3. tensor calculus jan 2013
DESCRIPTION
Continuum Mechanics,Linear Spaces,Prof OA Fakinlede,SSG 815,Tensor Analysis,UNILAG,University of Lagos Nigeria,Vectors,Tensor CalculusTRANSCRIPT
Tensor Calculus Differentials & Directional Derivatives
We are presently concerned with Inner Product spaces in our treatment of the Mechanics of Continua. Consider a map,
๐ญ: V โ W This maps from the domainV to W โ 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
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 2
Let ๐0 โ V and ๐0 โ W, as usual we can say that the limit
lim๐โ ๐0
๐ญ ๐ = ๐0
if for any pre-assigned real number ๐ > 0, no matter how small, we can always find a real number ๐ฟ > 0 such that ๐ญ ๐ โ ๐0 โค ๐ whenever ๐ โ ๐0 < ๐ฟ.
The function is said to be continuous at ๐0 if ๐ญ ๐0 exists and ๐ญ ๐0 = ๐0
The Gateaux Differential
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 3
Specifically, for ๐ผ โ โ let this map be:
๐ท๐ญ ๐, ๐ โก lim๐ผโ0
๐ญ ๐ + ๐ผ๐ โ ๐ญ ๐
๐ผ=
๐
๐๐ผ๐ญ ๐ + ๐ผ๐
๐ผ=0
We focus attention on the second variable โ while we allow the dependency on ๐ to be as general as possible. We shall show that while the above function can be any given function of ๐ (linear or nonlinear), the above map is always linear in ๐ irrespective of what kind of Euclidean space we are mapping from or into. It is called the Gateaux Differential.
The Gateaux Differential
[email protected] 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 โ โ ๐๐ฅ and we obtain,
๐ท๐น(๐ฅ, ๐๐ฅ) = lim๐ผโ0
๐น ๐ฅ + ๐ผ๐๐ฅ โ ๐น ๐ฅ
๐ผ=
๐
๐๐ผ๐น ๐ฅ + ๐ผ๐๐ฅ
๐ผ=0
And let ๐ผ๐๐ฅ โ ฮ๐ฅ, the middle term becomes,
limฮ๐ฅโ0
๐น ๐ฅ + ฮ๐ฅ โ ๐น ๐ฅ
ฮ๐ฅ๐๐ฅ =
๐๐น
๐๐ฅ๐๐ฅ
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
[email protected] 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 ๐ผ โ R
๐ท๐ญ ๐, ๐ผ๐ = ๐ผ๐ท๐ญ ๐, ๐
Furthermore, ๐ท๐ญ ๐, ๐ + ๐ = ๐ท๐ญ ๐, ๐ + ๐ท๐ญ ๐, ๐
and that for ๐ผ, ๐ฝ โ R ๐ท๐ญ ๐, ๐ผ๐ + ๐ฝ๐ = ๐ผ๐ท๐ญ ๐, ๐ + ๐ฝ๐ท๐ญ ๐, ๐
The Gateaux Differential
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 6
The Frechรฉt derivative or gradient of a differentiable function is the linear operator, grad ๐ญ(๐) such that, for ๐๐ โ V,
๐๐ญ ๐
๐๐๐๐ โก grad ๐ญ ๐ ๐๐ โก ๐ท๐ญ ๐, ๐๐
Obviously, grad ๐ญ(๐) 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 Frechรฉt Derivative
[email protected] 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:
๐: V โ R The Gateaux differential in this case takes two vector arguments and maps to a real value:
๐ท๐: V ร V โ R This is the classical definition of the inner product so that we can define the differential as,
๐๐(๐)
๐๐โ ๐๐ โก ๐ท๐ ๐, ๐๐
This quantity, ๐๐(๐)
๐๐
defined by the above equation, is clearly a vector.
Differentiable Real-Valued Function
[email protected] 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 ๐ ๐ โ V. On such a basis, the function,
๐ = ๐ฃ๐๐ ๐
and,
๐ ๐ = ๐ ๐ฃ๐๐ ๐
we may also express the independent vector ๐๐ = ๐๐ฃ๐๐ ๐
on this basis.
Real-Valued Function
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 9
The Gateaux differential in the direction of the basis vector ๐ ๐ is,
๐ท๐ ๐, ๐ ๐ = lim๐ผโ0
๐ ๐ + ๐ผ๐ ๐ โ ๐ ๐
๐ผ= lim
๐ผโ0
๐ ๐ฃ๐๐ ๐ + ๐ผ๐ ๐ โ ๐ ๐ฃ๐๐ ๐
๐ผ
= lim๐ผโ0
๐ ๐ฃ๐ + ๐ผ๐ฟ๐๐
๐ ๐ โ ๐ ๐ฃ๐๐ ๐
๐ผ
As the function ๐ does not depend on the vector basis, we can substitute the vector function by the real function of the components ๐ฃ1, ๐ฃ2, ๐ฃ3 so that,
๐ ๐ฃ๐๐ ๐ = ๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
in which case, the above differential, similar to the real case discussed earlier becomes,
๐ท๐ ๐, ๐ ๐ =๐๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐
Real-Valued Function
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 10
Of course, it is now obvious that the gradient ๐๐(๐)
๐๐ of
the scalar valued vector function, expressed in its dual basis must be,
๐๐(๐)
๐๐=
๐๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐
so that we can recover,
๐ท๐ ๐, ๐ ๐ =๐๐ ๐
๐๐โ ๐๐ =
๐๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ โ ๐ ๐
=๐๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ฟ๐
๐=
๐๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐
Hence we obtain the well-known result that
grad ๐ ๐ =๐๐(๐)
๐๐=
๐๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐
which defines the Frechรฉt derivative of a scalar valued function with respect to its vector argument.
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 11
The Gradient of a Differentiable Vector valued Function
The definition of the Frechรฉt 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 ๐ ๐ โ V. On such a basis, the function,
๐ญ ๐ = ๐น๐ ๐ ๐ ๐
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,
๐ญ ๐ = ๐น๐ ๐ฃ1, ๐ฃ2, ๐ฃ3 ๐ ๐
Vector valued Function
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 12
๐ท๐ญ ๐, ๐๐ = ๐ท๐ญ ๐ฃ๐๐ ๐ , ๐๐ฃ๐๐ ๐
= ๐ท๐ญ ๐ฃ๐๐ ๐ , ๐ ๐ ๐๐ฃ๐
= ๐ท๐น๐ ๐ฃ๐๐ ๐ , ๐ ๐ ๐ ๐๐๐ฃ๐
Again, upon noting that the functions ๐ท๐น๐ ๐ฃ๐๐ ๐ , ๐ ๐ ,
๐ = 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,
๐ท๐ญ ๐, ๐๐ = ๐ท๐น๐ ๐ฃ๐๐ ๐ , ๐ ๐ ๐ ๐๐๐ฃ๐ = ๐ท๐น๐ ๐, ๐ ๐ ๐ ๐๐๐ฃ๐
Vector Derivatives
[email protected] 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,
๐ท๐ญ ๐, ๐๐ =๐๐น๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ โ ๐ ๐ ๐ ๐๐๐ฃ๐
=๐๐น๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ โ ๐ ๐ ๐ ๐๐๐ฃ๐
=๐๐น๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ โ ๐ ๐ ๐๐
= ๐๐ญ ๐
๐๐๐๐ โก grad ๐ญ ๐ ๐๐
Clearly, the tensor gradient of the vector-valued vector function is,
๐๐ญ ๐
๐๐= grad ๐ญ ๐ =
๐๐น๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ โ ๐ ๐
Where due attention should be paid to the covariance of the quotient indices in contrast to the contravariance of the non quotient indices. [email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 14
The trace of this gradient is called the divergence of the function:
div ๐ญ ๐ = tr๐๐น๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ โ ๐ ๐
=๐๐น๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ โ ๐ ๐
=๐๐น๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ฟ๐
๐
=๐๐น๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐
The Trace & Divergence
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 15
Consider the mapping, ๐ป: V โ 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:
๐ท๐ป: V ร V โ T With the usual notation, we may write,
๐ท๐ป ๐, ๐๐ = ๐ท๐ป ๐ฃ๐๐ ๐ , ๐๐ฃ๐๐ ๐ = ๐ท๐ป ๐ฃ๐๐ ๐ , ๐๐ฃ๐๐ ๐
= ๐๐ป ๐
๐๐๐๐ โก grad ๐ป ๐ ๐๐
= ๐ท๐๐ผ๐ฝ ๐ฃ๐๐ ๐, ๐๐ฃ๐๐ ๐ ๐ ๐ผ โ ๐ ๐ฝ
= ๐ท๐๐ผ๐ฝ ๐, ๐ ๐ ๐ ๐ผ โ ๐ ๐ฝ ๐๐ฃ๐
Each of these nine functions look like the real differential of several variables we considered earlier.
Tensor-Valued Vector Function
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 16
The independence of the basis vectors imply, as usual, that
๐ท๐๐ผ๐ฝ ๐, ๐ ๐ =๐๐๐ผ๐ฝ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐
so that,
๐ท๐ป ๐, ๐๐ = ๐ท๐๐ผ๐ฝ ๐, ๐ ๐ ๐ ๐ผ โ ๐ ๐ฝ ๐๐ฃ๐
= ๐๐๐ผ๐ฝ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ผ โ ๐ ๐ฝ ๐๐ฃ๐
= ๐๐๐ผ๐ฝ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ผ โ ๐ ๐ฝ ๐ ๐ โ ๐๐
= ๐๐๐ผ๐ฝ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ผ โ ๐ ๐ฝ โ ๐ ๐ ๐๐ =
๐๐ป ๐
๐๐๐๐
โก grad ๐ป ๐ ๐๐
Which defines the third-order tensor,
๐๐ป ๐
๐๐= grad ๐ป ๐ =
๐๐๐ผ๐ฝ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ผ โ ๐ ๐ฝ โ ๐ ๐
and with no further ado, we can see that a third-order tensor transforms a vector into a second order tensor.
[email protected] 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 ๐ป ๐ =๐๐๐ผ๐ฝ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ผ โ ๐ ๐ฝ โ ๐ ๐
=๐๐๐ผ๐ฝ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ผ ๐ ๐ฝ โ ๐ ๐
=๐๐๐ผ๐ฝ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ผ๐ฟ๐ฝ
๐ =๐๐๐ผ๐ ๐ฃ1, ๐ฃ2, ๐ฃ3
๐๐ฃ๐๐ ๐ผ
It can easily be shown that this is the particular vector that gives, for all constant vectors ๐,
div ๐ป ๐ โก div ๐ปT๐
The Divergence of a Tensor Function
[email protected] 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 Frechรฉt derivatives will be extracted once we can obtain the Gateaux differentials.
Real-Valued Tensor Functions
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 19
Consider the Map: ๐: T โ R
The Gateaux differential in this case takes two vector arguments and maps to a real value:
๐ท๐: T ร T โ R Which is, as usual, linear in the second argument. The Frechรฉt derivative can be expressed as the first component of the following scalar product:
๐ท๐ ๐ป, ๐๐ป =๐๐(๐ป)
๐๐ป: ๐๐ป
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.
Frechรฉt Derivative
[email protected] 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,
๐๐(๐ป)
๐๐ป=
๐๐(๐11, ๐12, โฆ , ๐33)
๐๐๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ
In the dual to the same basis, we can write,
๐๐ป = ๐๐๐๐๐ ๐ โ ๐ ๐
Clearly, ๐๐(๐ป)
๐๐ป: ๐๐ป =
๐๐(๐11, ๐12, โฆ , ๐33)
๐๐๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ : ๐๐๐๐๐ ๐ โ ๐ ๐
=๐๐(๐11, ๐12, โฆ , ๐33)
๐๐๐๐๐๐๐๐
Again, note the covariance of the quotient indices.
