12—Tensors 368The generalization of this statement to an arbitrary rotation should now be obvious. If the components of
the tensor areTijin the basisˆei, and if the components are desired in another basiseˆ′i, you need only to solve
the geometrical problem of expressing the vectorsˆe′iin terms of the vectorsˆei
Say ˆe′i=αjiˆej,where theα’s are a set of numbers. Then,
Tij′ = 20 T(ˆe′i,ˆe′j) =^02 T(αkiˆek, α`jˆe`) =αkiα`jTk` (24)and you have the solution.
Don’t worry about properties of theα’s for the moment. In most problems you realistically encounter, it’s
simpler to use the definitions of the relationships and some plane or solid geometry.
12.3 Relations between Tensors
Drop the clumsy extra indices on the tensors. You should be able to tell from context whether you’re dealing
with^11 Tor^02 T, and I will simply call itT.
Go back to the fundamental representation theorem for linear functionals and see what it looks like in
component form. Evaluatef(~v), where~v=viˆei. (The linear functional has one vector argument and a scalar
output.)
f(~v) =f(viˆei) =vif(ˆei). (25)
Denote the set of numbersf(ˆei)(i= 1, 2 , 3 ) byAi=f(ˆei), in which case,
f(ˆv) =Aivi=A 1 v 1 +A 2 v 2 +A 3 v 3.Now it is clear that the vectorA~of the theorem is just
A~=A 1 ˆe 1 +A 2 ˆe 2 +A 3 ˆe 3. (26)Again, examine the problem of starting from a bilinear functional and splitting off one of the two arguments
in order to obtain a vector valued function of a vector. I want to say
T(~u, ~v) =~u.T(~v)