
English: 
Hello welcome to my talk, All about Fluids. This talk it is the first part on
an interesting and important topic, 'Is the Navier-Stokes equation correct?'
In this talk the focus would be on the inconsistencies of Navier-Stokes
equation, the well-known universal governing equation for fluid motion
However some recent research work and the effort for understanding the
physics involved in the Navier-Stokes equation. I will show the inconsistencies
behind the Navier-Stokes equation. For instance, the Cauchy's symmetric stress tensor
on solids had been fully adopted to fluids by Cauchy himself

English: 
Hello welcome to my talk, All about Fluids. This talk it is the first part on
an interesting and important topic, 'Is the Navier-Stokes equation correct?'
In this talk the focus would be on the inconsistencies of Navier-Stokes
equation, the well-known universal governing equation for fluid motion
However some recent research work and the effort for understanding the
physics involved in the Navier-Stokes equation. I will show the inconsistencies
behind the Navier-Stokes equation. For instance, the Cauchy's symmetric stress tensor
on solids had been fully adopted to fluids by Cauchy himself

English: 
nearly 200 years ago. But this requirement may not be a
pre-requisite for fluids as that for solids, since fluids are fundamentally different from
the solids: the fluids flow under an applied shear stress,
but solids do not. In this talk I will show you what are the
inconsistencies behind the Navier-Stokes equation, based on the direct physical
analysis of fluid motion and the examples from the famous textbooks
is Navier-Stokes equation correct? this is quite a scary question, nobody
would expect an answer 'NO', because the Navier-Stokes equation had been
accepted and practically validated for nearly 200 years, since Navier first

English: 
nearly 200 years ago. But this requirement may not be a
pre-requisite for fluids as that for solids, since fluids are fundamentally different from
the solids: the fluids flow under an applied shear stress,
but solids do not. In this talk I will show you what are the
inconsistencies behind the Navier-Stokes equation, based on the direct physical
analysis of fluid motion and the examples from the famous textbooks
is Navier-Stokes equation correct? this is quite a scary question, nobody
would expect an answer 'NO',  because the Navier-Stokes equation had been
accepted and practically validated for nearly 200 years, since Navier first

English: 
formulated the full fluid dynamic equation in 1822, Stokes presented an
elegant derivation of the fluid dynamics equation in 1845
But I would say here the answer for this question is 'YES' and 'NO'. I will tell you
the reason why I say this.
For YES, that's because the Navier-Stocks equation
as the governing equation for fluid motion has been proven by numerous examples:
experimental data, analytical solutions and the numerical solutions.
for NO, it is because we have already seen some inconsistencies behind the Navier-Stokes
equation, specifically on the symmetrical stress tensor for fluids. It is well

English: 
formulated the full fluid dynamic equation in 1822, Stokes presented an
elegant derivation of the fluid dynamics equation in 1845.
But I would say here the answer for this question is 'YES' and 'NO'. I will tell you
the reason why I say this.
For YES, that's because the Navier-Stocks equation
as the governing equation for fluid motion has been proven by numerous examples:
experimental data, analytical solutions and the numerical solutions.
for NO, it is because we have already seen some inconsistencies behind the Navier-Stokes
equation, specifically on the symmetrical stress tensor for fluids. It is well

English: 
known the Cauchy's symmetric stress tensor was originally developed for
elastic materials, the solids, and then fully adopted to the fluids by Cauchy himself
in 1827. However, I will show the inconsistencies regarding the
symmetrical tensor of fluids.
For instance, how we can explain the vertical stress tensor in the Couette flow?
how to explain the viscous shear stress in the pure rotating motion of fluid?
why the factor 2 is added in formulating the symmetric stress tensor?
but without the physical basis. And in some textbooks it is clearly shown

English: 
known the Cauchy's symmetric stress tensor was originally developed for
elastic materials, the solids, and then fully adopted to the fluids by Cauchy himself
in 1827. However, I will show the inconsistencies regarding the
symmetrical tensor of fluids.
For instance, how we can explain the vertical stress tensor in the Couette flow?
how to explain the viscous shear stress in the pure rotating motion of fluid?
why the factor 2 is added in formulating the symmetric stress tensor?
but without the physical basis. And in some textbooks it is clearly shown