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 21
Let ๐ be a symmetric and positive definite tensor and let ๐ผ1 ๐บ , ๐ผ2 ๐บ and ๐ผ3 ๐บ be the three principal invariants of
๐ show that (a) ๐๐ฐ1 ๐
๐๐= ๐ the identity tensor, (b)
๐๐ฐ2 ๐
๐๐= ๐ผ1 ๐ ๐ โ ๐ and (c)
๐๐ผ3 ๐
๐๐= ๐ผ3 ๐ ๐โ1
๐๐ฐ1 ๐
๐๐ can be written in the invariant component form
as, ๐๐ผ1 ๐
๐๐=
๐๐ผ1 ๐
๐๐๐๐
๐ ๐ โ ๐ ๐
Examples
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 22
Recall that ๐ผ1 ๐ = tr ๐ = ๐ฮฑฮฑ hence
๐๐ผ1 ๐
๐๐=
๐๐ผ1 ๐
๐๐๐๐
๐ ๐ โ ๐ ๐
=๐๐ฮฑ
ฮฑ
๐๐๐๐๐ ๐ โ ๐ ๐
= ๐ฟ๐ผ๐ ๐ฟ๐
๐ผ๐ ๐โ ๐ ๐
= ๐ฟ๐๐๐ ๐โ ๐ ๐ = ๐
which is the identity tensor as expected.
(a) Continued
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 23
๐๐ผ2 ๐
๐๐ in a similar way can be written in the invariant component form
as, ๐๐ผ2 ๐
๐๐=
1
2
๐๐ผ1 ๐
๐๐๐๐
๐ฮฑฮฑ๐๐ฝ
๐ฝโ ๐๐ฝ
ฮฑ๐ฮฑ๐ฝ
๐ ๐ โ ๐ ๐
where we have utilized the fact that ๐ผ2 ๐ =1
2tr2 ๐ โ tr(๐2) .
Consequently, ๐๐ผ2 ๐
๐๐=
1
2
๐
๐๐๐๐
๐ฮฑฮฑ๐๐ฝ
๐ฝโ ๐๐ฝ
ฮฑ๐ฮฑ๐ฝ
๐ ๐ โ ๐ ๐
=1
2๐ฟ๐ผ
๐ ๐ฟ๐๐ผ๐๐ฝ
๐ฝ+ ๐ฟ๐ฝ
๐ ๐ฟ๐๐ฝ๐ฮฑ
ฮฑ โ ๐ฟ๐ฝ๐ ๐ฟ๐
๐ผ๐ฮฑ๐ฝ
โ ๐ฟ๐ผ๐ ๐ฟ๐
๐ฝ๐๐ฝ
ฮฑ ๐ ๐ โ ๐ ๐
=1
2๐ฟ๐
๐๐๐ฝ๐ฝ
+ ๐ฟ๐๐๐ฮฑ
ฮฑ โ ๐๐๐โ ๐๐
๐๐ ๐ โ ๐ ๐
= ๐ฟ๐๐๐ฮฑ
ฮฑ โ ๐๐๐
๐ ๐ โ ๐ ๐ = ๐ผ1 ๐ ๐ โ ๐
(b) ๐๐ผ2 ๐
๐๐= ๐ผ1 ๐ ๐ โ ๐
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 24
๐๐ฐ3 ๐
๐๐=
๐det ๐
๐๐= ๐๐
the cofactor of ๐. Clearly ๐๐ = det ๐ ๐โ1 = ๐ฐ3 ๐ ๐โ1 as required. Details of this for the contravariant components of a tensor is presented below. Let
๐ = ๐ =1
3!๐๐๐๐๐๐๐ ๐ก๐๐๐๐๐๐ ๐๐๐ก
Differentiating wrt ๐๐ผ๐ฝ, we obtain,
๐๐
๐๐๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ=
1
3!๐๐๐๐๐๐๐ ๐ก
๐๐๐๐
๐๐๐ผ๐ฝ๐๐๐ ๐๐๐ก + ๐๐๐
๐๐๐๐
๐๐๐ผ๐ฝ๐๐๐ก + ๐๐๐๐๐๐
๐๐๐๐ก
๐๐๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ
=1
3!๐๐๐๐๐๐๐ ๐ก ๐ฟ๐
๐ผ๐ฟ๐๐ฝ๐๐๐ ๐๐๐ก + ๐๐๐๐ฟ๐
๐ผ๐ฟ๐ ๐ฝ๐๐๐ก + ๐๐๐๐๐๐ ๐ฟ๐
๐ผ๐ฟ๐ก๐ฝ
๐ ๐ผ โ ๐ ๐ฝ
=1
3!๐๐ผ๐๐๐๐ฝ๐ ๐ก ๐๐๐ ๐๐๐ก + ๐๐๐ ๐๐๐ก + ๐๐๐ ๐๐๐ก ๐ ๐ผ โ ๐ ๐ฝ
=1
2!๐๐ผ๐๐๐๐ฝ๐ ๐ก๐๐๐ ๐๐๐ก๐ ๐ผ โ ๐ ๐ฝโก ๐c ๐ผ๐ฝ๐ ๐ผ โ ๐ ๐ฝ
Which is the cofactor of ๐๐ผ๐ฝ
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 25
Consider the function ๐(๐บ) = tr ๐บ๐ where ๐ โ โ, and
๐บ is a tensor.
๐ ๐บ = tr ๐บ๐ = ๐บ๐: ๐
To be specific, let ๐ = 3.
Real Tensor Functions
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 26
The Gateaux differential in this case,
๐ท๐ ๐บ, ๐๐บ =๐
๐๐ผ๐ ๐บ + ๐ผ๐๐บ
๐ผ=0
=๐
๐๐ผtr ๐บ + ๐ผ๐๐บ 3
๐ผ=0
=๐
๐๐ผtr ๐บ + ๐ผ๐๐บ ๐บ + ๐ผ๐๐บ ๐บ + ๐ผ๐๐บ
๐ผ=0
= tr๐
๐๐ผ๐บ + ๐ผ๐๐บ ๐บ + ๐ผ๐๐บ ๐บ + ๐ผ๐๐บ
๐ผ=0
= tr ๐๐บ ๐บ + ๐ผ๐๐บ ๐บ + ๐ผ๐๐บ+ ๐บ + ๐ผ๐๐บ ๐๐บ ๐บ + ๐ผ๐๐บ
+ ๐บ + ๐ผ๐๐บ ๐บ + ๐ผ๐๐บ ๐๐บ ๐ผ=0
= tr ๐๐บ ๐บ ๐บ + ๐บ ๐๐บ ๐บ + ๐บ ๐บ ๐ ๐บ = 3 ๐บ2 T: ๐๐บ
Real Tensor Functions
[email protected] 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 ๐ > 0, we have,
๐ท๐ ๐บ, ๐๐บ = ๐ ๐บ๐โ1 T: ๐๐บ
Clearly, ๐
๐๐บ tr ๐บ๐ = ๐ ๐บ๐โ1
T
Real Tensor Functions
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 28
When ๐ = 1,
๐ท๐ ๐บ, ๐๐บ =๐
๐๐ผ๐ ๐บ + ๐ผ๐๐บ
๐ผ=0
=๐
๐๐ผtr ๐บ + ๐ผ๐๐บ
๐ผ=0
= tr ๐ ๐๐บ = ๐: ๐๐บ
Or that, ๐
๐๐บ tr ๐บ = ๐.
Real Tensor Functions
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 29
Derivatives of the other two invariants of the tensor ๐บ can be found as follows:
๐
๐๐บ ๐ผ2(๐บ)
=1
2
๐
๐๐บtr2 ๐บ โ tr ๐บ2
=1
22tr ๐บ ๐ โ 2 ๐บT = tr ๐บ ๐ โ ๐บT
.
Real Tensor Functions
[email protected] 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 ๐บ: tr ๐บ3 โ ๐ผ1๐บ
2 + ๐ผ2๐บ โ ๐ผ3๐ฐ = tr(๐บ3) โ ๐ผ1tr ๐บ2 + ๐ผ2tr ๐บโ 3๐ผ3 = 0
Therefore, 3๐ผ3 = tr(๐บ3) โ ๐ผ1tr ๐บ2 + ๐ผ2tr ๐บ
๐ผ2 ๐บ =1
2tr2 ๐บ โ tr ๐บ2
.
Real Tensor Functions
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 31
We can therefore write, 3๐ผ3 ๐บ = tr(๐บ3) โ tr ๐บ tr ๐บ2
+1
2tr2 ๐บ โ tr ๐บ2 tr ๐บ
so that, in terms of traces only,
๐ผ3 ๐บ =1
6tr3 ๐บ โ 3tr ๐บ tr ๐บ2 + 2tr(๐บ3)
Differentiating the Invariants
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 32
Clearly, ๐๐ผ3(๐บ)
๐๐บ=
1
63tr2 ๐บ ๐ โ 3tr ๐บ2 โ 3tr ๐บ 2๐บ๐ + 2 ร 3 ๐บ2 ๐
= ๐ผ2๐ โ tr ๐บ ๐บ๐ + ๐บ2๐
Real Tensor Functions
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 33
.
Real Tensor Functions
[email protected] 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
[email protected] 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
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 36
The Euclidean Point Space โฐ is a set of elements called points. For each pair of points ๐, ๐ โ โฐ, โ ๐ ๐, ๐ โ E with the following two properties:
1. ๐ ๐, ๐ = ๐ ๐, ๐ + ๐ ๐, ๐ โ๐, ๐, ๐ โ โฐ
2. ๐ ๐, ๐ = ๐ ๐, ๐ โ ๐ = ๐
Based on these two, we proceed to show that, ๐ ๐, ๐ = ๐
And that, ๐ ๐, ๐ = โ๐ ๐, ๐
The Euclidean Point Space
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 37
From property 1, let ๐ โ ๐ it is clear that, ๐ ๐, ๐ = ๐ ๐, ๐ + ๐ ๐, ๐
And it we further allow ๐ โ ๐, we find that, ๐ ๐, ๐ = ๐ ๐, ๐ + ๐ ๐, ๐ = 2๐ ๐, ๐
Which clearly shows that ๐ ๐, ๐ = ๐ the zero vector.
Similarly, from the above, we find that, ๐ ๐, ๐ = ๐ = ๐ ๐, ๐ + ๐ ๐, ๐
So that ๐ ๐, ๐ = โ๐ ๐, ๐
The Euclidean Point Space
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 38
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 39
Coordinate System
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 40
Coordinate System
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 41
Note that โฐ is NOT a vector space. In our discussion, the vector space E to which ๐ 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 โฐ. We adopt the convention that ๐ ๐ โก ๐ ๐, ๐ referring to the vector ๐. If therefore we choose an arbitrarily fixed point ๐ โ โฐ, we are associating ๐ ๐ , ๐ ๐ and ๐ ๐ respectively with ๐ ๐, ๐ , ๐ ๐, ๐ and ๐ ๐, ๐ .
These are vectors based on the points ๐, ๐ and ๐ with reference to the origin chosen. To emphasize the association with both points of โฐ as well as members of E they are called Position Vectors.
Position Vectors
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 42
Recall that by property 1, ๐ ๐ โก ๐ ๐, ๐ = ๐ ๐, ๐ + ๐ ๐, ๐
Furthermore, we have deduced that ๐ ๐, ๐ = โ๐(๐, ๐)
We may therefore write that, ๐ ๐ = ๐ ๐ โ ๐(๐)
Which, when there is no ambiguity concerning the chosen origin, we can write as,
๐ ๐ = ๐ โ ๐
And the distance between the two is,
๐ ๐ ๐ = ๐ ๐ โ ๐ = ๐ โ ๐
Length in the Point Space
[email protected] 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 ๐ and ๐ on a set of independent vectors are ๐ฅ๐ and ๐ฆ๐, then the distance we seek is,
๐ ๐ โ ๐ = ๐ โ ๐ = ๐๐๐(๐ฅ๐ โ ๐ฆ๐)(๐ฅ๐ โ ๐ฆ๐)
The more familiar Pythagorean form occurring only when ๐๐๐ = 1.
