rewrite of opening Secton 15 chunk
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A Word document of Phil's working notes rewriting the opening of Section 15 of his new frames document, with a note that it was installed on 3.1.17 without archiving the old version. It relates Cartesian and cylindrical unit vectors of frames S and S' by z-rotations Rz(θ), Rz(φ) and Rz(φ+θ). It uses the Basis Theorem (1.1.30) to write the vector relations in matrix form, with basis vectors as matrix rows, and works two examples giving the linear combinations (15.8) and (15.9). Equations are partly lost in extraction.
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This was installed on 3.1.17 without archiving the old.
Within Frame S we have ( each of these equations has the (1.1.30) template form e'n = R-1en) :
i = Rz(θ)ei . for example = Rz(θ) (a)
A corresponding equation applies in Frame S' ,
'i = Rz(θ')e'i . for example ' = Rz(θ') ' (b)
The relation between the Frame S and Frame S' Cartesian unit vectors is
e'i = Rz(φ)ei . for example ' = Rz(φ) (c)
The relation between the Frame S and Frame S' cylindrical unit vectors is
'i = Rz(φ)i . for example ' = Rz(φ) (d)
Relations (d) and (a) can be combined to get
'i = Rz(φ+θ)ei . for example ' = Rz(φ+θ) (e) (15.4)
Writing the basis vector relations in matrix notation
Recall the Basis Theorem of (1.1.30) : (we use dummy basis vector names an and a'n ) ,
a'n = R-1 an an = Σm(R-1)nm a'm (15.5)
On the left, we rotate vector an by R-1 to get vector a'n.
On the right, we express an as a linear combination of the basis vectors a'n.
Remember that the subscripts on the a and a' are labels, not components!
Suppose we take the kth component of the equation on the right of (15.5),
[an]k = Σm(R-1)nm [a'm]k . (15.6a)
One can write this as
Ank = Σm(R-1)nm A'mk where Ank = [an]k and A'nk = [a'n]k . (15.6b)
For a matrix Ank one knows that n is the row index and k is the column index. Therefore, saying Ank = [an]k is the same as saying that the vector an is the nth row of matrix A. Thus we can write (15.6a) in this manner,
= R-1 an = R-1 a'n n = 1,2,3 . (15.7a)
Inverting,
= R . (15.7b)
We have thus provided an interpretation for the "alternative notation" shown in (1.1.32) : the three vectors in a column can be regards as rows of a matrix.
All our rotations of interest in (15.4) are z-rotations which, from (A.1), have the form
Rz(ψ) = . (A.1)z
Example 1: Apply (15.7b) to (15.4e) which says 'i = Rz(φ+θ)ei = R-1ei so R = Rz(-θ-φ) :
= Rz(-θ-φ) = // see (A.1)
or
= .
Writing out the linear combinations, one gets
' = cos(θ+φ) + sin(θ+φ)
' = - sin(θ+φ) + cos(θ+φ)
' = . (15.8)
Example 2: Apply (15.7b) to (15.4c) which says e'i = Rz(φ)ei = R-1ei so R = Rz(-φ) :
= Rz(-φ) =
or
= .
Writing out the linear combinations, one gets
' = cosφ + sinφ
' = -sinφ + cosφ // e'2 = - sinφ e1 + cosφ e2
' = . (15.9)
We could reduce our 3x3 matrix work to 2x2 for the turntable examples, but other problems require the full 3x3 notation so we maintain it throughout.