English: 
that there are examples where the symmetrical stress tensor is not
satisfied in some real flows. How we can explain these inconsistencies.
In this talk I will show you the details of all these inconsistencies
In the next two slides I will show you how the Cauchy's symmetric stress
tensor has been derived.
It must be noted this is basically for solids,
not the necessary for fluids. Consider a small element as shown in the figure,
and the stress tensor pairs, which could produce the moment in the
x direction as shown in the figure. and the the stress pair are the

English: 
that there are examples where the symmetrical stress tensor is not
satisfied in some real flows. How we can explain these inconsistencies.
In this talk I will show you the details of all these inconsistencies
In the next two slides I will show you how the Cauchy's symmetric stress
tensor has been derived.
It must be noted this is basically for solids,
not the necessary for fluids. Consider a small element as shown in the figure,
and the stress tensor pairs, which could produce the moment in the
x direction as shown in the figure. and the the stress pair are the

English: 
TAU_23 and TAU_32, TAU' _23 and TAU'_32, if the element is very small, the
stress tensor components on both sides are equal. For instance
TAU_23 = TAU'_23;  TAU_32 = TAU'_32,
the moment due to the stress couple TAO_32 and TAO'_32
is in x direction, so delta M_x1 equals to this
because of the stress pair TAU_32 and TAU'_32
Similarly the moment due to the stress couple TAU_23 and TAU'_23 would be

English: 
TAU_23 and TAU_32, TAU' _23 and TAU'_32, if the element is very small, the
stress tensor components on both sides are equal. For instance,
TAU_23 = TAU'_23; TAU_32 = TAU'_32,
the moment due to the stress couple TAO_32 and TAO'_32
is in x direction, so delta M_x1 equals to this
because of the stress pair TAU_32 and TAU'_32
Similarly the moment due to the stress couple TAU_23 and TAU'_23 would be

English: 
in negative x direction. So we have delta M_x2 given by this
Put these together, we get the total moment in x direction delta M_x given by this
because of the moment acting on the element so its rotation motion around
the x-axis would be given as this.
For this element the moment of inertia for
the element given delta I_xx as this
so put this into the equation we can
calculate the rotation acceleration Omega x dot as this
so if the element becomes infinitesimal, that is, delta y and delta z

English: 
in negative x direction. So we have delta M_x2 given by this
Put these together, we get the the total moment in x direction delta M_x given by this
because of the moment acting on the element so its rotation motion around
the x-axis would be given as this.
For this element the moment of inertia for
the element given delta I_xx as this
so put this into the equation we can
calculate the rotation acceleration Omega x dot as this
so if the element becomes infinitesimal, that is, delta y and delta z

English: 
are both close to zero, so the rotating acceleration would have
become infinite unless TAU_32= TAU_23.
Similarly we can prove TAU_12 =TAU_21; TAU_31 = TAU_13
this is the Cauchy's second law of rotation motion for solids: the stress
tensor must be symmetric.
And this law is then fully adopted to fluids by Cauchy himself.
The concept of the symmetric stress
tensor for fluids has been widely accepted ever since
However, consider the fundamental difference between the fluid motion and
the solid motion, can we ask you a question:

English: 
are both close to zero, so the rotating acceleration would have
become infinite unless TAU_32= TAU_23.
Similarly we can prove TAU_12 =TAU_21; TAU_31 = TAU_13
this is the Cauchy's second law of rotation motion for solids: the stress
tensor must be symmetric.
And this law is then fully adopted to fluids by Cauchy himself.
The concept of the symmetric stress
tensor for fluids has been widely accepted ever since
However, consider the fundamental difference between the fluid motion and
the solid motion, Can we ask you a question:

English: 
Is this symmetric stress tensor a prerequisite for fluids?
in this slide the difference of a shear stress applied on a solid
element and a fluid element is shown. Consider a small cube and apply a
shear stress TAU_23 on the element as shown in this figure
if the element is solid, an elastic material, under the applied shear
stress, the element would deform and thus a shear stress TAU_32 would be
produced, which could balance the applied shear stress due to the
elasticity of the material. This is the requirement from the Cauchy symmetric

