3. tensor calculus jan 2013
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Tensor Calculus Differentials & Directional Derivatives
We are presently concerned with Inner Product spaces in our treatment of the Mechanics of Continua. Consider a map,
ð: V â W This maps from the domainV to W â both of which are Euclidean vector spaces. The concepts of limit and continuity carries naturally from the real space to any Euclidean vector space.
The Gateaux Differential
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 2
Let ð0 â V and ð0 â W, as usual we can say that the limit
limðâ ð0
ð ð = ð0
if for any pre-assigned real number ð > 0, no matter how small, we can always find a real number ð¿ > 0 such that ð ð â ð0 †ð whenever ð â ð0 < ð¿.
The function is said to be continuous at ð0 if ð ð0 exists and ð ð0 = ð0
The Gateaux Differential
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 3
Specifically, for ðŒ â â let this map be:
ð·ð ð, ð â¡ limðŒâ0
ð ð + ðŒð â ð ð
ðŒ=
ð
ððŒð ð + ðŒð
ðŒ=0
We focus attention on the second variable â while we allow the dependency on ð to be as general as possible. We shall show that while the above function can be any given function of ð (linear or nonlinear), the above map is always linear in ð irrespective of what kind of Euclidean space we are mapping from or into. It is called the Gateaux Differential.
The Gateaux Differential
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 4
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 â â ðð¥ and we obtain,
ð·ð¹(ð¥, ðð¥) = limðŒâ0
ð¹ ð¥ + ðŒðð¥ â ð¹ ð¥
ðŒ=
ð
ððŒð¹ ð¥ + ðŒðð¥
ðŒ=0
And let ðŒðð¥ â Îð¥, the middle term becomes,
limÎð¥â0
ð¹ ð¥ + Îð¥ â ð¹ ð¥
Îð¥ðð¥ =
ðð¹
ðð¥ðð¥
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).
The Gateaux Differential
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 5
It is easily shown that the Gateaux differential is linear in its second argument, ie, for ðŒ â R
ð·ð ð, ðŒð = ðŒð·ð ð, ð
Furthermore, ð·ð ð, ð + ð = ð·ð ð, ð + ð·ð ð, ð
and that for ðŒ, ðœ â R ð·ð ð, ðŒð + ðœð = ðŒð·ð ð, ð + ðœð·ð ð, ð
The Gateaux Differential
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 6
The Frechét derivative or gradient of a differentiable function is the linear operator, grad ð(ð) such that, for ðð â V,
ðð ð
ðððð â¡ grad ð ð ðð â¡ ð·ð ð, ðð
Obviously, grad ð(ð) 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.
The Frechét Derivative
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 7
The Gradient of a Differentiable Real-Valued Function For a real vector function, we have the map:
ð: V â R The Gateaux differential in this case takes two vector arguments and maps to a real value:
ð·ð: V Ã V â R This is the classical definition of the inner product so that we can define the differential as,
ðð(ð)
ððâ ðð â¡ ð·ð ð, ðð
This quantity, ðð(ð)
ðð
defined by the above equation, is clearly a vector.
Differentiable Real-Valued Function
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 8
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 ð ð â V. On such a basis, the function,
ð = ð£ðð ð
and,
ð ð = ð ð£ðð ð
we may also express the independent vector ðð = ðð£ðð ð
on this basis.
Real-Valued Function
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 9
The Gateaux differential in the direction of the basis vector ð ð is,
ð·ð ð, ð ð = limðŒâ0
ð ð + ðŒð ð â ð ð
ðŒ= lim
ðŒâ0
ð ð£ðð ð + ðŒð ð â ð ð£ðð ð
ðŒ
= limðŒâ0
ð ð£ð + ðŒð¿ðð
ð ð â ð ð£ðð ð
ðŒ
As the function ð does not depend on the vector basis, we can substitute the vector function by the real function of the components ð£1, ð£2, ð£3 so that,
ð ð£ðð ð = ð ð£1, ð£2, ð£3
in which case, the above differential, similar to the real case discussed earlier becomes,
ð·ð ð, ð ð =ðð ð£1, ð£2, ð£3
ðð£ð
Real-Valued Function
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 10
Of course, it is now obvious that the gradient ðð(ð)
ðð of
the scalar valued vector function, expressed in its dual basis must be,
ðð(ð)
ðð=
ðð ð£1, ð£2, ð£3
ðð£ðð ð
so that we can recover,
ð·ð ð, ð ð =ðð ð
ððâ ðð =
ðð ð£1, ð£2, ð£3
ðð£ðð ð â ð ð
=ðð ð£1, ð£2, ð£3
ðð£ðð¿ð
ð=
ðð ð£1, ð£2, ð£3
ðð£ð
Hence we obtain the well-known result that
grad ð ð =ðð(ð)
ðð=
ðð ð£1, ð£2, ð£3
ðð£ðð ð
which defines the Frechét derivative of a scalar valued function with respect to its vector argument.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 11
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 ð ð â V. On such a basis, the function,
ð ð = ð¹ð ð ð ð
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,
ð ð = ð¹ð ð£1, ð£2, ð£3 ð ð
Vector valued Function
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 12
ð·ð ð, ðð = ð·ð ð£ðð ð , ðð£ðð ð
= ð·ð ð£ðð ð , ð ð ðð£ð
= ð·ð¹ð ð£ðð ð , ð ð ð ððð£ð
Again, upon noting that the functions ð·ð¹ð ð£ðð ð , ð ð ,
ð = 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,
ð·ð ð, ðð = ð·ð¹ð ð£ðð ð , ð ð ð ððð£ð = ð·ð¹ð ð, ð ð ð ððð£ð
Vector Derivatives
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 13
Which we can compare to the earlier case of scalar valued function and easily obtain,
ð·ð ð, ðð =ðð¹ð ð£1, ð£2, ð£3
ðð£ðð ð â ð ð ð ððð£ð
=ðð¹ð ð£1, ð£2, ð£3
ðð£ðð ð â ð ð ð ððð£ð
=ðð¹ð ð£1, ð£2, ð£3
ðð£ðð ð â ð ð ðð
= ðð ð
ðððð â¡ grad ð ð ðð
Clearly, the tensor gradient of the vector-valued vector function is,
ðð ð
ðð= grad ð ð =
ðð¹ð ð£1, ð£2, ð£3
ðð£ðð ð â ð ð
Where due attention should be paid to the covariance of the quotient indices in contrast to the contravariance of the non quotient indices. oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 14
The trace of this gradient is called the divergence of the function:
div ð ð = trðð¹ð ð£1, ð£2, ð£3
ðð£ðð ð â ð ð
=ðð¹ð ð£1, ð£2, ð£3
ðð£ðð ð â ð ð
=ðð¹ð ð£1, ð£2, ð£3
ðð£ðð¿ð
ð
=ðð¹ð ð£1, ð£2, ð£3
ðð£ð
The Trace & Divergence
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 15
Consider the mapping, ð»: V â 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:
ð·ð»: V Ã V â T With the usual notation, we may write,
ð·ð» ð, ðð = ð·ð» ð£ðð ð , ðð£ðð ð = ð·ð» ð£ðð ð , ðð£ðð ð
= ðð» ð
ðððð â¡ grad ð» ð ðð
= ð·ððŒðœ ð£ðð ð, ðð£ðð ð ð ðŒ â ð ðœ
= ð·ððŒðœ ð, ð ð ð ðŒ â ð ðœ ðð£ð
Each of these nine functions look like the real differential of several variables we considered earlier.
Tensor-Valued Vector Function
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 16
The independence of the basis vectors imply, as usual, that
ð·ððŒðœ ð, ð ð =ðððŒðœ ð£1, ð£2, ð£3
ðð£ð
so that,
ð·ð» ð, ðð = ð·ððŒðœ ð, ð ð ð ðŒ â ð ðœ ðð£ð
= ðððŒðœ ð£1, ð£2, ð£3
ðð£ðð ðŒ â ð ðœ ðð£ð
= ðððŒðœ ð£1, ð£2, ð£3
ðð£ðð ðŒ â ð ðœ ð ð â ðð
= ðððŒðœ ð£1, ð£2, ð£3
ðð£ðð ðŒ â ð ðœ â ð ð ðð =
ðð» ð
ðððð
â¡ grad ð» ð ðð
Which defines the third-order tensor,
ðð» ð
ðð= grad ð» ð =
ðððŒðœ ð£1, ð£2, ð£3
ðð£ðð ðŒ â ð ðœ â ð ð
and with no further ado, we can see that a third-order tensor transforms a vector into a second order tensor.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 17
The divergence operation can be defined in several ways. Most common is achieved by the contraction of the last two basis so that,
div ð» ð =ðððŒðœ ð£1, ð£2, ð£3
ðð£ðð ðŒ â ð ðœ â ð ð
=ðððŒðœ ð£1, ð£2, ð£3
ðð£ðð ðŒ ð ðœ â ð ð
=ðððŒðœ ð£1, ð£2, ð£3
ðð£ðð ðŒð¿ðœ
ð =ðððŒð ð£1, ð£2, ð£3
ðð£ðð ðŒ
It can easily be shown that this is the particular vector that gives, for all constant vectors ð,
div ð» ð â¡ div ð»Tð
The Divergence of a Tensor Function
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 18
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.
