ADDITIONAL TOPICS
DDITIONAL TOPICS
A
PROOF OF THE LAMB-GROMEKO FORMULA
PROOF OF THE LAMB-GROMEKO FORMULA
ś
We have =ijk śx vk ei and v2 = vivi.
j
Then
śv śv
j
v =kbg wkvbeg =kbgijk j vb eg = (dibdjg -digdjb ) vb eg =
śxi śxi
ć
ć śvj śv śvb
śvg
j
= dibd vb -digd vb eg = vb - vb eg =
jg jb
śxi śxi śxb śxg
Ł ł
Ł ł
ć
śvg
ś
= vb - (1 vb vb ) eg = (vŃ)v - Ń(1 v2)
2
śxb śxg 2
Ł ł
Thus
(vŃ)v = Ń(1 v2) + v
2
and the Lamb-Gromeko formula follows immediately. c&
REYNOLDS TRANSPORT THEOREM (1)
REYNOLDS TRANSPORT THEOREM (1)
We will prove the mathematical result known as the
x3,x3
Reynolds Transport Theorem, which plays the
fundamental role in derivation of differential forms
t>0
of the conservation principles in Continuum
W(t)
Mechanics.
x(t,x)
Consider any sufficiently regular scalar field t=0
f = f (t,x). Consider the integral of f calculated over
x
0
an arbitrary material volume (t).
W0
C(t) = f (t,x)dx.
x2,x2
x1,x1
W (t)
d
We need to compute the time derivative Có(t) = f (t,x)dx.
dt
W (t)
NOTE: This task is nontrivial since the integration domain is itself time dependent!
REYNOLDS TRANSPORT THEOREM (2)
REYNOLDS TRANSPORT THEOREM (2)
To calculate the derivative, we will switch from Eulerian variables x = [x1,x2,x3] to
Lagrangian variables = [1,2,3]. The integral C(t) can be written as
C(t) = f (t,x)dx = f[t,x(t,)]J(t,)d = f0(t,)J(t,)d.
W (t) W0 W0
In the above formula we have used the composite function f0
f0 = f0(t,) = f[t,x(t,)],
and also the Jacobi determinant (Jacobian) defined as
śx1 śx1 śx1
ł
śx1 śx2 śx3
ę ś
ęśx śx2 śx2 ś
2
J(t,) = det (t,).
śx1 śx2 śx3
ę ś
śx3 śx3 śx3
ę ś
śx1 śx2 śx3
ę ś
REYNOLDS TRANSPORT THEOREM (3)
REYNOLDS TRANSPORT THEOREM (3)
Since the domain 0 is time-independent (it is actually the initial form of the material volume
(t) at the time t = 0), we can move the differentiation operator under the sign of the integral
and get
śf0 (t,)J(t,)d + f0(t,) (t,)d
d śJ
Có(t) = f0(t,)J(t,)d =
dt śt śt
W0 W0 W0
Note that time differentiation of the composite function f0 yields
ś d ś ś ś
f0(t,) = f[t,x(t,)] = f[t,x(t,)]+ f[t,x(t,)] xi(t,) =
śt dt śt śxi śt
1424
3
=Vi(t,)=vi[t,x(t,)]
ś
= f + vŃf [t,x(t,)]
( )
śt
This part was easy! We need to calculate the time derivative of the Jacobian which has
appeared in the second integral in the formula for Có(t). This is much more complicated & .
Basically, we have two methods.
Method A
We write the Jacobian using the alternating symbol J(t,) =ijk śx1 śx2 śx3
śxi śxj śxk
Note that partial derivatives with respect to time and Lagrangian variables commute, hence
śx1 ś śx1 śV1 śx2 ś śx2 śV2 śx3 ś śx3 śV3
ś ś ś
= = , = = , = =
śt śxi śxi śt śxi śt śxj śxj śt śxj śt śxk śxk śt śxk
The time derivative
ś
J =ijk śV1 śx2 śx3 +ijk śx1 śV2 śx3 +ijk śx1 śx2 śV3 =
śt śxi śxj śxk śxi śxj śxk śxi śxj śxk
śV1 śV1 śV1 śx1 śx1 śx1 śx1 śx1 śx1
śx1 śx2 śx3 śx1 śx2 śx3 śx1 śx2 śx3
śx2 śx2 śx2 śV2 śV2 śV2 śx2 śx2 śx2
= + + =
śx1 śx2 śx3 śx1 śx2 śx3 śx1 śx2 śx3
śx3 śx3 śx3 śx3 śx3 śx3 śV3 śV3 śV3
śx1 śx2 śx3 śx1 śx2 śx3 śx1 śx2 śx3
3 3
ś
= Vi [cof J]
śxj
1 3
424ij
{
i=1 j=1
(Ńx V)ij cofactor (i,j) of J
Consider two square matrices A and B, and also the product C = ABT. It means that
cik =
a bkj aijbkj,
ij
j
so we conclude that
trC cii = aijbij (trace of the matrix C)
Moreover, from the construction of the inverse Jacobi matrix we have
1
J-1 = (cof J)T (cof J)T = detJ J-1 = J J-1
det J
Hence, the formula for the time derivative of the Jacobi determinant can be written as follows
ś
J(t,) = tr Ńx V(cof J)T ł(t,) = J(t,) tr Ńx VJ-1ł(t,)
śt
Finally, we need to get back to the Eulerian variables. To this end, we use the relation
between Lagrangian and Eulerian definitions of the fluid velocity
V(t,) = v[t,x(t,)]
124 14243
4 3
Lagrange Euler
and calculate the gradient operator with respect to the Lagrangian variables
3
ś ś k
Ńx Vł (t,) = Vi(t,) = vi[t,x(t,)]śx (t,).
