## You are here

# New theory of elasticity & deformation

## Primary tabs

Starting with a few questions which I asked in my introductory class 30 years ago at UC Davis, and which were never answered, I found enough reasons over the years to reject the current theory of elasticity, stress and continuum mechanics entirely.

One of the reasons was certainly that my first exposure to deformation was plastic simple shear. For years I searched for a way to understand the structures in shear zones, but did not get anywhere. Just as disconcerting were certain concepts in the theory which I had to accept due to grade pressure, but they were not at all logical to me. At the same time I spent six years studying applied mathematics, but there I had no such difficulties. Eventually it dawned on me that the conventional theory of continuum mechanics (henceforth **CM**) fails systematically and completely for simple shear – energetics, structures, kinematics, brittle, elastic, viscous and plastic.

However, asking new questions resulted in the stoniest of silence; if you folks think you are open to discussion, I know better. Just try this one: who proved that strain is a thermodynamic state function? The truth is, the question has never been discussed. The truth is also, strain is not a state function – that is: a term that fully describes the energetic state of the system – but displacement is. But the difference becomes obvious only in the study of deformation with less than orthorhombic properties, that is: simple shear.

In 1986 I began to go my own ways, in the fall of 1991 I worked out the theory. Published are now

(1) Koenemann FH (2001) Cauchy stress in mass distributions. Zeitschrift für angewandte Mathematik & Mechanik **81**, suppl.2, pp.S309-S310 (argument fully repeated in #3)

(2) Koenemann FH (2001) Unorthodox thoughts about deformation, elasticity, and stress. Zeitschrift für Naturforschung **56a**, 794-808

(3) Koenemann FH (2008) On the systematics of energetic terms in continuum mechanics, and a note on Gibbs (1877). International Journal of Modern Physics B **22**, 4863-4876, DOI 10.1142/S0217979208049078

(4) Koenemann FH (2008) An approach to deformation theory based on thermodynamic principles. International Journal of Modern Physics B **22**, 2617-2673, DOI 10.1142/S021797920803985X

(5) Koenemann FH (2008) Linear elasticity and potential theory: a comment on Gurtin (1972). International Journal of Modern Physics B **22**, 5035-5039, DOI 10.1142/S0217979208049224

Regarding the theory based on Euler & Cauchy, I maintain

- that the Cauchy stress tensor does not exist; its derivation is flawed because the limit in Cauchy's continuity approach does not exist (1, 3);

- that the theory of elasticity is profoundly incompatible with the theory of potentials (2, 3), but the latter is the theoretical backbone of all of classical physics;

- that the form of the First Law of thermodynamics as it is used in CM is in fact an emasculation of the First Law as it is commonly understood; it is invalid, it is not the First Law (3);

- that the current theory of elasticity is a perpetuum mobile theory: the predicted magnitude of the work done in a volume-neutral deformation is always zero (3);

- that the current theories of stress and deformation cannot possibly be right because there is no mention of bonds in the theory; hence half the acting forces have been left out of the equilibrium condition (3).

- For all these reasons, conventional continuum mechanics is not a field theory in the mathematical sense.

It is my firm conviction that the foundations of CM are hopelessly obsolete for at least 150 years. The entire field should have been started all over again after 1847 on the basis of the First Law, because elasticity (and hence stress) is by nature a change of state in the sense of the First Law, and the old concepts by Euler & Cauchy cannot be reconciled with it. To name just one example: it has escaped the community entirely that Clausius (1870) published a law that can reasonably be considered a modern counter proposition to the Navier-Stokes equations, the virial law. Ignoring that law has left CM on a dead track ever since.

My theory (4) has nothing in common with the theory used so far. It starts with the First Law. The equation of state

- is transformed from scalar form into vector form (which is, in fact, Clausius' virial law);

- is generalized to be applicable to all solids, (akin to the theory of Grueneisen 1908);

- is expanded to consider both work done by normal forces and by shear forces: the work term in scalar thermodynamics is PdV; in my approach, it becomes fdr where f stands for force, and r stands for the radius of the thermodynamic system.

My theory makes no a priori assumptions such as "incompressibility" (which does not exist in nature), but volume-constancy is derived as a result.

There is a large number of so far enigmatic phenomena for which I believe to have found a solution, especially the properties of elastic and plastic simple shear, including the Poynting effect. At the transition from elastic-reversible to irreversible behavior my approach predicts the existence of a bifurcation: the elastically loaded state is irreversibly relaxed into one of two possible lower states, but they are handed. This bifurcation gives very exactly the orientation of joints, and it is in my eyes an excellent candidate for the origin of turbulence in viscous flow.

Since mid-Feb 09 I am aware of a discussion here on iMechanica of my papers eight months ago, but nobody had bothered to let me know. So, to turn up the heat I open this blog. It is high time for public discussion.

On my homepage <www.elastic-plastic.de > are ready for download: preprint PDFs of all papers listed above; links to the journal sites if available; and English translations of Clausius (1870) and Grüneisen (1908).

Falk H. Koenemann

10 March 09

- Falk H. Koenemann's blog
- Log in or register to post comments
- 87999 reads

## Comments

## Dear Sir, You

Dear Sir,

You definitely have a lot of information and disagreement with how continuum mechanics has been developed and taught. Since I'm a young researcher (and perhaps you know more than I) let's just keep things simple and start with one single issue you have in your papers. The following is a quote extracted from one of your papers.

"I have published mathematical proof showing that the derivation of the stress tensor is so deeply

flawed that this term in effect does not exist (Koenemann 2001a), plus an in-depth critique of

deformation theory (Koenemann 2001b). But then I ran into resistance. All attempts to publish

my own approach failed – editors suddenly realized that their journal was "unfortunately

2

unsuited" for my topic, or they "could not find a reviewer". One reviewer took 18 months to

reject my paper – without review. When the president of the society that runs the journal asked

him to provide reasons, he claimed not to understand the relevance of a particular mathematical

argument.

That argument is the Gauss divergence theorem which is known to anyone who survived

Calculus 101. Incompatibility with this fundamental identity is commonly accepted as necessary

and sufficient proof of an error. I use the theorem to show that the stress tensor does not exist. No

argued objection has ever been offered. The answer is silence."

I would be interested in seeing the proof that the stress tensor does not exist. Please keep it simple, since this should be self evident from "Calculus 101" as you say. How well you illustrate this result will help me decide if further reading of your work is worthwhile

Thanks for your consideration to respond,

Louie

## New theory of elasticity & deformation

You may all wish to look at the discussion (July 08) on this, that Falk Koenemann refers to, which is here:

http://www.imechanica.org/node/3570

Charles

## Cauchy's tensor

Louie,

whether you read my papers or not is up to you. However, the argument re the non-existence of the Cauchy stress tensor has now passed review twice, in 2001 (ZAMM) and in 2008 (IJMPB).

I cannot spare you reading the original literature, that's your job; but an argument of max 1/2 page should not be too much. In non-mathematical language, I can express it this way:

In his argument involving the tetrahedron, Cauchy assumed that the correct equilibrium condition is Newton's 3rd Law, and that Newton's definition of pressure P = f/A applies. Thus he implicitly assumed that the planes A are free surfaces (surfaces that divide the universe into left and right). But this is not the case. Cauchy's tetrahedron is a system surrounded by a surrounding, and the surface separating them is closed. On closed surfaces, the ratio f/A must approach infinity as the system vanishes. The limit does not exist. However, the intrnal energy U of the system (an extensive parameter) and the volume V (ditto) must also vanish, but the ratio U/V approaches a finite value. Hence we cannot calculate with P = f/A in a continuum, but we can calculate with P = U/V.

Consider the distinction of system and surrounding. Cauchy inadvertently let the system vanish identically because he let his limit operation reach zero, an he had to do this in order to use Newton's 3rd Law as equilibrium condition. He never considered the thermodynamic equilibrium condition according to which system and surrounding are in equilibrium with one another.

Consider a section along the horizontal line through a system subjected to isotropic pressure:

a-->|<--b.c-->|<--dwhere

aanddare external forces acting on the system,bandcare the forces exerted by the system at the surrounding, the vertical bars are the boundaries of the system, and the period is the system center.Newton's equilibrium condition is the external condition - that

aanddmust balance; they point left and right. If they do not balance, the system is accelerated externally.The thermodynamic equilibrium condition takes a different point of view. Forces

aanddare oriented inward, hence they have the same sign (think in polar or spherical coordinates). Say, they are positive, because they tend to increase the pressure. Thus forcesbandcare oriented outward, they are negative. You can integrate the external forces like you would sweep up matches with your hand around the system; the sum of all forces acting on the system is positive and non-zero. Ditto for the forces exerted by system at surrounding, the sum is non-zero negative. The thermodynamic equilibrium condition Psyst + Psurr = 0 then states that the sum of all inward- and outward-pointing forces is zero. If they do not balance, the system would spontaneously, contract/expand.Newton's 3rd Law and the thermodynamic equilibrium condition cannot be transformed into one another. In order to do sound physics we must consider and observe both equilibrium conditions. I have never seen any mention of the thermodynamic equilibrium condition anywhere in CM; this is due to the fact that Cauchy let the system reach zero. I have explained this in "Unorthodox Thoughts" Chapter 4.

There (Chapter 8) I have also shown that the Cauchy postulate

f-x = -fxviolates elementary rules for vector spaces. These rules ensure that one notation is assigned to only one object (here: vector), and one object can be assigned only one notation, such that the description is unique; and the objects u and -u must be two different objects. But in CM, the vector

nand -n(above: x and -x) both describe the same object, the plane on which -fandfact. The Hesse notation for planes in space works for planes at all points in space, except for the coordinate origin Q which is associated with a zero object, here a null vector. That rule is violated.Falk

## Dear Mr Koenemann, I

Dear Mr Koenemann,

I respectfully suggest the following:

1. If you have claimed that current continuum mechanics (CM) as it is presented is wrong and we know that it is often accepted in journal articles, then

2. You cannot use acceptance of your journal articles as proof that they are correct. Because according to you, incorrect journal articles (regarding CM) are accepted all the time.

3. The limit of the Cauchy stress does not approach infinity. If you have pressure P(x,y) on the xy plane. And you integrate it over an area you get force, f=integral (P(x,y) dA). Hence, f is a function of area, f(A). Now take the limit of f(A)/A as A goes to zero. You will get in the limit 0/0. Whenever this happens elementary calculus tells us to use l'Hospital's rule. To do this take the derivative of the top divided by the derivative of the bottom. Doing this at a unique point (xo,yo), to which the area vanishes, we get df/dA=P(xo,yo) and dA/dA=1. Therefore, in the limit you get P(xo,yo)/1, which is the unique stress at point (xo,yo). This is the stress σ, in the limit, to which Cauchy refers. The stress does not approach infinity (unless P(xo,yo)=∞).

For the above reasons it is difficult to see where you are coming from with your statement regarding the nonexistence of the Cauchy stress.

sincerely,

Louie

## Gauss divergence theorem

Louie,

it would help if you look at the argument in question, see www.elastic-plastic.de/gibbs.pdf , chapter 5 "Refutation of Cauchy Stress"; and it would help if you understand the Gauss divergence theorem which is commonly taught in intro calculus. (At least that's where I learned it.) What you say is correct on free planes, and it is correct on closed planes if a part DA (D stands for Delta) approaches zero - while the dimension of the volume enclosed by the closed plane is not itself subject to the limit operation. But in Cauchy's theory the volume, the surface, and the effect exerted on its surrounding all approach zero. If you are familiar with the Gauss divergence theorem, you can derive the laws f ~ r inside masses and f ~ 1/r2 outside masses from the same principle.

Falk

## on the definition of "continuum"

Dear Falk,

.

In your first paper you seem to disagree with the assumption "that the force magnitude |f| is independent of scale." Is this not the very basic assumption of CM which has been experimentally verified? It is certainly not absolutely correct in the sense that, taken a small enough sample on the microscale level, in addition to accounting for size scaling, the force will be a function of absolute sample size, but for large enough samples, this holds experimentally. If you pull one tensile specimen and record the force vs displacement curve, and pull another specimen with double the thickness and record the force vs displacement curve the ratio of forces will be equal to the ratio of thicknesses. This has been well established for sufficiently large samples in which the grain size is much smaller than any other dimension. In other words, the proportionality of mass and potential is valid for sufficiently large systems.

.

There are enhancements of CM which include surface energy terms which are in some manner related to the curvature of the surface. In these enhanced theories, the stress field does indeed become unbounded as the volume vanishes. Thus, it they are in some way consistent with your views. But the influence of the surface energy term vanishes when the samples are large, and we are left with the original continuum assumption.

.

I would be interested to hear your thoughts on the subject.

.

With regards to your claim that "The Cauchy lemma above assumes that a finite value is reached as V vanishes; however, the limit does not exist:" Your argument comparing continuum to a planetary system is tenuous, as planetary systems are not really a continuum. This coupled with the V /propto r^3 and A /propto r^2 arguments (here I ask: volume and area of *what* exactly?), then assuming the system response is linear wrt changes in r destroys your arguments. In this manner, you have assumed a constitutive equation, thus your results are only applicable for your particular choice of a constituve equation.

.

Nevertheless, there are potential functions which can yield a finite stress as the volume of the body vanishes. For example, the often-used Kirchoff-Saint Venant model W = 1/2*E:EE:E is such a beast. For zero volume (J=0), E assumes a finite value. Since EE is bounded, this obviously predicts a finite potential W for zero volume. The problem here is not with CM but with the particular constitutive equation. It is simply not useful for large deformation. An alternative one (only considering volume changes) could be W = ln(J)*(J+1). In this equation, the potential is finite both as J goes to 0 and as J goes to infinity. Thus, these two states are never reached. The same argument applies to your derivation. If your constitutive equation allows a J=0 state to exist, it's a bad constitutive equation, and it's your fault for using it. But because it doesn't work is no way a reflection on whether or not CM is valid.

.

Regards,

Tim Kostka

## Continuum and boundary conditions

Dear Tim,

"in your first paper": You are right: the proportionality of mass and potential is scale-independent in a given state. But if so, the forces exerted by the system must be proportional to the radius. The magnitude of the force of gravity within the Earth is the oldest example of this law (Poisson 1813).

If I have problems with your first paragraph, it is something else. You think of the bar that is stretched, and its dimensions, you change the thickness, but not the length; you think in terms that you can measure. I have come to the conclusion that in CM we have two sets of external boundary conditions which we must deal with separately: (a) the force field that acts uon some small volume of solid within a larger body, exerted by its surrounding; and (b) the shape of the discrete body within which the system is, because the spatial extension of bonded matter causes gradients for the boundary conditions in (a). It matters if your bars have square or rectangular cross section.

Frankly, I think that so far the analysis of deformation has been incredibly sloppy. Consider an infinite continuum (2D for simplicity). Within this continuum a system of spherical shape is shortened in Y. If boundary conditions permit, it will bulge in X. Apart from that, the surrounding also pulls in X such that the system surface point on the X axis is under external tension, causing a further extension in X. At the exterior surface of the sample the second component must reach zero, but not the first. That is, there must be a deformation gradient near free surfaces. In conventional CM the standard question asks for the relation of stress to strain. I don't know what to do with this question; the stress tensor does not exist, and strain - which strain? The one in the interior, or on the free surface, or bulk? And why is it impossible to raise that question?

For that reason I have started with an infinite continuum, subject to gradient-free external boundary conditions, and no free-surface-effects are felt; at the end of my paper I show how the latter are to be considered in the deformation of a discrete body with surfaces to freespace.

"With regard to": I have trouble with your shortcuts (wrt?). So I don't know if I answer your question correctly. Your Q: volume and area of *what* exactly? A: Volume and surface of the thermodynamic system in the solid. Clearly, system shape is not free to choose in this case. I assume that the shape of the system represents the material properties. - I do not understand what you mean by linear response - the scale dependence of f with r at constant state, or Hooke's Law (at changing state)? In the second case: I think Hooke's law is phenomenological. Elasticity is a genuine part of thermodynamics; all thermodynamic potentials are logarithmic, cf. PdV; hence an elastic law must be logarithmic by nature. (To realize this at the short range that is available to the experimenter before failure may be hard. But it is very important because it tremendously affects the mathematical structure of the theory.) See "Unorthodox Thoughts ".

"Nevertheless": a theory that assumes a finite value for stress for zero volume is by definition wrong. It violates the most fundamental existence theorem of potential theory (eqn.10 in my ZAMM paper; www:elastic-plastic.de/kellogg147.jpg gives you a page of Kellogg's book, see Lemma 1) according to which a system with volume 0 cannot work on its surrounding and vice versa.

I consider my theory a generalization of thermodynamics. Thermodynamics is commonly taught in scalar form, I consider it valid for isotropic external and material property conditions. CM so far ignored the virial law of Clausius. With its help it is possible - this is my proposition - to develop a theory of thermodynamics in vector field form such that anisotropic loading and material conditions can be properly considered.

Falk

## Regarding the existence of Cauchy stress tensor

I apologize for my previous post. Somehow it

was left blank. Here are my thoughts

regarding the existence of Cauchy stress and continuum mechanics.

First I assume the discussion is just

philosophical since otherwise arguing whether Cauchy stress tensor exists or

not is pointless. Based on that I

certainly agree with Mr Koenemann that stress does not exist and therefore

somebody could argue that continuum mechanics is as a consequence an

illed-conditioned model.

From my engineering point of view I have always

understood and treated stress as a

mathematicalwhich cannot even be directly measured in the lab, it is as “dirty”concept

as the concept of mathematical limit of a function and perhaps it even violates

physics at some scales. However it is

still very useful when dealing with systems composed of an infinite number of

particles. Therefore for us engineers,

CM´s is just a model that works pretty well in most scales of interest and like

any other model proposed to understand nature it has its bounds in accuracy and

range of applicability. Therefore I

consider CM´s as a very efficient (this is proved by facts) engineering tool

although it is based on a non-existing mathematical concept.

To clarify my ideas let me present a simple

example. We all know that with respect

to classical CM´s theory there are more refined proposals. For instance, and just to mention a very

popular one there is the so called Cosserat and Cosserat (1904) couple stress

theory. This theory essentially maintains

within the formulation the couples per unit surface even at the level of a

material point as opposed to Cauchy who assumed this value to go to zero in the

limit. Of course someone could argue

that Cosserat theory gives more accurate results. This may be true but probably an unnecessary

burden in most engineering applications.

Consider for instance the problem of a typical beam found in a civil

engineering application and where the rotational modes are very strong. If this problem is tackled using a refined

Cosserat model and a classical Cauchy stress model there will be no difference

in the results. However if this same

beam exists for instance in a microelectronics package at a very small

size-scale probably the accuracy of the classical model will start to collapse and

more refined certainly phenomenological model is justified. So finally everything comes up to be a matter

of recognizing CM´s as a model which works well for the engineering community.

I will greatly appreciate if I receive your

input with regard to these ideas.

Thanks

JUAN GOMEZ

PhD Computational Mechanics

Professor

Applied Mechanics Group

EAFIT University

Medellin,Colombia

## Existence & Cosserat

Dear Juan,

there may be misunderstandings regarding "existence". It has nothing philosophical about it. "Proof of existence" means in mathematcis and physics to show that a certain proposition, term or argument is in logical connection with the rest of physics, such that contradictory situations cannot arise. If Isay the Cauchy stress tensor does not exist, I say that the argument by which it is derived is invalid, because it violates more profound logical-physical concepts. The concept that is being violated here is the First Law of thermodynamics, and the distinction of conservative and non-conservative physics.

Most people understand under "stress" the mechanical or physical state in which a solid under load is, forces etc. This is certainly real. But the tool by which we so far tried to understand it - the Cauchy tensor - is not valid. We must find new concepts. I have developed a vector field theory.

Re Cosserat: One of those who introduced me to CM had his PhD from Eringen. The Cosserat theory shares with the standard theory the disregard of the thermodynamic system, and it is in conflict with the First Law of thermodynamics and the virial law of Clausius . These are the questions that must be adressed. Besides, there simply are no free rotations in a bonded continuum (this side of the yield point), and theories that imply free rotations cannot be right.

Before anything else, elasticity is a genuine part of thermodynamics, in the sense that elastic work is akin to PdV-work. This aspect of the nature of elasticity is not at all apparent in the common CM theory.

Falk

## Applying the new theory to discrete bodies

Being a mechanical engineer, the most interesting aspect of Koenemann's new theory to me is his profound critique against the absence of zero potential distance in Euler-Cauchy theory and its consequences (see 'An Approach to Deformation Theory based on Thermodynamic Principles' published in IJMPB for full account of this critique).The reduction of a volume element with surface facets into Euler's group of planes passing through one common point is a fundamental principle of Cauchy's continuity approach and what is ignored here is the measure of spatial extent of the volume element i.e. zero potential distance. The whole idea of FEM in a way is to re-invent this distance and feed it back to theory by using a grid of predetermined nodes for which the solution can be found. The distance between the nodes now acts as a local unit distance. Koenemann's theory explains how reference to this distance was lost in the process of deriving the Cauchy Stress Tensor resulting in an inherent inconsistency between classic continuum mechanics and potential theory and the reason why we need to use mass body based formulation at predetermined discrete nodes to extract results in the absence of a proper field theory which can give a closed form analytic solution for the whole domain. Indeed, he has proposed a new solution to deformation based on field theory.

