Tensor Analysis - University of Colorado Boulder · show that for the spherical system, the tensor...

259

For orthogonal curvilinear coordinates,

(98)

Expanding the derivative, we have,

Now expanding (some of the details are left to the reader),

Tensor Analysis

ˆ ˆˆgrad ( ).ij ji

i

ah q

∂≡ ∇ =

∂ea a e

ˆˆˆ ˆ ˆ

ˆˆ ˆ1 ˆ ˆ ˆ .

j jij ji i

i

j j ji j ii i

i i

aa

h q qa a

h q h q

∂ ∂⎛ ⎞∇ = +⎜ ⎟∂ ∂⎝ ⎠

∂ ∂= +

∂ ∂

eea e

ee e e

ˆ / ij q∂ ∂e

260

Tensor Analysis

ˆ ˆ

1 ˆ ˆ

1 ˆ .

j ji i

i

jk k j i

j

jk jk ki

j

q q h

hkh

i jh q

hkh

i jh qδ

∂ ⎛ ⎞∂= ⎜ ⎟∂ ∂ ⎝ ⎠

∂⎡ ⎤⎧ ⎫= −⎨ ⎬⎢ ⎥∂⎩ ⎭⎣ ⎦

∂⎡ ⎤⎧ ⎫= −⎨ ⎬⎢ ⎥∂⎩ ⎭⎣ ⎦

e e

e e

e

No summation on hi thus must insert to factor .ˆ ke

261

After much more work, employing eq. (18) and

one obtains,

(99)

with no summation with the hj and hk and finally,

(100)

Again with no summation of the scale factors.

Tensor Analysis

ˆˆ .j jk ij ji

ki j kj k

hhq h q h q

δ δ⎛ ⎞∂ ∂∂= −⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠

ee

22, . (no summation on )ijij

ij i ij ii

g h g hhδ

δ= =

ˆ ˆ1 ˆ ˆ .k k i i iik i ki j k

i j k

a a h a hh q h q h q

δ⎛ ⎞∂ ∂ ∂

∇ = + −⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠a e e

262

Example: Tensor gradient in cylindrical and spherical coordinates.

For each case we will first write the terms in eq. (100), incrementing the index i then combine them for the final result.cylindrical: q1 = R, q2 = φ, q3 = z

h1 = 1, h2 = R, h3 = 1

Tensor Analysis

263

i = 1:

Tensor Analysis1 2 2

1 1 1 2 1 31 1 11

31 1 2 1 11 2 3

1 2 3

11 1 1 12 1 2 13 1 3

1 1 1 1 1 11 1 1 2 1 31 2 3

1 2 3

ˆ ˆ ˆ1 ˆ ˆ ˆ ˆ ˆ ˆ

ˆˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ( )

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ

ˆˆ ˆ ˆRR R

a a ah q q q

aa h a h hh q h q h q

a h a h a hh q h q h q

aR

δ δ δ

⎡∂ ∂ ∂+ +⎢∂ ∂ ∂⎣

⎛ ⎞∂ ∂ ∂+ + +⎜ ⎟∂ ∂ ∂⎝ ⎠× + +

⎤⎛ ⎞∂ ∂ ∂− + + ⎥⎜ ⎟∂ ∂ ∂⎝ ⎠⎦

∂∂= +∂

e e e e e e

e e e e e e

e e e e e e

e eˆˆ ˆ ˆ ˆZ

R R z

a aR Rφ

φ∂

+∂ ∂

e e e e

0 0 0

0 0 000

264

i = 2:

Tensor Analysis

31 22 1 2 2 2 32 2 2

2

31 2 2 2 21 2 3

1 2 3

21 2 1 22 2 2 23 2 3

2 2 2 2 2 22 1 2 2 2 31 2 3

1 2 3

ˆˆ ˆ1 ˆ ˆ ˆ ˆ ˆ ˆ

ˆˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ( )

