This document provides notes on advanced fluid mechanics. It introduces fluid particle kinematics, including definitions of the particle position, velocity, and acceleration vectors in both Lagrangian and Eulerian descriptions. It defines the material derivative and discusses its use in describing how a fluid property changes as it moves with the flow. Pathlines and streamlines are also defined. Two examples of fluid flow in a convergent duct are provided to illustrate the concepts.

1. Azzeddine Soulaijmani
M¶ecanique des °uides avanc¶ee
SYS 860
Notes de cours
21 septembre 2006
¶Ecole de technologie sup¶erieure
Montr¶eal

3. 1
Introduction
In °uid mechanics analysis the °uid is considered as a continuum material. Any in¯nitesimal
volume in space contains a large number of molecules in mutual interactions and only global
e®ects such as pressure, temperature or velocity are analyzed. In the sequel, a °uid particle
refers to such an in¯nitesimal volume of °uid with a measurable mass and containing a large
number of molecules.
1.1 Fluid particle kinematics
Fluid kinematics is the study of the particles motion without considering the forces that
cause the motion. It describes the time evolution of particles properties (position, velocity,
acceleration, density, pressure, temperature, etc) during the motion as function of time and
space position.
1.1.1 Particle position vector
Consider a Cartesian reference system with origin O and an orthonormal basis (i1; i2; i3).
For any °uid particle, the vector position is denoted by x(t) and is function of time t :
x(t) = x1(t)i1 + x2(t)i2 + x3(t)i3 (1.1)
with x1(t); x2(t); x3(t) are the position coordinates.
In the following we will use the so-called Einstein's notation convention, i.e. when an index
is repeated in a mathematical expression that means a summation over the repeated index.
For example, equation (1.1) can be rewritten in a more compact form as
x(t) = x1(t)i1 + x2(t)i2 + x3(t)i3 =
X
k
xkik ´ xkik:

4. 2 1 Introduction
1.1.2 Particle velocity and acceleration vectors
Consider a °uid particle at time t. During a small time duration ¢t, the vector position
becomes x(t + ¢t) and the displacement vector of the particle is
¢x = x(t + ¢t) ¡ x(t):
x3
x1
O
i1
Particle at time t + ¢t
x2
Particle at time t
x(t + ¢t)
x(t)
i3
i2
Vector displacement after ¢t
¢x = x(t + ¢t) ¡ x(t)
Fig. 1.1. Particle position vector
The particle velocity vector is de¯ned as the rate of change of the vector position :
u(t) = lim
¢t!0
¢x
¢t
= dx
dt
(1.2)
The particle acceleration vector is de¯ned as the rate of change of the velocity vector :
a(t) = lim
¢t!0
¢u
¢t
= du
dt
= d2x
dt2 (1.3)
1.1.3 Lagrangian and Eulerian Kinematic Descriptions
To de¯ne the velocity and acceleration vectors, we need actually to specify or to identify
which particle is under study. One convenient way is to label each particle by its position » at a
certain initial time t0. Then the motion of the whole °uid can be described by two independent

5. 1.1 Fluid particle kinematics 3
variables » and t. For a particle with label », equation x = x(»; t) gives the particle trajectory
or patheline. The velocity and acceleration vectors are then given more precisely by
u(»; t) =
·
@x(»; t)
@t
¸
»
; (1.4)
and
a(»; t) =
·
@u(»; t)
@t
¸
»
: (1.5)
The use of the independent variables » and t is called the Lagrangian or the material
description. Another description very often used in °uid kinematics is to use as independent
variables x and t. That is, we are interested to describe the motion of the particle which passes
at a ¯xed position x in space and at the time instant t. This description is called the Eulerian
kinematic description and is very often used in Fluid Mechanics. Solving equation » = »(x; t)
gives the labels of all particles that occupied the ¯xed position x. The velocity and acceleration
vectors can now be described as functions of x and t,
u(x; t) = u(x(»; t); t) (1.6)
and
a(x; t) = a(x(»; t); t) (1.7)
1.1.4 Material derivative
In the following we will obtain the derivative with respect to time when using the Eulerian
description. Consider ¯rst a scalar material property µ(x; t) such as pressure or temperature.
The rate of change of this property felt by the particle labeled by » as it moves is de¯ned by
Dµ
Dt
´
·
dµ(x(»; t); t)
dt
¸
»
: (1.8)
Using the di®erentiation chain rule, equation (1.8) is developed into
Dµ
Dt
=
·
@µ(x; t)
@t
¸
t
+ @µ
@x1
@x1(»; t)
@t
+ @µ
@x2
@x2(»; t)
@t
+ @µ
@x3
@x3(»; t)
@t
: (1.9)
The ¯rst term on the right hand side of equation (1.9) is the rate of change of the property
µ at the ¯xed position x. This term is called the local derivative, and is non zero when, for
instance, the boundary conditions of the °uid domain change in time. The other terms in (1.9)
give the rate of change when the particle moves from x to x + ¢x. It is called the convective
derivative. In fact the terms @xk
@t
are the velocity components uk which convect or transport
the material property. The convective derivative is non zero when, for instance, there is a
variation in the domain geometry that causes a spatial variation @µ
@xk
. An illustration of the