Today, FEM, CFD and other numerical techniques applied to solve continuum mechanics problems dominate our academic departments to the extent that you can hardly find any scholar bothering themselves to go back to basics and check the fundamentals with a critical mind. If you review the state of the art papers published in this field they are mainly numerical or experimental papers. Surprisingly, it's sometimes even difficult to find people with proper background research and sound understanding of the theoretical principles of continuum mechanics among the academic staff of mechanical engineering departments. I have come across funny titles in the curriculum of so called world class universities such as "computational continuum mechanics" without any module to cover continuum theory itself! Computational continuum has become the theory itself and nobody cares about the origins. They keep saying as long as it works that's fine! No mention of the complexities, CPU time, likelihood of new developments with a proper field theory and total ignorance of the history of science. After all, Newtonian Physics is still capable of solving most of our problems. At least this was certainly the case 80 years ago. So, as long as it works that's fine! No need for Albert Einstein; no time for him actually...

It's a shame to say we are so much obsessed with what we call "state of the art research", number of publications, grants & funds that we are willing to totally ignore any new brave idea in our fields and would rather stick to our traditional ways of thought. I would like to take this opportunity to say thank you to Falk Koenemann for his breakthrough research and invite every mechanical engineer to check his work for themselves and enjoy the immense intellectual and scientific challenge these papers offer.

## Stress vs. potential theory

Dear Ajit,

to other readers: this is a discussion that started on Ajit's blog .

Regarding your questions:

(2.1.) He says in his paper that "div f is a measure of the work done by/upon a system."

See here, p.52 in Kellogg (1929) Foundations of potential theory, Springer Verlag. I understand your reservations farther down, but I did not say, and Kellogg did not say that div

fis the work term itself, just a measure of work. divfis the source density or charge density, and int divfdV is the charge (int = integral sign).Having said that, I think Kellogg's statement must be taken with caution. To make things simple, I prefer to restrict the discussion for the moment to isotropic conditions. I think that Kellogg's statement is not universally valid within bonded mass distributions, but this is not the right moment yet to discuss details. And do not take me as simple-minded as you appear to have taken me in the July 08 discussion (which I became aware of only in Feb 09 because no one bothered to contact me, including you).

(3.0) In another paper: "General solutions for the Poisson equation exist only for reversible processes, e.g. the Helmholtz equation." [sic]

That's what I learned. Apart from that I have my own opinion. The thermodynamic equilibrium condition states that two sources of forces interact, the external and the internal one, such that P_syst + P_surr = 0. This implies in my view that both system and surrounding act as a sources of forces, and they serve to one another as sinks; hence there are two Poisson equations in one equilibrium equation, because this eqn can also be written div

f_syst + divf_surr = 0 (for isotropic conditions, n.b.). They balance only if there are no inequality terms, i.e. Second-Law effects, a.k.a. entropy. Hence the eqn is reversible. That's elasticity.Very much apart from that, your distaste for my writing is misplaced, whether I am right or wrong. I have fought for 20 years for discussion, and have experienced very literally the most massive and long-lasting excommunication from the moment on I let it be known that I differ. The reasons never mattered, it has simply been impossible to discuss the foundations of CM. In my experience,

CM is like a faith or a church.While I disagree with your assessment of my work in the discussion on Biswajit's blog, I gathered that you are a thoughtful man. That's why I placed my comment on your blog.(i) Why do you say that a characterization of mechanical response of solids from a thermodynamic perspective is absent in the traditional theory (i.e. all solid mechanics than what you say of it)? I can clearly see that such a description exists. What makes you say that it doesn't?

I cannot see the faintest trace of thermodynamic thinking in continuum mechanics, starting with the assumption that elasticity is linear (Hookean). Thermodynamic potentials are logarithmic. Thermodynamics distinguishes system and surrounding, Cauchy let the system vanish identically. I have never ever seen any mention of the thermodynamic equilibrium condition in CM. Just the idea to start a theory of elasticity with an equation of motion is un-thermodynamic. I could go on ranting, but it is all published . What I do see in the literature is a mix of conservative and nonconservative terms as if they belong together. They don't. They are wolds apart.

I have shown - in Systematics & Gibbs , Chapter 2 - that the form of the First Law of thermodynamics as it is used in CM is invalid, and that as a consequence, the work done in a volume-neutral elastic deformation is always zero (Chapter 4). The path from E_kin to E_pot = const to dU = dw + dq is the virial law of Clausius who interpreted

E_kin + E_pot = const

as

mv^2 + fr = PV

Note that only the first term LHS is accessible for an equation of motion, but Clausius interpreted this as the _heat_: it is the free vibration of the individual atoms. The second term LHS are the bonds in a solid. The term RHS is the total energy of the system in the standard state. In case of a change of state, evidently the RHS becomes a variable, and the LHS is no longer interesting because now we consider exchange of energy between system and surrounding. Thus I see no way to start an elastic deformation with an equation of motion under the rule of E_kin + E_pot = const. The key to elasticity is PV = nRT and dU = dq + dq. For the rest, see my theory.

(ii) What part of Cauchy's definition lacks thermodynamic implications?

Cauchy's tetrahedron is effectively a thermodynamic system. But then he lets it vanish identically. There was no mention ever of the thermodynamic equilibrium condition in any of the many derivations of the Cauchy tensor I have seen; it was never considered. In potential theory it is stated that for logarithmic problems (such as thermodynamics) one must define a zero potential state by convention because they do not have a natural zero point. That means - as it is done in thermodynamics - that a system with unit mass and unit size is chosen (n and V are finite in PV = nRT). It cannot vanish. Its dimension can be taken as the zero potential distance d_0 of potential theory. It was Cauchy's error to let d_0 vanish; it is required for the definition of work. - In fact, this d_0 is in Hooke's law the length of the spring. Cauchy let it vanish.

(iii) In your theory, do you believe that stress remains a 2nd rank tensor quantity?

No.

If not, what are the physical units or dimensions of stress according to your?

I have written a force field theory (in [Newton], and not in [Pascal]). There are two fields: the external one that is controlled by the external boundary conditions; and the internal one which is subject to the material properties (compressibility, anisotropy, orientation thereof, etc.). The two fields are independent of one another (in origin, and in properties, but yes, they are controlled by field property tensors); but they are brought into equilibrium by the condition that system and surrounding are solidly bonded to one another, hence disequilibrium within a solid (below the yield point) is impossible. The two independent force fields result in a third one that combines their properties. To calculate the displacement the work function fdr is used in the same way as PdV is used in standard thermodynamics (f and r as in the virial law above). The details of that are explained in my paper, but be assured that no Joules vanish.

Can you give component-wise derivation as to how such a quantity is defined after starting with the thermodynamic laws?

Very simple: f = dU/dx. That's the only definition of a force field known to me. The fact that conventional CM does _not_ use this definition is enough evidence to show that conventional CM is not a proper field theory. (in fact, it uses

f= maexclusively, which is by nature _not_ a field force.)And, why should anyone begin using such a "stress"?

I do not use the term "stress", simply because everybody would instantly take it as an orthogonal tensor in the sense of Cauchy. My theory is a vector field theory. I use the term "properties of the loaded state". Its eigendirections may be non-orthogonal. The reason for using it is: the rationale of Cauchy's tensor is simply invalid, the term does not exist; if you disagree I expect you to refute my argument in Systematics & Gibbs , Chapter 5.

Or better, everybody is invited.Hint: no one, from my first discussion partners in 1986 to the last one two weeks ago , could do it. The argument is of a quality that cannot be refuted. It is time to acknowledge this. Thus, whatever you do, you are in need of a new theory.Apart from that, I suggest you read my Approach paper. There are enough demonstrations that my theory delivers excellent results. - Of course, satisfactory predictions do not prove that a theory is right. But I offer solutions to problems that have been considered intractable using the conventional theory, - such as turbulence in viscous flow - so it might be worthwhile looking at it.

Falk H. Koenemann

## My final reply to Falk

Dear Falk,

(1)If you do believe that stress as a second rank tensor is not the right quantity to address solid-mechanical problems, I believe that almost 95% (or more) of what you have to say regarding solid mechanics isn't going to be helpful to anyone---either to you or to others.The issue isn't so much what is true or false, primarily, here. The issue now is: the causally mandated hierarchical order ...

... If you do not agree with the very basics, and know this for yourself, you are only confusing your reader if you do not directly tell him this basic fact at the outset, and instead, tell him a mix of many advanced notions, apparently jumping from one topic to another in no apparent order.

All this is, of course, assuming that you have something valid to say in the first place. Assuming that, a lack of communication would occur primarily because the disagreement on the fundamentals had not been spelt out forthright, and also the difference indicated at each step in a

systematic and hierarchicalmanner.In the light of your denial of the idea that solid mechanics requires a second rank tensor for a quantity of stress, frankly, I don't think that you have anything of importance to offer me.(2)

Nevertheless, some of your statements tell me that I should explain in greater detail my earlier advice. This, I do in point (3) below:

(3)In this point, I am not necessarily assuming that you have something valid to say. What I am actually doing here is to put down some (really) off-the-cuff remarks about how someone with a new and valid theory to replace SM or CM should go about doing the job of explaining his theory to others.(3.i)Bring some order to your fundamental or basic concepts and state them directly. This would be necessary first and foremost. For example, tell the world that they would be better off using a vector field theory for stress analysis---if that is what you want to tell them (and them, here, includes me).(3.ii)Do not bother writing journal articles. Do not bother addressing advanced cases. And do not at all bother trying to convince other people. All this can follow later, if you so wish. Instead, first do the following.(3.iii)Take a good undergraduate text-book generally followed in good universities. Note, the emphasis is ontextbooks, not reference books, research monographs, etc. Also, the emphasis is on the undergraduate curricula, not the graduate ones. Actually, you want books on the introductory or the very first courses. For example, you might take, say, Timoshenko's elasticity; or better still (because they are easier on the absolutely beginning student), Shames/Popov/Beer & Johnston. Or, any other text. Perhaps, Timoshenko itself, if you are like some who claim that way. Even a title from Schaum's Series would do.(3.iv)Carefully isolate a sequence of about 10 or so problems that illustrate the progression of ideas from the simple to the more complex, using these text books. At least 5-6 problems, and not more than 15-20 or so. (Not for the first shot anyway.)(3.v)Write brief descriptions of what physics or engineering situations these illustrative problems address, using, as they do, Cauchy's (shall we say, "religious"?) theory.I mean to say, separate out the actual physical or engineering problem from the approach taken or the theory used to solve them.

For instance, consider the 1D stress, the experiment performed by Hooke. That's where it all began (to my mind, thinking offhand). So, you might take that as one of the typical problems.

(3.vi)Next, solve each of these problems using your, shall we say, "theory"?Please do not tell your reader that you have experimental validation to offer anyway or that you have given it already. Keep it all aside.

What you are doing here is simply to write a self-study book to learn your "theory". You must not engage in polemics---in any way. A tutorial, it will have to be, so to speak....

(3.vii)Thus, apply your theory to a set of representative or core problems from a typical introductory university course in SM (solid mechanics).For instance, take the problem of the tension of a 1D bar---the abstraction for Hooke's experiment. The "Church of Cauchy" solves it by definining 1D stress as sigma = F/A in the appropriate limiting process. We all know it, and presumably you do, too. Yet, do not mention the "Church of Cauchy" at all in your writing. Simply because you are going to replace it with your "better theory," right?. Instead, just state how to model the physically concrete situation of Hooke's experiment using your ideas.

If you do the above, you will be able to show how to obtain a quantitative result that helps solve, at least to the first-order or baseline estimate, some practical design problem---say, of being able to design, a wire-rope holding a weight.

You are going to solve this practical problem using your own theory. If your theory is valid, you should be able to do it. If so, why not do it?

You might want to give the solution in mathematical terms, and then, also illustrate it using an actual example.

(3.viii)Go on doing this for all those carefully selected 10-odd illustrative problems.The problem set might include, say, a few of bending, one or two of torsion, one or two of 2D elasticity, etc. It could be any composition of problems---but they must cover the entire text in a fairly representative way.

Also, make sure to have a variety of boundary conditions---displacements and loads. (Those are the two things that CM takes for granted from basic physics and doesn't define on its own. So, you can start with them rather than with the stress or strain boundary conditions.)

(3.ix)Include ample diagrams, photographs, etc. In terms of such illustrations and the simplicity of the accompanying text, personally, I find Beer and Johnston's introductory texts to be one of the best. (Though, their selection of topics, their order, emphasis etc. doesn't always fully meet my other criteria.)(3.x)Finally, also include a brief paragraph showing conceptual interconnections or parallels between your theoretical constructs/concepts/methods/algorithms and those of solid mechanics as it exists today.Remember, this last step also is necessary---though, it can be of secondary importance.

And, remember, it can always be done without keeping polemics at the fore-front.

(3.xi)Once you have prepared such a tutorial, let others know, say, by posting it on your Web site.(3.xii)You can then let us know here at iMechanica that you have updated your site.(4)If you do the above, I am sure that you will get far more positive response than what, apparently, your course of 20 years has brought you. (And I wonder what you have done or do for living, though you need not answer me.)In any case, I believe something like what I have given in point no. (3) above is what is going to be necessary for anyone before you begin to publish research articles in journals about it...

(5)Chances are very high that the availability of such a tutorial would, first of all, interest not the research community but the practical engineers. The latter might not immediately use your theory. But they have no tenure-track etc. considerations clouding/hampering/tunnelling their visions or judgments concerning a new theory. So, they might be---just might be---more open to your "theory." Even if they can't tell in precise words why they support you. (Though, never mistake their support for validation of your theory.)This is what routinely happens in computer science in general and software in particular.

Note, the practical engineers may not pay you. But they will listen to you if you do have valid results (which I think you don't). If so, some will put you in touch with their class-mates in academia. (Often times, practical engineers are actually smarter than even their most awarded/honored academic class-mates are.) Some of these professors might then be a little more sympathetic to your ideas (because, despite all their awards and honors, in the heart of their hearts, these professors also respect the practical engineer if the latter is a smart guy.) Some of these professors might be influential in research and academia. And *that* might begin your research career.

Provided, you have that tutorial to offer anytime anyone asks you for one.

Since this is a public posting, accessed by people who aren't mechanicians/engineers/physical scientists, I want to note: This is not necessarily the only pattern available to someone who has a (valid) new theory to offer. Or the one I would suggest everyone.

But for Falk's case, I cannot think of any route whatsoever other than his putting on the table a tutorial giving parallel examples of most all of Shames/Popov/Beer and Johnston/Etc. Including the treatment of the energy principles contained in the introductory text books such as these.

(6)Let me close now. I think I have already written far beyond what is objectively required of me for a case like this.Only one more point.

Falk, the reason I didn't think it necessary to bother contacting you itself was reasoned. It was based both on my judgment of the contents of what you have written, as well as that picture of the naked hippie-like guy writing equations on the board which you have put up prominently on your site... (Just so media doesn't begin something after reading this: The issue isn't nudity per say; Michelangelo's David, too, is nude. The main point, here, is that hippie-like thing about it.... Falk may have his freedom of expression; but I too have my freedom of forming my own judgments. If he cannot care too much for the common civic standards of society, I cannot care too much for common etiquettes such as keeping him informed via an email.... Others may have their own reaons for not contacting Falk; this was mine... And, also, many other things---e.g., absence of any other information on the man such as his education, work background, etc.)

Not my idea of what a man capable of dwarfing men like Cauchy would do, especially given today's culture and times.

Indeed, there are many ways in which I believe a man dwarfing Cauchy would act like---and how he would not, in different circumstances... But all that is---and has been---rather secondary. The primary things have been the contents of what you have to offer, and the absence, rather conspicuous absence, of what you don't at all offer---e.g., that tutorial.

(7)Note, the tutorial must be sufficiently detailed, and must have been written with ease of learning as its first objective. It has to be at par with the better (if not the best) of the undergraduate text-books of solid mechanics as followed in universities today. Good examples of good tutorials abound; e.g. have a look at this one.If you create a comprehensive tutorial for your theory to address and solve the typical problems as are covered by the introductory courses in solid mechanics, and if you leave a message about the availability of such a tutorial at, say, iMechanica, then I will sure come to know about it, and I will sure take a look at it.

(8)I will not be interested in your work---or in having a correspondence with you---until then. I suppose I could go a step further and say that I do not think that you are going to come back with that tutorial---ever! (Go ahead, prove me wrong!)This is my final judgment about this entire matter, and I mean it.

Good bye.

## conservative and nonconservative physics

Dear Ajit,

what you do is bible-slapping. Physics is this:

1. processes that take place under the rule of E_kin + E_pot = const are called

conservativebecause the total energy of the kinetic system is "conserved", i.e. it is constant. Examples: revolution of planets in solar systems, most discrete body mechanical problems (friction-free). This is Newtonian mechanics. Such processes take place in an isolated system without any contribution of a surrounding.2. Processes that change the total energy of a system require exchange between system and surrounding, be it in the form of heat or work. They take place under the rule of the First Law, dU = dw + dq. Such processes are called

nonconservativebecause the energy is changed, from U_0 to U_1. Example: pressure increase due to volume change.The second group can be split into the subgroups of the

reversibleand theirreversibleprocesses.This is the outline of classical physics. Elasticity is a

nonconservative reversibleprocess that takes place due to interaction of system and surrounding such that its internal energy is changed, and an elastic potential builds up. Hence it must be approached through an equation of state and the First Law. It means that the most primitive elastic law known to us is PV = const - for Boyle's "spring of the air". For solids we must find a better equation of state (I have done that).The entire mathematical structure of conventional CM is only and exclusively in line with a conservative theory in the sense of E_kin + E_pot = const. By nature, elasticity falls into the category of the nonconservative, yet reversible class. Nothing in the current theory suggests that it is so. Any theory that violates the systematics above is by definition wrong. Ignoring this outline does not change the physical reality.

I have demonstrated threefold that standard textbooks on the conventional theory cogently lead to the conclusion that a volume-neutral elastic deformation does not cost physical work, including Love and Landau & Lifshitz. I notice that neither you nor anyone else made an attempt to argue.

Falk

P.S. The naked scientist: that's not a hippie. Have you never heard the Tale of the Naked Emperor? I took it from Scientific American.

## Mr Koenemann, Ajit has

Mr Koenemann,

Ajit has been very kind to you to bother to explain why it is that you are not being listened to in the mechanics community. I have thought for a while now that I should write something similar to what Ajit wrote. It would help your cause to demonstrate by some simple examples how your theory comes to the same conclusions as continuum mechanics. Then also, you should demonstrate by some simple examples how your theory concludes with some IMPORTANT differences than current continuum mechanics theory. This would help illustrate the validity and benefit of your approach.

This situtation sort of reminds me of the historical account of Einstein and his theory of relativity, which the physics community was resistant to accept (not that I'm comparing you to Einstein). I'm just suggesting by mentioning Einstein's situation that you need to do what he did. He sat himself down and thought about how he could convince people in a demonstrable fashion that his theory was valid and MEANINGFUL. In the end he came up with an astronomy experiment that would illustrate the bending of light as it passed the sun's gravitational field. You may sort of follow his example by doing what Ajit has suggested regarding the simple tutorials.

You complain because we are resistant to listening to your new ideas. And, we are complaining because you are resistant to explaining yourself in an orderly fashion. You would do well to note the many comments about how your papers jump around and the organization is poor.

It might be that you have something useful to say. However, as I often tell my students, your work is worth nothing if no one understands it because of your failure to communicate and lack of organization. In conclusion, it might be that you are on to something but it's difficult to tell and you are not being successful because your public relations skills (or presentation of your work in your papers) are not so good

good luck

Louie

p.s. Later I'm going to follow up on a quote from your comments "I cannot see the faintest trace of thermodynamic thinking in continuum mechanics..." This seems to be an odd statement. I will look for some references on this, because my memory suggests that I have seen many continuum mechanics texts talk about thermodynamics.

## Mr Koenemann, Regarding

Mr Koenemann,

Regarding the following quote from your comments above.

"I cannot see the faintest trace of thermodynamic thinking in continuum

mechanics, starting with the assumption that elasticity is linear

(Hookean). Thermodynamic potentials are logarithmic. Thermodynamics

distinguishes system and surrounding, Cauchy let the system vanish

identically. I have never ever seen any mention of the thermodynamic

equilibrium condition in CM."

It is odd that you have not seen the "faintest trace" of thermodynamic thinking in continuum mechanics. Also, seems to me that the assumption of elasticty being linear is just that, an assumption for very small deformations and is a special case of the overall thermodynamic view and hence probably doesn't warrant a preliminary deduction from thermodynamics since it is a self evident subset from a thermodynamic starting point. I encourage you to look at the following literature.

See the following for continuum mechanics and thermodynamics

1. Eringen, A. C., Mechanics of Continua, Krieger Publishing, 2nd ed, 1980

(see chapter 4, "Thermodynamics of continuous media")

2. Truesdell, Clifford et al, "The non-linear field theories of mechanics"

(see page 294)

3. See the journal "Continuum Mechanics and Thermodynamics", by Springer.

4. See symposium book "Advances in continuum mechanics and thermodynamics of material behavior"

5. Green and Adkins, "Large Elastic Deformations (and nonlinear continuum mechanics",

Oxford, Clarendon Press, 1960. (see chapter VIII "Thermodynamics of Deformation")

6. Bazant, Zedenek and Cedolin, Luigi, "Stability of structures", Dover, 2003

(see chapter 10"Stability of Inelastic Structures, Bifurcation and Thermodynamic Basis")

7. Malvern, Lawrence, "Introduction to the mechanics of a continuous medium", Prentic-Hall, 1969.

(see page 226 onwards)

8. Holzapfel, Gerhard, "Nonlinear Solid Mechanics", Wiley, 2000

(see chapter 7, Thermodynamics of Materials)

Also, regarding your comment "I have never ever seen any mention of the thermodynamic

equilibrium condition in CM." see item 7 above starting on page 226, where it talks about energy balance; first law of thermodynamices; energy equation, if you want to see a basic discussion of equilibrium in continuum mechanics from a thermodynamic point of view.