Metric Properties
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 44
Consider a Cartesian coordinate system ๐ฅ, ๐ฆ, ๐ง or ๐ฆ1, ๐ฆ2 and ๐ฆ3with an orthogonal basis. Let us now have the possibility of transforming to another coordinate system of an arbitrary nature: ๐ฅ1, ๐ฅ2, ๐ฅ3. We can represent the transformation and its inverse in the equations:
๐ฆ๐ = ๐ฆ๐ ๐ฅ1, ๐ฅ2, ๐ฅ3 , ๐ฅ๐ = ๐ฅ๐(๐ฆ1, ๐ฆ2, ๐ฆ3)
And if the Jacobian of transformation,
Coordinate Transformations
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 45
๐๐ฆ๐
๐๐ฅ๐โก
๐ ๐ฆ1, ๐ฆ2, ๐ฆ3
๐ ๐ฅ1, ๐ฅ2, ๐ฅ3 โก
๐๐ฆ1
๐๐ฅ1
๐๐ฆ1
๐๐ฅ2
๐๐ฆ1
๐๐ฅ3
๐๐ฆ2
๐๐ฅ1
๐๐ฆ2
๐๐ฅ2
๐๐ฆ2
๐๐ฅ3
๐๐ฆ3
๐๐ฅ1
๐๐ฆ3
๐๐ฅ2
๐๐ฆ3
๐๐ฅ3
does not vanish, then the inverse transformation will exist. So that,
๐ฅ๐ = ๐ฅ๐(๐ฆ1, ๐ฆ2, ๐ฆ3)
Jacobian Determinant
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 46
Given a position vector ๐ซ = ๐ซ(๐ฅ1, ๐ฅ2, ๐ฅ3)
In the new coordinate system, we can form a set of three vectors,
๐ ๐ =๐๐ซ
๐๐ฅ๐, ๐ = 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. ๐ซ = ๐ซ ๐ฅ1, ๐ฅ2, ๐ฅ3 = ๐ฆ1๐ข +๐ฆ2๐ฃ + ๐ฆ3๐ค So that
๐ข =๐๐ซ
๐๐ฆ1 , ๐ฃ =๐๐ซ
๐๐ฆ2 and ๐ค =๐๐ซ
๐๐ฆ3
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,
๐ซ = ๐ซ ๐ฅ1, ๐ฅ2, ๐ฅ3 The total differential of ๐ซ is simply,
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 47
๐๐ซ =๐๐ซ
๐๐ฅ1๐๐ฅ1 +
๐๐ซ
๐๐ฅ2๐๐ฅ2 +
๐๐ซ
๐๐ฅ3๐๐ฅ3
โก ๐๐ฅ1๐ 1 + ๐๐ฅ2๐ 2 + ๐๐ฅ3๐ 3
With ๐ ๐ ๐ฅ1, ๐ฅ2, ๐ฅ3 , ๐ = 1,2,3 now depending in general on ๐ฅ1, ๐ฅ2 and ๐ฅ3 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, ๐ ๐ , ๐ = 1,2,3 are constants, normalized and orthogonal. They are our familiar
๐ข =๐๐ซ
๐๐ฆ1, ๐ฃ =
๐๐ซ
๐๐ฆ2 and ๐ค =
๐๐ซ
๐๐ฆ3.
Natural Dual Bases
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 48
We now proceed to show that the set of basis vectors, ๐ ๐ โก ๐ป๐ฅ๐
is reciprocal to ๐ ๐. The total differential ๐๐ซ can be written in terms of partial differentials as,
๐๐ซ =๐๐ซ
๐๐ฆ๐๐๐ฆ๐
Which when expressed in component form yields,
๐๐ฅ1 =๐๐ฅ1
๐๐ฆ1๐๐ฆ1 +
๐๐ฅ1
๐๐ฆ2๐๐ฆ2 +
๐๐ฅ1
๐๐ฆ3๐๐ฆ3
๐๐ฅ2 =๐๐ฅ2
๐๐ฆ1๐๐ฆ1 +
๐๐ฅ2
๐๐ฆ2๐๐ฆ2 +
๐๐ฅ2
๐๐ฆ3๐๐ฆ3
๐๐ฅ3 =๐๐ฅ3
๐๐ฆ1๐๐ฆ1 +
๐๐ฅ3
๐๐ฆ2๐๐ฆ2 +
๐๐ฅ3
๐๐ฆ3๐๐ฆ3
Or, more compactly that,
๐๐ฅ๐ =๐๐ฅ๐
๐๐ฆ๐๐๐ฆ๐ = ๐ป๐ฅ๐ โ ๐๐ซ
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 49
In Cartesian coordinates, ๐ฆ๐ , ๐ = 1, . . , 3 for any scalar field ๐ (๐ฆ1, ๐ฆ2, ๐ฆ3),
๐๐
๐๐ฆ๐๐๐ฆ๐ =
๐๐
๐๐ฅ๐๐ฅ +
๐๐
๐๐ฆ๐๐ฆ +
๐๐
๐๐ง๐๐ง = grad ๐ โ ๐๐ซ
We are treating each curvilinear coordinate as a scalar field, ๐ฅ๐ = ๐ฅ๐(๐ฆ1, ๐ฆ2, ๐ฆ3).
๐๐ฅ๐ = ๐ป๐ฅ๐ โ ๐๐ซ
๐๐ฅ๐ ๐๐ฅ๐
= ๐ ๐ โ ๐ ๐๐๐ฅ๐
= ๐ฟ๐๐๐๐ฅ๐.
The last equality arises from the fact that this is the only way one component of the coordinate differential can equal another.
โ ๐ ๐ โ ๐ ๐ = ๐ฟ๐๐
Which recovers for us the reciprocity relationship and shows that ๐ ๐ and ๐ ๐ are reciprocal systems.
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 50
The position vector in Cartesian coordinates is ๐ = ๐๐๐๐. Show that (a) ๐๐ข๐ฏ ๐ = ๐, (b) ๐๐ข๐ฏ ๐ โ ๐ = ๐๐, (c) ๐๐ข๐ฏ ๐ = ๐, and (d) ๐ ๐ซ๐๐ ๐ = ๐ and (e) ๐๐ฎ๐ซ๐ฅ ๐ โ ๐ = โ๐ ร
grad ๐ = ๐ฅ๐ ,๐ ๐๐ โ ๐๐ = ๐ฟ๐๐๐๐ โ ๐๐ = ๐
div ๐ = ๐ฅ๐ ,๐ ๐๐ โ ๐๐ = ๐ฟ๐๐๐ฟ๐๐ = ๐ฟ๐๐ = 3.
๐ โ ๐ = ๐ฅ๐๐๐ โ ๐ฅ๐๐๐ = ๐ฅ๐๐ฅ๐๐๐ โ ๐๐
grad ๐ โ ๐ = ๐ฅ๐๐ฅ๐ ,๐ ๐๐ โ ๐๐ โ ๐๐
div ๐ โ ๐ = ๐ฅ๐ ,๐ ๐ฅ๐ + ๐ฅ๐๐ฅ๐ ,๐ ๐๐ โ ๐๐ โ ๐๐
= ๐ฟ๐๐๐๐ + ๐๐๐ฟ๐๐ ๐ฟ๐๐๐๐ = ๐ฟ๐๐๐๐ + ๐๐๐ฟ๐๐ ๐๐ = 4๐ฅ๐๐๐ = 4๐
curl ๐ โ ๐ = ๐๐ผ๐ฝ๐พ ๐ฅ๐๐ฅ๐พ ,๐ฝ ๐๐ผ โ ๐๐
= ๐๐ผ๐ฝ๐พ ๐ฅ๐ ,๐ฝ ๐ฅ๐พ + ๐ฅ๐๐ฅ๐พ,๐ฝ ๐๐ผ โ ๐๐
= ๐๐ผ๐ฝ๐พ ๐ฟ๐๐ฝ๐ฅ๐พ + ๐ฅ๐๐ฟ๐พ๐ฝ ๐๐ผ โ ๐๐ = ๐๐ผ๐๐พ ๐ฅ๐พ๐๐ผ โ ๐๐ + ๐๐ผ๐ฝ๐ฝ๐ฅ๐๐๐ผ โ ๐๐
= โ๐๐ผ๐พ๐ ๐ฅ๐พ๐๐ผ โ ๐๐ = โ๐ ร
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 51
For a scalar field ๐ and a tensor field ๐ show that ๐ ๐ซ๐๐ ๐๐ = ๐๐ ๐ซ๐๐ ๐ + ๐ โ ๐ ๐ซ๐๐๐. Also show that ๐๐ข๐ฏ ๐๐ = ๐ ๐๐ข๐ฏ ๐ + ๐๐ ๐ซ๐๐๐.
grad ๐๐ = ๐๐๐๐ ,๐ ๐ ๐ โ ๐ ๐ โ ๐ ๐
= ๐,๐ ๐๐๐ + ๐๐๐๐ ,๐ ๐ ๐ โ ๐ ๐ โ ๐ ๐ = ๐ โ grad๐ + ๐grad ๐
Furthermore, we can contract the last two bases and obtain,
div ๐๐ = ๐,๐ ๐๐๐ + ๐๐๐๐ ,๐ ๐ ๐ โ ๐ ๐ โ ๐ ๐
= ๐,๐ ๐๐๐ + ๐๐๐๐ ,๐ ๐ ๐๐ฟ๐๐
= ๐๐๐ ๐,๐ ๐ ๐ + ๐๐๐๐ ,๐ ๐ ๐ = ๐grad๐ + ๐ div ๐
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 52
For two arbitrary vectors, ๐ and ๐, show that ๐ ๐ซ๐๐ ๐ โ ๐ = ๐ ร ๐ ๐ซ๐๐๐ โ ๐ ร ๐ ๐ซ๐๐๐
grad ๐ โ ๐ = ๐๐๐๐๐ข๐๐ฃ๐ ,๐ ๐ ๐ โ ๐ ๐
= ๐๐๐๐๐ข๐ ,๐ ๐ฃ๐ + ๐๐๐๐๐ข๐๐ฃ๐,๐ ๐ ๐ โ ๐ ๐
= ๐ข๐ ,๐ ๐๐๐๐๐ฃ๐ + ๐ฃ๐ ,๐ ๐๐๐๐๐ข๐ ๐ ๐ โ ๐ ๐
= โ ๐ ร grad๐ + ๐ ร grad๐
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 53
For any two tensor fields ๐ฎ and ๐ฏ, Show that, ๐ฎ ร : grad ๐ฏ = ๐ฎ โ curl ๐ฏ
๐ฎ ร : grad ๐ฏ = ๐๐๐๐๐ข๐๐ ๐ โ ๐ ๐ : ๐ฃ๐ผ,๐ฝ ๐ ๐ผ โ ๐ ๐ฝ
= ๐๐๐๐๐ข๐๐ฃ๐ผ,๐ฝ ๐ ๐ โ ๐ ๐ผ ๐ ๐ โ ๐ ๐ฝ
= ๐๐๐๐๐ข๐๐ฃ๐ผ,๐ฝ ๐ฟ๐๐ผ๐ฟ๐
๐ฝ= ๐๐๐๐๐ข๐๐ฃ๐ ,๐
= ๐ฎ โ curl ๐ฏ
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 54
For a vector field ๐, show that ๐ ๐ซ๐๐ ๐ ร is a third ranked tensor. Hence or otherwise show that ๐๐ข๐ฏ ๐ ร = โ๐๐ฎ๐ซ๐ฅ ๐.
