Page 56 - Intro to Tensor Calculus

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Operations Using Tensors

The following are some important tensor operations which are used to derive special equations and to
prove various identities.

Addition and Subtraction
Tensors of the same type and weight can be added or subtracted. For example, two third order mixed
tensors, when added, produce another third order mixed tensor. Let A i and B i denote two third order
jk jk
mixed tensors. Their sum is denoted
i
C i = A i + B .
jk jk jk
That is, like components are added. The sum is also a mixed tensor as we now verify. By hypothesis A i jk
and B i are third order mixed tensors and hence must obey the transformation laws
jk
n
i
i m ∂x ∂x ∂x p
A jk = A np j
∂x m ∂x ∂x k
n
i
i m ∂x ∂x ∂x p
B = B .
jk np m j k
∂x ∂x ∂x
i i i
We let C jk = A jk + B jk denote the sum in the transformed coordinates. Then the addition of the above
transformation equations produces
i
i
n
n
i i i m m ∂x ∂x ∂x p m ∂x ∂x ∂x p
C jk = A jk + B jk = A np + B np j = C np j .
∂x m ∂x ∂x k ∂x m ∂x ∂x k
Consequently, the sum transforms as a mixed third order tensor.
Multiplication (Outer Product)
The product of two tensors is also a tensor. The rank or order of the resulting tensor is the sum of
the ranks of the tensors occurring in the multiplication. As an example, let A i denote a mixed third order
jk
tensor and let B l m denote a mixed second order tensor. The outer product of these two tensors is the fifth
order tensor
i
l
C il = A B , i,j,k,l,m =1, 2,...,N.
m
jk
jkm
i l
Here all indices are free indices as i, j, k, l, m take on any of the integer values 1, 2,... ,N. Let A and B
jk m
il
denote the components of the given tensors in the barred system of coordinates. We define C as the
jkm
outer product of these components. Observe that C il is a tensor for by hypothesis A i and B l are tensors
jkm jk m
and hence obey the transformation laws
α
j
α i ∂x ∂x ∂x k
A βγ = A jk γ
β
i
∂x ∂x ∂x (1.2.55)
δ
δ l ∂x ∂x m
B = B m .

l
∂x ∂x
The outer product of these components produces
δ
α
j
k
αδ α δ i l ∂x ∂x ∂x ∂x ∂x m
C βγ = A βγ B = A B m γ
jk

β
l
i
∂x ∂x ∂x ∂x ∂x (1.2.56)
δ
α
j
k
∂x ∂x ∂x ∂x ∂x m
il
= C jkm γ
i
l
β
∂x ∂x ∂x ∂x ∂x
il
which demonstrates that C jkm transforms as a mixed fifth order absolute tensor. Other outer products are
analyzed in a similar way.