Page 19 - Intro to Tensor Calculus

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In this expression the indices s and m are dummy summation indices and can be replaced by any other
letters. We replace s by k and m by j to obtain

D i = e ikj A j B k = −e ijk A j B k = −C i .

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Consequently, we find that D = −C or B × A = −A × B. That is, D = D i b e i = −C i b e i = −C.
Note 1. The expressions
and
C i = e ijk A j B k C m = e mnp A n B p
with all indices having the range 1, 2, 3, appear to be different because different letters are used as sub-
scripts. It must be remembered that certain indices are summed according to the summation convention
and the other indices are free indices and can take on any values from the assigned range. Thus, after
summation, when numerical values are substituted for the indices involved, none of the dummy letters
used to represent the components appear in the answer.
Note 2. A second important point is that when one is working with expressions involving the index notation,
the indices can be changed directly. For example, in the above expression for D i we could have replaced
j by k and k by j simultaneously (so that no index repeats itself more than twice) to obtain

D i = e ijk B j A k = e ikj B k A j = −e ijk A j B k = −C i .

Note 3. Be careful in switching back and forth between the vector notation and index notation. Observe that a
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vector A can be represented
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A = A i b e i
or its components can be represented

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A · b e i = A i , i =1, 2, 3.

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Do not set a vector equal to a scalar. That is, do not make the mistake of writing A = A i as this is a
misuse of the equal sign. It is not possible for a vector to equal a scalar because they are two entirely
different quantities. A vector has both magnitude and direction while a scalar has only magnitude.

EXAMPLE 1.1-15. Verify the vector identity

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A · (B × C)= B · (C × A)
Solution: Let
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B × C = D = D i b e i where D i = e ijk B j C k and let
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C × A = F = F i b e i where F i = e ijk C j A k
where all indices have the range 1, 2, 3. To prove the above identity, we have
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A · (B × C)= A · D = A i D i = A i e ijk B j C k
= B j (e ijk A i C k )
= B j (e jki C k A i )