6. 4 1 Introduction
Lateral boundary heated in
a non uniform manner
Incomming flow with
time varying
temperature
Temperature measurement
A(x1) x1
x1 = 0
u1(x1; t)
£(0; t)
£(x1; t)
x1
Fig. 1.2. An illustration for temperature material derivative measurement
local and convective rate of change is given in ¯gure (1.2). A °uid °ows in a duct with a variable
cross section A(x). For simplicity, we assume that the °ow is unidirectional with velocity u1.
The incoming °uid has a time dependent temperature µ(0; t), that is the boundary condition
at the inlet is time varying. A thermometer is placed at a ¯xed position x1. By taking two
successive temperature observations µ(x1; t) and µ(x1; t + ¢t) in a small time interval ¢t we
can indirectly measure the local temperature derivative, i.e.
·
@µ(x1; t)
@t
¸
x1
¼
µ(x1; t + ¢t) ¡ µ(x1; t)
¢t
:
Another thermometer is placed at a close position x1 + ¢x1, so that the rate of temperature
change observed by the particle in its travel of the distance ¢x1 (with ¢x1 ¼ u1(x1; t)¢t) can
be approximated by
u1(x1; t) µ(x1 + ¢x1; t) ¡ µ(x1; t)
¢x1
:
The temperature gradient can be caused by heating the lateral boundary in a nonuniform
manner. Thus the material derivative of the particle passing at position x1 at time t can be
indirectly measured without actually moving the thermometer with this particle. Therefore,
only the °ow ¯eld u1(x1; t) and the temperature ¯eld µ(x1; t) have to be measured at di®erent
time intervals to be able to compute the material derivative.
Using indicial notation, the material derivative can be rewritten as
Dµ
Dt
(x; t) = @µ(x; t)
@t
+ uk
@µ
@xk
(1.10)
Equation (1.10) can also be written in a matrix form as

7. 1.1 Fluid particle kinematics 5
Dµ
Dt
= @µ(x; t)
@t
+ [u1; u2; u3] ¢
8>>>>>>>>>>><
>>>>>>>>>>>:
@µ
@x1
@µ
@x2
@µ
@x3
9>>>>>>>>>>>=
>>>>>>>>>>>;
; (1.11)
or
Dµ
Dt
= @µ(x; t)
@t
+ [u1; u2; u3] ¢
8>>>>>>>>>>><
>>>>>>>>>>>:
@
@x1
@
@x2
@
@x3
9>>>>>>>>>>>=
>>>>>>>>>>>;
µ (1.12)
or in a more compact form, using tensorial notation, as
Dµ
Dt
= @µ(x; t)
@t
+ (u ¢ r)µ: (1.13)
Now, we consider the expression of the rate of change in the Eulerian description of a
material vector property £ = µkik (Here, we use Eisntein's notation). Since the operator D()
Dt
is linear then
D£
Dt
= Dµk
Dt
ik + µk
Dik
Dt
(1.14)
If the reference frame is chosen ¯xed then there is no change in its basis vectors as the °uid
particles move, i.e. Dik
Dt
= 0, therefore
D£
Dt
= Dµk
Dt
ik: (1.15)
The expression of the particle acceleration in the Eulerian description can be found by
taking £ as the velocity vector. The kth acceleration component is obtained as
ak(x; t) = Duk
Dt
= @uk(x; t)
@t
+ (u ¢ r)uk = @uk
@t
+ ui
@uk
@xi
; (1.16)
and the acceleration vector is then written using tensorial notation as

