Biomedical Engineering Reference
In-Depth Information
a result that may be verified by expanding it to show that it coincides with (A.113).
If c
¼
a
b denotes the result of the vector cross product, then from (A.114),
c
¼
e ijk a i b j e k ; ð
c k ¼
e ijk a i b j Þ:
(A.115)
b a vector or a tensor? It is called a vector, but a second
order tensorial character is suggested by the fact that the components of a
Is the vector c
¼
a
b
coincide with the components of the skew-symmetric part of 2( a
b ), see (A.32).
The answer is that the vector c
b is unusual. Although it is, basically, a
second order tensor, it can be treated as a vector as long as the transformations are
between coordinate systems of the same handedness. In that case equation (A.113)
shows that the alternator transforms as a proper tensor of order three, thus there is
no ambiguity in the representation (A.114) for a
¼
a
b . When students first learn
about the vector cross product they are admonished (generally without explanation)
to always use right handed coordinate systems. This handedness problem is the
reason for that admonishment. The “vector” that is the result of the cross product of
two vectors has names like “axial vector” or “pseudo vector” to indicate its special
character. Typical axial vectors are moments in mechanics and the vector curl
operator (Sect. A.11).
Example A.8.3
Prove that a
b
¼
b
a .
Solution: In the formula (A.114) let i
!
j and j
!
i , thus
a
b
¼
e jik a j b i e k ;
Next change e jik to
e ijk and rearrange the order of a j and b i , then the result is
proved:
a b ¼
e ijk b i a j e k ¼b a:
The scalar triple product of three vectors is a scalar formed from three vectors,
a
ð
b
c
Þ
and the triple vector product is a vector formed from three vectors, ( r
( p
q )). An expression for the scalar triple product is obtained by taking the dot
product of the vector c with the cross product in the representation (A.114) for a
b ,
thus
c
ð
a
b
Þ¼
e jik a j b i c k :
(A.116)
From the properties of the alternator it follows that
c
ð
a
b
Þ¼
a
ð
b
c
Þ¼
b
ð
c
a
Þ¼
a
ð
c
b
Þ¼
b
ð
a
c
Þ¼
c
ð
Þ:
(A.117)
b
a
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