Vector Calculus & General Coordinate Systems Orthogonal curvilinear coordinates For orthogonal...

175

Derivatives of a General Unitary Basis We have examined the computation of derivatives in an orthonormal curvilinear system. With the exception of the Cartesian system, these are the easiest to compute. Now we will develop more general procedures and formulae for an arbitrary curvilinear system with a unitary basis. Although thedevelopment is straightforward, we will introduce symbols and nomenclature that is probably unfamiliar. You must pay close attention to the definitions. Computational mechanics (fluid, structures, etc.) when the physical domain is mapped onto a curvilinear grid is an example where the basis vectors change with respect to location in the grid, thus partial derivatives such as must be evaluated.

Vector Calculus & General Coordinate Systems

/ ji q∂ ∂e

176

Christoffel symbols and contravariant derivativesRecall, for an arbitrary vector a,

If , we can write

(11)

For a more compact notation, let’s define the Christoffelsymbol of the second kind,

(12)


( )ii= ⋅a a e e

/ ji q= ∂ ∂a e

.ki ikj jq q

⎛ ⎞∂ ∂= ⋅⎜ ⎟∂ ∂⎝ ⎠

e e e e

.ki

j

ki j q

⎧ ⎫ ∂= ⋅⎨ ⎬ ∂⎩ ⎭

e e

177

Now, the contravariant derivative of the covariant basis is,

(13)

Note that the christoffel symbol is the contravariant component of the ek basis. In eq. (13) note the summation over the dummy k. Reference to the dual basis (ek) is eliminated by introducing the components of the fundamental metric,

(14)


,

( ),

.

k krr

k krirj

kr ij

g

gqk

gi j q

=∂

⋅ =∂

⎧ ⎫ ∂=⎨ ⎬ ∂⎩ ⎭

e ee e e

e

.i

kj

ki jq

⎧ ⎫∂= ⎨ ⎬∂ ⎩ ⎭

e e

178

A few more manipulations will enable us to write the Christoffel symbol completely in terms of components of the fundamental metric. First, rewrite eq. (14),

(15)

Now use the fact that r(qi) is a continuous function then,

(16)

This result might be unexpected. The equality of mixed second-partial derivatives, however, should not surprise us if we recall from the differential calculus,


1 .2

kr i ir rj j

kg

i j q q⎧ ⎫ ⎛ ⎞∂ ∂

= ⋅ + ⋅⎨ ⎬ ⎜ ⎟∂ ∂⎩ ⎭ ⎝ ⎠

e ee e

.jij j i i j iq q q q q q

∂⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂= = =⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

ee r r

179

iff f (x,y) is continuous. Substituting eq. (16) into eq. (15), we have,

(17)

For the final steps of this development, we now write the dot products in terms of the fundamental metric components,


1 .2

jkr ir ri j

kg

i j q q∂⎛ ⎞⎧ ⎫ ∂

= ⋅ + ⋅⎨ ⎬ ⎜ ⎟∂ ∂⎩ ⎭ ⎝ ⎠

e ee e

( ) ,j rj i r ji i iq q q

∂∂ ∂⋅ = ⋅ + ⋅

∂ ∂ ∂

e ee e e e

2 2f fx y y x

∂ ∂=

∂ ∂ ∂ ∂

180

or

Then,

or finally,

(18)


1 ( ) ( ) ,2

1 ( ) ( ) ( ) ,2

kr r rj r j i r ii i j j

krj r i r i ji j r

kg

i j q q q q

kg

i j q q q

⎧ ⎫ ⎡ ⎤∂ ∂ ∂ ∂= ⋅ − ⋅ + ⋅ − ⋅⎨ ⎬ ⎢ ⎥∂ ∂ ∂ ∂⎩ ⎭ ⎣ ⎦

⎧ ⎫ ⎡ ⎤∂ ∂ ∂= ⋅ + ⋅ − ⋅⎨ ⎬ ⎢ ⎥∂ ∂ ∂⎩ ⎭ ⎣ ⎦

e ee e e e e e

e e e e e e

( ) .j rr j r ji i iq q q

∂ ∂ ∂⋅ = ⋅ − ⋅

∂ ∂ ∂

e ee e e e

1 .2

jr ijkr iri j r

g gk ggi j q q q

∂ ∂⎡ ⎤⎧ ⎫ ∂= + −⎨ ⎬ ⎢ ⎥∂ ∂ ∂⎩ ⎭ ⎣ ⎦

181

Thus, eq. (18) gives the scalar components of the covariant derivative in eq. (13).Now we develop a similar relation for the contravariantderivative of the contravariant basis. Start with,

(19)

Again employ the relation


( ) 0,

0,

.

k ki ij j

kki

ij j

kki

i j j

q q

q qk

i jq q

δ∂ ∂⋅ = =

∂ ∂

∂ ∂⋅ + ⋅ =

∂ ∂

⎧ ⎫∂∂⋅ = − ⋅ = −⎨ ⎬∂ ∂ ⎩ ⎭

e e

e ee e

eee e

( ) .ii= ⋅a a e e

182

If we can write

or, using eq. (19),

(20)

Again, eq. (20) is the contravariant derivative of the contravariant basis. Finally, without going through the details, one can show,

(21)


.