8. 6 1 Introduction
a(x; t) = Du(x; t)
Dt
= @u(x; t)
@t
+ (u ¢ r)u: (1.17)
The ¯rst term on the right hand side of equation (1.17) represents the local acceleration. The
last term represents the convective acceleration which is nonlinear since a product of u by its
gradient appears.
1.1.5 Pathlines and streamlines
The pathline is de¯ned by the successive positions occupied by the particle ». If the ve-
locity ¯eld u(x; t) is known then the pathline can be obtained by solving for x the following
di®erential equation :
dx
dt
= u(x; t): (1.18)
The streamlines are de¯ned as the lines tangent to the velocity vector. Any vector displacement
Streamline
Pathline
Particule position at time
u(x; t)
t
x(»; t)
Fig. 1.3. Figure 1.3. Pathline and streamline
dx along the streamlines is parallel to the velocity vector :
dx £ u = 0; (1.19)
which can be developed in the orthonnormal basis as
¯¯¯¯¯¯
dx1 u1 i1
dx2 u2 i2
dx3 u3 i3
¯¯¯¯¯¯
= (u3dx2 ¡ u2dx3)i1 + (u1dx3 ¡ u3dx1)i2 + (u2dx1 ¡ u1dx2)i3 = 0 (1.20)
This leads to solving a set of di®erential equations :
dx1
u1
= dx2
u2
= dx3
u3
: (1.21)
Example 1.1 : Steady °ow in a convergent duct
An incompressible °uid °ows in a duct having a convergent cross section A(x1) = A0
1 + x1
l
,

9. 1.1 Fluid particle kinematics 7
with l a positive constant. The velocity at the inlet section x = 0 is u0. Calculate the velocity
and the acceleration in the Eulerian and Lagragian descriptions.
"
Ax

10. 0
u u
x 0 x
Fig. 1.4. Flow in a convergent duct
Solution
Since the °ow is assumed one-dimensional u = (u1; 0; 0). The °uid is also assumed incompres-
sible so that the °ow rate (see chapter 2) is constant, i.e. u1(x1)A(x1) ³
= const = u0A0. Hence,
we get the velocity in the Eulerian description as u1(x1) = u0
1 + x1
l
´
. As the °ow is time
independent (steady °ow), the local acceleration is zero @u1
@t = 0. The convective acceleration
is calculated as
u1(x1)@u1
@x1
= u0u1(x1)=l:
Let »1 be the initial position of the °uid particle. The particle patheline is expressed by the
function x1(»1; t) and is calculated by solving
·
@x1
@t
¸
»1
= u1:
Inserting the expression of u1(x1) and using a separation of variables, we get u0dt = dx1
1 + x1
l
.
Integrating the left hand side from 0 to t and the right hand side from »1 to x1 leads to
u0t = l ln
0
B@1 + x1
l
1 + »1
l
1
CA
:
The pathline function is then described by
x1(»1; t) = l
0
B@
(1 + »1
l
Ã
u0t
)e
l
!
¡ 1
1
CA
:

11. 8 1 Introduction
On the other hand, given a ¯xed position x1 we can also obtain the Lagrangian variable
»1(x1; t) of the particles passing by x1. Hence,
»1(x1; t) = l
0
@(1 + x1
l
)e
(
¡u0t
l
)
¡ 1
1
A:
The velocity in the Lagrangian description is de¯ned by u1(»; t) =
·
@x1
@t
¸
»1
and thus by a
simple di®erentiation we obtain
u1(»1; t) = u0(1 + »1
l
)e( u0t
l
):
Since (1 + »1
l
)e
(
u0t
l
) = 1 + x1
l
we can verify that the we obtain the same result for the velocity
in the Lagranian and Eulerian descriptions.
The Lagrangian description of the acceleration is
a1(»1; t) =
·
@u1
@t
¸
»1
= u20
l
(1 + »1
l
)e( u0t
l
):
Example 1.2 : Unsteady °ow in a convergent duct
Repeat example (1.1) with a time dependent boundary condition u1(0; t) = u0(t) = Ue¡®t
with U and ® positive constants.
Solution
The velocity ¯eld is time dependent because of the time varying nature of the in°ow. As before,
invoking the principle of mass conservation ³
for an incompressible °uid the velocity at any
space position is obtained u1(x1; t) = u0(t)
1 + x1
l
´
. The local and convective accelerations
are respectively @u1(x; t)
@t
= ¡®Ue¡®t
³
1 + x1
l
´
and u0(t)u1(x1; t)=l.
Again, we obtain the particle pathline by solving the di®erential equation
·
@x1
@t
¸
»1
= u1(x1; t):
Using the method of separation of variables, we obtain the pathline equation
x1(t) = l
µ
(1 + »1
l
¶
;
)ef(t) ¡ 1
with
f(t) = ¡
U
l®
³
e(¡®t) ¡ 1
´
= ln
0
B@
1 + x1
l
1 + »1
l
1
CA
:

12. 1.1 Fluid particle kinematics 9
The Lagrangian variable at time t corresponding to a ¯xed spatial position is
»1(x1; t) = l
³
(1 + x1
l
)e¡f(t) ¡ 1
´
:
The velocity in the Lagrangian description is u1(»; t) =
·
@x1
@t
¸
»1
= lf0(t)(1 + 1
l
)ef(t), which
can be veri¯ed to be equal to Ue¡®t(1 + x1
l
) = u1(x1; t). When the velocity u1(»; t) is di®eren-
tiated with respect to time this gives the acceleration a1(»; t) in the Lagrangian description.
Example 1.3 : Steady two-dimensional °ow close to a stagnation point
The motion of a °uid is described by its velocity ¯eld given in the Eulerian description by :
u = ®(x1;¡x2)
with ® a positive constant.
a) Determine the velocity ¯eld and acceleration in the Lagrangian and Eulerian descriptions.
Find the pathline equation of a particle which occupied position » = (»1; »2) at time 0.
b) Find the streamline equation of a particle which occupied position x at time t.
Solution
a) The velocity components in the Eulerian description are : u1 = ®x1 and u2 = ¡®x2. As a
result, the acceleration components are determined using equation (1.16), a1(x; t) = ®2x1 and
a2(x; t) = ®2x2.
The pathline di®erential equations are :
ui(x; t) =
·
@xi
@t
¸
»
:
By integration, we obtain the pathline equations :
x1(»; t) = »1 exp(®t)
and
x2(»; t) = »2 exp(¡®t):
In these equations t represents the curve parameter of the pathline, and »i are the paramete-
rized families. By eliminating the curve parameter t,
t = (
1
®
) ln(x1
»1
) = ¡(
1
®
) ln(x2
»2
)
we obtain the explicit form of the pathline equation
x2 = »1»2
x1
;
which de¯nes a family of hyperboles.
The velocity components in the Lagrangian description are u1(»; t) = ®»1 exp(®t) and
u2(»; t) = ¡®»2 exp(¡®t). The acceleration components in the Lagrangian description are

13. 10 1 Introduction
a1(»; t) = ®2»1 exp(®t) and a2(»; t) = ®2»2 exp(¡®t).
b)The streamline di®erential equations are given by equation (1.21) which leads in the case of
a two-dimensional °ow to solving the di®erential equation
@x2
@x1
= u2
u1
:
Inserting the expression of the velocity components we obtain
@x2
@x1
= ¡
x2
x1
:
And by integration, we obtain x2 = C1C2
1
x1
with C1 and C2 two constants of integration.
Therefore, the streamlines de¯ne families of hyperboles identical to those de¯ned by the path-
lines.
Example 1.4 : Material derivative in cylindrical coordinates
a) Develop the divergence r ¢ v of the vector v using cylindrical coordinates (r; µ; z) and the
polar orthonormal basis (er; eµ; ez).
b) Derive the material derivative for a scalar function T in cylindrical coordinates.
c) Derive the material derivative for a vector function v in cylindrical coordinates.
Solution
a)Recall that the Cartesian basis vectors are related to the basis vectors (er; eµ; ez) by :
i1 = cos µ er ¡ sin µ eµ;
i2 = sin µ er + cos µ eµ;
i3 = ez:
The velocity in the cylindrical system is :
u = urer + uµeµ + uzez
The unit vectors er and eµ vary with the angle. We have the following relations @er
@µ
= eµ
and @eµ
@µ
= ¡er.
The gradient operator in cylindrical polar coordinates can be veri¯ed as (do it as an exer-
cise) :
r = er
@
@r
+ eµ
1
r
@
@µ
+ ez
@
@z
:
Let consider a vector v = vrer + vµeµ + vzez. The divergence is de¯ned as the scalar product
of the gradient vector with the vector v. Thus,
r¢v = (er
@
@r
+eµ
1
r
@
@µ
+ez
@
@z
)¢(vrer+vµeµ+vzez) = @vr
@r
+eµ
1
r
µ
vr
¢
@er
@µ
+ @vµ
@µ
eµ + vµ
@eµ
@µ
¶
+@vz
@z