Also, in the last century Truesdell did lots of work on bringing continuum mechanics up to date thermodynamically and the use of the Clausius Duhem inequality.

It remains a mystery how you can make such statements given the small list I've provided above. I'm certain that there is much much more literature available.

regards,

Louie

## conservative-nonconservative

Louie,

thanks for your comments. You ask for evidence that my theory can deliver better results than the old approach, and there is quite a bit, see below. However, I also call conventional CM in question for purely theoretical reasons because a theory without sound foundation isn't worth much.

I asked for the relation between Cauchy stress and the Gauss divergence theorem in my first year in grad school 1980, and to this day it takes only one hand to count those who actually answered. It is the pivotal question. Spare me to talk about the many people who did everything in their power in order to avoid answering, I had literally doors slammed in my face. Now, simply from a technical point: is that request so unreasonable? I asked you. Would you please address that question?

- Is Cauchy's tetrahedron a volume element with a surface?

- Is that surface a free surface or a closed surface?

- So what happens if V approaches zero? Where is the flaw in my argument?

There is none. Cauchy's argument is not valid. (See also Kellogg p.147, Lemma 1.)

You see, the point is that you have to step back farther than you are so far willing to do, in order to gain sufficient overview. The temptation to slip back into all-too familiar paths is very great. The farthest point of view I could find is the distinction of conservative and non-conservative physics (see my answer to Ajit, and let me say: it took me years to find out just how far I had to step back); and suddenly it is evident that elasticity (and thus stress) belongs into the nonconservative class - but that class has been discovered only 70 years after Euler postulated the stress tensor. Only then they learned how to handle this. However, in CM they did not throw the old theory away, but they adapted the First Law to fit on conservative concepts. That was a mistake. Hence the mathematical structure of CM is conservative: it must cogently lead to the conclusion that a volume-neutral deformation does not cost work, as I have shown in three ways in Systematics & Gibbs. I have yet to get a comment on that from you, or anyone else.

You can prove a theory wrong by showing that its basics are insufficient, that it leads to spurious results. Demonstrating that a theory is a perpetuum mobile theory is necessary and sufficient.

Re the literature you mention:

Malvern is a good example: no mention of thermodynamic concepts in the chapter on stress; and then you turn the page to the chapter on strain, and you just ask yourself how he got from one page to the next. The truth is, there is no connection. If you are given a particular state of stress, you cannot say which deformation will arise from it, and vice versa; the theories of stress and strain stand side by side (not just in Malvern, but always). There should be a cause-effect relation, but there is none. This is certainly different in my theory. - Yes, I know that thermodynamic concepts are mentioned later on in the book, but that is the classical case of mixing apples and oranges. Your remark "thermodynamics starting on p.226" is telling: the First Law should be on p.1. I insist that the theory of stress - as given in Malvern or any other book - should be derived from thermodynamic principles, starting with the First Law, and it isn't.

Eringen: I read a book by E. because one of his students was my advisor. He was the first who would not answer why there is no equation of state in the stress theory, why the latter is incompatible with the divergence theorem (see above), and why bonds in solids are never mentioned in CM. Instead of answering, he broke off contact. In Unorthodox Thoughts, Chapter 4, I quote Eringen directly: no mention of bonds, no mention of system and surrounding, and the quoted argument is simply wrong (see there). There is a point where you just stop reading.

Truesdell: at a time when I still had faith in conventional CM, I read Truesdell & Toupin, the entire 600 pages. The absolutely only time the term "field" is used is in the page header on page 1 above the title - never again. This is simply pretense. CM is not a field theory.

"Continuum Mechanics and Thermodynamics": in 1993 its editor Kolumban Hutter wiped the divergence theorem off the blackboard, saying "A discussion of the Gauss divergence theorem is below my intellectual level". Thank you very much, see above. It just means that he does not like to be proven wrong. He simply refused to let me submit to his journal, no reasons given.

The fact remains that we do have a perfectly good deformation theory, tried and tested, well established, carefully founded, 100% compatible with potential theory: it is called thermodynamics. So far it has been developed only for isotropic loading conditions: if you know Delta P and the equation of state, you do not need a strain theory because you can simply calculate the strain from the work function PdV. It is that simple. I propose a theory that does the same for anisotropic conditions, strictly following the mathematical structure of thermodynamics.

"It remains a mystery how you can make such statements given the small list I've provided above." I hope it is clearer now. I don't want thermodynamics to be poured over the theory when all the damage has already been done, but I want it on page 1. And I want to see system and surrounding treated as separately existent and finite, just as in thermodynamics.

The simple examples you want are all there - in my paper. No one can make better predictions for the properties of simple shear - elastic and plastic:

- Poynting effect (that an isotropic material subjected to elastic simple shear should dilate);

- energetics of elastic and plastic, simple and pure shear deformation: it is known that in the elastic field simple shear requires more Joule per chosen strain (strictly in the sense of the strain tensor) than in pure shear (my prediction is +10%), whereas in the plastic field simple shear requires substantially less than pure shear (my prediction of -30% is in line with experiments);

- the entire set of properties of plastic simple shear: geometric structures, kinematics, orientation of cracks and joints, orientation of microfabrics for simple materials (ice, mica, olivine) - it is not just one item, but a coherent overall picture.

My approach predicts a bifurcation at the transition from reversible to irreversible behavior. the geometric and energetic properties of this bifurcation give me

- brittle: precisely the orientation of cracks in 3D,

plastic: a wonderful mechanism why folds in shear zones nucleate with their fold axis in Y, perpendicular to the shear direction X, but then they devevelop into sheath folds which are so characteristic of large shear zones (sheath folds: tubular folds along X, but closed on one side like the finger of a glove; in a YZ section they form elliptic cross sections),

- viscous: everything that is required to understand the generation, and the known geometric properties, of turbulent flow in Couette flow - especially the property that due to flow along an XY plane the eddies develop with their rotation axis close to Z.

Is that enough? - Again, I ask you to shoot the divergence theorem arguement down, and to comment on the zero work predictions for isochoric deformation that follow from the old theory.

Finally, you mention your students, so you are a professor. Let your students read my paper and thrash it if they can.

Cheers,

Falk

## The mythical Helmholtz partition

The standard understanding of simple shear is strongly based on the Helmholtz partition: "a simple shear can be decomposed into a strain and an external rotation". Sorry, no – neither geometrically, nor energetically. Nor would Helmholtz agree to the way the partition is used today.

The partition works only once.It cannot be used in an iterative way. If a strain step is chosen, and the required rotation is found, ok that's a simple shear. But if the new points are fed back into the same equations it does not work any more. If one plays with other points iteratively, it becomes clear that the points follow a transformation matrix in the sense ofAx=b. It is possible to findAfrom the chosen strain and rotation, hence it is possible to find its eigenvectors. The field is incompatible with simple shear.The partition works only if one starts all over again – again and again and again. This is in incompatible with any attempt to develop a physical relation between energy spent and effect achieved.

Furthermore, if other strain steps are chosen, the eigenvectors vary. To anyone who is familiar with eigenvector problems this is an absolute No-No; one cannot manipulate the eigendirections as one pleases – except that it is still impossible to get the eigendirections for simple shear. Eigendirections are non-rotating directions, they are highly characteristic for the field to which they belong.

The eigendirections are non-orthogonal. The smaller the strain step, the more orthogonal the eigendirections become – and the closer they are to the eigendirections of the initial strain step. In the limit the non-orthogonality should vanish – great for those who have

faith in this concept, impossible from a physical point of view. A rotating eigendirection – that's like a roasted icicle.

The partition cannot be expressed in analytic form.The partition of the influence of strain and rotation is non-unique. It is therefore not possible to write a differential approach for simple shear that can be integrated to get the finite deformation. The concept isentirely phenomenological.

The partition is energetically wrong.A rigid-body rotation should not cost extra energy, hence all identical strain states, achieved through pure or simple shear, should cost the same amount of Joules. From the rubber experiments done in the mid-20th C it is perfectly clear that this is not the case – simple shear costs more energy than pure shear.Strong energetic differences are also observed in the plastic field, but there simple shear costs substantially less than pure shear.

This finding has the most profound significance:

strain cannot be a thermodynamic state function.(Elastic field: the data by Rivlin & Treloar are hard to evaluate because they are not given in raw form, but worked through a "stored energy function" of which it is not clear what it really means. But the fact that simple shear costs more energy is clearly apparent. Plastic field: thesis Franssen, Utrecht 1993.)

Helmholtz wrote the paper explicitly for an ideal Newtonian fluid– friction-free, conservative in the sense of E_kin + E_pot = const. Whether this is an acceptable thought today is not the question here (viscous flow is irreversible); but carrying this partition out of blind faith into the mechanics of solids is preposterous. Helmholtz never thought of a solid with strong internal bonds.Apart from that: would please anyone explain to me how a free rotation is supposed to come about in a continuously bonded solid? Isn't this so ridiculously out of the question that it is not worth thinking twice about this partition? A free rotation requires free surfaces around the volume to be rotated. Why is such nonsense being taught to this day?

All these inconsistencies are a clear sign that elastic deformation as a physical process is simply not understood. In Approach Chapter 7 I give more reasons why the current ideas about simple shear cannot be right, and I develop a theory which delivers far more realistic results.

To those who can read German: Helmholtz' original article can be downloaded here , the ref is:

Helmholtz H (1858) Über die Integrale der hydrodynamischen Gleichungen. Crelles Journal/Journal für die reine und angewandte Mathematik

55, 25-55Falk H. Koenemann

## Experimental data

Chad,

I did not make a mistake, you can't read. The quoted paper is entirely correct.

The data fully support my predictions, both in the elastic and in the plastic field: download the experimental data to compare for yourself.

Alright. Please proceed to shoot down my refutation of the Cauchy stress tensor, including a reason why in your opinion Kellogg's Lemma 1 does not apply. Or do you need any help to remember the other points you dodged? I'd be happy to assist you.

Cheers,

Falk

## Pure shear and simple shear are very difficult to impose.

Your interpretation of the experiments in this paper is rather strange. Very few researchers would claim that uniaxial extension is the same as pure shear or that the shearing set-up in these experiments is simple shear. If this is what you wish to claim then we can agree to disagree. However, you will not be able to convince any other scientist of this interpretation either. Simple shear is an especially difficult state of deformation to impose in an experiment on a stiff material.

As for your refutation of the Cauchy stress tensor, your "proof" is simple enough to refute by a counterexample.

You write that the integral over the area of f.n is equal to k. Then you claim that k is proportional to the volume of the region. From what you have written in your posts it seems that f.n is supposed to be the net inward directed force.

Consider a spherical region of a uniformly pressurized gas. Then it is very simple to show that the integral of f.n over the area is 4 pi R^2 p. So clearly the integral is proportional to the area for this example and not the volume. It seems that your proof is flawed.

If you wish to claim that there is something wrong with this counterexample, then simply show us how to compute your integral of f.n over the area for a uniformly pressurized gas. I am sure that we would all like to know the answer to this very basic problem.

## divergence theorem

Chad,

At constant scale you are right. The point is: how does f change if the scale is changed, i.e. if V approaches zero. In the two integrals

int f.n dA = int del.f dV = k,

(int = integral sign)

del.f is a scalar, an intensity term, the source/sink density. That is, both integrals are proportional to mass (here measured in V). If V approaches zero, we know that V ~ r^3, but A ~ r^2; for the entire LHS integral to be proportional to r^3 it follows that f ~ r.

If that sounds weird, there is no choice, and there is an example you will surely accept. If it were not so – that f ~ r – the LHS cannot reach zero if V reaches zero; but it must vanish with V (and mass) because if not, work could be done on a system with magnitude zero, or a system with zero mass could do work on its surrounding which would violate basic physics. You saw Kellogg's lemma according to which both integrals and k must reach zero with the max chord of V.

I am sure that you will not object to the same principle being applied to the force of gravity. The above principle tells us that within the Earth, the force of gravity approaches zero linearly with r such that at the center of the Earth the force of gravity is zero.

(The same principle applied to V larger than the earth will tell different things. In that case, k is constant because it is always proportional to mass, V changes; but the space added to V does no longer contain mass. For int del.f dV to be constant it follows that del.f ~ 1/r^3. In the LHS integral which is also constant, it follows then that since A ~ r^2, for int f.n dA to be constant it follows that f ~ 1/r^2. This law should look familiar to you. All I mean to say is this: f ~ r inside planets is the flipside of f ~ 1/r^2 outside planets. They are both valid within their respective context.)

In scalar terms: the 1D relation f ~ r and the 3D relation U ~ V are equivalent. All I mean to say is: we do have a pressure term, U/V, but we cannot work with f/A.

BTW, your comment shows a way how to measure f – it works if we assume an unit mass in the system. This thought is not at all weird, in thermodynamics it is one mol (of thermodynamic mass counted in atoms, not inertial mass counted in kg).

Falk

## Simple direct question, but no direct answer.

Falk,

I asked you to provide the result for your integral of f.n for a homogeneously pressurized gas, but no answer from you was forthcoming. If you refuse to answer simple direct questions, then how do you expect anyone to accept your theories? Let me restate the problem. You have a spherical region of a gas of radius R under uniform pressure. For this system, please identify f with an equality, f.n with an equality, and the integral of f.n over the surface with an equality.

Furthermore, it seems to me that the LHS is 4 pi R^2 p, so as the volume goes to zero, R goes to zero and hence the LHS goes to zero, so your response does not seem to make any sense.

Thank you,

Chad

## Consider units

Chad,

the integration of forces over the surface of a sphere at fixed scale which you request is a side-tracker. We don't need any red herrings here. Cauchy's continuity approach considers changing scales, hence we must find out what happens to f if V approaches zero.

Your term 4 pi R^2 p above contains the term p. Thus you take f to have the unit [Pascal]. That misconception is very characteristic of CM.

In the divergence theorem

int f.n dA = int del.f dV = k

k is the charge, its unit is C^2 [Coulomb squared] = [N m^2] or [J m]. It follows that f in the first member is in [Newton], and not in [Pascal].

We are dealing, after all, with electromagnetic forces which are defined as f = c QQ/r^2 (Coulomb's Law, where f = force, Q = one of two interacting point charges, r = the distance between them, and c is a constant = 1/(4 pi e_0) where e_0 is the electric field constant). If you do not understand this I suggest you consult an introductory physics textbook.

Note that Coulomb's law has the same mathematical structure as Newton's law of gravitation. As shown above, within distributed mass homogeneously filling space V, Coulomb's law -- or any other law of the form f ~ 1/r^2 -- transforms into f ~ r.

I am well aware of the fact that in CM, force f is often implicitly taken as f/A -- such that people in CM talk about forces, but they mean it in the sense of f/A = P for stress. This A is a fudge factor with an unit, and an indirect reference to Cauchy; it is just another indication that CM is catastrophically out of touch with the rest of physics. Force has the unit [Newton].

I have never -- repeat: never ever -- seen any other definition of force in CM than f = ma. This ignores the entire development of physics since AD 1800 -- Laplace, Coulomb, Lagrange, Poisson, Gauss, Green, Joule, Helmholtz, Clausius. Congrats. Thus a few plain truths:

f = ma is a single vector force. It acts

at the moment of collisionof two discrete bodies with their respective mass m and velocity v, and it acts only and exclusivelyat the point of collision. Nowhere else. f = ma is very useful in the understanding of the physics of discrete bodies in freespace. But f = ma is not and cannot be a field force. f = ma cannot be derived. The fact alone that CM uses this definition only and exclusively shows that CM is still at the level of knowledge of Leonhard Euler who died in 1783.f = dU/dx is a derived force. It was first proposed by Lagrange in 1784. Thus Euler is excused for not knowing about it -- but no one else after him. f = dU/dx is the definition of a field force, it is the f in Coulomb's law and in Newton's law of gravitation; it is the force type we are dealing with in any other physical situation except mechanics of discrete bodies in freespace.

Field forces must be derived from a potential energy term.Thus, with all due respect for Euler, it is time for CM to enter the 19th C.

Falk

## Still no answer?

So now k is charge, and charge has units of coulombs squared? And C^2 = N m^2? By my calculation C^2 = N^2 m^2 / V^2.

You still have not answered the question. What are f, f.n, and the integral for a uniformly pressurized spherical region of a gas?

## Are we wasting time here?

Hello all:

Perhaps it is time to realize that this discussion is not going to go anywhere?

Falk says:

"Your term 4 pi R^2 p above contains the term p. Thus you take f to

have the unit [Pascal]. That misconception is very characteristic of

CM."

The force does not have the units of Pascal: 4 pi R^2 p is area times pressure, that is [Newton]. The f is force density! Force per unit area. Misconception?

Shouldn't we stop wasting everybody's time and bandwidth?

Regards,

Petr

## Newton or Pascal?

Petr,

what are you getting impatient about? you and I agree. If you take the terms in the divergence theorem at face value, f is in [Newton], A is in [m^2], hence the integral int f . n dA has the unit [Jm] and that's how it should be. The p term [J/m^3] has no place in the equation.

Falk

## period

Yes, k has unit C^2. The Coulomb is kg^(1/2) m^(3/2) / sec, and not what you think. k is the charge, see e.g. the Feynman lectures. I hope this furthers your enlightenment.

I don't know why you keep asking questions that have nothing to do with the subject here. I explained on March 12 to Louie – who had a decent way of asking questions – that f ~ A on free planes (dividing the universe into left & right) and on closed planes (around a volume V) if Delta A approaches zero while the dimension of V is not itself subject to the limit operation. But in Cauchy's limit operation V approaches zero.

F.

## Yet, still no answer to a direct question.

How does the question have nothing to do with the subject? This is your "proof" that we are talking about. In your proof you have quantities like f, f.n, and the integral of f.n over the bounding area of a volume. You claim that these quantities have relevance to physical systems. I am simply asking you to tell us all what these quantities are for the physical system of a spherical region of gas. Why can't you answer the question?

By the way, you can go to http://physics.nist.gov/cuu/Units/units.html to learn more about SI units.

So I see you have read the Feynman lectures. What do you think of Volume 2, Chapters 38 and 39?

## comments

Mr. Koenemann,

I guess a blog is not the best place to solve these issues, but I think if some people in imechanica are sincerely trying to understand your claims, it doesn't help that you seem to be upset by them.

Even if publication of a few papers is to some extent an

"acceptance" of your claims, this is not final and rigorous assessment,

particularly in terms of implications. Especially considering that

there is an entire literature based on Cauchy stress, entire Design in

engineering is based on this, and people have worked with this for many

centuries. So you have to be prepared for people to be skeptical in the

first place by your statements, the harder you claim everything else is

wrong, the more skeptical people will be. If you continue this way,

most people will disregard some more your papers, and think you are

afraid to discover if your statements are wrong, or not very relevant.

The attempt to set the scene in some simple and clear terms, including the units, wherer the standard results are wrong, some examples, seems entirely reasonable.

It seems your main concern is about the "limit" at zero volume, so some attempt to think there is some link with "structure", or the need to think in discrete terms. What is your minimum lenght scale to which Cauchy stress works?

Regards

mike c.

## some answers

Mike,

I am perfectly aware of the long history of Cauchy stress, and of the long tradition of design. And to say this right away, I do not think that all these engineers are brain-amputated. They are very careful and conscious of the fact that most of their work has safety implications, and before they trust a prediction they do lots of testing to be sure it works. On the other hand, I have observed a most profound and massive distrust in theory especially among engineers, to the extent that they are not aware of it any more, but it exists nonetheless. I live in Aachen with its large technical university. People never admit they don't trust theory - in public. In private, a large number of profs told me they don't care about theory at all, "We don't work with thermodynamics, we run our projects" is a very recent quip. A Rektor of this RWTH told me once "Nobody wants new theories, this would create far too much unrest in the universities." A mechanics prof here sent my work to a colleague to evaluate it. When the answer was straightforwardly positive and endorsing, the prof here ceased to communicate with me. That was 15 years ago. The colleague helped me get a paper in print.

I am not at all worried about people being sceptic, that is their natural right. But I find it disconcerting if you or others ask for examples of errors - I have done that, I have shown in three different ways that the conventional theory always predicts that the work done in an isochoric elastic deformation is zero. At the same time, I do consider my published work to be part of this discussion, I don't have to repeat it here. The predictions for simple shear cannot be better.

I am not concerned about the limit of Cauchy stress, but about the fact that the Cauchy theory does not distinguish between system and surrounding, the most basic theoretical concept of thermodynamics. Also, I consider it incompatible to approach a change of state with an equation of motion. A theory of deformation of solids that does not take the existence of bonds in solids into account is nonsense, excuse me, no matter how well it works. I cannot agree with the way the First Law of thermodynamics has been turned upside down in CM. And I cannot live with the fact that it was so far simply impossible to discuss these matters. As said someplace above: potential theory says that for logarithmic problems in distributed mass one needs to take a unit volume in some standard state (to be defined by convention), and work is done if the state of that unit volume is changed (at constant mass). This is precisely as it is done in thermodynamics. I see no way to reconcile Cauchy with the basics of thermodynamics. Cauchy let the system vanish identically.