Real-Valued Tensor Functions
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 19
Consider the Map: ð: T â R
The Gateaux differential in this case takes two vector arguments and maps to a real value:
ð·ð: T à T â 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:
ð·ð ð», ðð» =ðð(ð»)
ðð»: ðð»
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.
Frechét Derivative
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 20
Guided by the previous result of the gradient of the scalar valued function of a vector, it is not difficult to see that,
ðð(ð»)
ðð»=
ðð(ð11, ð12, ⊠, ð33)
ðððŒðœð ðŒ â ð ðœ
In the dual to the same basis, we can write,
ðð» = ððððð ð â ð ð
Clearly, ðð(ð»)
ðð»: ðð» =
ðð(ð11, ð12, ⊠, ð33)
ðððŒðœð ðŒ â ð ðœ : ððððð ð â ð ð
=ðð(ð11, ð12, ⊠, ð33)
ðððððððð
Again, note the covariance of the quotient indices.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 21
Let ð be a symmetric and positive definite tensor and let ðŒ1 ðº , ðŒ2 ðº and ðŒ3 ðº be the three principal invariants of
ð show that (a) ðð°1 ð
ðð= ð the identity tensor, (b)
ðð°2 ð
ðð= ðŒ1 ð ð â ð and (c)
ððŒ3 ð
ðð= ðŒ3 ð ðâ1
ðð°1 ð
ðð can be written in the invariant component form
as, ððŒ1 ð
ðð=
ððŒ1 ð
ðððð
ð ð â ð ð
Examples
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 22
Recall that ðŒ1 ð = tr ð = ðαα hence
ððŒ1 ð
ðð=
ððŒ1 ð
ðððð
ð ð â ð ð
=ððα
α
ððððð ð â ð ð
= ð¿ðŒð ð¿ð
ðŒð ðâ ð ð
= ð¿ððð ðâ ð ð = ð
which is the identity tensor as expected.
(a) Continued
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 23
ððŒ2 ð
ðð in a similar way can be written in the invariant component form
as, ððŒ2 ð
ðð=
1
2
ððŒ1 ð
ðððð
ðααððœ
ðœâ ððœ
αðαðœ
ð ð â ð ð
where we have utilized the fact that ðŒ2 ð =1
2tr2 ð â tr(ð2) .
Consequently, ððŒ2 ð
ðð=
1
2
ð
ðððð
ðααððœ
ðœâ ððœ
αðαðœ
ð ð â ð ð
=1
2ð¿ðŒ
ð ð¿ððŒððœ
ðœ+ ð¿ðœ
ð ð¿ððœðα
α â ð¿ðœð ð¿ð
ðŒðαðœ
â ð¿ðŒð ð¿ð
ðœððœ
α ð ð â ð ð
=1
2ð¿ð
ðððœðœ
+ ð¿ðððα
α â ðððâ ðð
ðð ð â ð ð
= ð¿ðððα
α â ððð
ð ð â ð ð = ðŒ1 ð ð â ð
(b) ððŒ2 ð
ðð= ðŒ1 ð ð â ð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 24
ðð°3 ð
ðð=
ðdet ð
ðð= ðð
the cofactor of ð. Clearly ðð = det ð ðâ1 = ð°3 ð ðâ1 as required. Details of this for the contravariant components of a tensor is presented below. Let
ð = ð =1
3!ððððððð ð¡ðððððð ððð¡
Differentiating wrt ððŒðœ, we obtain,
ðð
ðððŒðœð ðŒ â ð ðœ=
1
3!ððððððð ð¡
ðððð
ðððŒðœððð ððð¡ + ððð
ðððð
ðððŒðœððð¡ + ðððððð
ðððð¡
ðððŒðœð ðŒ â ð ðœ
=1
3!ððððððð ð¡ ð¿ð
ðŒð¿ððœððð ððð¡ + ðððð¿ð
ðŒð¿ð ðœððð¡ + ðððððð ð¿ð
ðŒð¿ð¡ðœ
ð ðŒ â ð ðœ
=1
3!ððŒððððœð ð¡ ððð ððð¡ + ððð ððð¡ + ððð ððð¡ ð ðŒ â ð ðœ
=1
2!ððŒððððœð ð¡ððð ððð¡ð ðŒ â ð ðœâ¡ ðc ðŒðœð ðŒ â ð ðœ
Which is the cofactor of ððŒðœ
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 25
Consider the function ð(ðº) = tr ðºð where ð â â, and
ðº is a tensor.
ð ðº = tr ðºð = ðºð: ð
To be specific, let ð = 3.
Real Tensor Functions
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 26
The Gateaux differential in this case,
ð·ð ðº, ððº =ð
ððŒð ðº + ðŒððº
ðŒ=0
=ð
ððŒtr ðº + ðŒððº 3
ðŒ=0
=ð
ððŒtr ðº + ðŒððº ðº + ðŒððº ðº + ðŒððº
ðŒ=0
= trð
ððŒðº + ðŒððº ðº + ðŒððº ðº + ðŒððº
ðŒ=0
= tr ððº ðº + ðŒððº ðº + ðŒððº+ ðº + ðŒððº ððº ðº + ðŒððº
+ ðº + ðŒððº ðº + ðŒððº ððº ðŒ=0
= tr ððº ðº ðº + ðº ððº ðº + ðº ðº ð ðº = 3 ðº2 T: ððº
Real Tensor Functions
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 27
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 ð > 0, we have,
ð·ð ðº, ððº = ð ðºðâ1 T: ððº
Clearly, ð
ððº tr ðºð = ð ðºðâ1
T
Real Tensor Functions
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 28
When ð = 1,
ð·ð ðº, ððº =ð
ððŒð ðº + ðŒððº
ðŒ=0
=ð
ððŒtr ðº + ðŒððº
ðŒ=0
= tr ð ððº = ð: ððº
Or that, ð
ððº tr ðº = ð.
Real Tensor Functions
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 29
Derivatives of the other two invariants of the tensor ðº can be found as follows:
ð
ððº ðŒ2(ðº)
=1
2
ð
ððºtr2 ðº â tr ðº2
=1
22tr ðº ð â 2 ðºT = tr ðº ð â ðºT
.
Real Tensor Functions
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 30
To Determine the derivative of the third invariant, we begin with the trace of the Cayley-Hamilton for ðº: tr ðº3 â ðŒ1ðº
2 + ðŒ2ðº â ðŒ3ð° = tr(ðº3) â ðŒ1tr ðº2 + ðŒ2tr ðºâ 3ðŒ3 = 0
Therefore, 3ðŒ3 = tr(ðº3) â ðŒ1tr ðº2 + ðŒ2tr ðº
ðŒ2 ðº =1
2tr2 ðº â tr ðº2
.
Real Tensor Functions
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 31
We can therefore write, 3ðŒ3 ðº = tr(ðº3) â tr ðº tr ðº2
+1
2tr2 ðº â tr ðº2 tr ðº
so that, in terms of traces only,
ðŒ3 ðº =1
6tr3 ðº â 3tr ðº tr ðº2 + 2tr(ðº3)
Differentiating the Invariants
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 32
Clearly, ððŒ3(ðº)
ððº=
1
63tr2 ðº ð â 3tr ðº2 â 3tr ðº 2ðºð + 2 à 3 ðº2 ð
= ðŒ2ð â tr ðº ðºð + ðº2ð
Real Tensor Functions
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 33
.
Real Tensor Functions
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 34
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
The Euclidean Point Space
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 35
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.
The Euclidean Point Space
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 36
The Euclidean Point Space â° is a set of elements called points. For each pair of points ð, ð â â°, â ð ð, ð â E with the following two properties:
1. ð ð, ð = ð ð, ð + ð ð, ð âð, ð, ð â â°
2. ð ð, ð = ð ð, ð â ð = ð
Based on these two, we proceed to show that, ð ð, ð = ð
And that, ð ð, ð = âð ð, ð
The Euclidean Point Space
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 37
From property 1, let ð â ð it is clear that, ð ð, ð = ð ð, ð + ð ð, ð
And it we further allow ð â ð, we find that, ð ð, ð = ð ð, ð + ð ð, ð = 2ð ð, ð
Which clearly shows that ð ð, ð = ð the zero vector.
Similarly, from the above, we find that, ð ð, ð = ð = ð ð, ð + ð ð, ð
So that ð ð, ð = âð ð, ð
The Euclidean Point Space
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 38
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 39
Coordinate System
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 40
Coordinate System
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 41
Note that â° is NOT a vector space. In our discussion, the vector space E to which ð 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 â°. We adopt the convention that ð ð â¡ ð ð, ð referring to the vector ð. If therefore we choose an arbitrarily fixed point ð â â°, we are associating ð ð , ð ð and ð ð respectively with ð ð, ð , ð ð, ð and ð ð, ð .
These are vectors based on the points ð, ð and ð with reference to the origin chosen. To emphasize the association with both points of â° as well as members of E they are called Position Vectors.
