c3, b1 c2 - b2c1)T . (C.2.8)

As the determinant of a matrix with linearly dependent columns vanishes we further have a x a = 0, (C.2.9)

Finally, we need the distributive law for the cross product, i.e.,

(V1 + V2) X (V3 + V4) = Vi X V3 + Vi X V4 + V2 X V3 + V2 X V4. (C.2.11)

Let us now consider a surface embedded in a three-dimensional vector space parametrized by two coordinates 0 and <, i.e., fx (0, <)\

The normal vector, n, at a surface point, r, is defined as r0 x r<

where r0 and r< denote the partial derivatives of r w.r.t. 0 and <. If the surface encloses a finite region of the three-dimensional space, the normal vector n points into the region external to this volume, and the vectors r0, r<, and n establish the unit axis of a right-handed coordinate system. The surface element, da, is defined as da := |r0 x r< | d0d<. (C.2.14)

If we multiply (C.2.13) by er, we can eliminate |r0 x r< | from (C.2.14) by using cos p in definition (3.1.3), i.e., r cos p := - • n = er • n. (C.2.15)

If we do so, we get the expression da := er • (r0 x r<) d0d<. (C.2.16)

We now need to evaluate the term er • (r0 x r<) (C.2.17)

for the special case of interest

In this case we have r0 = re er + r^ r® = r® er + r® (C.2.19)

where rS and r^ denote the derivatives for the unit-sphere case, i.e.,

rS = r(0,®)| sin ® cos 0 , r® = r(0, ®) cos ® sin0 cos 0 / \ cos 0

If we substitute (C.2.19) into (C.2.17) and apply (C.2.11), we get er ■ (r0 x r®) = er ■ (rS x r^). (C.2.20)

The other three terms vanish due to (C.2.7) and (C.2.9). If we evaluate (C.2.20) we get cos < sin 0 r cos < cos 0 — r sin < sin 0 r ■ (rS x r<) = det I sin< sin0 r sin< cos0 r cos< sin0 1= r2 sin0.

So, finally, we get the surface element

cos P

For readers who prefer graphical demonstrations, we also provide the following: we can apply the sine rule to the infinitesimal small triangle ABC (left part of Fig. C.2):

Considering that d0 is an infinitesimal small angle, i.e., sin d0 « d0 and cos(P + d0) « cos P, we get ds r r

d0 cos P cos P

The right part of Fig. C.2 shows the projection of r onto the x-y plane, yielding

and an infinitesimal small triangle containing the angle d® and the arc dl. The arc dl follows as ds M n

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