I have found out why the Cauchy theory works to some degree, but at the same time I have found why I ran into such massive inconsistencies between prediction and observation in the first place. The explanation may be a bit out of context for you, but anyway:

when Euler reduced Newton's fxr to fxn, he effectively reduced the shape of the system to being spherical. This is an unrecognized boundary condition. In my theory, the shape of the system is a function of the material properties and the equilibrium conditions. The conventional theory is not completely out of whack with reality as long as this unrecognized boundary condition is in fact observed by the physical setup, namely that the system has spherical shape; if not, the disparity between prediction and observation is massive and profound. For simple shear the system must have elliptic shape, a sphere does not work.

Another reason why the error is not obvious in most cases is this: in the elastic field, Nature prefers pure shear deformation for energetic reasons. This is - independent of one another - my repeated observation in the field, and my prediction. But for pure shear (or higher symmetric situations) the shape of the system is indeed a sphere . The flipside is: simple shear is by far the most common deformation type in the plastic field (observation). I found out that it is strongly energetically favored (prediction). It cannot be understood by conventional means. The list of phenomena which I think to have given an explanation for, relating to simple shear, is quite long. - To be sure, it is at this point still a matter of contention if my theory is acceptable. A student has written a thesis on it, and enough people have checked it, so it is a serious proposition. The rest is up to public discussion. But this we need badly.

Falk

## Mike, did that clear everything up for you?

So Falk, have you come up with an answer for f, f.n, and the int f.n dA for the uniformly pressurized gas? If you don't like spherical regions then we can look at something else, say a cube or a parallelepiped.

## common language?

Chad

please don't be agressive in turn, I did find some effort from Falk, he repeats the problem is about the shape involved in doing the limit to zero. Falk says he proves isochoric deformations are work free. Then, he says something about plastic deformations and I am not sure why this concerns elastic media. In the end, I also see some confusion.

Additionally, there is obviously a problem of language here, in that Falk seems to insist with his notation, and also uses many coloured expressions like "massive and profound differences" which are difficult to beleive, and perhaps upset many people and not just the "Rectors" that clearly have many practical problems to solve...

Of course, no progress can be made until Falk finds ennoying to explain the questions in different terms, or at least we find some language mid way. A useful excercise for some "non-Rectors" in imechanica!

Mike

## My apologies.

My apologies Mike, and to you Falk.

I should take Mike's example and be more level-headed.

I am simply trying to set some common ground so that the discussion can progress. Without a knowledge of f, f.n, and the resulting integral, this appears to be impossible. Since, as most of us understand the quantity of force to be active at a single point, I can only understand int f.n dA as a summation of forces. Otherwise, what distribution are we supposed to be integrating over the area?

## constraints

time constraints due to my breadn'butter job, will answer Fri eve

Falk

## spherical calculus

Chad,

you can download your answer here . I fail to understand what's so difficult about this.

Falk

## You need to define f.

Falk,

This is getting silly. You still claim that you have no need to define f, but clearly you do. What is the vector f in your note for a homogeneously pressurized gas?

Chad

## funny questions

Chad,

I have not the faintest idea what you want. The unit of force is commonly N = kg m / sec^2. I thought you know this for yourself. If not – then what are you doing on this blog? In my papers and in my comments above I make it clear that I do not use the the force definition f = ma, but f = dU/dx, where U is some potential with unit [J], in my case the internal energy. What else is there to know? If you ask a serious question I have failed to understand this. And I think I am not alone.

Falk

## All I want is for you to define, unambiguously, the vector f.

Falk,

Just define the vector f. In my counterexample, my interpretation of the vector f for a homogeneously pressurized gas has magnitude p and is in the opposite direction of the unti normal to the sphere. This then means that the scalar quantity f.n is -p. Finally the integral of f.n over the area of the sphere is -4 pi R^2 p (the minus sign was left out previously).

All I am asking you to do is give an unambiguous definition of f, i.e what is its magnitude and what is its direction for a uniformly pressurized gas. If it is so easy for you to do, then just take two lines to give the magnitude and direction.

If you don't want to define f, then define your new quantity U for the uniformly pressurized gas. This should take less that two lines.

If you are not alone, please give me the other person's name. Perhaps they will answer my questions.

Chad

## force or pressure

From Falk's answer above: "Now consider forces. An external force acts inward on every point on the surface. Say, magnitude of force in some given condition is f1, from every direction. Thus the effect of all forces is proportional to A, say f1 A."

This seems to be a simple statement, but I find it difficult to understand. Perhaps I just cannot read. To help me out, please consider the following questions.

My first question is: If the external force acts on every point on the surface, how many points (or forces) are there for a given spherical volume V with area A? Are we talking about a discrete system or a continuum?

Second, how can an external force act inward on every point of the surface and meanwhile the magnitude is f1 (or whatever) from every direction? If an external force acts inward on the surface, I tend to believe that the force is a vector with a particular direction (normal to the surface) that varies from point to point. This applies for a pressuried sphere as well, i.e., the "external" pressure acts on the surface only in the direction normal to the surface at a particular point.

Third, what is the effect of all forces? Is it summation of all forces? By vector form or by magnitude only?

Considering a pressurized sphere, the external forces acting on the surface of the sphere may be considered as uniformly distrubuted as a pressure (force per unit area of the surface) in the continuum sense or may be considered as a discrete force system with point forces acting on individual points (particles? atoms?) of the surface. In any case, the external force (distributed pressure or point forces) at each point of the surface ought to be balanced by internal forces of the sphere at the same point. In continuum mechanics, such internal forces are expressed in terms of the local stress state and the surface normal, and the balance between the internal force and external force gives the boundary condition. If one simply adds up all the external forces by the vector form or equivalently integrates over the surface area the distributed pressure (with direction normal to the surface), the resultant force is always zero at equilibrium for the pressurized sphere. In other words, the external forces are self-balanced. For this reason, I do not understand what f.n is in the first integration of the divergence theorem. It appears to be the magnitude of the force f (external?) in the normal direction of the surface. The integration over area seems to imply that the force is continuously distributed (like pressure?) on the surface. Then, what is the physical meaning of adding up (or intgerating) all the pressure (as a scalar) on the surface?

I have to stop here, hoping to get my questions cleared before I try again to understand the next statement from Falk.

Thanks.

RH

## go on!

Rui,

"Now consider forces. An external force acts inward on every point on the surface. Say, magnitude of force in some given condition is f1, from every direction. Thus the effect of all forces is proportional to A, say f1 A." - This seems to be a simple statement, but I find it difficult to understand. Perhaps I just cannot read. To help me out, please consider the following questions.

My first question is: If the external force acts on every point on the surface, how many points (or forces) are there for a given spherical volume V with area A? Are we talking about a discrete system or a continuum?

Second question first: the system is a subvolume in a larger volume. It is defined by its mass n (in mol, not m in kg). It is situated in an external reference frame X, its location is determined by its center of mass.

First question: there's an infinite number of points on any plane, from the differential dA to km^2. If we want to know the sum of forces over a plane, we make use of the known fact that on free surfaces force and area are proportional, so we can integrate. The system surface is a closed surface, but as long as we consider the scale to be fixed, it works.

Second, how can an external force act inward on every point of the surface and meanwhile the magnitude is f1 (or whatever) from every direction? If an external force acts inward on the surface, I tend to believe that the force is a vector with a particular direction (normal to the surface) that varies from point to point. This applies for a pressured sphere as well, i.e., the "external" pressure acts on the surface only in the direction normal to the surface at a particular point.

That's how I would describe it too. I do not see where your problem is. - Note that in the case you imply, the force is perpendicular to the surface at its point of action P, and it is antiparallel to the radius vector of the point P relative to the center of mass of the system Q. (The force exerted by system at surrounding at point P is parallel to r.) The intent of this remark is to ease you back to Newton's understanding who considered the angular relation of radius to force, and away from Euler's way who considered the angular relation of surface to force. In the isotropic case there is no difference, but from the systematic point of view this is important nonetheless. I think Newton was right, Euler was not. In Newton's sense, a compressive normal force is r-antiparallel, a tensional normal force is r-parallel, and a shear force is r-normal.

There is the possibility to understand your question in the sense that you think: at the point P (surface point of the system whose center of mass is at Q, such that r = Q -> P) forces act from all sides. No, the point of reference is the center of gravity Q of the system, all forces act indirectly on Q, and the system interacts at P only with one single external force.

Third, what is the effect of all forces? Is it summation of all forces? By vector form or by magnitude only?

The effect of all forces in this case is an isotropic compression. If the external forces are expressed in polar coordinates and then integrated around the system, you get an answer in vectors; all vectors are oriented inward, hence they have the same sign. The sum is non-zero, and they are balanced by the forces exerted by system at surrounding.

Considering a pressurized sphere, the external forces acting on the surface of the sphere may be considered as uniformly distributed as a pressure (force per unit area of the surface) in the continuum sense or may be considered as a discrete force system with point forces acting on individual points (particles? atoms?) of the surface.

The former. What you call a discrete force I have often called a single force, your expression is better. The entire intent of my theory is to develop a field theory which assigns a force to every point in space (within the region of interest). Force per unit area accepted if the system size is fixed.

In any case, the external force (distributed pressure or point forces) at each point of the surface ought to be balanced by internal forces of the sphere at the same point. In continuum mechanics, such internal forces are expressed in terms of the local stress state and the surface normal, and the balance between the internal force and external force gives the boundary condition.

No. Continuum mechanics does not distinguish between system and surrounding. Here we are right at the core of the issue, I am glad you ask that question.

I am absolutely and positively sure that the thermodynamic equilibrium condition has never been asked in CM. If you distinguish system and surrounding, you perceive the forces exerted by system and by surrounding as being different in origin, hence they have different sources. The absolutely only equilibrium condition I have seen in CM is Newton's Third Law. The only energy conservation law possible in the context of that law is E_kin + E_pot = const. That is, processes under this law take place _within_ the system, that is: forces interacting in Newton's Third Law originate within the system. I discussed that matter in

Unorthodox Thoughts(see below), chapter 4.Continuum mechanics considers the sample and the forces acting on its surface; these are the external forces. Forces acting inside the sample are called internal forces, but they are thought to be the continuations of the external forces into the sample. They are not perceived as different in origin. (If that sounds weird, I quote here from memory; I don't know any more where I read this. But I spent a lot of time on literature research to make sure that this is indeed the only interpretation possible of the current understanding.) Continuum mechanics does not, repeat NOT, distinguish system and surrounding.

If one simply adds up all the external forces by the vector form or equivalently integrates over the surface area the distributed pressure (with direction normal to the surface), the resultant force is always zero at equilibrium for the pressurized sphere.

No. You mix up the two equilibrium conditions.

Consider a system of spherical shape subjected to an isotropic loading. Then we have two equilibrium conditions.

1. The Newtonian equilibrium condition (Newton's Third Law): the external forces must balance. It turns out that for every force on one side of the system pointing this way (e.g. left, negative) there is another force on the other side of the system pointing that way (e.g. right, positive). Hence the sum of forces is zero. But note that you implicitly considered the sign of the forces in a Cartesian coordinate system, with the center of mass Q of the system at the origin.

2. The thermodynamic equilibrium condition: external and internal forces must balance. If you express the vectors in polar coordinates, the external forces all have the same sign in the sense that they are directed inward (positive). Their sum over the surface is non-zero. If you do the same to the forces exterted by system at surrounding, they are all directed outward (negative). The sum of all system forces exerted at surrounding, and the sum of all surrounding forces, exerted at system, balances.

To be sure, the question is not one of coordinate systems, but the Newtonian eq.cond. comes very naturally in Cartesian coords, the thermodynamic eq.cond. comes just as naturally in polar coords. In either case, the questions can both be asked in the respectively other coordinate system, but they become algebraically complex. That's not the point. It suffices to realize that the questions answered in the respective coordinate systems is not the same.

In other words, the external forces are self-balanced.

In the Newtonian eq.cond. where left & right balance: yes. In the thermodynamic eq.cond. where inward and outward balance: no, because the inward-directed external forces can only be balanced by outward-directed system forces. The latter are not a part of the equation in Newton's Third Law.

For this reason, I do not understand what f.n is in the first integration of the divergence theorem. It appears to be the magnitude of the force f (external?) in the normal direction of the surface.

Yes, _ON_ the system.

The integration over area seems to imply that the force is continuously distributed (like pressure?) on the surface. Then, what is the physical meaning of adding up (or integrating) all the pressure (as a scalar) on the surface?

so far we haven't integrated pressures, but forces. The physical meaning is the effect exerted by the surrounding upon the system.

Rui, I know much of this sounds trivial, but it is fundamental. You ask good questions. Go on.

Falk

P.S. To everybody: the iMechanica software keeps changing my entries to the link creator, and I have no idea how to stop this, I don't even realize that it happens as long as I have not finally submitted my contribution. Thus, the following URLs are not links, but URLs which you please feed by cutn'paste into your address line in IE or FiFo to find my hp:

My homepage:

http:/www.elastic-plastic.deUnorthodox Thoughts

http://www.elastic-plastic.de/thoughts.pdfSystematics & Gibbs

http://www.elastic-plastic.de/gibbs.pdfApproach

http://www.elastic-plastic.de/theory-co.pdfGurtin

http://www.elastic-plastic.de/gurtin.pdfClausius

http://www.elastic-plastic.de/clausius1870.pdfExperimental data (answer to Landis)

http://www.elastic-plastic.de/experimentaldata.pdfKellogg p.147

http://www.elastic-plastic.de/kellogg147.jpgKellogg p.52

http://www.elastic-plastic.de/kellogg052.jpgTable 1

http://www.elastic-plastic.de/table_01.jpgSpherical (answer to Landis)

http://www.elastic-plastic.de/spherical.pdfHelmholtz:

http://www.digizeitschriften.de/index.php?id=166&ID=511628

## encouraged, but disagree...

Falk,

Thank you for your encouraging comments, but I disagree with your statements.

(1) It is not necessarily true that "on free surfaces force and area are proportional". It is only true when the force is uniformly distributed (same magnitude, same direction) over the surface.

(2) If "an external force acts inward on every point of the surface", the direction of the individual force (acting on individual point) is normal to the surface, which has nothing to do with the center of mass or gravity. Only in the case of a homogeneous sphere, the force points toward the center.

(3) If you agree that the external forces are vectors, summation of all the external forces over a closed surface would always be zero at equilibrium, independent of which coordinate system (Cartesian or polar) you use. Note that in polar coordinates, the unit vector (e_r) is not a constant vector on a spherical surface and thus cannot be taken out of the integrant.

(4) So far as I have been taught and have learnt by myself, continuum mechanics does and always distinguish between system and surrounding. The "external" forces from the surrounding act on the system and define the boundary conditions for the system, while the system responds with deformation or other change of states. The origins of the external forces and internal forces do not have to be the same, and they are often different.

(5) For the two equilibrium conditions you mentioned, if I understand correctly, the first (Newtonian equilibrium) considers sum of external forces and the second (thermodynamic equilibrium) considers balance between external and internal forces. As I have noted above in (3), the sum of external forces is zero in both Cartesian and polar coordinates. After all, coordinates are just mathematical tools and they do not change physics. As I have also mentioned in the previous comment, the balance between external and internal forces is required at each point of the surface, which sets the boundary condition in continuum mechanics. Therefore, no problem here in continuum mechanics.

(6) Again focusing on the case of pressurized sphere, the effect of the external forces (surface pressure) is that it induces an internal pressure in the sphere by changing the volume of the sphere or the distances between atoms/particles. The sum of all external forces is zero (Newtonian equilibrium) such that the center of the sphere remains static. The balance between the external force and the internal force (thermodynamic equilibrium) requires that the internal pressure equals the external pressure at the surface, which is the boundary condition that leads to uniform pressure everywhere in the sphere. This is a trivial problem in continuum mechanics, and I do not see any fundamental problem here.

Thanks for your patience.

RH

## good questions

Rui,

(1) It is not necessarily true that "on free surfaces force and area are proportional". It is only true when the force is uniformly distributed (same magnitude, same direction) over the surface.

Yes. You accepted these conditions when you quoted my file

Spherical, in which I make the assumption that the volume element and the external loading are isotropic, see figure in the file. Under these conditions the statement is correct.(2) If "an external force acts inward on every point of the surface", the direction of the individual force (acting on individual point) is normal to the surface, which has nothing to do with the center of mass or gravity. Only in the case of a homogeneous sphere, the force points toward the center.

Again, that was the starting condition, see figure. – In case you are tempted to consider other shapes, I'd prefer if you wait, for a number of reasons: (1) as I am going to use the divergence theorem, the shape of V is arbitrary only if no point of A runs through mass. The condition holds in freespace, but not within distributed mass. (2) There is to my knowledge no theory yet that considers forces not only in distributed mass, but in bonded distributed mass. The concept that must also be accommodated is the lever laws; within solids, shear forces do work too. The isotropic case circumvents that situation because shear forces (r-normal) do not occur. (3) In my theory, the shape of the system represents the material properties. It is goo to start with simple conditions before we go on. (4) Geometrically, the sphere is the body with the smallest A/V ratio. That is going to be important much later.

(3) If you agree that the external forces are vectors, summation of all the external forces over a closed surface would always be zero at equilibrium, independent of which coordinate system (Cartesian or polar) you use. Note that in polar coordinates, the unit vector (e_r) is not a constant vector on a spherical surface and thus cannot be taken out of the integrant.

You are right, I made a mistake. But only in my remark above, not in my theory. There I systematically explain everything in two eqns: in vector form what is being done, and in algebraic scalar form how it is done; the latter is always the magnitude of the former as a function of direction. (I got very used to that, but this could not be clear to you.)

(4) So far as I have been taught and have learnt by myself, continuum mechanics does and always distinguish between system and surrounding. The "external" forces from the surrounding act on the system and define the boundary conditions for the system, while the system responds with deformation or other change of states.

That's how it should be, but it is not so. Cauchy let the system vanish identically. I am well aware of a number of concepts that seem to be the equivalent of the system – often a small cube is assumed etc; but these are auxiliary concepts that are not logically built into the theory. In

Unorthodox ThoughtsChapter 4 I quote Eringen for precisely this reason, to show that he does not use the thermodynamic equilibrium condition. The thermodynamic system is finite, Cauchy's limit operation reaches zero.(5) For the two equilibrium conditions you mentioned, if I understand correctly, the first (Newtonian equilibrium) considers sum of external forces and the second (thermodynamic equilibrium) considers balance between external and internal forces. As I have noted above in (3), the sum of external forces is zero in both Cartesian and polar coordinates. After all, coordinates are just mathematical tools and they do not change physics.

Correct, see above.

As I have also mentioned in the previous comment, the balance between external and internal forces is required at each point of the surface, which sets the boundary condition in continuum mechanics. Therefore, no problem here in continuum mechanics.

I have not found it, and I must have checked dozens of textbooks. I have only found evidence to the contrary. Let me give you a few thermodynamic concepts that cannot be reconciled with CM.

(a) Consider two points in a continuum; consider them to be at opposite sides of a thermodynamic system. It follows quite naturally that these points can be made to coincide only by reducing the system to a point, i.e. by doing infinite work. The compatibility problem in conventional CM – that two points in the original state must not coincide after the transformation – cannot occur in thermodynamics. Instead, from this thought alone it comes very natural that the work function must be similar to PdV, that is, logarithmic. No such thing in CM.

(b) A solid has an internal pressure. That is: a mol of an ideal gas in the standard state has a V = 22,4 litres. A mol of solid potassium has a volume of 6 cm^3. If the ideal gas is compressed to 6 cm^3, an external pressure of ca.0.5kbar is required. In the solid, that pressure is internally balanced by the bonds. In order to consider the internal pressure, one must therefore acknowledge that bonds exist. I have not found mention of bonds in CM. Furthermore, the thought above suggests that all solids have the same compressibility if the external pressure is scaled to the internal pressure; that is, if you have the molar volume, all solids should have the same compressibility. At least for the alkalies I have shown this to be true (

ApproachFig.1, or download theslides below). Try to reconcile that concept with CM. You cannot even reconcile it with the P = f/A pressure definition.(c) Cauchy let the system reach zero. Hence there is no radius. If you leave the system finite, normalize, and you have r = 1 for the standard state. If work is done, r is changed (in 3D it is V). Hooke's spring has unit length, works ditto. Without a basal length term r you cannot define work. In potential theory this is the "zero potential distance". In CM it is completely awol.

I cancel your point 6 because it has been addressed above. CM cannot be compatible with thermodynamics because CM has a conservative Newtonian theoretical structure which contrasts in all possible points with thermodynamics, see

Table 1. It does not help at all to start with a conservative theory and then suddenly switch to thermodynamic terms. This is mixing apples and oranges. The two theories do not even use the same First Law. The First Law as used in CM is not the First Law.Links (please use Cut&Paste):

Unorthodox Thoughts http://www.elastic-plastic.de/thoughts.pdf

Approach http://www.elastic-plastic.de/theory-co.pdf

Table 1 http://www.elastic-plastic.de/table_01.jpg

Spherical http://www.elastic-plastic.de/spherical.pdf

Slide 39 http://www.elastic-plastic.de/slide39.jpg

Slide 40 http://www.elastic-plastic.de/slide40.jpg (this is also Fig.1 in

Approach)Falk

## sum of forces?

Falk,

I will be focusing on point (3) above because I am confused by your answers. I would be interested to see the two equations you mentioned there, one in vector form and one in scalar form.

