T - V,
where T is the kinetic energy density of the fluid, and V is the pressure of the fluid. The pressure acts as the potential energy density.
The viscosity of the fluid has to be zero. Otherwise, friction would drain energy, and the traditional lagrangian method would not work.
^ ^
| | flow of fluid
| |
Suppose that we have a nice smooth solution for the equations. Let us transform it infinitesimally, so that we make the "parcels" of fluid to bump into each other:
^ ^
\ /
/ \ "bumping" flows of fluid
\ /
/ \
The slight variation in the path of the parcels contributes negligibly to T, but the bumping contributes significantly to V.
We proved that the nice smooth solution is not a stationary point of the action?
This might be a negative solution to one of the Millennium Problems:
Our thought experiment also explains turbulence: the action seeks an extremal point, and makes the paths of fluid parcels meandering, so that they bump into each other.
Let us check if we really can modify the total, integrated pressure with this trick. Maybe the pressure of the flow decreases somewhere else, and compensates for the pressure in the bumps?
An incompressible fluid is very unnatural. Can we make a stable physical system if one can create pressure without expending energy? If not, then the Navier-Stokes equations might have no solutions, except in some trivial cases.
In our example, we have a uniform laminar flow, which we replace with a "bumping" flow. Does this mean that even trivial solutions of the equations are unstable if the fluid is incompressible?
Can we make a meandering flow without affecting the kinetic energy T much? Yes
Let us add walls like this to the laminar flow:
^ ^ ^
| | | flow out
/ \ / \
\ / \ / "meandering" walls
/ \ / \
\ / \ /
^ ^ ^
| | | flow in
Each wall guides the flow which is close to the wall, making the flow meandering.
We can imagine parcels of the fluid moving along the walls. The parcels bump to each other. Each bump involves an increased pressure.
If the walls differ from a vertical line by an angle ε, then the kinetic energy T may grow by ~ ε², but the pressure grows by ~ ε. We found an infinitesimal variation which changes the value of the action. The original laminar flow was not an extremal point of the action.
There is a problem, however. If the horizontal pressure is increased in the meandering zone, the zone will start to expand outward. Does this spoil the variation? It makes the end state different from the one we started from. The end state should stay the same in the variation.
Preventing the meandering flow from expanding: put it inside a pipe
If the flow is infinitely wide in the horizontal dimension, then the outward pressure would not be a problem. Another way is to enclose the flow into a pipe.
The Millennium Problem does not allow walls, though. It is stated in an infinite pool of water, filling the entire ℝ³.
Let us have an infinitelty rigid pipe in the fluid, preventing the fluid from spreading from the zone where we make the flow meandering.
meandering
flow
■ ^ ^ ■
■ / \ ■
■ \ / ■
■ / \ ■
solid solid
wall wall
Does the variation above really increase the pressure V, while the kinetic energy T stays essentially the same?
If the original configuration already contains pressure, what happens?
Pressure must be the same in all directions: this spoils our simple idea
The pressure cannot be just horizontal in an ideal fluid. The pressure has to be the same in all directions. This observation spoils our simple idea.
pipe
/ /
/ /
\ \ bend
\ \
^
\ flow
In the bend, a difference in the horizontal pressure must accelerate a fluid parcel to the right. But the average pressure in the bend must be the same as in the pipe overall. Otherwise, the fluid would slow down in the bend, and its density would grow. That is not allowed as the fluid is incompressible.
This means that our meandering flows are not able to change the average pressure of the flow. The action integral over the pressure V is unchanged.
Turbulence must be a second order phenomenon?
Suppose that the system looks for an extremal point of the action. The analysis above showed that the value of the action cannot change much if we add meandering. However, the action could change a little, and that may be enough to create turbulence.
Hypothesis. Adding turbulence at some length scale L optimizes the action. Adding it at ever shorter scales, L / 2ⁿ, will optimize the action further. The optimization will not stop at any length scale. There is no smooth solution for the Navier-Stokes equations, in most cases.
The potential is created by the motion: this explains turbulence?
Suppose that the potential V in the lagrangian density is an external potential, say, gravity:
T - V.