14. 1.1 Fluid particle kinematics 11
=)
r ¢ v = @vr
@r
+ vr
r
+
1
r
@vµ
@µ
+ @vz
@z
:
b)The material derivative of a function T(r; µ; z; t) is obtained as :
DT
Dt
= @T
@t
+ @T
@r
@r
@t
+ @T
@µ
@µ
@t
+ @T
@z
@z
@t
:
The velocity components in the polar system are ur = @r
@t
, uµ = r
@µ
@t
and uz = @z
@t
. Then, the
material derivative in cylindrical coordinates is :
DT
Dt
= @T
@t
+ ur
@T
@r
+ uµ
r
@T
@µ
+ uz
@T
@z
:
It can be veri¯ed that the convection operator is given in cylindrical coordinates by
u ¢ r = (urer + uµeµ + uzez) ¢ (er
@
@r
+ eµ
1
r
@
@µ
+ ez
@
@z
:) = ur
@
@r
+ uµ
r
@
@µ
+ uz
@T
@z
:
Therefore, the material derivative can be written as
DT
Dt
= @T
@t
+ (u ¢ r)T
which is the same tensorial expression as in the Cartesian coordinates system.
c)The material derivative for a vector ¯eld v(r; µ; z; t) = vrer + vµeµ + vzez is obtained as
Dv
Dt
= Dvr
Dt
er + vr
Der
Dt
+ Dvµ
Dt
eµ + vµ
Deµ
Dt
+ Dvz
Dt
ez + vz
Dez
Dt
:
Noting that
Der
Dt
= @er
@µ
@µ
@t
= uµ
r
eµ
Deµ
Dt
= @eµ
@µ
@µ
@t
= ¡
uµ
r
er
then
Dv
Dt
=
µ
Dvr
Dt
¡
vµuµ
r
¶
er +
µ
Dvµ
Dt
+ vruµ
r
¶
eµ + Dvz
Dt
ez:
If we replace v by the velocity then the acceleration a has in cylindrical coordinates system
the components :
ar = Dur
Dt
¡
2
r
uµ
= @ur
@t
+ (u ¢ r)ur ¡
2
r
uµ
;
aµ = Duµ
Dt
+ uµur
r
= @uµ
@t
+ (u ¢ r)uµ + uµur
r

15. 12 1 Introduction
and
az = Duz
Dt
= @uz
@t
+ (u ¢ r)uz:
Example 1.5 : Material derivative in natural coordinates
a) The unit tangent vector to the pathline is :
¡!¿
=
dx
dx
and its unit normal vector in the two dimensional space is
n = R
d¡!¿
ds
with s the coordinate in the direction of ¡!¿
and n is the coordinate in the direction of n. (¡!¿
; n)
de¯ne a local orthonormal basis. Derive the acceleration components on this basis.
Solution
The velocity components on the local basis are u¿ and un = 0. The material derivative of the
velocity is
Du
Dt
= Du
Dt
¡!¿
+ u
D¡!¿
dt
Now we develop the following terms :
Du(s; n; t)
Dt
= @u
@t
+ u
@u
@s
+ un
@u
@n
;
and
D¡!¿
Dt
= u
R
n:
Since un = 0 and @un
@s = 0 then
Du
Dt
= @u
@t
+ u
@u
@s
:
It follows that the acceleration in the local basis is
Du
Dt
= (@u
@t
+ u
@u
@s
)¡!¿
+ u2
R
n:
1.2 Velocity gradients and Deformations
1.2.1 Linear Deformations
The simplest motion that a °uid particle can undergo is translation. Consider a small °uid
element with a brick shape. The edges lengths are in¯nitesimal and are denoted by ±x1, ±x2
and ±x3. The original volume is ±# = ±x1±x2±x3.

16. 1.2 Velocity gradients and Deformations 13
B C C’
O A A’
±x2
u1
u1 +
@u1
@x1
±x1
±x1
@u1
@x1
±x1
u1 +
@u1
@x1
±x1
u1
@u1
@x1
±x1
Fig. 1.5. Linear deformation for a °uid element
Let us ¯rst analyze the e®ect of a translation motion in the i1 direction.
Particularly, we are interested to study the deformations as the °uid element moves during
a small time interval ±t. The edge OB is translated into O0B0 by the velocity component u1 and
the edge AC is translated into A0C0 by the velocity component u1(x1 + ±x1) ¼ u1 + @u1
@x1
±x1.
Because of this velocity di®erence, edges' OA and BC lengths are changed by @u1
@x1
±x1±t.
Therefore, the particle deforms and its volume changes by ±# = @u1
@x1
±x1±x2±x3±t. The rate at
which the original volume is changing per unit volume is
lim
±t!0
1
±#
±#
±t
= lim
±t!0
@u1
@x1
±x1±x2±x3±t
±x1±x2±x3±t
= @u1
@x1
(1.22)
When there is no velocity gradient, the °uid element undergoes a pure translation without
deformation.
The total rate of change of the °uid particle volume per unit volume for a three-dimensional
translation can be readily obtained
lim
±t!0
1
±#
±#
±t
= @u1
@x1
+ @u2
@x2
+ @u3
@x3
= @uk
@xk
= r:u = div(u) (1.23)
The velocity gradient components @uk
@xk
are responsible for the volumetric dilatation. The
other components of the velocity gradient (i.e., @ui
@xk
for i6= k) cause rotations and angular
deformations. By de¯nition an incompressible °uid has no volumetric dilation therefore in this
case div(u) = 0.
1.2.2 Angular deformation
For illustration, we will consider motion in the x1¡x2 plane. The velocity gradient compo-
nents that cause rotation and angular deformation are illustrated in ¯gure ?. During a small