śxj śxk śxj
ij
k=1
The above formula can be written shortly as
Ńx V(t,) = Ńv[t,x(t,)]J(t,)
Thus, the time derivative of the Jacobian can be re-written in the following form
ś
J(t,) = J(t,) trŃv [t,x(t,)].
( )
śt
ś
Taking into account that trŃv = vi = divv Ńv
śxi
ś
we finally get the formula J(t,) = J(t,)Ńv[t,x(t,)]
śt
Method B
This method is based upon the group property of the transformation of the material volume
at initial time t = 0 to the volume (consisting of the same fluid particles) at some later time
t > 0. We can write x(t + s,) = x[ t,x(s,)] or
xi(t + s,x1,x2,x3) = xi [t,x1(s,x1,x2,x3),x2(s,x1,x2,x3),x3(s,x1,x2,x3)] , i = 1,2,3.
Let s differentiate the above formula with respect to the Lagrangian coordinate j:
śxi śxi śxk
(t + s,) = [t,x(s,)] (s,),
śxj śxk śxj
which can also be written as [J]ij(t +s,) = [J]ik[t,x(s,)] [J]kj(s,),
which is equivalent to J(t +s,) = J[t,x(s,)] J(s,).
From the fundamental property of determinant
J(t +s,) = J[t,x(s,)] J(s,).
We need to calculate the derivative
ś J(t + Dt,) - J(t,) J(t,) J[Dt,x(t,)]- J(t,)
J(t,):= lim = lim =
Dt0 Dt0
śt Dt Dt
= J(t,)Dt0 J[Dt,x(t,)]-1
lim
Dt
Note that J[Dt,x(t,)] is the Jacobian of the nearly identical transformation
x(t,) a x(t + Dt,), which can be written shortly as x aY (x).
Dt
The explicit form of this transformation is
[Y (x)]i = xi + vi(t,x1,x2,x3)Dt + O(Dt2) , i = 1, 2, 3,
Dt
This, the Jacobi matrix can be calculated as follows
śvi
ś
[J]ij(Dt,x) = [Y (x)]i = dij + (t,x)Dt + O(Dt2)
śxj Dt śx
j
or simply J(Dt,x) = I + Ńv(t,x)Dt + O(Dt2).
Now, it is not difficult to show (do it!) that
ć
śv1 + śv2 + śv3 (t,x)Dt + O(Dt2) =1+ Ńv(t,x)Dt + O(Dt2)
J(Dt,x) =1+
śx1 śx2 śx3
Ł ł
14442444
3
div v
J(Dt,x) -1
Thus, we get lim = Ńv(t,x)
Dt0
Dt
and after returning back to the Lagrangian variables - the formula for the time derivative of
the Jacobian is obtained
ś
J(t,):= J(t,) Ńv [t,x(t,)].
( )
śt
REYNOLDS TRANSPORT THEOREM (4)
REYNOLDS TRANSPORT THEOREM (4)
The time derivative Có(t) can be now evaluated as follows
ś
Có(t) = f + vŃf + f Ńv [t,x(t,)]J(t,)d =
( )
śt
W0
ś ś
śt
= f + vŃf + f Ńv (t,x)dx = f +Ń(f v)ł(t,x)dx =
( )
śt
W (t) W (t)
ś ś
= f dx + Ń(f v) dx = f dx + f vn ds
śt śt
{
Ż
W (t) W (t) W (t) śW (t)
vn
GGO
normal
Theorem
velocity
Note that the last equality has been obtained by the use of the Green-Gauss-Ostrogradsky
(GGO) Theorem. We see that the rate of change of C(t) is the sum of two components. The
first component appears due to the local time variation of the integrated function f and it
appears even if the fluid is in rest (no motion). In contrast, the second term is entirely due to
ś
the fluid motion and it assumes nonzero value even if the field f is stationary (i.e. śt f 0).
TIME RATE OF CHANGE OF AN EXTENSIVE QUANTITY
TIME RATE OF CHANGE OF AN EXTENSIVE QUANTITY
Consider an extensive physical quantity, characterized by its mass-specific density
H = H(t,x). The amount of this quantity contained in the material volume (t) is
expressed by the following volume integral
h(t) = r Hdx
W (t)
The examples are: the Cartesian components of the linear momentum, kinetic and internal
energy. We need to know how to evaluate the time derivative of such integrals.
Using the Reynolds theorem and the differential equation of mass conservation we can
write
d d ś
h(t) = r Hdx = [śt (r H) + Ń(r H v)]dx =
dt dt
Ż
W (t) W (t)
Reynolds
Trans.Th.
ś ś D
= H r + Ń(r v)łdx + r H + vŃH dx = r Hdx
( )
Dt
śt śt
144424443 1442443
W (t) W (t) W (t)
= 0! DH
=
Dt
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