Position Vectors
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 42
Recall that by property 1, ð ð â¡ ð ð, ð = ð ð, ð + ð ð, ð
Furthermore, we have deduced that ð ð, ð = âð(ð, ð)
We may therefore write that, ð ð = ð ð â ð(ð)
Which, when there is no ambiguity concerning the chosen origin, we can write as,
ð ð = ð â ð
And the distance between the two is,
ð ð ð = ð ð â ð = ð â ð
Length in the Point Space
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 43
The norm of a vector is the square root of the inner product of the vector with itself. If the coordinates of ð and ð on a set of independent vectors are ð¥ð and ðŠð, then the distance we seek is,
ð ð â ð = ð â ð = ððð(ð¥ð â ðŠð)(ð¥ð â ðŠð)
The more familiar Pythagorean form occurring only when ððð = 1.
Metric Properties
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 44
Consider a Cartesian coordinate system ð¥, ðŠ, ð§ or ðŠ1, ðŠ2 and ðŠ3with an orthogonal basis. Let us now have the possibility of transforming to another coordinate system of an arbitrary nature: ð¥1, ð¥2, ð¥3. We can represent the transformation and its inverse in the equations:
ðŠð = ðŠð ð¥1, ð¥2, ð¥3 , ð¥ð = ð¥ð(ðŠ1, ðŠ2, ðŠ3)
And if the Jacobian of transformation,
Coordinate Transformations
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 45
ððŠð
ðð¥ðâ¡
ð ðŠ1, ðŠ2, ðŠ3
ð ð¥1, ð¥2, ð¥3 â¡
ððŠ1
ðð¥1
ððŠ1
ðð¥2
ððŠ1
ðð¥3
ððŠ2
ðð¥1
ððŠ2
ðð¥2
ððŠ2
ðð¥3
ððŠ3
ðð¥1
ððŠ3
ðð¥2
ððŠ3
ðð¥3
does not vanish, then the inverse transformation will exist. So that,
ð¥ð = ð¥ð(ðŠ1, ðŠ2, ðŠ3)
Jacobian Determinant
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 46
Given a position vector ð« = ð«(ð¥1, ð¥2, ð¥3)
In the new coordinate system, we can form a set of three vectors,
ð ð =ðð«
ðð¥ð, ð = 1,2,3
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. ð« = ð« ð¥1, ð¥2, ð¥3 = ðŠ1ð¢ +ðŠ2ð£ + ðŠ3ð€ So that
ð¢ =ðð«
ððŠ1 , ð£ =ðð«
ððŠ2 and ð€ =ðð«
ððŠ3
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,
ð« = ð« ð¥1, ð¥2, ð¥3 The total differential of ð« is simply,
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 47
ðð« =ðð«
ðð¥1ðð¥1 +
ðð«
ðð¥2ðð¥2 +
ðð«
ðð¥3ðð¥3
â¡ ðð¥1ð 1 + ðð¥2ð 2 + ðð¥3ð 3
With ð ð ð¥1, ð¥2, ð¥3 , ð = 1,2,3 now depending in general on ð¥1, ð¥2 and ð¥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, ð ð , ð = 1,2,3 are constants, normalized and orthogonal. They are our familiar
ð¢ =ðð«
ððŠ1, ð£ =
ðð«
ððŠ2 and ð€ =
ðð«
ððŠ3.
Natural Dual Bases
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 48
We now proceed to show that the set of basis vectors, ð ð â¡ ð»ð¥ð
is reciprocal to ð ð. The total differential ðð« can be written in terms of partial differentials as,
ðð« =ðð«
ððŠðððŠð
Which when expressed in component form yields,
ðð¥1 =ðð¥1
ððŠ1ððŠ1 +
ðð¥1
ððŠ2ððŠ2 +
ðð¥1
ððŠ3ððŠ3
ðð¥2 =ðð¥2
ððŠ1ððŠ1 +
ðð¥2
ððŠ2ððŠ2 +
ðð¥2
ððŠ3ððŠ3
ðð¥3 =ðð¥3
ððŠ1ððŠ1 +
ðð¥3
ððŠ2ððŠ2 +
ðð¥3
ððŠ3ððŠ3
Or, more compactly that,
ðð¥ð =ðð¥ð
ððŠðððŠð = ð»ð¥ð â ðð«
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 49
In Cartesian coordinates, ðŠð , ð = 1, . . , 3 for any scalar field ð (ðŠ1, ðŠ2, ðŠ3),
ðð
ððŠðððŠð =
ðð
ðð¥ðð¥ +
ðð
ððŠððŠ +
ðð
ðð§ðð§ = grad ð â ðð«
We are treating each curvilinear coordinate as a scalar field, ð¥ð = ð¥ð(ðŠ1, ðŠ2, ðŠ3).
ðð¥ð = ð»ð¥ð â ðð«
ðð¥ð ðð¥ð
= ð ð â ð ððð¥ð
= ð¿ðððð¥ð.
The last equality arises from the fact that this is the only way one component of the coordinate differential can equal another.
â ð ð â ð ð = ð¿ðð
Which recovers for us the reciprocity relationship and shows that ð ð and ð ð are reciprocal systems.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 50
The position vector in Cartesian coordinates is ð = ðððð. Show that (a) ðð¢ð¯ ð = ð, (b) ðð¢ð¯ ð â ð = ðð, (c) ðð¢ð¯ ð = ð, and (d) ð ð«ðð ð = ð and (e) ðð®ð«ð¥ ð â ð = âð Ã
grad ð = ð¥ð ,ð ðð â ðð = ð¿ðððð â ðð = ð
div ð = ð¥ð ,ð ðð â ðð = ð¿ððð¿ðð = ð¿ðð = 3.
ð â ð = ð¥ððð â ð¥ððð = ð¥ðð¥ððð â ðð
grad ð â ð = ð¥ðð¥ð ,ð ðð â ðð â ðð
div ð â ð = ð¥ð ,ð ð¥ð + ð¥ðð¥ð ,ð ðð â ðð â ðð
= ð¿ðððð + ððð¿ðð ð¿ðððð = ð¿ðððð + ððð¿ðð ðð = 4ð¥ððð = 4ð
curl ð â ð = ððŒðœðŸ ð¥ðð¥ðŸ ,ðœ ððŒ â ðð
= ððŒðœðŸ ð¥ð ,ðœ ð¥ðŸ + ð¥ðð¥ðŸ,ðœ ððŒ â ðð
= ððŒðœðŸ ð¿ððœð¥ðŸ + ð¥ðð¿ðŸðœ ððŒ â ðð = ððŒððŸ ð¥ðŸððŒ â ðð + ððŒðœðœð¥ðððŒ â ðð
= âððŒðŸð ð¥ðŸððŒ â ðð = âð Ã
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 51
For a scalar field ð and a tensor field ð show that ð ð«ðð ðð = ðð ð«ðð ð + ð â ð ð«ððð. Also show that ðð¢ð¯ ðð = ð ðð¢ð¯ ð + ðð ð«ððð.
grad ðð = ðððð ,ð ð ð â ð ð â ð ð
= ð,ð ððð + ðððð ,ð ð ð â ð ð â ð ð = ð â gradð + ðgrad ð
Furthermore, we can contract the last two bases and obtain,
div ðð = ð,ð ððð + ðððð ,ð ð ð â ð ð â ð ð
= ð,ð ððð + ðððð ,ð ð ðð¿ðð
= ððð ð,ð ð ð + ðððð ,ð ð ð = ðgradð + ð div ð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 52
For two arbitrary vectors, ð and ð, show that ð ð«ðð ð â ð = ð Ã ð ð«ððð â ð Ã ð ð«ððð
grad ð â ð = ððððð¢ðð£ð ,ð ð ð â ð ð
= ððððð¢ð ,ð ð£ð + ððððð¢ðð£ð,ð ð ð â ð ð
= ð¢ð ,ð ððððð£ð + ð£ð ,ð ððððð¢ð ð ð â ð ð
= â ð Ã gradð + ð Ã gradð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 53
For any two tensor fields ð® and ð¯, Show that, ð® à : grad ð¯ = ð® â curl ð¯
ð® à : grad ð¯ = ððððð¢ðð ð â ð ð : ð£ðŒ,ðœ ð ðŒ â ð ðœ
= ððððð¢ðð£ðŒ,ðœ ð ð â ð ðŒ ð ð â ð ðœ
= ððððð¢ðð£ðŒ,ðœ ð¿ððŒð¿ð
ðœ= ððððð¢ðð£ð ,ð
= ð® â curl ð¯
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 54
For a vector field ð, show that ð ð«ðð ð à is a third ranked tensor. Hence or otherwise show that ðð¢ð¯ ð à = âðð®ð«ð¥ ð.
