1. Tensor Calculus
Differentials & Directional Derivatives

2. The Gateaux Differential
We are presently concerned with Inner Product spaces
in our treatment of the Mechanics of Continua.
Consider a map,
𝑭: V → W
This maps from the domainV to W – both of which are
Euclidean vector spaces. The concepts of limit and
continuity carries naturally from the real space to any
Euclidean vector space.
Dept of Systems Engineering, University of Lagos 2 oafak@unilag.edu.ng 12/27/2012

3. The Gateaux Differential
Let 𝒗0 ∈ V and 𝒘0 ∈ W, as usual we can say that the
limit
lim 𝑭 𝒗 = 𝒘0
𝒗→ 𝒗0
if for any pre-assigned real number 𝜖 > 0, no matter
how small, we can always find a real number 𝛿 > 0
such that 𝑭 𝒗 − 𝒘0 ≤ 𝜖 whenever 𝒗 − 𝒗0 < 𝛿.
The function is said to be continuous at 𝒗0 if 𝑭 𝒗0
exists and 𝑭 𝒗0 = 𝒘0
Dept of Systems Engineering, University of Lagos 3 oafak@unilag.edu.ng 12/27/2012

4. The Gateaux Differential
Specifically, for 𝛼 ∈ ℛ let this map be:
𝑭 𝒙 + 𝛼𝒉 − 𝑭 𝒙 𝑑
𝐷𝑭 𝒙, 𝒉 ≡ lim = 𝑭 𝒙 + 𝛼𝒉
𝛼→0 𝛼 𝑑𝛼 𝛼=0
We focus attention on the second variable ℎ while we allow
the dependency on 𝒙 to be as general as possible. We shall
show that while the above function can be any given function
of 𝒙 (linear or nonlinear), the above map is always linear in 𝒉
irrespective of what kind of Euclidean space we are mapping
from or into. It is called the Gateaux Differential.
Dept of Systems Engineering, University of Lagos 4 oafak@unilag.edu.ng 12/27/2012

5. The Gateaux Differential
Let us make the Gateaux differential a little more familiar in
real space in two steps: First, we move to the real space and
allow ℎ → 𝑑𝑥 and we obtain,
𝐹 𝑥 + 𝛼𝑑𝑥 − 𝐹 𝑥 𝑑
𝐷𝐹(𝑥, 𝑑𝑥) = lim = 𝐹 𝑥 + 𝛼𝑑𝑥
𝛼→0 𝛼 𝑑𝛼 𝛼=0
And let 𝛼𝑑𝑥 → Δ𝑥, the middle term becomes,
𝐹 𝑥 + Δ𝑥 − 𝐹 𝑥 𝑑𝐹
lim 𝑑𝑥 = 𝑑𝑥
Δ𝑥→0 Δ𝑥 𝑑𝑥
from which it is obvious that the Gateaux derivative is a
generalization of the well-known differential from
elementary calculus. The Gateaux differential helps to
compute a local linear approximation of any function (linear
or nonlinear).
Dept of Systems Engineering, University of Lagos 5 oafak@unilag.edu.ng 12/27/2012

6. The Gateaux Differential
It is easily shown that the Gateaux differential is linear
in its second argument, ie, for 𝛼 ∈ R
𝐷𝑭 𝒙, 𝛼𝒉 = 𝛼𝐷𝑭 𝒙, 𝒉
Furthermore,
𝐷𝑭 𝒙, 𝒈 + 𝒉 = 𝐷𝑭 𝒙, 𝒈 + 𝐷𝑭 𝒙, 𝒉
and that for 𝛼, 𝛽 ∈ R
𝐷𝑭 𝒙, 𝛼𝒈 + 𝛽𝒉 = 𝛼𝐷𝑭 𝒙, 𝒈 + 𝛽𝐷𝑭 𝒙, 𝒉
Dept of Systems Engineering, University of Lagos 6 oafak@unilag.edu.ng 12/27/2012

7. The Frechét Derivative
The Frechét derivative or gradient of a differentiable
function is the linear operator, grad 𝑭(𝒗) such that, for
𝑑𝒗 ∈ V,
𝑑𝑭 𝒗
𝑑𝒗 ≡ grad 𝑭 𝒗 𝑑𝒗 ≡ 𝐷𝑭 𝒗, 𝑑𝒗
𝑑𝒗
Obviously, grad 𝑭(𝒗) is a tensor because it is a linear
transformation from one Euclidean vector space to
another. Its linearity derives from the second argument
of the RHS.
Dept of Systems Engineering, University of Lagos 7 oafak@unilag.edu.ng 12/27/2012

8. Differentiable Real-Valued Function
The Gradient of a Differentiable Real-Valued Function
For a real vector function, we have the map:
𝑓: V → R
The Gateaux differential in this case takes two vector
arguments and maps to a real value:
𝐷𝑓: V × V → R
This is the classical definition of the inner product so that we
can define the differential as,
𝑑𝑓(𝒗)
⋅ 𝑑𝒗 ≡ 𝐷𝑓 𝒗, 𝑑𝒗
𝑑𝒗
This quantity,
𝑑𝑓(𝒗)
𝑑𝒗
defined by the above equation, is clearly a vector. oafak@unilag.edu.ng 12/27/2012
Dept of Systems Engineering, University of Lagos 8

9. Real-Valued Function
It is the gradient of the scalar valued function of the
vector variable. We now proceed to find its components
in general coordinates. To do this, we choose a basis
𝐠 𝑖 ⊂ V. On such a basis, the function,
𝒗 = 𝑣𝑖 𝐠𝑖
and,
𝑓 𝒗 = 𝑓 𝑣𝑖 𝐠𝑖
we may also express the independent vector
𝑑𝒗 = 𝑑𝑣 𝑖 𝐠 𝑖
on this basis.
Dept of Systems Engineering, University of Lagos 9 oafak@unilag.edu.ng 12/27/2012

10. Real-Valued Function
The Gateaux differential in the direction of the basis vector 𝐠 𝑖 is,
𝑓 𝒗 + 𝛼𝐠 𝑖 − 𝑓 𝒗 𝑓 𝑣 𝑗 𝐠 𝑗 + 𝛼𝐠 𝑖 − 𝑓 𝑣 𝑖 𝐠 𝑖
𝐷𝑓 𝒗, 𝐠 𝑖 = lim = lim
𝛼→0 𝛼 𝛼→0 𝛼
𝑗
𝑓 𝑣 𝑗 + 𝛼𝛿 𝑖 𝐠 𝑗 − 𝑓 𝑣 𝑖 𝐠 𝑖
= lim
𝛼→0 𝛼
As the function 𝑓 does not depend on the vector basis, we can
substitute the vector function by the real function of the components
𝑣 1 , 𝑣 2 , 𝑣 3 so that,
𝑓 𝑣 𝑖 𝐠 𝑖 = 𝑓 𝑣1, 𝑣 2, 𝑣 3
in which case, the above differential, similar to the real case discussed
earlier becomes,
𝜕𝑓 𝑣 1 , 𝑣 2 , 𝑣 3
𝐷𝑓 𝒗, 𝐠 𝑖 =
𝜕𝑣 𝑖
Dept of Systems Engineering, University of Lagos 10 oafak@unilag.edu.ng 12/27/2012

11. 𝑑𝑓(𝒗)
Of course, it is now obvious that the gradient of
𝑑𝒗
the scalar valued vector function, expressed in its dual
basis must be,
𝑑𝑓(𝒗) 𝜕𝑓 𝑣 1 , 𝑣 2 , 𝑣 3 𝑖
= 𝑖
𝐠
𝑑𝒗 𝜕𝑣
so that we can recover,
𝑑𝑓 𝒗 𝜕𝑓 𝑣 1 , 𝑣 2 , 𝑣 3 𝑗
𝐷𝑓 𝒗, 𝐠 𝑖 = ⋅ 𝑑𝒗 = 𝑗
𝐠 ⋅ 𝐠𝑖
𝑑𝒗 𝜕𝑣
𝜕𝑓 𝑣 1 , 𝑣 2 , 𝑣 3 𝑗 𝜕𝑓 𝑣 1 , 𝑣 2 , 𝑣 3
= 𝑗
𝛿𝑖 =
𝜕𝑣 𝜕𝑣 𝑖
Hence we obtain the well-known result that
𝑑𝑓(𝒗) 𝜕𝑓 𝑣 1 , 𝑣 2 , 𝑣 3 𝑖
grad 𝑓 𝒗 = = 𝑖
𝐠
𝑑𝒗 𝜕𝑣
which defines the Frechét derivative of a scalar valued
function with respect to its vector argument.
Dept of Systems Engineering, University of Lagos 11 oafak@unilag.edu.ng 12/27/2012

12. Vector valued Function
 The Gradient of a Differentiable Vector valued Function
The definition of the Frechét clearly shows that the gradient
of a vector-valued function is itself a second order tensor. We
now find the components of this tensor in general
coordinates. To do this, we choose a basis 𝐠 𝑖 ⊂ V. On such
a basis, the function,
𝑭 𝒗 = 𝐹𝑘 𝒗 𝐠𝑘
The functional dependency on the basis vectors are ignorable
on account of the fact that the components themselves are
fixed with respect to the basis. We can therefore write,
𝑭 𝒗 = 𝐹 𝑘 𝑣1 , 𝑣 2 , 𝑣 3 𝐠 𝑘
Dept of Systems Engineering, University of Lagos 12 oafak@unilag.edu.ng 12/27/2012

13. Vector Derivatives
𝐷𝑭 𝒗, 𝑑𝒗 = 𝐷𝑭 𝑣 𝑖 𝐠 𝑖 , 𝑑𝑣 𝑖 𝐠 𝑖
= 𝐷𝑭 𝑣 𝑖 𝐠 𝑖 , 𝐠 𝑖 𝑑𝑣 𝑖
= 𝐷𝐹 𝑘 𝑣 𝑖 𝐠 𝑖 , 𝐠 𝑖 𝐠 𝑘 𝑑𝑣 𝑖
Again, upon noting that the functions 𝐷𝐹 𝑘 𝑣 𝑖 𝐠 𝑖 , 𝐠 𝑖 ,
𝑘 = 1,2,3 are not functions of the vector basis, they can
be written as functions of the scalar components alone
so that we have, as before,
𝐷𝑭 𝒗, 𝑑𝒗 = 𝐷𝐹 𝑘 𝑣 𝑖 𝐠 𝑖 , 𝐠 𝑖 𝐠 𝑘 𝑑𝑣 𝑖
= 𝐷𝐹 𝑘 𝒗, 𝐠 𝑖 𝐠 𝑘 𝑑𝑣 𝑖
Dept of Systems Engineering, University of Lagos 13 oafak@unilag.edu.ng 12/27/2012

14. Which we can compare to the earlier case of scalar valued
function and easily obtain,
𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3 𝑗
𝐷𝑭 𝒗, 𝑑𝒗 = 𝐠 ⋅ 𝐠 𝑖 𝐠 𝑘 𝑑𝑣 𝑖
𝜕𝑣 𝑗
𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3
= 𝐠 𝑘 ⊗ 𝐠 𝑗 𝐠 𝑖 𝑑𝑣 𝑖
𝜕𝑣 𝑗
𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3
= 𝐠 𝑘 ⊗ 𝐠 𝑗 𝑑𝒗
𝜕𝑣 𝑗
𝑑𝑭 𝒗
= 𝑑𝒗 ≡ grad 𝑭 𝒗 𝑑𝒗
𝑑𝒗
Clearly, the tensor gradient of the vector-valued vector function
is,
𝑑𝑭 𝒗 𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3
= grad 𝑭 𝒗 = 𝐠 𝑘 ⊗ 𝐠𝑗
𝑑𝒗 𝜕𝑣 𝑗
Where due attention should be paid to the covariance of the
quotient indices in contrast to the contravariance of the non
quotient indices.
Dept of Systems Engineering, University of Lagos 14 oafak@unilag.edu.ng 12/27/2012

15. The Trace & Divergence
The trace of this gradient is called the divergence of the
function:
𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3
div 𝑭 𝒗 = tr 𝐠 𝑘 ⊗ 𝐠𝑗
𝜕𝑣 𝑗
𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3
= 𝐠 𝑘 ⋅ 𝐠𝑗
𝜕𝑣 𝑗
𝜕𝐹 𝑘 𝑣 1 , 𝑣 2 , 𝑣 3 𝑗
= 𝑗
𝛿𝑘
𝜕𝑣
𝜕𝐹 𝑗 𝑣 1 , 𝑣 2 , 𝑣 3
=
𝜕𝑣 𝑗
Dept of Systems Engineering, University of Lagos 15 oafak@unilag.edu.ng 12/27/2012

