3. tensor calculus jan 2013

Tensor Calculus Differentials & Directional Derivatives

We are presently concerned with Inner Product spaces in our treatment of the Mechanics of Continua. Consider a map,

𝑭: V → W This maps from the domainV to W – both of which are Euclidean vector spaces. The concepts of limit and continuity carries naturally from the real space to any Euclidean vector space.

The Gateaux Differential

[email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 2

Let 𝒗0 ∈ V and 𝒘0 ∈ W, as usual we can say that the limit

lim𝒗→ 𝒗0

𝑭 𝒗 = 𝒘0

if for any pre-assigned real number 𝜖 > 0, no matter how small, we can always find a real number 𝛿 > 0 such that 𝑭 𝒗 − 𝒘0 ≤ 𝜖 whenever 𝒗 − 𝒗0 < 𝛿.

The function is said to be continuous at 𝒗0 if 𝑭 𝒗0 exists and 𝑭 𝒗0 = 𝒘0



Specifically, for 𝛼 ∈ ℛ let this map be:

𝐷𝑭 𝒙, 𝒉 ≡ lim𝛼→0

𝑭 𝒙 + 𝛼𝒉 − 𝑭 𝒙

𝛼=

𝑑

𝑑𝛼𝑭 𝒙 + 𝛼𝒉

𝛼=0

We focus attention on the second variable ℎ while we allow the dependency on 𝒙 to be as general as possible. We shall show that while the above function can be any given function of 𝒙 (linear or nonlinear), the above map is always linear in 𝒉 irrespective of what kind of Euclidean space we are mapping from or into. It is called the Gateaux Differential.



Let us make the Gateaux differential a little more familiar in real space in two steps: First, we move to the real space and allow ℎ → 𝑑𝑥 and we obtain,

𝐷𝐹(𝑥, 𝑑𝑥) = lim𝛼→0

𝐹 𝑥 + 𝛼𝑑𝑥 − 𝐹 𝑥

𝛼=

𝑑

𝑑𝛼𝐹 𝑥 + 𝛼𝑑𝑥

𝛼=0

And let 𝛼𝑑𝑥 → Δ𝑥, the middle term becomes,

limΔ𝑥→0

𝐹 𝑥 + Δ𝑥 − 𝐹 𝑥

Δ𝑥𝑑𝑥 =

𝑑𝐹

𝑑𝑥𝑑𝑥

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).



It is easily shown that the Gateaux differential is linear in its second argument, ie, for 𝛼 ∈ R

𝐷𝑭 𝒙, 𝛼𝒉 = 𝛼𝐷𝑭 𝒙, 𝒉

Furthermore, 𝐷𝑭 𝒙, 𝒈 + 𝒉 = 𝐷𝑭 𝒙, 𝒈 + 𝐷𝑭 𝒙, 𝒉

and that for 𝛼, 𝛽 ∈ R 𝐷𝑭 𝒙, 𝛼𝒈 + 𝛽𝒉 = 𝛼𝐷𝑭 𝒙, 𝒈 + 𝛽𝐷𝑭 𝒙, 𝒉



The Frechét derivative or gradient of a differentiable function is the linear operator, grad 𝑭(𝒗) such that, for 𝑑𝒗 ∈ V,

𝑑𝑭 𝒗

𝑑𝒗𝑑𝒗 ≡ grad 𝑭 𝒗 𝑑𝒗 ≡ 𝐷𝑭 𝒗, 𝑑𝒗

Obviously, grad 𝑭(𝒗) is a tensor because it is a linear transformation from one Euclidean vector space to another. Its linearity derives from the second argument of the RHS.

The Frechét Derivative


The Gradient of a Differentiable Real-Valued Function For a real vector function, we have the map:

𝑓: V → R The Gateaux differential in this case takes two vector arguments and maps to a real value:

𝐷𝑓: V × V → R This is the classical definition of the inner product so that we can define the differential as,

𝑑𝑓(𝒗)

𝑑𝒗⋅ 𝑑𝒗 ≡ 𝐷𝑓 𝒗, 𝑑𝒗

This quantity, 𝑑𝑓(𝒗)

𝑑𝒗

defined by the above equation, is clearly a vector.

Differentiable Real-Valued Function


It is the gradient of the scalar valued function of the vector variable. We now proceed to find its components in general coordinates. To do this, we choose a basis 𝐠𝑖 ⊂ V. On such a basis, the function,

𝒗 = 𝑣𝑖𝐠𝑖

and,

𝑓 𝒗 = 𝑓 𝑣𝑖𝐠𝑖

we may also express the independent vector 𝑑𝒗 = 𝑑𝑣𝑖𝐠𝑖

on this basis.

Real-Valued Function


The Gateaux differential in the direction of the basis vector 𝐠𝑖 is,

𝐷𝑓 𝒗, 𝐠𝑖 = lim𝛼→0

𝑓 𝒗 + 𝛼𝐠𝑖 − 𝑓 𝒗

𝛼= lim

𝛼→0

𝑓 𝑣𝑗𝐠𝑗 + 𝛼𝐠𝑖 − 𝑓 𝑣𝑖𝐠𝑖

𝛼

= lim𝛼→0

𝑓 𝑣𝑗 + 𝛼𝛿𝑖𝑗

𝐠𝑗 − 𝑓 𝑣𝑖𝐠𝑖

𝛼

As the function 𝑓 does not depend on the vector basis, we can substitute the vector function by the real function of the components 𝑣1, 𝑣2, 𝑣3 so that,

𝑓 𝑣𝑖𝐠𝑖 = 𝑓 𝑣1, 𝑣2, 𝑣3

in which case, the above differential, similar to the real case discussed earlier becomes,

𝐷𝑓 𝒗, 𝐠𝑖 =𝜕𝑓 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖

Real-Valued Function


Of course, it is now obvious that the gradient 𝑑𝑓(𝒗)

𝑑𝒗 of

the scalar valued vector function, expressed in its dual basis must be,

𝑑𝑓(𝒗)

𝑑𝒗=

𝜕𝑓 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖𝐠𝑖

so that we can recover,

𝐷𝑓 𝒗, 𝐠𝑖 =𝑑𝑓 𝒗

𝑑𝒗⋅ 𝑑𝒗 =

𝜕𝑓 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑗𝐠𝑗 ⋅ 𝐠𝑖

=𝜕𝑓 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑗𝛿𝑖

𝑗=

𝜕𝑓 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖

Hence we obtain the well-known result that

grad 𝑓 𝒗 =𝑑𝑓(𝒗)

𝑑𝒗=

𝜕𝑓 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖𝐠𝑖

which defines the Frechét derivative of a scalar valued function with respect to its vector argument.


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 𝐠𝑖 ⊂ V. On such a basis, the function,

𝑭 𝒗 = 𝐹𝑘 𝒗 𝐠𝑘

The functional dependency on the basis vectors are ignorable on account of the fact that the components themselves are fixed with respect to the basis. We can therefore write,

𝑭 𝒗 = 𝐹𝑘 𝑣1, 𝑣2, 𝑣3 𝐠𝑘

Vector valued Function


𝐷𝑭 𝒗, 𝑑𝒗 = 𝐷𝑭 𝑣𝑖𝐠𝑖 , 𝑑𝑣𝑖𝐠𝑖

= 𝐷𝑭 𝑣𝑖𝐠𝑖 , 𝐠𝑖 𝑑𝑣𝑖

= 𝐷𝐹𝑘 𝑣𝑖𝐠𝑖 , 𝐠𝑖 𝐠𝑘𝑑𝑣𝑖

Again, upon noting that the functions 𝐷𝐹𝑘 𝑣𝑖𝐠𝑖 , 𝐠𝑖 ,

𝑘 = 1,2,3 are not functions of the vector basis, they can be written as functions of the scalar components alone so that we have, as before,

𝐷𝑭 𝒗, 𝑑𝒗 = 𝐷𝐹𝑘 𝑣𝑖𝐠𝑖 , 𝐠𝑖 𝐠𝑘𝑑𝑣𝑖 = 𝐷𝐹𝑘 𝒗, 𝐠𝑖 𝐠𝑘𝑑𝑣𝑖

Vector Derivatives


Which we can compare to the earlier case of scalar valued function and easily obtain,

𝐷𝑭 𝒗, 𝑑𝒗 =𝜕𝐹𝑘 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑗𝐠𝑗 ⋅ 𝐠𝑖 𝐠𝑘𝑑𝑣𝑖

=𝜕𝐹𝑘 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑗𝐠𝑘 ⊗ 𝐠𝑗 𝐠𝑖𝑑𝑣𝑖

=𝜕𝐹𝑘 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑗𝐠𝑘 ⊗ 𝐠𝑗 𝑑𝒗

= 𝑑𝑭 𝒗

𝑑𝒗𝑑𝒗 ≡ grad 𝑭 𝒗 𝑑𝒗

Clearly, the tensor gradient of the vector-valued vector function is,

𝑑𝑭 𝒗

𝑑𝒗= grad 𝑭 𝒗 =

𝜕𝐹𝑘 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑗𝐠𝑘 ⊗ 𝐠𝑗

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 𝑭 𝒗 = tr𝜕𝐹𝑘 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑗𝐠𝑘 ⊗ 𝐠𝑗

=𝜕𝐹𝑘 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑗𝐠𝑘 ⋅ 𝐠𝑗

=𝜕𝐹𝑘 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑗𝛿𝑘

𝑗

=𝜕𝐹𝑗 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑗

The Trace & Divergence


Consider the mapping, 𝑻: V → T

Which maps from an inner product space to a tensor space. The Gateaux differential in this case now takes two vectors and produces a tensor:

𝐷𝑻: V × V → T With the usual notation, we may write,

𝐷𝑻 𝒗, 𝑑𝒗 = 𝐷𝑻 𝑣𝑖𝐠𝑖 , 𝑑𝑣𝑖𝐠𝑖 = 𝐷𝑻 𝑣𝑖𝐠𝑖 , 𝑑𝑣𝑖𝐠𝑖

= 𝑑𝑻 𝒗

𝑑𝒗𝑑𝒗 ≡ grad 𝑻 𝒗 𝑑𝒗

= 𝐷𝑇𝛼𝛽 𝑣𝑖𝐠𝑖, 𝑑𝑣𝑖𝐠𝑖 𝐠𝛼 ⊗ 𝐠𝛽

= 𝐷𝑇𝛼𝛽 𝒗, 𝐠𝑖 𝐠𝛼 ⊗ 𝐠𝛽 𝑑𝑣𝑖

Each of these nine functions look like the real differential of several variables we considered earlier.

Tensor-Valued Vector Function


The independence of the basis vectors imply, as usual, that

𝐷𝑇𝛼𝛽 𝒗, 𝐠𝑖 =𝜕𝑇𝛼𝛽 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖

so that,

𝐷𝑻 𝒗, 𝑑𝒗 = 𝐷𝑇𝛼𝛽 𝒗, 𝐠𝑖 𝐠𝛼 ⊗ 𝐠𝛽 𝑑𝑣𝑖

= 𝜕𝑇𝛼𝛽 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖𝐠𝛼 ⊗ 𝐠𝛽 𝑑𝑣𝑖

= 𝜕𝑇𝛼𝛽 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖𝐠𝛼 ⊗ 𝐠𝛽 𝐠𝑖 ⋅ 𝑑𝒗

= 𝜕𝑇𝛼𝛽 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖𝐠𝛼 ⊗ 𝐠𝛽 ⊗ 𝐠𝑖 𝑑𝒗 =

𝑑𝑻 𝒗

𝑑𝒗𝑑𝒗

≡ grad 𝑻 𝒗 𝑑𝒗

Which defines the third-order tensor,

𝑑𝑻 𝒗

𝑑𝒗= grad 𝑻 𝒗 =

𝜕𝑇𝛼𝛽 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖𝐠𝛼 ⊗ 𝐠𝛽 ⊗ 𝐠𝑖

and with no further ado, we can see that a third-order tensor transforms a vector into a second order tensor.


The divergence operation can be defined in several ways. Most common is achieved by the contraction of the last two basis so that,

div 𝑻 𝒗 =𝜕𝑇𝛼𝛽 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖𝐠𝛼 ⊗ 𝐠𝛽 ⋅ 𝐠𝑖

=𝜕𝑇𝛼𝛽 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖𝐠𝛼 𝐠𝛽 ⋅ 𝐠𝑖

=𝜕𝑇𝛼𝛽 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖𝐠𝛼𝛿𝛽

𝑖 =𝜕𝑇𝛼𝑖 𝑣1, 𝑣2, 𝑣3

𝜕𝑣𝑖𝐠𝛼

It can easily be shown that this is the particular vector that gives, for all constant vectors 𝒂,

div 𝑻 𝒂 ≡ div 𝑻T𝒂

The Divergence of a Tensor Function


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


Consider the Map: 𝑓: T → R

The Gateaux differential in this case takes two vector arguments and maps to a real value:

𝐷𝑓: T × T → R Which is, as usual, linear in the second argument. The Frechét derivative can be expressed as the first component of the following scalar product:

𝐷𝑓 𝑻, 𝑑𝑻 =𝑑𝑓(𝑻)

𝑑𝑻: 𝑑𝑻

which, we recall, is the trace of the contraction of one tensor with the transpose of the other second-order tensors. This is a scalar quantity.

Frechét Derivative


Guided by the previous result of the gradient of the scalar valued function of a vector, it is not difficult to see that,

𝑑𝑓(𝑻)

𝑑𝑻=

𝜕𝑓(𝑇11, 𝑇12, … , 𝑇33)

𝜕𝑇𝛼𝛽𝐠𝛼 ⊗ 𝐠𝛽

In the dual to the same basis, we can write,

𝑑𝑻 = 𝑑𝑇𝑖𝑗𝐠𝑖 ⊗ 𝐠𝑗

Clearly, 𝑑𝑓(𝑻)

𝑑𝑻: 𝑑𝑻 =

𝜕𝑓(𝑇11, 𝑇12, … , 𝑇33)

𝜕𝑇𝛼𝛽𝐠𝛼 ⊗ 𝐠𝛽 : 𝑑𝑇𝑖𝑗𝐠𝑖 ⊗ 𝐠𝑗

=𝜕𝑓(𝑇11, 𝑇12, … , 𝑇33)

𝜕𝑇𝑖𝑗𝑑𝑇𝑖𝑗

Again, note the covariance of the quotient indices.