17. 14 1 Introduction
B C
±x2
O A
±x1
u2 +
@u2
@x1
±x1
u1 +
@u1
@x2
±x2
u1
u2
Fig. 1.6. Velocity components responsible for angular rotation and deformation
µ
@u1
@x2
±x2
B C
O
C’
B’
A’
A
±x2
U1 ±x1
U2
µ
@u2
@x1
±x1
¶
±® ±t
±¯
¶
±t
Fig. 1.7. Angular rotation and deformation for a °uid element
time interval ±t, edges OA and OB will rotate to O0A0 and O0B0 by angles ±® and ±¯. A
positive rotation is by convention counterclockwise. The angular velocity of edge OA is
!0A = lim
±t!0
±®
±t
(1.24)
For small angles :
±® ¼ tan(±®) =
@u2
@x1
±x1±t
±x1
= @u2
@x1
±t (1.25)
and (1.24) becomes
!0A = lim
±t!0
±®
±t
= @u2
@x1
(1.26)
Similarly, the angular velocity of edge OB is
!0B = ¡ lim
±t!0
±¯
±t
= ¡
@u1
@x2
(1.27)
The rotation of the °uid element about the x3 axis is de¯ned as the average of the angular
velocities !0A and !0B, it follows that

18. 1.2 Velocity gradients and Deformations 15
!3 = !0A + !0B
2
=
1
2
µ
@u2
@x1
¡
@u1
@x2
¶
(1.28)
The angular velocities about the other axis !1 and !2 can be obtained using a similar analysis,
and they are respectively
!1 =
1
2
µ
@u3
@x2
¡
@u2
@x3
¶
and !2 =
1
2
µ
@u1
@x3
¡
@u3
@x1
¶
(1.29)
The three angular velocities !1, !2 and !3 de¯ne the rotation vector ! as
! = !1i1 + !2i2 + !3i3 (1.30)
This vector represents actually half of the curl of the velocity vector,
! =
1
2curl(u) (1.31)
since by de¯nition of the vector operator r £ u
curl(u = r £ u =
¯¯¯¯¯¯¯
i1 i2 i3
@
@
@x1
@x2
@
@x3
u1 u2 u3
¯¯¯¯¯¯¯
= 2! (1.32)
To eliminate the factor 1=2, the vorticity is de¯ned as twice the rotation vector
³ = 2! = r £ u: (1.33)
It can be concluded at this point that the divergence of the velocity represents the linear
deformations while the curl represents rotations. On the other hand, there are also angular
deformations associated with the cross derivatives as can be observed from ¯gure (1.7). The
change in the original right angle formed by the edges OA and OB is a shearing deformation
±°12 = ±® + ±¯
where it is considered positive if the original right angle is decreased. The rate of angular
deformation (or shearing strain) is
°_12 = lim
±t!0
±°12
±t
= lim
±t!0
2
664
@u2
@x1
±t + @u1
@x2
±t
±t
3
775
= @u2
@x1
+ @u1
@x2
(1.34)
If @u2
@x1
= ¡
@u1
@x2
, the angular deformation is zero and this corresponds to a pure rotation. Using
a similar analysis, the other two rate shearing strains in the planes x1 ¡ x3 and x2 ¡ x3 can
be obtained and are respectively
°_13 = @u3
@x1
+ @u1
@x3
and °_23 = @u3
@x2
+ @u2
@x3
: (1.35)

