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Precise relation between quaternion and vectors

  1. Oct 25, 2009 #1
    I am reading "history of vector analysis":

    https://www.amazon.com/History-Vect...=sr_1_3?ie=UTF8&s=books&qid=1256515819&sr=8-3




    We have two quaternions [tex]\alpha , \alpha' [/tex] with "scalar" component equal to zero, these are their expressions:

    [tex]\alpha' = x' \vec{i} +y' \vec{j} + z' \vec{k}[/tex]

    [tex]\alpha = x \vec{i} +y \vec{j} + z \vec{k}[/tex]

    with i,j and k fulfilling these known relations :

    ij= k , jk=i , ik= -j ... and so on.


    Then Hamilton wrote:

    S(scalar, which is our -dot product) of [tex]\alpha*\alpha' = -(xx' +yy'+zz') [/tex] and

    V(vector, which is our cross product) of [tex]\alpha*\alpha'[/tex] = i(yz'-zy') -j(zx' -xz') + k(xy'-yx')


    Well, let's now compare this equations with the equations of our modern vector calculus, for example, if we have the vector position R of a particle, expressed in the base of the vectors i,j and k (orthogonal unit basis vectors) :

    R= xi +yj +zk

    R' =x'i + y'j +z'k

    RR'=xx'+yy'+zz'

    and R x R' =i(yz'-zy') -j(zx'-xz') + k(xy'-yx')

    But now it is clear that we are using a basis that is conmutative , because ij=ji .

    I don't understand this, if our vector calculus started with the calculations of Hamilton cited above ¿why can we use basis vectors that are conmutative? .

    Perhaps you think that is a stupid question, but I would like to understand exactly what is the relation between the vectors that I use in my physics classes and these numbers discovered by Hamilton.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Oct 26, 2009 #2
    Are you sure that it isn't

    [tex] i j = - j i [/tex]

    ?

    EDIT: please ignore. I used a different metric calculating the above.
     
    Last edited: Oct 26, 2009
  4. Oct 26, 2009 #3

    D H

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    How did you arrive at that conclusion?

    The cross product is anti-commutative.
     
  5. Oct 26, 2009 #4
    Yes you are right, the cross product is not conmutative, and is not associative.

    But I don't understand why the quaternion has these three elements i, j, k (the equivalent of the i part of the imaginary numbers) and when you use vectors you don't say anything of these i,j, k elements, you use real numbers. I don't see how do you arrive from the algebra of the vector part of the quaternions, to the algebra of the vectors, as where introduced by gibbs and heaviside.

    Perhaps I am not expressing the question clearly, if you want some extra explanation , please tell me.

    Thanks for the answers.
     
  6. Oct 26, 2009 #5

    D H

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    Sure you do! What is the difference between the components of a three vector the x, y, and z components, versus the x1, x2, and x3 components, versus something else?

    Besides, where do you think the notation [tex]\vec x = x\hat i + y\hat j + z \hat k[/tex] came from?

    One way to look at quaternions is that a quaternion comprises a real scalar part and an imaginary three vector part:

    [tex]\mathcal Q = \bmatrix q_s \\ \vec q_v\endbmatrix[/tex]

    With this notation, the product of two quaternions can be written as

    [tex]\mathcal Q_1 \mathcal Q_2 =
    \bmatrix q_{1,s} \\ \vec q_{1,v} \endbmatrix \,
    \bmatrix q_{2,s} \\ \vec q_{2,v} \endbmatrix =
    \bmatrix
    q_{1,s}q_{2,s}-\vec q_{1,v}\cdot \vec q_{2,v} \\
    q_{1,s}\vec q_{2,v} + q_{2,s}\vec q_{1,v} + \vec q_{1,v}\times \vec q_{2,v}
    \endbmatrix[/tex]

    That, however, puts the cart before the horse. The development of the quaternions preceded (and motivated) the development of vectors. One way to define the cross product given the quaternion product is

    [tex]\bmatrix 0 \\ \vec v_1 \times \vec v_2 \endbmatrix =
    \frac 1 2 \left (
    \bmatrix 0 \\ \vec v_1 \endbmatrix \, \bmatrix 0 \\ \vec v_2 \endbmatrix -
    \bmatrix 0 \\ \vec v_2 \endbmatrix \, \bmatrix 0 \\ \vec v_1 \endbmatrix
    \right)[/tex]
     
  7. Oct 27, 2009 #6

    Thanks, very apreciated.

    I start to undertand what is happening, Hamilton described imaginary numbers as pairs of real numbers, then he was searching a similar concept but with three components.

    I don't know why he found the quaternions, which have four components, but they have symetry for three of his components, so if you put the scalar part equal to zero, you have the three components that he was searching.

    So as a conclusion ¿could i considere vector calculus as the calculus of quaternions with scalar part equal to zero?

    If I learn quaterionic calculus and then particularize for scalar part=0 ¿will i find the vector algebra? ¿do you agree?
     
  8. Oct 27, 2009 #7
    You can go the other way around too, start with vector algebra, and end up with quaternions, utilizing a Clifford Algebra (or Geometric Algebra) product of vectors.

    The basic idea is defining a vector product so that the product of two colinear vectors is a scalar. For a unit vector [itex]u[/itex] in physics you'd typically want one of:

    [tex] u^2 = \pm 1[/tex]

    With linearity, and associativity, this implies among other things that perpendicular vectors are anticommutative. Adding a scalar to the product of two such perpendicular vectors (a bivector) recovers the complex or quaternion algebras.

    Some details can be found here for example:

    http://www.mrao.cam.ac.uk/~clifford/ptIIIcourse/index.html
     
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