L

L

E

E

C

C

T

T

U

U

R

R

E

E

4

4

L

L

I

I

N

N

E

E

A

A

R

R

M

M

O

O

M

M

E

E

N

N

T

T

U

U

M

M

P

P

R

R

I

I

N

N

C

C

I

I

P

P

L

L

E

E

A

A

N

N

D

D

G

G

E

E

N

N

E

E

R

R

A

A

L

L

E

E

Q

Q

U

U

A

A

T

T

I

I

O

O

N

N

O

O

F

F

M

M

O

O

T

T

I

I

O

O

N

N

.

.

T

T

H

H

E

E

A

A

N

N

G

G

U

U

L

L

A

A

R

R

M

M

O

O

M

M

E

E

N

N

T

T

U

U

M

M

P

P

R

R

I

I

N

N

C

C

I

I

P

P

L

L

E

E

.

.

D

D

E

E

R

R

I

I

V

V

A

A

T

T

I

I

O

O

N

N

O

O

F

F

E

E

Q

Q

U

U

A

A

T

T

I

I

O

O

N

N

O

O

F

F

F

F

L

L

U

U

I

I

D

D

M

M

O

O

T

T

I

I

O

O

N

N

We have already mentioned that the Linear Momentum Principle is formulated by setting

h  v

in the general conservation law introduced in the Lecture 3. Since the “sources” for

linear momentum are the acting external forces, we get the following equality

(

)(

)

S

V

d

dV

dS

dt













 



















v

v v n

F

F

As before, we can change order of time differentiation and integration. Then, after insertion of
the expressions for the forces, we get

(

)

(

)(

)

t

dV

dS

dS

dV



































v

v v n

σ

f

Later we will show that the surface stress can be expressed as the following matrix-vector
product



σ

Ξn

More precisely: the stress vector

σ

is obtained by the application of the stress tensor

Ξ

to the

normal vector n. We will discuss this problems in details in one of the next lectures.

It the case of idealized fluid with no viscosity (the Euler fluid) the stress vector has always
only normal component directed into the volume and expressed by the local value of pressure
by the same formula as in the statics, i.e.

p

 

σ

n

It means that the stress tensor is particularly simple, namely

p

 

Ξ

I

, where

I

stands for the

unary matrix (identity tensor).

Our next goal is to derive a general differential equation of motion. The procedure as usual
consists in transformation all surface integrals to volume ones. Then, thanks to arbitrariness of
the volume



, we conclude that the integrand vanishes everywhere in the flow domain.

The appropriate calculations are more laborious that in the case of mass conservation. We will
start with transforming the surface integral describing the flux of the linear momentum
through the boundary:





.

(

)

(

)(

)

(

)

(

)

ij

j

GGO
Theor

i

i

j

j

i

i

j

x

dS

n dS

dV

dV





 

 























  











v v

v v n

e

e

v

v

Note that we have obtained a volume integral from the vector field which is a tensorial
divergence from the momentum flux tensor







Π

v

v

, where the operator



denotes

the tensor (or dyadic) product of two vectors.

Next let us consider the surface integral representing the surface force. The transformation to
the volume integral is conducted as follows



.

j

GGO
Th

i

ij

j

i

ij

x

eor

dS

dS

n dS

dV

dS

























  











σ

Ξn

e

e

Ξ

Note that the volumetric integration is applied to the vector which is the tensorial divergence
of the stress tensor field

Ξ

. In particular case of the ideal (inviscid) fluid, we get



.

(

)

j

i

GGO
Theor

i

ij

j

i

ij

x

x

p

dS

p

dS

p

n dS

p dV

p dS



























 

 

  











σ

I n

e

e



Insertion of the volumetric integrals into the original equality yields in the general case





(

)

(

)

t

dV













  

 







v

v

v Ξ

f

0

Thus, we conclude that the following differential equation must be satisfied at each point in
fluid domain

(

)

(

)

t











  

 



v

v

v Ξ

f

We have obtained the general equation of motion in the conservative form.