16. Tensor-Valued Vector Function
Consider the mapping,
𝑻: V → T
Which maps from an inner product space to a tensor space. The
Gateaux differential in this case now takes two vectors and
produces a tensor:
𝐷𝑻: V × V → T
With the usual notation, we may write,
𝐷𝑻 𝒗, 𝑑𝒗 = 𝐷𝑻 𝑣 𝑖 𝐠 𝑖 , 𝑑𝑣 𝑖 𝐠 𝑖 = 𝐷𝑻 𝑣 𝑖 𝐠 𝑖 , 𝑑𝑣 𝑖 𝐠 𝑖
𝑑𝑻 𝒗
= 𝑑𝒗 ≡ grad 𝑻 𝒗 𝑑𝒗
𝑑𝒗
= 𝐷𝑇 𝛼𝛽 𝑣 𝑖 𝐠 𝑖 , 𝑑𝑣 𝑖 𝐠 𝑖 𝐠 𝛼 ⊗ 𝐠 𝛽
= 𝐷𝑇 𝛼𝛽 𝒗, 𝐠 𝑖 𝐠 𝛼 ⊗ 𝐠 𝛽 𝑑𝑣 𝑖
Each of these nine functions look like the real differential of several
variables we considered earlier.
Dept of Systems Engineering, University of Lagos 16 oafak@unilag.edu.ng 12/27/2012

17. The independence of the basis vectors imply, as usual, that
𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3
𝐷𝑇 𝛼𝛽 𝒗, 𝐠 𝑖 =
𝜕𝑣 𝑖
so that,
𝐷𝑻 𝒗, 𝑑𝒗 = 𝐷𝑇 𝛼𝛽 𝒗, 𝐠 𝑖 𝐠 𝛼 ⊗ 𝐠 𝛽 𝑑𝑣 𝑖
𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3
= 𝐠 𝛼 ⊗ 𝐠 𝛽 𝑑𝑣 𝑖
𝜕𝑣 𝑖
𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3
= 𝐠 𝛼 ⊗ 𝐠 𝛽 𝐠 𝑖 ⋅ 𝑑𝒗
𝜕𝑣 𝑖
𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3 𝑑𝑻 𝒗
= 𝐠 𝛼 ⊗ 𝐠 𝛽 ⊗ 𝐠 𝑖 𝑑𝒗 = 𝑑𝒗
𝜕𝑣 𝑖 𝑑𝒗
≡ grad 𝑻 𝒗 𝑑𝒗
Which defines the third-order tensor,
𝑑𝑻 𝒗 𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3
= grad 𝑻 𝒗 = 𝐠 𝛼 ⊗ 𝐠 𝛽 ⊗ 𝐠𝑖
𝑑𝒗 𝜕𝑣 𝑖
and with no further ado, we can see that a third-order tensor
transforms a vector into a second order tensor.
Dept of Systems Engineering, University of Lagos 17 oafak@unilag.edu.ng 12/27/2012

18. The Divergence of a Tensor Function
The divergence operation can be defined in several ways.
Most common is achieved by the contraction of the last two
basis so that,
𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3
div 𝑻 𝒗 = 𝐠 𝛼 ⊗ 𝐠 𝛽 ⋅ 𝐠𝑖
𝜕𝑣 𝑖
𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3
= 𝐠 𝛼 𝐠 𝛽 ⋅ 𝐠𝑖
𝜕𝑣 𝑖
𝜕𝑇 𝛼𝛽 𝑣 1 , 𝑣 2 , 𝑣 3 𝑖 𝜕𝑇 𝛼𝑖 𝑣 1 , 𝑣 2 , 𝑣 3
= 𝑖
𝐠 𝛼 𝛿𝛽 = 𝑖
𝐠𝛼
𝜕𝑣 𝜕𝑣
It can easily be shown that this is the particular vector that
gives, for all constant vectors 𝒂,
div 𝑻 𝒂 ≡ div 𝑻T 𝒂
Dept of Systems Engineering, University of Lagos 18 oafak@unilag.edu.ng 12/27/2012

19. Real-Valued Tensor Functions
Two other kinds of functions are critically important in
our study. These are real valued functions of tensors
and tensor-valued functions of tensors. Examples in the
first case are the invariants of the tensor function that
we have already seen. We can express stress in terms of
strains and vice versa. These are tensor-valued functions
of tensors. The derivatives of such real and tensor
functions arise in our analysis of continua. In this section
these are shown to result from the appropriate Gateaux
differentials. The gradients or Frechét derivatives will be
extracted once we can obtain the Gateaux differentials.
Dept of Systems Engineering, University of Lagos 19 oafak@unilag.edu.ng 12/27/2012

20. Frechét Derivative
Consider the Map:
𝑓: T → R
The Gateaux differential in this case takes two vector
arguments and maps to a real value:
𝐷𝑓: T × T → R
Which is, as usual, linear in the second argument. The Frechét
derivative can be expressed as the first component of the
following scalar product:
𝑑𝑓(𝑻)
𝐷𝑓 𝑻, 𝑑𝑻 = : 𝑑𝑻
𝑑𝑻
which, we recall, is the trace of the contraction of one tensor
with the transpose of the other second-order tensors. This is
a scalar quantity.
Dept of Systems Engineering, University of Lagos 20 oafak@unilag.edu.ng 12/27/2012

21. Guided by the previous result of the gradient of the scalar valued
function of a vector, it is not difficult to see that,
𝑑𝑓(𝑻) 𝜕𝑓(𝑇11 , 𝑇12 , … , 𝑇 33 ) 𝛼
= 𝐠 ⊗ 𝐠𝛽
𝑑𝑻 𝜕𝑇 𝛼𝛽
In the dual to the same basis, we can write,
𝑑𝑻 = 𝑑𝑇 𝑖𝑗 𝐠 𝑖 ⊗ 𝐠 𝑗
Clearly,
𝑑𝑓(𝑻) 𝜕𝑓(𝑇11 , 𝑇12 , … , 𝑇 33 ) 𝛼
: 𝑑𝑻 = 𝐠 ⊗ 𝐠 𝛽 : 𝑑𝑇 𝑖𝑗 𝐠 𝑖 ⊗ 𝐠 𝑗
𝑑𝑻 𝜕𝑇 𝛼𝛽
𝜕𝑓(𝑇11 , 𝑇12 , … , 𝑇 33 )
= 𝑑𝑇 𝑖𝑗
𝜕𝑇 𝑖𝑗
Again, note the covariance of the quotient indices.
Dept of Systems Engineering, University of Lagos 21 oafak@unilag.edu.ng 12/27/2012

22. Examples
Let 𝐒 be a symmetric and positive definite tensor and let
𝐼1 𝑺 , 𝐼2 𝑺 and 𝐼3 𝑺 be the three principal invariants of
𝜕𝑰1 𝐒
𝐒 show that (a) = 𝟏 the identity tensor, (b)
𝜕𝐒
𝜕𝑰2 𝐒 𝜕𝐼3 𝐒
= 𝐼1 𝐒 𝟏 − 𝐒 and (c) = 𝐼3 𝐒 𝐒 −1
𝜕𝐒 𝜕𝐒
𝜕𝑰1 𝐒
can be written in the invariant component form
𝜕𝐒
as,
𝜕𝐼1 𝐒 𝜕𝐼1 𝐒
= 𝑗
𝐠𝑖 ⊗ 𝐠𝑗
𝜕𝐒 𝜕𝑆 𝑖
Dept of Systems Engineering, University of Lagos 22 oafak@unilag.edu.ng 12/27/2012

23. (a) Continued
α
Recall that 𝐼1 𝐒 = tr 𝐒 = 𝑆α hence
𝜕𝐼1 𝐒 𝜕𝐼1 𝐒
= 𝑗
𝐠𝑖 ⊗ 𝐠𝑗
𝜕𝐒 𝜕𝑆 𝑖
α
𝜕𝑆α
= 𝑗
𝐠𝑖 ⊗ 𝐠𝑗
𝜕𝑆 𝑖
= 𝛿 𝑖𝛼 𝛿 𝑗 𝛼 𝐠 𝑖 ⊗ 𝐠 𝑗
= 𝛿 𝑗𝑖 𝐠 𝑖 ⊗ 𝐠 𝑗 = 𝟏
which is the identity tensor as expected.
Dept of Systems Engineering, University of Lagos 23 oafak@unilag.edu.ng 12/27/2012

24. 𝜕𝐼2 𝐒
(b) = 𝐼1 𝐒 𝟏 − 𝐒
𝜕𝐒
𝜕𝐼2 𝐒
in a similar way can be written in the invariant component form
𝜕𝐒
as,
𝜕𝐼2 𝐒 1 𝜕𝐼1 𝐒 α 𝛽 𝛽
= 𝑆α 𝑆 𝛽 − 𝑆 α 𝑆α 𝐠 𝑖 ⊗ 𝐠 𝑗
𝛽
𝜕𝐒 2 𝜕𝑆 𝑗
𝑖
1
where we have utilized the fact that 𝐼2 𝐒 = 2 tr 2 𝐒 − tr(𝐒 2 ) .
Consequently,
𝜕𝐼2 𝐒 1 𝜕 α 𝛽 𝛽
= 𝑆α 𝑆 𝛽 − 𝑆 α 𝑆α 𝐠 𝑖 ⊗ 𝐠 𝑗
𝛽
𝜕𝐒 2 𝜕𝑆 𝑗
𝑖
1 𝑖 𝛼 𝛽 𝑖 𝛽 α 𝛽 𝛽
= 𝛿 𝛼 𝛿 𝑗 𝑆 𝛽 + 𝛿 𝛽 𝛿 𝑗 𝑆α − 𝛿 𝛽 𝛿 𝑗 𝛼 𝑆α − 𝛿 𝑖𝛼 𝛿 𝑗 𝑆 α 𝐠 𝑖 ⊗ 𝐠 𝑗
𝑖
𝛽
2
1 𝑖 𝛽 𝑗 𝑗
= 𝛿 𝑗 𝑆 𝛽 + 𝛿 𝑗𝑖 𝑆α − 𝑆 𝑖 − 𝑆 𝑖 𝐠 𝑖 ⊗ 𝐠 𝑗
α
2
𝑗
= 𝛿 𝑗𝑖 𝑆α − 𝑆 𝑖 𝐠 𝑖 ⊗ 𝐠 𝑗 = 𝐼1 𝐒 𝟏 − 𝐒
α
Dept of Systems Engineering, University of Lagos 24 oafak@unilag.edu.ng 12/27/2012

25. 𝜕𝑰3 𝐒 𝜕det 𝐒
 = = 𝐒𝐜
𝜕𝐒 𝜕𝐒
 the cofactor of 𝐒. Clearly 𝐒 𝐜 = det 𝐒 𝐒−1 = 𝑰3 𝐒 𝐒−1 as required. Details of
this for the contravariant components of a tensor is presented below. Let
1 𝑖𝑗𝑘 𝑟𝑠𝑡
𝐒 = 𝑆= 𝑒 𝑒 𝑆 𝑖𝑟 𝑆 𝑗𝑠 𝑆 𝑘𝑡
3!
Differentiating wrt 𝑆 𝛼𝛽 , we obtain,
𝜕𝑆 1 𝑖𝑗𝑘 𝑟𝑠𝑡 𝜕𝑆 𝑖𝑟 𝜕𝑆 𝑗𝑠 𝜕𝑆 𝑘𝑡
𝐠 ⊗ 𝐠 𝛽= 𝑒 𝑒 𝑆 𝑆 + 𝑆 𝑖𝑟 𝑆 + 𝑆 𝑖𝑟 𝑆 𝑗𝑠 𝐠 ⊗ 𝐠𝛽
𝜕𝑆 𝛼𝛽 𝛼 3! 𝜕𝑆 𝛼𝛽 𝑗𝑠 𝑘𝑡 𝜕𝑆 𝛼𝛽 𝑘𝑡 𝜕𝑆 𝛼𝛽 𝛼
1 𝑖𝑗𝑘 𝑟𝑠𝑡 𝛼 𝛽 𝛽 𝛽
= 𝑒 𝑒 𝛿 𝑖 𝛿 𝑟 𝑆 𝑗𝑠 𝑆 𝑘𝑡 + 𝑆 𝑖𝑟 𝛿 𝑗 𝛼 𝛿 𝑠 𝑆 𝑘𝑡 + 𝑆 𝑖𝑟 𝑆 𝑗𝑠 𝛿 𝑘𝛼 𝛿 𝑡 𝐠 𝛼 ⊗ 𝐠 𝛽
3!
1 𝛼𝑗𝑘 𝛽𝑠𝑡
= 𝑒 𝑒 𝑆 𝑗𝑠 𝑆 𝑘𝑡 + 𝑆 𝑗𝑠 𝑆 𝑘𝑡 + 𝑆 𝑗𝑠 𝑆 𝑘𝑡 𝐠 𝛼 ⊗ 𝐠 𝛽
3!
1 𝛼𝑗𝑘 𝛽𝑠𝑡
= 𝑒 𝑒 𝑆 𝑗𝑠 𝑆 𝑘𝑡 𝐠 𝛼 ⊗ 𝐠 𝛽 ≡ 𝑆 c 𝛼𝛽 𝐠 𝛼 ⊗ 𝐠 𝛽
2!
Which is the cofactor of 𝑆 𝛼𝛽
Dept of Systems Engineering, University of Lagos 25 oafak@unilag.edu.ng 12/27/2012