Let 𝐒 be a symmetric and positive definite tensor and let 𝐼1 𝑺 , 𝐼2 𝑺 and 𝐼3 𝑺 be the three principal invariants of

𝐒 show that (a) 𝜕𝑰1 𝐒

𝜕𝐒= 𝟏 the identity tensor, (b)

𝜕𝑰2 𝐒

𝜕𝐒= 𝐼1 𝐒 𝟏 − 𝐒 and (c)

𝜕𝐼3 𝐒

𝜕𝐒= 𝐼3 𝐒 𝐒−1

𝜕𝑰1 𝐒

𝜕𝐒 can be written in the invariant component form

as, 𝜕𝐼1 𝐒

𝜕𝐒=

𝜕𝐼1 𝐒

𝜕𝑆𝑖𝑗

𝐠𝑖 ⊗ 𝐠𝑗

Examples


Recall that 𝐼1 𝐒 = tr 𝐒 = 𝑆αα hence

𝜕𝐼1 𝐒

𝜕𝐒=

𝜕𝐼1 𝐒

𝜕𝑆𝑖𝑗


=𝜕𝑆α

α

𝜕𝑆𝑖𝑗𝐠𝑖 ⊗ 𝐠𝑗

= 𝛿𝛼𝑖 𝛿𝑗

𝛼𝐠𝑖⊗ 𝐠𝑗

= 𝛿𝑗𝑖𝐠𝑖⊗ 𝐠𝑗 = 𝟏

which is the identity tensor as expected.

(a) Continued


𝜕𝐼2 𝐒

𝜕𝐒 in a similar way can be written in the invariant component form

as, 𝜕𝐼2 𝐒

𝜕𝐒=

1

2

𝜕𝐼1 𝐒

𝜕𝑆𝑖𝑗

𝑆αα𝑆𝛽

𝛽− 𝑆𝛽

α𝑆α𝛽


where we have utilized the fact that 𝐼2 𝐒 =1

2tr2 𝐒 − tr(𝐒2) .

Consequently, 𝜕𝐼2 𝐒

𝜕𝐒=

1

2

𝜕

𝜕𝑆𝑖𝑗

𝑆αα𝑆𝛽

𝛽− 𝑆𝛽

α𝑆α𝛽


=1

2𝛿𝛼

𝑖 𝛿𝑗𝛼𝑆𝛽

𝛽+ 𝛿𝛽

𝑖 𝛿𝑗𝛽𝑆α

α − 𝛿𝛽𝑖 𝛿𝑗

𝛼𝑆α𝛽

− 𝛿𝛼𝑖 𝛿𝑗

𝛽𝑆𝛽

α 𝐠𝑖 ⊗ 𝐠𝑗

=1

2𝛿𝑗

𝑖𝑆𝛽𝛽

+ 𝛿𝑗𝑖𝑆α

α − 𝑆𝑖𝑗− 𝑆𝑖

𝑗𝐠𝑖 ⊗ 𝐠𝑗

= 𝛿𝑗𝑖𝑆α

α − 𝑆𝑖𝑗

𝐠𝑖 ⊗ 𝐠𝑗 = 𝐼1 𝐒 𝟏 − 𝐒

(b) 𝜕𝐼2 𝐒

𝜕𝐒= 𝐼1 𝐒 𝟏 − 𝐒


𝜕𝑰3 𝐒

𝜕𝐒=

𝜕det 𝐒

𝜕𝐒= 𝐒𝐜

the cofactor of 𝐒. Clearly 𝐒𝐜 = det 𝐒 𝐒−1 = 𝑰3 𝐒 𝐒−1 as required. Details of this for the contravariant components of a tensor is presented below. Let

𝐒 = 𝑆 =1

3!𝑒𝑖𝑗𝑘𝑒𝑟𝑠𝑡𝑆𝑖𝑟𝑆𝑗𝑠𝑆𝑘𝑡

Differentiating wrt 𝑆𝛼𝛽, we obtain,

𝜕𝑆

𝜕𝑆𝛼𝛽𝐠𝛼 ⊗ 𝐠𝛽=

1

3!𝑒𝑖𝑗𝑘𝑒𝑟𝑠𝑡

𝜕𝑆𝑖𝑟

𝜕𝑆𝛼𝛽𝑆𝑗𝑠𝑆𝑘𝑡 + 𝑆𝑖𝑟

𝜕𝑆𝑗𝑠

𝜕𝑆𝛼𝛽𝑆𝑘𝑡 + 𝑆𝑖𝑟𝑆𝑗𝑠

𝜕𝑆𝑘𝑡

𝜕𝑆𝛼𝛽𝐠𝛼 ⊗ 𝐠𝛽

=1

3!𝑒𝑖𝑗𝑘𝑒𝑟𝑠𝑡 𝛿𝑖

𝛼𝛿𝑟𝛽𝑆𝑗𝑠𝑆𝑘𝑡 + 𝑆𝑖𝑟𝛿𝑗

𝛼𝛿𝑠𝛽𝑆𝑘𝑡 + 𝑆𝑖𝑟𝑆𝑗𝑠𝛿𝑘

𝛼𝛿𝑡𝛽

𝐠𝛼 ⊗ 𝐠𝛽

=1

3!𝑒𝛼𝑗𝑘𝑒𝛽𝑠𝑡 𝑆𝑗𝑠𝑆𝑘𝑡 + 𝑆𝑗𝑠𝑆𝑘𝑡 + 𝑆𝑗𝑠𝑆𝑘𝑡 𝐠𝛼 ⊗ 𝐠𝛽

=1

2!𝑒𝛼𝑗𝑘𝑒𝛽𝑠𝑡𝑆𝑗𝑠𝑆𝑘𝑡𝐠𝛼 ⊗ 𝐠𝛽≡ 𝑆c 𝛼𝛽𝐠𝛼 ⊗ 𝐠𝛽

Which is the cofactor of 𝑆𝛼𝛽


Consider the function 𝑓(𝑺) = tr 𝑺𝑘 where 𝑘 ∈ ℛ, and

𝑺 is a tensor.

𝑓 𝑺 = tr 𝑺𝑘 = 𝑺𝑘: 𝟏

To be specific, let 𝑘 = 3.

Real Tensor Functions


The Gateaux differential in this case,

𝐷𝑓 𝑺, 𝑑𝑺 =𝑑

𝑑𝛼𝑓 𝑺 + 𝛼𝑑𝑺

𝛼=0

=𝑑

𝑑𝛼tr 𝑺 + 𝛼𝑑𝑺 3

𝛼=0

=𝑑

𝑑𝛼tr 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺

𝛼=0

= tr𝑑

𝑑𝛼𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺

𝛼=0

= tr 𝑑𝑺 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺+ 𝑺 + 𝛼𝑑𝑺 𝑑𝑺 𝑺 + 𝛼𝑑𝑺

+ 𝑺 + 𝛼𝑑𝑺 𝑺 + 𝛼𝑑𝑺 𝑑𝑺 𝛼=0

= tr 𝑑𝑺 𝑺 𝑺 + 𝑺 𝑑𝑺 𝑺 + 𝑺 𝑺 𝒅𝑺 = 3 𝑺2 T: 𝑑𝑺



With the last equality coming from the definition of the Inner product while noting that a circular permutation does not alter the value of the trace. It is easy to establish inductively that in the most general case, for 𝑘 > 0, we have,

𝐷𝑓 𝑺, 𝑑𝑺 = 𝑘 𝑺𝑘−1 T: 𝑑𝑺

Clearly, 𝑑

𝑑𝑺 tr 𝑺𝑘 = 𝑘 𝑺𝑘−1

T



When 𝑘 = 1,

𝐷𝑓 𝑺, 𝑑𝑺 =𝑑

𝑑𝛼𝑓 𝑺 + 𝛼𝑑𝑺

𝛼=0

=𝑑

𝑑𝛼tr 𝑺 + 𝛼𝑑𝑺

𝛼=0

= tr 𝟏 𝑑𝑺 = 𝟏: 𝑑𝑺

Or that, 𝑑

𝑑𝑺 tr 𝑺 = 𝟏.



Derivatives of the other two invariants of the tensor 𝑺 can be found as follows:

𝑑

𝑑𝑺 𝐼2(𝑺)

=1

2

𝑑

𝑑𝑺tr2 𝑺 − tr 𝑺2

=1

22tr 𝑺 𝟏 − 2 𝑺T = tr 𝑺 𝟏 − 𝑺T

.



To Determine the derivative of the third invariant, we begin with the trace of the Cayley-Hamilton for 𝑺: tr 𝑺3 − 𝐼1𝑺

2 + 𝐼2𝑺 − 𝐼3𝑰 = tr(𝑺3) − 𝐼1tr 𝑺2 + 𝐼2tr 𝑺− 3𝐼3 = 0

Therefore, 3𝐼3 = tr(𝑺3) − 𝐼1tr 𝑺2 + 𝐼2tr 𝑺

𝐼2 𝑺 =1

2tr2 𝑺 − tr 𝑺2

.



We can therefore write, 3𝐼3 𝑺 = tr(𝑺3) − tr 𝑺 tr 𝑺2

+1

2tr2 𝑺 − tr 𝑺2 tr 𝑺

so that, in terms of traces only,

𝐼3 𝑺 =1

6tr3 𝑺 − 3tr 𝑺 tr 𝑺2 + 2tr(𝑺3)

Differentiating the Invariants


Clearly, 𝑑𝐼3(𝑺)

𝑑𝑺=

1

63tr2 𝑺 𝟏 − 3tr 𝑺2 − 3tr 𝑺 2𝑺𝑇 + 2 × 3 𝑺2 𝑇

= 𝐼2𝟏 − tr 𝑺 𝑺𝑇 + 𝑺2𝑇



.



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


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 ℰ is a set of elements called points. For each pair of points 𝒙, 𝒚 ∈ ℰ, ∃ 𝒖 𝒙, 𝒚 ∈ E with the following two properties:

1. 𝒖 𝒙, 𝒚 = 𝒖 𝒙, 𝒛 + 𝒖 𝒛, 𝒚 ∀𝒙, 𝒚, 𝒛 ∈ ℰ

2. 𝒖 𝒙, 𝒚 = 𝒖 𝒙, 𝒛 ⇔ 𝒚 = 𝒛

Based on these two, we proceed to show that, 𝒖 𝒙, 𝒙 = 𝟎

And that, 𝒖 𝒙, 𝒛 = −𝒖 𝒛, 𝒙



From property 1, let 𝒚 → 𝒙 it is clear that, 𝒖 𝒙, 𝒙 = 𝒖 𝒙, 𝒛 + 𝒖 𝒛, 𝒙

And it we further allow 𝒛 → 𝒙, we find that, 𝒖 𝒙, 𝒙 = 𝒖 𝒙, 𝒙 + 𝒖 𝒙, 𝒙 = 2𝒖 𝒙, 𝒙

Which clearly shows that 𝒖 𝒙, 𝒙 = 𝒐 the zero vector.

Similarly, from the above, we find that, 𝒖 𝒙, 𝒙 = 𝒐 = 𝒖 𝒙, 𝒛 + 𝒖 𝒛, 𝒙

So that 𝒖 𝒙, 𝒛 = −𝒖 𝒛, 𝒙



Coordinate System


Note that ℰ is NOT a vector space. In our discussion, the vector space E to which 𝒖 belongs, is associated as we have shown. It is customary to oscillate between these two spaces. When we are talking about the vectors, they are in E while the points are in ℰ. We adopt the convention that 𝒙 𝒚 ≡ 𝒖 𝒙, 𝒚 referring to the vector 𝒖. If therefore we choose an arbitrarily fixed point 𝟎 ∈ ℰ, we are associating 𝒙 𝟎 , 𝒚 𝟎 and 𝒛 𝟎 respectively with 𝒖 𝒙, 𝟎 , 𝒖 𝒚, 𝟎 and 𝒖 𝒛, 𝟎 .

These are vectors based on the points 𝒙, 𝒚 and 𝒛 with reference to the origin chosen. To emphasize the association with both points of ℰ as well as members of E they are called Position Vectors.

Position Vectors


Recall that by property 1, 𝒙 𝒚 ≡ 𝒖 𝒙, 𝒚 = 𝒖 𝒙, 𝟎 + 𝒖 𝟎, 𝒚

Furthermore, we have deduced that 𝒖 𝟎, 𝒚 = −𝒖(𝒚, 𝟎)

We may therefore write that, 𝒙 𝒚 = 𝒙 𝟎 − 𝒚(𝟎)

Which, when there is no ambiguity concerning the chosen origin, we can write as,

𝒙 𝒚 = 𝒙 − 𝒚

And the distance between the two is,

𝒅 𝒙 𝒚 = 𝒅 𝒙 − 𝒚 = 𝒙 − 𝒚

Length in the Point Space


The norm of a vector is the square root of the inner product of the vector with itself. If the coordinates of 𝒙 and 𝒚 on a set of independent vectors are 𝑥𝑖 and 𝑦𝑖, then the distance we seek is,

𝒅 𝒙 − 𝒚 = 𝒙 − 𝒚 = 𝑔𝑖𝑗(𝑥𝑖 − 𝑦𝑖)(𝑥𝑗 − 𝑦𝑗)

The more familiar Pythagorean form occurring only when 𝑔𝑖𝑗 = 1.