For the ideal fluid

p

 

Ξ

I

, hence the equation of motion reads

(

)

(

)

t

p











  

 



v

v

v

I

f

The conservative form of the equation of motion is not the only one possible. There exists a
simpler form. It is worth noting that this simpler form can be obtained directly from an
alternative method of derivation, which is based on the usage of material rather than control
volumes. Here, to obtain simplified form of the equation of motion we perform the following
calculations

0!!! (

)

(

)

(

)

(

)

(

)

[

(

) ]

]

(

)

[

(

)

j

j

j

i

i

i

j

t

t

x

i

i

j

i

t

x

t

mass conserva

i

i

j

t

x

t

tion







 

 





























































  































 

  





 



v

v

v

e

e

v

v

v

v

v

a







Note that the expression in the second square bracket is nothing else but the fluid
acceleration

(

)

t









 

a

v

v

v

The general equation of motion can be now written as

[

(

) ]

t













   

v

v

v

Ξ

f

For the ideal (inviscid) fluid the general equation reduces to the Euler Equation

[

(

) ]

t

p













  

v

v

v

f

which is often delivered in the formed divided by density ρ

1

(

)

t

p











   

v

v

v

f

Later, we will consider also consider the case of a viscous fluid.

A

A

N

N

G

G

U

U

L

L

A

A

R

R

M

M

O

O

M

M

E

E

N

N

T

T

U

U

M

M

P

P

R

R

I

I

N

N

C

C

I

I

P

P

L

L

E

E

.

.

S

S

Y

Y

M

M

M

M

E

E

T

T

R

R

Y

Y

O

O

F

F

S

S

T

T

R

R

E

E

S

S

S

S

T

T

E

E

N

N

S

S

O

O

R

R

.

.

The angular momentum of the fluid contained in the volume



(computed with respect to the

origin of the reference frame) is equal

dV











K

x

v

Following the general rule, the net rate of time variation of K is expressed by the formula

(

)(

)

d

dt

production

d

dV

dS

dt























K

x

v

x

v v n

The Angular Momentum Principle (AMP)

S

V

production

d

dt





K

M

M

takes the following form

(

)(

)

d

dt

dV

dS

dS

dV







































x

v

x

v v n

x Ξn

x

f

As usual, we would like to derive a differential form of this principle. Again, the time
differentiation operation can be done under the sign of integral, i.e.

(

)

d

dt

t

dV

dV























x

v

x

v

As usual, we need to transform surface integral into the volume integrals. Note that the
integral in the left–hand side of the above equation can be write as

(

)

(

)(

)

[(

) ]

matrix tensor

dS

dS























x

v v n

x

v

v n



Thus, both surface integrals in the AMP are particular cases of

dS









j

x An

Transformation of such integral to the volume one is a good opportunity to practice the index
formalism.









.

(

)

(

)

32

23

1

13

31

2

21

12

3

(

)

(

)

(

)

(

)

(

)

(

)

m

k

j

km

m

m

jm

k

i i

i

i

i

ijk

j

km

m

i

ijk

j

km

x

GGO Th

x

a

i

ijk

km

i

ijk

j

i

ijk

kj

i

x

x

i

j

dS

x a n

dS

x a

dV

a dV

x

dV

a dV

dV

a

a

a

a

a

a

































































  

























  















An

A

j

e

e

x An

e

e

e

e

e

x

A

e

e

e

e

x

dV





A

Note that if the matrix

A

is symmetric (

T



A

A

) then three first terms are zero. This is the

case for the first of our integrals, where

ij

i

j

a



 









A

v

v

and hence

,

,

1,2,3

ij

ji

a

a

i j





. Thus

(

)

(

)

dS

dV















  







x

v

v n

x

v

v

In the second surface integral we have



A

Ξ

and

1

32

23

2

13

31

3

21

12

(

)

(

)

(

)

dS

dV

dV

dV

dV















  



  



  





  











x Ξn

x

Ξ

e

e

e

Insertion to the AMP integral formula brings the equality

1

32

23

2

13

31

3

21

12

[ (

)

(

)

]

(

)

(

)

(

)

0

t

dV

dV

dV

dV





















  

 





  





  



  











x

v

v

v Ξ

f

e

e

e

But

(

)

(

)

t











  

 





v

v

v Ξ

f

0

(equation of motion)! We finally conclude that

32

23



 

,

12

21

  

,

13

31

  

i.e., that the stress tensor

Ξ

is symmetric

T



Ξ

Ξ

We conclude that if the stress tensor is symmetric then the validity of the Angular
Momentum Principle follows from the Linear Momentum Principle. In other words: in
the fluid model where symmetry of the stress tensor is guaranteed, only Linear
Momentum Principle is necessary to describe the fluid motion.