Sect. 4.2: Orthogonal Transformations• For convenience, change of notation:

x x1, y x2, z x3, x´ x1, y´ x2, z´ x3

Also: aij cosθij

• In new notation, transformation eqtns between primed & unprimed coords become:

x1 = a11 x1+a12 x2 +a13 x3

x2 = a21 x1+a22 x2 +a23 x3

x3 = a31 x1+a32 x2 +a33 x3

Or: xi = ∑j aij xj (i,j = 1,2,3) (1)

• (1) = An example of what mathematicians call a

Linear (or Vector) Transformation.

• For convenience, another change of notation: If the index is repeated, summation over it is implied.

xi = ∑j aij xj (i,j = 1,2,3)

xi = aij xj (i,j = 1,2,3)

Einstein summation convention• To avoid possible ambiguity when powers of an

indexed quantity occur: ∑i(xi)2 xixi

• For the rest of the course, summation convention is automatically assumed, unless stated otherwise.

• Linear Transformation: xi = aij xj (i,j = 1,2,3) (1)

• With aij cosθij as derived, (1) is only a special case of a general linear transformation, since, as already discussed, the direction cosines cosθij are not all independent. – Re-derive connections between them, use new notation.

• Both coord systems are Cartesian: Square of magnitude of vector = sum of squares of

components.

Magnitude is invariant on transformation of coords:

xixi = xixi

Using (1), this becomes: aijaikxjxk = xixi (i,j,k = 1,2,3)

• aijaikxjxk = xixi (i,j,k = 1,2,3)

Can be valid if & only if

aijaik = δj,k (j,k = 1,2,3)

Identical previous results for orthogonality of direction cosines.

• Any Linear Transformation:

xi = aij xj (i,j = 1,2,3) (1)

Orthogonal Transformation

aijaik = δj,k Orthogonality Condition

• Linear (or Vector) Transformation.

xi = aijxj (i,j = 1,2,3) (1)• Can arrange direction cosines into a square matrix:

a11 a12 a13

A a21 a22 a23

a31 a32 a33

• Consider coordinate axes as column vector components:

x1 x1

r = x2 r = x2

x3 x2 Coordinate transformation reln can be written:

r = Ar with A Transformation matrix or rotation matrix (or tensor)

Example: 2d Coordinate Rotation• Application to 2d rotation. See figure:

• Easy to show that: x3 = x3

x1 = x1cos + x2sin = x1cos + x2cos( - π/2)

x2 = -x1sin + x2cos = x1cos( + π/2) + x2cos

• 2d rotation. See fig:

aij cosθij

a33 = cosθ33 = 1

a11 = cosθ11 = cos

a22 = cosθ22 = cos

a12 = cosθ12 = cos( - π/2) = sin

a21 = cosθ21 = cos( + π/2) = -sin

a31 = cosθ31 = cos(π/2) = 0, a32 = cosθ32 = cos(π/2) = 0

Transformation matrix has form:

a11 a12 0 cos sin 0

A = a21 a22 0 = -sin cos 0

0 0 1 0 0 1

• 2d rotation. See fig:

aij cosθij

Orthogonality Condition:

aijaik = δj,k

a11a11 + a21a21 = 1

a12a12 + a22a22 = 1 , a11a12 + a21a22 = 0

Use expressions for aij & get: cos2 + sin2 =1

sin2 + cos2 =1, cossin - sincos = 0

Need only one angle to specify a 2d rotation.

• Transformation matrix A Math operator that, acting on unprimed system, transforms it to primed system. Symbolically:

r = Ar (1) Matrix A, acting on components of the r in unprimed system yields

components of r in the primed system.

• Assumption: Vector r itself is unchanged (in length & direction) on operation with A. (r2 = (r)2)

• NOTE: Same formal mathematics results from another interpretation of (1): A acts on r & changes it into r . Components of 2 vectors related by (1).

• Which interpretation depends on context of problem. Usually, for rigid body motion, use 1st interpretation.

• For general transformation (1), nature of A depends on which interpretation is used. A acting on coords: Passive transformation. A acting on vector: Active transformation

Example from Marion• In the unprimed system, point P is represented

as (x1, x2, x3) = (2,1,3).

In the primed system, x2 has been rotated from x2, towards x3 by

a 30º angle as in the

figure. Find the

rotation matrix A &

the representation of

P = (x1, x2, x3) in the

primed system.

• From figure, using aij cosθij

a11 = cosθ11 = cos(0º) =1

a12 = cosθ12 = cos(90º) = 0

a13 = cosθ13 = cos(90º) = 0

a21 = cosθ21 = cos(90º) = 0

a22 = cosθ22 = cos(30º) = 0.866

a23 = cosθ23 = cos(90º-30º) = cos(60º) = 0.5

a31 = cosθ31 = cos(90º) = 0

a32 = cosθ32 = cos(90º+30º) = -0.5

a33 = cosθ33 = cos(30º) = 0.866

1 0 0

A = 0 0.866 0.5

0 -0.5 0.866

• To find new representation of P, apply r = Ar or

x1 = a11 x1+a12 x2 +a13 x3

x2 = a21 x1+a22 x2 +a23 x3

x3 = a31 x1+a32 x2 +a33 x3

Using (x1, x2, x3) = (2,1,3)

x1 = x1 = 2

x2 = 0.866x2 +0.5x3 = 2.37

x3 = -0.5 x2 + 0.866x3 = 2.10

(x1, x2, x3) = (2,2.37,2.10)

Useful Relations • Consider a general line segment, as in the figure:

Angles α, β, γ between the segment & x1, x2, x3

Direction cosines of line cosα, cosβ, cosγManipulation, using orthogonality relns from before:

cos2α + cos2β + cos2γ = 1

• Consider 2 line segments, direction cosines, as in the figure:

cosα, cosβ, cosγ, & cosα , cos β, cosγ• Angle θ between

segments:• Manipulation (trig):

cosθ

= cosαcosα +cosβcosβ +cosγcosγ

Sect. 4.3: Formal (math) Properties of the Transformation Matrix

• For a while (almost) pure math!