ki

j

ki jq

⎧ ⎫∂= − ⎨ ⎬∂ ⎩ ⎭

e e

/ ,k jq= ∂ ∂a e

,k k

iij jq q

⎛ ⎞∂ ∂= ⋅⎜ ⎟∂ ∂⎝ ⎠

e e e e

1ln . 2i i

k ggi k q g q

⎧ ⎫ ∂ ∂= =⎨ ⎬ ∂ ∂⎩ ⎭

183

Christoffel symbol of the first kind The Christoffel symbol of the first kind, is defined in terms of derivatives of the transformation relating the curvilinear system to the original Cartesian system,

(22)

Now with

(23)or

(24)


[ , ] .imm i j jij m

q q q q∂∂ ∂

= ⋅ = ⋅∂ ∂ ∂ ∂

er r e

( ) ,mm= ⋅a a e e

,mi imj jq q

⎛ ⎞∂ ∂= ⋅⎜ ⎟∂ ∂⎝ ⎠

e e e e

[ , ] .mij ij m

q∂

=∂

e e

184

The difference between eqs. (23) and (24) and eqs. (12) and (13) is that eqs. (12) and (13 ) give the contravariant derivative of the covariant basis in terms of contravariant scalar components and the covariant basis. Equations (23) and (24) give the contravariant derivative in terms of covariant scalar components and the contravariant (dual) basis. To summarize and show the relation of the two Christoffelsymbols,


Expansion in terms of

[ , ] Expansion in terms of

ik kj

k kij

ki jq

ij kq

⎧ ⎫∂= ⎨ ⎬∂ ⎩ ⎭

∂=

∂

e e e

e e e

185

The symbols can be directly related with ek = gkmem,

(25)

Now we can write [ij,m] in terms of the covariant components of the fundamental metric,

(26)

Example: Orthogonal curvilinear coordinates For orthogonal curvilinear coordinates, recall,


[ , ] .

m mi

kmj

kij m g

i jq⎧ ⎫∂

= = ⎨ ⎬∂ ⎩ ⎭

e e e

1[ , ] .2

jm ijimkm j i m

g gk gij m gi j q q q

∂ ∂⎡ ⎤⎧ ⎫ ∂= = + −⎨ ⎬ ⎢ ⎥∂ ∂ ∂⎩ ⎭ ⎣ ⎦

10, (no summation)ij i iiii

g i j h gg

= ≠ =

186


11 1211 11 111 1 1

1 11 1 2

g g gg gq q q

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

121

gq

∂∂

122

gq

∂+

∂112

13

gq

g

⎡ ⎛ ⎞∂−⎢ ⎜ ⎟⎜ ⎟∂⎢ ⎝ ⎠⎣

+ 131

gq

∂∂

132

gq

∂+

∂113

211 11 11 1 2 1

11 1

11

1

1 1 1 1 ( )22 21 .

gq

g g hg q q h q

hh q

⎤⎛ ⎞∂− ⎥⎜ ⎟⎜ ⎟∂ ⎥⎝ ⎠⎦

⎛ ⎞∂ ∂ ∂= − =⎜ ⎟∂ ∂ ∂⎝ ⎠

∂=

∂

187


These results of course agree with eqs. (25) and (26).

Before concluding this discussion, it is worth mentioning that the Christoffel symbol of second kind is most often used in the literature. It is often denoted with the symbol, . i

jkΓ

188

Directional Derivative, Gradient, and Divergence Directional derivative and gradientImagine a day in the Boulder area with a temperature distribution (°F) shown in the contour schematic. With With Boulder at the origin of the 2-D coordinate system, the temperature distribution is an example of a scalar field, i.e., a scalar function of a vector (in this case, the position).


Lyons

Ned

SupLou

Laf

Nwt Lmt

Wmr

Bfd

Bldr

87

90

89

88

US-36

90

89

87

189

Directional Derivative, Gradient, and Divergence In general,φ = φ(r) = φ(q1, q2, q3) (φ = temperature) . Suppose we want to determine the rate of change of φ roughly along US-36 towards Denver, indicated by the direction. To do this, we develop the directional derivative that gives the rate of change of φ at a given r, in the direction of . Start with,

Recall, for a differential displacement,


1 2 31 2 3 .i

id dq dq dq dqq q q qφ φ φ φφ ∂ ∂ ∂ ∂

= + + =∂ ∂ ∂ ∂

e

e

.iid dq=r e

190

Then,

Now with ds = |dr|, we can define the direction by the unit vector,

The directional derivative is defined by

(27)


ˆ .dds

=re

1 2 31 2 3

j iijd dq

q

dq q q

φφ

φ φ φ

∂= ⋅

∂

⎛ ⎞∂ ∂ ∂= + + ⋅⎜ ⎟∂ ∂ ∂⎝ ⎠

e e

e e e r

ˆ

ˆ,jj

dds qφ φ∂⎛ ⎞ = ⋅⎜ ⎟ ∂⎝ ⎠e

e e

191

and is the rate of change of φ with respect to s in the direction of . Define the vector,

(28)as the gradient vector that points in the direction of maximum change of φ. Note that grad φ is defined in terms of covariant components scalar components and the contravariant basis.For a general representation of level surfaces of φ, i.e., surfaces for which φ = const, the unit normal vector is the unit vector locally perpendicular to a level surface,


e

grad jjq

φφ ∂≡

∂e

192

For the surface, the unit normal is (29)


e n

n

r

gradφ

1Cφ =

2Cφ =

ˆ gradddsφ φ= ⋅e

gradˆ .|grad |

φφ

= ±n

193

For a closed surface, it is the convention to choose pointing outward.Now define the del operator, ∇,

