L
L
E
E
C
C
T
T
U
U
R
R
E
E
6
6
K
K
I
I
N
N
E
E
M
M
A
A
T
T
I
I
C
C
S
S
O
O
F
F
F
F
L
L
U
U
I
I
D
D
S
S
–
–
P
P
A
A
R
R
T
T
2
2
L
L
E
E
C
C
T
T
U
U
R
R
E
E
6
6
K
K
I
I
N
N
E
E
M
M
A
A
T
T
I
I
C
C
S
S
O
O
F
F
F
F
L
L
U
U
I
I
D
D
S
S
–
–
P
P
A
A
R
R
T
T
2
2
R
R
E
E
L
L
A
A
T
T
I
I
V
V
E
E
M
M
O
O
T
T
I
I
O
O
N
N
O
O
F
F
F
F
L
L
U
U
I
I
D
D
E
E
L
L
E
E
M
M
E
E
N
N
T
T
S
S
Consider two fluid elements located instantaneously at the close points A and B. We ask what
happens to the relative position of these fluid elements after a short time interval
t
.
The location of the first fluid element after the time
t
can
be expressed as follows
2
A
A
A
(t,
)Δt O(Δt )
x
x
v
x
Since
B
A
x
x
ρ
then analogously we have
2
B
A
A
(t,
)Δt O(Δt )
x
x
ρ v
x
ρ
,
where the vector
ρ
describes the relative position of the
fluid elements at the time
t
.
During a short time interval
t
this vector has changed and can be expressed as
2
2
B
A
A
A
2
(t
t)
(t) [ (t,
)
(t,
)] t O( t )
(t)
(t, )
t O( t , t | | )
ρ
x
x
ρ
v
x
ρ
v
x
ρ
v
x ρ
ρ
In the above, we have dropped the lower index “A” at the location vector corresponding to the
first element.
t
t+t
x
1
x
3
x
2
0
A
A'
B
B'
=x
B
-x
A
'=x
B'
-x
A'
'
The rate of change of the vector describing the relative position of two close fluid elements
can be calculated
2
t
0
d
(t
t)
(t)
lim
(t, )
O(| | )
dt
t
ρ
ρ
ρ
v
x ρ
ρ
.
We have introduced the matrix (tensor) called the velocity gradient
j
x
i
ij
v
v
.
The velocity gradient
v
can be written as a sum of two tensors
v
D R
, where
T
1
2
[
(
) ]
D
v
v
or
j
i
ij
j
i
v
v
1
d
2
x
x
- symmetric tensor,
and
T
1
2
[
(
) ]
R
v
v
or
j
i
ij
j
i
v
v
1
r
2
x
x
- skew-symmetric tensor
We will show that the change of the relative position of the fluid elements due to the
action of the antysymmetric tensor
R
corresponds to the local “rigid” rotation of the
fluid.
Next, we will show that the action of the symmetric part
D
corresponds to the “real”
deformation, i.e. it is responsible of the change in shape and volume.
To this end, we note that
1
2
ij
ijk
k
r
, where
k
are the Cartesian components of the
vorticity vector
k
k
ijk
i
j
v
rot
x
ω
v
e
.
Indeed, we have
j
j
i
i
1
1
2
2
ijk
k
i
j
i
j
ij
i
j
j
i
v
v
v
v
v
v
1
1
(
)
r
x
x
2
x
x
2
x
x
Thus, we can write
1
1
1
2
2
2
ij
j
i
ijk
j
k
i
r
R ρ
e
e
ρ ω
ω ρ
.
Moreover, we get
2
d
d
d
dt
dt
dt
| |
( , )
2( ,
) 2( ,
)
(
)
ρ
ρ ρ
ρ
ρ
ρ Rρ
ρ ω ρ
0
i.e., the distance between two (arbitrary) fluid elements is fixed and there is no shape
deformation.
The skew-symmetric part of the velocity gradient describes
pure rigid rotation
of the
fluid and the
local angular velocity is equal
1
2
ω
.