You seem to have agreed that sum of all external forces for a system in equilibrium is zero no matter which coordinate (Cartesian or polar) one uses. Thus, the external forces are self-balanced, which is what you called "Newtonian equilibrium", and there is no mix up here with the other equilibrium (which you called "thermodynamic equilibrium").

You have also agreed that f.n in the first integral of the divergence theorem (see your note on spherical calculus above) is the magnitude of the external forces acting on the system in the normal direction of the surface. Now if we go back to my original question about the effect of all external forces, apparently what you meant is not sum of all the forces, not in the vector form and not by the magnitude either, becasue the integral (or f1 times A) would have a unit in Newton times meter squared. I simply cannot understand the physical meaning of this integral.

If I look one step further in your note on spherical calculus, I see that you have assumed that f.n is a continuous function on the surface and f is a continous function in the volume. So in essense, you are considering a contimuum system. In such a case, the mathematical form of the divergence theorem can be understood. However, the phyiscal meaning of f in the volume remains puzzling to me. Is it internal force? Why does del.f have to be a constant everywhere in the volume? What do you mean by "source density"?

Thanks.

RH

## Source density

Rui,

I would be interested to see the two equations you mentioned there, one in vector form and one in scalar form.

I suggest you look up my paper

Approach, chapter 5 to 7. There are plenty of them. - I mean, on the one hand this blog makes it hard to discuss equations without symbols. On the other hand, my papers are there for the reading, and they are part of the discussion.You seem to have agreed that sum of all external forces for a system in equilibrium is zero no matter which coordinate (Cartesian or polar) one uses. Thus, the external forces are self-balanced, which is what you called "Newtonian equilibrium", and there is no mix up here with the other equilibrium (which you called "thermodynamic equilibrium").

Yes. Certainly it is coordinate-independent. But the fact remains that we have two different equilibrium conditions to consider. The first one only says: the system is not accelerated externally. The second one permits us to ask: what does the surrounding do with the system? - I would go one step further. The external equilibrium, or Newtonian equilibrium, is observed by definition in solids because system and surrounding are bonded to one another, thus the disequilibrium case cannot occur (below yield). So let's concentrate on the second one.

You have also agreed that f.n in the first integral of the divergence theorem (see your note on spherical calculus above) is the magnitude of the external forces acting on the system in the normal direction of the surface. Now if we go back to my original question about the effect of all external forces, apparently what you meant is not sum of all the forces, not in the vector form and not by the magnitude either, because the integral (or f1 times A) would have a unit in Newton times meter squared. I simply cannot understand the physical meaning of this integral.

I mean the sum of all forces, and since f.n is a scalar, the entire integral is a scalar. Regarding the unit of the integral I cannot help you; do what I have done: take Gauss literally.

The physical meaning of f in the volume remains puzzling to me. Is it internal force? Why does del.f have to be a constant everywhere in the volume? What do you mean by "source density"?

The thermodynamic equilibrium condition at the surface point P of the thermodynamic system situated at Q is f_syst + f_surr = 0. Thus in the divergence theorem, f is either f_syst or f_surr; it must hold for both, and you need both. Say del.f = phi; then we have

phi_syst + phi_surr = 0

which means that in the static loaded state (phi =/ 0), the surrounding and the system act as sources of forces, they serve as sinks to one another, and equilibrium is maintained. Only in the unloaded state phi = 0. Thus phi characterizes the energetic state in which the system is.

del.f is a constant in distributed mass (homogeneous conditions implied). It characterizes the ability of a region V to act as a source/sink of forces. I could bore you with examples from my intro calculus book; but you probably do not need them. I cannot say where I got the term "source density" from, it goes back to my earliest exposure to calculus.

But your question highlights a much larger problem: all these terms are in use in potential theory. Current CM has nothing whatsoever to do with potential theory. Please do not understand me personally, I have made this observation far too often: people in CM are nearly completely unaware of potential theory. But if we want to understand deformation, we must know about it. The condition phi = 0 says: there are no sources in V; if there are fluxes (e.g. water), they enter and leave a region. In the context of CM and thermodynamics this means: no work is done; because the elastic work is energy stored in the system, so phi must become non-zero during elastic loading.

You can express it even more radically: all of Newtonian mechanics takes place under the condition phi = 0 (called the Laplace condition). The statements phi = 0 and E_kin + E_pot = const are equivalent. If phi =/ 0 (called the Poisson condition), you enter the field of thermodynamics.

Links:

Approach http://www.elastic-plastic.de/theory-co.pdf

Falk

## my problem with units and more

Falk,

By "I mean the sum of all forces" did you suggest that the integration of f.n over the surface area result in the sum of external forces acting on the system? Then, should sum of all forces have the same unit as each individual force (f)? Since n is dimensionless unit vector, I can only see that the integral is a quantity with unit same as a force times area, thus not a force quantity or the sum of forces.

By "Regarding the unit of the integral I cannot help you" you seem to imply that it is trivial. However, it is only trivial if you would agree that the integral is not sum of all forces, but something else you yet to explain. Alternatively, the quantity f is actually force per unit area (like pressure), but you have pointed out several times this is not the case in your theory.

Speaking of units, I would remind you that you have made a mistake in your previous comment "Consider units" on March 25, where you said the integral equals a charge k with unit C^2 [Coulomb squared] = [N m^2] or [J m]. I believe someone has pointed out that this is wrong, at least in the standard SI unit system (see http://physics.nist.gov/cuu/Units/units.html). It may be helpful to note that, in the equation of Coulomb's Law for electrostatic forces, f = c QQ/r^2 as you wrote in the same comment, the constant c is not dimensionless as you assumed.

Well, apparently there are more disagreements than units, and I do not see our discussions are converging at any point. Perhaps it is best to leave it for each individual to decide what to believe.

RH

## units again

Rui,

By "I mean the sum of all forces" did you suggest that the integration of f.n over the surface area result in the sum of external forces acting on the system?

Yes. For example, if you want them to balance by themselves, consider a field with orthonormal eigendirections where the vectors point inside along x_2 and outward along x_1. In that case the sum is zero.

Then, should the sum of all forces have the same unit as each individual force (f)? Since n is dimensionless unit vector, I can only see that the integral is a quantity with unit same as a force times area, thus not a force quantity or the sum of forces.

Yes. For that reason I proceeded as follows: I accepted the result of the divergence theorem as it is; after all, it gives the right results in the sense that you can properly derive f ~ r inside bodies and f ~ 1/r^2 outside bodies. However, in order to deal with vectors with the unit [N] I defined the fields as described yesterday, but henceforth I considered all forces to be a function of direction relative to Q (the local coordinate origin). If you then integrate not over A, but over alpha, the unit is unchanged and in [N]. Or if you integrate radius-force interactions the result is in [J] (result must be a scalar). That makes a lot of sense in my eyes because we are talking about energy fluxes, after all.

By "Regarding the unit of the integral I cannot help you" you seem to imply that it is trivial. However, it is only trivial if you would agree that the integral is not sum of all forces, but something else you yet to explain. Alternatively, the quantity f is actually force per unit area (like pressure), but you have pointed out several times this is not the case in your theory.

No, I do not think it is trivial. I just think that there are concepts for which it is correct to use the divergence theorem as it is; for other purposes there may be equivalent concepts. I could not find a reason why integrating over A is different from integrating over alpha. But the latter makes it easier to understand force fields. (I kept track of units very carefully.)

But there is something else I ask you to consider. I know very well that my theory is a proposition. I have worked for a few years to develop it, others have read it carefully and found it sound, and when I offer it to the public I do so with a certain assuredness that it is airtight. The question you ask here require an understanding where not only you or Petr, but even I reach the limit. (For example, the divergence theorem is not restricted to force vectors alone. The vector quantity in question may be velocity or something else, and the theorem is still valid. This shows that the unit of the divergence theorem is not fixed. In the light of the unit disparity – see below – I can see where the confusion comes from, but I still cannot see that my interpretation is wrong.) I just had more time than you to ponder the problem, to get used to ideas that initially looked unfamiliar, to check in all directions available to me whether I create contradictions, and if not, I went on, in this particular case by choosing to use alpha and not A as the integrand. There are other items in my theory which I expect to create a good deal more fuzz than this one here. I make no excuses for errors if we find one, but I do say I went into the unknown, and to the best of my knowledge I found something new. But the only way we can find out if my theory really holds is by subjecting it to public discussion. This is what you provide, and I am grateful for this. We need more.

Having said that, please do not forget that the real incentive to develop a new theory were the flaws in the old one. My blog exists for three weeks now, there were >400 reads yesterday alone, and so far no hard objection to my critique of the Euler-Cauchy theory has surfaced. That would be the easiest way to shoot me down. Any single of the reasons taken by itself is sufficient to reject Euler-Cauchy. We have yet to clear whether my theory is valid; but it is absolutely certain that we do need a new one.

It may be helpful to note that, in the equation of Coulomb's Law for electrostatic forces, f = c QQ/r^2 as you wrote in the same comment, the constant c is not dimensionless as you assumed.

Yes. The unit I gave – kg^(1/2) m^(3/2) sec^(-1), with a dimensionless constant – is correct, and used in a textbook I have to explain Coulomb's Law. It is not wrong, but there are a number of definitions around which are subject to convention. The reason is that the electrical phenomena could not be fitted easily into the common mks-system which was not made for them. The unit above does not represent today's technical convention for which the Ampere is the most fundamental unit. Other definitions for the unit of the charge were adjusted accordingly, creating various constants, with various units. – Which means that indeed there are several right answers. If you look into Feynman V.1, 12.7, he makes a difference between the physical phenomenon and the units and treats the latter with a certain ease.

Falk

## Charade coming to a close.

Falk,

It seems your little charade is coming to a close. Your "proof" has been discounted with a trivial counterexample. If it was so simple to demonstrate the error in the counterexample, then any rational person would have done it quickly and concisely. Instead, you avoid the question because either you do not know the answer, or you have realized that the answer invalidates your "proof". In either case I will leave it for the readers of this blog to judge.

All the best,

Chad

## finally

If this means that you promise to finally vanish, I open a bottle of best Belgian brew.

F.

## Enjoy your beer.

Have you realized your error or not?

## the old approach

Michele,

to be precise: I did not prove that isochoric deformation is work-free. I only showed that the conventional theory always predicts that an isochoric deformation is always work-free. The math is in Systematics & Gibbs , take a pick, there are three examples with the same result. It shows that the conventional theory is a perpetuum mobile theory. The dissatisfaction with this work-free result was one of my incentives to look for better answers.

I'd be glad if we could become a little more serious now. If there are other questions regarding my critique of the Euler-Cauchy theory, I shall be happy to answer them. Especially if the tone is right.

Falk

## Cauchy's Original Writings

I find this thread both interesting and sad. It is sad to see the

old ways of treating others arising and discussions laced with contempt. It is interesting in

that a challenge to a "fact" that most take for granted has been called into question in anyway. In this regard I think that people should not be

so flip in saying what or what not Cauchy did or did not do. I

encourage all to read the original writings. To save time, I have set

up the following links with excerpts from Cauchy's collected works [Ouevers Complètes]:

1) Augustin Louis Cauchy 1823 (short discussion)

2) Augustin Louis Cauchy 1822/1826 (longer presentation)

3) The first few pages form an interesting article by Gurtin and Martins on trying to beat one of Cauchy's arbitrary assumptions. Note that I did not post the whole article since this is copyrighted material. The full reference is Archives for Rational Mechanics and Analysis, v 60, pp 305-324, 1976, if you have a decent library.

As a last bit of advice for such discussions, be precise, make your definitions clear, do not ASSuME (as a former physics teacher of mine liked to say, because it makes an ass out of you and me); and especially do not assume that what someone tells you about someone elses work is reflective of what the other person actually wrote. Give it a read youself and make up your own mind.

Prof. Dr. Sanjay Govindjee

University of California, Berkeley

## thanks to Sanjay

It is good for many of us to know where we stand: there is a literature about the problem of continuity in the Cauchy stress, and Gurtin and others well known researchers have worked on it (see the ref. below as well). So next step would be to identify where exactly Falk's contribution stands within this and where he refers to this (I haven't seen Falk's paper, nor I have been able to download anything from his web page). One distracting feature is that Gurtin says "Cauchy's contribution remains monumental in continuum mechanics", while Falk seems to suggest this alternative derivations of CM are crucial.

Šilhavý, M. (2008) Cauchy’s stress theorem for stresses represented by measures.

Continuum Mechanics and Thermodynamics[CrossRef]

Segev, Reuven (1986) Forces and the existence of stresses in invariant continuum mechanics.

Journal of Mathematical Physics27(1)[CrossRef]

Šilhavy, Miroslav (1985) The existence of the flux vector and the divergence theorem for general Cauchy fluxes.

Archive for Rational Mechanics and Analysis90(3)[CrossRef]

Šilhavý, M. (1991) Cauchy's stress theorem and tensor fields with divergences in Lp

.

Archive for Rational Mechanics and Analysis116(3)[CrossRef]

Schuricht, Friedemann (2006) A New Mathematical Foundation for Contact Interactions in Continuum Physics.

Archive for Rational Mechanics and Analysis[CrossRef]

Williams, W. O. (1979) On stresses in mixtures.

Archive for Rational Mechanics and Analysis70(3)[CrossRef]

Ziemer, William P. (1983) Cauchy flux and sets of finite perimeter.

Archive for Rational Mechanics and Analysis84(3)[CrossRef]

Degiovanni, Marco (2006) Edge-force densities and second-order powers.

Annali di Matematica Pura ed Applicata185(1)[CrossRef]

Chen, Gui-Qiang (2005) Divergence-Measure Fields, Sets of Finite Perimeter, and Conservation Laws.

Archive for Rational Mechanics and Analysis175(2)[CrossRef]

## no download?

Michele,

re Gurtin: I have a very solid opinion on him, see here , in reply to Rui Huang. I think Gurtin knows for nearly 40 years about the main flaw in CM, but chose to keep that to himself.

Unable to download: that is possible, I apologize. I found out the hard way that if I inserted an URL in the links on iMechanica and I started with "www", the software would add "http://www.imechanica.org/blog/" by itself in front of the URL I inserted. Then naturally the link leads nowhere. The URLs to be inserted in iMechanica must imperatively start with "http://[etc.]", or the system software has its own opinion as to what the URL should look like.

The more recent links should work, or you go to www.elastic-plastic.de and find your way forward.

Falk

## Koenemann hoax?

I'm sure some of us have already wondered if this whole new continuum mechanics discovery was a hoax. Consider how obviously false some of the statements are (no work for isochoric deformation? The fantastic units of some of the integrals!), and how impossibly hard it is to make sense of the writings in general. Mr. Koenemann makes the words mean what he wants them to mean, not exactly conducive to a scientific dialogue. It is all a big joke!

## Then laugh

Once again, even for the dyslectic: I did not say that an isochoric deformation does not cost work. I have only shown that this conclusion follows cogently from the conventional approach, documented threefold in

Systematics & Gibbs,http://www.elastic-plastic.de/gibbs.pdf

I suggest you look it up, and if you find a reason to disagree, come back here.

The reason for this zero result of the conventional theory is its mathematical structure. CM uses a form of the First Law of Thermodynamics that is invalid, and subordinate to E_kin + E_pot = const. It cannot come to another conclusion because in Newtonian mechanics mass and energy are proportional. The First Law, common understanding, separates mass and energy as independent variables. In

E_kin + E_pot = const

the first term LHS is the heat, the second term are the bonds in a solid; the RHS term is the total energy H = PV of the system (

Clausius' virial law). In Newtonian mechanics, H = const. In thermodynamics it is called U and a variable, but it is the same term. If U is changed at constant mass, the change dU = dw + dq.http://www.elastic-plastic.de/clausius1870.pdf

Falk

## It would be nice if that were true.

Petr,

Thanks for the comment. For a while there I was wondering if I had gone mad. At least now I know that I am not the only one who has trouble deciphering Mr. Koenemann's prose.

Chad

Falk,

I would love to get into the "no work for isochoric deformation" discussion, but before I can do that I really need to understand what you mean by your vector f. Can you give the magnitude and direction of the vector f for a uniformly pressurized gas? If we need to be more specific about the material, it can even be an ideal gas following the PV=nRT law.

Chad

## some linear algebra

Chad,

the zero work topic has nothing to do with your question re the force. (And, to tell the truth, I have little confidence in your good will, to put it gently.)

The ideal gas law has nothing to do with the force, it determines the material reaction. The force itself: at first I could not think of another answer to you than the one I gave last Sunday. I honestly do not understand what you mean. Right now it occurs to me that maybe you do not know how to build up a vector field. I think this is possible – honestly no insult intended – because CM is not a field theory, and you (like many others) may lack experience with this.

The basic equation in linear algebra is Ax = b where A is a square matrix, x is the location vector of a point P, and b is a vector at P. P is a point in a local coordinate set x_i situated at Q which itself is located in a global coordinate set X_i. (In vernacular: you study a material sample of dimensions XYZ; all points in the sample are given in the coordinate set X_i. Right now you have a particular interest in the point Q (X_i) deep inside the sample. The immediate environment of that point is then viewed from Q, relative to which all other points are P (x_i). If P = Q, x is a zero vector, hence b is a zero vector too.) – I reduce the setup to 2D for simplicity.

Consider g(Ax = b). g is a scaling factor, set it to -1, left is Ax = b. A is [1 0 | 0 1] (this is a 2x2 matrix, at | you switch from first to second row). For every point P whose position vector is x you get an inward-directed vector at P. Interpret this as the external field f_surr.

If g is 1, you get an outward-directed field. You can interpret this as the field f_syst. Naturally, x is the radius field.

If you increase the pressure, that will change g. The work function (here you need the EOS) then gives you the change of the length of x from the unloaded to the final loaded state.

Note that Ax = b gives you vectors at all points possible, but here we are only interested at the vectors at points on the system surface. Assume that the system is spherical (implying isotropic material properties). If we agree that the system has unit size such that r = 1, you need not worry about other points inside or outside V, because thermodynamics is not interested in processes within the system, only and exclusively in processes that require interaction of system and surrounding on the system surface. Thus you need f_syst and f_surr at all points P. This is – referring to the discussion with Rui – the thermodynamic equilibrium condition.

In case you wonder how I justify A: take a potential U. Take the derivative with respect to the coordinates, dU/dx = f gives you a vector field. The second derivative d^2U/dx^2 gives you A. The multiplication Ax is equivalent to integrating the 2nd derivative over an unit distance x in all directions. For the external field you can define A as you like, the matrix above is our choice. For the internal field you must consider the material properties.

Falk

## The question is quite direct.

Falk,

If someone asks you for the magnitude and direction of the velocity vector v for a car traveling east at 100 km/hr, you set up a coordinate system (let's say east is the x-direction) and report

v = 100 e_x, where e_x is the unit vector in the x-direction.

Where is your confusion? I am asking you to report the vector f in your equation int f.n dA for a homogeneously pressurized gas. Clearly from our discussion it appears that we agree that the direction of f is opposite to the normal direction to the surface of the sphere. So all I am asking you to do is fill in the ??? in the following equation:

f = - ??? n

Where is your confusion, and why does it take you a page of text to answer a simple question?

Chad

## Is that what you want?

Chad,

now you finally say precisely what you want. I really did not understand your question, it was too vague.

In thermodynamics, all terms are scaled per unit mass, that is: per mol. Hence, from PV = nRT follows that f = nRTA/V (after all, V is finite) or f = nRT/r. That is, if x J of work are done per mol, it is possible to calculate the force if the molar volume V is known. Or else, if (for a gas) a pressure of x Pa is exerted, it is a short calculation to find out how much work has been done; in this case you can work with P = f/A because the scale is fixed. Or you divide by the molar surface and get the force.

I foresee some problems here for other materials, because in the case of solids the internal pressure will have to be considered (see above) which introduces the Grueneisen exponent. In practice, one has to do it as in general in thermodynamics: a value in a measurement (technical problem) will have to be recalculated to be expressed for the material (natural property). But this is way too far ahead.

Falk

## Almost there.

So, since nRT/V = p you must be claiming:

f = -(p 4 pi r^2) e_r

where e_r is the unit vector in the radial direction. However, I am not entirely sure what you mean by A. This A is the area of what exactly? For now, I am assuming A = 4 pi r^2.

So, assuming my interpretation of your comments are correct,

int f.n dA = int -4 pi p R^2 dA = -16 pi^2 p R^4

where R is the radius of the region we are integrating over, and the r above is the radial coordinate. So, in the end, this result is not proportional to the volume of the region either.

We can apply the divergence theorem to the expression for f above as well and we will arrive at the same answer.

## f/A vs. U/V

No. This is almost an intentional attempt at misunderstanding.

A is the surface of the molar volume. If df/d alpha = 0, 4 f pi r^2 is the answer you seek.

The question you should consider seriously:

How does P = f/A relate to P = U/V at changing scale? This is the key.

Falk

## What?

Falk,

You must be kidding. What is the surface of the molar volume? Do you mean the surface of the sphere containing 1 mole of ideal gas at the pressure p? And how would I, or anyone for that matter, possibly know that this is what you meant by A, when this is the first time that you have mentioned it?

If this is what you mean then A is simply some constant value related to the square of the cubed root of RT/p. Then the result for the int f.n dA is simply this constant times p times 4 pi R^2. And we are back to my original claim that your integral is proportional to R^2 and again not the volume.

In order to evaluate your integrals you need to give the vector f in terms of the pressure and possibly the radial distance since we are considering a spherically symmetric system. So, if I am yet again "misunderstanding" you, then please be clear and unambiguous, define all of your terms, and give the magnitude of f in terms of the pressure p and radial distance.