(30)Note that ∇ is a vector differential operator, i.e., it has some properties of a vector. Since we have defined ∇ to operate from left to right, it does not share the commutativity property of a vector for the dot product, i.e.,

Vector Calculus & General Coordinate Systemsˆ+n

1 2 31 2 3grad ,

q q qφ φ φ

⎛ ⎞∂ ∂ ∂≡ ∇ + +⎜ ⎟∂ ∂ ∂⎝ ⎠

e e e

.iiq

∂∇ ≡

∂e

⋅∇ ≠ ∇ ⋅a a

194

In fact, the left side of the inequality is itself an operator. For orthogonal curvilinear coordinates,

(31)

For the Cartesian system,

(32)The divergence of a vector a is defined as,

(33)


1 2 3

1 2 31 2 3

ˆ (no summation),

ˆ ˆ ˆ.

ii

ih

h q h q h q

=

∂ ∂ ∂∇ = + +

∂ ∂ ∂

ee

e e e

ˆ .i ix∂

∇ =∂

i

div .≡ ∇ ⋅a a

195

As we will see, is related to the net efflux of some vector quantity a per unit volume, at a point in space. In general,

Now recall another relation you thought we’d never use, eq. (21),

Vector Calculus & General Coordinate Systems∇ ⋅a

div

.

ij i j

i kj

ii

j

ka ai jq

ia ai jq

δ δ⎧ ⎫∂

= + ⎨ ⎬∂ ⎩ ⎭⎧ ⎫∂

= + ⎨ ⎬∂ ⎩ ⎭

a

1ln . 2i i

k ggi k q g q

⎧ ⎫ ∂ ∂= =⎨ ⎬ ∂ ∂⎩ ⎭

196

Then,

So we have two alternate forms commonly seen in the literature,

(34)and

(35)

Where J is the Jacobian.


( )div .

i i

i i

ga aq qg

∂∂= +

∂ ∂a

1div ( ),ii ga

qg∂

=∂

a

1div ( ).ii Ja

J q∂

=∂

a

197

For orthogonal curvilinear coordinates,

Then,

(36)

(37)

Note we intentionally emphasize the physical components in the form of eq. (37). And finally, for the Cartesian system,

(39)


ˆ ˆˆ ˆ ˆ( no summation)i i ii i i i i ia h a a a h a= = = =a e e

1 2 31 2 3

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

1div ( ),

1 ˆ ˆ ˆ( ) ( ) ( ) .

ii h h h a

h h h q

h h a h h a h h ah h h q q q

∂=

∂

⎡ ⎤∂ ∂ ∂= + +⎢ ⎥∂ ∂ ∂⎣ ⎦

a

div .i

i

ax

∂=

∂a

198

Integral Relations (Theorem of Gauss) The following relations from vector integral calculus relate surface integrals to volume integrals. As we will see, these relations can be used to give a more physical interpretation to some of the vector differential relations defined in the previous sections. The vector integral relations form the basis of the integral and differential conservation laws. For development, we will focus on the conservation laws of continuum mechanics. For computational methods, they form the basis of finite-volume and finite-element methods.We begin with an arbitrary region R in space enclosed by a surface S,


199


dτ

n

ndS

S

Rr

200

A differential volume element is dτ, and dS is the magnitude of a differential surface element whose orientation is determined by the outward unit normal . Now introduce three vector integral theorems,

Gradient Theorem (49)

Divergence Theorem (50)

Curl Theorem (51)


n

ˆgradR S

d dSφ τ φ=∫∫∫ ∫∫ n

ˆdivR S

d dSτ = ⋅∫∫∫ ∫∫a n a

ˆcurlR S

d dSτ = ×∫∫∫ ∫∫a n a

201

Using the divergence theorem, we can create a derivative relation similar to the directional derivative. Set a = ∇φ,

(52)

Here, we identify the normal derivative,


2 ˆdiv grad gradR R S

d d dSφ τ φ τ φ= ∇ = ⋅∫∫∫ ∫∫∫ ∫∫ n

n φ∇

nφ∂

∂

(53)

is the rate of change of φ in the direction normal to the surface Senclosing region R.

ˆnφ φ∂

≡ ⋅∇∂

n

202

In Cartesian coordinates, the normal derivative is,

(54)

Because is a unit vector, the ni are direction cosines. Using the integral relations, we can develop more intuitive, alternative definitions for some of the vector differential relations introduced earlier. (Note that in the integral approach we make no references to any coordinate system.) We begin by examining the limit as the region R shrinks to a point, i.e.,


ˆ ˆi i j j

j j

n nn x xφ φ φ∂ ∂ ∂

= ⋅ =∂ ∂ ∂

i i

n

, .R S Sτ→ ∆ → ∆

203

Why does ∆τ → 0 but S → ∆S ≠ 0? The integral relations are now,

Again, since we have not referred to any coordinate system, and these relations involve vectors, the following are called the invariant forms,


ˆ ˆgrad gradR S S

d dS dSφ τ φ φ τ φ∆

= → ∆ ≅∫∫∫ ∫∫ ∫∫n n

ˆ ˆdiv divR S S

d dS dSτ τ∆

= ⋅ → ∆ ≅ ⋅∫∫∫ ∫∫ ∫∫a n a a n a

ˆ ˆcurl curlR S S

d dS dSτ τ= × → ∆ ≅ ×∫∫∫ ∫∫ ∫∫a n a a n a

204

(55)

(56)

(57)

Physical interpretations of integral definitionsNote that each of the invariant forms resembles the difference-quotient definition of a derivative from the differential calculus (1-D),

where u is some scalar function of x.