The secondâorder tensor ð à is defined as ððððð¢ðð ð â
ð ð. Taking the covariant derivative with an independent base, we have
grad ð à = ððððð¢ð ,ð ð ð â ð ð â ð ð
This gives a third order tensor as we have seen. Contracting on the last two bases,
div ð à = ððððð¢ð ,ð ð ð â ð ð â ð ð
= ððððð¢ð ,ð ð ðð¿ðð
= ððððð¢ð ,ð ð ð = âcurl ð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 55
Show that ðð®ð«ð¥ ðð = grad ð Ã
Note that ðð = ðððŒðœ ð ðŒ â ð ðœ, and that curl ð» =
ððððððŒð ,ð ð ð â ð ðŒ so that,
curl ðð = ðððð ðððŒð ,ð ð ð â ð ðŒ
= ðððð ð,ð ððŒð ð ð â ð ðŒ = ððððð,ð ð ð â ð ð = grad ð Ã
Show that curl ð Ã = div ð ð â grad ð
ð à = ððŒðœðð£ðœ ð ðŒ â ð ð
curl ð» = ððððððŒð ,ð ð ð â ð ðŒ
so that
curl ð à = ððððððŒðœðð£ðœ ,ð ð ð â ð ðŒ
= ðððŒðððœ â ðððœðððŒ ð£ðœ ,ð ð ð â ð ðŒ
= ð£ð ,ð ð ðŒ â ð ðŒ â ð£ð ,ð ð ð â ð ð = div ð ð â grad ð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 56
Show that ðð¢ð¯ ð à ð = ð â ðð®ð«ð¥ ð â ð â ðð®ð«ð¥ ð
div ð à ð = ððððð¢ðð£ð ,ð
Noting that the tensor ðððð behaves as a constant under a covariant differentiation, we can write,
div ð à ð = ððððð¢ðð£ð ,ð
= ððððð¢ð ,ð ð£ð + ððððð¢ðð£ð,ð = ð â curl ð â ð â curl ð
Given a scalar point function Ï and a vector field ð¯, show that ðð®ð«ð¥ ðð¯ = ð curl ð¯ + gradð à ð¯.
curl ðð¯ = ðððð ðð£ð ,ð ð ð
= ðððð ð,ð ð£ð + ðð£ð ,ð ð ð
= ððððð,ð ð£ðð ð + ðððððð£ð,ð ð ð = gradð à ð¯ + ð curl ð¯
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 57
Given that ð ð¡ = ð(ð¡) , Show that ð ð¡ =ð
ð ð¡: ð
ð2 â¡ ð: ð
Now, ð
ðð¡ð2 = 2ð
ðð
ðð¡=
ðð
ðð¡: ð + ð:
ðð
ðð¡= 2ð:
ðð
ðð¡
as inner product is commutative. We can therefore write that
ðð
ðð¡=
ð
ð:ðð
ðð¡=
ð
ð ð¡: ð
as required.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 58
Given a tensor field ð», obtain the vector ð â¡ ð»ðð¯ and show that its divergence is ð»: ðð¯ + ð¯ â ðð¢ð¯ ð»
The divergence of ð is the scalar sum , ðððð£ð ,ð.
Expanding the product covariant derivative we obtain,
div ð»Tð¯ = ðððð£ð ,ð = ððð ,ð ð£ð + ðððð£
ð ,ð
= div ð» â ð¯ + tr ð»Tgradð¯
= div ð» â ð¯ + ð»: grad ð¯
Recall that scalar product of two vectors is commutative so that
div ð»Tð¯ = ð»: gradð¯ + ð¯ â div ð»
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 59
For a second-order tensor ð» define
curl ð» â¡ ððððð»ðŒð ,ð ð ð â ð ð¶ show that for any constant
vector ð, curl ð» ð = curl ð»ð»ð
Express vector ð in the invariant form with
contravariant components as ð = ððœð ðœ. It follows that
curl ð» ð = ððððððŒð,ð ð ð â ð ðŒ ð
= ððððððŒð,ð ððœ ð ð â ð ðŒ ð ðœ
= ððððððŒð,ð ððœð ðð¿ðœðŒ
= ðððð ððŒð ,ð ð ðððŒ
= ðððð ððŒðððŒ ,ð ð ð
The last equality resulting from the fact that vector ð is a constant vector. Clearly,
curl ð» ð = curl ð»ðð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 60
For any two vectors ð® and ð¯, show that curl ð® â ð¯ =grad ð® ð¯ à ð» + curl ð¯ â ð where ð¯ à is the skew tensor
ððððð£ð ð ð â ð ð.
Recall that the curl of a tensor ð» is defined by curl ð» â¡
ððððððŒð ,ð ð ð â ð ðŒ. Clearly therefore,
curl ð â ð = ðððð ð¢ðŒð£ð ,ð ð ð â ð ðŒ
= ðððð ð¢ðŒ ,ð ð£ð + ð¢ðŒð£ð ,ð ð ð â ð ðŒ
= ððððð¢ðŒ ,ð ð£ð ð ð â ð ðŒ + ððððð¢ðŒð£ð ,ð ð ð â ð ðŒ
= ððððð£ð ð ð â ð¢ðŒ ,ð ð ðŒ + ððððð£ð ,ð ð ð â ð¢ðŒð ðŒ
= ððððð£ð ð ð â ð ð ð¢ðŒ ,ðœ ð ðœ â ð ðŒ + ððððð£ð ,ð ð ð
â ð¢ðŒð ðŒ = â ð à grad ð ð» + curl ð â ð= grad ð ð à ð» + curl ð â ð
upon noting that the vector cross is a skew tensor.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 61
Show that curl ð Ã ð = div ð â ð â ð â ð
The vector ð â¡ ð à ð = ð€ðð ð = ðððŒðœð¢ðŒð£ðœð ð and
curl ð = ððððð€ð ,ð ð ð. Therefore,
curl ð à ð = ððððð€ð ,ð ð ð
= ðððððððŒðœ ð¢ðŒð£ðœ ,ð ð ð
= ð¿ðŒð ð¿ðœ
ðâ ð¿ðœ
ð ð¿ðŒð
ð¢ðŒð£ðœ ,ð ð ð
= ð¿ðŒð ð¿ðœ
ðâ ð¿ðœ
ð ð¿ðŒð
ð¢ðŒ,ð ð£ðœ + ð¢ðŒð£ðœ,ð ð ð
= ð¢ð ,ð ð£ð + ð¢ðð£ð ,ð â ð¢ð ,ð ð£ð + ð¢ðð£ð ,ð ð ð
= ð¢ðð£ð ,ð â ð¢ðð£ð ,ð ð ð = div ð â ð â ð â ð
since div ð â ð = ð¢ðð£ð ,ðŒ ð ð â ð ð â ð ðŒ = ð¢ðð£ð ,ð ð ð.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 62
Given a scalar point function ð and a second-order tensor field ð»,
show that ðð®ð«ð¥ ðð» = ð ðð®ð«ð¥ ð» + gradð à ð»ð» where
gradð Ã is the skew tensor ððððð,ð ð ð â ð ð
curl ðð» â¡ ðððð ðððŒð ,ð ð ð â ð ðŒ
= ðððð ð,ð ððŒð + ðððŒð,ð ð ð â ð ðŒ
= ððððð,ð ððŒð ð ð â ð ðŒ + ðððððððŒð,ð ð ð â ð ðŒ
= ððððð,ð ð ð â ð ð ððŒðœð ðœ â ð ðŒ + ðððððððŒð,ð ð ð â ð ðŒ
= ð curl ð» + gradð à ð»ð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 63
For a second-order tensor field ð», show that
div curl ð» = curl div ð»ð
Define the second order tensor ð as
curl ð» â¡ ððððððŒð,ð ð ð â ð ðŒ = ð.ðŒð ð ð â ð ðŒ
The gradient of ðº is
ð.ðŒð ,ðœ ð ð â ð ðŒ â ð ðœ = ððððððŒð,ððœ ð ð â ð ðŒ â ð ðœ
Clearly,
div curl ð» = ððððððŒð,ððœ ð ð â ð ðŒ â ð ðœ
= ððððððŒð,ððœ ð ð ððŒðœ
= ððððððœð ,ððœ ð ð = curl div ð»T
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 64
If the volume ðœ is enclosed by the surface ðº, the position vector ð = ððð ð and ð is the external unit normal to each surface
element, show that ð
ð ð ð â ð â ðð ðºðº
equals the volume
contained in ðœ.
ð â ð = ð¥ðð¥ðð ð â ð ð = ð¥ðð¥ðððð
By the Divergence Theorem,
ð» ð â ð â ðððð
= ð» â ð» ð â ð ððð
= ðð ðð ð¥ðð¥ðððð ð ð â ð ð ððð
= ðð ððð ð¥ð ,ð ð¥ð + ð¥ðð¥ð ,ð ð ð â ð ð ððð
= ðððððð ð¿ð
ð ð¥ð + ð¥ðð¿ðð
,ð ððð
= 2ððððððð¥ð,ð ððð
= 2ð¿ððð¿ð
ð ððð
= 6 ððð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 65
For any Euclidean coordinate system, show that div ð® à ð¯ =ð¯ curl ð® â ð® curl ð¯
Given the contravariant vector ð¢ð and ð£ð with their associated vectors ð¢ð and ð£ð, the contravariant component of the above cross product is ððððð¢ðð£ð .The required divergence is simply the contraction of the covariant ð¥ð derivative of this quantity:
ððððð¢ðð£ð ,ð= ððððð¢ð,ðð£ð + ððððð¢ðð£ð,ð
where we have treated the tensor ðððð as a constant under the covariant derivative.
Cyclically rearranging the RHS we obtain,
ððððð¢ðð£ð ,ð= ð£ðððððð¢ð,ð + ð¢ðð
ðððð£ð,ð = ð£ðððððð¢ð,ð + ð¢ðððððð£ð,ð
where we have used the anti-symmetric property of the tensor ðððð. The last expression shows clearly that
div ð® à ð¯ = ð¯ curl ð® â ð® curl ð¯
as required.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 66
For a scalar variable ðŒ, if the tensor ð = ð(ðŒ) and ð â¡ðð
ððŒ, Show that
ð
ððŒdet ð =
det ð tr(ð ðâð) Proof:
Let ð â¡ ð ðâ1 so that, ð = ðð. In component form, we have ð ðð = ðŽð
ð ððð.