D
D
E
E
F
F
O
O
R
R
M
M
A
A
T
T
I
I
O
O
N
N
O
O
F
F
F
F
L
L
U
U
I
I
D
D
E
E
L
L
E
E
M
M
E
E
N
N
T
T
S
S
The rate of change of the relative position vector (or – equivalently – the velocity of the
relative motion of two infinitely close fluid elements) can be expressed by the formula
deformation
rigid rotation
d
1
2
dt
ρ
Dρ
R ρ
Dρ
ω ρ
.
The first terms consists the symmetric tensor
D
, called the deformation rate tensor.
The tensor
D
can be expressed as the sum of the spherical part
D
SPH
and the deviatoric part
D
DEV
DEV
SPH
D
D
D
The spherical part
D
SPH
describes pure volumetric deformation (uniform expansion or
contraction without any shape changes) and it defined as
trace
of
k
SPH
SPH ij
ij
k
tr
v
1
1
1
(
)
(
)
3
3
3 x
D
D
D
I
v I
D
,
Note that
SPH
1
tr
(
) tr
(
)
div
3
D
v
I
v
v
.
The second part
D
DEV
describes shape changes which preserve the volume.
We have
j
i
k
DEV
DEV ij
ij
j
i
k
v
v
v
1
1
1
div
(
)
3
2
x
x
3 x
D
D
v I
D
and
DEV
SPH
tr
tr
tr
0
D
D
D
To explain the geometric interpretation of both parts of the deformation rate tensor, consider
the deformation of a small, initially rectangular portion of a fluid in two dimensions. Assume
there is no rotation part and thus we can write
d
DEV
SPH
dt
ρ Dρ D
ρ D
ρ
.
For a short time interval
t
the above relation yields
2
1
2
DEV
SPH
(t
t)
(t)
t
t
O( t )
ρ
ρ
ρ
ρ
D
ρ
D
ρ
Consider the 2D case when only volumetric part of the deformation exists (see picture).
We have
SPH
d 0
0 d
D
,
tr
2d
D
The relative position vector at the time
instant
t
t
is expressed as
2
(t
t)
(1 d t) (t) O( t )
ρ
ρ
.
The shape of the volume is preserved because the above formula describes the isotropic
expansion/contraction. The volume of the region
1
2
Vol (t)
L L
has been changed to
2
2
1
2
Vol (t
t)
L L (1 d t)
Vol (t)(1 2d t) O( t )
,
and
t
0
Vol (t
t) Vol (t)
1
lim
2d
tr
Vol (t)
t
D
v
O
x
1
x
2
A
B
C
(t)
B'
C'
A'
{
{
dt L
2
dt L
1
(t+t)
O
x
1
x
2
A
B
C
(t)
Assume now that the spherical part of the deformation rate tensor is absent. The deviatoric
part of this tensor in a 2D flow can be written as follows
1
1
2
2
11
12
11
22
12
DEV
1
1
2
2
12
22
12
22
11
d
d
d
(d
d )
d
d
d
d
d
(d
d )
D
.
The fluid deformation during the short time
interval can be now expressed as
2
DEV
O( t )
(t
t)
(
t
) (t)
ρ
I
D
ρ
or in the explicit form as
1
1
2
2
1
2
x (t
t)
(1
t) x (t)
t x (t)
x (t
t)
t x (t) (1
t) x (t)
Note the presence of shear, which manifests in the change of the angles between the position
vectors corresponding to different fluid elements in the deforming region.
O
x
1
x
2
A
B
C
(t)
B'
C'
A'
{
{
t L
2
t L
1
(t+t)
{
{
t L
2
t L
1
O
x
1
x
2
A
B
C
(t)
Let’s compute again the change of the volume of the fluid region during such deformation.
We get
2
2
2
2
1
2
1
2
2
1
t
t
Vol (t
t)
L L
L L (1
t
t )
t
1
t
Vol (t) O( t )
,
so
t
0
Vol (t
t) Vol (t)
1
lim
0
Vol (t)
t
.
We conclude that this time the instantaneous rate of the volume change is zero.
Thus, instantaneously, the deviatoric part of the deformation describes pure shear (no
expansion/contraction).