Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
Scarica il documento per vederlo tutto.
vuoi
o PayPal
tutte le volte che vuoi
Understanding the Denominator in Versor Equations
To understand this, we have to start by realizing that the denominator is always a scalar because:
† A12 + A22 + ... + Ak2 = |A1|2 + |A2|2 + ... + |Ak|2
And since a scalar divided by itself equals one, this means that:
†† A-1 A = 1
Furthermore, it also proves that the left and right inverse are the same. Division by a scalar α is multiplication by 1/α, which is, according to equation (3.3) commutative, proving that the left and right inverses of a versor are indeed equal.
This means that for a versor A, we have A-1 = A-1 = A-1 and therefore the following:
A-1 A = AA-1 = A-1 A = AA-1 = 1
It is important to notice that in the case of vectors, the scalar represents the squared magnitude of the vector. As a consequence, the inverse of a unit vector is equal to itself.
Not many people are comfortable with the idea of...
Il tuo compito è formattare il testo fornito utilizzando tag html.
ATTENZIONE: non modificare il testo in altro modo, NON aggiungere commenti, NON utilizzare tag h1;
Il testo formattato con i tag html è il seguente:
represent planes. Geometricalgebra introduces bivectors which can be used for the same purpose. By usingthe pseudoscalar we can get an understanding of the relationship between thetwo representations.∗The dual A of a multivector A is defined as follows:∗ −1A = AI (4.3)where I represents the pseudoscalar of the geometric algebra that is being used.The pseudoscalar is a blade (the blade with highest grade) and therefore its leftand right inverse are the same, and hence the above formula is not ambiguous.Let us consider a simple example in C` , calculating the dual of the basis3bivector e . The pseudoscalar is e . Pseudoscalars are blades and thus ver-12 123sors. You can check yourself that its inverse is e e e . We’ll use this to calculate3 2 1the dual of e :12 ∗e = e e e e12 3 2 112 = e e e e e1 2 3 2 1= −e e e e e1 3 2 2 1= −e e e1 3 1= e e e1 1 3= e 3 (4.4)Thus, the dual is basis vector e , which is exactly the normal vector of
basis3bivector e . In fact, this is true for all bivectors. If we have two arbitrary12vectors a and b ∈ C` :
a = α e1 + α e2 + α e3
b = β e1 + β e2 + β e3
According to equation (2.7) their outer product is:
a ∧ b = (α β - α β )e12 + (α β - α β )e13 + (α β - α β )e23
And its dual (a ∧ b) becomes:
-1*(a ∧ b) =(a ∧ b)e123 =(a ∧ b)e1 e2 e3
=((α β - α β )e12 + (α β - α β )e13 + (α β - α β )e23)e1 e2 e3
=(α β - α β )e1 e2 e3 +1(α β - α β )e12 e3 e1 +1(α β - α β )e13 e3 e1 +1(α β - α β )e23 e3 e2
=(α β - α β )e1 e2 e3 +1(α β - α β )e12 e3 e1 +1(α β - α β )e13 e3 e1 +1(α β - α β )e23 e3 e2
- − α β )e e e e e +1 3 3 1 1 3 3 2 1(α β − α β )e e e e e2 3 3 2 2 3 3 2 1=(α β − α β )e − (α β − α β )e + (α β − α β )e1 2 2 1 3 1 3 3 1 2 2 3 3 2 1=(α β − α β )e + (α β − α β )e + (α β − α β )e (4.5)2 3 3 2 1 3 1 1 3 2 1 2 2 1 3
- Which is exactly the traditional cross product. We conclude that in three dimensions, the dual of a bivector is its normal. The dual can be used to convert between bivector and normal representations. But the dual is even more, because it is defined for any multivector.