26. Real Tensor Functions
Consider the function 𝑓(𝑺) = tr 𝑺 𝑘 where 𝑘 ∈ ℛ, and
𝑺 is a tensor.
𝑓 𝑺 = tr 𝑺 𝑘 = 𝑺 𝑘 : 𝟏
To be specific, let 𝑘 = 3.
Dept of Systems Engineering, University of Lagos 26 oafak@unilag.edu.ng 12/27/2012

27. Real Tensor Functions
The Gateaux differential in this case,
𝑑 𝑑 3
𝐷𝑓 𝑺, 𝑑𝑺 = 𝑓 𝑺 + 𝛼𝑑𝑺 = tr 𝑺 + 𝛼𝑑𝑺
𝑑𝛼 𝛼=0
𝑑𝛼 𝛼=0
𝑑
= tr 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺
𝑑𝛼 𝛼=0
𝑑
= tr 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺
𝑑𝛼 𝛼=0
= tr 𝑑𝑺 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺
+ 𝑺 + 𝛼𝑑𝑺 𝑑𝑺 𝑺 + 𝛼𝑑𝑺
+ 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺 𝑑𝑺
𝛼=0
= tr 𝑑𝑺 𝑺 𝑺 + 𝑺 𝑑𝑺 𝑺 + 𝑺 𝑺 𝒅𝑺 = 3 𝑺2 T : 𝑑𝑺
Dept of Systems Engineering, University of Lagos 27 oafak@unilag.edu.ng 12/27/2012

28. Real Tensor Functions
With the last equality coming from the definition of the
Inner product while noting that a circular permutation
does not alter the value of the trace. It is easy to
establish inductively that in the most general case, for
𝑘 > 0, we have,
𝑘−1 T
𝐷𝑓 𝑺, 𝑑𝑺 = 𝑘 𝑺 : 𝑑𝑺
Clearly,
𝑑 T
tr 𝑺 𝑘 = 𝑘 𝑺 𝑘−1
𝑑𝑺
Dept of Systems Engineering, University of Lagos 28 oafak@unilag.edu.ng 12/27/2012

29. Real Tensor Functions
When 𝑘 = 1,
𝑑
𝐷𝑓 𝑺, 𝑑𝑺 = 𝑓 𝑺 + 𝛼𝑑𝑺
𝑑𝛼 𝛼=0
𝑑
= tr 𝑺 + 𝛼𝑑𝑺
𝑑𝛼 𝛼=0
= tr 𝟏 𝑑𝑺 = 𝟏: 𝑑𝑺
Or that,
𝑑
tr 𝑺 = 𝟏.
𝑑𝑺
Dept of Systems Engineering, University of Lagos 29 oafak@unilag.edu.ng 12/27/2012

30. Real Tensor Functions
Derivatives of the other two invariants of the tensor 𝑺
can be found as follows:
𝑑
𝐼2 (𝑺)
𝑑𝑺
1 𝑑
= tr 2 𝑺 − tr 𝑺2
2 𝑑𝑺
1
= 2tr 𝑺 𝟏 − 2 𝑺T = tr 𝑺 𝟏 − 𝑺T
2
.
Dept of Systems Engineering, University of Lagos 30 oafak@unilag.edu.ng 12/27/2012

31. Real Tensor Functions
To Determine the derivative of the third invariant, we
begin with the trace of the Cayley-Hamilton for 𝑺:
tr 𝑺3 − 𝐼1 𝑺2 + 𝐼2 𝑺 − 𝐼3 𝑰 = tr(𝑺3 ) − 𝐼1 tr 𝑺2 + 𝐼2 tr 𝑺
− 3𝐼3 = 0
Therefore,
3𝐼3 = tr(𝑺3 ) − 𝐼1 tr 𝑺2 + 𝐼2 tr 𝑺
1 2
𝐼2 𝑺 = tr 𝑺 − tr 𝑺2
2
.
Dept of Systems Engineering, University of Lagos 31 oafak@unilag.edu.ng 12/27/2012

32. Differentiating the Invariants
We can therefore write,
3𝐼3 𝑺 = tr(𝑺3 ) − tr 𝑺 tr 𝑺2
1 2
+ tr 𝑺 − tr 𝑺2 tr 𝑺
2
so that, in terms of traces only,
1 3
𝐼3 𝑺 = tr 𝑺 − 3tr 𝑺 tr 𝑺2 + 2tr(𝑺3 )
6
Dept of Systems Engineering, University of Lagos 32 oafak@unilag.edu.ng 12/27/2012

33. Real Tensor Functions
Clearly,
𝑑𝐼3 (𝑺) 1
= 3tr 2 𝑺 𝟏 − 3tr 𝑺2 − 3tr 𝑺 2𝑺 𝑇 + 2 × 3 𝑺2 𝑇
𝑑𝑺 6
= 𝐼2 𝟏 − tr 𝑺 𝑺 𝑇 + 𝑺2𝑇
Dept of Systems Engineering, University of Lagos 33 oafak@unilag.edu.ng 12/27/2012

34. Real Tensor Functions
.
Dept of Systems Engineering, University of Lagos 34 oafak@unilag.edu.ng 12/27/2012

35. The Euclidean Point Space
In Classical Theory, the world is a Euclidean Point Space
of dimension three. We shall define this concept now
and consequently give specific meanings to related
concepts such as
 Frames of Reference,
 Coordinate Systems and
 Global Charts
Dept of Systems Engineering, University of Lagos 35 oafak@unilag.edu.ng 12/27/2012

36. The Euclidean Point Space
 Scalars, Vectors and Tensors we will deal with in
general vary from point to point in the material. They
are therefore to be regarded as functions of position
in the physical space occupied.
 Such functions, associated with positions in the
Euclidean point space, are called fields.
 We will therefore be dealing with scalar, vector and
tensor fields.
Dept of Systems Engineering, University of Lagos 36 oafak@unilag.edu.ng 12/27/2012

37. The Euclidean Point Space
 The Euclidean Point Space ℰ is a set of elements
called points. For each pair of points 𝒙, 𝒚 ∈ ℰ,
∃ 𝒖 𝒙, 𝒚 ∈ E with the following two properties:
1. 𝒖 𝒙, 𝒚 = 𝒖 𝒙, 𝒛 + 𝒖 𝒛, 𝒚 ∀𝒙, 𝒚, 𝒛 ∈ ℰ
2. 𝒖 𝒙, 𝒚 = 𝒖 𝒙, 𝒛 ⇔ 𝒚 = 𝒛
Based on these two, we proceed to show that,
𝒖 𝒙, 𝒙 = 𝟎
And that,
𝒖 𝒙, 𝒛 = −𝒖 𝒛, 𝒙
Dept of Systems Engineering, University of Lagos 37 oafak@unilag.edu.ng 12/27/2012

38. The Euclidean Point Space
From property 1, let 𝒚 → 𝒙 it is clear that,
𝒖 𝒙, 𝒙 = 𝒖 𝒙, 𝒛 + 𝒖 𝒛, 𝒙
And it we further allow 𝒛 → 𝒙, we find that,
𝒖 𝒙, 𝒙 = 𝒖 𝒙, 𝒙 + 𝒖 𝒙, 𝒙 = 2𝒖 𝒙, 𝒙
Which clearly shows that 𝒖 𝒙, 𝒙 = 𝒐 the zero vector.
Similarly, from the above, we find that,
𝒖 𝒙, 𝒙 = 𝒐 = 𝒖 𝒙, 𝒛 + 𝒖 𝒛, 𝒙
So that 𝒖 𝒙, 𝒛 = −𝒖 𝒛, 𝒙
Dept of Systems Engineering, University of Lagos 38 oafak@unilag.edu.ng 12/27/2012

39. Dept of Systems Engineering, University of Lagos 39 oafak@unilag.edu.ng 12/27/2012

40. Coordinate System
Dept of Systems Engineering, University of Lagos 40 oafak@unilag.edu.ng 12/27/2012

41. Coordinate System
Dept of Systems Engineering, University of Lagos 41 oafak@unilag.edu.ng 12/27/2012

42. Position Vectors
Note that ℰ is NOT a vector space. In our discussion, the
vector space E to which 𝒖 belongs, is associated as we
have shown. It is customary to oscillate between these
two spaces. When we are talking about the vectors,
they are in E while the points are in ℰ.
We adopt the convention that 𝒙 𝒚 ≡ 𝒖 𝒙, 𝒚 referring
to the vector 𝒖. If therefore we choose an arbitrarily
fixed point 𝟎 ∈ ℰ, we are associating 𝒙 𝟎 , 𝒚 𝟎 and
𝒛 𝟎 respectively with 𝒖 𝒙, 𝟎 , 𝒖 𝒚, 𝟎 and 𝒖 𝒛, 𝟎 .
These are vectors based on the points 𝒙, 𝒚 and 𝒛 with
reference to the origin chosen. To emphasize the
association with both points of ℰ as well as members of
E they are called Position Vectors.
Dept of Systems Engineering, University of Lagos 42 oafak@unilag.edu.ng 12/27/2012

43. Length in the Point Space
Recall that by property 1,
𝒙 𝒚 ≡ 𝒖 𝒙, 𝒚 = 𝒖 𝒙, 𝟎 + 𝒖 𝟎, 𝒚
Furthermore, we have deduced that 𝒖 𝟎, 𝒚 = −𝒖(𝒚, 𝟎)
We may therefore write that,
𝒙 𝒚 = 𝒙 𝟎 − 𝒚(𝟎)
Which, when there is no ambiguity concerning the
chosen origin, we can write as,
𝒙 𝒚 = 𝒙− 𝒚
And the distance between the two is,
𝒅 𝒙 𝒚 = 𝒅 𝒙− 𝒚 = 𝒙− 𝒚
Dept of Systems Engineering, University of Lagos 43 oafak@unilag.edu.ng 12/27/2012

44. Metric Properties
 The norm of a vector is the square root of the inner
product of the vector with itself. If the coordinates of
𝒙 and 𝒚 on a set of independent vectors are 𝑥 𝑖 and 𝑦 𝑖 ,
then the distance we seek is,
𝒅 𝒙− 𝒚 = 𝒙− 𝒚 = 𝑔 𝑖𝑗 (𝑥 𝑖 − 𝑦 𝑖 )(𝑥 𝑗 − 𝑦 𝑗 )
The more familiar Pythagorean form occurring only
when 𝑔 𝑖𝑗 = 1.
Dept of Systems Engineering, University of Lagos 44 oafak@unilag.edu.ng 12/27/2012

45. Coordinate Transformations
 Consider a Cartesian coordinate system
𝑥, 𝑦, 𝑧 or 𝑦1 , 𝑦 2 and 𝑦 3 with an orthogonal basis. Let us
now have the possibility of transforming to another
coordinate system of an arbitrary nature: 𝑥 1 , 𝑥 2 , 𝑥 3 .
We can represent the transformation and its inverse
in the equations:
 𝑦 𝑖 = 𝑦 𝑖 𝑥 1 , 𝑥 2 , 𝑥 3 , 𝑥 𝑖 = 𝑥 𝑖 (𝑦1 , 𝑦 2 , 𝑦 3 )
 And if the Jacobian of transformation,
Dept of Systems Engineering, University of Lagos 45 oafak@unilag.edu.ng 12/27/2012