Metric Properties


Consider a Cartesian coordinate system 𝑥, 𝑦, 𝑧 or 𝑦1, 𝑦2 and 𝑦3with an orthogonal basis. Let us now have the possibility of transforming to another coordinate system of an arbitrary nature: 𝑥1, 𝑥2, 𝑥3. We can represent the transformation and its inverse in the equations:

𝑦𝑖 = 𝑦𝑖 𝑥1, 𝑥2, 𝑥3 , 𝑥𝑖 = 𝑥𝑖(𝑦1, 𝑦2, 𝑦3)

And if the Jacobian of transformation,

Coordinate Transformations


𝜕𝑦𝑖

𝜕𝑥𝑗≡

𝜕 𝑦1, 𝑦2, 𝑦3

𝜕 𝑥1, 𝑥2, 𝑥3 ≡

𝜕𝑦1

𝜕𝑥1

𝜕𝑦1

𝜕𝑥2

𝜕𝑦1

𝜕𝑥3

𝜕𝑦2

𝜕𝑥1

𝜕𝑦2

𝜕𝑥2

𝜕𝑦2

𝜕𝑥3

𝜕𝑦3

𝜕𝑥1

𝜕𝑦3

𝜕𝑥2

𝜕𝑦3

𝜕𝑥3

does not vanish, then the inverse transformation will exist. So that,

𝑥𝑖 = 𝑥𝑖(𝑦1, 𝑦2, 𝑦3)

Jacobian Determinant


Given a position vector 𝐫 = 𝐫(𝑥1, 𝑥2, 𝑥3)

In the new coordinate system, we can form a set of three vectors,

𝐠𝑖 =𝜕𝐫

𝜕𝑥𝑖, 𝑖 = 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. 𝐫 = 𝐫 𝑥1, 𝑥2, 𝑥3 = 𝑦1𝐢 +𝑦2𝐣 + 𝑦3𝐤 So that

𝐢 =𝜕𝐫

𝜕𝑦1 , 𝐣 =𝜕𝐫

𝜕𝑦2 and 𝐤 =𝜕𝐫

𝜕𝑦3

The fact that this is true in the general case is easily seen when we consider that along those lines, only the variable we are differentiating with respect to varies. Consider the general case where,

𝐫 = 𝐫 𝑥1, 𝑥2, 𝑥3 The total differential of 𝐫 is simply,


𝑑𝐫 =𝜕𝐫

𝜕𝑥1𝑑𝑥1 +

𝜕𝐫

𝜕𝑥2𝑑𝑥2 +

𝜕𝐫

𝜕𝑥3𝑑𝑥3

≡ 𝑑𝑥1𝐠1 + 𝑑𝑥2𝐠2 + 𝑑𝑥3𝐠3

With 𝐠𝑖 𝑥1, 𝑥2, 𝑥3 , 𝑖 = 1,2,3 now depending in general on 𝑥1, 𝑥2 and 𝑥3 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, 𝐠𝑖 , 𝑖 = 1,2,3 are constants, normalized and orthogonal. They are our familiar

𝐢 =𝜕𝐫

𝜕𝑦1, 𝐣 =

𝜕𝐫

𝜕𝑦2 and 𝐤 =

𝜕𝐫

𝜕𝑦3.

Natural Dual Bases


We now proceed to show that the set of basis vectors, 𝐠𝑖 ≡ 𝛻𝑥𝑖

is reciprocal to 𝐠𝑖. The total differential 𝑑𝐫 can be written in terms of partial differentials as,

𝑑𝐫 =𝜕𝐫

𝜕𝑦𝑖𝑑𝑦𝑖

Which when expressed in component form yields,

𝑑𝑥1 =𝜕𝑥1

𝜕𝑦1𝑑𝑦1 +

𝜕𝑥1

𝜕𝑦2𝑑𝑦2 +

𝜕𝑥1

𝜕𝑦3𝑑𝑦3

𝑑𝑥2 =𝜕𝑥2

𝜕𝑦1𝑑𝑦1 +

𝜕𝑥2

𝜕𝑦2𝑑𝑦2 +

𝜕𝑥2

𝜕𝑦3𝑑𝑦3

𝑑𝑥3 =𝜕𝑥3

𝜕𝑦1𝑑𝑦1 +

𝜕𝑥3

𝜕𝑦2𝑑𝑦2 +

𝜕𝑥3

𝜕𝑦3𝑑𝑦3

Or, more compactly that,

𝑑𝑥𝑗 =𝜕𝑥𝑗

𝜕𝑦𝑖𝑑𝑦𝑖 = 𝛻𝑥𝑗 ⋅ 𝑑𝐫


In Cartesian coordinates, 𝑦𝑖 , 𝑖 = 1, . . , 3 for any scalar field 𝜙 (𝑦1, 𝑦2, 𝑦3),

𝜕𝜙

𝜕𝑦𝑖𝑑𝑦𝑖 =

𝜕𝜙

𝜕𝑥𝑑𝑥 +

𝜕𝜙

𝜕𝑦𝑑𝑦 +

𝜕𝜙

𝜕𝑧𝑑𝑧 = grad 𝜙 ⋅ 𝑑𝐫

We are treating each curvilinear coordinate as a scalar field, 𝑥𝑗 = 𝑥𝑗(𝑦1, 𝑦2, 𝑦3).

𝑑𝑥𝑗 = 𝛻𝑥𝑗 ⋅𝜕𝐫

𝜕𝑥𝑚 𝑑𝑥𝑚

= 𝐠𝑗 ⋅ 𝐠𝑚𝑑𝑥𝑚

= 𝛿𝑚𝑗𝑑𝑥𝑚.

The last equality arises from the fact that this is the only way one component of the coordinate differential can equal another.

⇒ 𝐠𝑗 ⋅ 𝐠𝑚 = 𝛿𝑚𝑗

Which recovers for us the reciprocity relationship and shows that 𝐠𝑗 and 𝐠𝑚 are reciprocal systems.


The position vector in Cartesian coordinates is 𝒓 = 𝒙𝒊𝒆𝒊. Show that (a) 𝐝𝐢𝐯 𝒓 = 𝟑, (b) 𝐝𝐢𝐯 𝒓 ⊗ 𝒓 = 𝟒𝒓, (c) 𝐝𝐢𝐯 𝒓 = 𝟑, and (d) 𝐠𝐫𝐚𝐝 𝒓 = 𝟏 and (e) 𝐜𝐮𝐫𝐥 𝒓 ⊗ 𝒓 = −𝒓 ×

grad 𝒓 = 𝑥𝑖 ,𝒋 𝒆𝑖 ⊗ 𝒆𝒋 = 𝛿𝑖𝑗𝒆𝑖 ⊗ 𝒆𝒋 = 𝟏

div 𝒓 = 𝑥𝑖 ,𝒋 𝒆𝑖 ⋅ 𝒆𝒋 = 𝛿𝑖𝑗𝛿𝑖𝑗 = 𝛿𝑗𝑗 = 3.

𝒓 ⊗ 𝒓 = 𝑥𝑖𝒆𝑖 ⊗ 𝑥𝑗𝒆𝑗 = 𝑥𝑖𝑥𝑗𝒆𝑖 ⊗ 𝒆𝒋

grad 𝒓 ⊗ 𝒓 = 𝑥𝑖𝑥𝑗 ,𝑘 𝒆𝑖 ⊗ 𝒆𝒋 ⊗ 𝒆𝒌

div 𝒓 ⊗ 𝒓 = 𝑥𝑖 ,𝑘 𝑥𝑗 + 𝑥𝑖𝑥𝑗 ,𝑘 𝒆𝑖 ⊗ 𝒆𝒋 ⋅ 𝒆𝒌

= 𝛿𝑖𝑘𝒙𝒋 + 𝒙𝒊𝛿𝑗𝑘 𝛿𝑗𝑘𝒆𝑖 = 𝛿𝑖𝑘𝒙𝒌 + 𝒙𝒊𝛿𝑗𝑗 𝒆𝑖 = 4𝑥𝑖𝒆𝑖 = 4𝒓

curl 𝒓 ⊗ 𝒓 = 𝜖𝛼𝛽𝛾 𝑥𝑖𝑥𝛾 ,𝛽 𝒆𝛼 ⊗ 𝒆𝒊

= 𝜖𝛼𝛽𝛾 𝑥𝑖 ,𝛽 𝑥𝛾 + 𝑥𝑖𝑥𝛾,𝛽 𝒆𝛼 ⊗ 𝒆𝒊

= 𝜖𝛼𝛽𝛾 𝛿𝑖𝛽𝑥𝛾 + 𝑥𝑖𝛿𝛾𝛽 𝒆𝛼 ⊗ 𝒆𝒊 = 𝜖𝛼𝑖𝛾 𝑥𝛾𝒆𝛼 ⊗ 𝒆𝒊 + 𝜖𝛼𝛽𝛽𝑥𝑖𝒆𝛼 ⊗ 𝒆𝒊

= −𝜖𝛼𝛾𝑖 𝑥𝛾𝒆𝛼 ⊗ 𝒆𝒊 = −𝒓 ×


For a scalar field 𝝓 and a tensor field 𝐓 show that 𝐠𝐫𝐚𝐝 𝝓𝐓 = 𝝓𝐠𝐫𝐚𝐝 𝐓 + 𝐓 ⊗ 𝐠𝐫𝐚𝐝𝝓. Also show that 𝐝𝐢𝐯 𝝓𝐓 = 𝝓 𝐝𝐢𝐯 𝐓 + 𝐓𝐠𝐫𝐚𝐝𝝓.

grad 𝜙𝐓 = 𝜙𝑇𝑖𝑗 ,𝑘 𝐠𝑖 ⊗ 𝐠𝑗 ⊗ 𝐠𝑘

= 𝜙,𝑘 𝑇𝑖𝑗 + 𝜙𝑇𝑖𝑗 ,𝑘 𝐠𝑖 ⊗ 𝐠𝑗 ⊗ 𝐠𝑘 = 𝐓 ⊗ grad𝜙 + 𝜙grad 𝐓

Furthermore, we can contract the last two bases and obtain,

div 𝜙𝐓 = 𝜙,𝑘 𝑇𝑖𝑗 + 𝜙𝑇𝑖𝑗 ,𝑘 𝐠𝑖 ⊗ 𝐠𝑗 ⋅ 𝐠𝑘

= 𝜙,𝑘 𝑇𝑖𝑗 + 𝜙𝑇𝑖𝑗 ,𝑘 𝐠𝑖𝛿𝑗𝑘

= 𝑇𝑖𝑘 𝜙,𝑘 𝐠𝑖 + 𝜙𝑇𝑖𝑘 ,𝑘 𝐠𝑖 = 𝐓grad𝜙 + 𝜙 div 𝐓


For two arbitrary vectors, 𝒖 and 𝒗, show that 𝐠𝐫𝐚𝐝 𝒖 ⊗ 𝒗 = 𝒖 × 𝐠𝐫𝐚𝐝𝒗 − 𝒗 × 𝐠𝐫𝐚𝐝𝒖

grad 𝒖 ⊗ 𝒗 = 𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝑘 ,𝑙 𝐠𝑖 ⊗ 𝐠𝑙

= 𝜖𝑖𝑗𝑘𝑢𝑗 ,𝑙 𝑣𝑘 + 𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝑘,𝑙 𝐠𝑖 ⊗ 𝐠𝑙

= 𝑢𝑗 ,𝑙 𝜖𝑖𝑗𝑘𝑣𝑘 + 𝑣𝑘 ,𝑙 𝜖𝑖𝑗𝑘𝑢𝑗 𝐠𝑖 ⊗ 𝐠𝑙

= − 𝒗 × grad𝒖 + 𝒖 × grad𝒗


For any two tensor fields 𝐮 and 𝐯, Show that, 𝐮 × : grad 𝐯 = 𝐮 ⋅ curl 𝐯

𝐮 × : grad 𝐯 = 𝜖𝑖𝑗𝑘𝑢𝑗𝐠𝑖 ⊗ 𝐠𝑘 : 𝑣𝛼,𝛽 𝐠𝛼 ⊗ 𝐠𝛽

= 𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝛼,𝛽 𝐠𝑖 ⋅ 𝐠𝛼 𝐠𝑘 ⋅ 𝐠𝛽

= 𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝛼,𝛽 𝛿𝑖𝛼𝛿𝑘

𝛽= 𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝑖 ,𝑘

= 𝐮 ⋅ curl 𝐯


For a vector field 𝒖, show that 𝐠𝐫𝐚𝐝 𝒖 × is a third ranked tensor. Hence or otherwise show that 𝐝𝐢𝐯 𝒖 × = −𝐜𝐮𝐫𝐥 𝒖.