The secondโorder tensor ๐ ร is defined as ๐๐๐๐๐ข๐๐ ๐ โ
๐ ๐. Taking the covariant derivative with an independent base, we have
grad ๐ ร = ๐๐๐๐๐ข๐ ,๐ ๐ ๐ โ ๐ ๐ โ ๐ ๐
This gives a third order tensor as we have seen. Contracting on the last two bases,
div ๐ ร = ๐๐๐๐๐ข๐ ,๐ ๐ ๐ โ ๐ ๐ โ ๐ ๐
= ๐๐๐๐๐ข๐ ,๐ ๐ ๐๐ฟ๐๐
= ๐๐๐๐๐ข๐ ,๐ ๐ ๐ = โcurl ๐
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 55
Show that ๐๐ฎ๐ซ๐ฅ ๐๐ = grad ๐ ร
Note that ๐๐ = ๐๐๐ผ๐ฝ ๐ ๐ผ โ ๐ ๐ฝ, and that curl ๐ป =
๐๐๐๐๐๐ผ๐ ,๐ ๐ ๐ โ ๐ ๐ผ so that,
curl ๐๐ = ๐๐๐๐ ๐๐๐ผ๐ ,๐ ๐ ๐ โ ๐ ๐ผ
= ๐๐๐๐ ๐,๐ ๐๐ผ๐ ๐ ๐ โ ๐ ๐ผ = ๐๐๐๐๐,๐ ๐ ๐ โ ๐ ๐ = grad ๐ ร
Show that curl ๐ ร = div ๐ ๐ โ grad ๐
๐ ร = ๐๐ผ๐ฝ๐๐ฃ๐ฝ ๐ ๐ผ โ ๐ ๐
curl ๐ป = ๐๐๐๐๐๐ผ๐ ,๐ ๐ ๐ โ ๐ ๐ผ
so that
curl ๐ ร = ๐๐๐๐๐๐ผ๐ฝ๐๐ฃ๐ฝ ,๐ ๐ ๐ โ ๐ ๐ผ
= ๐๐๐ผ๐๐๐ฝ โ ๐๐๐ฝ๐๐๐ผ ๐ฃ๐ฝ ,๐ ๐ ๐ โ ๐ ๐ผ
= ๐ฃ๐ ,๐ ๐ ๐ผ โ ๐ ๐ผ โ ๐ฃ๐ ,๐ ๐ ๐ โ ๐ ๐ = div ๐ ๐ โ grad ๐
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 56
Show that ๐๐ข๐ฏ ๐ ร ๐ = ๐ โ ๐๐ฎ๐ซ๐ฅ ๐ โ ๐ โ ๐๐ฎ๐ซ๐ฅ ๐
div ๐ ร ๐ = ๐๐๐๐๐ข๐๐ฃ๐ ,๐
Noting that the tensor ๐๐๐๐ behaves as a constant under a covariant differentiation, we can write,
div ๐ ร ๐ = ๐๐๐๐๐ข๐๐ฃ๐ ,๐
= ๐๐๐๐๐ข๐ ,๐ ๐ฃ๐ + ๐๐๐๐๐ข๐๐ฃ๐,๐ = ๐ โ curl ๐ โ ๐ โ curl ๐
Given a scalar point function ฯ and a vector field ๐ฏ, show that ๐๐ฎ๐ซ๐ฅ ๐๐ฏ = ๐ curl ๐ฏ + grad๐ ร ๐ฏ.
curl ๐๐ฏ = ๐๐๐๐ ๐๐ฃ๐ ,๐ ๐ ๐
= ๐๐๐๐ ๐,๐ ๐ฃ๐ + ๐๐ฃ๐ ,๐ ๐ ๐
= ๐๐๐๐๐,๐ ๐ฃ๐๐ ๐ + ๐๐๐๐๐๐ฃ๐,๐ ๐ ๐ = grad๐ ร ๐ฏ + ๐ curl ๐ฏ
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 57
Given that ๐ ๐ก = ๐(๐ก) , Show that ๐ ๐ก =๐
๐ ๐ก: ๐
๐2 โก ๐: ๐
Now, ๐
๐๐ก๐2 = 2๐
๐๐
๐๐ก=
๐๐
๐๐ก: ๐ + ๐:
๐๐
๐๐ก= 2๐:
๐๐
๐๐ก
as inner product is commutative. We can therefore write that
๐๐
๐๐ก=
๐
๐:๐๐
๐๐ก=
๐
๐ ๐ก: ๐
as required.
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 58
Given a tensor field ๐ป, obtain the vector ๐ โก ๐ป๐๐ฏ and show that its divergence is ๐ป: ๐๐ฏ + ๐ฏ โ ๐๐ข๐ฏ ๐ป
The divergence of ๐ is the scalar sum , ๐๐๐๐ฃ๐ ,๐.
Expanding the product covariant derivative we obtain,
div ๐ปT๐ฏ = ๐๐๐๐ฃ๐ ,๐ = ๐๐๐ ,๐ ๐ฃ๐ + ๐๐๐๐ฃ
๐ ,๐
= div ๐ป โ ๐ฏ + tr ๐ปTgrad๐ฏ
= div ๐ป โ ๐ฏ + ๐ป: grad ๐ฏ
Recall that scalar product of two vectors is commutative so that
div ๐ปT๐ฏ = ๐ป: grad๐ฏ + ๐ฏ โ div ๐ป
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 59
For a second-order tensor ๐ป define
curl ๐ป โก ๐๐๐๐๐ป๐ผ๐ ,๐ ๐ ๐ โ ๐ ๐ถ show that for any constant
vector ๐, curl ๐ป ๐ = curl ๐ป๐ป๐
Express vector ๐ in the invariant form with
contravariant components as ๐ = ๐๐ฝ๐ ๐ฝ. It follows that
curl ๐ป ๐ = ๐๐๐๐๐๐ผ๐,๐ ๐ ๐ โ ๐ ๐ผ ๐
= ๐๐๐๐๐๐ผ๐,๐ ๐๐ฝ ๐ ๐ โ ๐ ๐ผ ๐ ๐ฝ
= ๐๐๐๐๐๐ผ๐,๐ ๐๐ฝ๐ ๐๐ฟ๐ฝ๐ผ
= ๐๐๐๐ ๐๐ผ๐ ,๐ ๐ ๐๐๐ผ
= ๐๐๐๐ ๐๐ผ๐๐๐ผ ,๐ ๐ ๐
The last equality resulting from the fact that vector ๐ is a constant vector. Clearly,
curl ๐ป ๐ = curl ๐ป๐๐
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 60
For any two vectors ๐ฎ and ๐ฏ, show that curl ๐ฎ โ ๐ฏ =grad ๐ฎ ๐ฏ ร ๐ป + curl ๐ฏ โ ๐ where ๐ฏ ร is the skew tensor
๐๐๐๐๐ฃ๐ ๐ ๐ โ ๐ ๐.
Recall that the curl of a tensor ๐ป is defined by curl ๐ป โก
๐๐๐๐๐๐ผ๐ ,๐ ๐ ๐ โ ๐ ๐ผ. Clearly therefore,
curl ๐ โ ๐ = ๐๐๐๐ ๐ข๐ผ๐ฃ๐ ,๐ ๐ ๐ โ ๐ ๐ผ
= ๐๐๐๐ ๐ข๐ผ ,๐ ๐ฃ๐ + ๐ข๐ผ๐ฃ๐ ,๐ ๐ ๐ โ ๐ ๐ผ
= ๐๐๐๐๐ข๐ผ ,๐ ๐ฃ๐ ๐ ๐ โ ๐ ๐ผ + ๐๐๐๐๐ข๐ผ๐ฃ๐ ,๐ ๐ ๐ โ ๐ ๐ผ
= ๐๐๐๐๐ฃ๐ ๐ ๐ โ ๐ข๐ผ ,๐ ๐ ๐ผ + ๐๐๐๐๐ฃ๐ ,๐ ๐ ๐ โ ๐ข๐ผ๐ ๐ผ
= ๐๐๐๐๐ฃ๐ ๐ ๐ โ ๐ ๐ ๐ข๐ผ ,๐ฝ ๐ ๐ฝ โ ๐ ๐ผ + ๐๐๐๐๐ฃ๐ ,๐ ๐ ๐
โ ๐ข๐ผ๐ ๐ผ = โ ๐ ร grad ๐ ๐ป + curl ๐ โ ๐= grad ๐ ๐ ร ๐ป + curl ๐ โ ๐
upon noting that the vector cross is a skew tensor.
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 61
Show that curl ๐ ร ๐ = div ๐ โ ๐ โ ๐ โ ๐
The vector ๐ โก ๐ ร ๐ = ๐ค๐๐ ๐ = ๐๐๐ผ๐ฝ๐ข๐ผ๐ฃ๐ฝ๐ ๐ and
curl ๐ = ๐๐๐๐๐ค๐ ,๐ ๐ ๐. Therefore,
curl ๐ ร ๐ = ๐๐๐๐๐ค๐ ,๐ ๐ ๐
= ๐๐๐๐๐๐๐ผ๐ฝ ๐ข๐ผ๐ฃ๐ฝ ,๐ ๐ ๐
= ๐ฟ๐ผ๐ ๐ฟ๐ฝ
๐โ ๐ฟ๐ฝ
๐ ๐ฟ๐ผ๐
๐ข๐ผ๐ฃ๐ฝ ,๐ ๐ ๐
= ๐ฟ๐ผ๐ ๐ฟ๐ฝ
๐โ ๐ฟ๐ฝ
๐ ๐ฟ๐ผ๐
๐ข๐ผ,๐ ๐ฃ๐ฝ + ๐ข๐ผ๐ฃ๐ฝ,๐ ๐ ๐
= ๐ข๐ ,๐ ๐ฃ๐ + ๐ข๐๐ฃ๐ ,๐ โ ๐ข๐ ,๐ ๐ฃ๐ + ๐ข๐๐ฃ๐ ,๐ ๐ ๐
= ๐ข๐๐ฃ๐ ,๐ โ ๐ข๐๐ฃ๐ ,๐ ๐ ๐ = div ๐ โ ๐ โ ๐ โ ๐
since div ๐ โ ๐ = ๐ข๐๐ฃ๐ ,๐ผ ๐ ๐ โ ๐ ๐ โ ๐ ๐ผ = ๐ข๐๐ฃ๐ ,๐ ๐ ๐.
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 62
Given a scalar point function ๐ and a second-order tensor field ๐ป,
show that ๐๐ฎ๐ซ๐ฅ ๐๐ป = ๐ ๐๐ฎ๐ซ๐ฅ ๐ป + grad๐ ร ๐ป๐ป where
grad๐ ร is the skew tensor ๐๐๐๐๐,๐ ๐ ๐ โ ๐ ๐
curl ๐๐ป โก ๐๐๐๐ ๐๐๐ผ๐ ,๐ ๐ ๐ โ ๐ ๐ผ
= ๐๐๐๐ ๐,๐ ๐๐ผ๐ + ๐๐๐ผ๐,๐ ๐ ๐ โ ๐ ๐ผ
= ๐๐๐๐๐,๐ ๐๐ผ๐ ๐ ๐ โ ๐ ๐ผ + ๐๐๐๐๐๐๐ผ๐,๐ ๐ ๐ โ ๐ ๐ผ
= ๐๐๐๐๐,๐ ๐ ๐ โ ๐ ๐ ๐๐ผ๐ฝ๐ ๐ฝ โ ๐ ๐ผ + ๐๐๐๐๐๐๐ผ๐,๐ ๐ ๐ โ ๐ ๐ผ
= ๐ curl ๐ป + grad๐ ร ๐ป๐
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 63
For a second-order tensor field ๐ป, show that
div curl ๐ป = curl div ๐ป๐
Define the second order tensor ๐ as
curl ๐ป โก ๐๐๐๐๐๐ผ๐,๐ ๐ ๐ โ ๐ ๐ผ = ๐.๐ผ๐ ๐ ๐ โ ๐ ๐ผ
The gradient of ๐บ is
๐.๐ผ๐ ,๐ฝ ๐ ๐ โ ๐ ๐ผ โ ๐ ๐ฝ = ๐๐๐๐๐๐ผ๐,๐๐ฝ ๐ ๐ โ ๐ ๐ผ โ ๐ ๐ฝ
Clearly,
div curl ๐ป = ๐๐๐๐๐๐ผ๐,๐๐ฝ ๐ ๐ โ ๐ ๐ผ โ ๐ ๐ฝ
= ๐๐๐๐๐๐ผ๐,๐๐ฝ ๐ ๐ ๐๐ผ๐ฝ
= ๐๐๐๐๐๐ฝ๐ ,๐๐ฝ ๐ ๐ = curl div ๐ปT
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 64
If the volume ๐ฝ is enclosed by the surface ๐บ, the position vector ๐ = ๐๐๐ ๐ and ๐ is the external unit normal to each surface
element, show that ๐
๐ ๐ ๐ โ ๐ โ ๐๐ ๐บ๐บ
equals the volume
contained in ๐ฝ.