English: 
Is this symmetric stress tensor a prerequisite for fluids?
in this slide the difference of a shear stress applied on a solid
element and a fluid element is shown. Consider a small cube and apply a
shear stress TAU_23 on the element as shown in this figure
if the element is solid, an elastic material, under the applied shear
stress, the element would deform and thus a shear stress TAU_32 would be
produced, which could balance the applied shear stress due to the
elasticity of the material. This is the requirement from the Cauchy symmetric

English: 
stress tensor, as TAU_23 must equal to TAU_32.
However, if the element is a fluid, and the applied shear stress will cause a flow
in the shear stress direction as seen here. And due to the flow in the fluid a
friction is induced which could balance the applied shear stress in a steady flow.
Therefore there will be no additional shear stress to balance the
applied shear stress as seen in the solids. This fundamental difference
can be understood a fluid could have a translation or rotation motion or both
as those for a solid, but the fluid may also have a flowing motion, while the solid does not.

English: 
stress tensor, as TAU_23 must equal to TAU_32.
however if the element is a fluid, and the applied shear stress will cause a flow
in the shear stress direction as seen here. And due to the flow in the fluid a
friction is induced which could balance the applied shear stress in a steady flow.
Therefore there will be no additional shear stress to balance the
applied shear stress as seen in the solids. This fundamental difference
can be understood a fluid could have a translation or rotation motion or both
as those for a solid, but the fluid may also have a flowing motion, while the solid

English: 
does not
Now we look at the first inconsistency Q1 in the simple classic Couette flow.
The Couette flow is confined between two large parallel plates of a distance h
the bottom plate is fixed while the upper plate moves at a constant velocity U0
the moving velocity of the upper plate is small
hence, the flow between two plates is assumed to be laminar. Such an
assumption is for an analytical solution for the Couette flow. Based on the
analysis of the flow we can see there should be no fluid velocity in y and

English: 
Now we look at the first inconsistency Q1 in the simple classic Couette flow.
The Couette flow is confined between two large parallel plates of a distance h
the bottom plate is fixed while the upper plate moves at a constant velocity U0
the moving velocity of the upper plate is small
hence, the flow between two plates is assumed to be laminar. Such an
assumption is for an analytical solution for the Couette flow. Based on the
analysis of the flow we can see there should be no fluid velocity in y and

English: 
z directions, both v and w are zero. And u is the function of y only.
As such the Navier-Stokes equation can be simplified into this partial
differential equation, and apply the no-slip boundary
condition at both plates, the solution is given as this and the
plot of the velocity profile is seen as this: the flow velocity is
increasing linearly with y. This classical flow solution can be found in many
textbooks and the correctness of this solution has been validated using experiment data.
Now we can calculate the shear stress in the Couette flow, we can draw a small

English: 
z directions, both v and w are zero. And u is the function of y only.
As such the Navier-Stokes equation can be simplified into this partial
differential equation, and apply the no-slip boundary
condition at both plates, the solution is given as this and the
plot of the velocity profile is seen as this: the flow velocity is
increasing linearly with y. This classical flow solution can be found in many
textbooks and the correctness of this solution has been validated using
experiment data
Now we can calculate the shear stress in the Couette flow, we can draw a small

English: 
rectangle and there will be 4 stress components as seen in here
and based on the requirement of the Cauchy's symmetric stress tensor,
we have TAU_12 equaling to TAU_21, and TAU'_12 equaling to TAU'_21.
and the horizontal shear stress
based on the Cauchy's symmetric stress tensor is given as this.
This non-zero horizontal stress can be  easily understood: it acts in x direction on the
surface of a constant y and it causes a horizontal flow
based on the requirement of the Symmetric stress tensor, TAU_21 equals to TAU_12,

English: 
rectangle and there will be 4 stress components as seen in here
and based on the requirement of the Cauchy's symmetric stress tensor,
we have TAU_12 equaling to TAU_21, and TAU'_12 equaling to TAU'_21.
and the horizontal shear stress
based on the Cauchy's symmetric stress tensor is given as this.
This non-zero horizontal stress can be easily understood: it acts in x direction on the
surface of a constant y and it causes a horizontal flow
based on the requirement of the Symmetric stress tensor, TAU_21 equals to TAU_12,