0

1 ˆgrad limS

dSτ

φ φτ ∆∆ →

≅∆ ∫∫ n

0

1 ˆdiv limS

dSτ τ ∆∆ →

≅ ⋅∆ ∫∫a n a

0

1 ˆcurl limS

dSτ τ∆ →

≅ ×∆ ∫∫a n a

0

( ) ( )limx

du u x x u xdx x∆ →

+ ∆ −=

∆

205

grad φ:

In eq. (55) the function φ appears as a weighting function for, i.e., larger values of contribute more to the total

integral than smaller (in magnitude) values. As ∆τ → 0, the integral is weighted most in the direction of greatest increase of φ, i.e., the term in the direction of grad φ is the dominant term in the integral, thus the integral definition corresponds with the original gradient definition, eq. (28).


φ = C1

φ = C2φ = C3

φ = C4

φ = C5

R

S

ˆφn

ˆφn

ˆdSn ˆdSφn

206

div a:

For illustration, set a = v in eq. (56), where v is the velocity of some quantity flowing from region R through the surface S.Note that R and S define a bounded region in space, not necessarily a physical boundary. Then,


dSv

vn

v

R

vt

S

nn

207

As ∆τ → 0, the integral, eq. (56), becomes,

(58)


ˆ outflow (efflux) through .nv dS dS dS= ⋅ ≡v n

net outflowdiv unit volume

≡v

208

curl a: We use Stokes’ theorem to show the curl is related to the circulation of a vector field:


ds

a

curl a

dS

C

capping surface n

209

Shrink the region until the capping surface just covers a differential plane area bounded by the curve C, i.e.,


Cn∆S

|curl a|ncurl a

ds

a

n

C → Cn: bounds the infinitesimal plane surface ∆S.

circulation of about

ˆ(curl )S C

C

dS⋅ = ⋅∫∫ ∫a

a n a ds

210

Let ∆S → 0,


0

1ˆ ˆ(curl ) (curl ) lim

circulation around ˆ ˆ(curl ) (curl ) .area enclosed by

n n

n

CS SC

C

∆ →⋅ = ⋅ = ⋅

∆

⋅ = ⋅ ≡

∫n a n a a ds

n a n a

ˆ curl ,C

⋅ ≅ ⋅∫n a a ds

211

Example: Show that in a fluid, the bouyancy force f is equal and opposite to the local gravitational acceleration g and equal to the weight of the displaced fluid. Solution:Pressure is defined as the inward normal stress on a surface, i.e., it is the (force)/(unit area) at a point on a surface S, oriented in direction opposite the outward unit normal to thesurface. For a body immersed in a static pressure field p(r), the total force is

The magnitude and direction of the greatest increase in pressure is grad p = ρg, where ρ is the mass density. From the gradient theorem eq. (49),


n

ˆ .S

p dS= − ∫∫f n

212

Now with the scalar field set to φ(r) = p(r),

Thus,

Eureka! This is Archimedes’ principal!


ˆ grad

Q.E.D.

RS

R R

p dS p d

d d m

τ

ρ τ ρ τ

= − = −

= − = − = − ⇐

∫∫ ∫∫∫

∫∫∫ ∫∫∫

f n

g g g

ˆgradR S

d dSφ τ φ=∫∫∫ ∫∫ n

ˆgradR S

p d p dSτ =∫∫∫ ∫∫ n

213

Example: Show that the continuity equation

is the local mass conservation relation. Where t is time, ρ is density, and v is velocity. Solution:Begin with a fixed region in space with, ρ = ρ(r,t), v = v(r,t), and


div 0tρ ρ∂

+ =∂

v

massmass flux= .(area)(time)

ρ ≡v

214

The mass conservation law is, (with no mass sources)


S

R fixed in space

ρv

r

n

rate of change of net flux of massmass in region across surface

ˆR S

R Sd d dSdt

ρ τ ρ

⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

= ⋅∫∫∫ ∫∫ v n

215

Since R is fixed in space,

Applying the divergence theorem, the surface integral is converted to a volume integral, and

Since this relation must hold for any arbitrary region R, the only way the integral is always zero is if the integrand is zero, thus


ˆ 0.R S

d dStρ τ ρ∂

+ ⋅ =∂∫∫∫ ∫∫ v n

div 0.R

dtρ ρ τ∂⎛ ⎞+ =⎜ ⎟∂⎝ ⎠∫∫∫ v

div 0. Q.E.D.tρ ρ∂

+ = ⇐∂

v

216

Reynolds Transport TheoremThe Reynolds transport theorem allows us to develop general transport relations. For example, in some medium (gas, liquid, solid, vacuum), we can write relations that describe the evolution of the field distribution of some vector (tensor) or scalar quantity due to the transport or convection of medium quantities (e.g., mass, momentum, energy, etc.) Recall, in 1-D integro-differential calculus, the derivative of an integral function I(t) is given by the Liebniz rule,


( )

( )( ) ( ) ,

( ( )) ( ( )).

b t

a tI t f x dx

dI db daf b t f a tdt dt dt

=

= −

∫

217

For a function of 1-D space and time,

Now generalize this to a 3-D system,


( )

( )

( )

( )

( ) ( , ) ,

( ( ), ) ( ( ), ) .

b t

a t

b t

a t

I t f x t dx

dI db da ff b t t f a t t dxdt dt dt t

=

∂= − +

∂

∫

∫

S(t)

R(t)

vS

dS

n Let Q(r,t) represent some flow quantity, e.g., density. In the general case, each point on the surface S(t) has a velocity vS.

