12—Tensors 372Proof: Consider the functionΛ−αΛ′whereαis a scalar. Pick any three independent vectors~v 10 ,~v 20 ,
~v 30 as long asΛ′on this set is non-zero. Let
α=Λ(~v 10 , ~v 20 , ~v 30 )
Λ′(~v 10 , ~v 20 , ~v 30 )(30)
(IfΛ′gives zero for every set of~vs then it’s a trivial tensor, zero.) This guarantees thatΛ−αΛ′will vanish for
at least one set of values of the arguments. Now take a general set of three vectors~v 1 ,~v 2 , and~v 3 and ask for
the effect ofΛ−αΛ′on them. ~v 1 ,~v 2 , and~v 3 can be expressed as linear combinations of the original~v 10 ,~v 20 ,
and~v 30. Do so. Substitute intoΛ−αΛ′, use linearity and notice that all possible terms give zero.
The above argument is unchanged in a higher number of dimensions. It is also easy to see that you cannot
have a totally antisymmetric tensor of rankn+ 1 inndimensions. In this case, one of then+ 1 variables
would have to be a linear combination of the othern. Use linearity, and note that when any two variables equal
each other, antisymmetry forces the result to be zero. These observations imply that the function must vanish
identically. See also problem 19.12.4 Non-Orthogonal Bases
The next topic is the introduction of more general computational techniques. These will lift the restriction on the
type of basis that can be used for computing components of various tensors. Until now, the basis vectors have
formed an orthonormal set
|ˆei|= 1, eˆi.ˆej= 0ifi 6 =j
Consider instead a more general set of vectors~ei. These must be independent. That is, in three dimensions they
are not coplanar. Other than this there is no restriction. Since by assumption the vectors~eispan the space you
can write
~v=viˆei.
with the numbersvibeing as before the components of the vector~v.NOTE: Here is a change in notation. Before, every index
was a subscript. (It could as easily have been a super-
script.) Now, I want to make a careful distinction between
sub- and superscripts. They will have different meanings.