English: 
hence in the Couette flow, the vertical shear stress component, TAU_21, is
given by this, same as TAU_12, both are nonzero.
Now we look back at the definition for fluid. A fluid cannot bear any shear stress
at rest and under a shear stress the fluid flows in the direction
of the applied shear stress. Therefore it is easily understood that
the non-zero horizontal shear stress TAU_12 causes a horizontal flow.
Upon the requirement of the symmetric stress tensor, the vertical shear stress
TAU_21 is non-zero, same as TAU_12 in
the Couette flow.

English: 
hence in the Couette flow, the vertical shear stress component, TAU_21, is
given by this, same as TAU_12, both are nonzero.
Now we look back at the definition for fluid. A fluid cannot bear any shear stress
at rest and under a shear stress the fluid flows in the direction
of the applied shear stress. Therefore it is easily understood that
the non-zero horizontal shear stress TAU_12 causes a horizontal flow.
Upon the requirement of the symmetric stress tensor, the vertical shear stress
TAU_21 is non-zero, same as TAU_12 in the Couette flow.

English: 
But why this vertical shear stress TAU_21 does not cause a vertical
flow in the Couette flow? if based on the fluid definition in the
Couette flow, we have a non zero horizontal
shear stress TAU_12 - due to the
horizontal flow, but the vertical shear stress TAU_21 must be zero because there's
no flow in the vertical direction in the Couette flow.
Therefore, TAU_12  should not equal to TAU_21. That is, this classical Couette
flow has shown that the fluid stress tensor can be asymmetric.

English: 
But why this vertical shear stress TAU_21 does not cause a vertical
flow in the Couette flow? if based on the fluid definition in the
Couette flow, we have a non zero horizontal shear stress TAU_12 - due to the
horizontal flow, but the vertical shear stress TAU_21 must be zero because there's
no flow in the vertical direction in the Couette flow.
Therefore, TAU_12 should not equal to TAU_21. That is, this classical Couette
flow has shown that the fluid stress tensor can be asymmetric.

English: 
It may be argued that there may be a vertical stress tensor but the wall,
the upper plate has stopped a vertical flow due to TAU_21
we can imagine that there may be a flow coming out
if we make an opening on the upper plate like this. Is this true?
The answer for this is NO, because whether there will be a flow coming out
from the opening on the wall, totally depend on the fluid pressure, not the vertical shear stress.
If the flow velocity is large enough, based on Bernoulli's principle,
the pressure in the flow would be low and hence there might be a suction of the flow
through the opening, rather than an outflow through the opening

English: 
It may be argued that there may be a vertical stress tensor but the wall,
the upper plate has stopped a vertical flow due to TAU_21
we can imagine that there may be a flow coming out
if we make an opening on the upper plate like this. Is this true?
The answer for this is NO,  because whether there will be a flow coming out
from the opening on the wall, totally depend on the fluid pressure, not the vertical shear stress.
If the flow velocity is large enough, based on Bernoulli's principle,
the pressure in the flow would be low and hence there might be a suction of the flow
through the opening, rather than an outflow through the opening

English: 
we can exam the flow in this figure, If the non-zero vertical shear stress
exists in the horizontal pipe, it will cause a continuous flow in the small
tubes, regardless of how high of the small tubes. In reality, the water levels in 2 small
tubes can never be higher than the water level h0.
and the heights in the small tubes only depend on the pressures in the pipes.
Obviously, in such an opening, the so-called vertical shear stress does not exist.
thus this proves that there is no vertical shear stress in the Couette flow.
This conclusion can be also applied to the Hagen-Poiseuille flow, a laminar flow

English: 
we can exam the flow in this figure, If the non-zero vertical shear stress
exists in the horizontal pipe, it will cause a continuous flow in the small
tubes, regardless of how high of the small tubes. In reality, the water levels in 2 small
tubes can never be higher than the water level h0.
and the heights in the small tubes only depend on the pressures in the pipes.
Obviously, in such an opening, the so called vertical shear stress does not exist.
thus this proves that there is no vertical shear stress in the Couette flow.
This conclusion can be also applied to the Hagen-Poiseuille flow, a laminar flow