46. Jacobian Determinant
𝜕𝑦1 𝜕𝑦1 𝜕𝑦1
𝜕𝑥 1 𝜕𝑥 2 𝜕𝑥 3
𝜕𝑦 𝑖 𝜕 𝑦1 , 𝑦 2 , 𝑦 3 𝜕𝑦 2 𝜕𝑦 2 𝜕𝑦 2
𝑗
≡ 1, 𝑥2, 𝑥3 ≡
𝜕𝑥 𝜕 𝑥 𝜕𝑥 1 𝜕𝑥 2 𝜕𝑥 3
𝜕𝑦 3 𝜕𝑦 3 𝜕𝑦 3
𝜕𝑥 1 𝜕𝑥 2 𝜕𝑥 3
does not vanish, then the inverse transformation will exist.
So that,
𝑥 𝑖 = 𝑥 𝑖 (𝑦1 , 𝑦 2 , 𝑦 3 )
Dept of Systems Engineering, University of Lagos 46 oafak@unilag.edu.ng 12/27/2012

47. Given a position vector
𝐫 = 𝐫(𝑥 1 , 𝑥 2 , 𝑥 3 )
 In the new coordinate system, we can form a set of three
vectors,
𝜕𝐫
𝐠𝑖 = 𝑖
, 𝑖 = 1,2,3
𝜕𝑥
Which represent vectors along the tangents to the
coordinate lines. (This is easily established for the Cartesian
system and it is true in all systems. 𝐫 = 𝐫 𝑥 1 , 𝑥 2 , 𝑥 3 = 𝑦1 𝐢 +
𝑦2 𝐣 + 𝑦3 𝐤
So that
𝜕𝐫 𝜕𝐫 𝜕𝐫
𝐢 = 1 , 𝐣 = 2 and 𝐤 = 3
𝜕𝑦 𝜕𝑦 𝜕𝑦
The fact that this is true in the general case is easily seen
when we consider that along those lines, only the variable we
are differentiating with respect to varies. Consider the
general case where,
𝐫 = 𝐫 𝑥1 , 𝑥 2 , 𝑥 3
The total differential of 𝐫 is simply,
Dept of Systems Engineering, University of Lagos 47 oafak@unilag.edu.ng 12/27/2012

48. Natural Dual Bases
𝜕𝐫 1+
𝜕𝐫 2+
𝜕𝐫
𝑑𝐫 = 1 𝑑𝑥 𝑑𝑥 𝑑𝑥 3
𝜕𝑥 𝜕𝑥 2 𝜕𝑥 3
≡ 𝑑𝑥 1 𝐠1 + 𝑑𝑥 2 𝐠 2 + 𝑑𝑥 3 𝐠 3
With 𝐠 𝑖 𝑥 1 , 𝑥 2 , 𝑥 3 , 𝑖 = 1,2,3 now depending in general
on 𝑥 1 , 𝑥 2 and 𝑥 3 now forming a basis on which we can
describe other vectors in the coordinate system. We
have no guarantees that this vectors are unit in length
nor that they are orthogonal to one another. In the
Cartesian case, 𝐠 𝑖 , 𝑖 = 1,2,3 are constants, normalized
and orthogonal. They are our familiar
𝜕𝐫 𝜕𝐫 𝜕𝐫
𝐢 = 1 , 𝐣 = 2 and 𝐤 = 3 .
𝜕𝑦 𝜕𝑦 𝜕𝑦
Dept of Systems Engineering, University of Lagos 48 oafak@unilag.edu.ng 12/27/2012

49. We now proceed to show that the set of basis vectors,
𝐠 𝑖 ≡ 𝛻𝑥 𝑖
is reciprocal to 𝐠 𝑖 . The total differential 𝑑𝐫 can be written in terms
of partial differentials as,
𝜕𝐫
𝑑𝐫 = 𝑑𝑦 𝑖
𝜕𝑦 𝑖
Which when expressed in component form yields,
𝜕𝑥 1 1 𝜕𝑥 1 2 𝜕𝑥 1 3
𝑑𝑥 1 = 1 𝑑𝑦 + 2 𝑑𝑦 + 3 𝑑𝑦
𝜕𝑦 𝜕𝑦 𝜕𝑦
𝜕𝑥 2 1 𝜕𝑥 2 2 𝜕𝑥 2 3
𝑑𝑥 2 = 1 𝑑𝑦 + 2 𝑑𝑦 + 3 𝑑𝑦
𝜕𝑦 𝜕𝑦 𝜕𝑦
𝜕𝑥 3 1 𝜕𝑥 3 2 𝜕𝑥 3 3
𝑑𝑥 3 = 1 𝑑𝑦 + 2 𝑑𝑦 + 3 𝑑𝑦
𝜕𝑦 𝜕𝑦 𝜕𝑦
Or, more compactly that,
𝑗 =
𝜕𝑥 𝑗
𝑑𝑥 𝑑𝑦 𝑖 = 𝛻𝑥 𝑗 ⋅ 𝑑𝐫
𝜕𝑦 𝑖
Dept of Systems Engineering, University of Lagos 49 oafak@unilag.edu.ng 12/27/2012

50. In Cartesian coordinates, 𝑦 𝑖 , 𝑖 = 1, . . , 3 for any scalar field
𝜙 (𝑦1 , 𝑦 2 , 𝑦 3 ),
𝜕𝜙 𝑖 =
𝜕𝜙 𝜕𝜙 𝜕𝜙
𝑑𝑦 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧 = grad 𝜙 ⋅ 𝑑𝐫
𝜕𝑦 𝑖 𝜕𝑥 𝜕𝑦 𝜕𝑧
We are treating each curvilinear coordinate as a scalar field,
𝑥 𝑗 = 𝑥 𝑗 (𝑦1 , 𝑦 2 , 𝑦 3 ).
𝑗 = 𝛻𝑥 𝑗 ⋅
𝜕𝐫
𝑑𝑥 𝑑𝑥 𝑚
𝜕𝑥 𝑚
= 𝐠 𝑗 ⋅ 𝐠 𝑚 𝑑𝑥 𝑚
𝑗
= 𝛿 𝑚 𝑑𝑥 𝑚 .
The last equality arises from the fact that this is the only way
one component of the coordinate differential can equal
another.
𝑗
⇒ 𝐠𝑗 ⋅ 𝐠 𝑚 = 𝛿 𝑚
Which recovers for us the reciprocity relationship and shows
that 𝐠 𝑗 and 𝐠 𝑚 are reciprocal systems.
Dept of Systems Engineering, University of Lagos 50 oafak@unilag.edu.ng 12/27/2012

51. The position vector in Cartesian coordinates is 𝒓 = 𝒙 𝒊 𝒆 𝒊. Show
that (a) 𝐝𝐢𝐯 𝒓 = 𝟑, (b) 𝐝𝐢𝐯 𝒓 ⊗ 𝒓 = 𝟒𝒓, (c) 𝐝𝐢𝐯 𝒓 = 𝟑, and (d)
𝐠𝐫𝐚𝐝 𝒓 = 𝟏 and (e) 𝐜𝐮𝐫𝐥 𝒓 ⊗ 𝒓 = −𝒓 ×
grad 𝒓 = 𝑥 𝑖 , 𝒋 𝒆 𝑖 ⊗ 𝒆 𝒋
= 𝛿 𝑖𝑗 𝒆 𝑖 ⊗ 𝒆 𝒋 = 𝟏
div 𝒓 = 𝑥 𝑖 , 𝒋 𝒆 𝑖 ⋅ 𝒆 𝒋
= 𝛿 𝑖𝑗 𝛿 𝑖𝑗 = 𝛿 𝑗𝑗 = 3.
𝒓 ⊗ 𝒓 = 𝑥 𝑖 𝒆 𝑖 ⊗ 𝑥𝑗 𝒆𝑗 = 𝑥 𝑖 𝑥𝑗 𝒆 𝑖 ⊗ 𝒆 𝒋
grad 𝒓 ⊗ 𝒓 = 𝑥 𝑖 𝑥 𝑗 , 𝑘 𝒆 𝑖 ⊗ 𝒆 𝒋 ⊗ 𝒆 𝒌
div 𝒓 ⊗ 𝒓 = 𝑥 𝑖 , 𝑘 𝑥 𝑗 + 𝑥 𝑖 𝑥 𝑗 , 𝑘 𝒆 𝑖 ⊗ 𝒆 𝒋 ⋅ 𝒆 𝒌
= 𝛿 𝑖𝑘 𝒙 𝒋 + 𝒙 𝒊 𝛿 𝑗𝑘 𝛿 𝑗𝑘 𝒆 𝑖 = 𝛿 𝑖𝑘 𝒙 𝒌 + 𝒙 𝒊 𝛿 𝑗𝑗 𝒆 𝑖
= 4𝑥 𝑖 𝒆 𝑖 = 4𝒓
curl 𝒓 ⊗ 𝒓 = 𝜖 𝛼𝛽𝛾 𝑥 𝑖 𝑥 𝛾 , 𝛽 𝒆 𝛼 ⊗ 𝒆 𝒊
= 𝜖 𝛼𝛽𝛾 𝑥 𝑖 , 𝛽 𝑥 𝛾 + 𝑥 𝑖 𝑥 𝛾 , 𝛽 𝒆 𝛼 ⊗ 𝒆 𝒊
= 𝜖 𝛼𝛽𝛾 𝛿 𝑖𝛽 𝑥 𝛾 + 𝑥 𝑖 𝛿 𝛾𝛽 𝒆 𝛼 ⊗ 𝒆 𝒊
= 𝜖 𝛼𝑖𝛾 𝑥 𝛾 𝒆 𝛼 ⊗ 𝒆 𝒊 + 𝜖 𝛼𝛽𝛽 𝑥 𝑖 𝒆 𝛼 ⊗ 𝒆 𝒊
= −𝜖 𝛼𝛾𝑖 𝑥 𝛾 𝒆 𝛼 ⊗ 𝒆 𝒊 = −𝒓 ×
Dept of Systems Engineering, University of Lagos 51 oafak@unilag.edu.ng 12/27/2012

52. For a scalar field 𝝓 and a tensor field 𝐓 show that 𝐠𝐫𝐚𝐝
𝝓𝐓 = 𝝓𝐠𝐫𝐚𝐝 𝐓 + 𝐓 ⊗ 𝐠𝐫𝐚𝐝𝝓. Also show that 𝐝𝐢𝐯
𝝓𝐓 = 𝝓 𝐝𝐢𝐯 𝐓 + 𝐓𝐠𝐫𝐚𝐝𝝓.
grad 𝜙𝐓 = 𝜙𝑇 𝑖𝑗 , 𝑘 𝐠 𝑖 ⊗ 𝐠 𝑗 ⊗ 𝐠 𝑘
= 𝜙, 𝑘 𝑇 𝑖𝑗 + 𝜙𝑇 𝑖𝑗 , 𝑘 𝐠 𝑖 ⊗ 𝐠 𝑗 ⊗ 𝐠 𝑘
= 𝐓 ⊗ grad𝜙 + 𝜙grad 𝐓
Furthermore, we can contract the last two bases and
obtain,
div 𝜙𝐓 = 𝜙, 𝑘 𝑇 𝑖𝑗 + 𝜙𝑇 𝑖𝑗 , 𝑘 𝐠 𝑖 ⊗ 𝐠 𝑗 ⋅ 𝐠 𝑘
= 𝜙, 𝑘 𝑇 𝑖𝑗 + 𝜙𝑇 𝑖𝑗 , 𝑘 𝐠 𝑖 𝛿 𝑗 𝑘
= 𝑇 𝑖𝑘 𝜙, 𝑘 𝐠 𝑖 + 𝜙𝑇 𝑖𝑘 , 𝑘 𝐠 𝑖
= 𝐓grad𝜙 + 𝜙 div 𝐓
Dept of Systems Engineering, University of Lagos 52 oafak@unilag.edu.ng 12/27/2012