Therefore, ð
ððŒdet ð =
ð
ððŒðððððð
1ðð2ðð
3 = ðððð ð ð1ðð
2ðð3 + ðð
1ð ð2ðð
3 + ðð1ðð
2ð ð3
= ðððð ðŽð1ðð
ððð2ðð
3 + ðð1ðŽð
2 ððððð
3 + ðð1ðð
2ðŽð3ðð
ð
= ðððð ðŽ11ðð
1 + ðŽ21ðð
2 + ðŽ31ðð
3 ðð2ðð
3
+ ðð1 ðŽ1
2ðð1 + ðŽ2
2ðð2 + ðŽ3
2ðð3 ðð
3 + ðð1ðð
2 ðŽ13ðð
1 + ðŽ23ðð
2 + ðŽ33ðð
3
All the boxed terms in the above equation vanish on account of the contraction of a symmetric tensor with an antisymmetric one.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 67
Liouville Formula
(For example, the first boxed term yields, ðððððŽ21ðð
2ðð2ðð
3
Which is symmetric as well as antisymmetric in ð and ð. It
therefore vanishes. The same is true for all other such terms.)
ð
ððŒdet ð = ðððð ðŽ1
1ðð1 ðð
2ðð3 + ðð
1 ðŽ22ðð
2 ðð3 + ðð
1ðð2 ðŽ3
3ðð3
= ðŽðð ðððððð
1ðð2ðð
3
= tr ð ðâ1 det ð
as required.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 68
Liouville Theorem Contâd
For a general tensor field ð» show that, ðð®ð«ð¥ ðð®ð«ð¥ ð» =ðð ðð« ð» â ðð¢ð¯ ðð¢ð¯ ð» ð° + ð ð«ðð ðð¢ð¯ ð» + ð ð«ðð ðð¢ð¯ ð»
ðâ
ð ð«ðð ð ð«ðð ðð«ð» â ððð»ð
curl ð» = ððŒð ð¡ððœð¡ ,ð ð ðŒ â ð ðœ
= ð .ðœðŒ ð ðŒ â ð ðœ
curl ðº = ððððð .ððŒ ,ð ð ð â ð ðŒ
so that
curl ðº = curl curl ð» = ððððððŒð ð¡ððð¡,ð ð ð ð â ð ðŒ
=
ðððŒ ððð ððð¡
ðððŒ ððð ððð¡
ðððŒ ððð ððð¡
ððð¡ ,ð ð ð ð â ð ðŒ
=ðððŒ ððð ððð¡ â ððð¡ððð + ððð ððð¡ðððŒ â ðððŒððð¡
+ððð¡ ðððŒððð â ððð ðððŒððð¡ ,ð ð ð ð â ð ðŒ
= ððð ð .ð¡ð¡ ,ð ð âð ..
ð ð ,ð ð ð ðŒ â ð ðŒ + ð ..ðŒð ,ð ð âðððŒð .ð¡
ð¡ ,ð ð ð ð â ð ðŒ
+ ðððŒð .ð¡ð .,ð ð âððð ð .ð¡
ðŒ.,ð ð ð ð¡ â ð ðŒ
= ð»2 tr ð» â div div ð» ð° + grad div ð»T
â grad grad trð»
+ grad div ð» â ð»2ð»T oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 69
When ð» is symmetric, show that tr(curl ð») vanishes.
curl ð» = ððððððœð ,ð ð ð â ð ðœ
tr curl ð» = ððððððœð ,ð ð ð â ð ðœ
= ððððððœð ,ð ð¿ððœ
= ððððððð ,ð
which obviously vanishes on account of the symmetry and antisymmetry in ð and ð. In this case, curl curl ð»
= ð»2 tr ð» â div div ð» ð° â grad grad trð»
+ 2 grad div ð» â ð»2ð»
as grad div ð»T
= grad div ð» if the order of
differentiation is immaterial and T is symmetric.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 70
For a scalar function Ί and a vector ð£ð show that the divergence of the vector ð£ðΊ is equal to, ð¯ â gradΊ +Ί div ð¯
ð£ðΊ,ð
= Ίð£ð,ð + ð£ðΊ,i
Hence the result.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 71
Show that ðððð ð® à ð¯ = ð¯ â ðð® + ð® â div ð¯ â ð¯ â div ð® âð® â ð ð¯
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,
ðððð ððððð¢ðð£ð,ð
Expanding in the usual way, noting the relation between the alternating tensors and the Kronecker deltas,
ðððð ððððð¢ðð£ð,ð
= ð¿ðððððð ð¢ð
,ðð£ð â ð¢ðð£ð,ð
= ð¿ðððð ð¢ð
,ðð£ð â ð¢ðð£ð,ð =
ð¿ðð ð¿ð
ð
ð¿ðð ð¿ð
ðð¢ð
,ðð£ð â ð¢ðð£ð,ð
= ð¿ððð¿ð
ð â ð¿ðð ð¿ð
ð ð¢ð,ðð£ð â ð¢ðð£ð
,ð
= ð¿ððð¿ð
ðð¢ð,ðð£ð â ð¿ð
ðð¿ððð¢ðð£ð
,ð + ð¿ðð ð¿ð
ðð¢ð,ðð£ð
â ð¿ðð ð¿ð
ðð¢ðð£ð,ð
= ð¢ð,ðð£ð â ð¢ð
,ðð£ð + ð¢ðð£ð,ð â ð¢ðð£ð
,ð
Which is the result we seek in indicial notation.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 72
Given a vector point function ð ð¥1, ð¥2, ð¥3 and its covariant components, ð¢ðŒ ð¥1, ð¥2, ð¥3 , ðŒ = 1,2,3, with ð ðŒas the reciprocal basis vectors, Let
ð = ð¢ðŒð ðŒ , then ðð =ð
ðð¥ðð¢ðŒð ðŒ ðð¥ð =
ðð¢ðŒ
ðð¥ðð ðŒ +
ðð ðŒ
ðð¥ðð¢ðŒ ðð¥ð
Clearly, ðð
ðð¥ð=
ðð¢ðŒ
ðð¥ðð ðŒ +
ðð ðŒ
ðð¥ðð¢ðŒ
And the projection of this quantity on the ð ðdirection is, ðð
ðð¥ðâ ð ð =
ðð¢ðŒ
ðð¥ðð ðŒ +
ðð ðŒ
ðð¥ðð¢ðŒ â ð ð =
ðð¢ðŒ
ðð¥ðð ðŒ â ð ð +
ðð ðŒ
ðð¥ðâ ð ðð¢ðŒ
=ðð¢ðŒ
ðð¥ðð¿ð
ðŒ +ðð ðŒ
ðð¥ðâ ð ðð¢ðŒ
Differentiation of Fields
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 73
Now, ð ð â ð ð = ð¿ðð so that
ðð ð
ðð¥ðâ ð ð + ð ð â
ðð ð
ðð¥ð= 0.
ðð ð
ðð¥ðâ ð ð = âð ð â
ðð ð
ðð¥ðâ¡ â
ððð
.
This important quantity, necessary to quantify the derivative of a tensor in general coordinates, is called the Christoffel Symbol of the second kind.
Christoffel Symbols
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 74
Using this, we can now write that, ðð
ðð¥ð â ð ð =ðð¢ðŒ
ðð¥ð ð¿ððŒ +
ðð ðŒ
ðð¥ð â ð ðð¢ðŒ =ðð¢ð
ðð¥ð âðŒðð
ð¢ðŒ
The quantity on the RHS is the component of the derivative of vector ð along the ð ð direction using covariant components. It is the covariant derivative of ð. Using contravariant components, we could write,
ðð =ðð¢ð
ðð¥ð ð ð +ðð ð
ðð¥ð ð¢ð ðð¥ð =ðð¢ð
ðð¥ð ð ð +ðŒðð
ð ðŒð¢ð ðð¥ð
So that, ðð
ðð¥ð =ðð¢ðŒ
ðð¥ð ð ðŒ +ðð ðŒ
ðð¥ð ð¢ðŒ
The components of this in the direction of ð ð can be obtained by taking a dot product as before:
ðð
ðð¥ð â ð ð =ðð¢ðŒ
ðð¥ð ð ðŒ +ðð ðŒ
ðð¥ð ð¢ðŒ â ð ð =ðð¢ðŒ
ðð¥ð ð¿ðŒð +
ðð ðŒ
ðð¥ð â ð ðð¢ðŒ =ðð¢ð
ðð¥ð +ð
ðŒðð¢ðŒ
Derivatives in General Coordinates
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 75
The two results above are represented symbolically as,
ð¢ð,ð =ðð¢ð
ðð¥ð âðŒðð
ð¢ðŒ and ð¢,ðð =
ðð¢ðŒ
ðð¥ð +ð
ðŒðð¢ðŒ
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.