The second–order tensor 𝒖 × is defined as 𝜖𝑖𝑗𝑘𝑢𝑗𝐠𝑖 ⊗

𝐠𝑘. Taking the covariant derivative with an independent base, we have

grad 𝒖 × = 𝜖𝑖𝑗𝑘𝑢𝑗 ,𝑙 𝐠𝑖 ⊗ 𝐠𝑘 ⊗ 𝐠𝑙

This gives a third order tensor as we have seen. Contracting on the last two bases,

div 𝒖 × = 𝜖𝑖𝑗𝑘𝑢𝑗 ,𝑙 𝐠𝑖 ⊗ 𝐠𝑘 ⋅ 𝐠𝑙

= 𝜖𝑖𝑗𝑘𝑢𝑗 ,𝑙 𝐠𝑖𝛿𝑘𝑙

= 𝜖𝑖𝑗𝑘𝑢𝑗 ,𝑘 𝐠𝑖 = −curl 𝒖


Show that 𝐜𝐮𝐫𝐥 𝜙𝟏 = grad 𝜙 ×

Note that 𝜙𝟏 = 𝜙𝑔𝛼𝛽 𝐠𝛼 ⊗ 𝐠𝛽, and that curl 𝑻 =

𝜖𝑖𝑗𝑘𝑇𝛼𝑘 ,𝑗 𝐠𝑖 ⊗ 𝐠𝛼 so that,

curl 𝜙𝟏 = 𝜖𝑖𝑗𝑘 𝜙𝑔𝛼𝑘 ,𝒋 𝐠𝑖 ⊗ 𝐠𝛼

= 𝜖𝑖𝑗𝑘 𝜙,𝒋 𝑔𝛼𝑘 𝐠𝑖 ⊗ 𝐠𝛼 = 𝜖𝑖𝑗𝑘𝜙,𝒋 𝐠𝑖 ⊗ 𝐠𝑘 = grad 𝜙 ×

Show that curl 𝒗 × = div 𝒗 𝟏 − grad 𝒗

𝒗 × = 𝜖𝛼𝛽𝑘𝑣𝛽 𝐠𝛼 ⊗ 𝐠𝑘

curl 𝑻 = 𝜖𝑖𝑗𝑘𝑇𝛼𝑘 ,𝑗 𝐠𝑖 ⊗ 𝐠𝛼

so that

curl 𝒗 × = 𝜖𝑖𝑗𝑘𝜖𝛼𝛽𝑘𝑣𝛽 ,𝑗 𝐠𝑖 ⊗ 𝐠𝛼

= 𝑔𝑖𝛼𝑔𝑗𝛽 − 𝑔𝑖𝛽𝑔𝑗𝛼 𝑣𝛽 ,𝑗 𝐠𝑖 ⊗ 𝐠𝛼

= 𝑣𝑗 ,𝑗 𝐠𝛼 ⊗ 𝐠𝛼 − 𝑣𝑖 ,𝑗 𝐠𝑖 ⊗ 𝐠𝑗 = div 𝒗 𝟏 − grad 𝒗


Show that 𝐝𝐢𝐯 𝒖 × 𝒗 = 𝒗 ⋅ 𝐜𝐮𝐫𝐥 𝒖 − 𝒖 ⋅ 𝐜𝐮𝐫𝐥 𝒗

div 𝒖 × 𝒗 = 𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝑘 ,𝑖

Noting that the tensor 𝜖𝑖𝑗𝑘 behaves as a constant under a covariant differentiation, we can write,

div 𝒖 × 𝒗 = 𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝑘 ,𝑖

= 𝜖𝑖𝑗𝑘𝑢𝑗 ,𝑖 𝑣𝑘 + 𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝑘,𝑖 = 𝒗 ⋅ curl 𝒖 − 𝒖 ⋅ curl 𝒗

Given a scalar point function ϕ and a vector field 𝐯, show that 𝐜𝐮𝐫𝐥 𝜙𝐯 = 𝜙 curl 𝐯 + grad𝜙 × 𝐯.

curl 𝜙𝐯 = 𝜖𝑖𝑗𝑘 𝜙𝑣𝑘 ,𝑗 𝐠𝑖

= 𝜖𝑖𝑗𝑘 𝜙,𝑗 𝑣𝑘 + 𝜙𝑣𝑘 ,𝑗 𝐠𝑖

= 𝜖𝑖𝑗𝑘𝜙,𝑗 𝑣𝑘𝐠𝑖 + 𝜖𝑖𝑗𝑘𝜙𝑣𝑘,𝑗 𝐠𝑖 = grad𝜙 × 𝐯 + 𝜙 curl 𝐯


Given that 𝜑 𝑡 = 𝐀(𝑡) , Show that 𝜑 𝑡 =𝐀

𝐀 𝑡: 𝐀

𝜑2 ≡ 𝐀: 𝐀

Now, 𝑑

𝑑𝑡𝜑2 = 2𝜑

𝑑𝜑

𝑑𝑡=

𝑑𝐀

𝑑𝑡: 𝐀 + 𝐀:

𝑑𝐀

𝑑𝑡= 2𝐀:

𝑑𝐀

𝑑𝑡

as inner product is commutative. We can therefore write that

𝑑𝜑

𝑑𝑡=

𝐀

𝜑:𝑑𝐀

𝑑𝑡=

𝐀

𝐀 𝑡: 𝐀

as required.


Given a tensor field 𝑻, obtain the vector 𝒘 ≡ 𝑻𝐓𝐯 and show that its divergence is 𝑻: 𝛁𝐯 + 𝐯 ⋅ 𝐝𝐢𝐯 𝑻

The divergence of 𝒘 is the scalar sum , 𝑇𝑗𝑖𝑣𝑗 ,𝑖.

Expanding the product covariant derivative we obtain,

div 𝑻T𝐯 = 𝑇𝑗𝑖𝑣𝑗 ,𝑖 = 𝑇𝑗𝑖 ,𝑖 𝑣𝑗 + 𝑇𝑗𝑖𝑣

𝑗 ,𝑖

= div 𝑻 ⋅ 𝐯 + tr 𝑻Tgrad𝐯

= div 𝑻 ⋅ 𝐯 + 𝑻: grad 𝐯

Recall that scalar product of two vectors is commutative so that

div 𝑻T𝐯 = 𝑻: grad𝐯 + 𝐯 ⋅ div 𝑻


For a second-order tensor 𝑻 define

curl 𝑻 ≡ 𝜖𝑖𝑗𝑘𝑻𝛼𝑘 ,𝑗 𝐠𝒊 ⊗ 𝐠𝜶 show that for any constant

vector 𝒂, curl 𝑻 𝒂 = curl 𝑻𝑻𝒂

Express vector 𝒂 in the invariant form with

contravariant components as 𝒂 = 𝑎𝛽𝐠𝛽. It follows that

curl 𝑻 𝒂 = 𝜖𝑖𝑗𝑘𝑇𝛼𝑘,𝑗 𝐠𝑖 ⊗ 𝐠𝛼 𝒂

= 𝜖𝑖𝑗𝑘𝑇𝛼𝑘,𝑗 𝑎𝛽 𝐠𝑖 ⊗ 𝐠𝛼 𝐠𝛽

= 𝜖𝑖𝑗𝑘𝑇𝛼𝑘,𝑗 𝑎𝛽𝐠𝑖𝛿𝛽𝛼

= 𝜖𝑖𝑗𝑘 𝑇𝛼𝑘 ,𝑗 𝐠𝑖𝑎𝛼

= 𝜖𝑖𝑗𝑘 𝑇𝛼𝑘𝑎𝛼 ,𝑗 𝐠𝑖

The last equality resulting from the fact that vector 𝒂 is a constant vector. Clearly,

curl 𝑻 𝒂 = curl 𝑻𝑇𝒂


For any two vectors 𝐮 and 𝐯, show that curl 𝐮 ⊗ 𝐯 =grad 𝐮 𝐯 × 𝑻 + curl 𝐯 ⊗ 𝒖 where 𝐯 × is the skew tensor

𝜖𝑖𝑘𝑗𝑣𝑘 𝐠𝒊 ⊗ 𝐠𝒋.

Recall that the curl of a tensor 𝑻 is defined by curl 𝑻 ≡

𝜖𝑖𝑗𝑘𝑇𝛼𝑘 ,𝑗 𝐠𝑖 ⊗ 𝐠𝛼. Clearly therefore,

curl 𝒖 ⊗ 𝒗 = 𝜖𝑖𝑗𝑘 𝑢𝛼𝑣𝑘 ,𝑗 𝐠𝑖 ⊗ 𝐠𝛼

= 𝜖𝑖𝑗𝑘 𝑢𝛼 ,𝑗 𝑣𝑘 + 𝑢𝛼𝑣𝑘 ,𝑗 𝐠𝑖 ⊗ 𝐠𝛼

= 𝜖𝑖𝑗𝑘𝑢𝛼 ,𝑗 𝑣𝑘 𝐠𝑖 ⊗ 𝐠𝛼 + 𝜖𝑖𝑗𝑘𝑢𝛼𝑣𝑘 ,𝑗 𝐠𝑖 ⊗ 𝐠𝛼

= 𝜖𝑖𝑗𝑘𝑣𝑘 𝐠𝑖 ⊗ 𝑢𝛼 ,𝑗 𝐠𝛼 + 𝜖𝑖𝑗𝑘𝑣𝑘 ,𝑗 𝐠𝑖 ⊗ 𝑢𝛼𝐠𝛼

= 𝜖𝑖𝑗𝑘𝑣𝑘 𝐠𝑖 ⊗ 𝐠𝑗 𝑢𝛼 ,𝛽 𝐠𝛽 ⊗ 𝐠𝛼 + 𝜖𝑖𝑗𝑘𝑣𝑘 ,𝑗 𝐠𝑖

⊗ 𝑢𝛼𝐠𝛼 = − 𝒗 × grad 𝒖 𝑻 + curl 𝒗 ⊗ 𝒖= grad 𝒖 𝒗 × 𝑻 + curl 𝒗 ⊗ 𝒖

upon noting that the vector cross is a skew tensor.


Show that curl 𝒖 × 𝒗 = div 𝒖 ⊗ 𝒗 − 𝒗 ⊗ 𝒖

The vector 𝒘 ≡ 𝒖 × 𝒗 = 𝑤𝒌𝐠𝒌 = 𝜖𝑘𝛼𝛽𝑢𝛼𝑣𝛽𝐠𝒌 and

curl 𝒘 = 𝜖𝑖𝑗𝑘𝑤𝑘 ,𝑗 𝐠𝑖. Therefore,

curl 𝒖 × 𝒗 = 𝜖𝑖𝑗𝑘𝑤𝑘 ,𝑗 𝐠𝑖

= 𝜖𝑖𝑗𝑘𝜖𝑘𝛼𝛽 𝑢𝛼𝑣𝛽 ,𝑗 𝐠𝑖

= 𝛿𝛼𝑖 𝛿𝛽

𝑗− 𝛿𝛽

𝑖 𝛿𝛼𝑗

𝑢𝛼𝑣𝛽 ,𝑗 𝐠𝑖

= 𝛿𝛼𝑖 𝛿𝛽

𝑗− 𝛿𝛽

𝑖 𝛿𝛼𝑗

𝑢𝛼,𝑗 𝑣𝛽 + 𝑢𝛼𝑣𝛽,𝑗 𝐠𝑖

= 𝑢𝑖 ,𝑗 𝑣𝑗 + 𝑢𝑖𝑣𝑗 ,𝑗 − 𝑢𝑗 ,𝑗 𝑣𝑖 + 𝑢𝑗𝑣𝑖 ,𝑗 𝐠𝑖

= 𝑢𝑖𝑣𝑗 ,𝑗 − 𝑢𝑗𝑣𝑖 ,𝑗 𝐠𝑖 = div 𝒖 ⊗ 𝒗 − 𝒗 ⊗ 𝒖

since div 𝒖 ⊗ 𝒗 = 𝑢𝑖𝑣𝑗 ,𝛼 𝐠𝑖 ⊗ 𝐠𝑗 ⋅ 𝐠𝛼 = 𝑢𝑖𝑣𝑗 ,𝑗 𝐠𝑖.


Given a scalar point function 𝜙 and a second-order tensor field 𝑻,

show that 𝐜𝐮𝐫𝐥 𝜙𝑻 = 𝜙 𝐜𝐮𝐫𝐥 𝑻 + grad𝜙 × 𝑻𝑻 where

grad𝜙 × is the skew tensor 𝜖𝑖𝑗𝑘𝜙,𝑗 𝐠𝒊 ⊗ 𝐠𝒌

curl 𝜙𝑻 ≡ 𝜖𝑖𝑗𝑘 𝜙𝑇𝛼𝑘 ,𝑗 𝐠𝑖 ⊗ 𝐠𝛼

= 𝜖𝑖𝑗𝑘 𝜙,𝑗 𝑇𝛼𝑘 + 𝜙𝑇𝛼𝑘,𝑗 𝐠𝑖 ⊗ 𝐠𝛼

= 𝜖𝑖𝑗𝑘𝜙,𝑗 𝑇𝛼𝑘 𝐠𝑖 ⊗ 𝐠𝛼 + 𝜙𝜖𝑖𝑗𝑘𝑇𝛼𝑘,𝑗 𝐠𝑖 ⊗ 𝐠𝛼

= 𝜖𝑖𝑗𝑘𝜙,𝑗 𝐠𝑖 ⊗ 𝐠𝑘 𝑇𝛼𝛽𝐠𝛽 ⊗ 𝐠𝛼 + 𝜙𝜖𝑖𝑗𝑘𝑇𝛼𝑘,𝑗 𝐠𝑖 ⊗ 𝐠𝛼

= 𝜙 curl 𝑻 + grad𝜙 × 𝑻𝑇


For a second-order tensor field 𝑻, show that

div curl 𝑻 = curl div 𝑻𝐓

Define the second order tensor 𝑆 as

curl 𝑻 ≡ 𝜖𝑖𝑗𝑘𝑇𝛼𝑘,𝑗 𝐠𝑖 ⊗ 𝐠𝛼 = 𝑆.𝛼𝑖 𝐠𝑖 ⊗ 𝐠𝛼

The gradient of 𝑺 is

𝑆.𝛼𝑖 ,𝛽 𝐠𝑖 ⊗ 𝐠𝛼 ⊗ 𝐠𝛽 = 𝜖𝑖𝑗𝑘𝑇𝛼𝑘,𝑗𝛽 𝐠𝑖 ⊗ 𝐠𝛼 ⊗ 𝐠𝛽

Clearly,

div curl 𝑻 = 𝜖𝑖𝑗𝑘𝑇𝛼𝑘,𝑗𝛽 𝐠𝑖 ⊗ 𝐠𝛼 ⋅ 𝐠𝛽

= 𝜖𝑖𝑗𝑘𝑇𝛼𝑘,𝑗𝛽 𝐠𝑖 𝑔𝛼𝛽

= 𝜖𝑖𝑗𝑘𝑇𝛽𝑘 ,𝑗𝛽 𝐠𝑖 = curl div 𝑻T


If the volume 𝑽 is enclosed by the surface 𝑺, the position vector 𝒓 = 𝒙𝒊𝐠𝒊 and 𝒏 is the external unit normal to each surface

element, show that 𝟏

𝟔 𝛁 𝒓 ⋅ 𝒓 ⋅ 𝒏𝒅𝑺𝑺

equals the volume

contained in 𝑽.