53.  For two arbitrary vectors, 𝒖 and 𝒗, show that
𝐠𝐫𝐚𝐝 𝒖 ⊗ 𝒗 = 𝒖 × 𝐠𝐫𝐚𝐝𝒗 − 𝒗 × 𝐠𝐫𝐚𝐝𝒖
 grad 𝒖 ⊗ 𝒗 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘 , 𝑙 𝐠 𝑖 ⊗ 𝐠 𝑙
= 𝜖 𝑖𝑗𝑘 𝑢 𝑗 , 𝑙 𝑣 𝑘 + 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘 , 𝑙 𝐠 𝑖 ⊗ 𝐠 𝑙
= 𝑢 𝑗 , 𝑙 𝜖 𝑖𝑗𝑘 𝑣 𝑘 + 𝑣 𝑘 , 𝑙 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝐠 𝑖 ⊗ 𝐠 𝑙
= − 𝒗 × grad𝒖 + 𝒖 × grad𝒗
Dept of Systems Engineering, University of Lagos 53 oafak@unilag.edu.ng 12/27/2012

54. For any two tensor fields 𝐮 and 𝐯, Show that,
𝐮 × : grad 𝐯 = 𝐮 ⋅ curl 𝐯
𝐮 × : grad 𝐯 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝐠 𝑖 ⊗ 𝐠 𝑘 : 𝑣 𝛼 , 𝛽 𝐠 𝛼 ⊗ 𝐠 𝛽
= 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝛼 , 𝛽 𝐠 𝑖 ⋅ 𝐠 𝛼 𝐠 𝑘 ⋅ 𝐠 𝛽
𝛽
= 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝛼 , 𝛽 𝛿 𝑖 𝛼 𝛿 𝑘 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑖 , 𝑘
= 𝐮 ⋅ curl 𝐯
Dept of Systems Engineering, University of Lagos 54 oafak@unilag.edu.ng 12/27/2012

55. For a vector field 𝒖, show that 𝐠𝐫𝐚𝐝 𝒖 × is a third
ranked tensor. Hence or otherwise show that
𝐝𝐢𝐯 𝒖 × = −𝐜𝐮𝐫𝐥 𝒖.
The second–order tensor 𝒖 × is defined as 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝐠 𝑖 ⊗
𝐠 𝑘 . Taking the covariant derivative with an independent
base, we have
grad 𝒖 × = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 , 𝑙 𝐠 𝑖 ⊗ 𝐠 𝑘 ⊗ 𝐠 𝑙
This gives a third order tensor as we have seen.
Contracting on the last two bases,
div 𝒖 × = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 , 𝑙 𝐠 𝑖 ⊗ 𝐠 𝑘 ⋅ 𝐠 𝑙
= 𝜖 𝑖𝑗𝑘 𝑢 𝑗 , 𝑙 𝐠 𝑖 𝛿 𝑙𝑘
= 𝜖 𝑖𝑗𝑘 𝑢 𝑗 , 𝑘 𝐠 𝑖
= −curl 𝒖
Dept of Systems Engineering, University of Lagos 55 oafak@unilag.edu.ng 12/27/2012

56. Show that 𝐜𝐮𝐫𝐥 𝜙𝟏 = grad 𝜙 ×
Note that 𝜙𝟏 = 𝜙𝑔 𝛼𝛽 𝐠 𝛼 ⊗ 𝐠 𝛽 , and that curl 𝑻 =
𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼 so that,
curl 𝜙𝟏 = 𝜖 𝑖𝑗𝑘 𝜙𝑔 𝛼𝑘 , 𝒋 𝐠 𝑖 ⊗ 𝐠 𝛼
= 𝜖 𝑖𝑗𝑘 𝜙, 𝒋 𝑔 𝛼𝑘 𝐠 𝑖 ⊗ 𝐠 𝛼 = 𝜖 𝑖𝑗𝑘 𝜙, 𝒋 𝐠 𝑖 ⊗ 𝐠 𝑘
= grad 𝜙 ×
Show that curl 𝒗 × = div 𝒗 𝟏 − grad 𝒗
𝒗 × = 𝜖 𝛼𝛽𝑘 𝑣 𝛽 𝐠 𝛼 ⊗ 𝐠 𝑘
curl 𝑻 = 𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
so that
curl 𝒗 × = 𝜖 𝑖𝑗𝑘 𝜖 𝛼𝛽𝑘 𝑣 𝛽 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
= 𝑔 𝑖𝛼 𝑔 𝑗𝛽 − 𝑔 𝑖𝛽 𝑔 𝑗𝛼 𝑣 𝛽 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
= 𝑣 𝑗, 𝑗 𝐠 𝛼 ⊗ 𝐠 𝛼 − 𝑣 𝑖, 𝑗 𝐠 𝑖 ⊗ 𝐠 𝑗
= div 𝒗 𝟏 − grad 𝒗
Dept of Systems Engineering, University of Lagos 56 oafak@unilag.edu.ng 12/27/2012

57. Show that 𝐝𝐢𝐯 𝒖 × 𝒗 = 𝒗 ⋅ 𝐜𝐮𝐫𝐥 𝒖 − 𝒖 ⋅ 𝐜𝐮𝐫𝐥 𝒗
div 𝒖 × 𝒗 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘 , 𝑖
Noting that the tensor 𝜖 𝑖𝑗𝑘 behaves as a constant under
a covariant differentiation, we can write,
div 𝒖 × 𝒗 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘 , 𝑖
= 𝜖 𝑖𝑗𝑘 𝑢 𝑗 , 𝑖 𝑣 𝑘 + 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘 , 𝑖
= 𝒗 ⋅ curl 𝒖 − 𝒖 ⋅ curl 𝒗
Given a scalar point function ϕ and a vector field 𝐯,
show that 𝐜𝐮𝐫𝐥 𝜙𝐯 = 𝜙 curl 𝐯 + grad𝜙 × 𝐯.
curl 𝜙𝐯 = 𝜖 𝑖𝑗𝑘 𝜙𝑣 𝑘 , 𝑗 𝐠 𝑖
= 𝜖 𝑖𝑗𝑘 𝜙, 𝑗 𝑣 𝑘 + 𝜙𝑣 𝑘 , 𝑗 𝐠 𝑖
= 𝜖 𝑖𝑗𝑘 𝜙, 𝑗 𝑣 𝑘 𝐠 𝑖 + 𝜖 𝑖𝑗𝑘 𝜙𝑣 𝑘 , 𝑗 𝐠 𝑖
= grad𝜙 × 𝐯 + 𝜙 curl 𝐯
Dept of Systems Engineering, University of Lagos 57 oafak@unilag.edu.ng 12/27/2012

58. 𝐀
Given that 𝜑 𝑡 = 𝐀(𝑡) , Show that 𝜑 𝑡 = : 𝐀
𝐀 𝑡
𝜑2 ≡ 𝐀: 𝐀
Now,
𝑑 2 = 2𝜑
𝑑𝜑 𝑑𝐀 𝑑𝐀 𝑑𝐀
𝜑 = : 𝐀 + 𝐀: = 2𝐀:
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡
as inner product is commutative. We can therefore
write that
𝑑𝜑 𝐀 𝑑𝐀 𝐀
= : = : 𝐀
𝑑𝑡 𝜑 𝑑𝑡 𝐀 𝑡
as required.
Dept of Systems Engineering, University of Lagos 58 oafak@unilag.edu.ng 12/27/2012

59. Given a tensor field 𝑻, obtain the vector 𝒘 ≡ 𝑻 𝐓 𝐯 and
show that its divergence is 𝑻: 𝛁𝐯 + 𝐯 ⋅ 𝐝𝐢𝐯 𝑻
The divergence of 𝒘 is the scalar sum , 𝑇𝑗𝑖 𝑣 𝑗 , 𝑖 .
Expanding the product covariant derivative we obtain,
div 𝑻T 𝐯 = 𝑇𝑗𝑖 𝑣 𝑗 , 𝑖 = 𝑇𝑗𝑖 , 𝑖 𝑣 𝑗 + 𝑇𝑗𝑖 𝑣 𝑗 , 𝑖
= div 𝑻 ⋅ 𝐯 + tr 𝑻T grad𝐯
= div 𝑻 ⋅ 𝐯 + 𝑻: grad 𝐯
Recall that scalar product of two vectors is
commutative so that
div 𝑻T 𝐯 = 𝑻: grad𝐯 + 𝐯 ⋅ div 𝑻
Dept of Systems Engineering, University of Lagos 59 oafak@unilag.edu.ng 12/27/2012

60. For a second-order tensor 𝑻 define
curl 𝑻 ≡ 𝜖 𝑖𝑗𝑘 𝑻 𝛼𝑘 , 𝑗 𝐠 𝒊 ⊗ 𝐠 𝜶 show that for any constant
vector 𝒂, curl 𝑻 𝒂 = curl 𝑻 𝑻 𝒂
Express vector 𝒂 in the invariant form with
contravariant components as 𝒂 = 𝑎 𝛽 𝐠 𝛽 . It follows that
curl 𝑻 𝒂 = 𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼 𝒂
= 𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗 𝑎 𝛽 𝐠 𝑖 ⊗ 𝐠 𝛼 𝐠 𝛽
= 𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗 𝑎 𝛽 𝐠 𝑖 𝛿 𝛽𝛼
= 𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗 𝐠 𝑖 𝑎 𝛼
= 𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 𝑎 𝛼 , 𝑗 𝐠 𝑖
The last equality resulting from the fact that vector 𝒂 is
a constant vector. Clearly,
curl 𝑻 𝒂 = curl 𝑻 𝑇 𝒂
Dept of Systems Engineering, University of Lagos 60 oafak@unilag.edu.ng 12/27/2012

61. For any two vectors 𝐮 and 𝐯, show that curl 𝐮 ⊗ 𝐯 =
grad 𝐮 𝐯 × 𝑻 + curl 𝐯 ⊗ 𝒖 where 𝐯 × is the skew tensor
𝜖 𝑖𝑘𝑗 𝑣 𝑘 𝐠 𝒊 ⊗ 𝐠 𝒋 .
Recall that the curl of a tensor 𝑻 is defined by curl 𝑻 ≡
𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼 . Clearly therefore,
curl 𝒖 ⊗ 𝒗 = 𝜖 𝑖𝑗𝑘 𝑢 𝛼 𝑣 𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
= 𝜖 𝑖𝑗𝑘 𝑢 𝛼 , 𝑗 𝑣 𝑘 + 𝑢 𝛼 𝑣 𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
= 𝜖 𝑖𝑗𝑘 𝑢 𝛼 , 𝑗 𝑣 𝑘 𝐠 𝑖 ⊗ 𝐠 𝛼 + 𝜖 𝑖𝑗𝑘 𝑢 𝛼 𝑣 𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
= 𝜖 𝑖𝑗𝑘 𝑣 𝑘 𝐠 𝑖 ⊗ 𝑢 𝛼 , 𝑗 𝐠 𝛼 + 𝜖 𝑖𝑗𝑘 𝑣 𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝑢 𝛼 𝐠 𝛼
= 𝜖 𝑖𝑗𝑘 𝑣 𝑘 𝐠 𝑖 ⊗ 𝐠 𝑗 𝑢 𝛼 , 𝛽 𝐠 𝛽 ⊗ 𝐠 𝛼 + 𝜖 𝑖𝑗𝑘 𝑣 𝑘 , 𝑗 𝐠 𝑖
⊗ 𝑢 𝛼 𝐠 𝛼 = − 𝒗 × grad 𝒖 𝑻 + curl 𝒗 ⊗ 𝒖
= grad 𝒖 𝒗 × 𝑻 + curl 𝒗 ⊗ 𝒖
upon noting that the vector cross is a skew tensor.
Dept of Systems Engineering, University of Lagos 61 oafak@unilag.edu.ng 12/27/2012