Covariant Derivatives
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 76
We observe that the derivative of the covariant basis, ð ð =ðð«
ðð¥ð ,
ðð ð
ðð¥ð=
ð2ð«
ðð¥ððð¥ð=
ð2ð«
ðð¥ððð¥ð=
ðð ð
ðð¥ð
Taking the dot product with ð ð, ðð ð
ðð¥ðâ ð ð =
1
2
ðð ð
ðð¥ðâ ð ð +
ðð ð
ðð¥ðâ ð ð
=1
2
ð
ðð¥ðð ð â ð ð +
ð
ðð¥ðð ð â ð ð â ð ð â
ðð ð
ðð¥ðâ ð ð â
ðð ð
ðð¥ð
=1
2
ð
ðð¥ðð ð â ð ð +
ð
ðð¥ðð ð â ð ð â ð ð â
ðð ð
ðð¥ðâ ð ð â
ðð ð
ðð¥ð
=1
2
ð
ðð¥ðð ð â ð ð +
ð
ðð¥ðð ð â ð ð â
ð
ðð¥ðð ð â ð ð
=1
2
ðððð
ðð¥ð+
ðððð
ðð¥ðâ
ðððð
ðð¥ð
Christoffel Symbols
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 77
Which is the quantity defined as the Christoffel symbols of the first kind in the textbooks. It is therefore possible for us to write,
ðð, ð â¡ ðð ð
ðð¥ðâ ð ð =
ðð ð
ðð¥ðâ ð ð =
1
2
ðððð
ðð¥ð+
ðððð
ðð¥ðâ
ðððð
ðð¥ð
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 ð and ð. The extension of this definition to the Christoffel symbols of the second kind is immediate:
The First Kind
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 78
Contract the above equation with the conjugate metric tensor, we have,
ðððŒ ðð, ðŒ â¡ ðððŒ ðð ð
ðð¥ðâ ð ðŒ = ðððŒ
ðð ð
ðð¥ðâ ð ðŒ =
ðð ð
ðð¥ðâ ð ð =
ððð
= âðð ð
ðð¥ðâ ð ð
Which connects the common definition of the second Christoffel symbol with the one defined in the above derivation. The relationship,
ðððŒ ðð, ðŒ =ððð
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.
The Second Kind
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 79
We contract the above equation with ðððœand obtain,
ðððœðððŒ ðð, ðŒ = ðððœððð
ð¿ðœðŒ ðð, ðŒ = ðð, ðœ = ðððœ
ððð
so that,
ðððŒ
ðŒðð = ðð, ð
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.
Two Christoffel Symbols Related
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 80
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 ð, we can express the components in dyadic form along the product basis as follows:
ð = ðððð ð â ð ð= ðððð ðâšð ð = ð.ð
ð ð ð â ð ð= ðð.ðð ðâšð ð
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.
Higher Order Tensors
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 81
We have:
ðð
ðð¥ð=
ðððð
ðð¥ðð ðâšð ð + ððð
ðð ð
ðð¥ðâšð ð + ðððð ðâš
ðð ð
ðð¥ð
Recall that, ðð ð
ðð¥ð â ð ð =ððð
. It follows therefore that,
ðð ð
ðð¥ðâ ð ð â
ðŒðð
ð¿ðŒð
=ðð ð
ðð¥ðâ ð ð â
ðŒðð
ð ðŒ â ð ð
=ðð ð
ðð¥ðâ
ðŒðð
ð ðŒ â ð ð = 0.
Clearly, ðð ð
ðð¥ð =ðŒðð
ð ðŒ
(Obviously since ð ð is a basis vector it cannot vanish)
Higher Order Tensors
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 82
ðð
ðð¥ð=
ðððð
ðð¥ðð ðâšð ð + ððð
ðð ð
ðð¥ðâšð ð + ðððð ðâš
ðð ð
ðð¥ð
=ðððð
ðð¥ð ð ðâšð ð + ððð ðŒðð
ð ðŒ âšð ð + ðððð ðâšðŒðð ð ðŒ
=ðððð
ðð¥ð ð ðâšð ð + ððŒð ððŒð
ð ð âšð ð + ðððŒð ðâšð
ðŒðð ð
=ðððð
ðð¥ð +ððŒð ððŒð
+ ðððŒ ððŒð
ð ðâšð ð = ð ,ððð
ð ðâšð ð
Where
ð ,ð
ðð=
ðððð
ðð¥ð +ððŒð ððŒð
+ ðððŒ ððŒð
or ðððð
ðð¥ð +ððŒð ÎðŒðð + ðððŒÎðŒð
ð
are the components of the covariant derivative of the tensor ð in terms of contravariant components on the product covariant bases as shown.
Higher Order Tensors
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 83
In the same way, by taking the tensor expression in the dyadic form of its contravariant product bases, we can write,
ðð
ðð¥ð =ðððð
ðð¥ð ð ð â ð ð +ððð
ðð ð
ðð¥ð â ð ð +ðððð ð â
ðð ð
ðð¥ð
=ðððð
ðð¥ð ð ð â ð ð +ðððÎðŒðð â ð ð +ðððð
ð âðð ð
ðð¥ð
Again, notice from previous derivation above, ððð
= âðð ð
ðð¥ð â ð ð so that, ðð ð
ðð¥ð = âð
ðŒðð ðŒ = âÎðŒð
ð ð ðŒ Therefore,
ðð
ðð¥ð=
ðððð
ðð¥ðð ð â ð ð +ððð
ðð ð
ðð¥ðâ ð ð +ðððð
ð âðð ð
ðð¥ð
=ðððð
ðð¥ðð ð â ð ð âðððÎðŒð
ð ð ðŒ â ð ð + ðððð ð â ÎðŒð
ðð ðŒ
=ðððð
ðð¥ðâ ððŒðÎðð
ðŒ â ðððŒÎðððŒ ð ð â ð ð= ððð,ðð ð â ð ð
So that
ððð,ð =ðððð
ðð¥ð â ððŒðÎðððŒ â ðððŒÎðð
ðŒ
Higher Order Tensors
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 84
Two other expressions can be found for the covariant derivative components in terms of the mixed tensor components using the mixed product bases defined above. It is a good exercise to derive these.
The formula for covariant differentiation of higher order tensors follow the same kind of logic as the above definitions. Each covariant index will produce an additional term similar to that in 3 with a dummy index supplanting the appropriate covariant index. In the same way, each contravariant index produces an additional term like that in 3 with a dummy index supplanting an appropriate contravariant index.
Higher Order Tensors
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 85
The covariant derivative of the mixed tensor, ðŽð1 ,ð2 ,âŠ,ðð
ð1 ,ð2 ,âŠ,ðð is the
most general case for the covariant derivative of an absolute tensor:
ðŽð1 ,ð2 ,âŠ,ðð
ð1 ,ð2 ,âŠ,ðð
,ð
= ððŽð1 ,ð2 ,âŠ,ðð
ð1 ,ð2 ,âŠ,ðð
ðð¥ðâ
ðŒð1ð
ðŽðŒ ,ð2 ,âŠ,ðð
ð1 ,ð2 ,âŠ,ðð â ðŒð2ð
ðŽð1,ðŒ,âŠ,ðð
ð1 ,ð2 ,âŠ,ðð â ⯠âðŒððð ðŽð1 ,ð2 ,âŠ,ðŒ
ð1 ,ð2 ,âŠ,ðð
+ ð1ðœð
ðŽð1 ,ð2 ,âŠ,ðð
ðœ ,ð2 ,âŠ,ðð + ð2ðœð
ðŽð1 ,ð2 ,âŠ,ðð
ð1 ,ðœ,âŠ,ðð + â¯+ ðððœð
ðŽð1 ,ð2 ,âŠ,ðð
ð1 ,ð2,âŠ,ðœ
Higher Order Mixed Tensors
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 86
The metric tensor components behave like constants under a covariant differentiation. The proof that this is so is due to Ricci:
ððð ,ð =ðððð
ðð¥ðâ
ðŒðð
ððŒð âðœðð
ðððœ
=ðððð
ðð¥ðâ ðð, ð â ðð, ð
=ðððð
ðð¥ðâ
1
2
ðððð
ðð¥ð+
ðððð
ðð¥ðâ
ðððð
ðð¥ðâ
1
2
ðððð
ðð¥ð+
ðððð
ðð¥ðâ
ðððð
ðð¥ð
= 0.
Ricciâs Theorem
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 87
The conjugate metric tensor behaves the same way as can be seen from the relationship,
ðððððð = ð¿ð
ð
The above can be differentiated covariantly with respect to ð¥ð to obtain
ððð ,ð ððð + ðððððð ,ð = ð¿ð
ð,ð
0 + ðððððð ,ð =
ðð¿ðð
ðð¥ð âðŒðð
ð¿ðŒð
+ð
ðŒðð¿ð
ðŒ
ðððððð ,ð = 0 â
ððð
+ððð
= 0
The contraction of ððð with ððð ,ð vanishes. Since we know that the metric tensor cannot vanish in general, we can only conclude that
ððð ,ð = 0
Ricciâs Theorem
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 88
Which is the second part of the Ricci theorem. With these two, we can treat the metric tensor as well as its conjugate as constants in covariant differentiation. Notice that along with this proof, we also obtained the result that the Kronecker Delta vanishes under a partial derivative (an obvious fact since it is a constant), as well as under the covariant derivative. We summarize these results as follows:
The metric tensors ðððand ððð as well as the alternating tensors ðððð , ðððð are all constants under a covariant
differention. The Kronecker delta ð¿ðð is a constant under
both covariant as well as the regular partial differentiation
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 89
The divergence theorem is central to several other results in Continuum Mechanics. We present here a generalized form [Ogden] which states that,
Gauss Divergence Theorem
For a tensor field ðµ, The volume integral in the region Ω â â°,
grad ðµÎ©
ðð£ = ðµ â ð§ðΩ
ðð
where ð§ is the outward drawn normal to ðΩ â the boundary of Ω.