218

(59)

For the special case of a material region,vS = v ≡ velocity of the medium, i.e., each point on the surface S(t) moves with the local velocity of the medium. For this case, it is standard to denotethe time derivative with a special notation,


( ) ( , )I t Q t dτ= ∫∫∫ r

( ) ( )ˆ( , ) SR t S t

dI Q d Q t dSdt t

τ∂= + ⋅

∂∫∫∫ ∫∫ r v n

the total amount of Q in region R at time t.

substantial (material) derivative.d Ddt Dt

→ ≡

219

Then,

(60)

Conservation EquationsUsing the Reynolds transport theorem, we can develop the physical conservation laws in the form used in study of classical tensor, vector, scalar fields. MassThe mass dm in a differential volume element dτ is ρdτ. Then the total mass in the region R(t) is,


( ) ( )ˆ( , )

R t S t

DI Q d Q t dSDt t

τ∂= + ⋅

∂∫∫∫ ∫∫ r v n

( ).

R tdρ τ∫∫∫

220

The statement of mass conservation is that the mass contained in a region R(t) that is moving at the medium velocity v is conserved,

(61)

Alternatively, using the Reynolds transport relation eq. (60),

(62)


( )( , ) 0.

R t

D t dDt

ρ τ =∫∫∫ r

( ) ( )ˆ( , ) 0.

R t S td t dS

tρ τ ρ∂

+ ⋅ =∂∫∫∫ ∫∫ r v n

221

MomentumNewton’s second law is:

Now write this for a material region,

Momentum conservation is then written as

(63)


( )( , ) .

R t

D t dDt

ρ τ = ∑∫∫∫ v r f

sum of external time rate of changeforces of momentum

( ).d mdt

⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

=∑f v

( )

momentum of massmomentum in ( ) is

is R tR t d

dm dρ τ

ρ τ⎫

→⎬⎭

∫∫∫ vv

222


(64)

EnergyThe first law of thermodynamics is


( ) ( )

( ) ˆ .R t S t

d dSt

ρ τ ρ∂+ ⋅ =

∂ ∑∫∫∫ ∫∫v vv n f

heat transfer acrosschange in system work on the

the boundary intototal energy system boundary

the systemE Q Wδ δ

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

∆ = +

223

The total energy in dτ is

where u is the specific internal energy and v is speed. We then write the energy equation as

(65)



( ).

2R t

D Q Wu dDt dt dt

δ δρ τ⋅⎛ ⎞+ = +⎜ ⎟⎝ ⎠∫∫∫

v v

2

potentialkinetic

total energy in 2vd u dτ ρ ρ τ

⎛ ⎞⎜ ⎟= +⎜ ⎟⎜ ⎟⎝ ⎠

224


(64)

Note that we have explicitly used a ‘δ’ to designate differential of heat Q and work W. Why?


( ) ( )ˆ .

2 2R t S t

Q Wu d u dSt dt dt

δ δρ τ ρ⎡ ⎤∂ ⋅ ⋅⎛ ⎞ ⎛ ⎞+ + + ⋅ = +⎜ ⎟ ⎜ ⎟⎢ ⎥∂ ⎝ ⎠ ⎝ ⎠⎣ ⎦∫∫∫ ∫∫

v v v v v n

225

Dyadics and Tensors We have discussed the general notion that a scalar represents a zero-order tensor and a vector is a first-order tensor. Now introduce the notion of a second-order tensor and a dyadic. Begin with an example (Reddy & Rasmussen), Example: Angular momentum of a rigid body.

Tensor Analysis

r

ω

O

dτ

0H0 ( )d dρ τ= ×H r v

( ) = ×v r ω r

= angular momentum about O for differential mass element dm = ρ dτ.

226

Now write HO in a form to show the explicit dependence on the angular velocity vector ω,

(65)Recall the triple vector product,

Then eq. (65) becomes,

(66)

Tensor Analysis

0 ( ) angular momentum vector.R

dρ τ= × =∫∫∫H r v

0 ( ) .R

dρ τ= × ×∫∫∫H r ω r

( ) ( ) ( ).× × = ⋅ − ⋅a b c b a c c a b

20 ( ) .

Rr dρ τ⎡ ⎤= − ⋅⎣ ⎦∫∫∫H ω r ω r

227

For rigid-body rotation, ω is not a function of position, i.e., ω ≠ω(r), so we should be able to factor it. To do this, we introduce the dyad rr.A dyad is two vectors, side by side, that act as one entity. A dyadic is a sum of dyads. The unit dyadic is

(67)Like the identity matrix of linear algebra, the unit dyadic changes a vector into itself, e.g.,

Tensor Analysis

1 2 31 2 3 .= + +I e e e e e e

1 2 31 2 3

1 2 31 2 3

( ) ( ) ( )

.ω ω ω

⋅ = ⋅ + ⋅ + ⋅

= + + =

I ω e e ω e e ω e e ω

e e e ω

228

The unit tensor is symmetric, thus,(68)

and(69)

Returning to the angular momentum example, with the aid of the unit dyadic, we can factor ω in eq. (65),

(70)With this form, we define the moment of inertia tensor,

(71)and

(72)

Tensor Analysis

1 2 31 2 3 .= + +I e e e e e e

.⋅ = ⋅ =I ω ω I ω

20 .

Rr dρ τ⎡ ⎤= − ⋅⎣ ⎦∫∫∫H I rr ω

20 .