62. Show that curl 𝒖 × 𝒗 = div 𝒖 ⊗ 𝒗 − 𝒗 ⊗ 𝒖
The vector 𝒘 ≡ 𝒖 × 𝒗 = 𝑤 𝒌 𝐠 𝒌 = 𝜖 𝑘𝛼𝛽 𝑢 𝛼 𝑣 𝛽 𝐠 𝒌 and
curl 𝒘 = 𝜖 𝑖𝑗𝑘 𝑤 𝑘 , 𝑗 𝐠 𝑖 . Therefore,
curl 𝒖 × 𝒗 = 𝜖 𝑖𝑗𝑘 𝑤 𝑘 , 𝑗 𝐠 𝑖
= 𝜖 𝑖𝑗𝑘 𝜖 𝑘𝛼𝛽 𝑢 𝛼 𝑣 𝛽 , 𝑗 𝐠 𝑖
𝑗 𝑖 𝑗
= 𝛿 𝑖𝛼 𝛿 𝛽 − 𝛿 𝛽 𝛿 𝛼 𝑢𝛼𝑣 𝛽 ,𝑗 𝐠𝑖
𝑗 𝑖 𝑗
= 𝛿 𝑖𝛼 𝛿 𝛽 − 𝛿 𝛽 𝛿 𝛼 𝑢 𝛼, 𝑗 𝑣 𝛽 + 𝑢 𝛼 𝑣 𝛽, 𝑗 𝐠 𝑖
= 𝑢 𝑖, 𝑗 𝑣 𝑗 + 𝑢 𝑖 𝑣 𝑗, 𝑗 − 𝑢 𝑗, 𝑗 𝑣 𝑖 + 𝑢 𝑗 𝑣 𝑖, 𝑗 𝐠𝑖
= 𝑢𝑖 𝑣 𝑗 ,𝑗 − 𝑢 𝑗 𝑣𝑖 ,𝑗 𝐠𝑖
= div 𝒖 ⊗ 𝒗 − 𝒗 ⊗ 𝒖
since div 𝒖 ⊗ 𝒗 = 𝑢𝑖 𝑣 𝑗 , 𝛼 𝐠𝑖 ⊗ 𝐠 𝑗 ⋅ 𝐠 𝛼 = 𝑢 𝑖 𝑣 𝑗 , 𝑗 𝐠 𝑖.
Dept of Systems Engineering, University of Lagos 62 oafak@unilag.edu.ng 12/27/2012

63. Given a scalar point function 𝜙 and a second-order tensor field 𝑻,
show that 𝐜𝐮𝐫𝐥 𝜙𝑻 = 𝜙 𝐜𝐮𝐫𝐥 𝑻 + grad𝜙 × 𝑻 𝑻 where
grad𝜙 × is the skew tensor 𝜖 𝑖𝑗𝑘 𝜙, 𝑗 𝐠 𝒊 ⊗ 𝐠 𝒌
curl 𝜙𝑻 ≡ 𝜖 𝑖𝑗𝑘 𝜙𝑇 𝛼𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
= 𝜖 𝑖𝑗𝑘 𝜙, 𝑗 𝑇 𝛼𝑘 + 𝜙𝑇 𝛼𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
= 𝜖 𝑖𝑗𝑘 𝜙, 𝑗 𝑇 𝛼𝑘 𝐠 𝑖 ⊗ 𝐠 𝛼 + 𝜙𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
= 𝜖 𝑖𝑗𝑘 𝜙, 𝑗 𝐠 𝑖 ⊗ 𝐠 𝑘 𝑇 𝛼𝛽 𝐠 𝛽 ⊗ 𝐠 𝛼 + 𝜙𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
= 𝜙 curl 𝑻 + grad𝜙 × 𝑻 𝑇
Dept of Systems Engineering, University of Lagos 63 oafak@unilag.edu.ng 12/27/2012

64. For a second-order tensor field 𝑻, show that
div curl 𝑻 = curl div 𝑻 𝐓
Define the second order tensor 𝑆 as
curl 𝑻 ≡ 𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼 = 𝑆.𝛼 𝐠 𝑖 ⊗ 𝐠 𝛼
𝑖
The gradient of 𝑺 is
𝑆.𝛼 , 𝛽 𝐠 𝑖 ⊗ 𝐠 𝛼 ⊗ 𝐠 𝛽 = 𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗𝛽 𝐠 𝑖 ⊗ 𝐠 𝛼 ⊗ 𝐠 𝛽
𝑖
Clearly,
div curl 𝑻 = 𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗𝛽 𝐠 𝑖 ⊗ 𝐠 𝛼 ⋅ 𝐠 𝛽
= 𝜖 𝑖𝑗𝑘 𝑇 𝛼𝑘 , 𝑗𝛽 𝐠 𝑖 𝑔 𝛼𝛽
= 𝜖 𝑖𝑗𝑘 𝑇 𝛽 𝑘 , 𝑗𝛽 𝐠 𝑖 = curl div 𝑻T
Dept of Systems Engineering, University of Lagos 64 oafak@unilag.edu.ng 12/27/2012

65. If the volume 𝑽 is enclosed by the surface 𝑺, the position vector
𝒓 = 𝒙 𝒊 𝐠 𝒊 and 𝒏 is the external unit normal to each surface
𝟏
element, show that 𝑺 𝛁 𝒓 ⋅ 𝒓 ⋅ 𝒏𝒅𝑺 equals the volume
𝟔
contained in 𝑽.
𝒓 ⋅ 𝒓 = 𝑥 𝑖 𝑥 𝑗 𝐠 𝑖 ⋅ 𝐠 𝑗 = 𝑥 𝑖 𝑥 𝑗 𝑔 𝑖𝑗
By the Divergence Theorem,
𝛻 𝒓 ⋅ 𝒓 ⋅ 𝒏𝑑𝑆 = 𝛻⋅ 𝛻 𝒓⋅ 𝒓 𝑑𝑉
𝑆 𝑉
= 𝜕 𝑙 𝜕 𝑘 𝑥 𝑖 𝑥 𝑗 𝑔 𝑖𝑗 𝐠 𝑙 ⋅ 𝐠 𝑘 𝑑𝑉
𝑉
= 𝜕 𝑙 𝑔 𝑖𝑗 𝑥 𝑖 , 𝑘 𝑥 𝑗 + 𝑥 𝑖 𝑥 𝑗 , 𝑘 𝐠 𝑙 ⋅ 𝐠 𝑘 𝑑𝑉
𝑉
𝑗
= 𝑔 𝑖𝑗 𝑔 𝑙𝑘 𝛿 𝑖𝑘 𝑥 𝑗 + 𝑥 𝑖 𝛿 𝑘 , 𝑙 𝑑𝑉 = 2𝑔 𝑖𝑘 𝑔 𝑙𝑘 𝑥 𝑖 , 𝑙 𝑑𝑉
𝑉 𝑉
= 2𝛿 𝑖𝑙 𝛿 𝑙𝑖 𝑑𝑉 = 6 𝑑𝑉
𝑉 𝑉
Dept of Systems Engineering, University of Lagos 65 oafak@unilag.edu.ng 12/27/2012

66. For any Euclidean coordinate system, show that div 𝐮 × 𝐯 =
𝐯 curl 𝐮 − 𝐮 curl 𝐯
Given the contravariant vector 𝑢 𝑖 and 𝑣 𝑖 with their associated
vectors 𝑢 𝑖 and 𝑣 𝑖 , the contravariant component of the above cross
product is 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘 .The required divergence is simply the
contraction of the covariant 𝑥 𝑖 derivative of this quantity:
𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘 = 𝜖 𝑖𝑗𝑘 𝑢 𝑗,𝑖 𝑣 𝑘 + 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘,𝑖
,𝑖
where we have treated the tensor 𝜖 𝑖𝑗𝑘 as a constant under the
covariant derivative.
Cyclically rearranging the RHS we obtain,
𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘 = 𝑣 𝑘 𝜖 𝑘𝑖𝑗 𝑢 𝑗,𝑖 + 𝑢 𝑗 𝜖 𝑗𝑘𝑖 𝑣 𝑘,𝑖 = 𝑣 𝑘 𝜖 𝑘𝑖𝑗 𝑢 𝑗,𝑖 + 𝑢 𝑗 𝜖 𝑗𝑖𝑘 𝑣 𝑘,𝑖
,𝑖
where we have used the anti-symmetric property of the tensor
𝜖 𝑖𝑗𝑘 . The last expression shows clearly that
div 𝐮 × 𝐯 = 𝐯 curl 𝐮 − 𝐮 curl 𝐯
as required.
Dept of Systems Engineering, University of Lagos 66 oafak@unilag.edu.ng 12/27/2012

67. Liouville Formula
𝑑𝐓 𝑑
For a scalar variable 𝛼, if the tensor 𝐓 = 𝐓(𝛼) and 𝐓 ≡ 𝑑𝛼, Show that 𝑑𝛼 det 𝐓 =
det 𝐓 tr(𝐓 𝐓 −𝟏 )
Proof:
Let 𝐀 ≡ 𝐓 𝐓 −1 so that, 𝐓 = 𝐀𝐓. In component form, we have 𝑇𝑗 𝑖 = 𝐴 𝑖 𝑚 𝑇𝑗 𝑚 .
Therefore,
𝑑 𝑑
det 𝐓 = 𝜖 𝑖𝑗𝑘 𝑇𝑖1 𝑇𝑗2 𝑇 3 = 𝜖 𝑖𝑗𝑘 𝑇𝑖1 𝑇𝑗2 𝑇 3 + 𝑇𝑖1 𝑇𝑗2 𝑇 3 + 𝑇𝑖1 𝑇𝑗2 𝑇 3
𝑘 𝑘 𝑘 𝑘
𝑑𝛼 𝑑𝛼
= 𝜖 𝑖𝑗𝑘 𝐴1 𝑇𝑖 𝑙 𝑇𝑗2 𝑇 3 + 𝑇𝑖1 𝐴2𝑚 𝑇𝑗 𝑚 𝑇 3 + 𝑇𝑖1 𝑇𝑗2 𝐴3𝑛 𝑇 𝑘𝑛
𝑙 𝑘 𝑘
= 𝜖 𝑖𝑗𝑘 𝐴1 𝑇𝑖1 + 𝐴1 𝑇𝑖2 + 𝐴1 𝑇𝑖3
1 2 3 𝑇𝑗2 𝑇 3
𝑘
+ 𝑇𝑖1 𝐴1 𝑇𝑗1 + 𝐴2 𝑇𝑗2 + 𝐴2 𝑇𝑗3
2
2 3 𝑇 3 + 𝑇𝑖1 𝑇𝑗2
𝑘 𝐴1 𝑇 1 + 𝐴3 𝑇 2 + 𝐴3 𝑇 3
3
𝑘 2 𝑘 3 𝑘
 All the boxed terms in the above equation vanish on account of the contraction
of a symmetric tensor with an antisymmetric one.
Dept of Systems Engineering, University of Lagos 67 oafak@unilag.edu.ng 12/27/2012

68. Liouville Theorem Cont’d
(For example, the first boxed term yields, 𝜖 𝑖𝑗𝑘 𝐴1 𝑇𝑖2 𝑇𝑗2 𝑇 3
2 𝑘
Which is symmetric as well as antisymmetric in 𝑖 and 𝑗. It
therefore vanishes. The same is true for all other such terms.)
𝑑
det 𝐓 = 𝜖 𝑖𝑗𝑘 𝐴1 𝑇𝑖1 𝑇𝑗2 𝑇 3 + 𝑇𝑖1 𝐴2 𝑇𝑗2 𝑇 3 + 𝑇𝑖1 𝑇𝑗2 𝐴3 𝑇 3
1 𝑘 2 𝑘 3 𝑘
𝑑𝛼
= 𝐴 𝑚 𝜖 𝑖𝑗𝑘 𝑇𝑖1 𝑇𝑗2 𝑇 3
𝑚 𝑘
= tr 𝐓 𝐓 −1 det 𝐓
as required.
Dept of Systems Engineering, University of Lagos 68 oafak@unilag.edu.ng 12/27/2012