๐ โ ๐ = ๐ฅ๐๐ฅ๐๐ ๐ โ ๐ ๐ = ๐ฅ๐๐ฅ๐๐๐๐
By the Divergence Theorem,
๐ป ๐ โ ๐ โ ๐๐๐๐
= ๐ป โ ๐ป ๐ โ ๐ ๐๐๐
= ๐๐ ๐๐ ๐ฅ๐๐ฅ๐๐๐๐ ๐ ๐ โ ๐ ๐ ๐๐๐
= ๐๐ ๐๐๐ ๐ฅ๐ ,๐ ๐ฅ๐ + ๐ฅ๐๐ฅ๐ ,๐ ๐ ๐ โ ๐ ๐ ๐๐๐
= ๐๐๐๐๐๐ ๐ฟ๐
๐ ๐ฅ๐ + ๐ฅ๐๐ฟ๐๐
,๐ ๐๐๐
= 2๐๐๐๐๐๐๐ฅ๐,๐ ๐๐๐
= 2๐ฟ๐๐๐ฟ๐
๐ ๐๐๐
= 6 ๐๐๐
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 65
For any Euclidean coordinate system, show that div ๐ฎ ร ๐ฏ =๐ฏ curl ๐ฎ โ ๐ฎ curl ๐ฏ
Given the contravariant vector ๐ข๐ and ๐ฃ๐ with their associated vectors ๐ข๐ and ๐ฃ๐, the contravariant component of the above cross product is ๐๐๐๐๐ข๐๐ฃ๐ .The required divergence is simply the contraction of the covariant ๐ฅ๐ derivative of this quantity:
๐๐๐๐๐ข๐๐ฃ๐ ,๐= ๐๐๐๐๐ข๐,๐๐ฃ๐ + ๐๐๐๐๐ข๐๐ฃ๐,๐
where we have treated the tensor ๐๐๐๐ as a constant under the covariant derivative.
Cyclically rearranging the RHS we obtain,
๐๐๐๐๐ข๐๐ฃ๐ ,๐= ๐ฃ๐๐๐๐๐๐ข๐,๐ + ๐ข๐๐
๐๐๐๐ฃ๐,๐ = ๐ฃ๐๐๐๐๐๐ข๐,๐ + ๐ข๐๐๐๐๐๐ฃ๐,๐
where we have used the anti-symmetric property of the tensor ๐๐๐๐. The last expression shows clearly that
div ๐ฎ ร ๐ฏ = ๐ฏ curl ๐ฎ โ ๐ฎ curl ๐ฏ
as required.
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 66
For a scalar variable ๐ผ, if the tensor ๐ = ๐(๐ผ) and ๐ โก๐๐
๐๐ผ, Show that
๐
๐๐ผdet ๐ =
det ๐ tr(๐ ๐โ๐) Proof:
Let ๐ โก ๐ ๐โ1 so that, ๐ = ๐๐. In component form, we have ๐ ๐๐ = ๐ด๐
๐ ๐๐๐.
Therefore, ๐
๐๐ผdet ๐ =
๐
๐๐ผ๐๐๐๐๐๐
1๐๐2๐๐
3 = ๐๐๐๐ ๐ ๐1๐๐
2๐๐3 + ๐๐
1๐ ๐2๐๐
3 + ๐๐1๐๐
2๐ ๐3
= ๐๐๐๐ ๐ด๐1๐๐
๐๐๐2๐๐
3 + ๐๐1๐ด๐
2 ๐๐๐๐๐
3 + ๐๐1๐๐
2๐ด๐3๐๐
๐
= ๐๐๐๐ ๐ด11๐๐
1 + ๐ด21๐๐
2 + ๐ด31๐๐
3 ๐๐2๐๐
3
+ ๐๐1 ๐ด1
2๐๐1 + ๐ด2
2๐๐2 + ๐ด3
2๐๐3 ๐๐
3 + ๐๐1๐๐
2 ๐ด13๐๐
1 + ๐ด23๐๐
2 + ๐ด33๐๐
3
All the boxed terms in the above equation vanish on account of the contraction of a symmetric tensor with an antisymmetric one.
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 67
Liouville Formula
(For example, the first boxed term yields, ๐๐๐๐๐ด21๐๐
2๐๐2๐๐
3
Which is symmetric as well as antisymmetric in ๐ and ๐. It
therefore vanishes. The same is true for all other such terms.)
๐
๐๐ผdet ๐ = ๐๐๐๐ ๐ด1
1๐๐1 ๐๐
2๐๐3 + ๐๐
1 ๐ด22๐๐
2 ๐๐3 + ๐๐
1๐๐2 ๐ด3
3๐๐3
= ๐ด๐๐ ๐๐๐๐๐๐
1๐๐2๐๐
3
= tr ๐ ๐โ1 det ๐
as required.
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 68
Liouville Theorem Contโd
For a general tensor field ๐ป show that, ๐๐ฎ๐ซ๐ฅ ๐๐ฎ๐ซ๐ฅ ๐ป =๐๐ ๐ญ๐ซ ๐ป โ ๐๐ข๐ฏ ๐๐ข๐ฏ ๐ป ๐ฐ + ๐ ๐ซ๐๐ ๐๐ข๐ฏ ๐ป + ๐ ๐ซ๐๐ ๐๐ข๐ฏ ๐ป
๐โ
๐ ๐ซ๐๐ ๐ ๐ซ๐๐ ๐ญ๐ซ๐ป โ ๐๐๐ป๐
curl ๐ป = ๐๐ผ๐ ๐ก๐๐ฝ๐ก ,๐ ๐ ๐ผ โ ๐ ๐ฝ
= ๐ .๐ฝ๐ผ ๐ ๐ผ โ ๐ ๐ฝ
curl ๐บ = ๐๐๐๐๐ .๐๐ผ ,๐ ๐ ๐ โ ๐ ๐ผ
so that
curl ๐บ = curl curl ๐ป = ๐๐๐๐๐๐ผ๐ ๐ก๐๐๐ก,๐ ๐ ๐ ๐ โ ๐ ๐ผ
=
๐๐๐ผ ๐๐๐ ๐๐๐ก
๐๐๐ผ ๐๐๐ ๐๐๐ก
๐๐๐ผ ๐๐๐ ๐๐๐ก
๐๐๐ก ,๐ ๐ ๐ ๐ โ ๐ ๐ผ
=๐๐๐ผ ๐๐๐ ๐๐๐ก โ ๐๐๐ก๐๐๐ + ๐๐๐ ๐๐๐ก๐๐๐ผ โ ๐๐๐ผ๐๐๐ก
+๐๐๐ก ๐๐๐ผ๐๐๐ โ ๐๐๐ ๐๐๐ผ๐๐๐ก ,๐ ๐ ๐ ๐ โ ๐ ๐ผ
= ๐๐๐ ๐ .๐ก๐ก ,๐ ๐ โ๐ ..
๐ ๐ ,๐ ๐ ๐ ๐ผ โ ๐ ๐ผ + ๐ ..๐ผ๐ ,๐ ๐ โ๐๐๐ผ๐ .๐ก
๐ก ,๐ ๐ ๐ ๐ โ ๐ ๐ผ
+ ๐๐๐ผ๐ .๐ก๐ .,๐ ๐ โ๐๐๐ ๐ .๐ก
๐ผ.,๐ ๐ ๐ ๐ก โ ๐ ๐ผ
= ๐ป2 tr ๐ป โ div div ๐ป ๐ฐ + grad div ๐ปT
โ grad grad tr๐ป
+ grad div ๐ป โ ๐ป2๐ปT [email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 69
When ๐ป is symmetric, show that tr(curl ๐ป) vanishes.
curl ๐ป = ๐๐๐๐๐๐ฝ๐ ,๐ ๐ ๐ โ ๐ ๐ฝ
tr curl ๐ป = ๐๐๐๐๐๐ฝ๐ ,๐ ๐ ๐ โ ๐ ๐ฝ
= ๐๐๐๐๐๐ฝ๐ ,๐ ๐ฟ๐๐ฝ
= ๐๐๐๐๐๐๐ ,๐
which obviously vanishes on account of the symmetry and antisymmetry in ๐ and ๐. In this case, curl curl ๐ป
= ๐ป2 tr ๐ป โ div div ๐ป ๐ฐ โ grad grad tr๐ป
+ 2 grad div ๐ป โ ๐ป2๐ป
as grad div ๐ปT
= grad div ๐ป if the order of
differentiation is immaterial and T is symmetric.
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 70
For a scalar function ฮฆ and a vector ๐ฃ๐ show that the divergence of the vector ๐ฃ๐ฮฆ is equal to, ๐ฏ โ gradฮฆ +ฮฆ div ๐ฏ
๐ฃ๐ฮฆ,๐
= ฮฆ๐ฃ๐,๐ + ๐ฃ๐ฮฆ,i
Hence the result.
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 71
Show that ๐๐๐๐ ๐ฎ ร ๐ฏ = ๐ฏ โ ๐๐ฎ + ๐ฎ โ div ๐ฏ โ ๐ฏ โ div ๐ฎ โ๐ฎ โ ๐ ๐ฏ
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,
๐๐๐๐ ๐๐๐๐๐ข๐๐ฃ๐,๐
Expanding in the usual way, noting the relation between the alternating tensors and the Kronecker deltas,
๐๐๐๐ ๐๐๐๐๐ข๐๐ฃ๐,๐
= ๐ฟ๐๐๐๐๐๐ ๐ข๐
,๐๐ฃ๐ โ ๐ข๐๐ฃ๐,๐
= ๐ฟ๐๐๐๐ ๐ข๐
,๐๐ฃ๐ โ ๐ข๐๐ฃ๐,๐ =
๐ฟ๐๐ ๐ฟ๐
๐
๐ฟ๐๐ ๐ฟ๐
๐๐ข๐
,๐๐ฃ๐ โ ๐ข๐๐ฃ๐,๐
= ๐ฟ๐๐๐ฟ๐
๐ โ ๐ฟ๐๐ ๐ฟ๐
๐ ๐ข๐,๐๐ฃ๐ โ ๐ข๐๐ฃ๐
,๐
= ๐ฟ๐๐๐ฟ๐
๐๐ข๐,๐๐ฃ๐ โ ๐ฟ๐
๐๐ฟ๐๐๐ข๐๐ฃ๐
,๐ + ๐ฟ๐๐ ๐ฟ๐
๐๐ข๐,๐๐ฃ๐
โ ๐ฟ๐๐ ๐ฟ๐
๐๐ข๐๐ฃ๐,๐
= ๐ข๐,๐๐ฃ๐ โ ๐ข๐
,๐๐ฃ๐ + ๐ข๐๐ฃ๐,๐ โ ๐ข๐๐ฃ๐
,๐
Which is the result we seek in indicial notation.