𝒓 ⋅ 𝒓 = 𝑥𝑖𝑥𝑗𝐠𝑖 ⋅ 𝐠𝑗 = 𝑥𝑖𝑥𝑗𝑔𝑖𝑗

By the Divergence Theorem,

𝛻 𝒓 ⋅ 𝒓 ⋅ 𝒏𝑑𝑆𝑆

= 𝛻 ⋅ 𝛻 𝒓 ⋅ 𝒓 𝑑𝑉𝑉

= 𝜕𝑙 𝜕𝑘 𝑥𝑖𝑥𝑗𝑔𝑖𝑗 𝐠𝑙 ⋅ 𝐠𝑘 𝑑𝑉𝑉

= 𝜕𝑙 𝑔𝑖𝑗 𝑥𝑖 ,𝑘 𝑥𝑗 + 𝑥𝑖𝑥𝑗 ,𝑘 𝐠𝑙 ⋅ 𝐠𝑘 𝑑𝑉𝑉

= 𝑔𝑖𝑗𝑔𝑙𝑘 𝛿𝑘

𝑖 𝑥𝑗 + 𝑥𝑖𝛿𝑘𝑗

,𝑙 𝑑𝑉𝑉

= 2𝑔𝑖𝑘𝑔𝑙𝑘𝑥𝑖,𝑙 𝑑𝑉𝑉

= 2𝛿𝑖𝑙𝛿𝑙

𝑖 𝑑𝑉𝑉

= 6 𝑑𝑉𝑉


For any Euclidean coordinate system, show that div 𝐮 × 𝐯 =𝐯 curl 𝐮 − 𝐮 curl 𝐯

Given the contravariant vector 𝑢𝑖 and 𝑣𝑖 with their associated vectors 𝑢𝑖 and 𝑣𝑖, the contravariant component of the above cross product is 𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝑘 .The required divergence is simply the contraction of the covariant 𝑥𝑖 derivative of this quantity:

𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝑘 ,𝑖= 𝜖𝑖𝑗𝑘𝑢𝑗,𝑖𝑣𝑘 + 𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝑘,𝑖

where we have treated the tensor 𝜖𝑖𝑗𝑘 as a constant under the covariant derivative.

Cyclically rearranging the RHS we obtain,

𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝑘 ,𝑖= 𝑣𝑘𝜖𝑘𝑖𝑗𝑢𝑗,𝑖 + 𝑢𝑗𝜖

𝑗𝑘𝑖𝑣𝑘,𝑖 = 𝑣𝑘𝜖𝑘𝑖𝑗𝑢𝑗,𝑖 + 𝑢𝑗𝜖𝑗𝑖𝑘𝑣𝑘,𝑖

where we have used the anti-symmetric property of the tensor 𝜖𝑖𝑗𝑘. The last expression shows clearly that

div 𝐮 × 𝐯 = 𝐯 curl 𝐮 − 𝐮 curl 𝐯

as required.


For a scalar variable 𝛼, if the tensor 𝐓 = 𝐓(𝛼) and 𝐓 ≡𝑑𝐓

𝑑𝛼, Show that

𝑑

𝑑𝛼det 𝐓 =

det 𝐓 tr(𝐓 𝐓−𝟏) Proof:

Let 𝐀 ≡ 𝐓 𝐓−1 so that, 𝐓 = 𝐀𝐓. In component form, we have 𝑇 𝑗𝑖 = 𝐴𝑚

𝑖 𝑇𝑗𝑚.

Therefore, 𝑑

𝑑𝛼det 𝐓 =

𝑑

𝑑𝛼𝜖𝑖𝑗𝑘𝑇𝑖

1𝑇𝑗2𝑇𝑘

3 = 𝜖𝑖𝑗𝑘 𝑇 𝑖1𝑇𝑗

2𝑇𝑘3 + 𝑇𝑖

1𝑇 𝑗2𝑇𝑘

3 + 𝑇𝑖1𝑇𝑗

2𝑇 𝑘3

= 𝜖𝑖𝑗𝑘 𝐴𝑙1𝑇𝑖

𝑙𝑇𝑗2𝑇𝑘

3 + 𝑇𝑖1𝐴𝑚

2 𝑇𝑗𝑚𝑇𝑘


2𝐴𝑛3𝑇𝑘

𝑛

= 𝜖𝑖𝑗𝑘 𝐴11𝑇𝑖

1 + 𝐴21𝑇𝑖

2 + 𝐴31𝑇𝑖

3 𝑇𝑗2𝑇𝑘

3

+ 𝑇𝑖1 𝐴1

2𝑇𝑗1 + 𝐴2

2𝑇𝑗2 + 𝐴3

2𝑇𝑗3 𝑇𝑘


2 𝐴13𝑇𝑘

1 + 𝐴23𝑇𝑘

2 + 𝐴33𝑇𝑘

3

All the boxed terms in the above equation vanish on account of the contraction of a symmetric tensor with an antisymmetric one.


Liouville Formula

(For example, the first boxed term yields, 𝜖𝑖𝑗𝑘𝐴21𝑇𝑖

2𝑇𝑗2𝑇𝑘

3

Which is symmetric as well as antisymmetric in 𝑖 and 𝑗. It

therefore vanishes. The same is true for all other such terms.)

𝑑

𝑑𝛼det 𝐓 = 𝜖𝑖𝑗𝑘 𝐴1

1𝑇𝑖1 𝑇𝑗

2𝑇𝑘3 + 𝑇𝑖

1 𝐴22𝑇𝑗

2 𝑇𝑘3 + 𝑇𝑖

1𝑇𝑗2 𝐴3

3𝑇𝑘3

= 𝐴𝑚𝑚 𝜖𝑖𝑗𝑘𝑇𝑖

1𝑇𝑗2𝑇𝑘

3

= tr 𝐓 𝐓−1 det 𝐓

as required.


Liouville Theorem Cont’d

For a general tensor field 𝑻 show that, 𝐜𝐮𝐫𝐥 𝐜𝐮𝐫𝐥 𝑻 =𝛁𝟐 𝐭𝐫 𝑻 − 𝐝𝐢𝐯 𝐝𝐢𝐯 𝑻 𝑰 + 𝐠𝐫𝐚𝐝 𝐝𝐢𝐯 𝑻 + 𝐠𝐫𝐚𝐝 𝐝𝐢𝐯 𝑻

𝐓−

𝐠𝐫𝐚𝐝 𝐠𝐫𝐚𝐝 𝐭𝐫𝑻 − 𝛁𝟐𝑻𝐓

curl 𝑻 = 𝝐𝛼𝑠𝑡𝑇𝛽𝑡 ,𝑠 𝐠𝛼 ⊗ 𝐠𝛽

= 𝑆 .𝛽𝛼 𝐠𝛼 ⊗ 𝐠𝛽

curl 𝑺 = 𝜖𝑖𝑗𝑘𝑆 .𝑘𝛼 ,𝑗 𝐠𝑖 ⊗ 𝐠𝛼

so that

curl 𝑺 = curl curl 𝑻 = 𝜖𝑖𝑗𝑘𝜖𝛼𝑠𝑡𝑇𝑘𝑡,𝑠𝑗 𝐠𝑖 ⊗ 𝐠𝛼

=

𝑔𝑖𝛼 𝑔𝑖𝑠 𝑔𝑖𝑡

𝑔𝑗𝛼 𝑔𝑗𝑠 𝑔𝑗𝑡

𝑔𝑘𝛼 𝑔𝑘𝑠 𝑔𝑘𝑡

𝑇𝑘𝑡 ,𝑠𝑗 𝐠𝑖 ⊗ 𝐠𝛼

=𝑔𝑖𝛼 𝑔𝑗𝑠𝑔𝑘𝑡 − 𝑔𝑗𝑡𝑔𝑘𝑠 + 𝑔𝑖𝑠 𝑔𝑗𝑡𝑔𝑘𝛼 − 𝑔𝑗𝛼𝑔𝑘𝑡

+𝑔𝑖𝑡 𝑔𝑗𝛼𝑔𝑘𝑠 − 𝑔𝑗𝑠𝑔𝑘𝛼𝑇𝑘𝑡 ,𝑠𝑗 𝐠𝑖 ⊗ 𝐠𝛼

= 𝑔𝑗𝑠𝑇 .𝑡𝑡 ,𝑠𝑗 −𝑇 ..

𝑠𝑗 ,𝑠𝑗 𝐠𝛼 ⊗ 𝐠𝛼 + 𝑇 ..𝛼𝑗 ,𝑠𝑗 −𝑔𝑗𝛼𝑇 .𝑡

𝑡 ,𝑠𝑗 𝐠𝑠 ⊗ 𝐠𝛼

+ 𝑔𝑗𝛼𝑇 .𝑡𝑠.,𝑠𝑗 −𝑔𝑗𝑠𝑇 .𝑡

𝛼.,𝑠𝑗 𝐠𝑡 ⊗ 𝐠𝛼

= 𝛻2 tr 𝑻 − div div 𝑻 𝑰 + grad div 𝑻T

− grad grad tr𝑻

+ grad div 𝑻 − 𝛻2𝑻T [email protected] 12/27/2012 Dept of Systems Engineering, University of Lagos 69

When 𝑻 is symmetric, show that tr(curl 𝑻) vanishes.

curl 𝑻 = 𝝐𝑖𝑗𝑘𝑇𝛽𝑘 ,𝑗 𝐠𝑖 ⊗ 𝐠𝛽

tr curl 𝑻 = 𝜖𝑖𝑗𝑘𝑇𝛽𝑘 ,𝑗 𝐠𝑖 ⋅ 𝐠𝛽

= 𝜖𝑖𝑗𝑘𝑇𝛽𝑘 ,𝑗 𝛿𝑖𝛽

= 𝜖𝑖𝑗𝑘𝑇𝑖𝑘 ,𝑗

which obviously vanishes on account of the symmetry and antisymmetry in 𝑖 and 𝑘. In this case, curl curl 𝑻

= 𝛻2 tr 𝑻 − div div 𝑻 𝑰 − grad grad tr𝑻

+ 2 grad div 𝑻 − 𝛻2𝑻

as grad div 𝑻T

= grad div 𝑻 if the order of

differentiation is immaterial and T is symmetric.


For a scalar function Φ and a vector 𝑣𝑖 show that the divergence of the vector 𝑣𝑖Φ is equal to, 𝐯 ⋅ gradΦ +Φ div 𝐯

𝑣𝑖Φ,𝑖

= Φ𝑣𝑖,𝑖 + 𝑣𝑖Φ,i

Hence the result.


Show that 𝒄𝒖𝒓𝒍 𝐮 × 𝐯 = 𝐯 ∙ 𝛁𝐮 + 𝐮 ⋅ div 𝐯 − 𝐯 ⋅ div 𝐮 −𝐮 ∙ 𝛁 𝐯

Taking the associated (covariant) vector of the expression for the cross product in the last example, it is straightforward to see that the LHS in indicial notation is,

𝜖𝑙𝑚𝑖 𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝑘,𝑚

Expanding in the usual way, noting the relation between the alternating tensors and the Kronecker deltas,

𝜖𝑙𝑚𝑖 𝜖𝑖𝑗𝑘𝑢𝑗𝑣𝑘,𝑚

= 𝛿𝑗𝑘𝑖𝑙𝑚𝑖 𝑢𝑗

,𝑚𝑣𝑘 − 𝑢𝑗𝑣𝑘,𝑚

= 𝛿𝑗𝑘𝑙𝑚 𝑢𝑗

,𝑚𝑣𝑘 − 𝑢𝑗𝑣𝑘,𝑚 =

𝛿𝑗𝑙 𝛿𝑗

𝑚

𝛿𝑘𝑙 𝛿𝑘

𝑚𝑢𝑗

,𝑚𝑣𝑘 − 𝑢𝑗𝑣𝑘,𝑚

= 𝛿𝑗𝑙𝛿𝑘

𝑚 − 𝛿𝑘𝑙 𝛿𝑗

𝑚 𝑢𝑗,𝑚𝑣𝑘 − 𝑢𝑗𝑣𝑘

,𝑚

= 𝛿𝑗𝑙𝛿𝑘

𝑚𝑢𝑗,𝑚𝑣𝑘 − 𝛿𝑗

𝑙𝛿𝑘𝑚𝑢𝑗𝑣𝑘

,𝑚 + 𝛿𝑘𝑙 𝛿𝑗

𝑚𝑢𝑗,𝑚𝑣𝑘

− 𝛿𝑘𝑙 𝛿𝑗

𝑚𝑢𝑗𝑣𝑘,𝑚

= 𝑢𝑙,𝑚𝑣𝑚 − 𝑢𝑚

,𝑚𝑣𝑙 + 𝑢𝑙𝑣𝑚,𝑚 − 𝑢𝑚𝑣𝑙

,𝑚

Which is the result we seek in indicial notation.


Given a vector point function 𝒖 𝑥1, 𝑥2, 𝑥3 and its covariant components, 𝑢𝛼 𝑥1, 𝑥2, 𝑥3 , 𝛼 = 1,2,3, with 𝐠𝛼as the reciprocal basis vectors, Let

𝒖 = 𝑢𝛼𝐠𝛼 , then 𝑑𝒖 =𝜕

𝜕𝑥𝑘𝑢𝛼𝐠𝛼 𝑑𝑥𝑘 =

𝜕𝑢𝛼

𝜕𝑥𝑘𝐠𝛼 +

𝜕𝐠𝛼

𝜕𝑥𝑘𝑢𝛼 𝑑𝑥𝑘

Clearly, 𝜕𝒖

𝜕𝑥𝑘=

𝜕𝑢𝛼


𝜕𝐠𝛼

𝜕𝑥𝑘𝑢𝛼

And the projection of this quantity on the 𝐠𝑖direction is, 𝜕𝒖

𝜕𝑥𝑘∙ 𝐠𝑖 =

𝜕𝑢𝛼


𝜕𝐠𝛼

𝜕𝑥𝑘𝑢𝛼 ∙ 𝐠𝑖 =

𝜕𝑢𝛼

𝜕𝑥𝑘𝐠𝛼 ∙ 𝐠𝑖 +

𝜕𝐠𝛼

𝜕𝑥𝑘∙ 𝐠𝑖𝑢𝛼

=𝜕𝑢𝛼

𝜕𝑥𝑘𝛿𝑖

𝛼 +𝜕𝐠𝛼

𝜕𝑥𝑘∙ 𝐠𝑖𝑢𝛼

Differentiation of Fields


Now, 𝐠𝑖 ∙ 𝐠𝑗 = 𝛿𝑖𝑗 so that

𝜕𝐠𝑖

𝜕𝑥𝑘∙ 𝐠𝑗 + 𝐠𝑖 ∙

𝜕𝐠𝑗

𝜕𝑥𝑘= 0.