English: 
in a long horizontal pipe. Therefore the requirement of the
symmetrical stress tensor is not a pre-requisite for fluids.
the second inconsistency Q2 would be whether there will be
viscous stresses in the rotation motion of fluid. Based on the classic book, 'Marine
Hydrodynamics', it states: there will be no viscous force, if the flow
moves as a rigid body without deformation, or fully follow the velocity
description as this. Here vector V0 and the vector Omega are the
vectors for translation and rotation velocities and the vector R is the

English: 
in a long horizontal pipe. Therefore the requirement of the
symmetrical stress tensor is not a pre-requisite for fluids.
the second inconsistency Q2 would be whether there will be
viscous stresses in the rotation motion of fluid. Based on the classic book, 'Marine
Hydrodynamics', it states: there will be no viscous force, if the flow
moves as a rigid body without deformation, or fully follow the velocity
description as this. Here vector V0 and the vector Omega are the
vectors for translation and rotation velocities and the vector R is the

English: 
position vector from the origin of rotation
viscous stresses would be produced only
when the fluid motion differs from the above equation, which would be the
deformation or the flow of the fluid.  Consider a rotating solid cylinder as
shown in this figure, which rotates evenly around its axis. For such a rotating
solid, there will be no shear stress acting on
the solid. This is why the rotating motion is excluded in the calculations
of the shear stress in solids.
For the rate of the strain in fluids, it can be generally decomposed into two parts as

English: 
position vector from the origin of rotation
viscous stresses would be produced only
when the fluid motion differs from the above equation, which would be the
deformation or the flow of the fluid.  Consider a rotating solid cylinder as
shown in this figure, which rotates evenly around its axis. For such a rotating
solid, there will be no shear stress acting on
the solid. This is why the rotating motion is excluded in the calculations
of the shear stress in solids.
For the rate of the strain in fluids, it can be generally decomposed into two parts as

English: 
this: Stokes himself did the same decomposition in 1845
The symmetric part is the shear or angular deformation and the
anti-symmetric part is the rigid-body rotation
Based on the principle mentioned in the previous slide,
the rigid-body rotation would follow the velocity description by this
formula. Obviously the shear deformation would
differ from the velocity description by this.
as such the shear deformation could contribute to viscous shear stress
but the rotational term would not contribute to the viscous stress,
this is the why it is excluded in deriving Navier-Stokes equation
Is this true?
that the fluid has no viscous stresses if it rotates

English: 
this: Stokes himself did the same decomposition in 1845.
The symmetric part is the shear or angular deformation and the
anti-symmetric part is the rigid-body rotation
Based on the principle mentioned in the previous slide,
the rigid-body rotation would follow the velocity description by this
formula. Obviously the shear deformation would
differ from the velocity description by this.
as such the shear deformation could contribute to viscous shear stress
but the rotational term would not contribute to the viscous stress,
this is the why it is excluded in deriving Navier-Stokes equation
Is this true?
that the fluid has no viscous stresses if it rotates

English: 
like a rigid body.
let's look at a special Taylor-Couette flow
the Taylor-Couette flow is confined between two coaxial cylinders, with radii of
r1 and r2. And these two cylinders could rotate at different speeds Omega 1
and Omega 2. This is a setup Couette used to
measure the fluid viscosity. We can assume the cylinder rotation speeds
are small, thus the flow between two cylinders would
be laminar and hence we can find an analytical solution for such a flow.

English: 
like a rigid body.
let's look at a special Taylor-Couette flow
the Taylor-Couette flow is confined between two coaxial cylinders, with radii of
r1 and r2. And these two cylinders could rotate at different speeds Omega 1
and Omega 2. This is a setup Couette used to
measure the fluid viscosity. We can assume the cylinder rotation speeds
are small, thus the flow between two cylinders would
be laminar and hence we can find an analytical solution for such a flow.