Rr dρ τ⎡ ⎤= −⎣ ⎦∫∫∫ I rrI

0 0 .= ⋅H ωI

229

Note, similar to the square matrix in linear algebra, is an operator that transforms a vector (column or row matrix) into another vector (or into itself if the square matrix is the identity matrix). We finish this example by developing a relation for rotational kinetic energy E of the total mass m,

Tensor Analysis0I

0

2

0

1 12 21 1( ) ( )2 21 1( )2 2

rotational kinetic energy in terms of the moment of inertia tensor.

R

R R

R

E mv d

d d

d

ρ τ

ρ τ ρ τ

ρ τ

= = ⋅

= × ⋅ = ⋅ ×

= ⋅ × = ⋅ ⋅

=

∫∫∫

∫∫∫ ∫∫∫

∫∫∫H

v v

ω r v ω r v

ω r v ω ωI

230

Note the use of the generic term “tensor” generally refers to a 2nd-order tensor or dyad. Before discussing detailed rules of the tensor algebra and calculus, one more illustrative example of how tensors are used. Example: In a continuous medium, the stress at a given location depends on the force f at that location and the orientation of the plane area ∆S on which the force acts (stress = force/area). The stress vector at position r is a function of two vectors, and .

Tensor Analysis

ˆ( )f n n

231

The stress tensor contains the information necessary to determine the state of stress at a point. Consider a tetrahedron shaped element in the continuum,

Tensor Analysis

ˆ( )f n

n

∆S

ft

fn

232

From Newton’s second law,

Tensor Analysis

q3

q2q1

2 2ˆ( )− = −π e σ

3 3ˆ( )− = −π e σ

1 1ˆ( )− = −π e σ

ˆ( )π nn

1ˆ−e

3ˆ−e

2ˆ−e

surface body ,m= + =∑f f f a

233

Then,

Tensor Analysis

S

S′

n

ˆ ′n

ˆ

ˆ ˆˆ

ˆ ˆproj

S SS

S

′

′ ′ ′= =′ ′= ⋅

′= ⋅= n

S n S nS n

n nS

1 1

2 2

3 3

ˆ ˆ( ) ,ˆ ˆ( ) ,ˆ ˆ( ) .

S SS SS S

∆ = ⋅ ∆∆ = ⋅ ∆∆ = ⋅ ∆

n en en e

234

Now employ the divergence theorem,

Tensor Analysis

n

q1

q2

q3

∆h

r∆τ

ˆdivR S

d dSτ = ⋅∫∫∫ ∫∫r r n

Shrinking R to an infinitesimal element,

.div 3S

dS h Sτ⋅ ∆ ∆

∆ ≅ =∫∫ r n

r

235

Newton’s second law then becomes,

Solving for the stress vector π,

Continuing to shrink R,

Using tensor notation we write this as,

(72)

Tensor Analysis

1 1 2 2 3 3ˆ ˆ ˆ ˆ ˆ ˆ[ ( ) ( ) ( )] .3 3

h hS S Sρ ρ∆ ∆∆ − ⋅ − ⋅ − ⋅ + ∆ = ∆π σ n e σ n e σ n e f a

1 1 2 2 3 3ˆ ˆ ˆ ˆ ˆ ˆ( ) ( ) ( )] ( ).3

hρ∆= ⋅ + ⋅ + ⋅ + −π σ n e σ n e σ n e a f

1 1 2 2 3 30ˆ ˆ ˆ ˆlim ( ).

h∆ →⇒ = ⋅ + +π n e σ e σ e σ

ˆ ˆ( ) .= ⋅π n n σ

236

Let’s make a careful distinction,

The stress vectors are the stresses on the area perpendicular tothe ith-coordinate, in the direction.

The stress vector π is on the area perpendicular to the arbitrary unit normal vector . The constitutive relation for the stress tensor depends upon the properties of the medium.

Tensor Analysis

ˆ stress vector,ˆ ˆ stress tensor.

i ij j

ij i j

σ

σ

=

=

σ eσ e e

ˆ je

n

237

Dyadic RulesAs defined earlier, a dyad is a second-order tensor represented by two vectors adjacent to one another, e.g., ab. A dyadic is composed of a linear combination of dyads, e.g.,

(73)Define the dot (scalar) product operation between a dyadic and a vector v,

(74)

Note the dot product is generally not commutative. (Can you think of a situation where it is?)

Tensor Analysis

1 1 2 2 .n n= + + +Φ a b a b a bΦ

1 1 2 2

1 1 2 2

( ) ( ) ( ),

( ) ( ) ( ) .n n

n n

⋅ = ⋅ + ⋅ + + ⋅

⋅ = ⋅ + ⋅ + + ⋅

Φ v a b v a b v a b v

v Φ v a b v a b v a b

238

Define the transpose (conjugate), (75)

Then,

(76)

Note the correspondence of these rules and results with some of the rules of matrix and vector multiplication in linear algebra.Using index notation, let’s explore this further. Each of the dyads in the dyadic is composed of two vectors, i.e.,

Tensor Analysis

T1 1 2 2 .n n= + + +Φ b a b a b a

1 1 2 2

1 1 2 2

T

( )( )

.

n n

n n

⋅ = ⋅ + + += + + + ⋅

= ⋅

v Φ v a b a b a bb a b a b a v

Φ v

239

(77)

In 3-space,

(78)

or, using matrix linear algebra notation,

Tensor Analysis

.