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 72
Given a vector point function ๐ ๐ฅ1, ๐ฅ2, ๐ฅ3 and its covariant components, ๐ข๐ผ ๐ฅ1, ๐ฅ2, ๐ฅ3 , ๐ผ = 1,2,3, with ๐ ๐ผas the reciprocal basis vectors, Let
๐ = ๐ข๐ผ๐ ๐ผ , then ๐๐ =๐
๐๐ฅ๐๐ข๐ผ๐ ๐ผ ๐๐ฅ๐ =
๐๐ข๐ผ
๐๐ฅ๐๐ ๐ผ +
๐๐ ๐ผ
๐๐ฅ๐๐ข๐ผ ๐๐ฅ๐
Clearly, ๐๐
๐๐ฅ๐=
๐๐ข๐ผ
๐๐ฅ๐๐ ๐ผ +
๐๐ ๐ผ
๐๐ฅ๐๐ข๐ผ
And the projection of this quantity on the ๐ ๐direction is, ๐๐
๐๐ฅ๐โ ๐ ๐ =
๐๐ข๐ผ
๐๐ฅ๐๐ ๐ผ +
๐๐ ๐ผ
๐๐ฅ๐๐ข๐ผ โ ๐ ๐ =
๐๐ข๐ผ
๐๐ฅ๐๐ ๐ผ โ ๐ ๐ +
๐๐ ๐ผ
๐๐ฅ๐โ ๐ ๐๐ข๐ผ
=๐๐ข๐ผ
๐๐ฅ๐๐ฟ๐
๐ผ +๐๐ ๐ผ
๐๐ฅ๐โ ๐ ๐๐ข๐ผ
Differentiation of Fields
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 73
Now, ๐ ๐ โ ๐ ๐ = ๐ฟ๐๐ so that
๐๐ ๐
๐๐ฅ๐โ ๐ ๐ + ๐ ๐ โ
๐๐ ๐
๐๐ฅ๐= 0.
๐๐ ๐
๐๐ฅ๐โ ๐ ๐ = โ๐ ๐ โ
๐๐ ๐
๐๐ฅ๐โก โ
๐๐๐
.
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
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 74
Using this, we can now write that, ๐๐
๐๐ฅ๐ โ ๐ ๐ =๐๐ข๐ผ
๐๐ฅ๐ ๐ฟ๐๐ผ +
๐๐ ๐ผ
๐๐ฅ๐ โ ๐ ๐๐ข๐ผ =๐๐ข๐
๐๐ฅ๐ โ๐ผ๐๐
๐ข๐ผ
The quantity on the RHS is the component of the derivative of vector ๐ along the ๐ ๐ direction using covariant components. It is the covariant derivative of ๐. Using contravariant components, we could write,
๐๐ =๐๐ข๐
๐๐ฅ๐ ๐ ๐ +๐๐ ๐
๐๐ฅ๐ ๐ข๐ ๐๐ฅ๐ =๐๐ข๐
๐๐ฅ๐ ๐ ๐ +๐ผ๐๐
๐ ๐ผ๐ข๐ ๐๐ฅ๐
So that, ๐๐
๐๐ฅ๐ =๐๐ข๐ผ
๐๐ฅ๐ ๐ ๐ผ +๐๐ ๐ผ
๐๐ฅ๐ ๐ข๐ผ
The components of this in the direction of ๐ ๐ can be obtained by taking a dot product as before:
๐๐
๐๐ฅ๐ โ ๐ ๐ =๐๐ข๐ผ
๐๐ฅ๐ ๐ ๐ผ +๐๐ ๐ผ
๐๐ฅ๐ ๐ข๐ผ โ ๐ ๐ =๐๐ข๐ผ
๐๐ฅ๐ ๐ฟ๐ผ๐ +
๐๐ ๐ผ
๐๐ฅ๐ โ ๐ ๐๐ข๐ผ =๐๐ข๐
๐๐ฅ๐ +๐
๐ผ๐๐ข๐ผ
Derivatives in General Coordinates
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 75
The two results above are represented symbolically as,
๐ข๐,๐ =๐๐ข๐
๐๐ฅ๐ โ๐ผ๐๐
๐ข๐ผ and ๐ข,๐๐ =
๐๐ข๐ผ
๐๐ฅ๐ +๐
๐ผ๐๐ข๐ผ
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
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 76
We observe that the derivative of the covariant basis, ๐ ๐ =๐๐ซ
๐๐ฅ๐ ,
๐๐ ๐
๐๐ฅ๐=
๐2๐ซ
๐๐ฅ๐๐๐ฅ๐=
๐2๐ซ
๐๐ฅ๐๐๐ฅ๐=
๐๐ ๐
๐๐ฅ๐
Taking the dot product with ๐ ๐, ๐๐ ๐
๐๐ฅ๐โ ๐ ๐ =
1
2
๐๐ ๐
๐๐ฅ๐โ ๐ ๐ +
๐๐ ๐
๐๐ฅ๐โ ๐ ๐
=1
2
๐
๐๐ฅ๐๐ ๐ โ ๐ ๐ +
๐
๐๐ฅ๐๐ ๐ โ ๐ ๐ โ ๐ ๐ โ
๐๐ ๐
๐๐ฅ๐โ ๐ ๐ โ
๐๐ ๐
๐๐ฅ๐
=1
2
๐
๐๐ฅ๐๐ ๐ โ ๐ ๐ +
๐
๐๐ฅ๐๐ ๐ โ ๐ ๐ โ ๐ ๐ โ
๐๐ ๐
๐๐ฅ๐โ ๐ ๐ โ
๐๐ ๐
๐๐ฅ๐
=1
2
๐
๐๐ฅ๐๐ ๐ โ ๐ ๐ +
๐
๐๐ฅ๐๐ ๐ โ ๐ ๐ โ
๐
๐๐ฅ๐๐ ๐ โ ๐ ๐
=1
2
๐๐๐๐
๐๐ฅ๐+
๐๐๐๐
๐๐ฅ๐โ
๐๐๐๐
๐๐ฅ๐
Christoffel Symbols
[email protected] 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,
๐๐, ๐ โก ๐๐ ๐
๐๐ฅ๐โ ๐ ๐ =
๐๐ ๐
๐๐ฅ๐โ ๐ ๐ =
1
2
๐๐๐๐
๐๐ฅ๐+
๐๐๐๐
๐๐ฅ๐โ
๐๐๐๐
๐๐ฅ๐
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 ๐ and ๐. The extension of this definition to the Christoffel symbols of the second kind is immediate:
The First Kind
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 78
Contract the above equation with the conjugate metric tensor, we have,
๐๐๐ผ ๐๐, ๐ผ โก ๐๐๐ผ ๐๐ ๐
๐๐ฅ๐โ ๐ ๐ผ = ๐๐๐ผ
๐๐ ๐
๐๐ฅ๐โ ๐ ๐ผ =
๐๐ ๐
๐๐ฅ๐โ ๐ ๐ =
๐๐๐
= โ๐๐ ๐
๐๐ฅ๐โ ๐ ๐
Which connects the common definition of the second Christoffel symbol with the one defined in the above derivation. The relationship,
๐๐๐ผ ๐๐, ๐ผ =๐๐๐
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
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 79
We contract the above equation with ๐๐๐ฝand obtain,
๐๐๐ฝ๐๐๐ผ ๐๐, ๐ผ = ๐๐๐ฝ๐๐๐
๐ฟ๐ฝ๐ผ ๐๐, ๐ผ = ๐๐, ๐ฝ = ๐๐๐ฝ
๐๐๐
so that,
๐๐๐ผ
๐ผ๐๐ = ๐๐, ๐
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
[email protected] 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 ๐, we can express the components in dyadic form along the product basis as follows:
๐ = ๐๐๐๐ ๐ โ ๐ ๐= ๐๐๐๐ ๐โจ๐ ๐ = ๐.๐
๐ ๐ ๐ โ ๐ ๐= ๐๐.๐๐ ๐โจ๐ ๐
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
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 81
We have:
๐๐
๐๐ฅ๐=
๐๐๐๐
๐๐ฅ๐๐ ๐โจ๐ ๐ + ๐๐๐
๐๐ ๐
๐๐ฅ๐โจ๐ ๐ + ๐๐๐๐ ๐โจ
๐๐ ๐
๐๐ฅ๐
Recall that, ๐๐ ๐
๐๐ฅ๐ โ ๐ ๐ =๐๐๐
. It follows therefore that,
๐๐ ๐
๐๐ฅ๐โ ๐ ๐ โ
๐ผ๐๐
๐ฟ๐ผ๐
=๐๐ ๐
๐๐ฅ๐โ ๐ ๐ โ
๐ผ๐๐
๐ ๐ผ โ ๐ ๐
=๐๐ ๐
๐๐ฅ๐โ
๐ผ๐๐
๐ ๐ผ โ ๐ ๐ = 0.
Clearly, ๐๐ ๐
๐๐ฅ๐ =๐ผ๐๐
๐ ๐ผ
(Obviously since ๐ ๐ is a basis vector it cannot vanish)
Higher Order Tensors
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 82
๐๐
๐๐ฅ๐=
๐๐๐๐
๐๐ฅ๐๐ ๐โจ๐ ๐ + ๐๐๐
๐๐ ๐
๐๐ฅ๐โจ๐ ๐ + ๐๐๐๐ ๐โจ
๐๐ ๐
๐๐ฅ๐
=๐๐๐๐
๐๐ฅ๐ ๐ ๐โจ๐ ๐ + ๐๐๐ ๐ผ๐๐
๐ ๐ผ โจ๐ ๐ + ๐๐๐๐ ๐โจ๐ผ๐๐ ๐ ๐ผ
=๐๐๐๐
๐๐ฅ๐ ๐ ๐โจ๐ ๐ + ๐๐ผ๐ ๐๐ผ๐
๐ ๐ โจ๐ ๐ + ๐๐๐ผ๐ ๐โจ๐
๐ผ๐๐ ๐
=๐๐๐๐
๐๐ฅ๐ +๐๐ผ๐ ๐๐ผ๐
+ ๐๐๐ผ ๐๐ผ๐
๐ ๐โจ๐ ๐ = ๐ ,๐๐๐
๐ ๐โจ๐ ๐
Where
๐ ,๐
๐๐=
๐๐๐๐
๐๐ฅ๐ +๐๐ผ๐ ๐๐ผ๐
+ ๐๐๐ผ ๐๐ผ๐
or ๐๐๐๐
๐๐ฅ๐ +๐๐ผ๐ ฮ๐ผ๐๐ + ๐๐๐ผฮ๐ผ๐
๐
are the components of the covariant derivative of the tensor ๐ in terms of contravariant components on the product covariant bases as shown.
