Page 58 - Intro to Tensor Calculus

54

Quotient Law

s
Assume B qs and C are arbitrary absolute tensors. Further assume we have a quantity A(ijk) which
r p
i
we think might be a third order mixed tensor A . By showing that the equation
jk
r
A B qs = C s
qp r p
is satisfied, then it follows that A r must be a tensor. This is an example of the quotient law. Obviously,
qp
this result can be generalized to apply to tensors of any order or rank. To prove the above assertion we shall
i
i
show from the above equation that A i is a tensor. Let x and x denote a barred and unbarred system of
jk
coordinates which are related by transformations of the form defined by equation (1.2.30). In the barred
system, we assume that
r qs s
A B r = C p (1.2.59)
qp
ij l
where by hypothesis B and C are arbitrary absolute tensors and therefore must satisfy the transformation
k m
equations
s
q
qs ij ∂x ∂x ∂x k
B = B
r k i j r
∂x ∂x ∂x
s
s l ∂x ∂x m
C = C m p .
p
l
∂x ∂x
qs s
We substitute for B and C in the equation (1.2.59) and obtain the equation
r p
q s k s m
r ij ∂x ∂x ∂x l ∂x ∂x
A B = C
qp k i j r m l p
∂x ∂x ∂x ∂x ∂x
s
∂x ∂x m
= A r B ql .
qm r l p
∂x ∂x
Since the summation indices are dummy indices they can be replaced by other symbols. We change l to j,
q to i and r to k and write the above equation as
q
∂x s r ∂x ∂x k k ∂x m ij
A qp − A im B =0.
i
∂x j ∂x ∂x r ∂x p k
n
∂x
Use inner multiplication by s and simplify this equation to the form
∂x
r ∂x ∂x k ∂x m
q
δ n A − A k B ij =0 or
j qp i r im p k
∂x ∂x ∂x
q k m
r ∂x ∂x k ∂x in
A qp r − A im p B k =0.
i
∂x ∂x ∂x
Because B in is an arbitrary tensor, the quantity inside the brackets is zero and therefore
k
q
r ∂x ∂x k k ∂x m
A qp r − A im p =0.
i
∂x ∂x ∂x
i
This equation is simplified by inner multiplication by ∂x ∂x l
∂x ∂x k to obtain
j
i
r k ∂x m ∂x ∂x l
q l
δ δ A qp − A im p j =0 or
j r
∂x ∂x ∂x k
i
l ∂x m ∂x ∂x l
A = A k
jp im p j k
∂x ∂x ∂x
which is the transformation law for a third order mixed tensor.