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ

ˆ1 ˆ ˆRR

aa ah q q q

aa h a h hh q h q h q

a h a h a hh q h q h q

aR φ

δ δ δ

φ

⎡ ∂∂ ∂+ +⎢∂ ∂ ∂⎣

⎛ ⎞∂ ∂ ∂+ + +⎜ ⎟∂ ∂ ∂⎝ ⎠× + +

⎤⎛ ⎞∂ ∂ ∂− + + ⎥⎜ ⎟∂ ∂ ∂⎝ ⎠⎦

∂= +

∂

e e e e e e

e e e e e e

e e e e e e

e eˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ

ˆˆ ˆ1 1 1ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ

Zz R R

R zR R z

a a a a

aa aa aR R R

φφ φ φ φ φ φ φ

φφ φ φ φ φ

φ φ

φ φ φ

∂⎡ ⎤∂+ + −⎢ ⎥∂ ∂⎣ ⎦∂⎛ ⎞⎛ ⎞∂ ∂

= − + + +⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠

e e e e e e e e

e e e e e e

0 0

0 00 0

265

i = 3:

Tensor Analysis

31 23 1 3 2 3 33 3 3

3

3 3 3 31 21 2 3

1 2 3

31 3 1 32 3 2 33 3 3

3 3 3 3 3 33 1 3 2 3 31 2 3

1 2 3

ˆˆ ˆ1 ˆ ˆ ˆ ˆ ˆ ˆ

ˆˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆ( )

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ

ˆˆ ˆ ˆRz R

aa ah q q q

h h a ha ah q h q h q

a h a h a hh q h q h q

az

δ δ δ

⎡ ∂∂ ∂+ +⎢∂ ∂ ∂⎣

⎛ ⎞∂ ∂ ∂+ + +⎜ ⎟∂ ∂ ∂⎝ ⎠× + +

⎤⎛ ⎞∂ ∂ ∂− + + ⎥⎜ ⎟∂ ∂ ∂⎝ ⎠⎦

∂∂= +

∂

e e e e e e

e e e e e e

e e e e e e

e eˆˆ ˆ ˆ ˆz

z z z

a az zφ

φ∂

+∂ ∂

e e e e

0 0

0 00 0

0

0

266

Combining, we have the cylindrical system,

(101)

Tensor Analysis

ˆˆ ˆˆ ˆ ˆ ˆ ˆ ˆ

ˆˆ ˆ1 1 1ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ

ˆˆ ˆˆ ˆ ˆ ˆ ˆ ˆ

R ZR R R R z

R zR R z

R zz R z z z

aa aR R R

aa aa aR R R

aa az z z

φφ

φφ φ φ φ φ

φφ

φ φ φ

∂∂ ∂∇ = + +

∂ ∂ ∂∂⎛ ⎞⎛ ⎞∂ ∂

+ − + + +⎜ ⎟⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠∂∂ ∂

+ + +∂ ∂ ∂

a e e e e e e

e e e e e e

e e e e e e

267

Following this example for the cylindrical system, you can show that for the spherical system, the tensor gradient is,

(102)

Tensor Analysis

ˆˆˆ ˆ ˆ ˆ ˆ ˆ ˆ

ˆˆˆ1 1 1ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ

ˆˆ ˆ ˆsin cosˆ ˆ ˆ ˆ

sin sinˆ

ˆ ˆsin cosˆ ˆ

sin

rr r r r

rr r

r

r

r

aaar r r

aaa a ar r r

aa a a

r ra

a a

r

φθθ φ

φθθ θ θ θ θ φ

θφ φ

φ φ θ

φθ

φ φ

θ θ φ

θ θφ φ

θ θ

θ θφ

θ

∂∂∂∇ = + +

∂ ∂ ∂∂∂∂ ⎛ ⎞⎛ ⎞+ − + + +⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠

∂∂− −

∂ ∂+ +

∂− +

∂+

a e e e e e e

e e e e e e

e e e e

e e

268

Divergence of a Second-Order TensorWe define the divergence operation for a second-order tensor,

(103)

Now relabel the dummy indices and factor,

Tensor Analysis

div ( )i jkj kiq

φ∂≡ ∇⋅ = ⋅

∂Φ Φ e e e

.

jkji jk jk k

j k k ji i i

jki jk jk

j k m k j ni

q q q

m ni j i kq

φ φ φ

φ φ φ

∂⎛ ⎞∂∂∇ ⋅ = ⋅ + +⎜ ⎟∂ ∂ ∂⎝ ⎠

⎛ ⎞⎧ ⎫ ⎧ ⎫∂= ⋅ + +⎨ ⎬ ⎨ ⎬⎜ ⎟∂ ⎩ ⎭ ⎩ ⎭⎝ ⎠

e eΦ e e e e e

e e e e e e e

269

(104)