The orbit of an object in a newtonian gravity field is very regular, for example, an ellipse. It is the very opposite of the chaos of turbulence.
In the Navier-Stokes lagrangian density, the potential V is created by the flow itself, as it is the pressure. This opens the door for a chaotic process.
Adding viscosity creates turbulence
A large viscosity creates turbulence. What happens to our meandering flows if there is a large viscosity?
Viscosity means that there is a shear stress. The fluid is more like a solid substance then. The pressure still is the same in all directions, though.
The lagrangian density cannot be as simple as given above, if viscosity is present.
D'Alembert's paradox
D'Alembert's paradox states that, in the absence of viscosity, a laminar, time-independent, flow is possible, and there is no turbulence and no drag.
The pressure of the fluid depends on the speed of the fluid. In the diagram, the flow is symmetric on the left and on the right. Consequently, the pressure is the same on the left and on the right. The pressure does not exert any force on the circular cylinder, i.e., there is no drag.
The physicist Ludwig Prandtl suggested in 1904 that there is a "no slip" condition of the fluid on the surface. The viscosity of the fluid is large close to the surface, even if the viscosity would be close to zero in the bulk of the fluid. The hypothesis can explain many empirical phenomena.
From where does the energy for turbulence come?
Let us assume that the fluid has a zero viscosity and is incompressible. In the diagram above, the red cylinder is static. It cannot do any work.
Since the fluid is incompressible, the average horizontal speed component of the parcels of fluid must be constant. If the fluid arrives straight from the left, it cannot lose any of its original kinetic energy.
If there is turbulence on the right, then parcels have a vertical velocity component. They gained kinetic energy. From where did they obtain that energy? The explanation has to be that the pressure is larger on the left than on the right. A parcel slides down this pressure potential and gains kinetic energy.
Bernoulli's principle
For a time-independent (potential) flow, a larger velocity means a lower pressure.
Bernoulli's principle states that for a steady (= time-independent) flow with a zero viscosity, the energy of a parcel of fluid is constant. The pressure serves as the "potential" of the parcel.
Thus, the lagrangian density which we mentioned at the beginning of this blog post, only works for a time-independent flow.
Suppose that the flow is time-independent and we have a flow which is the stationary point of the action
∫ T - V dt,
where V is the pressure. For a time-independent flow, Bernoulli's principle holds. Let us assume that we "feed" the flow at a constant pressure p and kinetic energy density E. Then
T + V = E + p.
Is the stationary point also a stationary point of the action
∫ T dt?
Probably, yes.
Constructing a singularity: can we solve the nonviscous Navier-Stokes starting from a singular flow? We are not able to construct a singularity
If there is no friction, then fluid physics is time-symmetric. Suppose that we are able to construct a solution which contains a singularity at the time
t = 0,
but not at later times. If we run time backwards, we can let a smooth flow to develop into a singularity.
^ y
|
___ | ____ line of flow
\|/
---------------------> x
_____/|\_____ line of flow
t = 0
We probably can construct a flow such that the speed of the flow approaches infinity at x = 0, for y close to zero, when t = 0.
The total kinetic energy is finite because the amount of fluid running very fast is very small.
The solution is smooth and defined for all t > 0, but we cannot define it at t = 0 in the origin.
If Navier-Stokes has solutions for all smooth initial states, then we can solve it starting from any t > 0. But there is no guarantee that we can combine these solutions to obtain one solution which is valid for all t > 0.
Since Navier-Stokes is nonlinear, an interesting question is if nonlinear equations generally develop singularities.
We cannot use a Green's function to construct a singularity because the Navier-Stokes equations are nonlinear. We cannot construct new solutions through summation.
Nonlinear differential equations decomposed into linear equations plus a perturbation
The tools used for linear differential equations do not work for nonlinear equations. If the nonlinear equations are "close" to linear ones, then we might be able to construct approximate – or even exact – solutions for the nonlinear equations by perturbing solutions of the linear equations.
Let
G(f) = 0
be the approximate linear equation for a function f. Let
G(f) = H(f)
be the nonlinear equation, where H is "small". The function H(f) is the perturbation which acts as the source for the homogeneous linear equation G(f) = 0.