69. For a general tensor field 𝑻 show that,
𝐜𝐮𝐫𝐥 𝐜𝐮𝐫𝐥 𝑻 =
𝟐 𝐭𝐫 𝑻 − 𝐝𝐢𝐯 𝐝𝐢𝐯 𝑻 𝑰 + 𝐠𝐫𝐚𝐝 𝐝𝐢𝐯 𝑻 + 𝐓
𝛁 𝐠𝐫𝐚𝐝 𝐝𝐢𝐯 𝑻 −
𝐠𝐫𝐚𝐝 𝐠𝐫𝐚𝐝 𝐭𝐫𝑻 − 𝛁 𝟐 𝑻 𝐓
curl 𝑻 = 𝝐 𝛼𝑠𝑡 𝑇 𝛽𝑡 , 𝑠 𝐠 𝛼 ⊗ 𝐠 𝛽
𝛼
= 𝑆 .𝛽 𝐠 𝛼 ⊗ 𝐠 𝛽
𝛼
curl 𝑺 = 𝜖 𝑖𝑗𝑘 𝑆 .𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
so that
curl 𝑺 = curl curl 𝑻 = 𝜖 𝑖𝑗𝑘 𝜖 𝛼𝑠𝑡 𝑇 𝑘𝑡 , 𝑠𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
𝑔 𝑖𝛼 𝑔 𝑖𝑠 𝑔 𝑖𝑡
= 𝑔 𝑗𝛼 𝑔 𝑗𝑠 𝑔 𝑗𝑡 𝑇 𝑘𝑡 , 𝑠𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
𝑔 𝑘𝛼 𝑔 𝑘𝑠 𝑔 𝑘𝑡
𝑔 𝑖𝛼 𝑔 𝑗𝑠 𝑔 𝑘𝑡 − 𝑔 𝑗𝑡 𝑔 𝑘𝑠 + 𝑔 𝑖𝑠 𝑔 𝑗𝑡 𝑔 𝑘𝛼 − 𝑔 𝑗𝛼 𝑔 𝑘𝑡
= 𝑖𝑡 𝑗𝛼 𝑘𝑠 𝑗𝑠 𝑘𝛼
𝑇 𝑘𝑡 , 𝑠𝑗 𝐠 𝑖 ⊗ 𝐠 𝛼
+𝑔 𝑔 𝑔 − 𝑔 𝑔
𝑡 𝑡
= 𝑔 𝑗𝑠 𝑇 .𝑡 , 𝑠𝑗 −𝑇..𝑠𝑗 , 𝑠𝑗 𝐠 𝛼 ⊗ 𝐠 𝛼 + 𝑇 ..𝛼𝑗 , 𝑠𝑗 −𝑔 𝑗𝛼 𝑇 .𝑡 , 𝑠𝑗 𝐠 𝑠 ⊗ 𝐠 𝛼
𝑠. 𝛼.
+ 𝑔 𝑗𝛼 𝑇 .𝑡 , 𝑠𝑗 −𝑔 𝑗𝑠 𝑇 .𝑡 , 𝑠𝑗 𝐠 𝑡 ⊗ 𝐠 𝛼
T
= 𝛻 2 tr 𝑻 − div div 𝑻 𝑰 + grad div 𝑻 − grad grad tr𝑻
+ grad div 𝑻 − 𝛻 2 𝑻T
Dept of Systems Engineering, University of Lagos 69 oafak@unilag.edu.ng 12/27/2012

70. When 𝑻 is symmetric, show that tr(curl 𝑻) vanishes.
curl 𝑻 = 𝝐 𝑖𝑗𝑘 𝑇 𝛽𝑘 , 𝑗 𝐠 𝑖 ⊗ 𝐠 𝛽
tr curl 𝑻 = 𝜖 𝑖𝑗𝑘 𝑇 𝛽𝑘 , 𝑗 𝐠 𝑖 ⋅ 𝐠 𝛽
𝛽
= 𝜖 𝑖𝑗𝑘 𝑇 𝛽𝑘 , 𝑗 𝛿 𝑖 = 𝜖 𝑖𝑗𝑘 𝑇𝑖𝑘 , 𝑗
which obviously vanishes on account of the symmetry
and antisymmetry in 𝑖 and 𝑘. In this case,
curl curl 𝑻
= 𝛻 2 tr 𝑻 − div div 𝑻 𝑰 − grad grad tr𝑻
+ 2 grad div 𝑻 − 𝛻 2 𝑻
T
as grad div 𝑻 = grad div 𝑻 if the order of
differentiation is immaterial and T is symmetric.
Dept of Systems Engineering, University of Lagos 70 oafak@unilag.edu.ng 12/27/2012

71. For a scalar function Φ and a vector 𝑣 𝑖 show that the
divergence of the vector 𝑣 𝑖 Φ is equal to, 𝐯 ⋅ gradΦ +
Φ div 𝐯
𝑣 𝑖 Φ ,𝑖 = Φ𝑣 𝑖 ,𝑖 + 𝑣 𝑖 Φ,i
Hence the result.
Dept of Systems Engineering, University of Lagos 71 oafak@unilag.edu.ng 12/27/2012

72. Show that 𝒄𝒖𝒓𝒍 𝐮 × 𝐯 = 𝐯 ∙ 𝛁𝐮 + 𝐮 ⋅ div 𝐯 − 𝐯 ⋅ div 𝐮 −
𝐮∙ 𝛁 𝐯
Taking the associated (covariant) vector of the expression for
the cross product in the last example, it is straightforward to
see that the LHS in indicial notation is,
𝜖 𝑙𝑚𝑖 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘
,𝑚
Expanding in the usual way, noting the relation between the
alternating tensors and the Kronecker deltas,
𝑙𝑚𝑖
𝜖 𝑙𝑚𝑖 𝜖 𝑖𝑗𝑘 𝑢 𝑗 𝑣 𝑘 = 𝛿 𝑗𝑘𝑖 𝑢 𝑗 ,𝑚 𝑣 𝑘 − 𝑢 𝑗 𝑣 𝑘 ,𝑚
,𝑚
𝑙𝑚 𝛿 𝑗𝑙 𝛿𝑗 𝑚
= 𝛿 𝑗𝑘 𝑢 𝑗 ,𝑚 𝑣 𝑘 − 𝑢 𝑗 𝑣 𝑘 ,𝑚 = 𝑢 𝑗 ,𝑚 𝑣 𝑘 − 𝑢 𝑗 𝑣 𝑘 ,𝑚
𝛿 𝑙𝑘 𝛿 𝑘𝑚
= 𝛿 𝑗𝑙 𝛿 𝑘𝑚 − 𝛿 𝑙𝑘 𝛿 𝑗 𝑚 𝑢 𝑗 ,𝑚 𝑣 𝑘 − 𝑢 𝑗 𝑣 𝑘 ,𝑚
= 𝛿 𝑗𝑙 𝛿 𝑘𝑚 𝑢 𝑗 ,𝑚 𝑣 𝑘 − 𝛿 𝑗𝑙 𝛿 𝑘𝑚 𝑢 𝑗 𝑣 𝑘 ,𝑚 + 𝛿 𝑙𝑘 𝛿 𝑗 𝑚 𝑢 𝑗 ,𝑚 𝑣 𝑘
− 𝛿 𝑙𝑘 𝛿 𝑗 𝑚 𝑢 𝑗 𝑣 𝑘 ,𝑚
= 𝑢 𝑙 ,𝑚 𝑣 𝑚 − 𝑢 𝑚 ,𝑚 𝑣 𝑙 + 𝑢 𝑙 𝑣 𝑚 ,𝑚 − 𝑢 𝑚 𝑣 𝑙 ,𝑚
Which is the result we seek in indicial notation.
Dept of Systems Engineering, University of Lagos 72 oafak@unilag.edu.ng 12/27/2012

73. Differentiation of Fields
Given a vector point function 𝒖 𝑥 1 , 𝑥 2 , 𝑥 3 and its covariant
components, 𝑢 𝛼 𝑥 1 , 𝑥 2 , 𝑥 3 , 𝛼 = 1,2,3, with 𝐠 𝛼 as the reciprocal
basis vectors, Let
𝜕 𝜕𝑢 𝛼 𝛼 𝜕𝐠 𝛼
𝒖 = 𝑢 𝛼 𝐠 𝛼 , then 𝑑𝒖 = 𝑢 𝛼 𝐠 𝛼 𝑑𝑥 𝑘 = 𝐠 + 𝑢 𝛼 𝑑𝑥 𝑘
𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘
Clearly,
𝜕𝒖 𝜕𝑢 𝛼 𝛼 𝜕𝐠 𝛼
𝑘
= 𝑘
𝐠 + 𝑢
𝑘 𝛼
𝜕𝑥 𝜕𝑥 𝜕𝑥
And the projection of this quantity on the 𝐠 𝑖 direction is,
𝜕𝒖 𝜕𝑢 𝛼 𝛼 𝜕𝐠 𝛼 𝜕𝑢 𝛼 𝛼 𝜕𝐠 𝛼
∙ 𝐠 = 𝐠 + 𝑢 ∙ 𝐠𝑖 = 𝐠 ∙ 𝐠𝑖 + ∙ 𝐠 𝑢
𝜕𝑥 𝑘 𝑖 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝛼 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝑖 𝛼
𝜕𝑢 𝛼 𝛼 𝜕𝐠 𝛼
= 𝛿 +
𝑘 𝑖 𝑘
∙ 𝐠𝑖 𝑢 𝛼
𝜕𝑥 𝜕𝑥
Dept of Systems Engineering, University of Lagos 73 oafak@unilag.edu.ng 12/27/2012

74. Christoffel Symbols
𝑗
 Now, 𝐠 𝑖 ∙ 𝐠 𝑗 = 𝛿 𝑖 so that
𝜕𝐠 𝑖 𝑖∙
𝜕𝐠 𝑗
𝑘
∙ 𝐠𝑗 + 𝐠 𝑘
= 0.
𝜕𝑥 𝜕𝑥
𝜕𝐠 𝑖 𝑖
𝜕𝐠 𝑗 𝑖
∙ 𝐠 𝑗 = −𝐠 ∙ ≡− .
𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝑗𝑘
This important quantity, necessary to quantify the
derivative of a tensor in general coordinates, is called
the Christoffel Symbol of the second kind.
Dept of Systems Engineering, University of Lagos 74 oafak@unilag.edu.ng 12/27/2012

75. Derivatives in General Coordinates
Using this, we can now write that,
𝜕𝒖 𝜕𝑢 𝛼 𝛼 𝜕𝐠 𝛼 𝜕𝑢 𝑖 𝛼
∙ 𝐠𝑖 = 𝛿𝑖 + ∙ 𝐠𝑖 𝑢 𝛼 = − 𝑢
𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝑖𝑘 𝛼
 The quantity on the RHS is the component of the derivative of vector 𝒖
along the 𝐠 𝑖 direction using covariant components. It is the covariant
derivative of 𝒖. Using contravariant components, we could write,
𝜕𝑢 𝑖 𝜕𝐠 𝑖 𝑖 𝜕𝑢 𝑖 𝛼
𝑑𝒖 = 𝐠𝑖 + 𝑢 𝑑𝑥 𝑘 = 𝐠𝑖 + 𝐠 𝑢 𝑖 𝑑𝑥 𝑘
𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝑖𝑘 𝛼
 So that,
𝜕𝒖 𝜕𝑢 𝛼 𝜕𝐠 𝛼 𝛼
= 𝐠𝛼+ 𝑢
𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘
 The components of this in the direction of 𝐠 𝑖 can be obtained by taking a
dot product as before:
𝜕𝒖 𝑖 =
𝜕𝑢 𝛼 𝜕𝐠 𝛼 𝛼 𝑖 =
𝜕𝑢 𝛼 𝑖 𝜕𝐠 𝛼 𝑖 𝑢𝛼 =
𝜕𝑢 𝑖 𝑖
∙ 𝐠 𝐠𝛼+ 𝑢 ∙ 𝐠 𝛿𝛼 + ∙ 𝐠 + 𝑢𝛼
𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝛼𝑘
Dept of Systems Engineering, University of Lagos 75 oafak@unilag.edu.ng 12/27/2012

76. Covariant Derivatives
The two results above are represented symbolically as,
𝜕𝑢 𝑖 𝛼 𝑖 𝜕𝑢 𝛼 𝑖
𝑢 𝑖,𝑘 = − 𝑢 𝛼 and 𝑢,𝑘 = + 𝑢𝛼
𝜕𝑥 𝑘 𝑖𝑘 𝜕𝑥 𝑘 𝛼𝑘
 Which are the components of the covariant derivatives in
terms of the covariant and contravariant components
respectively.
 It now becomes useful to establish the fact that our
definition of the Christoffel symbols here conforms to the
definition you find in the books using the transformation
rules to define the tensor quantities.
Dept of Systems Engineering, University of Lagos 76 oafak@unilag.edu.ng 12/27/2012