In the Cartesian system this is

(105)

Tensor Analysis

.

rsi js rk

r s r s r si

rsjs rk i

r si

r ri j i jq

r ri j i jq

φ φ φ

φ φ φ δ

⎛ ⎞⎧ ⎫ ⎧ ⎫∂∇ ⋅ = ⋅ + +⎨ ⎬ ⎨ ⎬⎜ ⎟∂ ⎩ ⎭ ⎩ ⎭⎝ ⎠

⎛ ⎞⎧ ⎫ ⎧ ⎫∂= + +⎨ ⎬ ⎨ ⎬⎜ ⎟∂ ⎩ ⎭ ⎩ ⎭⎝ ⎠

Φ e e e e e e e

e

.rs

js rksi

i si j i kq

φ φ φ⎛ ⎞⎧ ⎫ ⎧ ⎫∂

∇ ⋅ = + +⎨ ⎬ ⎨ ⎬⎜ ⎟∂ ⎩ ⎭ ⎩ ⎭⎝ ⎠Φ e

ˆ .ijj

ixφ∂

∇ ⋅ =∂

Φ i

270

Tensor Gradient RevisitedThere are two tensor operations that we have not defined, but now will because the result of these operations may shed more light on the physical interpretation of the tensor gradient. The contraction of a second-order tensor or dyadic is the result of placing the dot operation between the vectors of each dyad,

This operation reduces the order of the tensor by two. The vector of a second-order tensor or dyadic is the result of placing the cross product between the vectors of each dyad,

Tensor Analysis

ij ijs i j ij

i j ijij ij

g

g

φ φ φ

φ φ

≡ ⋅ =

≡ ⋅ =

e e

e e

j i is i j i

i j ij i i

φ φ φ

φ φ

≡ ⋅ =

≡ ⋅ =

e e

e e

271

This operation reduces the order of the tensor by one. One can show the the following two identities,

(106)(107)

Tensor Analysis

.

ij ij kV i j ijk

i j im jn kij ij mnk

j i j im ki j i mjk

i j i jm kj i j imk

g

g g g

g g

g g

φ φ ε

φ φ ε

φ φ ε

φ φ ε

≡ × =

≡ × =

≡ × =

≡ × =

Φ e e e

e e e

e e e

e e e

T

T

[ ] ,

[ ( ) ] ( ) .V⋅ − ≡ ×

⋅ ∇ − ∇ ≡ ∇× ×

a Φ Φ Φ a

a b b b a

272

Equation (107) was briefly discussed as eq. (97). Equation (106) shows how the antisymmetric part of , which has only three independent components, is related to the vector of , ΦV. So for the gradient of vector a, ∇a,

(108)Note that div a involves the diagonal terms ∇a thus is associated with the symmetric part of ∇a (since the diagonal terms of the antisymmetric part are always zero). The vector of ∇a, (∇a)V is,

(109)

Tensor Analysis

ΦΦ

, ,( ) div .k j kS j k ka a∇ = ⋅ = =a e e a

, ,( ) curl .k j ji k mV j k j ikma g g a ε∇ = × = =a e e e a

273

Therefore, (∇a)V and the antisymmetric part of ∇a are related to the rotation of the vector field as previously presented.So we can conclude that the tensor gradient contains information about the dilitation and rotation of a vector field. This information is extracted with the internal operations just defined in eqs. (108) and (109) or when ∇a is used to transform a vector into another vector, (e.g., ∇b in eq. (107) is used to transform a into another vector).

IN-CLASS EXAMPLES

Tensor Analysis

274

Integral Theorems for Dyadics and Second-Order TensorsThe vector integral relations (49)-(51) are increased by one order,

(110)

(111)

(112)

(113)

Tensor Analysis

ˆgrad R S

d dSτ =∫∫∫ ∫∫a na

ˆdiv R S

d dSτ = ⋅∫∫∫ ∫∫Φ n Φ

ˆcurl R S

d dSτ = ×∫∫∫ ∫∫Φ n Φ

ˆ (curl )CS C

dS⋅ = ⋅∫∫ ∫n Φ ds Φ

275

The invariant forms can also be shown as before in eqs. (55)-(57)

(110)

(111)

(112)

(113)

Note that (113) and (117) are added for completeness.