𝜕𝐠𝑖

𝜕𝑥𝑘∙ 𝐠𝑗 = −𝐠𝑖 ∙

𝜕𝐠𝑗

𝜕𝑥𝑘≡ −

𝑖𝑗𝑘

.

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


Using this, we can now write that, 𝜕𝒖

𝜕𝑥𝑘 ∙ 𝐠𝑖 =𝜕𝑢𝛼

𝜕𝑥𝑘 𝛿𝑖𝛼 +

𝜕𝐠𝛼

𝜕𝑥𝑘 ∙ 𝐠𝑖𝑢𝛼 =𝜕𝑢𝑖

𝜕𝑥𝑘 −𝛼𝑖𝑘

𝑢𝛼

The quantity on the RHS is the component of the derivative of vector 𝒖 along the 𝐠𝑖 direction using covariant components. It is the covariant derivative of 𝒖. Using contravariant components, we could write,

𝑑𝒖 =𝜕𝑢𝑖

𝜕𝑥𝑘 𝐠𝑖 +𝜕𝐠𝑖

𝜕𝑥𝑘 𝑢𝑖 𝑑𝑥𝑘 =𝜕𝑢𝑖

𝜕𝑥𝑘 𝐠𝑖 +𝛼𝑖𝑘

𝐠𝛼𝑢𝑖 𝑑𝑥𝑘

So that, 𝜕𝒖

𝜕𝑥𝑘 =𝜕𝑢𝛼

𝜕𝑥𝑘 𝐠𝛼 +𝜕𝐠𝛼

𝜕𝑥𝑘 𝑢𝛼

The components of this in the direction of 𝐠𝑖 can be obtained by taking a dot product as before:

𝜕𝒖

𝜕𝑥𝑘 ∙ 𝐠𝑖 =𝜕𝑢𝛼

𝜕𝑥𝑘 𝐠𝛼 +𝜕𝐠𝛼

𝜕𝑥𝑘 𝑢𝛼 ∙ 𝐠𝑖 =𝜕𝑢𝛼

𝜕𝑥𝑘 𝛿𝛼𝑖 +

𝜕𝐠𝛼

𝜕𝑥𝑘 ∙ 𝐠𝑖𝑢𝛼 =𝜕𝑢𝑖

𝜕𝑥𝑘 +𝑖

𝛼𝑘𝑢𝛼

Derivatives in General Coordinates


The two results above are represented symbolically as,

𝑢𝑖,𝑘 =𝜕𝑢𝑖


𝑢𝛼 and 𝑢,𝑘𝑖 =

𝜕𝑢𝛼

𝜕𝑥𝑘 +𝑖

𝛼𝑘𝑢𝛼

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


We observe that the derivative of the covariant basis, 𝐠𝑖 =𝜕𝐫

𝜕𝑥𝑖 ,

𝜕𝐠𝑖

𝜕𝑥𝑗=

𝜕2𝐫

𝜕𝑥𝑗𝜕𝑥𝑖=

𝜕2𝐫

𝜕𝑥𝑖𝜕𝑥𝑗=

𝜕𝐠𝑗

𝜕𝑥𝑖

Taking the dot product with 𝐠𝑘, 𝜕𝐠𝑖

𝜕𝑥𝑗∙ 𝐠𝑘 =

1

2

𝜕𝐠𝑗

𝜕𝑥𝑖∙ 𝐠𝑘 +

𝜕𝐠𝑖

𝜕𝑥𝑗∙ 𝐠𝑘

=1

2

𝜕

𝜕𝑥𝑖𝐠𝑗 ∙ 𝐠𝑘 +

𝜕

𝜕𝑥𝑗𝐠𝑖 ∙ 𝐠𝑘 − 𝐠𝑗 ∙

𝜕𝐠𝑘

𝜕𝑥𝑖− 𝐠𝑖 ∙

𝜕𝐠𝑘

𝜕𝑥𝑗

=1

2

𝜕


𝜕

𝜕𝑥𝑗𝐠𝑖 ∙ 𝐠𝑘 − 𝐠𝑗 ∙

𝜕𝐠𝑖

𝜕𝑥𝑘− 𝐠𝑖 ∙

𝜕𝐠𝑗

𝜕𝑥𝑘

=1

2

𝜕


𝜕

𝜕𝑥𝑗𝐠𝑖 ∙ 𝐠𝑘 −

𝜕

𝜕𝑥𝑘𝐠𝑖 ∙ 𝐠𝑗

=1

2

𝜕𝑔𝑗𝑘

𝜕𝑥𝑖+

𝜕𝑔𝑘𝑖

𝜕𝑥𝑗−

𝜕𝑔𝑖𝑗

𝜕𝑥𝑘

Christoffel Symbols


Which is the quantity defined as the Christoffel symbols of the first kind in the textbooks. It is therefore possible for us to write,

𝑖𝑗, 𝑘 ≡ 𝜕𝐠𝑖


𝜕𝐠𝑗

𝜕𝑥𝑖∙ 𝐠𝑘 =

1

2

𝜕𝑔𝑗𝑘

𝜕𝑥𝑖+

𝜕𝑔𝑘𝑖

𝜕𝑥𝑗−

𝜕𝑔𝑖𝑗

𝜕𝑥𝑘

It should be emphasized that the Christoffel symbols, even though the play a critical role in several tensor relationships, are themselves NOT tensor quantities. (Prove this). However, notice their symmetry in the 𝑖 and 𝑗. The extension of this definition to the Christoffel symbols of the second kind is immediate:

The First Kind


Contract the above equation with the conjugate metric tensor, we have,

𝑔𝑘𝛼 𝑖𝑗, 𝛼 ≡ 𝑔𝑘𝛼 𝜕𝐠𝑖

𝜕𝑥𝑗∙ 𝐠𝛼 = 𝑔𝑘𝛼

𝜕𝐠𝑗

𝜕𝑥𝑖∙ 𝐠𝛼 =

𝜕𝐠𝑖


𝑘𝑖𝑗

= −𝜕𝐠𝑘

𝜕𝑥𝑗∙ 𝐠𝑖

Which connects the common definition of the second Christoffel symbol with the one defined in the above derivation. The relationship,

𝑔𝑘𝛼 𝑖𝑗, 𝛼 =𝑘𝑖𝑗

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


We contract the above equation with 𝑔𝑘𝛽and obtain,

𝑔𝑘𝛽𝑔𝑘𝛼 𝑖𝑗, 𝛼 = 𝑔𝑘𝛽𝑘𝑖𝑗

𝛿𝛽𝛼 𝑖𝑗, 𝛼 = 𝑖𝑗, 𝛽 = 𝑔𝑘𝛽

𝑘𝑖𝑗

so that,

𝑔𝑘𝛼

𝛼𝑖𝑗 = 𝑖𝑗, 𝑘

Showing that the metric tensor can be used to lower the contravariant index of the Christoffel symbol of the second kind to obtain the Christoffel symbol of the first kind.

Two Christoffel Symbols Related


We are now in a position to express the derivatives of higher order tensor fields in terms of the Christoffel symbols.

For a second-order tensor 𝐓, we can express the components in dyadic form along the product basis as follows:

𝐓 = 𝑇𝑖𝑗𝐠𝑖 ⊗ 𝐠𝑗= 𝑇𝑖𝑗𝐠𝑖⨂𝐠𝑗 = 𝑇.𝑗

𝑖 𝐠𝑖 ⊗ 𝐠𝑗= 𝑇𝑗.𝑖𝐠𝑗⨂𝐠𝑖

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


We have:

𝜕𝐓

𝜕𝑥𝑘=

𝜕𝑇𝑖𝑗

𝜕𝑥𝑘𝐠𝑖⨂𝐠𝑗 + 𝑇𝑖𝑗

𝜕𝐠𝑖

𝜕𝑥𝑘⨂𝐠𝑗 + 𝑇𝑖𝑗𝐠𝑖⨂

𝜕𝐠𝑗

𝜕𝑥𝑘

Recall that, 𝜕𝐠𝑖

𝜕𝑥𝑘 ∙ 𝐠𝑗 =𝑗𝑖𝑘

. It follows therefore that,

𝜕𝐠𝑖

𝜕𝑥𝑘∙ 𝐠𝑗 −

𝛼𝑖𝑘

𝛿𝛼𝑗

=𝜕𝐠𝑖

𝜕𝑥𝑘∙ 𝐠𝑗 −

𝛼𝑖𝑘

𝐠𝛼 ∙ 𝐠𝑗

=𝜕𝐠𝑖

𝜕𝑥𝑘−

𝛼𝑖𝑘

𝐠𝛼 ∙ 𝐠𝑗 = 0.

Clearly, 𝜕𝐠𝑖

𝜕𝑥𝑘 =𝛼𝑖𝑘

𝐠𝛼

(Obviously since 𝐠𝑗 is a basis vector it cannot vanish)



𝜕𝐓

𝜕𝑥𝑘=

𝜕𝑇𝑖𝑗

𝜕𝑥𝑘𝐠𝑖⨂𝐠𝑗 + 𝑇𝑖𝑗

𝜕𝐠𝑖

𝜕𝑥𝑘⨂𝐠𝑗 + 𝑇𝑖𝑗𝐠𝑖⨂

𝜕𝐠𝑗

𝜕𝑥𝑘

=𝜕𝑇𝑖𝑗

𝜕𝑥𝑘 𝐠𝑖⨂𝐠𝑗 + 𝑇𝑖𝑗 𝛼𝑖𝑘

𝐠𝛼 ⨂𝐠𝑗 + 𝑇𝑖𝑗𝐠𝑖⨂𝛼𝑗𝑘 𝐠𝛼

=𝜕𝑇𝑖𝑗

𝜕𝑥𝑘 𝐠𝑖⨂𝐠𝑗 + 𝑇𝛼𝑗 𝑖𝛼𝑘

𝐠𝑖 ⨂𝐠𝑗 + 𝑇𝑖𝛼𝐠𝑖⨂𝑗

𝛼𝑘𝐠𝑗

=𝜕𝑇𝑖𝑗

𝜕𝑥𝑘 +𝑇𝛼𝑗 𝑖𝛼𝑘

+ 𝑇𝑖𝛼 𝑗𝛼𝑘

𝐠𝑖⨂𝐠𝑗 = 𝑇 ,𝑘𝑖𝑗

𝐠𝑖⨂𝐠𝑗

Where

𝑇 ,𝑘

𝑖𝑗=

𝜕𝑇𝑖𝑗

𝜕𝑥𝑘 +𝑇𝛼𝑗 𝑖𝛼𝑘

+ 𝑇𝑖𝛼 𝑗𝛼𝑘

or 𝜕𝑇𝑖𝑗

𝜕𝑥𝑘 +𝑇𝛼𝑗 Γ𝛼𝑘𝑖 + 𝑇𝑖𝛼Γ𝛼𝑘

𝑗

are the components of the covariant derivative of the tensor 𝐓 in terms of contravariant components on the product covariant bases as shown.



In the same way, by taking the tensor expression in the dyadic form of its contravariant product bases, we can write,

𝜕𝐓

𝜕𝑥𝑘 =𝜕𝑇𝑖𝑗

𝜕𝑥𝑘 𝐠𝑖 ⊗ 𝐠𝑗 +𝑇𝑖𝑗

𝜕𝐠𝑖

𝜕𝑥𝑘 ⊗ 𝐠𝑗 +𝑇𝑖𝑗𝐠𝑖 ⊗

𝜕𝐠𝑗

𝜕𝑥𝑘

=𝜕𝑇𝑖𝑗

𝜕𝑥𝑘 𝐠𝑖 ⊗ 𝐠𝑗 +𝑇𝑖𝑗Γ𝛼𝑘𝑖 ⊗ 𝐠𝑗 +𝑇𝑖𝑗𝐠

𝑖 ⊗𝜕𝐠𝑗

𝜕𝑥𝑘

Again, notice from previous derivation above, 𝑖𝑗𝑘

= −𝜕𝐠𝑖

𝜕𝑥𝑘 ∙ 𝐠𝑗 so that, 𝜕𝐠𝑖

𝜕𝑥𝑘 = −𝑖

𝛼𝑘𝐠𝛼 = −Γ𝛼𝑘

𝑖 𝐠𝛼 Therefore,

𝜕𝐓

𝜕𝑥𝑘=

𝜕𝑇𝑖𝑗

𝜕𝑥𝑘𝐠𝑖 ⊗ 𝐠𝑗 +𝑇𝑖𝑗

𝜕𝐠𝑖

𝜕𝑥𝑘⊗ 𝐠𝑗 +𝑇𝑖𝑗𝐠

𝑖 ⊗𝜕𝐠𝑗

𝜕𝑥𝑘

=𝜕𝑇𝑖𝑗

𝜕𝑥𝑘𝐠𝑖 ⊗ 𝐠𝑗 −𝑇𝑖𝑗Γ𝛼𝑘

𝑖 𝐠𝛼 ⊗ 𝐠𝑗 + 𝑇𝑖𝑗𝐠𝑖 ⊗ Γ𝛼𝑘

𝑗𝐠𝛼

=𝜕𝑇𝑖𝑗

𝜕𝑥𝑘− 𝑇𝛼𝑗Γ𝑖𝑘

𝛼 − 𝑇𝑖𝛼Γ𝑗𝑘𝛼 𝐠𝑖 ⊗ 𝐠𝑗= 𝑇𝑖𝑗,𝑘𝐠𝑖 ⊗ 𝐠𝑗

So that

𝑇𝑖𝑗,𝑘 =𝜕𝑇𝑖𝑗

𝜕𝑥𝑘 − 𝑇𝛼𝑗Γ𝑖𝑘𝛼 − 𝑇𝑖𝛼Γ𝑗𝑘

𝛼



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.