Integral Theorems
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 90
Vector field. Replacing the tensor with the vector field ð and contracting, we have,
div ðΩ
ðð£ = ð â ð§ðΩ
ðð
Which is the usual form of the Gauss theorem.
Special Cases: Vector Field
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 91
For a scalar field ð, the divergence becomes a gradient and the scalar product on the RHS becomes a simple multiplication. Hence the divergence theorem becomes,
grad ðΩ
ðð£ = ðð§ðΩ
ðð
The procedure here is valid and will become obvious if we write, ð = ðð where ð is an arbitrary constant vector.
Scalar Field
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 92
div ððΩ
ðð£ = ðð â ð§ðΩ
ðð = ð â ðð§ðΩ
ðð
For the LHS, note that, div ðð = tr grad ðð
grad ðð = ððð ,ð ð ð â ð ð
= ððð,ð ð ð â ð ð
The trace of which is,
ððð,ð ð ð â ð ð = ððð,ð ð¿ðð
= ððð,ð = ð â grad ð
For the arbitrary constant vector ð, we therefore have that,
div ððΩ
ðð£ = ð â grad ð Ω
ðð£ = ð â ðð§ðΩ
ðð
grad ð Ω
ðð£ = ðð§ðΩ
ðð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 93
For a second-order tensor ð, the Gauss Theorem becomes,
divðΩ
ðð£ = ðð§ðΩ
ðð
The original outer product under the integral can be expressed in dyadic form:
grad ðΩ
ðð£ = ððð ,ð ð ð â ð ð â ð ð
Ω
ðð£
= ð â ð§ðΩ
ðð
= ðððð ð â ð ð â ððð ð
ðΩ
ðð
Second-Order Tensor field
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 94
Or
ððð ,ð ð ð â ð ð âΩ
ð ððð£ = ðððððð ð â ð ð â ð ð
ðΩ
ðð
Contracting, we have
ððð ,ðΩ
ð ð â ð ð ð ððð£ = ððððð ð ð â ð ð ð ð
ðΩ
ðð
ððð ,ð ð¿ððð ð
Ω
ðð£ = ðððððð¿ððð ð
ðΩ
ðð
ððð ,ð ð ðΩ
ðð£ = ðððððð ððΩ
ðð
Which is the same as,
divðΩ
ðð£ = ðð§ðΩ
ðð
Second-Order Tensor field
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 95
The components of a vector or tensor in a Cartesian system are projections of the vector on directions that have no dimensions and a value of unity.
These components therefore have the same units as the vectors themselves.
It is natural therefore to expect that the components of a tensor have the same dimensions.
In general, this is not so. In curvilinear coordinates, components of tensors do not necessarily have a direct physical meaning. This comes from the fact that base vectors are not guaranteed to have unit values (âð â 1 in general).
Physical Components of Tensors
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 96
They may not be dimensionless. For example, in orthogonal spherical polar, the base vectors are ð 1, ð 2 and ð 3.
These can be expressed in terms of dimensionless unit vectors as, ððð, ð sin ð ðð , and ðð since the
magnitudes of the basis vectors are ð, ð sinð , and 1
or ð11, ð22, ð33 respectively.
As an example consider a force with the contravariant components ð¹1, ð¹2 and ð¹3,
ð = ð¹1ð 1 + ð¹2ð 2 + ð¹3 ð 3
= ðð¹1ðð + ð sin ð ð¹2ðð + ð¹3ðð
Physical Components
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 97
Which may also be expressed in terms of physical components,
ð = ð¹ððð + ð¹ððð + ð¹ððð.
While these physical components {ð¹ð, ð¹ð , ð¹ð} have the
dimensions of force, for the contravariant components normalized by the in terms of the unit vectors along these axes to be consistent, {ðð¹1, ð sin ð ð¹2, ð¹3} must each be in the units of a force. Hence, ð¹1, ð¹2 and ð¹3 may not themselves be in force units. The consistency requirement implies, ð¹ð = ðð¹1, ð¹ð = ð sin ð ð¹2, and ð¹ð = ð¹3
Physical Components
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 98
For the same reasons, if we had used covariant components, the relationships,
ð¹ð = ð¹1
ð, ð¹ð =
ð¹2
ð sin ð, and ð¹ð = ð¹3
The magnitudes of the reciprocal base vectors are 1
ð,
1
ð sin ð, and 1. While the physical components have the
dimensions of force, ð¹1 = ðð¹ð and ð¹2 = ð sin ð ð¹ð have the
dimensions of moment, while ð¹1 =ð¹ð
ð and ð¹2 =
ð¹ð
ð sin ð are
in dimensions of force per unit length. Only the third components in both cases are given in the dimensions of force.
Physical Components
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 99
The physical components of a vector or tensor are components that have physically meaningful units and magnitudes. Often it is convenient to derive the governing equations for a problem in terms of the tensor components but to solve the problem in physical components.
In an orthogonal system of coordinates, to obtain a physical component from a tensor component we must divide by the magnitude of the relevant coordinate for each covariant index and multiply by each contravariant index. To illustrate this point, consider the evaluation of the physical
components of the symmetric tensor components ððð or ððð or
ððð in spherical polar coordinates. Here, as we have seen, the
magnitudes of the base vectors ð11, ð22, ð33 or â1, â2 and â3 are ð, ð sin ð and 1. Using the rule specified above, the table below computes the physical components from the three associated tensors as follows:
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 100
Physical
Component
Contravariant Covariant Mixed
ð(ðð) ð11â1â1 = ð11ð2 ð11
â1 â1=
ð11
ð2 ð1
1
â1â1 = ð1
1
ð(ððœ) ð12â1â2 = ð12ð2 sinð ð12
â1 â1=
ð12
ð2 sin ð ð2
1
â2â1 =
ð21
sin ð
ð(ðœðœ) ð22â2â2 = ð22ð2 sin2 ð ð22
â2 â2=
ð22
ð2 sin2 ð ð2
2
â2â2 = ð2
2
ð(ðœð) ð23â2â3 = ð23ð sin ð ð23
â2 â3=
ð23
ð sin ð ð3
2
â3â2 = ð3
2ð sin ð
ð(ðð) ð33â3â3 = ð33 ð33
â3 â3= ð33 ð3
3
â3â3 = ð3
3
ð(ðð) ð31â3â1 = ð31ð ð31
â3 â1=
ð31
ð ð1
3
â1â3 =
ð13
ð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 101
17. The transformation equations from the Cartesian to the oblate spheroidal coordinates ð, ðŒ and ð are: ð = ðððŒ ð¬ð¢ð§ð, ð = ð ðð â ð ð â ðŒð , and ð = ðððŒ ððšð¬ ð, where f is a constant representing the half the distance between the foci of a family of confocal ellipses. Find the components of the metric tensor in this system. The metric tensor components are:
ððð =ðð¥
ðð
2
+ððŠ
ðð
2
+ðð§
ðð
2
= ðð sinð 2 + ð2ð21 â ð2
ð2 â 1+ ðð cos ð 2 = ð2
ð2 â ð2
ð2 â 1
ððð =ðð¥
ðð
2
+ððŠ
ðð
2
+ðð§
ðð
2
= ð2ð2 â ð2
1 â ð2
ððð =ðð¥
ðð
2
+ððŠ
ðð
2
+ðð§
ðð
2
= ððð 2
ððð =ðð¥
ðð
ðð¥
ðð+
ððŠ
ðð
ððŠ
ðð+
ðð§
ðð
ðð§
ðð
= ðð sin ð ðð sinð â ððð2 â 1
1 â ð2ðð
1 â ð2
ð2 â 1+ ðð cos ð ðð cos ð
= 0 = ððð = ððð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 102
Find an expression for the divergence of a vector in orthogonal curvilinear coordinates.
The gradient of a vector ð = ð¹ðð ð is ð» â ð = ð ððð â ð¹ðð ð =ð¹ð
,ðð ð â ð ð. The divergence is the contraction of the gradient. While
we may use this to evaluate the divergence directly it is often easier to use the computation formula in equation Ex 15:
ð» â ð = ð¹ð,ðð
ð â ð ð = ð¹ð ,ð =1
ð
ð ð ð¹ð
ðð¥ð
=1
â1â2â3
ð
ðð¥1â1â2â3ð¹
1 +ð
ðð¥2â1â2â3ð¹
2 +ð
ðð¥3â1â2â3ð¹
3
Recall that the physical (components having the same units as the tensor in question) components of a contravariant tensor are not equal to the tensor components unless the coordinate system is Cartesian. The physical component ð¹(ð) = ð¹ðâð (no sum on i). In terms of the physical components therefore, the divergence becomes,
=1
â1â2â3
ð
ðð¥1â2â3ð¹(1) +
ð
ðð¥2â1â3ð¹(2) +
ð
ðð¥3â1â2ð¹(3)
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 103
Find an expression for the Laplacian operator in Orthogonal coordinates. For the contravariant component of a vector, ð¹ð,
ð¹ð,ð =
1
ð
ð ð ð¹ð
ðð¥ð.
