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Derivation of Vector Product

by john fairbanks
Tags: derivation, product, vector
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john fairbanks
#1
Oct17-05, 09:15 PM
P: 8
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
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mezarashi
#2
Oct18-05, 04:15 AM
HW Helper
P: 660
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.
dextercioby
#3
Oct18-05, 04:36 AM
Sci Advisor
HW Helper
P: 11,915
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.

lurflurf
#4
Oct18-05, 06:19 PM
HW Helper
P: 2,264
Derivation of Vector Product

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.
john fairbanks
#5
Oct18-05, 10:39 PM
P: 8
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!
lurflurf
#6
Oct21-05, 11:33 PM
HW Helper
P: 2,264
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


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