We can use Green's functions to calculate the "response" of f to the perturbation. If the response is very small, we might be able to get f to converge toward a function which solves the nonlinear equation. This might be the method which has previously been used to derive partial existence and smoothness solutions for Navier-Stokes.
Could it be that we can perform Turing machine computations through this iterative process? Then the existence of solutions to certain problems could be undecidable in the axioms of set theory.
Letting two balls of incompressible nonviscous fluid to collide: a singularity forms? No
Let us have the space filled with an incompressible nonviscous fluid. We interpret the diagram above in this way: the red circle is a ball of the fluid moving horizontally relative to the main mass of the fluid.
The fluid speed seems to be the fastest and the pressure the lowest on the equator of the ball, if the axis is horizontal. Thus, the ball starts to flatten in the horizontal direction.
What happens if we let two such balls of fluid collide head-on?
Since the fluid is incompressible, a head-on hard collision would create an infinite pressure.
R = 1
solid ball
\ r
| | plane
/
--> v ρ fluid
Let a rigid solid ball of a radius R = 1 collide at a constant speed v to a solid plane. They are immersed in the fluid of the density
ρ.
Let r << R be the distance from the contact point. The flow of the fluid to larger radii r at the contact moment is
v * π r²,
and the area through which it flows is
A = 1/2 r² * 2 π r
= π r³.
The velocity of the flowing fluid is
v / r,
and its kinetic energy density is
1/2 ρ v² / r².
The kinetic energy contained in the fluid in a thin ring dr around the contact point is
1/2 ρ v² / r²
* π r³ dr
= 1/2 π ρ v² r dr
The total kinetic energy E of the fluid at radii less than r is
E = 1/4 π ρ v² r².
Let us assume that the density of the ball is ρ. Then the kinetic energy of the part of the ball at radii < r is
E' = 1/2 v² * ρ 2 * π r².
= ρ π v² r².
The kinetic energy of the ball is 4X the kinetic energy of the fluid. A hard collision might happen.
In the collision between two moving balls of fluid, is there a mechanism which softens the collision?
The speed of the fluid at the contact point would become infinite. The pressure required to accelerate the fluid is there infinite. Does that pressure deform the balls so that they will not touch?
The pressure which makes the fluid between the balls to acquire great speeds, also affects the fluid at the surface of a ball, and accelerates that fluid. The surface of the ball "melts" and flows along with the fluid between the balls.
We conclude that there probably is no singularity in this case.
Why a Green's function does not create a singularity for a nonlinear equation?
Suppose that the fluid is compressible. Then we can form an approximate linear equation which describes sound waves in the fluid.
A Green's function is the response of a linear equation to a singularity impulse – the impulse is a Dirac delta function. The Green's function is obviously singular at the point of the impulse.
But the precise Navier-Stokes equation is not linear. If we imagine a sound wave with an infinite energy density, then the fluid close to the infinite energy density is obviously very much compressed. A simple linear sound wave equation fails if the density of the fluid varies greatly, according to the phase of the sound wave.
Thus, the perturbation of the linear equation becomes very large at a singular point, and we cannot expect an iterative solution method to converge. There is no guarantee that any function f satisfies the nonlinear equation and looks similar to a Green's function.
Is it so that any linear approximation of the Navier-Stokes equation has the property that the approximation is very bad close to a singularity?
Transverse waves in a viscous fluid or elastic solid
If the fluid has a huge viscosity, we might be able to create transverse waves. If the viscosity grows without bounds even at moderate shear speeds, then the fluid might behave much like an elastic solid.
Let us investigate transverse waves in a newtonian solid. Are we able to create a singularity with a Green's function?
A linear wave equation probably only works for small-amplitude waves. A singularity would have a large amplitude. Thus, the wave equation would not be linear. This approach is not going to help us.
The wrinkle in a carpet trick can make solutions nonexistent?
In the fall of 2023 we argued that the Einstein equations cannot have a solution because they behave like we would need to fit an oversized carpet in too small a room. Locally, we can press the carpet optimally to the floor, but a wrinkle will always persist somewhere in the carpet.
