# Fluid Dynamic: Stokeslet? Rotlet?

1. Mar 3, 2007

### hanson

Can anyone explain what stroketlet and rotlet physically mean?
Also, I find two forms of strokes equation in different literature, one with the partial derivative of velocity w.r.t. time while another did not, so which is the correct one? Or they refers to different scenario?

I am very confused...could someone kindly help?

By the way, I still don't catch the rationale behind non-dimensionalizing the variables...and why certain variables are non-dimensionalized in that way, like the pressure in Stokes' approximation, which is made dimensionaless by defining P* = {P-P(infinity)}/(viscosity*velocity/length), why it is P-P-infinity? why not just P/(that thing)? Please help....

2. Mar 3, 2007

### arildno

The rationale behind non-dimensionalizing is two-fold:
1. Assuming the scales has been "suitably chosen", the magnitude of a term is carried by the known dimensional factors, whereas the unknown factor comprised of, typically, derivatives of an unkown function has roughly the same magnitude as any other such unkown factor.
Hence, by comparing the KNOWN factors (i.e, the ratios between scales), we can see which terms overall are "dominant", and therefore which terms we may ignore when making an approximation.

2. As long as the RATIOS between scales remain the same, we are essentially solving the same mathematical problem, and getting the same solution.
However, ratio equality does not mean equality in scale!!
Therefore, by non-dimensionalizing the equations, you get an equation valid for large CLASSES of physical problems, each choice of the parameters involved marking one class.

Last edited: Mar 3, 2007
3. Mar 3, 2007

### hanson

some text says that non-dimensionalization renders a nondimensional varible having O(1) and hence the importance of each term in a differential equation can be compred etc...

I don't understand this statement...what is mean by order of 1? This fancy term keep on appearing for so many times but I still don't know what "order' is...and why suddenly it is 1 etc..

can you please kindly explain in detail?

4. Mar 4, 2007

### Clausius2

Yes, and you will, you will keep on seeing that stuff many times. That is the essence of fluid mechanics. You better understand it. Even though some people here and there claim to be engineers, I have never seen such a fantastic maneuver of practical engineering such as that proposed by former scientists on the field to non dimensionalize the equations and split up the physical phenomena into modes. The symbol $$O$$ is what is called a big O of Landau, and it is read as "of the same order-magnitude than..". For instance, consider the dimensional momentum equation:

$$\rho\overline{u}\cdot \nabla\overline{u}=-\nabla P+\mu\Delta\overline{u}$$

and assume that your problem has a characteristic velocity $$U$$ and length $$a$$, such as those stemming from the motion of a body in a fluid. The non dimensionalization of the velocity and spatial coordinates is clear: $$\overline{u}=\overline{u}/U$$ and $$\overline{x}=\overline{x}/a$$. The non dimensionalization of the pressure IS NOT trivial at all. Two characteristic pressures emerge from the body motion: $$\rho U^2$$ and $$\mu U/a$$. At first sight one may think it does not make sense to non dimensionalize pressure with a measure proportional to the kinetic energy in slow flows, such as in hydrodynamic lubrication, where the flow is slow but the overpressures are large. Similarly, it wouldn't make sense to non dimensionalize pressure with the viscous stress for a slightly viscous fluid. Both non dimensionalizations have a famous name:

1)$$P=P/\rho U^2$$ is called the Reynolds Scaling. The non dimensional momentum equation then reads:

$$\overline{u}\cdot \nabla\overline{u}=-\nabla P+\frac{1}{Re}\Delta\overline{u}$$

so that $$\nabla P\sim O(\overline{u}\cdot \nabla\overline{u})$$ if the Reynolds number $$Re\gg 1$$. If you calculate the Reynolds number of your flow and turns out to satisfy this condition, the above scaling is well posed. Moreover, the resultant equation is the Euler equation for inviscid flow:

$$\overline{u}\cdot \nabla\overline{u}=-\nabla P$$

2)$$P=P/\mu U/a$$ is called the Stokes Scaling. The non dimensional momentum equation then reads:

$$Re\overline{u}\cdot \nabla\overline{u}=-\nabla P+\Delta\overline{u}$$

so that $$\nabla P\sim O(\Delta\overline{u})$$ if $$Re\ll 1$$. If you calculate the Reynolds number of your flow and turns out to satisfy this condition, this scaling is well posed. Moreover, the resultant equation is the Stokes equation for viscous flow:

$$\nabla P=\Delta\overline{u}$$

I don't know what is a rotlet but I can help you with the Stokeslet. This stuff can be explained by using Green's functions of the stokes equation, but I'm avoiding such a mess and put an easy example. Imagine a sphere translating through a quiescent viscous fluid. The non dimensional stream function can be written as:

$$\psi=\frac{1}{4}\left(3r-\frac{1}{r}\right)sin^2\theta$$

The first and second terms are called Stokeslet and Doublet or Dipole respectively. Surprisingly, the Stokeslet is the only term that is rotational. Look at distances very far away from the sphere. The stream function can be written at first approximation as $$\psi\sim 3rsin^2\theta/4$$, so that the only term that matters in that region is the Stokeslet. That term represents the flow field caused by a point force acting in a viscous fluid. Therefore, the effect of the moving sphere can be approximated far away as the effect of a point force acting at the center of the sphere (it sounds like the definition of a Green's function, isn't it?).

Hanson, all the above analysis is INTIMATELY-CLOSELY related to ASYMPTOTIC ANALYSIS. In both Stokes and Reynolds Scalings, the Reynolds number or its inverse appears as a small parameter, multiplying the neglected terms. But nothing is absolutely negligible in nature. Asymptotic analysis takes care of those SMALL REMAINDERS carried when you neglect some effect. Turns out to be that those apparently small effects neglected DO HAVE sometimes an extraordinary effect on the fluid motion. It is because of that reason that the Reynolds number in the Reynolds and Stokes scalings is referred as a SINGULAR PERTURBATION PARAMETER. That is for instance, what happens with the Euler equation and the Stokes equation in a Boundary Layer flow and the Viscous flow around a sphere respectively, where to leading order in $$1/Re$$ and $$Re$$ the flow is described by the mentioned Euler and Stokes equations (respectively), leading to classical paradoxes in fluid mechanics (the boundary layer one solved by Prandtl and the sphere motion- The Whitehead's paradox- solved by O'seen).

I don't know if you get the picture, but to my understanding, a well posed non dimensionalization is the first step on the way of "expanding in Asymptotic Series" the equations of motion around the flow regime (Re) in study. That's what we do in physics of fluids, we work out the scales of the problem, we localize small parameters, and we expand the equations around those small parameters. Man, that's all the artifact needed to formulate any problem in this field, but unfortunately understanding that artifact and locating the point of expansion is something that is only in the range of people with a lot of experience on the field. If you plan to get deeper in Fluid Mechanics as a graduate student, maybe you luckily happen not to have a problem statement for your thesis but you have to set it up by yourself (like me). In case of that being an analytical problem, you'll be surely enforced to use these powerful analytic methods.

Hope this helped.

Last edited: Mar 4, 2007
5. Mar 4, 2007

### hanson

I have not yet read through it but I just want to express my appreciation in advance.
You know, I have been seeing your threads and replies for long, and I notice your theoretical approch towards mechanical engineering and fluid mechnics. And I have the same attitude, but, unlike you, I am not as smart as you. I think I will keep on asking many stupid questions and I hope you would not mind to help a fool like me continuously.

6. Mar 4, 2007

### Clausius2

Hey man, these questions that depart from typical engineering questions are more than welcome, and moreover if it is related to this stuff of neglecting things in fluid mechanics and sentences of the type "of the order of". That is what I am up to everyday on my desk. Maybe someday you'll be a graduate student and you happen to remember a 1% of my above answer. Don't count on me for working out friction factors though...

7. Mar 4, 2007

### hanson

It is always said that non-dimensionalization would usually leave the nondimensional variable with typical sclae of O(1), what exactly does this mean? If I non-dimensionalize the velocity with the upstream velocity(U), u* = u/U, so u* has the same order of magnitude as "1"? what does it mean, and what's the significance?