English: 
Now we can consider a special case Omega 1 equals to Omega 2.
the two cylinders rotate at the same rotating speed, thus in a steady state
the fluid would rotate following the two cylinders like a rigid body
and the flow azimuthal velocity would be given by this
here Omiga is the rotating velocity of the cylinders and r is the radial
position from the central axis. Obviously at the different radial
oppositions the fluid velocities would be different, see u1 and u2 here.
Thus there will be viscous stress, calculated as this. Obviously, it is
nonzero. This proves that in the pure rotational motion,

English: 
Now we can consider a special case Omega 1 equals to Omega 2.
the two cylinders rotate at the same rotating speed, thus in a steady state
the fluid would rotate following the two cylinders like a rigid body
and the flow azimuthal velocity would be given by this.
here Omega is the rotating velocity of the cylinders and r is the radial
position from the central axis. Obviously at the different radial
oppositions the fluid velocities would be different, see u1 and u2 here.
Thus there will be viscous stress, calculated as this. Obviously, it is
nonzero. This proves that in the pure rotational motion,

English: 
the fluid would produce the viscous stresses. This is very different from the
rotating solids.
the third consistency Q3 would be the formulation of the symmetric stress tensor.
Similar to Solid mechanics, the fluid
motion may generally describe as the follows or their combinations:
translation motion which means the fluid element moves from one position to
another without the dilatation or distortion.
Dilatation, the fluid is compressed or expanded but without distortion

English: 
the fluid would produce the viscous stresses. This is very different from the
rotating solids.
the third consistency Q3 would be the formulation of the symmetric stress tensor.
Similar to Solid mechanics, the fluid
motion may generally described as the follows or their combinations:
translation motion which means the fluid element moves from one position to
another without the dilatation or distortion.
Dilatation, the fluid is compressed or expanded but without distortion

English: 
Distortion, it is a combination of the rigid rotation and the shear deformation
of the fluid element
All these three types of motion can be find in the reference books: Boundary Layer Theory;
Aerodynamics for engineering students; and Fluid Mechanics
However,  for a fluid,
based on the definition of fluid, it flows if a shear stress is applied as seen
as this,  and the flow would flow as this.
and since the fluid flows, the fluid viscous force would
balance the applied shear stress in a steady flow, given by this

English: 
Distortion, it is a combination of the rigid rotation and the shear deformation
of the fluid element
All these three types of motion can be found in the reference books: ‘Boundary Layer Theory’;
‘Aerodynamics for engineering students’; and ‘Fluid Mechanics’.
However, for a fluid,
based on the definition of fluid, it flows if a shear stress is applied as seen
as this, and the flow would flow as this.
and since the fluid flows, the fluid viscous force would
balance the applied shear stress in a steady flow, given by this

English: 
In the original derivation of Navier-Stokes equation as it has shown
that the fluid distortion can be decomposed into the rigid rotation and
the shear deformation, but that the only symmetric rate of the
shear strain could contribute to the viscous shear stresses as the rates are
given as this. For the x and y directions, and these for other rates
of the strain. Based on the Newton's formulation for
fluid viscous stress, which would be proportional to the rate of strain,
therefore the stress tensor of correct physics would be calculated as this

English: 
In the original derivation of Navier-Stokes equation as it has shown
that the fluid distortion can be decomposed into the rigid rotation and
the shear deformation, but that the only symmetric rate of the
shear strain could contribute to the viscous shear stresses as the rates are
given as this. For the x and y directions, and these for other rates
of the strain. Based on the Newton's formulation for
fluid viscous stress, which would be proportional to the rate of strain,
Therefore, the stress tensor of correct physics would be calculated as this

English: 
but this stress tensor would not lead to the correct Navier-Stokes equation.
Instead the symmetric stress tensor has to be formulated as this. Stokes did the
same in 1845 in which the artificial factor 2 must be added to the formulation.
However, the question is the fact 2 had no physical foundation.
In the books, 'Aerodynamics for Engineering
Students' and the 'Boundary Layer Theory', both give a similar explanation: the
factor 2 is merely used for convenience so as to cancel out the
factor 1/2 in the expression of the rate of shear strain. But this

English: 
but this stress tensor would not lead to the correct Navier-Stokes equation.
Instead the symmetric stress tensor has to be formulated as this. Stokes did the
same in 1845 in which the artificial factor 2 must be added to the formulation.
However, the question is the fact 2 had no physical foundation.
In the books, 'Aerodynamics for Engineering
Students' and the 'Boundary Layer Theory', both give a similar explanation: the
factor 2 is merely used for convenience so as to cancel out the
factor 1/2 in the expression of the rate of shear strain. But this