So let me try to be as specific as possible with the question. What is the magnitude of the vector f for a uniformly pressurized gas in terms of the pressure of the gas p and the radial distance from the origin of your coordinate system?

Chad

## the EOS

Chad, and everybody:

what is the unit of PV = nRT?

What is the unit of fr = nRT?

What unit does f have here?

What is U/V?

If V --> 0, what happens to U if nRT is constant?

If r --> 0, what happens to f if nRT is constant?

If V --> 0, what happens to U/V?

If V --> 0, what happens to n?

What happens to PV = nRT when n = 0?

What happens to fr = nRT when n = 0?

Can we do thermodynamics if n = 0?

So what happens to f/A if V --> 0?

Fooling around with energy terms is dangerous. lim(x-->0) 1/x is not convergent, ditto for f(x)/f(x^2) (here f stands for function).

Do you accept P_syst + P_surr = 0?

Do you accept f_syst + f_surr = 0 on the surface of the system?

If n = 0, can the surrounding exert a pressure on the system? Or vice versa? Ditto for forces?

If Cauchy's continuity approach were right, both P_syst and P_surr should be finite values as V = 0. That is, a system of magnitude 0 exerts a pressure. Not in this world. But Cauchy never distinguished system and surrounding, it was simply still too early in the 19th C for that distinction to be understood, that came only after 1847 and the First Law.

You would do yourself a favor by considering P first of all as energy density U/V. It is a scalar, and a state function.

Falk

## Avoiding the question.

Falk,

Over and over again you avoid the question. Why? Is it because you do not know the answer? The question is simple and direct. What is the magnitude of f in unambiguous terms for a uniformly pressurized ideal gas? Your last answer was f = AnRT/V, which is just p*A, but you did not unambiguously define A. If A is the surface of a sphere occupied by 1 mole of gas, as you seem to have suggested, then A is just a constant that only depends on p for an ideal gas.

All of your other words and claims have no meaning if your first "proof" is not valid.

Chad

## you got it long ago

Chad,

if it satisfies you to make a fool of yourself in public by advertizing that you cannot handle the divergence theorem, I cannot stop you. If the force field is

g Ax = b,

where A is (in this case) the matrix [-1 0 | 0 -1] and x is the radius vector of the system, b is the force vector f; its magnitude is determined by the intensity parameter g which is externally controlled, you get a clear answer. There are other ways. The rest is algebra.

The question you keep dodging is what happens to f if we consider changing scales. I have now shown you in two entirely independent ways that f must vanish with r,

- once by means of the divergence theorem, showing that CM is in profound conflict with potential theory,

- once by showing that the EOS does not leave an alternative, showing that CM is in profound conflict with thermodynamics.

This – the scale dependence of the magnitude of f – is the key issue, because it is squarely incompatible with Cauchy who assumed that f reaches a constant finite value.

If you insist on using an f/A as the basic term, that's your business, in physics f = [N], except in the space ship CM. Also, you are very good at ignoring evidence you don't like to consider, such as the zero work prediction of the Euler-Cauchy-theory, and the still open question where in Cauchy's theory the thermodynamic system is. I suggest you concentrate on these items.

Finally, it seems to me that you know little or nothing about the potential theory. If so, you are not in a good position to argue because you do not know what you are up against. Your only claim is that you already know enough. This is not so.

Falk

## Other misconceptions.

Falk,

To answer your first question on V->0:

"If V --> 0, what happens to U if nRT is constant?"

V in this equation is the volume of n moles of gas at temperature T. So, in order to compress some finite number of moles of gas to zero volume, the ideal gas law states that an infinite pressure is required. The work required to do this to an ideal gas would be:

int p dV from V=Vo to V=0 which would be logarithmically singular, i.e. an infinite amount of work is required to do this.

This, however, has nothing, I repeat nothing to do with Cauchy's limiting process. Cauchy did not look at smaller and smaller volumes of material by taking a given mass of material and compressing it down to zero volume. Cauchy cut out smaller and smaller chunks of a material at a given state. So, as Cauchy's volume element was taken down to zero volume, so was the mass within this volume.

If this has been your misunderstanding all along, then I and relieved to finally understand where your statements have been coming from.

Chad

## sidetracking

No, Chad, you are just good at changing the subject. You understand quite well. If V --> 0 at constant state (don't play stupid, you knew that this is what I meant), U --> 0.

f/r is the 1D-equivalent to U/V in 3D; f/r must be scale-independent in order to assess the work done by f on e.g. a spring: a given f will do different amounts of work on a large and a small spring, but the work per spring length must be scale-independent such that a small and a large spring stretched by the same relative length difference must be in the same loading state, hence f/r is scale-independent. But if so, f/A approaches infinity if f and r --> 0 at constant state.

Falk

## The most mysterious problem mankind has ever encountered

I would like to use the words of an Amazon book reviewer. Falk is "making 1+1=2 looks like the most mysterious problem mankind has ever encountered."

I just find this thread very funny.

## 1+1=2

Actually, I think Falk is trying to convince us that 1+1=3.

## Koenemann close to a breakthrough?

Koenemann says above: "... so far no hard objection to my critique of the Euler-Cauchy theory has surfaced".

Falk,

You are wrong there. That's where Sanjay's and Chad's comments are supposed to be helpful for you to understand your fundamental misunderstanding of what Cauchy is saying about the use of divergence theorem as a statement of equilibrium. It appears that you do not understand that f in the divergence theorem is stress, not force!

But I'm hopeful that we are close to a breakthrough.

Petr

## put up

Petr,

if f in the divergence theorem were stress = f/A, the integral would be

int ( f.n / A ) dA

the result would be

f ln A

That's a road into mathematical never-never land. Stay on the carpet. f is [N], and nothing else.

Question to you: the gravitational force f ~ r the radius of a subvolume within the Earth. This is known since Poisson 1813. Gravity is the gravitational effect of mass on its surrounding. If the system vanishes identically, necessarily its gravitational force must vanish identically. By precisely the same logic, a force exerted by a system upon its surrounding is the mechanical effect of loaded mass on its surrounding. If we let the system approach zero, necessarily the effect by the system upon the surrounding must vanish, such that if the system vanishes identically, its effect must vanish identically too, or else we are left with an effect without an origin. I cannot imagine that you maintain this, this would be magic. Or do you claim extra physical laws for mechanics?

I understood Sanjay in a very different way than you.

Plus: I did not base my critique of conventional CM on the divergence theorem alone. How about the zero work prediction? How about the wrong First Law? Do you understand the terms "conservative" and "non-conservative"? Do you agree or disagree that the work done in accelerating a body in freespace, and the work done in a thermodynamic change of state are entirely different things that cannot be summed? If you agree, you are in line with standard physics, but CM falls. If you disagree, you bring yourself into a very unenviable position.

Sanjay offered links to Cauchy's original papers. I am presently translating one. More on that later.

Falk

## wrong integration!

int (f.n/A) dA does not necessarily equal f ln(A).

On the surface of a pressurized sphere, f/A is a constant p (pressure), and thus int (f/A) dA = p 4 pi R^2, with R being the radius of the sphere.

If you treat A as a variable on the surface, what is your definition of A at an arbitrary point of the surface? I suppose dA means the differential area on the surface.

RH

## This is quite entertaining

This is quite entertaining, but it would be even more so if Mr Koenemann would reply on this.

Maybe he meant that A is the integration area (constant)? Even so, seeing how you get the wrong units from integrating a stress over a surface.. well it's a mistake not even undergrads would make. He also insist that f is a force by dividing by A to start with.

## Koenemann force?

Re the enclosed pressurized gas in a sphere: You say f is force, please do tell us what is acted upon by the force, and what exerts the force? This is of some consequence since in the divergence theorem you would be claiming the existence of both the surface integral and the volume integral. What is the distribution of the force in the volume? It cannot be constant, since the integral over the surface is nonzero (never mind the weird units).

Until you understand what is wrong with your divergence theorem argument there is no point in addressing the other fallacies you mention above.

Petr

## All over again

Petr,

f_syst is the force exerted by the system at the surrounding, and f_surr is the force exerted by the surrounding at the system. The divergence theorem is applied to both force fields independently. They behave similarly, with opposite sign.

The distribution of the force in the volume is not of importance because thermodynamics only considers energetic interactions of system and surrounding across the system surface. (I am not talking about statistical mechanics here, I talk about the First Law.) We can only choose progressively smaller systems at constant state. In that case the energy density P = U/V is scale-independent: both U and V are extensive terms, P is intensive. From the divergence theorem it follows that the forces f_syst and f_surr must both approach zero with r, such that f_syst, f_surr, r, U and V reach zero together. The ratio U/V remains constant, but does not have a value for V = 0, ditto for f/r.

Falk

## What do you remember from the continuum mechanics course?

Now we are getting somewhere.

So you claim that f is the total force vector between the gas sphere and its surroundings. Okay, but then to speak of a distribution of the force along the surface A is not possible, and the integral int f.n dA has no meaning. Moreover, the volume integral cannot be evaluated either, since the force f has no distribution in the volume.

If I'm correct about the force f, your argument that you refer to as the "proof" is meaningless.

Let us on the other hand assume that you mean f is the distributed force (traction, force density) along the surface A. Then it would be (classically) proportional to the pressure in the gas and oriented along the outer normal to A. Such a force density would have no distribution in the volume. Now that is a problem for the divergence theorem...

If you still have the textbook on mechanics from UC Davis (who was the instructor by the way?) you might wish to refer to concepts of force, especially interaction force (forces acting on imaginary surfaces that separate a continuum into subbodies) and the force derived from a potential as a gradient (note well, such a force would be defined within the volume V and on its surface A). For the compressible gas, the gradient of the potential energy of the gas would be the pressure.

Regards,

Petr

## Roots

Petr,

the books I was "raised" with soon became insufficient, including Malvern. I wanted more profound derivations and explanations. There was a reason why I read Truesdell & Toupin from cover to cover.

You may be the wrong person to explain this to, but anyway: From the first day on I expected thermodynamic concepts, and could not find any. They were promised for later, but they never came. In the morning I took PChem, in the aft I took Stress&Def, and they were to one another like different solar systems.

It took me a long time to learn sufficient physical background to put myself into a position where I was not just guessing, but could judge. Today it is this:

You can divide classical physics (this side of Planck) into the conservative field and the non-conservative field.

A conservative process observes E_kin + E_pot = const.That is: the energy of the system is a constant. Any work done in a conservative process results in the transformation of E_kin into E_pot and vice versa. Exchanges of energy of the system with any surrounding are explicitly excluded.A non-conservative process changes the energy of the system.Hence U is a variable. The energy exchange requires a surrounding with which the system interacts. The exchange can only consist of work being done, or heat being exchanged.hencedU = dw + dq.You can subdivide the latter category into the reversible and irreversible field, but that would invoke the Second Law, and we don't need it in elasticity. It suffices to realize that elastic deformation is a change of state _by_nature_. At this basic level no misunderstanding is possible. The consequence is that any attempt to understand an elastic deformation by conservative means (equation of motion, Newton's 3rd Law, E_kin + E_pot = const, others) is simply nonsense. But, putting this into a historical context: we are all children of our own time. The non-conservative energy conservation law, aka the First Law, was discovered only in 1847 (discounting Mayer).

Surely you know that deformation is understood as a variation. Now, who invented the theory of variations? Euler – in order to find a deformation theory under the rule of E_kin + E_pot = const. Can we blame him for not knowing better, him, a man of the 18th C? Of course not. But this does not change the fact that Euler's theory was obsolete from 1847 on. Up to today, the intro texts to CM, see Malvern or any other book, are clinically free of any First Law concept.

Falk

## is stress, not force, that's my vote

Good job, Petr is correct. Wow, it was quite difficult to determine where Mr Koenemann was coming from, but now he has clearly said in the above comments that he thinks f is force in the equation.

For continuum mechanics, in the following equation, T is stress (Malvern uses T instead of f).

int T.n dA = int del.T dV

This can be seen from observing pages 213,214 in the book by Malvern (Introduction to the mechanics of a continuous medium).

All the continuum mechanics books that I've seen indicate T as stress in the context of the above equation.

It appears that Mr Koenemann did not realize that f as he was using it (in place of T in the above equation) is referring to stress in continuum mechanics texts.

Louie

## Malvern

Louie,

Malvern Ch.5.3: you have everything in a nutshell:

- an equation of motion,

- use of rho as mass density term

- no mention of bonds,

- use of Newton's Third Law as equilibrium condition,

- leading to the momentum balance.

Rho is the inertial mass density, measured in [kg]. In Cauchy's theory it is needed to define the force, which is Newton's f = ma. It is a discrete force, not a field force, and plainly incompatible with the physics of changes of state. It is a force that you can use only in the context of E_kin + E_pot = const, in order to change E_kin; that is, to do acceleration work. To write a force field you need to start with a potential energy term, and take the derivative dU/dx. I haven't seen that in CM.

What you need instead, in order to understand an elastic deformation as a change of state in the sense of the First Law is:

- an equation of state,

- use of n as the mass term (which is dimensionless)

- mention of bonds,

- use of the thermodynamic equilibrium condition involving system & surrounding,

- leading to a work equation for the deformation, in the isotropic case PdV.

You cannot blame Cauchy for not considering the First Law, it was too early. But this does not change the fact that this theory is obsolete since 1847. Read this against the intent, just read what's there literally: M. speaks of particles; what he means are mass differentials. But if they can move relative to one another, and bonds are not mentioned, it implies that a solid is treated like a bondless, frictionless dense gas. The entire setup as you have it there is only and exclusively compatible with the condition

int f.n dA (or, if you prefer, I don't) int t.n dA = 0

which is the most concise form of saying that no work has been done upon the system. You are not under the First Law, you are within E_kin + E_pot = const. You are probably happy that your views are in line with the books. But, as carefully they are written, they are useless, they serve as bibles, but they do not help to understand a solid.

Note that for slow flow of a natural gas (consisting of discrete particles) the condition w = 0 for an isochoric deformation is indeed maintained: the shortening work in x_2 and the stretching work in x_1 cancel, such that the sum is zero. AND of course no elastic potential has built up; if you let go it will not snap back. But this elasticpotential is the key property of an elastic deformation.

Changes of state take place in energy space (PV-space in 2D), not in Euclidean space. Time is not a parameter in reversible thermodynamics. See Table 1 for the conceptual differences. Make a test: try to derive from p.214 the term PdV for an isotropic deformation (volume change). You don't have a chance.

Falk

Table 1 http://www.elastic-plastic.de/table_01.jpg

## Falk, Do you understand

Falk,

Do you understand that T as it is used in the equation I indicated in Malvern is indeed a stress Tensor? Or do you still claim that it is a force as it is used in Malvern's book on the pages I indicated? Please see the following if you don't have Malvern's book close at hand http://www.scribd.com/doc/5987971/MALVERN-LE-Introduction-to-the-Mechani...

Also, I'm curious if you believe that PV=nRT is applicable to solids?

Now, I will discuss/ask about a few of your comments.

1. I feel it is unfair to say that CM does not consider bonds. For the reason that every continuum mechanics book that I have ever read very carefully specifies that continuum mechanics is a theory that is intended to be used at a scale large enough that bonds at the atomic level may be neglected. This then means that CM is a theory that intends to find macroscopic models for materials based on matching theory with material experiments via carefully constructed constitutive models. I feel that to say it neglects bonds is to miss the point of what continuum mechanics is trying to do. It seems like you are arguing a point that continuum mechanics has said openly right up front that it is addressing from a different approach.

2. I do not contest that thermodynamics is an alternative approach to solving problems. It seems that if it is done correctly it should lead to same results (I realize you are indicating as part of your argument that sometimes you believe it does not match CM).

3. The equation int T.n dA = 0, is not, as you say, a concise way of saying that no work has been done. Rather it is an equation of momentum balance (see page 214 of Malvern when no accelerations are present).

Well, we could go round and round on this I'm sure. Certain things you say make sense, but then others are uncertain. Particularly your belief that f as used in the equation I mentioned in Malvern is force. It is intended to be stress. This is quite clear in all CM books.

If nothing else it is healthy for us to review the fundamentals.

regards

Louie

## divergence theorem

Louie,

if at all possible I follow up. Malvern's book is on my desk.

The way the divergence theorem is used on the pages you indicated is erroneous, but it is an error which I would have easily committed if I had been in Malvern's shoes. This is a classical case of an honest error.

The divergence theorem is considered to be of such universal validity that it is often used as a mathematical transformation (from surface integral to volume integral) which is true as a matter of course, no questions asked. This is not entirely true. Let me explain this.

The divergence theorem is 100% valid if the surface A of the volume V does not run through mass at any point. That is, in freespace. (In solids this is not the case.)

The divergence theorem only considers fluxes in and out of V through A. That is, if you simplify the situation and let V be spherical, and you consider the angular relation between flux f and radius r, the divergence theorem only considers normal - radius-parallel or anti-parallel - fluxes. The divergence theorem in its classical form is an implicit statement that lateral fluxes - such as tangential heat flow – do not contribute work. In that case the divergence is a measure of the total work done upon the system, and the shape of V is free. (In solids, shear forces do work.)

Thus the attempt at using the divergence theorem in the present context is massively flawed. Please notice that at this point I have made a strong effort to restrict the discussion to isotropic conditions; this right here is the reason. In the case of an anisotropic deformation where shear forces do work, the divergence theorem can only give an incomplete answer. In clear words, it is entirely possible that a physical situation exists in which normal forces do not do work, only shear forces. In that case the divergence theorem gives precisely the wrongest answer possible.

Note that in the universally valid case – when A does not run through mass – the shear situations cannot arise. Thus it is possible to assign universal validity to the divergence theorem – but only there.

Malvern also uses a cross-product form of the divergence theorem. It is entirely meaningless in the context here: in the form he uses it, it would be valid only in the case of an unbonded continuum (a dense gas making eddies), but not in a bonded continuum where the length of the lever matters, and free motions are impossible. But vortices imply free motion. Free motion imply no internal bonds. Back to square one: bonds were never considered in CM.

It remains to say that Malvern's use of the divergence theorem is rather thoughtless and uncritical. Except that I would not blame him: if I had been raised in his environment I would have made the same mistake, with the same of-course-attitude. It is due to my off-the-common-track education, in order to find out how I can beat CM, that I found out much more about the divergence theorem. It may take a week or two, but reading Kellogg is not wasted effort.

Your question if I believe that PV = nRT is applicable to solids? straight answer: Yes.

But in the form P^k V = z with the condition that if k = 1, z = nRT. For the rest see my paper Approach, Fig.1, plus accompanying text. k is the Grueneisen exponent. For the way I defined k, the condition holds for solids and for gases.

http://www.elastic-plastic.de/theory-co.pdf

http://www.elastic-plastic.de/grueneisen1908.pdf

Your question 1: bonds in solids. Bonds become terribly important if you reach free surfaces. Consider a small region within a large volume of solid, it is shortened in x_2. Its reaction will be – boundary conditions permitting – that it will stretch by itself in x_1. So far so good, but within a solid our small region is not unloaded in x_1, but is under stretch caused by the surrounding. Thus there are two stretch components, one provided by the system itself, the other one caused by the surrounding. Naturally, near free surfaces, the second component must be zero. Which means that in any discrete sample which has loaded and unloaded faces, there must inevitably be strain gradients. I have calculated them in my paper, the differences are quite important. But you can discern the two components only if you consider bonds, absence of bonds, and system & surrounding, all of which is not done by conventional CM.

Your question 2: thermodynamics. I suggest you become familiar with my theory. I may have been a pretty strange duck for you initially, but right now I hope to be a duck that you don't know yet. The first question is a matter of binocs; the second one is a matter of Peterson. (Yes, I am a birdwatcher.)

Your question 3: see answer to question 1. I maintain my point that T.n dA = 0 is a concise way of saying that no work has been done – if only normal forces are considered, and this condition is solidly built into the divergence theorem.

f and stress: I take the liberty to disagree from the books. f = [N].

And now I am going to bed. Seeya tomorra.

Falk

## Eerily familiar.

http://en.wikipedia.org/wiki/Bogdanov_Affair

I especially identify with the comment "like trying to nail jello to a wall".

## Tres important (translate "interesting")

"...Some of the papers of the Bogdanof brothers are really painful and clearly silly...."

## Falk Koenemann: hoax perpetrator or just dense?

Evidently, I'm still undecided. There is really no way of knowing for sure.

As a hoax this new theory of elasticity would be a fairly obvious one (and hence not a very good one). And there is no point in arguing with a hoax perpetrator.

Then there is the possibility that this gentleman is just very uneducated (to put it mildly) as far as mechanics goes. With the latter possibility in mind I've tried, alongside a few patient imechanicians, to show FK his errors.

However, it is now clear that this man will not be educated, he simply refuses. So that's it for me, I'm afraid I've lost interest. A word of advice to anyone else tempted to try: just ignore this man, and disregard his garbage.

Petr

## Bon Voyage

Petr and Chad,

Let me be the first to wish you bon voyage. Your postings to this discussion often resemble the antics of the class clown who imagines his jokes are a schoolyard sensation.

It requires a good deal of courage to question the status quo. How much courage does it take to mock and ridicule such efforts?

It was my understanding that science advances on the basis of seeking answers to questions. Koenemann has done just that. You may disagree with his answer (and he has encouraged such disagreement), but you have no basis for calling into question his motives or his integrity.

