Page 15 - Intro to Tensor Calculus

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more general results. Consider (p, q, r) as some permutation of the integers (1, 2, 3), and observe that the
determinant can be expressed

a p1 a p2 a p3

∆= a q1 a q2 a q3 = e ijk a pi a qj a rk .

a r1 a r2 a r3
If (p, q, r) is an even permutation of (1, 2, 3) then ∆ = |A|
If (p, q, r) is an odd permutation of (1, 2, 3) then ∆ = −|A|
If (p, q, r) is not a permutation of (1, 2, 3) then ∆ = 0.
We can then write
e ijk a pi a qj a rk = e pqr |A|.

Each of the above results can be verified by performing the indicated summations. A more formal proof of
the above result is given in EXAMPLE 1.1-25, later in this section.

EXAMPLE 1.1-10. The expression e ijk B ij C i is meaningless since the index i repeats itself more than
twice and the summation convention does not allow this. If you really did want to sum over an index which
occurs more than twice, then one must use a summation sign. For example the above expression would be
n
X
written e ijk B ij C i .
i=1

EXAMPLE 1.1-11.
The cross product of the unit vectors b e 1 , b e 2 , b e 3 can be represented in the index notation by


if (i, j, k) is an even permutation of (1, 2, 3)
 b e k

b e i × b e j = − b e k if (i, j, k) is an odd permutation of (1, 2, 3)


0 in all other cases
This result can be written in the form b e i × b e j = e kij b e k . This later result can be verified by summing on the
index k and writing out all 9 possible combinations for i and j.

EXAMPLE 1.1-12. Given the vectors A p ,p =1, 2, 3and B p ,p =1, 2, 3 the cross product of these two
vectors is a vector C p ,p =1, 2, 3 with components

C i = e ijk A j B k , i,j,k =1, 2, 3. (1.1.2)

The quantities C i represent the components of the cross product vector

~
~
~
C = A × B = C 1 b e 1 + C 2 b e 2 + C 3 b e 3 .
~
The equation (1.1.2), which defines the components of C, is to be summed over each of the indices which
repeats itself. We have summing on the index k

C i = e ij1 A j B 1 + e ij2 A j B 2 + e ij3 A j B 3 . (1.1.3)