i ij k

i j i k

j k jki i j k j k

a b

a b φ

=

=

= =

Φ a b

e e

e e e e

11 12 131 1 1 2 1 3

21 22 232 1 2 2 2 3

31 32 333 1 3 2 3 3

φ φ φ

φ φ φ

φ φ φ

= + +

+ + +

+ + +

Φ e e e e e e

e e e e e e

e e e e e e

240

(79)

Again, there is considerable overlap between the current analysis and the formal linear algebra. There are differences, however. For instance, we do not distinguish between row and column vectors when writing

Using the linear algebra notation, this would be written as

Tensor Analysis

( )

11 12 131

21 22 231 2 3 2

31 32 333

T[ ] .ijj i

φ φ φφ φ φφ φ φ

φ

⎡ ⎤⎛ ⎞⎢ ⎥ ⎜ ⎟= ⎢ ⎥⎜ ⎟

⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦=

eΦ e e e e

e

e e

T .⋅ = ⋅v Φ Φ v

T T .=v Φ Φ v

241

In general, we can write

Example: Computing stress at a point. Given the Cartesian system (x1,x2,x3) and the stress tensor

Find: The stress vector at a point on a plane passing through the point and parallel to the plane

Tensor Analysis

(contravariant scalar components)

(covariant scalar components)

= (mixed scalar components)

jkj k

jkj k

j kk j

φ

φ

φ

=

=

Φ e e

e e

e e

200 400 300[ ] 400 0 0 psi

300 0 100ijσ

⎡ ⎤⎢ ⎥= ⎢ ⎥

−⎢ ⎥⎣ ⎦

1 2 32 2 6 0.x x x+ + − =

π

242

Find the normal and tangential components.Solution:First we find the normal to the plane,

Now the stress vector,

Tensor Analysis

1 2 3 1 2 3ˆ , ( , , ) 2 2 6| |

P P x x x x x xP

∇= = + + −

∇n

1 2 3 1 2 31 2 3

1 2 3

ˆ ˆ ˆ ( 2 2 6)

ˆ ˆ ˆ2 2ˆ3

P x x xx x x

⎛ ⎞∂ ∂ ∂∇ = + + + + −⎜ ⎟∂ ∂ ∂⎝ ⎠

+ +→ =

i i i

i i in

ˆ ˆ( ) = ⋅π n n σ

243

Note that we actually computed . We can do this because the stress tensor is symmetric.

For the normal component,

and the tangential component is

Tensor Analysis

1

2

3

200 400 300 1/ 3 1600 / 3400 0 0 2 / 3 400 / 3 psi.300 0 100 2 / 3 100 / 3

πππ

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

ˆ= ⋅π σ n

1 2 31 ˆ ˆ ˆˆ( ) (1600 400 100 ) psi3

= + + ⇐π n i i i

ˆ ˆ( ) 2600 / 9 psi,nσ = ⋅ = ⇐π n n

2 100| | 1781 psi.81t nσ σ= − = ⇐π

244

Symmetric and Antisymmetric DyadicsWe just showed that the stress tensor is symmetric. Here we show that the properties of symmetry and antisymmetry (skew-symmetry) are analogous to those of matrices in linear algebra,

For the antisymmetric case,

Therefore, only the j ≠ i components are independent, i.e.,

Tensor Analysis

T symmetric..ij ij ji ij ji

i j j i i jφ φ φ φ φ

= →

= = → =

Φ Φe e e e e e

T antisymmetric..ij ij ji ij ji

i j j i i jφ φ φ φ φ

= − →

= − = − → = −

Φ Φe e e e e e

(no summation) 0.ii ii iiφ φ φ= − → =

12 21 13 31 23 32, , .φ φ φ φ φ φ= − = − = −

245

Thus, there are only three independent components for an antisymmetric dyadic. Any dyadic can be separated into a symmetric and antisymmetric part,

or

Tensor Analysis

T T

symmetric antisymmetric

1 1[ ] [ ],2 2

= + + −Φ Φ Φ Φ Φ

1 1( ) ( ).2 2

ij ij ji ij jiφ φ φ φ φ= + + −

246

Example:

Tensor Analysis

5 4 3 5 2 1[ ] 2 2 2 , [ ] 4 2 3 ,

1 3 5 3 2 5

10 6 4 0 2 21 1[ ] 6 4 5 2 0 1 .2 2

4 5 10 2 1 0

ij ji

ij

φ φ

φ

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= + − −⎢ ⎥ ⎢ ⎥

−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

247

Transformation laws for second-order tensorsThe same rules of invariance apply to second-order tensors and dyadics that apply to vectors (first-order tensors). Again we use the transformation between two curvilinear systems, the barred and unbarred systems,

The following are possible representations of the generic tensor in the two coordinate systems,

Tensor Analysis

Φ

, ,

, .

j j

s j s js s

s ss j s j

j j

q qq qq qq q

∂ ∂= =

∂ ∂

∂ ∂= =

∂ ∂

e e e e

e e e e

248

Transforming the basis, invariance requires that,

It then follows that we can write

Tensor Analysis

iji j

i jij

i jj i

j ii j

φ

φ

φ

φ

=

=

=

=

Φ e e

e e

e e

e e

mnm nm n

mnm n

n m

n mm n

φ

φ

φ

φ

=

=

=

=

Φ e e

e e

e e

e e

.m n

ij iji j m ni j

q qq q

φ φ ∂ ∂= =

∂ ∂Φ e e e e

.mnm nφ=Φ e e

249

So, in general,

(81)