English: 
formulation is physically incorrect. The problem is actually caused by the
adoption of the symmetric stress tensor for fluids where the part
corresponding to rotation has been wrongly taken as the non-contributor to the
viscous stress. In fact the shear stress tensor can be asymmetric for fluids, and
it would not cause the fluid spinning as we see in the solids.
This is the illustration for a fluid element if a shear stress TAU_23
is applied on the element, the fluid flows simply like this and the
viscous forces would balance out the applied shear stress.

English: 
formulation is physically incorrect. The problem is actually caused by the
adoption of the symmetric stress tensor for fluids where the part
corresponding to rotation has been wrongly taken as the non-contributor to the
viscous stress. In fact the shear stress tensor can be asymmetric for fluids, and
it would not cause the fluid spinning as we see in the solids.
This is the illustration for a fluid element if a shear stress TAU_23
is applied on the element, the fluid flows simply like this and the
viscous forces would balance out the applied shear stress.

English: 
The fourth inconsistency Q4 would be the exceptions for the symmetric stress tensor
for Navier-Stokes equation as we had in real flows. In the book of Kundu,
Cohen and Dowling, 'Fluid Mechanics', on page 126, there is a statement as this:
the stress tensor symmetry is violated in the electrical field on
polarised molecules where anti- symmetrical stresses must be included in
the analysis. Another example is in Wilcox's book,
'Turbulence modeling for CFD', on page 39, it states the Symmetric stress tensor
is for simple viscous flows but not for some anisotropic flows

English: 
The fourth inconsistency Q4 would be the exceptions for the symmetric stress tensor
for Navier-Stokes equation as we had in real flows. In the book of Kundu,
Cohen and Dowling, 'Fluid Mechanics', on page 126, there is a statement as this:
the stress tensor symmetry is violated in the electrical field on
polarised molecules where anti- symmetrical stresses must be included in
the analysis.        Another example is in Wilcox's book,
'Turbulence modeling for CFD', on page 39, it states the Symmetric stress tensor
is for simple viscous flows but not for some anisotropic flows

English: 
However, there's no reason given for this claim. hence we can ask a
question: if the Cauchy's symmetrical stress tensor is universally correct
for fluids, then how we can explain the exceptions
we found the in the real flows.
After all, we never see any exceptions for Newton's laws of motion.
In this slide, the summaries would be given
The classical Couette flow has shown that the shear stresses can be asymmetric for fluids;
There will be viscous stresses in the pure rotation motion of fluids;
Therefore, ignorance of the viscous stress due to the rotation motion would
not be physically consistent with the facts; so original form of the symmetric

English: 
However, there's no reason given for this claim. hence we can ask a
question: if the Cauchy's symmetrical stress tensor is universally correct
for fluids,  then how we can explain the exceptions
we found the in the real flows.
After all,  we never see any exceptions for
Newton's laws of motion
In this slide, the summaries would be given
The classical Couette flow has shown that the shear stresses can be asymmetric for fluids;
There will be viscous stresses in 
the pure rotation motion of fluids;
therefore ignorance of the viscous stress due to the rotation motion would
not be physically consistent with the facts; so original form of the symmetric

English: 
stress tensor has a factor 2, which 
has no physical significance or
foundation; In the real fluids, there are exceptions
for the symmetric stress tensor. why have such exceptions for the universally
correct laws? Having talked about all these inconsistencies, the Navier-Stokes
equation itself is correct, especially for the incompressible flows
In the second part of the talk we will discuss the solution to all these
inconsistencies behind the Navier-Stokes equation.

English: 
stress tensor has a factor 2, which has no physical significance or
foundation; In the real fluids, there are exceptions
for the symmetric stress tensor. why have such exceptions for the universally
correct laws? Having talked about all these inconsistencies, the Navier-Stokes
equation itself is correct, especially for the incompressible flows
In the second part of the talk we will discuss the solution to all these
inconsistencies behind the Navier-Stokes equation.