Now the contravariant component of gradient ð¹ð = ðððð,ð. Using this in place of the vector ð¹ð, we can write,
ðððð,ðð =1
ð
ð ð ðððð,ð
ðð¥ð
given scalar ð, the Laplacian ð»2ð is defined as, ðððð,ðð so that,
ð»2ð = ðððð,ðð =1
ð
ð
ðð¥ðð ððð
ðð
ðð¥ð
When coordinates are orthogonal, ððð = ððð = 0 whenever ð â ð. Expanding the computation formula therefore, we can write,
ð»2ð =1
â1â2â3
ð
ðð¥1
â1â2â3
â1
ðð
ðð¥1 +ð
ðð¥2
â1â2â3
â2
ðð
ðð¥2 +ð
ðð¥3
â1â2â3
â3
ðð
ðð¥3
=1
â1â2â3
ð
ðð¥1 â2â3
ðð
ðð¥1 +ð
ðð¥2 â1â3
ðð
ðð¥2 +ð
ðð¥3 â1â2
ðð
ðð¥3
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 104
Show that the oblate spheroidal coordinate systems are orthogonal. Find an expression for the Laplacian of a scalar function in this system.
Example above shows that ððð = ððð = ððð = 0 . This is the required proof of orthogonality. Using the computation formula in example 11, we may write for the oblate spheroidal coordinates that,
ð»2Ί =ð2 â 1 1 â ð2
ð3ð2 ð2 â ð2 ð
ððððð
ð2 â 1
1 â ð2
ðΊ
ðð
+ð
ððððð
1 â ð2
ð2 â 1
ðΊ
ðð
+ð
ðð
ð ð2 â ð2
ðð ð2 â 1 1 â ð2
ðΊ
ðð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 105
For a vector field ð, show that grad ð Ã is a third ranked tensor. Hence or otherwise show that div ð Ã = âcurl ð.
The secondâorder tensor ð à is defined as ððððð¢ðð ð âð ð. Taking the covariant derivative with an independent base, we have
grad ð à = ððððð¢ð ,ð ð ð â ð ð â ð ð
This gives a third order tensor as we have seen. Contracting on the last two bases,
div ð à = ððððð¢ð ,ð ð ð â ð ð â ð ð = ððððð¢ð ,ð ð ðð¿ð
ð = ððððð¢ð ,ð ð ð = âcurl ð
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 106
Begin with our familiar Cartesian system of coordinates. We can represent the position of a point (position vector) with three coordinates ð¥, ðŠ, ð§ (â R) such that,
ð« = ð¥ð + ðŠð + ð§ð
That is, the choice of any three scalars can be used to locate a point. We now introduce a transformation (called a polar transformation) of ð¥, ðŠ â {ð, ð} such that, ð¥ = ð cos ð, and ðŠ = ð sin ð. Note also that this
transformation is invertible: ð = ð¥2 + ðŠ2,and
ð = tanâ1 ðŠ
ð¥
Coordinate transformations
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With such a transformation, we can locate any point in the 3-D space with three scalars {ð, ð, ð§} instead of our previous set ð¥, ðŠ, ð§ . Our position vector is now,
ð« = ð cos ð ð + ð sin ð ð + ð§ð = ððð + ð§ðð§
where we define ðð â¡ cos ð ð + sin ð ð, ðð§ is no different from ð. In order to complete our triad of basis vectors, we need a third vector, ðð. In selecting
ðð, we want it to be such that ðð , ðð, ðð§ can form an
orthonormal basis.
Curvilinear Coordinates
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Let ðð = ðð + ðð
To satisfy our conditions,
ðð â ðð = 0, ðð â ðð§ = 0, and ð2 + ð2 = 1.
It is easy to see that ðð â¡ âsin ð ð + cos ð ð satisfies
these requirements. ðð , ðð, ðð§ forms an orthonormal (that is, each member has unit magnitude and they are pairwise orthogonal) triad just like ð, ð, ð . The transformation we have just described
can be given a geometric interpretation. In either case, it is the definition of the Cylindrical Polar coordinate system.
Cylindrical Polar coordinate system
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 109
Unlike our Cartesian system, we note that
ðð(ð), ðð(ð), ðð§ as the first two of these are not
constants but vary with angular orientation. ðð§ remains a constant vector as in the Cartesian case.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 110
Continuing further with our transformation, we may again introduce two new scalars such that ð, ð§ â {ð, ð} in such a way that the position vector,
ð« = ððð + ð§ðð§ = ð sin ð ðð + ð cos ð ðð§ â¡ ððð
Here, ð = ð sin ð , ð§ = ð cos ð. As before, we can use three scalars, ð, ð, ð instead of {ð, ð, ð§}. In comparison to the original Cartesian system we began with, we have that,
ð« = ð¥ð + ðŠð + ð§ð = ð sin ð ðð + ð cos ð ðð§ = ð sin ð cos ð ð + sin ð ð + ð cos ð ð = ð sin ð cos ð ð +ð sin ð sinð ð + ð cos ð ð â¡ ððð
from which it is clear that the unit vector ðð â¡ sin ð cos ð ð + sin ð sin ð ð + cos ð ð.
Spherical Polar
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 111
Again, we introduce the unit vector, ðð â¡ cos ð cos ð ð + cos ð sinð ð â sin ð ð and retain ðð = â sinð ð + cos ð ð as before.
It is easy to demonstrate the fact that these vectors constitute another orthonormal set. Combining the two transformations, we can move from ð¥, ðŠ, ð§ system of coordinates to ð, ð, ð directly by the transformation equations, ð¥ = ð sin ð cos ð, ðŠ = ð sin ð sinð and ð§ =ð cos ð.
The orthonormal set of basis for the ð, ð, ð system is
ðð ð, ð , ðð ð, ð , ðð ð .
This is the Spherical Polar Coordinate System.
Spherical Polar
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 112
There two main points to note in transforming from Cartesian to Cylindrical or Spherical polar coordinate systems. The latter two (also called curvilinear systems) have unit basis vector sets that are dependent on location. Explicitly, we may write,
ðð ð, ð, ð§ = cos ð ð + sin ð ð ðð(ð, ð, ð§) = âsin ð ð + cos ð ð ðð§ ð, ð, ð§ = ð
Variable Bases
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 113
For Cylindrical Polar, and for Spherical Polar,
ðð(ð, ð, ð) = sin ð cos ð ð + sin ð sinð ð + cos ð ð
ðð ð, ð, ð = cos ð cos ð ð + cos ð sinð ð â sin ð ð ðð ð, ð, ð = â sinð ð + cos ð ð.
Of course, there are specific cases in which some of these basis vectors are constants as we can see. It is instructive to note that curvilinear (so called because coordinate lines are now curves rather than straight lines) coordinates generally have basis vectors that depend on the coordinate variables.
Unlike the Cartesian system, we will no longer be able to assume that the derivatives of the basis vectors vanish.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 114
The second point to note is more subtle. While it is true that a vector referred to a curvilinear basis will have coordinate projections in each of the basis so that, a typical vector
ð¯ = ð£1ð 1 + ð£2ð 2 + ð£3ð 3
when referred to the basis vectors {ð 1, ð 2, ð 3}, for position vectors, in order that the transformation refers to the same position vector we started with in the Cartesian system
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 115
We have for Cylindrical Polar, the vector, ð«(ð, ð, ð§) = ððð(ð) + ð§ðð§
and for Spherical Polar, ð«(ð, ð, ð) = ððð(ð, ð)
Expressing position vectors in Curvilinear coordinates must be done carefully. We do well to take into consideration the fact that in curvilinear systems, the basis vectors are not fully specified until they are expressly specified with their locational dependencies.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 116
It is only in this situation that we can express the position vector of a specific location Spherical and Cylindrical Polar coordinates are two more commonly used curvilinear systems. Others will be introduced as necessary in due course.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 117
Beginning from the position vector, given a system with coordinate variables ðŒ1, ðŒ2, ðŒ3 , it is easy to prove that the set of vectors,
ðð« ðŒ1, ðŒ2, ðŒ3
ððŒ1,ðð« ðŒ1, ðŒ2, ðŒ3
ððŒ1,ðð« ðŒ1, ðŒ2, ðŒ3
ððŒ1
Which we can write more compactly as
ð ð â¡ðð«
ððŒð, ð = 1,2,3
constitute an independent set. This set, with its dual set
of vectors, ð ð such that ð ð â ð ð= ð¿ðð, constitute the
âNatural basesâ for the coordinate system.
oafak@unilag.edu.ng 12/27/2012 Dept of Systems Engineering, University of Lagos 118
Natural Bases
We can only guarantee the independence of the bases for any arbitrary system. They may NOT be normalized neither are they guaranteed of mutual orthogonality.
HW. Show that the natural bases for Spherical and Cylindrical polar systems are orthogonal but not normalized. And that the dual natural bases for the Cartesian system coincide.
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Orthonormality of Natural Bases
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