Tensor Analysis

0

1 ˆgrad limS

dSτ τ ∆∆ →

=∆ ∫∫a na

0

1 ˆdiv limS

dSτ τ ∆∆ →

= ⋅∆ ∫∫Φ n Φ

0

1 ˆcurl limS

dSτ τ∆ →

= ×∆ ∫∫Φ n Φ

0

1ˆ curl limnCS S∆ →

⋅ = ⋅∆ ∫n Φ ds Φ

276

Eigenvalues and Eigenvectors of DyadicsWe have shown how a dyadic can be written in matrix notation and some of the matrix operations that apply to dyadics. Let us now extend this further by computing eigenvalues and eigenvectors of second-order tensors.The dyadic is an operator that changes a vector into anothervector, e.g.,

where λ is the eigenvalue that characterizes the change in a.

Tensor Analysis

There is a special case where only the magnitude and/or sense of the vector is changed, i.e.,

Φ

.⋅ =Φ a b

λ⋅ =Φ a a

b

λaa

277

This is the same relation we developed for the homogeneous system in linear algebra. We can write this equation in matrix form as,

From Cramer’s theorem, we know that to obtain nontrivial solutions,

Tensor Analysis

( )

λ

λ

⋅ − =

− ⋅ =

Φ a a 0

Φ I a 0

11 12 13 1

21 22 23 2

31 32 33 3

00 .0

aaa

φ λ φ φφ φ λ φφ φ φ λ

−⎡ ⎤⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎢ ⎥− =⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎜ ⎟ ⎜ ⎟−⎢ ⎥⎣ ⎦ ⎝ ⎠ ⎝ ⎠

det( ) 0.ij ijλ φ λδ− = − =Φ I

278

From Cramer’s theorem, we know that to obtain nontrivial solutions,

Note:i. Each eigenvalue has an associated eigenvector.ii. Since complex roots must appear in conjugate pairs,

at least one λ is real for n = 3.iii. For symmetric , all λ are real.

For most practical engineering problems is symmetric (Reddy & Rasmussen), so for the remainder of the analysis we assume is symmetric. Furthermore,

iv. The eigenvectors associated with distinct eigenvalues of a symmetric dyadic are orthogonal.

Tensor Analysis

det( ) 0.ij ijλ φ λδ− = − =Φ I

ΦΦ

Φ

279

Computing eigenvectors (symmetric dyadic):Since only the direction of the eigenvectors is of importance, we can choose to write the set of eigenvectors as unit vectors, i.e, if a is a given eigenvector, then the unit vector associated with a is

(119)Then,

(120)

Tensor Analysis

* 1 1 2 2 3 32 2 21 2 3

ˆ ˆ ˆˆ .

| |a a a

a a a+ +

= ± =+ +

i i iaea

*11 1

1

*22 2

2

*33 3

3

ˆ| |

ˆ| |

ˆ| |

λ

λ

λ

→ ± =

→ ± =

→ ± =

a eaa eaa ea

Choose ± to ensure a right-handed system.

280

The represent the principal directions of . Axes of a coordinate system aligned with the principal directions are the principal axes. The orthogonal set of eigenvectors forms a basis. Thus we can use

and write,

Then,

Tensor AnalysisΦ

* *ˆ ˆ ˆ( )j j j⋅i e e

* *ˆ ˆ ˆ ˆ .ij i j mn m nφ φ= =Φ i i e e

* * *

( )

ˆ ˆ( ) (no summation)j j

mn m n j j j

λ

φ λ

− ⋅ =

− ⋅ =

Φ I a 0

e e I a 0

281

or,

This result shows that a dyadic written in principal axis coordinates reduces to only components along the diagonal that are the eigenvalues.

Tensor Analysis* * *

* * *

* * * * *

ˆ ˆ

ˆ ˆ

ˆ ˆ ˆ( ) (no summation)

mn m nj j k kj

mj m j j

j mj m j j ij j ij j

φ δ λ δ

φ λ

φ λ φ λ δ

− =

− =

⋅ = → =

e e 0

e e 0

e e e

1*

2

3

0 00 0 .0 0

ij

λφ λ

λ

⎡ ⎤⎢ ⎥⎡ ⎤ =⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦

282

Problems with coordinate symmetry will result in a repeated root, e.g., λ1 distinct from λ2 = λ3. In this case, two directions are independent and the third is arbitrary. So one can choose the third eigenvalue such that the principal axes form a right-handed coordinate system, i.e., a3 = a1 × a2.