Higher Order Tensors
[email protected] 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,
๐๐
๐๐ฅ๐ =๐๐๐๐
๐๐ฅ๐ ๐ ๐ โ ๐ ๐ +๐๐๐
๐๐ ๐
๐๐ฅ๐ โ ๐ ๐ +๐๐๐๐ ๐ โ
๐๐ ๐
๐๐ฅ๐
=๐๐๐๐
๐๐ฅ๐ ๐ ๐ โ ๐ ๐ +๐๐๐ฮ๐ผ๐๐ โ ๐ ๐ +๐๐๐๐
๐ โ๐๐ ๐
๐๐ฅ๐
Again, notice from previous derivation above, ๐๐๐
= โ๐๐ ๐
๐๐ฅ๐ โ ๐ ๐ so that, ๐๐ ๐
๐๐ฅ๐ = โ๐
๐ผ๐๐ ๐ผ = โฮ๐ผ๐
๐ ๐ ๐ผ Therefore,
๐๐
๐๐ฅ๐=
๐๐๐๐
๐๐ฅ๐๐ ๐ โ ๐ ๐ +๐๐๐
๐๐ ๐
๐๐ฅ๐โ ๐ ๐ +๐๐๐๐
๐ โ๐๐ ๐
๐๐ฅ๐
=๐๐๐๐
๐๐ฅ๐๐ ๐ โ ๐ ๐ โ๐๐๐ฮ๐ผ๐
๐ ๐ ๐ผ โ ๐ ๐ + ๐๐๐๐ ๐ โ ฮ๐ผ๐
๐๐ ๐ผ
=๐๐๐๐
๐๐ฅ๐โ ๐๐ผ๐ฮ๐๐
๐ผ โ ๐๐๐ผฮ๐๐๐ผ ๐ ๐ โ ๐ ๐= ๐๐๐,๐๐ ๐ โ ๐ ๐
So that
๐๐๐,๐ =๐๐๐๐
๐๐ฅ๐ โ ๐๐ผ๐ฮ๐๐๐ผ โ ๐๐๐ผฮ๐๐
๐ผ
Higher Order Tensors
[email protected] 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
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 85
The covariant derivative of the mixed tensor, ๐ด๐1 ,๐2 ,โฆ,๐๐
๐1 ,๐2 ,โฆ,๐๐ is the
most general case for the covariant derivative of an absolute tensor:
๐ด๐1 ,๐2 ,โฆ,๐๐
๐1 ,๐2 ,โฆ,๐๐
,๐
= ๐๐ด๐1 ,๐2 ,โฆ,๐๐
๐1 ,๐2 ,โฆ,๐๐
๐๐ฅ๐โ
๐ผ๐1๐
๐ด๐ผ ,๐2 ,โฆ,๐๐
๐1 ,๐2 ,โฆ,๐๐ โ ๐ผ๐2๐
๐ด๐1,๐ผ,โฆ,๐๐
๐1 ,๐2 ,โฆ,๐๐ โ โฏ โ๐ผ๐๐๐ ๐ด๐1 ,๐2 ,โฆ,๐ผ
๐1 ,๐2 ,โฆ,๐๐
+ ๐1๐ฝ๐
๐ด๐1 ,๐2 ,โฆ,๐๐
๐ฝ ,๐2 ,โฆ,๐๐ + ๐2๐ฝ๐
๐ด๐1 ,๐2 ,โฆ,๐๐
๐1 ,๐ฝ,โฆ,๐๐ + โฏ+ ๐๐๐ฝ๐
๐ด๐1 ,๐2 ,โฆ,๐๐
๐1 ,๐2,โฆ,๐ฝ
Higher Order Mixed Tensors
[email protected] 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:
๐๐๐ ,๐ =๐๐๐๐
๐๐ฅ๐โ
๐ผ๐๐
๐๐ผ๐ โ๐ฝ๐๐
๐๐๐ฝ
=๐๐๐๐
๐๐ฅ๐โ ๐๐, ๐ โ ๐๐, ๐
=๐๐๐๐
๐๐ฅ๐โ
1
2
๐๐๐๐
๐๐ฅ๐+
๐๐๐๐
๐๐ฅ๐โ
๐๐๐๐
๐๐ฅ๐โ
1
2
๐๐๐๐
๐๐ฅ๐+
๐๐๐๐
๐๐ฅ๐โ
๐๐๐๐
๐๐ฅ๐
= 0.
Ricciโs Theorem
[email protected] 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,
๐๐๐๐๐๐ = ๐ฟ๐
๐
The above can be differentiated covariantly with respect to ๐ฅ๐ to obtain
๐๐๐ ,๐ ๐๐๐ + ๐๐๐๐๐๐ ,๐ = ๐ฟ๐
๐,๐
0 + ๐๐๐๐๐๐ ,๐ =
๐๐ฟ๐๐
๐๐ฅ๐ โ๐ผ๐๐
๐ฟ๐ผ๐
+๐
๐ผ๐๐ฟ๐
๐ผ
๐๐๐๐๐๐ ,๐ = 0 โ
๐๐๐
+๐๐๐
= 0
The contraction of ๐๐๐ with ๐๐๐ ,๐ vanishes. Since we know that the metric tensor cannot vanish in general, we can only conclude that
๐๐๐ ,๐ = 0
Ricciโs Theorem
[email protected] 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 ๐๐๐and ๐๐๐ as well as the alternating tensors ๐๐๐๐ , ๐๐๐๐ are all constants under a covariant
differention. The Kronecker delta ๐ฟ๐๐ is a constant under
both covariant as well as the regular partial differentiation
[email protected] 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 ๐ต, The volume integral in the region ฮฉ โ โฐ,
grad ๐ตฮฉ
๐๐ฃ = ๐ต โ ๐ง๐ฮฉ
๐๐
where ๐ง is the outward drawn normal to ๐ฮฉ โ the boundary of ฮฉ.
Integral Theorems
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 90
Vector field. Replacing the tensor with the vector field ๐ and contracting, we have,
div ๐ฮฉ
๐๐ฃ = ๐ โ ๐ง๐ฮฉ
๐๐
Which is the usual form of the Gauss theorem.
Special Cases: Vector Field
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 91
For a scalar field ๐, the divergence becomes a gradient and the scalar product on the RHS becomes a simple multiplication. Hence the divergence theorem becomes,
grad ๐ฮฉ
๐๐ฃ = ๐๐ง๐ฮฉ
๐๐
The procedure here is valid and will become obvious if we write, ๐ = ๐๐ where ๐ is an arbitrary constant vector.
Scalar Field
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 92
div ๐๐ฮฉ
๐๐ฃ = ๐๐ โ ๐ง๐ฮฉ
๐๐ = ๐ โ ๐๐ง๐ฮฉ
๐๐
For the LHS, note that, div ๐๐ = tr grad ๐๐
grad ๐๐ = ๐๐๐ ,๐ ๐ ๐ โ ๐ ๐
= ๐๐๐,๐ ๐ ๐ โ ๐ ๐
The trace of which is,
๐๐๐,๐ ๐ ๐ โ ๐ ๐ = ๐๐๐,๐ ๐ฟ๐๐
= ๐๐๐,๐ = ๐ โ grad ๐
For the arbitrary constant vector ๐, we therefore have that,
div ๐๐ฮฉ
๐๐ฃ = ๐ โ grad ๐ ฮฉ
๐๐ฃ = ๐ โ ๐๐ง๐ฮฉ
๐๐
grad ๐ ฮฉ
๐๐ฃ = ๐๐ง๐ฮฉ
๐๐
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 93
For a second-order tensor ๐, the Gauss Theorem becomes,
div๐ฮฉ
๐๐ฃ = ๐๐ง๐ฮฉ
๐๐
The original outer product under the integral can be expressed in dyadic form:
grad ๐ฮฉ
๐๐ฃ = ๐๐๐ ,๐ ๐ ๐ โ ๐ ๐ โ ๐ ๐
ฮฉ
๐๐ฃ
= ๐ โ ๐ง๐ฮฉ
๐๐
= ๐๐๐๐ ๐ โ ๐ ๐ โ ๐๐๐ ๐
๐ฮฉ
๐๐
Second-Order Tensor field
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 94
Or
๐๐๐ ,๐ ๐ ๐ โ ๐ ๐ โฮฉ
๐ ๐๐๐ฃ = ๐๐๐๐๐๐ ๐ โ ๐ ๐ โ ๐ ๐
๐ฮฉ
๐๐
Contracting, we have
๐๐๐ ,๐ฮฉ
๐ ๐ โ ๐ ๐ ๐ ๐๐๐ฃ = ๐๐๐๐๐ ๐ ๐ โ ๐ ๐ ๐ ๐
๐ฮฉ
๐๐
๐๐๐ ,๐ ๐ฟ๐๐๐ ๐
ฮฉ
๐๐ฃ = ๐๐๐๐๐๐ฟ๐๐๐ ๐
๐ฮฉ
๐๐
๐๐๐ ,๐ ๐ ๐ฮฉ
๐๐ฃ = ๐๐๐๐๐๐ ๐๐ฮฉ
๐๐
Which is the same as,
div๐ฮฉ
๐๐ฃ = ๐๐ง๐ฮฉ
๐๐
Second-Order Tensor field
[email protected] 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 (โ๐ โ 1 in general).
Physical Components of Tensors
[email protected] 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 ๐ 1, ๐ 2 and ๐ 3.
These can be expressed in terms of dimensionless unit vectors as, ๐๐๐, ๐ sin ๐ ๐๐ , and ๐๐ since the
magnitudes of the basis vectors are ๐, ๐ sin๐ , and 1
or ๐11, ๐22, ๐33 respectively.
As an example consider a force with the contravariant components ๐น1, ๐น2 and ๐น3,
๐ = ๐น1๐ 1 + ๐น2๐ 2 + ๐น3 ๐ 3
= ๐๐น1๐๐ + ๐ sin ๐ ๐น2๐๐ + ๐น3๐๐
Physical Components
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 97
Which may also be expressed in terms of physical components,
๐ = ๐น๐๐๐ + ๐น๐๐๐ + ๐น๐๐๐.
While these physical components {๐น๐, ๐น๐ , ๐น๐} have the
dimensions of force, for the contravariant components normalized by the in terms of the unit vectors along these axes to be consistent, {๐๐น1, ๐ sin ๐ ๐น2, ๐น3} must each be in the units of a force. Hence, ๐น1, ๐น2 and ๐น3 may not themselves be in force units. The consistency requirement implies, ๐น๐ = ๐๐น1, ๐น๐ = ๐ sin ๐ ๐น2, and ๐น๐ = ๐น3
Physical Components
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 98
For the same reasons, if we had used covariant components, the relationships,
๐น๐ = ๐น1
๐, ๐น๐ =
๐น2
๐ sin ๐, and ๐น๐ = ๐น3
The magnitudes of the reciprocal base vectors are 1
๐,
1
๐ sin ๐, and 1. While the physical components have the
dimensions of force, ๐น1 = ๐๐น๐ and ๐น2 = ๐ sin ๐ ๐น๐ have the
dimensions of moment, while ๐น1 =๐น๐
๐ and ๐น2 =
๐น๐
๐ sin ๐ are
in dimensions of force per unit length. Only the third components in both cases are given in the dimensions of force.
Physical Components
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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 ๐๐๐ or ๐๐๐ or
๐๐๐ in spherical polar coordinates. Here, as we have seen, the
magnitudes of the base vectors ๐11, ๐22, ๐33 or โ1, โ2 and โ3 are ๐, ๐ sin ๐ and 1. Using the rule specified above, the table below computes the physical components from the three associated tensors as follows:
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 100
Physical
Component
Contravariant Covariant Mixed
๐(๐๐) ๐11โ1โ1 = ๐11๐2 ๐11
โ1 โ1=
๐11
๐2 ๐1
1
โ1โ1 = ๐1
1
๐(๐๐ฝ) ๐12โ1โ2 = ๐12๐2 sin๐ ๐12
โ1 โ1=
๐12
๐2 sin ๐ ๐2
1
โ2โ1 =
๐21
sin ๐
๐(๐ฝ๐ฝ) ๐22โ2โ2 = ๐22๐2 sin2 ๐ ๐22
โ2 โ2=
๐22
๐2 sin2 ๐ ๐2
2
โ2โ2 = ๐2
2
๐(๐ฝ๐) ๐23โ2โ3 = ๐23๐ sin ๐ ๐23
โ2 โ3=
๐23
๐ sin ๐ ๐3
2
โ3โ2 = ๐3
2๐ sin ๐
๐(๐๐) ๐33โ3โ3 = ๐33 ๐33
โ3 โ3= ๐33 ๐3
3
โ3โ3 = ๐3
3
๐(๐๐) ๐31โ3โ1 = ๐31๐ ๐31
โ3 โ1=
๐31
๐ ๐1
3
โ1โ3 =
๐13
๐
[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 101
17. The transformation equations from the Cartesian to the oblate spheroidal coordinates ๐, ๐ผ and ๐ are: ๐ = ๐๐๐ผ ๐ฌ๐ข๐ง๐, ๐ = ๐ ๐๐ โ ๐ ๐ โ ๐ผ๐ , and ๐ = ๐๐๐ผ ๐๐จ๐ฌ ๐, 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:
๐๐๐ =๐๐ฅ
๐๐
2
+๐๐ฆ
๐๐
2
+๐๐ง
๐๐
2
= ๐๐ sin๐ 2 + ๐2๐21 โ ๐2
๐2 โ 1+ ๐๐ cos ๐ 2 = ๐2
๐2 โ ๐2
๐2 โ 1
๐๐๐ =๐๐ฅ
๐๐
2
+๐๐ฆ
๐๐
2
+๐๐ง
๐๐
2
= ๐2๐2 โ ๐2
1 โ ๐2
๐๐๐ =๐๐ฅ
๐๐
2
+๐๐ฆ
๐๐
2
+๐๐ง
๐๐
2
= ๐๐๐ 2
๐๐๐ =๐๐ฅ
๐๐
๐๐ฅ
๐๐+
๐๐ฆ
๐๐
๐๐ฆ
๐๐+
๐๐ง
๐๐
๐๐ง
๐๐
= ๐๐ sin ๐ ๐๐ sin๐ โ ๐๐๐2 โ 1
1 โ ๐2๐๐
1 โ ๐2
๐2 โ 1+ ๐๐ cos ๐ ๐๐ cos ๐
= 0 = ๐๐๐ = ๐๐๐
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Find an expression for the divergence of a vector in orthogonal curvilinear coordinates.
