Class 06 Handout

7/25/2019 Class 06 Handout

1/19

Fluid Mechanics AS102

Class Note No: 06

Wednesday, August 8, 2007


2/19


3/19

Review: C. C. S. & Tensor Analysis - differentiations

two important relations:

gk

ul =

m

k l

gm

g

k

ul =

kl ngn (2)

m

k l:=

1

2

gmn(gnk

ul

+gnl

uk

gkl

un

) (3)

the christoffel symbols of the second kind


4/19


scalars:

= (x) = (u) {= (u)} (4)

=

umgm

=;mgm

{=

umg

m

} (5)

um =

un

um

un (6)

1st order covariant tensor


5/19

Review: C. C. S. & Tensor Analysis - differentiationsvectors:

a= ak(u)gk=ak(u)gk

{=ak(u)gk =ak(u)gk} (7)

case 1:

a=am;lglgm , am;l := am

ul +

mk l

ak

2nd o.mixed t. Understand the rule (8)

case 2:

a= am;lglgm , am;l :=

am

ul

k

m l

ak

2nd o.covariant t. understand the rule (9)

am;l=g

mn

an;l (10)


6/19

Review: C. C. S. & Tensor Analysis - differentiationsuseful formula:

diva := a= gk

uk[algl] =g

k

uk[algl]

= gk al ;kgl=ak;k (11)

2 := =gl

ul[gm

um] =gl

ul[,mg

m]

= gl,m;lgm =glm(

2

ulum

k

m l

uk) (12)

curla := a= gk

uk(alg

l) =gk

uk(alg

l)

= gk(al;kgl) =al;kgkgl (13)

ba = (bmgm)gk

uka=bk

uk(algl)

= b

k

a

l

;kgl (14)


7/19

Curvilinear Coordinate Systems & Tensor Analysis

Todays topic:

# differentiation of

the 2nd order tensors

# example to apply the learned ruels equations of motion in curvilinear c. s.


8/19


2nd order tensors A= Aij(u)gigj ...:

A = gk uk

A= gk uk

(Aijgigj)

= gk

Aij

ukgigj+ A

ijgi

ukgj+ A

ijgigj

uk

= gk

Aij

ukgigj+ Aij

mi k

gmgj + Aijgi

m

j kgm

=

Amn

uk +

m

i k

Ain +

n

j k

Amj

gkgmgn

= Amn

;kg

k

gmgn (15)

Amn;k := Amn

uk +

m

i k

Ain +

n

j k

Amj (16)

third order tensor of mixed


9/19


find the pattern:

am;l= am

ul +

mk l

ak , am;l=

am

ul

k

m l

ak (17)

am;l=gmnan;l (18)

Amn;k = Amn

uk +

m

i k

Ain +

n

j k

Amj (19)

(17) as the starting point, covariant / contravariant;

orders, number of free indexes;

positions of & correspondences between the indexes

Amn;k= Amn

uk +

m

i k

Ain

j

n k

Amj (20)


10/19


useful formula:

gij;m=gij;m=0 (21) to show the 1st:

1 gij;ma third order mixed tensor (the consequence of diff.)

2 in a rectangular c.s.

gij ij, gij ;m ij;m=

ij

xm=0

3

gij;m= ui

xp

uj

xq

xr

umpq;r=0

the characteristic of tensors: if a tensor is zero in onespecific coordinate system, it is zero in other coordinatesystems. (roughly speaking)


11/19

Curvilinear Coordinate Systems & Tensor Analysisuseful formula:

examples of product rules

(Aij bk);m=Aij;mbk+ A

ij bk;m

(Aij bj);m=Aij;mbj+A

ij bj;m (22)

#

Aij;mbk+ A

ij bk;m=

Aij

um +

i

m l

Alj

+ j

m l

Ail

bk+ Aij bk

um l

m k

bl

=

um(Aij bk) +

i

m l

Alj bk+

j

m l

Ail bk

l

m kAij bl= (Aij bk);m


12/19


useful formula:

A = gk

ukA= gk

ukA=gk

uk(Aijgigj)

= gkAij

ukgigj+ A

ijgi

ukgj+ A

ijgigj

uk

= gk

Aij

ukgigj+ A

ij

m

i k

gmgj+ A

ijgi

m

j k

gm

= Amn

um

+ m

i mAin +

n

j mAmj gn

= Amn;mgn (23)


13/19


equations of motion (conservations of mass & linear momentum):

t + v+v= 0, (24)

vt + vv f= 0 (25)

# to apply the equations, we need the corresponding

component forms !!

= (um, t), v= vi(um, t)gi =vi(um, t)gi,

=ij(um, t)gigj, f=fi(um, t)gi

gi independent of timet


14/19


equations of motion (conservations of mass & linear momentum):

v= vk;k, v=vk;k,

v

t =

vi

t gi,

vv= vk vi;kgi, =ki;kgi (26)

t + v

k;k+vk;k =0, (27)

vit + v

kvi;k

ki

;k fi =0 (28)


15/19


equations of motion

(conservations of mass & linear momentum):

# another perspective to look at the issue

we have the following component forms of the equations inarectangularcoordinate system (from the conservations of

mass & linear momentum)

t

+vk,k+ vk,k=0, (29)

vit

+vkvi,k

ki,k fi=0 (30)

how to write the corresponding equations in acurvilinearc.s. ?

(if needed in application, for the sake of convenience)

C ili C di S T A l i


16/19


# another perspective to look at the issue

define

ai :=

vi

t +vkvi,k

ki,k fi in(xm)

ai :=vi

t + vkvi;kki;k fi in(um)

show that such a defined ai is a 1st order contravarianttensor

ai = ui

xmam=0, see equation (30)

vi

t + vkvi;k

ki;k f

i =0 (31)

C ili C di S & T A l i


17/19


equations of motion

(conservations of mass & linear momentum):

# example for ij, ij

for a Newtonian fluid,

ki=Pki+(vk,i+vi,k), in(xm) (32)

ki =Pgki +(vk;lgli + vi;lg

kl) (33)

ki;k=P;kgki +(v

k;l);kg

li + (vi;l);kgkl

(34)(vi;l);k=

ukvi;l+

i

m k

vm;l

m

l k

vi;m (35)

further simplfication in an orthogonal c.s.

C ili C di t S t & T A l i


18/19


This Fridays tutorial time

11:00am 11:50am

... the equations of motion in a cylindrical c.s. ...

& the concept of physical components

C ili C di t S t & T A l i


19/19


derive the specific expressions for (find the rules for orthogonal

c. s. first)

the cylindrical c. s.

2

1 2

=

2

2 1

=

1

r,

1

2 2

= r, other= 0 (36)

the spherical c. s.

1

2 2

=r,

1

3 3

=rsin2,

2

1 2

= ..=

1

r

23 3

=sin cos,

31 3

=..= 1

r 3

2 3

=..= cot, (37)

Class 06 Handout

Documents

Transcript of Class 06 Handout