Also, it is usually said that by scaling the variables, the terms in the dimensionless form of the equation can be meaningfully compared because the terms are dimensionaless. I don't quite get this.
Even with a dimensional equation, say the dimensionl X-momentum equation in the figure.

Though it is not non-dimensionalized, the terms in the equation have the same dimension still, so why can't these terms be compared? Say, if we think the viscou force is small compared to the inertia force, then we drop the viscous force term, can it be done it this way?

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8. Mar 4, 2007

### AlephZero

Yes you can. Since Reynolds number is the ratio of a typical inertia force to a typical viscous force, for a given physical situation, you get to the same place by a different route.

Re the wonders of engineering (or lack of) compared with the beauty of nondimensional physics, I've never yet seen a nondimensional aeroplane. The important word in the definition of Re is "the ratio of a typical inertia force to a typical viscous force". If you use nonsense parameters to estimate your nondimensional quantities, the results from a nondimensional model will also probably be nonsense.

Engineering is about dealing with how the real world works, not proving pretty theorems about idealised situations. That's why Engineering is hard.

Last edited: Mar 4, 2007
9. Mar 4, 2007

### hanson

10. Mar 4, 2007

### hanson

Why $$\rho U^2$$ and $$\mu U/a$$ came to your mind when you want to non-dimensionalize the pressure? I just don't know how these scales come out and what kind of different effects would using different scales bring aabout. Could you please explain again?

Also, what do you mean by a "well-posted" scaling in the Reynolds and Stokes scaling? What effects will a not-well-posted scaling bring about?

And I don't have a physical feel or perception of why using different scales to non-dimensionalize the pressure will produce two different equations, namely the Euler equation and the Stokes equation. Can you explain in some more layman terms? Also, do you have any books that include the discussion on Reynolds scaling and Stokes scaling? I can hardly find any books talking about that.

Last edited: Mar 4, 2007
11. Mar 4, 2007

### hanson

Something suddenly comes to my mind.
By a "well-posted" scale, do you mean it could scale the variable (the pressure) properly like normalizing it or to make it O(1) in the N-S equation?

Say, if we use Reynolds scaling, $$P=P/\rho U^2$$, we still not yet know whehter the term $$\rho U^2$$ could properly scale the pressure. However by substituting into the equation, it reads:

$$\overline{u}\cdot \nabla\overline{u}=-\nabla P+\frac{1}{Re}\Delta\overline{u}$$

And if we assume that the velocity $$\overline{u}$$ and its derivatives are well-scaled in make them O(1), then that means,
$$\overline{u}\cdot \nabla\overline{u}$$ and $$\Delta\overline{u}$$ are O(1).

So, if now, it happens that the Reynolds number is of order greater than 1, then the term $$\frac{1}{Re}\Delta\overline{u}$$can be neglected and hence $$\nabla P\sim O(\overline{u}\cdot \nabla\overline{u})$$. In other words, the pressure has been properly scaled to O(1), like what does to the velocity and derivatives of velocity.

Am I thinking alright?

Last edited: Mar 5, 2007
12. Mar 5, 2007

### Clausius2

That is damn right!. You answered yourself. I happen to not remember any special chapter of a book talking about this stuff deeply. Again I would recommend you to take a look at Batchelor's book. Non dimensionalization is only made via experience and common sense. Sometimes one only needs to reason as a layman and very basically to perform a good one. But therein lays the difficulty: one has to forget about the apparent complexity of the flow and think as a layman when non dimensionalizing, and that is not so easy believe me. We are prone to think things are more complex than they really are.

It is also true that people we work on this stuff, when non dimensionalizing and posing the problem we "place" it in such a way that we feel with some steps in advance how the solution is gonna come automatically. It's kind of cheating, but we do know how to set up the formulation in order to keep the next steps of resolution as easy and coherent as possible. And that is what the Stokes and Reynolds scalings are all about: they give us the problem ready to be expanded in PERTURBATION SERIES. Hhahaha!! It's so funny! (I'm going crazy).