- 4.5 Projection and Rejection
- If we have a vector a and bivector B we can decompose a in two parts. One part a that is collinear with B. We call this the projection of a onto B. The|| B 2 other part is a , and orthogonal to B. We call this the rejection of a from ⊥ BB.
vector a onto abivector B. By definition, the inner and outer product of respectively orthogonaland collinear blades are zero. In other words, the inner product of a vectororthogonal to a bivector is zero: a · B = 0 (4.8)⊥B. Likewise the outer product of a vector collinear with a bivector is zero: a ∧ B = 0 (4.9)|| B. Let’s see what happens if we multiply the orthogonal part of vector a withbivector B: a · B = a ∧ B⊥ ⊥ ⊥B B B = a ∧ B using equation (4.8)⊥B = a ∧ B + a ∧ B equation (4.9)⊥ ||B B = (a + a) ∧ B equation (2.5)⊥ ||B B = a ∧ B equation (4.6). Thus, the perpendicular part of vector a times bivector B is equal to theouter product of a and B. Now all we need to do is divide both sides of theequation by B to obtain the perpendicular part of a: a · B = a ∧ B⊥B−1 −1a B = (a ∧ B)B⊥ B −1a = (a ∧ B)B⊥ B 32. Notice that there is no ambiguity in using the inverse
because B is a blade orversor, and its left and right inverses are therefore the same. The conclusion is: −1a = (a ∧ B)B (4.10)⊥ B
Calculating the collinear part of vector a follows similar steps: a B = a · B + a ∧ B
|| || ||B B B= a · B|| B= a · B + a · B⊥
|| BB= (a + a ) · B⊥ ||B B= a · B
Again multiply both sides with the inverse bivector: −1 −1a BB = (a · B)B|| B
To conclude: −1a = (a · B)B (4.11)|| B
Using these definitions, we can now confirm that a + a = a⊥|| BB−1 −1 and (4.10)a + a = (a ∧ B)B + (a · B)B equation (4.11)⊥|| BB −1= (a ∧ B + a · B)B equation (3.3)−1= aBB equation (4.7)= a by definition of the inverse
334.6 Reflections
Armed with a way of decomposing blades in orthogonal and collinear parts we can take a look at reflections. We will get ahead of ourselves and take a3specific look at the geometric algebra of the Euclidian
space R denoted with∗C` . Suppose we have a bivector U . Its dual U will be the normal vector u.3What if we multiply a vector a with vector u, projecting and rejecting a ontoU at the same time: ua = u(a + a )⊥|| UU= ua + ua ⊥|| UUUsing (3.2) we write it in full:ua = (u · a + u ∧ a ) + (u · a + u ∧ a )⊥ ⊥|| || U UU UNote that (because u is the normal of U ) the vectors a and u are perpen-|| Udicular. This means that the inner product a · u equals zero. Likewise, the|| Uvectors a and u are collinear. This means that the outer product a ∧ u⊥ ||U Uequals zero. Removing these two 0 terms:ua = u ∧ a + u · a ⊥|| UU= u · a + u ∧ a⊥ ||U URecall that the inner product between two vectors is commutative, and the outerproduct is anticommutative, so we can write:ua = a · u − a ∧ u⊥ ||U UWe can now insert those 0-terms back in (putting in the form of equation (3.2)):ua = (a
· u + a ∧ u) − (a · u + a ∧ u)⊥ ⊥ || ||U U U U
Writing it as a geometric product now:ua = a u − a u⊥|| UU= (a − a )u⊥|| UU
Meaning that: −ua = −(a − a )u⊥|| UU= (a − a )u⊥ ||U U
Notice how we changed the addition of the perpendicular part into a subtractionby multiplying with −u. Now, if we add a multiplication with the inverse we4.2:obtain the following, depicted in figure−1 −1−uau = −u(a + a )u⊥|| UU −1= (a − a )uu⊥|| UU= a − a⊥|| UU34u aU a ⊥Ua|| U −1a − a = −uau⊥|| UUFigure 4.2: Reflection
In general, if we sandwich a vector a in between another vector −u and its−1