It is virtually a cliche that those who propose radically different solutions or explanations that challenge or threaten accepted wisdom will be dismissed as amateurs or kooks, or stand accused of intellectual dishonesty. One will have no trouble finding examples of such behavior in this blog.

This professional community should feel free to thoroughly examine and test Koenemann's proposals. The conceptual framework has been constructed, the tools sharpened, and the methods outlined. Now it's time to do the work and compare the results.

What is truly unfortunate is that many young people reading this blog will get the overwhelming impression that this field is dominated by orthodoxy, that bold thinking and aggressive questioning will trigger mob retribution.

This is not the sort of behavior one would expect to find in a field of intellectual inquiry. To be cowed into keeping silent for fear of being ridiculed is to risk losing the truth itself.

## I agree

I agree with Petr and Chad.

The logic goes like this:

1. Mr Koenemann says that continuum mechanics is invalid.

2. As far as I can tell Mr Koenemann demonstrates by his statements that he does not understand what continuum mechanics is saying.

3. Hence Mr Koenemann is not equipped to question the validity of continuum mechanics, since it appears that he does not understand what it is that he is criticizing.

Therefore, I choose to move on to different discussions. I sincerely wish many years of enlightenment for Mr. Koenemann. If he is indeed sincere, I appreciate his willingness to discuss the matter with me.

Louie

## Evidence

I have not seen hard arguments so far from anyone, disregarding Chad's various attempts to willfully misunderstand the divergence theorem. By hard argument I mean either mathematical or physical logic. So far, it has survived. The best contributions, by tone, attitude and quality, came from Rui.

I have far more reasons to question the Euler-Cauchy theory than just the divergence theorem argument. No one addressed them.

No onecommented on the distinction "conservative – non-conservative", the most profound distinction in classical physics. If you do not understand the difference between E_kin + E_pot = const and dU = dw + dq, you cannot claim to know much about physics altogether.No oneobjected to the zero work predictions of the Euler-Cauchy theory. This is simply ignoring straightforward proof.No oneseriously replied to my claim that bonds in solids are not contained in the theory. Instead, once again I am given reasons why bonds are not important. This is voodoo science.No onechallenged my claim that the foundations of CM are clinically free of any thought compatible with thermodynamics. Quite the opposite, Louie's references have confirmed this.And, as far as I can tell, none of those who contributed, bothered to read my papers. What do you think they are there for? Louie asked for the equation of state (EOS) that works for for gas and solids. I gave an illustration for the EOS I propose – it is possible to show that one and the same EOS can be used for different materials. So, Louie, if you withdraw, it is not because I did not answer your question. It is because you feel better in the crowd. Or do you have any objections to my discussion regarding the application and validity of the divergence theorem? Let me guess: you don't know enough, but you are not going to admit it.

Never mind. The gun is loose now. Among 7000 reads there must be a few thoughtful readers, and that's enough.

Falk H. Koenemann

http:/www.elastic-plastic.de

## my problem with divergence theorem

In my effort to understand Mr. Koenemann's theory, I have so far managed to read only one and a half of his papers besides most of the discussions here on imechanica. The first paper I read about a year ago was his "comment on Gurtin" (International Journal of Modern Physics B

22, 5035-5039), which was not impressive to me at all, by style, tone, and its technical content. More recently, I glanced over his 60-page article, An approach to deformation theory based on thermodynamic principles (International Journal of Modern Physics B22, 2617-2673). Apparently, this is a more serious paper. After Introduction, Terminology, and General Remarks in the first three sections, I ran into the discussions on thermodynamic system in Euclidean space along with the divergence theorem, which has been the focal point of many discussions above in this thread. It is clear to me that I must understand Mr. Koenemann's interpretation of the divergence theorem before I can move on with his papers, and that's where I stopped and that's where I started my comments above, After several rounds of discussions, I am still confused by how Mr. Koenemann interprets the divergence theorem, physically and mathematically, which I will try to summarize briefly as below. On the other hand, I must point out that there are many bigger questions in Mr. Koenemann's theory that could be discussed, but it would be even more difficult if we cannot reach any agreements at the starting point.Now, for the divergence theorem that has been discussed above, I maintain the following:

(1) Mathematically, the equality between a surface integral and a volume integral can be established for an arbitrary vector field f. The mathematic conditions for the divergence theorem to hold may be further discussed, with specific physical problems in mind.

(2) If the vector f is the external force acting on the surface of the system, the surface integral is not a force quantity. Its unit is [force(N)][length(m)]^2. The explanation above by Mr. Koenemann as to how the unit of this integral remains in [N] does not make sense to me. One may convert an integral over a spherical surface area to an integral over spherical angles, but the conversion does not change the unit.

(3) In vector form, summation of all external forces acting on a system equals zero at equilibrium, and this is the only correct way to sum up the forces.

(4) With (2) and (3), I do not understand the physical meaning of the surface integral of f.n (normal component of external forces) in the divergence theorem.

(5) It remains puzzling to me what f is in the system (V) and why the divergence of f has to be a constant everywhere in V. Mr. Koenemann’s answer, “f is either f_syst or f_surr”, seems to apply on the surface, but not in the volume. Say, at an arbitrary point in V, what is f_syst and what is f_surr? Does the "system" vary with the location of the point? Do we need a surface passing that particular point to separate system from surrounding? If so, there are infinite choices for such a surface at each point, and which one should be used?

(6) The purpose of the divergence theorem in Mr. Koenemann’s theory seems to show that f is proportional to r and thus f/A does not exist as r -> 0 (i.e., non-existence of the Cauchy stress). This would only make sense to me if I understand (5).

(7) In Mr. Koenemann’s paper (Approach), the force is also defined as the energy flux, f_i = dU/dx_i (partial differentiation). Such a definition for a force field seems to make sense in a discrete system (particles, atoms), but in a continuum the partial differentiation with respect to the coordinate of a particular point would inevitably lead to singularity.

RH

## discrete system, and others

Rui,

(1) Mathematically, the equality between a surface integral and a volume integral can be established for an arbitrary vector field f. The mathematic conditions for the divergence theorem to hold may be further discussed, with specific physical problems in mind.

OK. - The point of importance is this: the validity of the divergence theorem is unlimited in freespace, it is certainly limited in distributed mass, and it is further limited – this is not in Kellogg, but it is my conviction – in bonded distributed mass.

(2) If the vector f is the external force acting on the surface of the system, the surface integral is not a force quantity. Its unit is [force(N)][length(m)]^2.

Yes. As far as I could find out, [Jm] is the unit in which Coulomb's law was originally perceived. The definition was changed later when the Ampere was chosen to be the most basic unit in electricity.

The explanation above by Mr. Koenemann as to how the unit of this integral remains in [N] does not make sense to me. One may convert an integral over a spherical surface area to an integral over spherical angles, but the conversion does not change the unit.

int f.n dA is in [Jm] because it is an integration over a surface term. If you integrate over the space angles, you need a scaling factor to indicate what theta means in real distances. That scaling factor is the magnitude of the radius in question. It is always involved (e.g. in fxr). The unit is in [J]. This is convenient because [J]-terms are scalars by definition, so there's no ambiguity.

(3) In vector form, summation of all external forces acting on a system equals zero at equilibrium, and this is the only correct way to sum up the forces.

f.r and |fxr| are [J] terms, and scalars which you can integrate around 2pi.

(4) With (2) and (3), I do not understand the physical meaning of the surface integral of f.n (normal component of external forces) in the divergence theorem.

I can only repeat what I wrote further up: I accepted Gauss' view that [Jm] is a meaningful unit. In the first definition of Coulomb's law this unit was clearly implied. (It has been changed since, see above.) But I preferred integrals which led to [N] or [J], and integrating over the angles satisfied me.

(5) It remains puzzling to me what f is in the system (V) and why the divergence of f has to be a constant everywhere in V.

Excluding external gradients, the constancy of div f within distributed mass is not my contention, but that of Gauss. But I do not wish to defer responsibility to others. Treating del.f as a constant made sense to me. There must be some term that scales a force to a length term of the system involved (e.g. a spring) to make a relation of force to work possible, see answer to Chad, April 3, 23.49h. – del.f is not a constant if V varies in scale, but mass does not. This is the case of a system with a finite body in freespace.

Within distributed mass, the constancy of del.f follows in my view from the fact that int del.f dV must be linearly proportional to the limits of integration, or else the result is not proportional to mass. The latter would bring us into a real mess. consider k an extensive parameter, and del.f an intensive parameter.

Mr. Koenemann’s answer, “f is either f_syst or f_surr”, seems to apply on the surface, but not in the volume. Say, at an arbitrary point in V, what is f_syst and what is f_surr? Do we need a surface passing that particular point to separate system from surrounding? If so, there are infinite choices for such a surface at each point, and which one should be used?

On the surface of the system, two forces interact, f_syst and f_surr. The question what happens inside is not the subject of thermodynamics; this theory only considers exchanges between system and surrounding. – I am not trying to dodge your question. If you find that unsatisfying, we can only change the scale of the system, i.e. chose systems at varying scale.

In principle we do not need the surface as a planar or curviplanar geometrical element. But the sum of all radius vectors emanating from Q, indicating all points of action of external forces f_surr, necessarily forms a surface.

(6) The purpose of the divergence theorem in Mr. Koenemann’s theory seems to show that f is proportional to r and thus f/A does not exist as r -> 0 (i.e., non-existence of the Cauchy stress). This would only make sense to me if I understand (5).

The purpose of the div-theorem argument is (a) to show that the Cauchy theory does not work, (b) to show that we must establish an unit distance between Q and P, the radius, which can then serve as a zero potential distance. The latter can be used to define work; e.g., if V_0 is changed to V_1 by a pressure increase, the same holds for the radius r_0 to r_1.

(7) In Mr. Koenemann’s paper (Approach), the force is also defined as the energy flux, f_I = dU/dx_I (partial differentiation). Such a definition for a force field seems to make sense in a discrete system (particles, atoms), but in a continuum the partial differentiation with respect to the coordinate of a particular point would inevitably lead to a singularity.

Surprised, explain. The particular point is in my case always the point Q, the center of mass of the system, in the ideal case removed from the surface by one unit distance.

Perhaps the confusion is in the expression 'discrete system'. I have seen this expression used with a different meaning. A discrete body is a body bounded by sharp surfaces against some other medium or the vacuum such that a surface can be wrapped around it that does not run through mass. But a discrete system need not be a discrete body; a discrete system is a system defined by a given mass the identity of which is fixed. This system can be separated from its surrounding (which may consist of the same substance) by a sharp surface. This condition is implied here. Thermodynamicists use other types of systems, e.g. in the case of diffusion: a system of water may be open to mass exchange provided the mass in the system remains constant (in the same way as heat is exchanged between system and surrounding at constant temperature).

Falk

## Re: discrete system

In my language, a discrete system consists of a finite number of particles (e.g., atoms, molecules, etc.) enclosed by a closed surface. The particles interact with each other (not necessarily bonded) and the system interacts with its environment across its surface (e.g., mass transport, heat conduction, energy flux). Please see my post on "a point and a particle " for clarification.

RH

## points & distributions

Rui:

point 0: agreed.

point 1: agreed.

point 2: agreed.

point 3: sentence 1 & 2 agreed. Sentence 3 is in conflict with the thermodynamic theory because the P-terms in the thermodynamic equilibrium condition P_syst + P_surr = 0 are not zero in the loaded state. It is time to consider the nature of pressure itself. U/V is the energy density, a scalar by definition. f/A is a force density on a surface. They are not equivalent, and U/V is the more fundamental one.

point 4: disagree. You avoid the thermodynamic definition P = U/V at all cost. Area does not have vector nature, but the surface normal vector is a representative of the plane. If you want to use it, you can do it for planes at any point except the chosen reference point Q, because there is no notation for these planes. The use of planes in space in the Cauchy theory is incompatible with basic vector space properties, see

Unorthodox Thoughts, chapter 8.point 5: no contest, but irrelevant in the present context. Thermodynamic work is time-independent, and not an acceleration.

point 6: Sentence 1 & 2 agreed. Last sentence is foggy.

In my theory, bonds occur in two different ways:

(1) Bonded substances have an internal pressure. This pressure is internally balanced in the unloaded solid/fluid. Its nature is compatible with your sentences 6.1 & 2. See

Grueneisen 1908, and farther backClausius 1870.(2) The bonds between system and surrounding must be taken into account because without them the surrounding cannot do tensional work on the system. This has implications for the final deformation. Surface-bonding forces are constraint forces, i.e. they do not do work themselves, but without them the acting forces (here the tensional forces exerted by the surrounding) cannot do work.

Unorthodox Thoughts http://www.elastic-plastic.de/thoughts.pdf

Clausius http://www.elastic-plastic.de/clausius1870.pdf

Grueneisen http://www.elastic-plastic.de/grueneisen1908.pdf

CM suffers from the heritage it carries around from Newtonian mechanics of discrete bodies. The 'part' or 'particle' is required, among other reasons, for the definition of the force. The limit operation in Cauchy's theory is called the continuity approach because it is thought to dissolve the continuously distributed mass into mass points, mass differentials, parts or particles which move past another. The concept is in part influenced by the recognition that all matter is discrete, but the two concepts do not harmonize.

In potential theory a very different view is offered. A mass point is something entirely different: a body of mass in freespace can be considered a mass point with the condition that its mass is conserved, but concentrated in one point. In effect, only the volume of the mass is ignored. One condition for this assumption to be valid is that the mass distribution in the system is scale-independent, in clear English it is constant. The concept is of use in astronomy, for example. It must be given up whenever the dimension of the mass can no longer be ignored, e.g. if the shape of the body matters.

This concept of the mass point cannot be carried into continuum physics, however, because the mass considered in a continuum is necessarily scale-dependent. The logical alternative is then to consider an unit mass of finite extent.

This is what is done in thermodynamics, and this is what I propose. It would mean to give up the concept of the 'part' or 'particle' entirely and treat a solid as a perfect continuum of mass.

I submit this to your consideration.

Falk

## Work-free isochoric deformation

Dear Mr Koenemann,

I am fascinated by this exchange, but I too suspect that there are serious problems with your new theory. I will nevertheless try to keep an open mind and give you an opportunity to respond to queries. Since no one else has addressed your contention that continuum mechanics will always predict zero work for an isochoric elastic deformation, I will make an attempt to understand it. I have read your paper entitled "ON THE SYSTEMATICS OF ENERGETIC TERMS IN CONTINUUM MECHANICS, AND A NOTE ON GIBBS (1877)", in which this claim is detailed. You have asserted that you demonstrate three times in this paper that standard continuum mechanics erroneously evaluates the work necessary for a volume-neutral deformation, and hence I will comment on the three examples you provide.

1. You use the standard continuity equation, and make the statement (a) "that stretch displacements and contraction displacements sum to zero" and hence (b) work must be equal to zero. In a sense (a) is true: if one takes a bar of material with Poisson's ratio 0.5 and stretches it, volume is conserved because radial contractions occur in directions perpendicular to the stretching. In this case, using standard notation and stretching in the 11 direction, sigma_11 = sigma, while sigma_22 = sigma_33 = 0; the corresponding strains will be something like epsilon_11 = sigma_11 / E; epsilon_22 = epsilon_33 = - sigma_11 / 2E. Continuum mechanics would find that work has been performed because the extension is accompanied by a force in the same direction, while the radial contractions are work-free because no forces are associated with the direction of these displacements. Hence there is a net positive work, and your statement (b) that continuum mechanics would find zero net work is incorrect. Then you make the remarkable claim that "stretch work and contraction work have opposite sign". Typically, positive work is that which is done upon a system while negative work is done by a system (or vice versa, if that suits your tastes). If your statement were true I would be able to extract energy from a system merely by compressing it (or stretching it, depending upon your sign convention).

2. You note that for isochoric, isentropic deformations, continuum mechanics would say that the increment in internal energy dU is equal to the product of sigma_ij (the stress) and d epsilon_ij (the strain increment). You then state that there are two conditions for an isochoric deformation: sigma_ii = 0 and epsilon_ii = 0. The latter is true; the former is not (see the example in 1. above, which has an isochoric deformation). In any case, even if this were true, it would not make the product sigma_ij d epsilon_ij identical to zero. Take, for example, the tensors associated with the principal stresses sigma_i = [1 -1/2 -1/2] and the principal strains epsilon_i = [1 -1/2 -1/2]: the traces of both tensors would be zero but the product would not. Moreover, you state that Bachelor notes that "the deviatoric stress tensor [...] does not contribute energetically to the deformation". Bachelor, on page 142 in my edition (section 3.3), states that the deviatoric stress tensor is directly associated with the motion of a viscous fluid, which is a disspiative process requiring energy and hence work. This is the opposite to what you claim Bachelor wrote.

3. I don't have Landau & Lifshitz on my desk so I cannot adequately comment on point 3. There is also, I suspect, a typographical error in (10), which would further complicate discussion until clarified.

I also found curious your critiques of Love and Fung earlier in this paper. Are you objecting to the partitioning of the energy of an object into the kinetic energy of the object as a whole and the internal energy due to temperature, strain, etc? Or are you objecting to the integration of the kinetic energy of the differential elements composing the object to determine the kinetic energy of the entire object?

I appreciate your assistance with these issues. I am, however, increasingly convinced that the theory you propound is incorrect.

Regards,

Craig

## work, and the First Law

First of all, to general audience: sorry for posting the same twice, a technical misunderstanding.

Craig,

1. I separate your point 1 into two. But I must discuss the second point first.

Your argument contains a sign error. In thermodynamics, the sign convention is this: a contraction is due to a pressure increase and perceived as positive work, so that an expansion requires negative work. If you perform an isotropic contraction and then let go, the system will expand; the work of step 1 and 2 cancel such that U_2 = U_0. Your logic would imply that contraction work done by the surrounding and contraction work done by the system have opposite sign, or that the work done in a contraction starting from U_0 and an expansion starting from U_0 have the same sign. Not so.

Now the hard part. If you stretch a bar of length d in x by Delta d_x, it will attenuate in y and z, but only if you let it. If you prevent the system from contracting in y and z, this Delta d_x will cost you far more Joule because in effect you perform an expansion. Gibbs (explained at the end of my paper) concluded that if nothing happens in y and z, no work is done in these directions. This is to show that Gibbs' interpretation and yours are already in conflict: either no work is done when nothing happens, or no work is done when the thing attenuates. I think both views are too simplistic.

If you stretch in x while keeping y and z unchanged, and then relax the boundary conditions by making the walls soft, the material will relax energetically by pulling the walls in along y and z, and the energy released is the work done in y and z in the first place. What we have now is this:

- if the substance is a gas, and V = const, indeed no work would be done in the deformation. But it would not snap back upon release; no elastic potential has built up.

- If the substance is a solid, and V = const, a finite amount of work is observed: negative stretch work is done by the surrounding in x, and positive work is done by the system upon the surrounding in y and z – and there is an elastic potential.

Now you can conclude: the material contracts because it has this and this Poisson's ratio. Or you can conclude that the material attenuates because of the law of least work: the change of state is kept to a minimum. The former is phenomenological. The latter is energetic, a physical thought.

I think a proper deformation theory should not need the Poisson term. It is not a material property, but a result of the physical setup, subject to specific boundary conditions, the presence of the free surfaces. It is dependent on the shape of the bar, it matters if the bar has square or rectangular cross section.

This example, the stretching of a bar, is a nice example to show that the boundary conditions were never sufficiently examined in the beginning: it matters a lot if the bar is allowed to attenuate, i.e. if there are free surfaces nearby. But if you think of deformation in the Earth's mantle, the nearest free surface is very far away, and talking about Poisson's ratio down there is at best funny.

All this is to show that a change of state in the sense of the First Law is always work done upon a volume, not work done in a particular direction. Whatever happens, it is spatial, and the material reactions in the various directions are never independent of one another. A proper theory must take this into account.

2. In the light of the above, your sentence "The latter is true, the former is not" no longer holds. Besides, you can find the statement – that for an isochoric deformation the trace of the stress tensor is zero – in textbooks, e.g. Gurtin 1972, p.53, and I have seen it in Gurtin 1981, but there I cannot give you the page. Given the conservative mathematical structure of conventional CM, the zero condition for the stress tensor in isochoric deformation is in fact unavoidable: if the material paths cancel, the work done in the deformation must cancel. And if epsilon_ii = 0 and sigma_ii = 0: it is true, tensor products do not communicate; but the products of the invariants do communicate.

Bachelor: I refer to the entire chapter, specifically to the eqn. without number above eqn.3.3.8, p.144. But I disregard parts where Bachelor explicitly refers to viscous effects, because viscous flow is irreversible, and thus the Second Law is involved. Before we get into the irreversible field we should first clear the reversible field. Bachelor must have been aware of Poisson's partition of a viscous flow step into an elastic, time-independent, reversible loading step and a diffusion-controlled, time-dependent, irreversible relaxation step. He must have learned about this from the same source as myself: from Stokes' paper. I refer to the elastic part only.

3. There is no typo in my eqn.10, and the pages of L&L referring to stress are available at

www.elastic-plastic.de/landau-stress.pdf

Love, Fung et al: you must make a decision which process you wish to consider. If it is the acceleration of a body as a whole, it is done under E_kin + E_pot = const = H where H is the entire mechanical energy of the system. Note that up to here the system is isolated. If a change of state is to be considered, be it isotropic leading to PdV-work, or anisotropic as in elasticity, interaction of system and surrounding is required, and the energy conservation law is dU = dw + dq. U_0 and H have different names in CM and thermodynamics for historical conventions, but they are identical terms. If a change of state is to be studied, U = H is a variable, and Delta U is the elastic potential. In all the quoted books an attempt is made to treat E_kin and U as separate variables. This is physically impossible, E_kin is a subset of U, they cannot be summed.