The covariant derivative of the mixed tensor, 𝐴𝑖1 ,𝑖2 ,…,𝑖𝑛

𝑗1 ,𝑗2 ,…,𝑗𝑚 is the

most general case for the covariant derivative of an absolute tensor:

𝐴𝑖1 ,𝑖2 ,…,𝑖𝑛

𝑗1 ,𝑗2 ,…,𝑗𝑚

,𝑗

= 𝜕𝐴𝑖1 ,𝑖2 ,…,𝑖𝑛

𝑗1 ,𝑗2 ,…,𝑗𝑚

𝜕𝑥𝑗−

𝛼𝑖1𝑗

𝐴𝛼 ,𝑖2 ,…,𝑖𝑛

𝑗1 ,𝑗2 ,…,𝑗𝑚 − 𝛼𝑖2𝑗

𝐴𝑖1,𝛼,…,𝑖𝑛

𝑗1 ,𝑗2 ,…,𝑗𝑚 − ⋯ −𝛼𝑖𝑛𝑗 𝐴𝑖1 ,𝑖2 ,…,𝛼

𝑗1 ,𝑗2 ,…,𝑗𝑚

+ 𝑗1𝛽𝑗

𝐴𝑖1 ,𝑖2 ,…,𝑖𝑛

𝛽 ,𝑗2 ,…,𝑗𝑚 + 𝑗2𝛽𝑗

𝐴𝑖1 ,𝑖2 ,…,𝑖𝑛

𝑗1 ,𝛽,…,𝑗𝑚 + ⋯+ 𝑗𝑚𝛽𝑗

𝐴𝑖1 ,𝑖2 ,…,𝑖𝑛

𝑗1 ,𝑗2,…,𝛽

Higher Order Mixed Tensors


The metric tensor components behave like constants under a covariant differentiation. The proof that this is so is due to Ricci:

𝑔𝑖𝑗 ,𝑘 =𝜕𝑔𝑖𝑗

𝜕𝑥𝑘−

𝛼𝑖𝑘

𝑔𝛼𝑗 −𝛽𝑘𝑗

𝑔𝑖𝛽

=𝜕𝑔𝑖𝑗

𝜕𝑥𝑘− 𝑖𝑘, 𝑗 − 𝑘𝑗, 𝑖

=𝜕𝑔𝑖𝑗

𝜕𝑥𝑘−

1

2

𝜕𝑔𝑗𝑘

𝜕𝑥𝑖+

𝜕𝑔𝑗𝑖

𝜕𝑥𝑘−

𝜕𝑔𝑖𝑘

𝜕𝑥𝑗−

1

2

𝜕𝑔𝑘𝑗

𝜕𝑥𝑗+

𝜕𝑔𝑖𝑘

𝜕𝑥𝑗−

𝜕𝑔𝑗𝑘

𝜕𝑥𝑖

= 0.

Ricci’s Theorem


The conjugate metric tensor behaves the same way as can be seen from the relationship,

𝑔𝑖𝑙𝑔𝑙𝑗 = 𝛿𝑖

𝑗

The above can be differentiated covariantly with respect to 𝑥𝑘 to obtain

𝑔𝑖𝑙 ,𝑘 𝑔𝑙𝑗 + 𝑔𝑖𝑙𝑔𝑙𝑗 ,𝑘 = 𝛿𝑖

𝑗,𝑘

0 + 𝑔𝑖𝑙𝑔𝑙𝑗 ,𝑘 =

𝜕𝛿𝑖𝑗


𝛿𝛼𝑗

+𝑗

𝛼𝑘𝛿𝑖

𝛼

𝑔𝑖𝑙𝑔𝑙𝑗 ,𝑘 = 0 −

𝑗𝑖𝑘

+𝑗𝑖𝑘

= 0

The contraction of 𝑔𝑖𝑙 with 𝑔𝑙𝑗 ,𝑘 vanishes. Since we know that the metric tensor cannot vanish in general, we can only conclude that

𝑔𝑖𝑗 ,𝑘 = 0

Ricci’s Theorem


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 𝑔𝑖𝑗and 𝑔𝑖𝑗 as well as the alternating tensors 𝜖𝑖𝑗𝑘 , 𝜖𝑖𝑗𝑘 are all constants under a covariant

differention. The Kronecker delta 𝛿𝑖𝑗 is a constant under

both covariant as well as the regular partial differentiation


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 𝚵, The volume integral in the region Ω ⊂ ℰ,

grad 𝚵Ω

𝑑𝑣 = 𝚵 ⊗ 𝐧𝜕Ω

𝑑𝑠

where 𝐧 is the outward drawn normal to 𝜕Ω – the boundary of Ω.

Integral Theorems


Vector field. Replacing the tensor with the vector field 𝐟 and contracting, we have,

div 𝐟Ω

𝑑𝑣 = 𝐟 ⋅ 𝐧𝜕Ω

𝑑𝑠

Which is the usual form of the Gauss theorem.

Special Cases: Vector Field


For a scalar field 𝜙, the divergence becomes a gradient and the scalar product on the RHS becomes a simple multiplication. Hence the divergence theorem becomes,

grad 𝜙Ω

𝑑𝑣 = 𝜙𝐧𝜕Ω

𝑑𝑠

The procedure here is valid and will become obvious if we write, 𝐟 = 𝜙𝒂 where 𝒂 is an arbitrary constant vector.

Scalar Field


div 𝜙𝒂Ω

𝑑𝑣 = 𝜙𝒂 ⋅ 𝐧𝜕Ω

𝑑𝑠 = 𝒂 ⋅ 𝜙𝐧𝜕Ω

𝑑𝑠

For the LHS, note that, div 𝜙𝒂 = tr grad 𝜙𝒂

grad 𝜙𝒂 = 𝜙𝑎𝑖 ,𝑗 𝐠𝑖 ⊗ 𝐠𝑗

= 𝑎𝑖𝜙,𝑗 𝐠𝑖 ⊗ 𝐠𝑗

The trace of which is,

𝑎𝑖𝜙,𝑗 𝐠𝑖 ⋅ 𝐠𝑗 = 𝑎𝑖𝜙,𝑗 𝛿𝑖𝑗

= 𝑎𝑖𝜙,𝑖 = 𝒂 ⋅ grad 𝜙

For the arbitrary constant vector 𝒂, we therefore have that,

div 𝜙𝒂Ω

𝑑𝑣 = 𝒂 ⋅ grad 𝜙 Ω

𝑑𝑣 = 𝒂 ⋅ 𝜙𝐧𝜕Ω

𝑑𝑠

grad 𝜙 Ω

𝑑𝑣 = 𝜙𝐧𝜕Ω

𝑑𝑠


For a second-order tensor 𝐓, the Gauss Theorem becomes,

div𝐓Ω

𝑑𝑣 = 𝐓𝐧𝜕Ω

𝑑𝑠

The original outer product under the integral can be expressed in dyadic form:

grad 𝐓Ω

𝑑𝑣 = 𝑇𝑖𝑗 ,𝑘 𝐠𝑖 ⊗ 𝐠𝑗 ⊗ 𝐠𝑘

Ω

𝑑𝑣

= 𝐓 ⊗ 𝐧𝜕Ω

𝑑𝑠

= 𝑇𝑖𝑗𝐠𝑖 ⊗ 𝐠𝑗 ⊗ 𝑛𝑘𝐠𝑘

𝜕Ω

𝑑𝑠

Second-Order Tensor field


Or

𝑇𝑖𝑗 ,𝑘 𝐠𝑖 ⊗ 𝐠𝑗 ⊗Ω

𝐠𝑘𝑑𝑣 = 𝑇𝑖𝑗𝑛𝑘𝐠𝑖 ⊗ 𝐠𝑗 ⊗ 𝐠𝑘

𝜕Ω

𝑑𝑠

Contracting, we have

𝑇𝑖𝑗 ,𝑘Ω

𝐠𝑖 ⊗ 𝐠𝑗 𝐠𝑘𝑑𝑣 = 𝑇𝑖𝑗𝑛𝑘 𝐠𝑖 ⊗ 𝐠𝑗 𝐠𝑘

𝜕Ω

𝑑𝑠

𝑇𝑖𝑗 ,𝑘 𝛿𝑗𝑘𝐠𝑖

Ω

𝑑𝑣 = 𝑇𝑖𝑗𝑛𝑘𝛿𝑗𝑘𝐠𝑖

𝜕Ω

𝑑𝑠

𝑇𝑖𝑗 ,𝑗 𝐠𝑖Ω

𝑑𝑣 = 𝑇𝑖𝑗𝑛𝑗𝐠𝑖𝜕Ω

𝑑𝑠

Which is the same as,

div𝐓Ω

𝑑𝑣 = 𝐓𝐧𝜕Ω

𝑑𝑠

Second-Order Tensor field


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 (ℎ𝑖 ≠ 1 in general).

Physical Components of Tensors


They may not be dimensionless. For example, in orthogonal spherical polar, the base vectors are 𝐠1, 𝐠2 and 𝐠3.

These can be expressed in terms of dimensionless unit vectors as, 𝜌𝐞𝜌, 𝜌 sin 𝜙 𝐞𝜃 , and 𝐞𝜙 since the

magnitudes of the basis vectors are 𝜌, 𝜌 sin𝜙 , and 1

or 𝑔11, 𝑔22, 𝑔33 respectively.

As an example consider a force with the contravariant components 𝐹1, 𝐹2 and 𝐹3,

𝐅 = 𝐹1𝐠1 + 𝐹2𝐠2 + 𝐹3 𝐠3

= 𝜌𝐹1𝐞𝜌 + 𝜌 sin 𝜙 𝐹2𝐞𝜃 + 𝐹3𝐞𝜙

Physical Components


Which may also be expressed in terms of physical components,

𝐅 = 𝐹𝜌𝐞𝜌 + 𝐹𝜃𝐞𝜃 + 𝐹𝜙𝐞𝜙.

While these physical components {𝐹𝜌, 𝐹𝜃 , 𝐹𝜙} have the

dimensions of force, for the contravariant components normalized by the in terms of the unit vectors along these axes to be consistent, {𝜌𝐹1, 𝜌 sin 𝜃 𝐹2, 𝐹3} must each be in the units of a force. Hence, 𝐹1, 𝐹2 and 𝐹3 may not themselves be in force units. The consistency requirement implies, 𝐹𝜌 = 𝜌𝐹1, 𝐹𝜃 = 𝜌 sin 𝜃 𝐹2, and 𝐹𝜙 = 𝐹3

Physical Components


For the same reasons, if we had used covariant components, the relationships,

𝐹𝜌 = 𝐹1

𝜌, 𝐹𝜃 =

𝐹2

𝜌 sin 𝜃, and 𝐹𝜙 = 𝐹3

The magnitudes of the reciprocal base vectors are 1

𝜌,

1

𝜌 sin 𝜙, and 1. While the physical components have the

dimensions of force, 𝐹1 = 𝜌𝐹𝜌 and 𝐹2 = 𝜌 sin 𝜙 𝐹𝜃 have the

dimensions of moment, while 𝐹1 =𝐹𝜌

𝜌 and 𝐹2 =

𝐹𝜃

𝜌 sin 𝜙 are

in dimensions of force per unit length. Only the third components in both cases are given in the dimensions of force.

Physical Components


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 𝜏𝑖𝑗 or 𝜏𝑗𝑖 or

𝜏𝑖𝑗 in spherical polar coordinates. Here, as we have seen, the

magnitudes of the base vectors 𝑔11, 𝑔22, 𝑔33 or ℎ1, ℎ2 and ℎ3 are 𝜌, 𝜌 sin 𝜙 and 1. Using the rule specified above, the table below computes the physical components from the three associated tensors as follows:


Physical

Component

Contravariant Covariant Mixed

𝝉(𝝆𝝆) 𝜏11ℎ1ℎ1 = 𝜏11𝜌2 𝜏11

ℎ1 ℎ1=

𝜏11

𝜌2 𝜏1

1

ℎ1ℎ1 = 𝜏1

1

𝝉(𝝆𝜽) 𝜏12ℎ1ℎ2 = 𝜏12𝜌2 sin𝜙 𝜏12

ℎ1 ℎ1=

𝜏12

𝜌2 sin 𝜙 𝜏2

1

ℎ2ℎ1 =

𝜏21

sin 𝜙

𝝉(𝜽𝜽) 𝜏22ℎ2ℎ2 = 𝜏22𝜌2 sin2 𝜙 𝜏22

ℎ2 ℎ2=

𝜏22

𝜌2 sin2 𝜙 𝜏2

2

ℎ2ℎ2 = 𝜏2

2

𝝉(𝜽𝝓) 𝜏23ℎ2ℎ3 = 𝜏23𝜌 sin 𝜙 𝜏23

ℎ2 ℎ3=

𝜏23

𝜌 sin 𝜙 𝜏3

2

ℎ3ℎ2 = 𝜏3

2𝜌 sin 𝜙

𝝉(𝝓𝝓) 𝜏33ℎ3ℎ3 = 𝜏33 𝜏33

ℎ3 ℎ3= 𝜏33 𝜏3

3

ℎ3ℎ3 = 𝜏3

3

𝝉(𝝆𝝓) 𝜏31ℎ3ℎ1 = 𝜏31𝜌 𝜏31

ℎ3 ℎ1=

𝜏31

𝜌 𝜏1

3

ℎ1ℎ3 =

𝜏13

𝜌


17. The transformation equations from the Cartesian to the oblate spheroidal coordinates 𝝃, 𝜼 and 𝝋 are: 𝒙 = 𝒇𝝃𝜼 𝐬𝐢𝐧𝝋, 𝒚 = 𝒇 𝝃𝟐 − 𝟏 𝟏 − 𝜼𝟐 , and 𝒛 = 𝒇𝝃𝜼 𝐜𝐨𝐬 𝝋, 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:

𝑔𝜉𝜉 =𝜕𝑥

𝜕𝜉

2

+𝜕𝑦

𝜕𝜉

2

+𝜕𝑧

𝜕𝜉

2

= 𝑓𝜂 sin𝜑 2 + 𝑓2𝜉21 − 𝜂2

𝜉2 − 1+ 𝑓𝜂 cos 𝜑 2 = 𝑓2

𝜉2 − 𝜂2

𝜉2 − 1

𝑔𝜂𝜂 =𝜕𝑥

𝜕𝜂

2

+𝜕𝑦

𝜕𝜂

2

+𝜕𝑧

𝜕𝜂

2

= 𝑓2𝜉2 − 𝜂2

1 − 𝜉2

𝑔𝜑𝜑 =𝜕𝑥

𝜕𝜑

2

+𝜕𝑦

𝜕𝜑

2

+𝜕𝑧

𝜕𝜑

2

= 𝑓𝜉𝜂 2

𝑔𝜉𝜂 =𝜕𝑥

𝜕𝜉

𝜕𝑥

𝜕𝜂+

𝜕𝑦

𝜕𝜉

𝜕𝑦

𝜕𝜂+

𝜕𝑧

𝜕𝜉

𝜕𝑧

𝜕𝜂

= 𝑓𝜂 sin 𝜑 𝑓𝜉 sin𝜑 − 𝑓𝜂𝜉2 − 1

1 − 𝜂2𝑓𝜉

1 − 𝜂2

𝜉2 − 1+ 𝑓𝜂 cos 𝜑 𝑓𝜉 cos 𝜑

= 0 = 𝑔𝜂𝜑 = 𝑔𝜑𝜉


Find an expression for the divergence of a vector in orthogonal curvilinear coordinates.