• 2 successive orthogonal transformations B and A, acting on unprimed coordinates:

r = Br followed by r = Ar = ABr• In component form, application of B followed by A gives

(summation convention assumed, of course!):

xk = bkjxj , xi = aikxk = aikbkjxj (1)(i,j,k = 1,2,3)

Rewrite (1) as: xi = cijxj (2)

• (2) has the form of an orthogonal transformation C AB

with elements of the square matrix C given by cij aikbkj

Products Product of 2 orthogonal transformations B (matrix

elements bkj) & A (matrix elements aik) is another orthogonal transformation C = AB (matrix elements

cij aikbkj). – Proof that C is also orthogonal: See Prob. 1, p 180.

• Can show (student exercise!): Product of orthogonal transformations is not commutative: BA AB – Define: D BA (matrix elements dij bikakj). Find, in

general: dij cij.

Final coords depend on order of application of A & B.

• Can also show (student exercise!): Products of such transformations are associative: (AB)C = A(BC)

• Note: Text now begins to use vector r & vector x interchangeably! r = Ar x = Ax can be represented in terms of matrices, with coord vectors being column vectors: x = Ax

xi = aijxj or:

x1 a11 a12 a13 x1

x2 = a21 a22 a23 x2

x2 a31 a32 a33 x3

• Addition of 2 transformation matrices: C = A + B Matrix elements are: cij = aij + bij

Inverse• Define the inverse A-1 of transformation A:

x = Ax (1), x A-1 x (2)

In terms of matrix elements, these are:

xk = akixi (1 ), xi aij xj (2)

where aij are matrix elements of A-1

• Combining (1) & (2): xk = akiaij xj clearly, this can hold if & only if:

akiaij = δj,k (3)Define: Unit Matrix

1 0 0

1 0 1 0 akiaij = δj,k are clearly matrix 0 0 1 elements of 1

Transpose In terms of matrices, DEFINE A-1 by:

AA-1 A-1A 1– Proof that AA-1 A-1A : p 146 of text.

• 1 Identity transformation because:

x = 1 x and A = 1 A• Matrix elements of A-1 & of A are related by:

aij = aji (4)– Proof of this: p 146-147 of text.

• Define: Ã Transpose of A matrix obtained from A by interchanging rows & columns.

Clearly, (4) A-1 = Ã & thus: ÃA = AÃ = 1

A-1 = Ã For orthogonal matrices, the reciprocal is equal to the transpose.

• Combine aij = aji with akiaij = δj,k

akiaji = δj,k (5)

(5): A restatement of the orthogonality relns for the aki !

• Dimension of rectangular matrix, m rows, n columns m n. A, A-1, Ã : Square matrices with m = n. – Column vector (1 column matrix) x, dimension m 1.

Transpose x: dimension 1 m (one row matrix).– Matrix multiplication: Product AB exists only if # columns

of A = # rows of B: cij = aikbkj

– See text about multiplication of x & its transpose with A & Ã

Define:

• Symmetric Matrix A square matrix that is the same as its transpose: A = Ã aij = aji

• Antisymmetric Matrix A square matrix that is the negative of its transpose:

A = - Ã aij = - aji

– Obviously, diagonal elements in this case: aii = 0

• 2 interpretations of orthogonal transformation

Ax = x– 1) Transforming coords. 2) Transforming vector x.

• How does arbitrary vector F (column matrix) transform under transformation A? Obviously,

G AF (some other vector). • If also, the coord system is transformed under operation

B, components of G in new system are given by

G BG BAF

Rewrite (using B-1B = 1) as: G = BG = BAB-1BF

Also, components of F in new system are given by F BF

• Combining gives: G = BAB-1F where:F BF, G BG

If define operator BAB-1 A we have:

G A F (same form as G = AF, but expressed in transformed coords)

Transformation of operator A under coord transformation B is given as:

A BAB-1

Similarity transformation

• Properties of determinant formed from elements of an orthogonal transformation matrix:

det(A) |A|

Some identities (no proofs):• |AB| = |A||B|

From orthogonality reln ÃA = AÃ = 1 get

• |Ã||A| = |A||Ã| = 1

Determinant is unaffected by interchange of rows & columns:

• |Ã| = |A| Using this with above gives:

• |A|2 = 1 |A| = 1

• Value of determinant is invariant under a similarity transformation. Proof: A, B orthogonal transformations– Assumes 1) B-1 exists & 2) |B| 0

Similarity transformation: A BAB-1

Multiply from right by B: AB = BAB-1B = BA

Determinant: |A||B| = |B||A| (|B| = a number 0)

Divide by |B| on both sides & get

|A| = |A|