Can we utilize this idea if we assume an exotic fluid whose viscosity behaves in a strange way when the shear velocity varies?
Then we may locally have a nice solution to the fluid flow problem, but globally, an annoying "wrinkle" always persists?
Regarding the Millennium Problem, this is cheating. The purpose of the problem is to study turbulence – not oversized carpets.
Exotic viscosity which declines when the shear speed increases: this produces singularities
F shear force
<----
----------
---------- pile of papers
----------
---->
F shear force
Suppose that we have a pile of paper sheets on the table. The friction between sheets of paper becomes less if the speed between the sheets is larger. We exert a shear force. Exactly one pair of papers starts to slide relative to each other.
The sliding gap is a singularity. The same trick probably can be used for a viscous fluid. The speed of the fluid will be a non-smooth function at the gap.
Let us check if the Millennium Problem bans exotic viscosity.
The statement of the problem assumes a newtonian fluid, in which the "friction" is linear in the shear velocity. Thus, our trivial singularity is ruled out. The constant ν is the kinematic viscosity.
Viscous fluid and two opposite streams: explosion backward in time
^
| --------> v
0 -- 2 L transition zone
| <------- -v
x
Let us have two streams, between which there is a narrow transition zone where the speed of the fluid varies smoothly between -v and v. The velocity
v(x)
is a monotonously growing function of x, and symmetric around x = 0.
If we calculate the system backward in time, will the velocity field end up with a discontinuity at x = 0 in a finite time?
The differential equation is
dv / dt = ν d²v / dx².
Let v = 1 asymptotically when we go to large x, and v = -1 asymptotically when we go to large negative x. Let the viscosity ν be 1.
Let us assume that
v(t, x) = 1 - exp(t - x)
at t = 0. Then
d²v(t, x) / dx² = -exp(t - x),
and
dv(t, x) = -exp(t - x).
We found a solution? No. The value of v(t, 0) should be 0 for all t.
Let
v(t, x) = exp(-a² t) * cos(a x),
where a is any real number. Then
d²v / dx² = -a² exp(-a² t) * cos(a x),
dv / dt = -a² exp(-a² t) * cos(a x).
We found another solution.
Let us choose a sum of the form:
∞
v(t, x) = ∑ 1 / 2ⁿ * exp(-n² t) * cos(n x)
n = 0
The function v(t, x) is defined at t = 0, x =0, but is infinite for t = -1, x = 0.
Are all the derivatives of v(t, x) defined at t = 0? The derivatives contain polynomials of n. They do converge.
We found a function v(t, x) which explodes if we go backward in time. But we are interested in solutions forward in time, starting from a smooth initial condition.
Does the following work: v(t, x) is defined as as an alternating series where negative terms cancel the positive ones. Let us make negative terms such that they shrink when t > 0. The sum becomes infinite. But can we define it for all x at t = 0?
Our example has only two spatial dimensions. Wikipedia says that Navier-Stokes does have smooth solutions in that case. We should construct an example which uses three spatial dimensions, if we wish to prove that a smooth solution does not always exist.
Incompressibility leads to paradoxes?
If the fluid is compressible, we can, in principle, solve the Navier-Stokes equations locally. No change in the state of the fluid can propagate faster than the speed of sound.
pipe
________________________ ...
---> ________________________ ...
push
Suppose that we have an infinitely long rigid pipe filled with an incompressible fluid. Obviously, one is not able to get the fluid to move at all.
The same holds for fluid between two infinite, rigid planes. If the distance from the person pushing the fluid is r, then the velocity of the fluid would be roughly
~ 1 / r,
the kinetic energy density would be
~ 1 / r².
The integral
∞
∫ 1 / r² * 2 π r dr
0
diverges. Pushing the fluid would require infinite energy. In three spatial dimensions, the energy integral does converge.
Conclusions
We were not able to discover anything new about the Navier-Stokes equations. They are nonlinear equations. We do not know a nice action principle for them, so that we could utilize variational calculus to study the existence of a solution.
It might be that the answer to the Millennium Problem is affirmative: there always exists a smooth solution. Our efforts to disprove that failed.
We will next return to studying retardation effects in gravity. Can retardation explain "dark energy"?
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