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Unformatted text preview: show that in general Aii is not an invariant. Similarly, if Aij is a covariant tensor of rank two, A,; is not, in general, an invariant. 2.10. QUOTIENT RULE Consider a set of n3 functions A(lll),A(112),A(123), etc., or A(i,j,Ic) for short, with the indices i, j , Ic each ranging over 1 , 2 , . . . n. Although the set of functions A(i,j, I c ) has the right number of components, we do not know whether it is a tensor or not. Now, suppose that we know something about the nature of the product of A(i, j , k) with an arbitrary tensor. Then there is a theorem which enables us to establish whether A(i, j , k ) is a tensor without going to the trouble of determining the law of transformation directly. For example, let <"(x) be an arbitrary tensor of rank 1 (a vector). Let us suppose that the product A(a, j , k)<" (summation convention used over a) is known to yield a tensor of the type Ai(x), A(a, j , k , )Ea = A; . Then we can prove that A(i,j)k) is a tensor of the type A:k(x)....
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- Fall '05