19. 16 1 Introduction
1.2.3 Tensors in Fluid Kinematics
It follows from the previous section that the spatial derivatives of the velocity ¯eld de¯ne
three di®erent tensors :
¤ The velocity gradient :
F =
2
666664
@u1
@x1
@u1
@x2
@u1
@x3
@u2
@x1
@u2
@x2
@u2
@x3
@u3
@x1
@u3
@x2
@u3
@x3
3
777775
= ruT : (1.36)
¤ The rate of strain (or rate of deformation) :
D =
2
4
²11 ²12 ²13
²21 ²22 ²23
²31 ²32 ²33
3
5 =
F + FT
2
=
(ru + (ru)T )
2
(1.37)
where
²kl =
1
2
(@uk
@xl
+ @ul
@xk
) (1.38)
{ The strain tensor is symmetric and detailed more clearly as follows :
D =
2
666664
@u1
@x1
1
2
(@u1
@x2
+ @u2
@x1
)
1
2
(@u1
@x3
+ @u3
@x1
)
1
2
(@u2
@x1
+ @u1
@x2
) @u2
@x2
1
2
(@u2
@x3
+ @u3
@x2
)
1
2
(@u3
@x1
+ @u1
@x3
)
1
2
(@u3
@x2
+ @u2
@x3
) @u3
@x3
3
777775
(1.39)
{ ²kk = @uk
@xk
= tr(D) is the volumetric deformation.
{ 2²kl = °_kl is the rate of angular deformation.
¤ The spin tensor :
S = F ¡ D =
F ¡ FT
2
(1.40)
is the antisymmetric part of the velocity gradient and is detailed as follows

20. 1.2 Velocity gradients and Deformations 17
S =
2
666664
0
1
2
(@u1
@x2
¡
@u2
@x1
)
1
2
(@u1
@x3
¡
@u3
@x1
)
1
2
(@u2
@x1
¡
@u1
@x2
) 0
1
2
(@u2
@x3
¡
@u3
@x2
)
1
2
(@u3
@x1
¡
@u1
@x3
)
1
2
(@u3
@x2
¡
@u2
@x3
) 0
3
777775
=
2
4
0 ¡!3 !2
!3 0 ¡!1
¡!2 !1 0
3
5 (1.41)
{ Using index notation, the components of the above tensors can be written in a compact
form as follows :
Fij = @ui
@xj
(1.42)
Dij = Fij + Fji
2
=
1
2
( @ui
@xj
+ @uj
@xi
) (1.43)
Sij = Fij ¡ Fji
2
=
1
2
( @ui
@xj
¡
@uj
@xi
) (1.44)
Remark : The velocity ¯eld in the vicinity of a °uid particle at position x can be completely
described if the velocity gradient tensor is known. Using Taylor's expansion, the velocity at a
close position x + dx is approximated at the ¯rst order by
u(x + dx; t) = u(x; t) + ruT ¢ dx (1.45)
since the velocity gradient tensor is decomposed into a symmetric and an antisymmetric part
then
u(x + dx; t) = u(x; t) + S ¢ dx + D ¢ dx (1.46)
Furthermore, it can be easily veri¯ed that
S ¢ dx = ! £ dx
The motion of a °uid element located at position x with length dx and aligned along vector
dx can be decomposed into a pure translation by velocity u(x; t), a pure rotation at velocity
!, and deformations (linear and angular) at a rate D ¢ dx.
Example 1.6 : Deformations for a shearing °ow
Consider a simple two dimensional °ow whose velocity is given by
u1 = ax2
and
u2 = 0:

21. 18 1 Introduction
Derive all related tensors and give an interpretation for the °uid motion.
Solution
The velocity gradient tensor is readily obtained
ruT =
·
0 a
0 0
¸
:
Its symmetric and antisymmetric parts are respectively
D =
1
2
·
0 a
a 0
¸
and
S =
1
2
·
0 a
¡a 0
¸
:
The rotation vector has one non zero component
!3 = ¡
a
2 ;
hence the °ow is rotational. It can be seen that a square element translates horizontally by a
velocity u1 while its vertical edges rotate clockwise by an angle
d¯ = adt
so that its rate of rotation is a. The volume of the °uid element is unchanged, but the diagonal
line will rotate clockwise at the rate ¡1
2a. On the other hand, the angle originally at ¼
2 is
reduced by an angle adt, which shows that the °uid element is sheared at the rate °_ = a.
Example 1.7 : Deformations for a stagnation °ow
Consider a simple two dimensional °ow whose velocity is given by
u1 = ax1
and
u2 = ¡ax2:
Derive all related tensors and give an interpretation for the °uid motion.
Solution
The velocity gradient tensor is readily obtained
ruT =
·
a 0
0 ¡a
¸
:
Since it is a symmetric tensor then its symmetric and antisymmetric parts are respectively