Last edited: Mar 5, 2007
13. Mar 5, 2007

### Clausius2

I'm gonna clarify you what is a Perturbation series. Afterwards I have to go to sleep. Take Stokes scaling resultant momentum equation:

$$Re\overline{u}\cdot \nabla\overline{u}=-\nabla P+\Delta\overline{u}$$

This scaling is well posed if $$Re\ll 1$$. Imagine that I want to solve that equation defining the next series solution in terms of the velocity field and pressure:

$$\overline{u}=\sum_{k=0}^N \overline{u}_k Re^k$$
$$P=\sum_{k=0}^N P_k Re^k$$

That is called an Asymptotic Expansion of the Poincare type of the velocity, a power expansion in the small Re. By plugging in this series in the momentum equation and collecting terms of the same Re power, I obtain:

To leading order $$O(Re^0)$$:
$$\nabla P_o+\Delta\overline{u}_o=0$$ (Stokes Equation)

To second order $$O(Re^1)$$:
$$\overline{u}_o\cdot \nabla\overline{u}_o=-\nabla P_1+\Delta\overline{u}_1$$

And so on. This scheme is called a PERTURBATION SCHEME. Unfortunately things are not so easy, and even though $$Re\ll1$$ the above series solution is divergent and does not represent an uniform approximation of the solution. But it can give you an idea how things going on after a non dimensionalization from the point of view of an analytical approach to the problem.

14. Mar 6, 2007

### hanson

Clausius2, could you please explain in more detail about stokeslet? "That term represents the flow field caused by a point force acting in a viscous fluid." Well....I have a problem in imagining it physiclly. Do you mean a stokeslet is the flow field that caused by a point force? (suppose there is not any object in the flow, no sphere, just the point force?) What physically will this point force do to the flow?

And...what is the motivation of adding that point force?

15. Mar 6, 2007

### Clausius2

I'm not gonna come out with the mathematical theory here, but you might realise by yourself that any body, no matter its shape, is viewed from far far away as a point force when moving through a viscous fluid. From your studies on PDEs and Green's Functions you should be able to understand the implications of my statement. The point force is a fundamental solution of viscous flow.

16. Mar 7, 2007

### hanson

Clausius2, do you mean, in general, for an object of arbitary shapes, the force acting on the object or the force acting on the fluid by the object can be approximated by a point force if we are far enough away from the object? (so that it the force can be view as a point force?)

That means in general, when I have an arbitary-shaped object and I incorporate a stokeslet to take care of the drag force, the solution or the flow field I obtained by doing so is not exactly true when we are "near" the object but is "more exactly true" when we are far away from the object?

And, for arbitary-shaped sphere, if we want to have an exact solution that is true for all domain, we cannot just add a stokeslet at the center of the sphere? Rather, we shall something like a function that describe the profile of the object and integrate with the "point force solution"...? Is it like the relationship between delta function and green's function? (Sorry that I did not formally learn these things)

Also, for the special case when the object being concerned is a sphere, it happens that solution by incoporating a stokeslet at the center of the sphere is exactly true no matter near or far away from the sphere, is it? In this case, it is due to the perfect symmetry of the sphere so that we don't need to perform other treatment of the fundamental solution? Please help.

Last edited: Mar 7, 2007
17. Mar 8, 2007

### Clausius2

Basically yes, but only for axisymmetric shapes, like a sphere or an ellipsoid moving parallel to its major axis. It is not that the force can be viewed as a point force, the force is just a number or a vector. It is said that the flow field- caused by the motion of the body through a fluid at rest-at large distances far from the body can be considered asymptotically similar to that caused by the motion of a point force.

Well, take the example I gave to you with that Stream Function, go to the page before this one. At small distances, who is larger? The doublet or the stokeslet, the doublet is dominant isn't it?. The contrary happens at large distances. The solution is exact though, and exactly true over the entire domain.

That makes no sense. Nobody adds anything except in potential flow where everybody seems to know how to locate dipoles, sources and sinks for building a solution without solving the laplace equation. The solutions of viscous flows are not so trivial. The stokeslet will automatically emerge from your analysis in these kind of problems, in which an axisymmetric body moves through a fluid at rest at infinity.