Example: Find the eigenvalues and eigenvectors of

Tensor Analysis

4 4 04 0 0 .

0 0 3

−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

283

Taylor Series ExpansionFrom differential calculus, any analytic function can be expanded as an infinite power series, i.e., for a scalar function of a scalar, u(x),

Tensor Analysis

x − δ x + δx

δ

284

(121)

This is called the increment form of the Taylor series. For a scalar function of two scalar variables,

(122)

Tensor Analysis2 2

20 0 0

( ) ( ) .2! !

n n

n

du d u d uu x u xdx dx n dxδ δ δ

δ δδ δ= = =

+ = + + + +

( , ) ( , ) ( , )

1 ( , )!

n

u x h y k u x y h k u x yx y

h k u x yn x y

⎛ ⎞∂ ∂+ + = + + +⎜ ⎟∂ ∂⎝ ⎠

⎛ ⎞∂ ∂+ +⎜ ⎟∂ ∂⎝ ⎠

285

Tensor Analysis

x y y + hx + h

u(x,y)

Note, can you write aa sample term from the operator of eq. (122)?

2

( , ) ?h k u x yx y

⎛ ⎞∂ ∂+ =⎜ ⎟∂ ∂⎝ ⎠

286

We now generalize these results to a scalar function of a vector, φ(r), or a vector function of a vector, a(r),

Tensor Analysis

r

φ(r +)

φ(r)

287

(123)

(124)

Recall the Taylor series for the exponential function,

Noting the form of (121) - (124), we can define a convenient operator form of the Taylor series,

(125)

Tensor Analysis

( ) ( ) ( )2!

( ) ( ) ( )2!

φφ φ φ φ⋅∇+ = + ⋅∇ + ⋅∇ +

⋅∇+ = + ⋅∇ + ⋅∇ +

δr δ r δ δ

δ aa r δ a r δ a δ a

2

1 .2! !

nx x xe x

n= + + + +

2 2

21 ,2! !

d n ndx

n

d d dedx dx n dx

δ δ δδ= + + + +

288

Continuum Velocity Field: Relative MotionGiven the velocity at a given point in a continuum velocity field, we can use the Taylor series expansion to determine the velocity at a neighboring location. This motion can be decomposed into a combination of translational, rotational and shearing (deforming) motions,

Tensor Analysis

v(r)∆rr

r+∆rv(r+∆r)

The gradient of the velocity is written in its symmetric and antisymmetric parts,

( ) ( ) h.o.t.+ ∆ = + ∆ ⋅∇ +v r r v r r v

T T1 1[ ( ) ] [ ( ) ],2 2

∇ = ∇ + ∇ + ∇ − ∇v v v v v

289

where,

Thus,

Tensor Analysis

T

T

1 [ ( ) ] rate of strain tensor,2

1 1[ ( ) ] curl 2 2 rate of rigid-body rotation.

∇ = ∇ + ∇ ≡ =

∆ ⋅ ∇ − ∇ = ×∆

=

v v v ε

r v v v r

rate of straintranslationrate of rotation

1( ) ( ) curl .2

+ ∆ = + ∆ ⋅ + ×∆v r r v r r ε v r

290

The rate of strain tensor looks like

Here is an illustration of the combined straining motions.

Tensor Analysis

11 12 13

21 22 23

31 32 33

[ ]ij

ε ε εε ε ε ε

ε ε ε

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

3 1

1 3

12

v vx x

⎛ ⎞∂ ∂+⎜ ⎟∂ ∂⎝ ⎠

3 3 3

3 3 3

12

v v vx x x

⎛ ⎞∂ ∂ ∂+ =⎜ ⎟∂ ∂ ∂⎝ ⎠

+ +

shear(off-diagonal terms)extension

(diagonal terms)

291

Note:

This concludes our discussion of tensor analysis.

Tensor Analysis

1 2 3

1 2 3

trace[ ] div

rate of change of volume per unit volume (dilatation).

ij iiv v vx x x

ε ε ∂ ∂ ∂= = + + =

∂ ∂ ∂=

v