22. 1.2 Velocity gradients and Deformations 19
D = ruT
and
S = 0:
Since S = 0, the °ow is irrotational. The sum of the diagonal coe±cients of the deformation
tensor is zero, therefore the volumetric dilatation is zero (in other words the °ow is incompres-
sible). From ¯gure ( ) we can see that the °uid element can be decomposed into a translation
motion by the velocity vector u and a relative motion which causes deformations whose velo-
city is ( @u1
@x1
dx1; @u2
@x2
dx2) = (adx1;¡adx2). The angle variations d¯ and d® are zero. This shows
that the °uid elements does not rotate and has no shearing. It has only linear deformations
(for a positive value of a, it corresponds to an expansion in the horizontal direction and a
compression in the vertical one).
Example 1.8 : Rate of change of material line
Consider a material line element dx with length ds, i.e. ds2 = dx ¢ dx and l = ( dx
ds ). Show that
the rate of extension or stretching of the element de¯ned by 1
ds
D(ds)
Dt is given by
1
ds
D(ds)
Dt
= l ¢ D ¢ l = li²ij lj : (1.47)
Solution
First of all, one can verify the following identities (proof left as an exercise) :
1
ds
D(ds)
Dt
=
1
2ds2
D(ds2)
Dt
;
D(dx)
Dt
= du;
and for any antisymmetric tensor W and any vector v one has
v ¢W¢ v = 0:
Using these relations (prove them as an exercise) one gets :
1
2ds2
D(ds2)
Dt
=
1
ds2 dx ¢ du:
Using equation (1.45), we have du = ru ¢ dx.
We denote by l the unit vector parallel to dx, that is : l = ( dx
ds ): It follows that
1
ds
D(ds)
Dt
= l ¢ ru ¢ l:
Since the velocity gradient tensor can be decomposed into a symmetric and an antisymmetric
parts :

23. 20 1 Introduction
1
ds
D(ds)
Dt
= l ¢ (D + S) ¢ l:
Since the spin tensor S is antisymmetric, the product l ¢ S ¢ l is zero. We can conclude that,
1
ds
D(ds)
Dt
= l ¢ D ¢ l = li²ij lj : (1.48)
Equation (1.48) gives an interpretation of the coe±cients of the strain tensor. If the material
element is parallel to a basis vector, for instance i1, then 1
ds
D(ds)
Dt = l ¢ D ¢ l = ²11. Therefore,
the diagonal elements of D represents the stretching of the material element parallel to the
axes.
Example 1.9 : Rate of change of the angle between material vectors
Consider two material line elements a and b. Initially these vectors make a right angle. Show
that the rate of change of the angle is :
D(µ)
Dt µ=¼2
= ¡2a ¢ D ¢ b
1.3 Problems
1.1. Measurements of a one dimensional velocity ¯eld give the following results :
Time x=0m x=10m x=20m
t=0s v=0m/s v=0m/s v=0m/s
t=1s v=1m/s v=1.2m/s v=1.4m/s
t=2s v=1.7m/s v=1.8m/s v=1.9m/s
t=3s v=2.1m/s v=2.15m/s v=2.2m/s
Calculate the acceleration at t = 1s and x = 10m .
1.2. Consider the velocity components : u = x2 ¡ y2 et v = ¡2xy. Calculate the divergence
and vorticity.
1.3. Consider the velocity ¯eld whose components are : u1 = a(x1 + x2)
u2 = a(x1 ¡ x2)
u3 = w with a, w are constants. Determine the divergence , vorticity and the pathlines.
1.4. Consider the velocity ¯eld : 8
:
u1 = bx2
u2 = bx1
u3 = 0
9=
;
et
8
:
u1 = ¡bx2
u2 = bx1
u3 = 0
9= ;
Is the °ow incompressible ?. Is it rotational ?.
1.5. Consider the velocity ¯eld v=(u,À) with u = cx + 2!y + y0 et À = cy + À0. Compute the
vorticity and the strain tensor.

24. 1.3 Problems 21
1.6. Consider the velocity ¯eld u = ax2 + by and v = ¡2axy + ct. Is the °ow incompressible ?
Determine the streamlines.
1.7. Verify the following relations :
{ a) div(Áv) = Ádiv(v) + vgrad(Á)
{ b) div(u £ v) = v:curl(u) ¡ u:curl(v)
{ c) (¿ : rv) = r(¿v) ¡ vr¿ with A : B =
P
i
P
j AijBij
1.8. { Find the expanded forms of the quantities :
(u:r)u
(ru):u
¾:u
div((ru):u)
{ Show that the magnitude of the vorticity vector is :
j³j2 = @ui
@xj
( @ui
@xj
¡
@uj
@xi
)
1.9. Show that the vorticity vector in cylindrical coordinates is given by :
curlu = r £ u = (
1
r
@uz
@µ
¡
@uµ
@r
)er + (@ur
@z
¡
@uz
@r
)eµ + (
1
r
(@(ruµ)
@r
¡
@ur
@µ
))ez
Calculate the vorticity for the following velocity ¯eld (°uid in solid rotation) : u = uµ(r)eµ
where uµ(r) = !r