The gradient of a vector ๐ญ = ๐น๐๐ ๐ is ๐ป โ ๐ญ = ๐ ๐๐๐ โ ๐น๐๐ ๐ =๐น๐
,๐๐ ๐ โ ๐ ๐. 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:
๐ป โ ๐ = ๐น๐,๐๐
๐ โ ๐ ๐ = ๐น๐ ,๐ =1
๐
๐ ๐ ๐น๐
๐๐ฅ๐
=1
โ1โ2โ3
๐
๐๐ฅ1โ1โ2โ3๐น
1 +๐
๐๐ฅ2โ1โ2โ3๐น
2 +๐
๐๐ฅ3โ1โ2โ3๐น
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 ๐น(๐) = ๐น๐โ๐ (no sum on i). In terms of the physical components therefore, the divergence becomes,
=1
โ1โ2โ3
๐
๐๐ฅ1โ2โ3๐น(1) +
๐
๐๐ฅ2โ1โ3๐น(2) +
๐
๐๐ฅ3โ1โ2๐น(3)
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Find an expression for the Laplacian operator in Orthogonal coordinates. For the contravariant component of a vector, ๐น๐,
๐น๐,๐ =
1
๐
๐ ๐ ๐น๐
๐๐ฅ๐.
Now the contravariant component of gradient ๐น๐ = ๐๐๐๐,๐. Using this in place of the vector ๐น๐, we can write,
๐๐๐๐,๐๐ =1
๐
๐ ๐ ๐๐๐๐,๐
๐๐ฅ๐
given scalar ๐, the Laplacian ๐ป2๐ is defined as, ๐๐๐๐,๐๐ so that,
๐ป2๐ = ๐๐๐๐,๐๐ =1
๐
๐
๐๐ฅ๐๐ ๐๐๐
๐๐
๐๐ฅ๐
When coordinates are orthogonal, ๐๐๐ = ๐๐๐ = 0 whenever ๐ โ ๐. Expanding the computation formula therefore, we can write,
๐ป2๐ =1
โ1โ2โ3
๐
๐๐ฅ1
โ1โ2โ3
โ1
๐๐
๐๐ฅ1 +๐
๐๐ฅ2
โ1โ2โ3
โ2
๐๐
๐๐ฅ2 +๐
๐๐ฅ3
โ1โ2โ3
โ3
๐๐
๐๐ฅ3
=1
โ1โ2โ3
๐
๐๐ฅ1 โ2โ3
๐๐
๐๐ฅ1 +๐
๐๐ฅ2 โ1โ3
๐๐
๐๐ฅ2 +๐
๐๐ฅ3 โ1โ2
๐๐
๐๐ฅ3
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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 ๐๐๐ = ๐๐๐ = ๐๐๐ = 0 . This is the required proof of orthogonality. Using the computation formula in example 11, we may write for the oblate spheroidal coordinates that,
๐ป2ฮฆ =๐2 โ 1 1 โ ๐2
๐3๐2 ๐2 โ ๐2 ๐
๐๐๐๐๐
๐2 โ 1
1 โ ๐2
๐ฮฆ
๐๐
+๐
๐๐๐๐๐
1 โ ๐2
๐2 โ 1
๐ฮฆ
๐๐
+๐
๐๐
๐ ๐2 โ ๐2
๐๐ ๐2 โ 1 1 โ ๐2
๐ฮฆ
๐๐
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For a vector field ๐, show that grad ๐ ร is a third ranked tensor. Hence or otherwise show that div ๐ ร = โcurl ๐.
The secondโorder tensor ๐ ร is defined as ๐๐๐๐๐ข๐๐ ๐ โ๐ ๐. Taking the covariant derivative with an independent base, we have
grad ๐ ร = ๐๐๐๐๐ข๐ ,๐ ๐ ๐ โ ๐ ๐ โ ๐ ๐
This gives a third order tensor as we have seen. Contracting on the last two bases,
div ๐ ร = ๐๐๐๐๐ข๐ ,๐ ๐ ๐ โ ๐ ๐ โ ๐ ๐ = ๐๐๐๐๐ข๐ ,๐ ๐ ๐๐ฟ๐
๐ = ๐๐๐๐๐ข๐ ,๐ ๐ ๐ = โcurl ๐
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Begin with our familiar Cartesian system of coordinates. We can represent the position of a point (position vector) with three coordinates ๐ฅ, ๐ฆ, ๐ง (โ R) such that,
๐ซ = ๐ฅ๐ + ๐ฆ๐ + ๐ง๐
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 ๐ฅ, ๐ฆ โ {๐, ๐} such that, ๐ฅ = ๐ cos ๐, and ๐ฆ = ๐ sin ๐. Note also that this
transformation is invertible: ๐ = ๐ฅ2 + ๐ฆ2,and
๐ = tanโ1 ๐ฆ
๐ฅ
Coordinate transformations
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With such a transformation, we can locate any point in the 3-D space with three scalars {๐, ๐, ๐ง} instead of our previous set ๐ฅ, ๐ฆ, ๐ง . Our position vector is now,
๐ซ = ๐ cos ๐ ๐ + ๐ sin ๐ ๐ + ๐ง๐ = ๐๐๐ + ๐ง๐๐ง
where we define ๐๐ โก cos ๐ ๐ + sin ๐ ๐, ๐๐ง is no different from ๐. In order to complete our triad of basis vectors, we need a third vector, ๐๐. In selecting
๐๐, we want it to be such that ๐๐ , ๐๐, ๐๐ง can form an
orthonormal basis.
Curvilinear Coordinates
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Let ๐๐ = ๐๐ + ๐๐
To satisfy our conditions,
๐๐ โ ๐๐ = 0, ๐๐ โ ๐๐ง = 0, and ๐2 + ๐2 = 1.
It is easy to see that ๐๐ โก โsin ๐ ๐ + cos ๐ ๐ satisfies
these requirements. ๐๐ , ๐๐, ๐๐ง forms an orthonormal (that is, each member has unit magnitude and they are pairwise orthogonal) triad just like ๐, ๐, ๐ . 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
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Unlike our Cartesian system, we note that
๐๐(๐), ๐๐(๐), ๐๐ง as the first two of these are not
constants but vary with angular orientation. ๐๐ง remains a constant vector as in the Cartesian case.
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Continuing further with our transformation, we may again introduce two new scalars such that ๐, ๐ง โ {๐, ๐} in such a way that the position vector,
๐ซ = ๐๐๐ + ๐ง๐๐ง = ๐ sin ๐ ๐๐ + ๐ cos ๐ ๐๐ง โก ๐๐๐
Here, ๐ = ๐ sin ๐ , ๐ง = ๐ cos ๐. As before, we can use three scalars, ๐, ๐, ๐ instead of {๐, ๐, ๐ง}. In comparison to the original Cartesian system we began with, we have that,
๐ซ = ๐ฅ๐ + ๐ฆ๐ + ๐ง๐ = ๐ sin ๐ ๐๐ + ๐ cos ๐ ๐๐ง = ๐ sin ๐ cos ๐ ๐ + sin ๐ ๐ + ๐ cos ๐ ๐ = ๐ sin ๐ cos ๐ ๐ +๐ sin ๐ sin๐ ๐ + ๐ cos ๐ ๐ โก ๐๐๐
from which it is clear that the unit vector ๐๐ โก sin ๐ cos ๐ ๐ + sin ๐ sin ๐ ๐ + cos ๐ ๐.
Spherical Polar
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Again, we introduce the unit vector, ๐๐ โก cos ๐ cos ๐ ๐ + cos ๐ sin๐ ๐ โ sin ๐ ๐ and retain ๐๐ = โ sin๐ ๐ + cos ๐ ๐ as before.
It is easy to demonstrate the fact that these vectors constitute another orthonormal set. Combining the two transformations, we can move from ๐ฅ, ๐ฆ, ๐ง system of coordinates to ๐, ๐, ๐ directly by the transformation equations, ๐ฅ = ๐ sin ๐ cos ๐, ๐ฆ = ๐ sin ๐ sin๐ and ๐ง =๐ cos ๐.
The orthonormal set of basis for the ๐, ๐, ๐ system is
๐๐ ๐, ๐ , ๐๐ ๐, ๐ , ๐๐ ๐ .
This is the Spherical Polar Coordinate System.
Spherical Polar
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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,
๐๐ ๐, ๐, ๐ง = cos ๐ ๐ + sin ๐ ๐ ๐๐(๐, ๐, ๐ง) = โsin ๐ ๐ + cos ๐ ๐ ๐๐ง ๐, ๐, ๐ง = ๐
Variable Bases
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For Cylindrical Polar, and for Spherical Polar,
๐๐(๐, ๐, ๐) = sin ๐ cos ๐ ๐ + sin ๐ sin๐ ๐ + cos ๐ ๐
๐๐ ๐, ๐, ๐ = cos ๐ cos ๐ ๐ + cos ๐ sin๐ ๐ โ sin ๐ ๐ ๐๐ ๐, ๐, ๐ = โ sin๐ ๐ + cos ๐ ๐.
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.
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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
๐ฏ = ๐ฃ1๐ 1 + ๐ฃ2๐ 2 + ๐ฃ3๐ 3
when referred to the basis vectors {๐ 1, ๐ 2, ๐ 3}, for position vectors, in order that the transformation refers to the same position vector we started with in the Cartesian system
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We have for Cylindrical Polar, the vector, ๐ซ(๐, ๐, ๐ง) = ๐๐๐(๐) + ๐ง๐๐ง
and for Spherical Polar, ๐ซ(๐, ๐, ๐) = ๐๐๐(๐, ๐)
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.
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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.
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Beginning from the position vector, given a system with coordinate variables ๐ผ1, ๐ผ2, ๐ผ3 , it is easy to prove that the set of vectors,
๐๐ซ ๐ผ1, ๐ผ2, ๐ผ3
๐๐ผ1,๐๐ซ ๐ผ1, ๐ผ2, ๐ผ3
๐๐ผ1,๐๐ซ ๐ผ1, ๐ผ2, ๐ผ3
๐๐ผ1
Which we can write more compactly as
๐ ๐ โก๐๐ซ
๐๐ผ๐, ๐ = 1,2,3
constitute an independent set. This set, with its dual set
of vectors, ๐ ๐ such that ๐ ๐ โ ๐ ๐= ๐ฟ๐๐, constitute the
โNatural basesโ for the coordinate system.
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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