The gradient of a vector 𝑭 = 𝐹𝑖𝐠𝑖 is 𝛻 ⊗ 𝑭 = 𝐠𝑗𝜕𝑗 ⊗ 𝐹𝑖𝐠𝑖 =𝐹𝑖

,𝑗𝐠𝑗 ⊗ 𝐠𝑖. 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:

𝛻 ⋅ 𝐅 = 𝐹𝑖,𝑗𝐠

𝑗 ⋅ 𝐠𝑖 = 𝐹𝑖 ,𝑖 =1

𝑔

𝜕 𝑔 𝐹𝑖

𝜕𝑥𝑖

=1

ℎ1ℎ2ℎ3

𝜕

𝜕𝑥1ℎ1ℎ2ℎ3𝐹

1 +𝜕


2 +𝜕


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 𝐹(𝑖) = 𝐹𝑖ℎ𝑖 (no sum on i). In terms of the physical components therefore, the divergence becomes,

=1

ℎ1ℎ2ℎ3

𝜕

𝜕𝑥1ℎ2ℎ3𝐹(1) +

𝜕

𝜕𝑥2ℎ1ℎ3𝐹(2) +

𝜕

𝜕𝑥3ℎ1ℎ2𝐹(3)


Find an expression for the Laplacian operator in Orthogonal coordinates. For the contravariant component of a vector, 𝐹𝑗,

𝐹𝑗,𝑗 =

1

𝑔

𝜕 𝑔 𝐹𝑗

𝜕𝑥𝑗.

Now the contravariant component of gradient 𝐹𝑗 = 𝑔𝑖𝑗𝜑,𝑖. Using this in place of the vector 𝐹𝑗, we can write,

𝑔𝑖𝑗𝜑,𝑗𝑖 =1

𝑔

𝜕 𝑔 𝑔𝑖𝑗𝜑,𝑗

𝜕𝑥𝑖

given scalar 𝜑, the Laplacian 𝛻2𝜑 is defined as, 𝑔𝑖𝑗𝜑,𝑗𝑖 so that,

𝛻2𝜑 = 𝑔𝑖𝑗𝜑,𝑗𝑖 =1

𝑔

𝜕

𝜕𝑥𝑖𝑔 𝑔𝑖𝑗

𝜕𝜑

𝜕𝑥𝑗

When coordinates are orthogonal, 𝑔𝑖𝑗 = 𝑔𝑖𝑗 = 0 whenever 𝑖 ≠ 𝑗. Expanding the computation formula therefore, we can write,

𝛻2𝜑 =1

ℎ1ℎ2ℎ3

𝜕

𝜕𝑥1

ℎ1ℎ2ℎ3

ℎ1

𝜕𝜑

𝜕𝑥1 +𝜕

𝜕𝑥2

ℎ1ℎ2ℎ3

ℎ2

𝜕𝜑

𝜕𝑥2 +𝜕

𝜕𝑥3

ℎ1ℎ2ℎ3

ℎ3

𝜕𝜑

𝜕𝑥3

=1

ℎ1ℎ2ℎ3

𝜕

𝜕𝑥1 ℎ2ℎ3

𝜕𝜑

𝜕𝑥1 +𝜕

𝜕𝑥2 ℎ1ℎ3

𝜕𝜑

𝜕𝑥2 +𝜕

𝜕𝑥3 ℎ1ℎ2

𝜕𝜑

𝜕𝑥3


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 𝑔𝜉𝜂 = 𝑔𝜂𝜑 = 𝑔𝜑𝜉 = 0 . This is the required proof of orthogonality. Using the computation formula in example 11, we may write for the oblate spheroidal coordinates that,

𝛻2Φ =𝜉2 − 1 1 − 𝜂2

𝑓3𝜉2 𝜉2 − 𝜂2 𝜕

𝜕𝜉𝑓𝜉𝜂

𝜉2 − 1

1 − 𝜂2

𝜕Φ

𝜕𝜉

+𝜕

𝜕𝜂𝑓𝜉𝜂

1 − 𝜂2

𝜉2 − 1

𝜕Φ

𝜕𝜂

+𝜕

𝜕𝜂

𝑓 𝜉2 − 𝜂2

𝜉𝜂 𝜉2 − 1 1 − 𝜂2

𝜕Φ

𝜕𝜂


For a vector field 𝒖, show that grad 𝒖 × is a third ranked tensor. Hence or otherwise show that div 𝒖 × = −curl 𝒖.

The second–order tensor 𝒖 × is defined as 𝜖𝑖𝑗𝑘𝑢𝑗𝐠𝑖 ⊗𝐠𝑘. Taking the covariant derivative with an independent base, we have

grad 𝒖 × = 𝜖𝑖𝑗𝑘𝑢𝑗 ,𝑙 𝐠𝑖 ⊗ 𝐠𝑘 ⊗ 𝐠𝑙

This gives a third order tensor as we have seen. Contracting on the last two bases,

div 𝒖 × = 𝜖𝑖𝑗𝑘𝑢𝑗 ,𝑙 𝐠𝑖 ⊗ 𝐠𝑘 ⋅ 𝐠𝑙 = 𝜖𝑖𝑗𝑘𝑢𝑗 ,𝑙 𝐠𝑖𝛿𝑘

𝑙 = 𝜖𝑖𝑗𝑘𝑢𝑗 ,𝑘 𝐠𝑖 = −curl 𝒖


Begin with our familiar Cartesian system of coordinates. We can represent the position of a point (position vector) with three coordinates 𝑥, 𝑦, 𝑧 (∈ R) such that,

𝐫 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌

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 𝑥, 𝑦 → {𝑟, 𝜙} such that, 𝑥 = 𝑟 cos 𝜙, and 𝑦 = 𝑟 sin 𝜙. Note also that this

transformation is invertible: 𝑟 = 𝑥2 + 𝑦2,and

𝜙 = tan−1 𝑦

𝑥

Coordinate transformations


With such a transformation, we can locate any point in the 3-D space with three scalars {𝑟, 𝜙, 𝑧} instead of our previous set 𝑥, 𝑦, 𝑧 . Our position vector is now,

𝐫 = 𝑟 cos 𝜙 𝒊 + 𝑟 sin 𝜙 𝒋 + 𝑧𝒌 = 𝑟𝒆𝑟 + 𝑧𝒆𝑧

where we define 𝒆𝑟 ≡ cos 𝜙 𝒊 + sin 𝜙 𝒋, 𝒆𝑧 is no different from 𝒌. In order to complete our triad of basis vectors, we need a third vector, 𝑒𝜙. In selecting

𝒆𝜙, we want it to be such that 𝒆𝑟 , 𝒆𝜙, 𝒆𝑧 can form an

orthonormal basis.

Curvilinear Coordinates


Let 𝒆𝜙 = 𝜉𝒊 + 𝜂𝒋

To satisfy our conditions,

𝒆𝜙 ⋅ 𝒆𝑟 = 0, 𝒆𝜙 ⋅ 𝒆𝑧 = 0, and 𝜉2 + 𝜂2 = 1.

It is easy to see that 𝒆𝜙 ≡ −sin 𝜙 𝒊 + cos 𝜙 𝒋 satisfies

these requirements. 𝒆𝑟 , 𝒆𝜙, 𝒆𝑧 forms an orthonormal (that is, each member has unit magnitude and they are pairwise orthogonal) triad just like 𝒊, 𝒋, 𝒌 . 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


Unlike our Cartesian system, we note that

𝒆𝑟(𝜙), 𝒆𝜙(𝜙), 𝒆𝑧 as the first two of these are not

constants but vary with angular orientation. 𝒆𝑧 remains a constant vector as in the Cartesian case.


Continuing further with our transformation, we may again introduce two new scalars such that 𝑟, 𝑧 → {𝜌, 𝜃} in such a way that the position vector,

𝐫 = 𝑟𝒆𝑟 + 𝑧𝒆𝑧 = 𝜌 sin 𝜃 𝒆𝑟 + 𝜌 cos 𝜃 𝒆𝑧 ≡ 𝜌𝒆𝜌

Here, 𝑟 = 𝜌 sin 𝜃 , 𝑧 = 𝜌 cos 𝜃. As before, we can use three scalars, 𝜌, 𝜃, 𝜙 instead of {𝑟, 𝜙, 𝑧}. In comparison to the original Cartesian system we began with, we have that,

𝐫 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 = 𝜌 sin 𝜃 𝒆𝑟 + 𝜌 cos 𝜃 𝒆𝑧 = 𝜌 sin 𝜃 cos 𝜙 𝒊 + sin 𝜙 𝒋 + 𝜌 cos 𝜃 𝒌 = 𝜌 sin 𝜃 cos 𝜙 𝒊 +𝜌 sin 𝜃 sin𝜙 𝒋 + 𝜌 cos 𝜃 𝒌 ≡ 𝜌𝒆𝜌

from which it is clear that the unit vector 𝒆𝜌 ≡ sin 𝜃 cos 𝜙 𝒊 + sin 𝜃 sin 𝜙 𝒋 + cos 𝜃 𝒌.

Spherical Polar


Again, we introduce the unit vector, 𝒆𝜃 ≡ cos 𝜃 cos 𝜙 𝒊 + cos 𝜃 sin𝜙 𝒋 − sin 𝜃 𝒌 and retain 𝒆𝜙 = − sin𝜙 𝒊 + cos 𝜙 𝒋 as before.

It is easy to demonstrate the fact that these vectors constitute another orthonormal set. Combining the two transformations, we can move from 𝑥, 𝑦, 𝑧 system of coordinates to 𝜌, 𝜃, 𝜙 directly by the transformation equations, 𝑥 = 𝜌 sin 𝜃 cos 𝜙, 𝑦 = 𝜌 sin 𝜃 sin𝜙 and 𝑧 =𝜌 cos 𝜃.

The orthonormal set of basis for the 𝜌, 𝜃, 𝜙 system is

𝒆𝜌 𝜃, 𝜙 , 𝒆𝜃 𝜃, 𝜙 , 𝒆𝜙 𝜙 .

This is the Spherical Polar Coordinate System.

Spherical Polar


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,

𝒆𝑟 𝑟, 𝜙, 𝑧 = cos 𝜙 𝒊 + sin 𝜙 𝒋 𝒆𝜙(𝑟, 𝜙, 𝑧) = −sin 𝜙 𝒊 + cos 𝜙 𝒋 𝒆𝑧 𝑟, 𝜙, 𝑧 = 𝒌

Variable Bases


For Cylindrical Polar, and for Spherical Polar,

𝒆𝜌(𝜌, 𝜃, 𝜙) = sin 𝜃 cos 𝜙 𝒊 + sin 𝜃 sin𝜙 𝒋 + cos 𝜃 𝒌

𝒆𝜃 𝜌, 𝜃, 𝜙 = cos 𝜃 cos 𝜙 𝒊 + cos 𝜃 sin𝜙 𝒋 − sin 𝜃 𝒌 𝒆𝜙 𝜌, 𝜃, 𝜙 = − sin𝜙 𝒊 + cos 𝜙 𝒋.

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.


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

𝐯 = 𝑣1𝐠1 + 𝑣2𝐠2 + 𝑣3𝐠3

when referred to the basis vectors {𝐠1, 𝐠2, 𝐠3}, for position vectors, in order that the transformation refers to the same position vector we started with in the Cartesian system


We have for Cylindrical Polar, the vector, 𝐫(𝑟, 𝜙, 𝑧) = 𝑟𝒆𝑟(𝜙) + 𝑧𝒆𝑧

and for Spherical Polar, 𝐫(𝑟, 𝜃, 𝜙) = 𝜌𝒆𝜌(𝜃, 𝜙)

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.


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.


Beginning from the position vector, given a system with coordinate variables 𝛼1, 𝛼2, 𝛼3 , it is easy to prove that the set of vectors,

𝜕𝐫 𝛼1, 𝛼2, 𝛼3

𝜕𝛼1,𝜕𝐫 𝛼1, 𝛼2, 𝛼3

𝜕𝛼1,𝜕𝐫 𝛼1, 𝛼2, 𝛼3

𝜕𝛼1

Which we can write more compactly as

𝐠𝑖 ≡𝜕𝐫

𝜕𝛼𝑖, 𝑖 = 1,2,3

constitute an independent set. This set, with its dual set

of vectors, 𝐠𝑗 such that 𝐠𝑖 ⋅ 𝐠𝑗= 𝛿𝑗𝑖, constitute the

“Natural bases” for the coordinate system.


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.


Orthonormality of Natural Bases

3. tensor calculus jan 2013

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