(82)

(83)

Tensor Analysis

contravariant lawm n

mn iji j

q qq q

φ φ ∂ ∂=

∂ ∂

covariant lawi j

mn ij m n

q qq q

φ φ ∂ ∂=

∂ ∂ covariant law

i j

mn ij m n

q qq q

φ φ ∂ ∂=

∂ ∂

mixed laws

m jm i

n j i n

i nn j

m i m j

q qq qq qq q

φ φ

φ φ

⎫∂ ∂= ⎪∂ ∂ ⎪

⎬∂ ∂ ⎪= ⎪∂ ∂ ⎭

250

For the Cartesian system, all transformations are the same, and

where βij are the direction cosines (symmetric matrix). Then, comparing to eq. (81), we get

and

or in matrix format

Tensor Analysis

ˆ ˆ ˆ ,ji j ij j

i

xx

β∂

= =∂

i i i

ˆ ˆ ˆ ˆmn ijm n mi nj i jφ φ β β= =Φ i i i i

mn ijmi njφ φ β β=

T[ ] [ ][ ][ ] .φ β φ β=

251

The Unit TensorWe already introduced the unit tensor (dyadic) in eq. (68) as

or(85)

Note this is the invariant component form since the and components are either zero (i ≠ j) or one (i = j) in any coordinate system. Recall,

Tensor Analysis

i ji j= =I e e e e

.i j j ij i i jδ δ= =I e e e e

ijδ j

iδ

( ) ,

( ) .

m mi i m imi i m im

m m

g

g

= ⋅ =

= ⋅ =

e e e e e

e e e e e

252

Then,(86)(87)

(88)

Note, a tensor is always invariant when taken as a whole. What we are referring to here as invariant components is the fact that in eqs. (88) the scalar components do not change during a coordinate transformation as they do in eqs. (86) and (87), since the gij and gij are system dependent.

Tensor Analysis

contravariant components

covariant components

invariant (mixed) components

iji j

i jij

j ii j

i jj i

g

g

δ

δ

=

=

⎫= ⎪⎬

= ⎪⎭

I e e

e e

e e

e e

253

Invariants of a second-order tensorThere are three quantities associated with a second-order tensor that do not change when the tensor undergoes a coordinate transformation. They are,

Tensor Analysis

1

2

3

trace ,1 ( ),2det .

ii

j i i ji j i j

ji

I

I

I

φ

φ φ φ φ

φ

= =

= −

=

Φ

254

Double-dot ProductDefine the operation between two second-order tensors

(89)Alternatively, you may see the definition

(90)or

(91)Note that Reddy and Rasmussen mistakenly say that the form in eq. (91) is not commutative. One can easily see that it is. We will use the definition in eq. (89).

Tensor Analysis

: ( )( ).≡ ⋅ ⋅ab cd b c a d

( )( ).⋅ ⋅ ≡ ⋅ ⋅ab cd b c a d

: ( )( ).≡ ⋅ ⋅ab cd a c b d

255

For two tensor

and

Tensor Analysis

: ( ) : ( )

.

ij mni j m n

ij mnjm ing g

φ ψ

φ ψ

=

=

Φ Ψ e e e e

and ,Φ Ψ

: ( ) : ( )

.

ij m ni j mn

ij m nmn j i

ijij

φ ψ

φ ψ δ δ

φ ψ

=

=

=

Φ Ψ e e e e

256

Tensor GradientWe have developed the general gradient operation on a scalar quantity, now let us look at the gradient of a vector,

(92)

Since i and k are dummies,

Tensor Analysis

grad ( )

.

j iij

ij i i

ij j

ij i

i kj

aq

a aq q

ka ai jq

∂≡ ∇ =

∂

⎛ ⎞∂∂= +⎜ ⎟∂ ∂⎝ ⎠

⎛ ⎞⎧ ⎫∂= + ⎨ ⎬⎜ ⎟∂ ⎩ ⎭⎝ ⎠

a a e e

ee e

e e e

.k

j ikj

ka ai jq

⎛ ⎞⎧ ⎫∂∇ = + ⎨ ⎬⎜ ⎟∂ ⎩ ⎭⎝ ⎠

a e e

257

Using the notation for partial derivatives, we write using the covariant derivative of the contravariant component ak,

(94)or using,

(95)where,

(96)

is the covariant derivative of the covariant component ak. In the Cartesian system, we have

(97)

Tensor Analysis

, ,k jj ka∇ =a e e

, ,jk j ka∇ =a e e

,

ki

k j j

iaa ak jq

⎧ ⎫∂≡ − ⎨ ⎬∂ ⎩ ⎭

ˆ ˆ .ji ji

aq

∂∇ =

∂a i i

258

One can show that ∇a is related to the rotational part of a vector field, i.e.,

(97)

Note that this vector identity is #14 in the table (handout). Compare eq. (97) to the earlier result

relating the rotation, angular velocity, and linear velocity.

Tensor Analysis

antisymmetric part of

T

rotation of change in over distance from rotation of

1 1( ) (curl ) .2 2

∇

⎧ ⎫⎡ ⎤⋅ ∇ − ∇ ≡ ×⎨ ⎬⎣ ⎦⎩ ⎭

a

aab a

b a a a b

1 (curl )2

= ×v v r

Vector Calculus & General Coordinate Systems Orthogonal curvilinear coordinates For orthogonal...

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Transcript of Vector Calculus & General Coordinate Systems Orthogonal curvilinear coordinates For orthogonal...