2.6 Vector and Tensor Calculus
43
= da i δ ij
∂φ
∂a j
+ dT ij
∂φ
∂T kl
δ ik δ jl
=
∂φ
∂a i
da i +
∂φ
∂T ij
dT ij
= dφ.
The last line follows from the representation φ(a, T) = φ(a i , T ij ), with the
total differential involving partial derivatives taken with respect to all the scalar
components. Differentiating a vector or tensor function follows simply by replacing
φ in Eq. (2.56) with the vector or tensor.
Example 2.9 Compute ∂a/∂a and ∂T/∂T.
Solution
Substituting the component forms of a and T into Eq. (2.58) yields
∂a
∂a
=
e i
∂
∂a i
a j e j
=
∂a j
∂a i
e i e j = δ ij e i e j
= e i e i = I
∂T
∂T
=
e i e j
∂
∂T ij
(T kl e k e l ) =
∂T kl
∂T ij
e i e j e k e l = δ ik δ jl e i e j e k e l
= e i e j e i e j = ˆ
I.
Here, we set ∂a j /∂a i = δ ij because the derivative is unity for i = j and zero for
i = j . Similar considerations lead to ∂T kl /∂T ij = δ ik δ jl .
Please note the following. First, differentiating a vector with respect to a vector
gives a second-order tensor, while differentiating a second-order tensor with respect
to a second-order tensor yields a fourth-order tensor. Second, just as differentiating a
scalar variable with respect to itself yields unity, differentiating a vector with respect
to itself gives the second-order identity tensor, and differentiating a second-order
tensor with respect to itself gives the fourth-order identity tensor ˆ
I, which is defined
above. Similar to the expression a = a · I, it can be shown that
T = T : ˆ
I
(2.59)
for any second-order tensor T (see Problem 2.6).