Derivation of Vector Product

  1. In my calc book the derivation of the vector cross product is not derrived but rather just given. I've read in another book that William Rowan Hamilton, after years of work, came up with the basic form we momorize today and symbolize with determinants. Does anybody know how this vector cross product is derrived. Some might say "assume the determinat definition and the results follow". The question is how was this definition derrived. Thanks
     
  2. jcsd
  3. mezarashi

    mezarashi 660
    Homework Helper

    I personally believe that the cross product is a definition rather than any kind of result. On the same level as the rules of multiplication and division. We've defined a new operation we can do on vectors.
     
  4. dextercioby

    dextercioby 12,307
    Science Advisor
    Homework Helper

    Just follow some exterior algebra and some geometry and you'll find that the "vector product" is a particular (up to Hodge duality) example of exterior product.

    Daniel.
     
  5. lurflurf

    lurflurf 2,326
    Homework Helper

    Definitions are not really derived, they are not chosen at random either. It is a matter of seeing what is nice about the object in question and seeing how that leads to a definition. Often different rotes can be followed. Here is one for this case
    we desire a function
    z=f(x,y)
    f:R^3xR^3->R^3
    That is we want to map two 3-space vectors into one.
    We also would like
    -the function to linearly map products of componets of x and y to componets
    -the function to not depend upon the coordinate system used for x and y that is it should be invariant

    So one (tedious) way to procede would be to write down a function with the linear componet map property that would have 81 constants. Then the invarence property would give 80 independent equations. Thus the cross product would be determined up to a multiplicative constant.
     
  6. content not to know for now

    :bugeye: lurflurf you gave me enough to know that I will wait till my calc 3 undegraduate level of math is completed before I really follow your reply. For now I think I know that the need for the properties of a cross product drove a legit derrivation -- originally maybe the one by Irish Mathematician Hamiliton, and more recently the process you so kindly mentioned. Thanks!
     
  7. lurflurf

    lurflurf 2,326
    Homework Helper

    I will try again.

    We want to define
    a (pseudo)vector
    axb
    where a and b are vectors
    and a,b,axb are in 3-space
    so that
    1) it is bilinear
    that is
    (ka)xb=k(axb)
    ax(kb)=k(axb)
    (a+c)xb=axb+cxb
    ax(b+c)=axb+axc
    a,b,c vectors k a scalar
    2) it is invariant under rotation
    that is is v' is the vector v after a (proper) rotation
    (axb)'=a'xb'
    A proper rotation can be written with a (special orthogonal) matrix A
    v'=Av where det(A)=1 and inverse(A)=transpose(A)
    or thought of as a rotation geometrically

    The point of this is if two coordinate systems are used the results should be the same.

    The approuch
    -define the general product
    -reduce the possibilities by enforcing rotation invariance
    -reduce 27=3^3 variables to 1 (I said 81 before oops)
    -only rightangle rotations will be needed
    -we can visuallize right angle rotations as the 24 ways a cube can be possitioned

    so since we have bilinearity we will have defined the cross produce when we have defined the 9 quantities
    ixi,ixj,ixk,jxi,jxj,jxk,kxi,kxj,kxk
    each of which will have the form
    ixj=Ai+Bj+Ck
    thus the 27 values we need to specify
    consider the rotation
    i'=-i j'=-j k'=k
    so
    ixj=Ai+Bj+Ck
    Invarience requres the equation to hold when
    i'xj'=Ai'+Bj'+Ck'
    which becomes
    (-i)x(-j)=-Ai-Bj+Ck
    but by bilinearity
    (-i)x(-j)=ixj
    so
    Ai+Bj+Ck=-Ai-Bj+Ck
    hence
    A=-A->2A=0->A=0
    B=-B->2B=0->B=0
    so
    ixj=CK
    this is a big start we have directly eliminated 2 of our 27 and using this fact later we have went from 27 to 9 as similar resoning applies to all nine products
    ixi,ixj,ixk,jxi,jxj,jxk,kxi,kxj,kxk

    in particular if
    kxk=Di+Ej+Fk
    then
    k'xk'=Di'+Ej'+Fk'
    so
    kxk=-Di-Ej+Fk
    so d=E=0 and
    kxk=Fk
    now consider the rotation
    i'=j j'=i k'=-k

    i'xj'=Ck' (recall A=B=0)
    jxi=-Ck

    also consider
    kxk=Fk
    k'xk'=Fk'
    (-k)x(-k)=-Fk
    so F=0
    kxk=0
    The full effect of this line of reasoning reduces the 9 variables to 3
    so far we have
    ixj=Ck
    jxi=-Ck
    kxk=0

    now consider the rotation
    i'=k j'=i k'=j
    i'xj'=Ck'
    so
    kxi=Cj

    j'xi'=-Ck'
    so
    ixk=-Cj

    kxk=0
    so
    jxj=0

    now the rotation
    i'=j j'=k k'=i
    i'xj'=Ck'
    so
    jxk=Ci

    j'xi'=-Ck'
    kxj=-Ci

    k'xk'=0
    ixi=0

    collecting things up again
    ixi=0
    ixj=Ck
    ixk=-Cj
    jxi=-Ck
    jxj=0
    jxk=Ci
    kxi=Cj
    kxj=-Ci
    kxk=0

    So any such product is a multiple of the standard one in which C is chosen C=1 giving
    ixi=0
    ixj=k
    ixk=-j
    jxi=-k
    jxj=0
    jxk=i
    kxi=j
    kxj=-i
    kxk=0

    Similar reasoning applies to higher products
    a.bxc=axb.c is the only scalar triple product
    all vector triple products are of the form
    u(a.b)c+v(b.c)a+w(c.a)b
    and so on
     
    Last edited: Oct 22, 2005
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