77. Christoffel Symbols
𝜕𝐫
 We observe that the derivative of the covariant basis, 𝐠 𝑖 = 𝑖 ,
𝜕𝑥
𝜕𝐠 𝑖 𝜕 2𝐫 𝜕 2𝐫 𝜕𝐠 𝑗
= = =
𝜕𝑥 𝑗 𝜕𝑥 𝑗 𝜕𝑥 𝑖 𝜕𝑥 𝑖 𝜕𝑥 𝑗 𝜕𝑥 𝑖
 Taking the dot product with 𝐠 𝑘 ,
𝜕𝐠 𝑖 1 𝜕𝐠 𝑗 𝜕𝐠 𝑖
𝑗
∙ 𝐠𝑘= 𝑖
∙ 𝐠𝑘+ 𝑗
∙ 𝐠𝑘
𝜕𝑥 2 𝜕𝑥 𝜕𝑥
1 𝜕 𝜕 𝜕𝐠 𝑘 𝜕𝐠 𝑘
= 𝐠 ∙ 𝐠𝑘 + 𝐠 ∙ 𝐠𝑘 − 𝐠𝑗 ∙ − 𝐠𝑖 ∙
2 𝜕𝑥 𝑖 𝑗 𝜕𝑥 𝑗 𝑖 𝜕𝑥 𝑖 𝜕𝑥 𝑗
1 𝜕 𝜕 𝜕𝐠 𝑖 𝜕𝐠 𝑗
= 𝐠𝑗 ∙ 𝐠𝑘 + 𝐠𝑖 ∙ 𝐠 𝑘 − 𝐠𝑗 ∙ 𝑘
− 𝐠𝑖 ∙
2 𝜕𝑥 𝑖 𝜕𝑥 𝑗 𝜕𝑥 𝜕𝑥 𝑘
1 𝜕 𝜕 𝜕
= 𝐠 ∙ 𝐠𝑘 + 𝐠 ∙ 𝐠𝑘 − 𝐠 ∙ 𝐠𝑗
2 𝜕𝑥 𝑖 𝑗 𝜕𝑥 𝑗 𝑖 𝜕𝑥 𝑘 𝑖
1 𝜕𝑔 𝑗𝑘 𝜕𝑔 𝑘𝑖 𝜕𝑔 𝑖𝑗
= + −
Dept of Systems Engineering,
𝜕𝑥 𝑖
2 University of Lagos 𝜕𝑥 𝑗 𝜕𝑥 𝑘 77 oafak@unilag.edu.ng 12/27/2012

78. The First Kind
 Which is the quantity defined as the Christoffel
symbols of the first kind in the textbooks. It is
therefore possible for us to write,
𝜕𝐠 𝑖 𝜕𝐠 𝑗 1 𝜕𝑔 𝑗𝑘 𝜕𝑔 𝑘𝑖 𝜕𝑔 𝑖𝑗
𝑖𝑗, 𝑘 ≡ ∙ 𝐠𝑘 = ∙ 𝐠𝑘 = + −
𝜕𝑥 𝑗 𝜕𝑥 𝑖 2 𝜕𝑥 𝑖 𝜕𝑥 𝑗 𝜕𝑥 𝑘
It should be emphasized that the Christoffel symbols,
even though the play a critical role in several tensor
relationships, are themselves NOT tensor quantities.
(Prove this). However, notice their symmetry in the
𝑖 and 𝑗. The extension of this definition to the
Christoffel symbols of the second kind is immediate:
Dept of Systems Engineering, University of Lagos 78 oafak@unilag.edu.ng 12/27/2012

79. The Second Kind
 Contract the above equation with the conjugate metric
tensor, we have,
𝑘𝛼 𝑘𝛼
𝜕𝐠 𝑖 𝑘𝛼
𝜕𝐠 𝑗 𝜕𝐠 𝑖 𝑘
𝑔 𝑖𝑗, 𝛼 ≡ 𝑔 ∙ 𝐠𝛼= 𝑔 ∙ 𝐠𝛼= ∙ 𝐠𝑘=
𝜕𝑥 𝑗 𝜕𝑥 𝑖 𝜕𝑥 𝑗 𝑖𝑗
𝜕𝐠 𝑘
=− 𝑗
∙ 𝐠𝑖
𝜕𝑥
Which connects the common definition of the second
Christoffel symbol with the one defined in the above
derivation. The relationship,
𝑔 𝑘𝛼 𝑖𝑗, 𝛼 = 𝑘
𝑖𝑗
apart from defining the relationship between the Christoffel
symbols of the first kind and that second kind, also
highlights, once more, the index-raising property of the
conjugate metric tensor.
Dept of Systems Engineering, University of Lagos 79 oafak@unilag.edu.ng 12/27/2012

80. Two Christoffel Symbols Related
We contract the above equation with 𝑔 𝑘𝛽 and obtain,
𝑘𝛼 𝑖𝑗, 𝛼 = 𝑔 𝑘
𝑔 𝑘𝛽 𝑔 𝑘𝛽 𝑖𝑗
𝛼 𝑘
𝛿 𝛽 𝑖𝑗, 𝛼 = 𝑖𝑗, 𝛽 = 𝑔 𝑘𝛽
𝑖𝑗
so that,
𝛼
𝑔 𝑘𝛼 𝑖𝑗 = 𝑖𝑗, 𝑘
Showing that the metric tensor can be used to lower
the contravariant index of the Christoffel symbol of the
second kind to obtain the Christoffel symbol of the first
kind.
Dept of Systems Engineering, University of Lagos 80 oafak@unilag.edu.ng 12/27/2012

81. Higher Order Tensors
We are now in a position to express the derivatives of
higher order tensor fields in terms of the Christoffel
symbols.
For a second-order tensor 𝐓, we can express the
components in dyadic form along the product basis as
follows:
𝐓 = 𝑇𝑖𝑗 𝐠 𝑖 ⊗ 𝐠 𝑗 = 𝑇 𝑖𝑗 𝐠 𝑖 ⨂𝐠 𝑗 = 𝑇.𝑗 𝐠 𝑖 ⊗ 𝐠 𝑗 = 𝑇𝑗.𝑖 𝐠 𝑗 ⨂𝐠 𝑖
𝑖
This is perfectly analogous to our expanding vectors in
terms of basis and reciprocal bases. Derivatives of the
tensor may therefore be expressible in any of these
product bases. As an example, take the product
covariant bases.
Dept of Systems Engineering, University of Lagos 81 oafak@unilag.edu.ng 12/27/2012

82. Higher Order Tensors
We have:
𝜕𝐓 𝜕𝑇 𝑖𝑗 𝑖𝑗
𝜕𝐠 𝑖 𝑖𝑗
𝜕𝐠 𝑗
𝑘
= 𝑘 𝑖
𝐠 ⨂𝐠 𝑗 + 𝑇 𝑘
⨂𝐠 𝑗 + 𝑇 𝐠 𝑖 ⨂
𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝑘
𝜕𝐠 𝑗
Recall that, 𝑖𝑘 ∙ 𝐠 𝑗 = . It follows therefore that,
𝜕𝑥 𝑖𝑘
𝜕𝐠 𝑖 𝑗− 𝛼 𝛿𝑗 =
𝜕𝐠 𝑖 𝛼
∙ 𝐠 ∙ 𝐠𝑗 − 𝐠 𝛼 ∙ 𝐠𝑗
𝜕𝑥 𝑘 𝑖𝑘 𝛼 𝜕𝑥 𝑘 𝑖𝑘
𝜕𝐠 𝑖 𝛼
= − 𝐠 𝛼 ∙ 𝐠 𝑗 = 0.
𝜕𝑥 𝑘 𝑖𝑘
𝜕𝐠 𝑖 𝛼
Clearly, 𝑘 = 𝐠𝛼
𝜕𝑥 𝑖𝑘
(Obviously since 𝐠 𝑗 is a basis vector it cannot vanish)
Dept of Systems Engineering, University of Lagos 82 oafak@unilag.edu.ng 12/27/2012

83. Higher Order Tensors
𝜕𝐓 𝜕𝑇 𝑖𝑗 𝑖𝑗
𝜕𝐠 𝑖 𝑖𝑗 𝐠 ⨂
𝜕𝐠 𝑗
𝜕𝑥 𝑘 = 𝜕𝑥 𝑘 𝐠 𝑖 ⨂𝐠 𝑗 + 𝑇 𝜕𝑥 𝑘 ⨂𝐠 𝑗 + 𝑇 𝑖
𝜕𝑥 𝑘
𝜕𝑇 𝑖𝑗 𝑖𝑗 𝛼 𝑖𝑗 𝐠 ⨂
𝛼
= 𝐠 ⨂𝐠 𝑗 + 𝑇 𝐠 ⨂𝐠 𝑗 + 𝑇 𝑖 𝑗𝑘 𝐠 𝛼
𝜕𝑥 𝑘 𝑖 𝑖𝑘 𝛼
𝜕𝑇 𝑖𝑗 𝑖 𝑗
= 𝑘
𝐠 𝑖 ⨂𝐠 𝑗 + 𝑇 𝛼𝑗 𝐠 𝑖 ⨂𝐠 𝑗 + 𝑇 𝑖𝛼 𝐠 𝑖 ⨂ 𝐠𝑗
𝜕𝑥 𝛼𝑘 𝛼𝑘
𝜕𝑇 𝑖𝑗 𝑖 𝑗 𝑖𝑗
= +𝑇 𝛼𝑗 + 𝑇 𝑖𝛼 𝐠 𝑖 ⨂𝐠 𝑗 = 𝑇 ,𝑘 𝐠 𝑖 ⨂𝐠 𝑗
𝜕𝑥 𝑘 𝛼𝑘 𝛼𝑘
Where
𝜕𝑇 𝑖𝑗
𝑖𝑗 𝑖 𝑗 𝜕𝑇 𝑖𝑗 𝑖 𝑗
= 𝑇,𝑘 +𝑇 𝛼𝑗 + 𝑇 𝑖𝛼 or +𝑇 𝛼𝑗 Γ 𝛼𝑘 + 𝑇 𝑖𝛼 Γ 𝛼𝑘
𝜕𝑥 𝑘 𝛼𝑘 𝛼𝑘 𝜕𝑥 𝑘
are the components of the covariant derivative of the tensor 𝐓 in terms
of contravariant components on the product covariant bases as shown.
Dept of Systems Engineering, University of Lagos 83 oafak@unilag.edu.ng 12/27/2012

84. Higher Order Tensors
In the same way, by taking the tensor expression in the dyadic form of its
contravariant product bases, we can write,
𝜕𝐓 𝜕𝑇𝑖𝑗 𝑖 𝑗
𝜕𝐠 𝑖 𝑗 𝑖
𝜕𝐠 𝑗
𝜕𝑥 𝑘 = 𝜕𝑥 𝑘 𝐠 ⊗ 𝐠 +𝑇 𝑖𝑗 𝜕𝑥 𝑘 ⊗ 𝐠 +𝑇 𝑖𝑗 𝐠 ⊗ 𝜕𝑥 𝑘
𝜕𝑇𝑖𝑗 𝑖 𝑗 +𝑇 Γ 𝑖 ⊗ 𝐠 𝑗 +𝑇 𝐠 𝑖 ⊗
𝜕𝐠 𝑗
= 𝐠 ⊗ 𝐠 𝑖𝑗 𝛼𝑘 𝑖𝑗
𝜕𝑥 𝑘 𝜕𝑥 𝑘
𝑖 𝜕𝐠 𝑖
Again, notice from previous derivation above, = − 𝜕𝑥 𝑘 ∙ 𝐠 𝑗 so that,
𝑗𝑘
𝜕𝐠 𝑖 𝑖 𝑖
=− 𝐠 𝛼 = −Γ 𝛼𝑘 𝐠 𝛼 Therefore,
𝜕𝑥 𝑘
𝛼𝑘
𝜕𝐓 𝜕𝑇𝑖𝑗 𝑖 𝑗 +𝑇
𝜕𝐠 𝑖 𝑗 +𝑇 𝐠 𝑖 ⊗
𝜕𝐠 𝑗
= 𝐠 ⊗ 𝐠 𝑖𝑗 ⊗ 𝐠 𝑖𝑗
𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘 𝜕𝑥 𝑘
𝜕𝑇𝑖𝑗 𝑖 𝑖 𝑗
= 𝐠 ⊗ 𝐠 𝑗 −𝑇𝑖𝑗 Γ 𝛼𝑘 𝐠 𝛼 ⊗ 𝐠 𝑗 + 𝑇𝑖𝑗 𝐠 𝑖 ⊗ Γ 𝛼𝑘 𝐠 𝛼
𝜕𝑥 𝑘
𝜕𝑇𝑖𝑗 𝛼 𝛼 𝑖 𝑗 𝑖 𝑗
= 𝑘 − 𝑇 𝛼𝑗 Γ𝑖𝑘 − 𝑇 𝑖𝛼 Γ𝑗𝑘 𝐠 ⊗ 𝐠 = 𝑇 𝑖𝑗,𝑘 𝐠 ⊗ 𝐠
𝜕𝑥
So that
𝜕𝑇𝑖𝑗 𝛼 𝛼
𝑇𝑖𝑗,𝑘 = 𝑘 − 𝑇 𝛼𝑗 Γ𝑖𝑘 − 𝑇 𝑖𝛼 Γ𝑗𝑘
𝜕𝑥
Dept of Systems Engineering, University of Lagos 84 oafak@unilag.edu.ng 12/27/2012