An external acceleration of a body cannot change its internal energy. If a change of state is the subject, any possibly existing external acceleration is simply irrelevant, besides I think it is impossible, because in order to perform a change of state, external equilibrium must exist.

Clausius outlined it quite well in the virial law: in

E_kin + E_pot = PV

the first term is the heat, the random motion of the atoms which accelerate one another. If so, an equation of motion causing an acceleration is no longer available for further use. The second term LHS is the bonds. I have always missed them in conventional CM. The term RHS is the state of the system in the standard state. If that term is a variable, RHS = PV = nRT, we have the equation of state and get into standard thermodynamics, and the LHS is no longer of interest; in effect, E_pot interacts with the external forces. But it is clear that Clausius has been completely overlooked in CM.

I have not found out who wrote the form of the First Law as used in Love etc., I suspect it must have happened already in the 1850s. But the attempt to treat U and E_kin as variables that can be summed amounts to making the First Law, the energy conservation law of non-conservative physics, subordinate to E_kin + E_pot = const. This is the First Law upside-down.

Falk H. Koenemann

## Work-free isochoric deformation

Dear Mr Koenemann,

Thank you for your detailed response. I would like to suggest two things. First, since you have written a response twice as long as my original post, and because each of us could easily continue to do the same until we have used all of the electrons on the internet, I'd like to reply only to one aspect of your post: the original assertion from your Gibbs paper (which you have rephrased earlier in this thread). The second suggestion is that sometimes, in cases of disagreement such as this, it is helpful first to determine on which points we are in agreement, and then progress to where our views diverge.

The original claim you made in your Gibbs paper is: "The consequence of the attempt to interpret elastic deformation as a conservative process in the sense of eqn.1 is that it is not possible to derive a non-zero work term for a volume-constant elastic deformation." In this thread you have rephrased this in your first, summary post of 9 March 2009 as "the current theory of elasticity is a perpetuum mobile theory: the predicted magnitude of the work done in a volume-neutral deformation is always zero". I interpret these to mean what they say on their faces: that if continuum mechanics is applied as it is currently understood, then zero work will be calculated for all isochoric deformations. In my previous post I gave a trivial example of stretching a material with Poisson's ratio 0.5 for which continuum mechanics calculates net positive work (or negative, if that is your preferred sign convention, but in either case non-zero). Either this contradicts your original assertion, or your original assertion does not mean what it seems to state. Before taking this further, could we come to an agreement on what your original statement means? It would help very much to know exactly what we are discussing.

By the way (violating my intention above to address only one issue), how are you defining E_kin?

Craig

## Work

Craig,

work: I thought I had answered this. The trace of the stress tensor is sigma_ii = 1 – 1/2 – 1/2 = 0, and ditto for epsilon_ii. Hence the product of the terms relating to work – the traces of the respective tensors – is 0 x 0 = 0.

It is entirely sufficient to consider sigma_ii only. sigma_ii = 0 is a tatement to the effect that the inward-directed and the outward-directed forces balance. In the conventional theory work done by shear forces is not considered.

Let me show you how this is done in my theory (a little simplified): the total force field was split into (1) an isotropic component, and the deviatoric field; the latter was split again into (2) the normal force field and (3) the shear force field. I assumed that the ideal deformation is the one that requires the most work, which is an isotropic contraction, caused by the isotropic inward-directed force field of step 1. Then I considered the deviatoric force field, and here normal forces and shear forces separately. The volumetric and energetic net effect of the normal force field is indeed zero. The work done by he shear forces is a geometric expansion which indeed balanced the volume loss by the first step; and an energetic relaxation, but here the work done in the first step is only partly relaxed, a rest remains, so the net work is non-zero.

E_kin = m v^2

Falk

## Work-free isochoric deformation

Dear Mr Koenemann,

I remain confused by this, and I am afraid that we still haven't quite agreed upon what exactly you mean by your statement: "the current theory of elasticity is a perpetuum mobile theory: the predicted magnitude of the work done in a volume-neutral deformation is always zero". Are you stating that by applying standard continuum mechanics to an isochoric deformation, then it is impossible to calculate non-zero work? Or am I misunderstanding this? We are both reasonable people, so we should try to come to an agreement on this point before complicating matters further.

If I am understanding you correctly, then my first example of an isochoric deformation (I apologise for being insufficiently specific in my previous post; sometimes my writing lacks rigour. I meant the one involving the stretching of a material with 0.5 Poisson's ratio by an axial force.) has principal stresses [sigma 0 0], and the corresponding tensor does not have zero trace. (Parenthetically, I disagree with your claim that an isochoric deformation must be accompanied by a stress tensor with trace zero, and further with your claim that the traces of the stress and strain tensors are related to work. Work is related to sigma_ij epsilon_ij; the products of the elements of the two tensors are taken before the summation, but to keep life simple I don't think we should consider that yet.) Standard continuum mechanics clearly calculates non-zero work (for a linear elastic material it will be something akin to sigma^2/2E) for the isochoric deformation in this example, and hence if your statement means what it appears to mean, then your statement is incorrect. Am I interpreting your statement accurately? If not, how ought it be interpreted?

Thanks again for your help with this.

Regards,

Craig

## condition tr s = 0

Craig,

Yes, I imply that the condition "trace of stress tensor = 0" implies that no work is done. In fact, it is not an implication, but there cannot be a more explicit statement to that effect. And yes, I know it flies into the face of everything you and I have learned in grad school. Potential theory does not leave you another option.

If I am incredibly glad for having found iMechanica, it is because I can now make such statements in public, and they cannot be drowned in silence any more. This blog is intensely watched. The condition "trace of the tensor = 0" is an expression of the Laplace condition div v = 0 (v = general vector field), indicating that the net flux of the variable in question is zero (in this case energy flux). Of course I know that most of the people out there are unfamiliar with potential theory, and they are as incredulous as you are, but this is why I create all this noise.

The question as to what source density is, has come up a week ago (31 March). Assume that the matrix [-1 0 | 0 1] describes the flow of water in and out of a region. The trace is zero; as much water flows out as flows in, such that no extra water flows out (the region would be a source) or remains there (it would be a sink). With mass flows this appears natural, but the same matrix describing a heat flow tells you that the region does not contain an oven. Now think of energy U and its flux dU/dx = f. If the trace is zero, it means that the region remains energetically unchanged, i.e. no work is done upon the system. – To be fair to Euler & Cauchy, these systematics were understood only way after 1830.

Second question: my first answer to you was a bit long because I did my best to discredit the Poisson ratio as a physical term. Its observation depends on the existence of free surfaces nearby. Your statement that the principal stresses are [sigma 0 0] is in my view wrong too. – You force me to another long answer. Please download

http://www.elastic-plastic.de/slide25.jpg

This is a matrix of states. First column: the unloaded state; second column: the loaded state with the condition that system and surrounding are solidly bonded; third column: the loaded state with the condition that system and surrounding are not bonded. First row: isotropic compression; second row: isotropic dilation; third row: anisotropic deformation. In isotropic compression, it does not matter if system and surrounding are bonded; the system must contract. Conclusion: normal compressive forces can always do work.

In dilation the bonding condition matters; if system and surrounding are not bonded, a cavity will be created in which the unloaded system rests as a discrete body.

In an anisotropic loading configuration normal compressive forces can do work (in y), and as a result the system bulges (in x). It does so in order to minimize the change of state. But it depends on the bonding condition whether the surface point of the system on the x axis is under tension by the surrounding, or unloaded. If the system is under tension in x, there will be two stretch components in x, one supplied by the system itself, another by the surrounding. Of necessity, the first stretch component must exist from the interior of the sample all the way to a free surface; but the second stretch component must reach zero on free surfaces.

This example shows a bunch of things:

- Inevitably, there is a strain gradient in loaded bodies with free surfaces that fades towards the interior.

- The state of loading (I avoid "stress" because I do not want to refer to Cauchy's tensor) must also vary as a function of position within the sample. Inside a sample the stored energy must be larger than near free surfaces. (See the model calculations in Approach, last chapter.)

- It is impossible to ignore bonds. The bonding forces are constraint forces (see my answer to Rui Huang on April 7).

Altogether, the description you gave above – [sigma 0 0 ] – becomes too crude. And what does Poisson's ratio mean here?

To an engineer, it is all too natural to start with stretching a bar, i.e. a discrete body with free surfaces. However, I am a geologist, and the infinite continuum came very natural to me. Free surfaces do not exist at metamorphic depths, and I missed the second stretch component from the beginning. I am saying this because it shows that different backgrounds produce different perceptions, and we must be careful with what we take for granted.

Falk

## Work-free isochoric deformation

Dear Mr Koenemann,

Thank you again for your detailed explanation. I am, however, still confused about the very first point which we are discussing. Consequently our conversations cannot be very productive as I am not certain that we are in fact discussing the same thing. I return to your original statement: "the current theory of elasticity is a perpetuum mobile theory: the predicted magnitude of the work done in a volume-neutral deformation is always zero". We have not yet managed to agree even on the meaning of this (or if we have, I missed that, since I don't yet know quite what this means). Are you stating that, for all volume-neutral deformations, the standard application of continuum mechanics (with which you, admittedly, disagree) will calculate zero work?

Let's come to an agreement on what this means. Once we have cleared up this point then I will at least know what we are discussing. Until then, it is very difficult for me to make rigorous statements about something which seems rather undefined. If we work on one thing at a time, we will at least make that much progress. By discussing dozens of topics simultaneously we seem to be foundering on the complex interrelationships between these things.

Regards,

Craig

## work again

Craig,

the condition tr sigma = 0 is the only possible condition for a volume-neutral deformation, pure or simple shear or other. Since that condtion is compatible – in the light of the principles of potential theory – only with the conclusion that no work has been done, I must conclude that a volume-neutral deformation can be achieved without doing work, according to the conventional theory. This is the perpetuum mobile condition of the first kind, namely that we get an effect (a deformation) without doing work. I hope this is as unambiguous as words can be.

I think I have made it clear that I have a sufficient number of systematic objections to the conventional theory in non-straightforward cases to be able to deal with them as well. Among others, the L&L-textbook example demonstrates that the error is solidly built into the conventional theory.

This is as focussed as possible. I have given one example where the zero work condition for a deformation is actually correct, namely for the volume-neutral deformation of a gas when no elastic potential builds up. The point is, from the entire mathematical & philosophical structure of conventional CM, one can only conclude that the work done in a deformation is work done under the conservative energy conservation law E_kin + E_pot = const; whereas it has never been properly considered that elastic deformation work must be akin to change-of-state work, i.e. work done under the First Law, in the isotropic case this is PdV-work.

Falk

## Work-free isochoric deformation

Dear Mr Koenemann,

Yet again, thank you for your reply. I feel that it is very important to be completely clear about these preliminary points, otherwise we will end up speaking past each other due to differences in terminology or in the way that we are defining our concepts. So, what you are stating is that:

(1) Conventional continuum mechanics will calculate non-zero work for an isochoric deformation such as the uniaxial stretching of a material with Poisson's ratio 0.5 (as I have done above, using conventional continuum mechanics).

(2) However, you believe that conventional continuum mechanics is incorrect; as an example, you believe that conventional continuum mechanics wrongly allows the trace of the stress tensor to be non-zero during an isochoric deformation.

(3) When you apply your interpretation of continuum mechanics to an isochoric deformation, you produce absurd results, such as uniformly zero work for volume-neutral deformations.

(4) Therefore conventional continuum mechanics is invalid.

Have we agreed on this much, at least?

Regards,

Craig

## isochoric

Craig,

because I do not believe in the necessity of Poisson's ratio as a physical term, because I have made it clear that free surfaces nearby change the end result, and because I reject the proposition that the attenuation in y and z comes free, I prefer a perfect infinite continuum simply so we are both in unambiguous terrain.

Assume a situation in which stress s_11 = a (a is a free numerical, not an area),

s_22 = -a/2, s_33 = -a/3 such that tr s = 0; in the following the brackets are 3x3 matrices with zero terms left out

then s_ij de_ij = s [a -a/2 -a/2] [de_11 de_22 de_33]

= as de_11 - (as/2) de_22 - (as/2) de_33

= as (de_11 – 1/2 de_22 – 1/2 de_33/2)

= 0

which means you do not have to do the integration any more. Integrated over unit length the strains e will be proportional to s, so the deformation is isochoric. But since the paths cancel, the work cancels.

Falk

I shall be absent for a day or two.

## Work-free isochoric deformation?

Dear Mr Koenemann,

I appreciate your response, but I am afraid this raises further questions.

(1) Your original claim was, effectively, that conventional continuum mechanics was so flawed that it will produce absurd results for simple cases. In particular you claimed that for all cases of isochoric deformation, conventional continuum mechanics will calculate zero work. You then demonstrate this claim by using your own unusual, perhaps unique, interpretation of mechanics, with which most if not all workers in the field would disagree, to generate absurd results. Surely if you are going to criticise conventional continuum mechanics for generating absurd results, you should at least use conventional continuum mechanics to generate the absurd results.

(2) After assigning the stress field, your equation would be correct provided de_11 = 1/2 (de_22 + de_33). This could be, most obviously, an isotropic expansion or contraction such that de_11 = de_22 = de_33, but it is most certainly not isochoric, nor is e_ij proportional to the s_ij you set. What exactly is s, and why are you multiplying by it? And how is this connected to the stress field you propose? I am not familiar with stress fields of the sort you suggest which would produce isotropic expansions. Are you suggesting that you can arbitrarily assign stress and strain fields? This makes no sense at all to me, physically, mathematically or logically.

Regards,

Craig

## signs?

Craig,

sorry, there is a sign error in your point 2, line 1: if tr s = s_11 + s_22 + s_33 = 1 -1/2 -1/2 then tr s = 0. The s_ii (no sum) are the algebraic terms; only their value is positive or negative. The relation s de (s = sigma = stress, e = epsilon = strain) is the formula by which the strain is calculated as a function of stress. – That's how I learned it. I don't agree with it. But right now you ask me to argue in terms of the conventional theory.

If we calculate s de, the sign of the result comes from the sign of the s term. Hence if we integrate over unit magnitude of e, the result is s_11 e + s_22 e + s_33 e = 1 e – 1/2 e – 1/2 e. so this is not an isotropic expansion, but an isochoric deformation.

Again, please check the signs. I did not indicate an isotropic expansion. And, no, I do not think we can arbitrarily assign stress and strain fields. But if we choose a stress with magnitudes as I proposed such that tr s = 0, by means of s de we cannot come to another result than that the total is zero too. For strain e it means that it is isochoric; a non-zero value would indicate a volume change. For work s de it means that it is zero.

Thank you,

Falk

## Work-free isochoric deformation?

Dear Mr Koenemann,

This is, at best, bizarre and laughable. If you integrate de_ij and get a constant e for e_11, e_22 and e_33, you are explicitly stating that e_11 = e_22 = e_33 = e ie isotropic expansion. Moreover, if e_11 does not equal e_22 and e_33, then you cannot, algebraically, remove s_11, s_22 and s_33 from the equation s_11 e_11 + s_22 e_22 + s_33 e_33. Furthermore, you are also ignoring the functional dependence of s_ij on e_ij: what is your constitutive relation? We are now reduced to discussing what is possible and not possible in basic algebra and calculus. To me, thus far, it appears that every step of your "theory" is based upon similar examples of mathematical or logical mistakes and misunderstandings. At each stage, a variety of errors in this "theory" have been pointed out by a series of different people, all of whom seem to have reached the same two conclusions:

(1) your theory is rubbish.

(2) it is futile trying to point out the errors in your thinking.

With a certain amount of relief, I will have now to abandon this thread as I am going to be away for several days, but I am left wondering if this is a strange piece of modernist performance art in which I have been foolish enough to participate. Most importantly, though, I do not want any future readers to stumble across this thread and be led to believe that the Koenemann theory is valid or uncontested. It is not. I don't recommend that anyone else take up this Quixotic discussion, but I will certainly be amused to read any further posts.

Regards,

Craig

## stress, strain, and material

Craig,

I did not say that e_11 = e_22 = e_33, but I said that de_11 = de_22 = de_33. The numerical values of the e_ii are yet to be determined.

Since I had just had nearly the same discussion with a private correspondent, I repeat the entire argument here:

Tensor products do not communicate, AB ≠ BA

but the trace of a tensor is an invariant, and tr AB = tr BA communicates.

Also, tr A . tr B = tr B . tr A

but tr A . tr B =/= tr AB.

Let the off-diagonal terms be zero to make it simpler, then

tr AB = a_11 b_11 + a_22 b_22 + a_33 b_33

but tr A . tr B = (a_11 + a_22 + a_33) (b_11 + b_22 + b_33)

If A = B such that a_11 =/= a_22 =/= a_33 but tr A = 0, (at least one term must be negative)

tr AB = tr AA = (a_11)^2 + (a_22)^2 + (a_33)^2 > 0 (it cannot contain negative terms)

whereas tr A . tr A = 0 . 0 = 0

So, the question is, do I pick my results as they please me? No:

Let s_ij = [1 -1/2 -1/2]

you can write s de as

s [1 -1/2 -1/2] de [1 1 1]

Why so?

The s matrix determines the properties of stress, the de matrix must be an identity matrix I for an isotropic material. If you let s be isotropic, but de is not isotropic, the result is necessarily anisotropic. This is what you get if you subject a single crystal of some non-cubic material to a high hydrostatic pressure. So, if de = I, and s is as above, the properties of the finite e are those of s. Thus

tr se = (s_11)^2 + (s_22)^2 + (s_33)^2

which is sign-insensitive and physically meaningless, it no longer reflects the properties of s itself;

but tr s tr e = 0.

Regarding my own theory, you haven't even looked at it yet. All of the above is in terms of the conventional theory.

Greetings,

Falk

## Deformation work

The discussion has continued with a few correspondents by private emails. Two subjects came up with two correspondents simulaneously: the sign of deformation work, and the way to integrate the stress-strain equation.

Work: it was claimed that the equation for elastic work can only be positive: if s = [1 -1/2 -1/2], and e = [1 -1/2 -1/2], then necessarily the trace of the result is 1 + 1/4 + 1/4. I think this cannot be right. An elastic expansion and a contraction must produce work with opposite sign, for two reasons: (1) the relaxation of a compressional loading condition is an expansion, and the work of loading and unloading must cancel; if they were both positive, they should add. (2) The work done in an expansion or compression, starting from some standard state, is different in thermodynamics, as everybody can check for himself.

I have pointed out that there are two ways to interpret the trace of two tensors A and B: it may be tr (A.B), or it may be tr A . tr B. The product above would be tr (A.B), and I think it is physically meaningless, for the reasons given. The product tr A . tr B can only be zero if tr A is zero.

I maintain that the condition tr s = 0 is a statement to the effect that no work is done in a process (s = stress, tr s = s_11 + s_22 + s_33). Consider a region V, consider fluxes f in and out of the region. If the fluxes balance such that div f = 0 holds, the state of the region is unchanged. If the fluxes are flow of water, the region is neither a source nor a sink of water.

If the flux is heat, the temp of the region is constant. And if f are forces, no work is done upon the region. Since no one would maintain that indeed no work is done in an elastic deformation, it is clear that something has so far been overlooked. I think the power of this argument is so far vastly underestimated.

One correspondent claimed that the involvement of Poisson's ratio changes the zero result. I disagree. Poisson's ratio is the attempt to explain an observation by itself, it is phenomenological. Rather, one should ask how much work is saved by allowing a stretched bar to attenuate, and then focus on very carefully analyzing the energetics of deformation.

Two correspondents took issue with my claim that in the integration of s de, de is isotropic. I think that the result of int f(x) dx should be fully determined by the integrand f(x), and the magnitude by the limits of integration. Thus if the result is a tensor, it is not appropriate to let the limits of integration be anisotropic unless there are compelling reasons which have not been offered so far.

Characterizing the discussion so far, it is still completely from the basis of conventional CM. While the correspondence with some partners continues, and it is too early to come to a final

opinion, the actual desire to simply explore what else is offered, and where it may lead to, is nil. People would not admit it, but it is plain orthodoxy. Questions that do not fit into the known canon will not be discussed. Examples:

(1) the obvious conflict between standard elasticity (which holds that the attenuation of a bar happens without work being done) and Gibbs (who said that no work is done in a direction in which nothing happens) has not been approached by anyone.

(2) What is so wrong about "Why does the theory of elasticity start with an equation of motion, and not with an equation of state?"

(3) My claim that Clausius' virial law must be taken into consideration, has not found any response.

Falk H. Koenemann

## Re: deformation work

Falk,

If your arguments make any sense, we would have gone much further into those questions. The fact is, you do not understand the conventional continuum mechanics theory, or you have intentionally applied it incorrectly to get all kinds of wrong/strange conclusions, for which you simply balme on the theory rather than your own intelligenece or honesty. Through all these discussions I have followed on imechanica and via private emails, you have shown clearly how badly you understand the conventional continuum mechanics theory. You have used thermodynamics as a weapon to attack continuum mechanics, but really you do not understand thermodynamics that well either. You have also shown your weakness in basic calculus several times regarding integrals and vectors. It is unbelievable that you can still maintain yourself with all the errors and misinterpretations of math, physics, and continuum mechanics.

To all other readers: I am one of the correspondents trying to understand Falk. Unfortunately, I am afraid it was a bad idea to waste time on this.

RH

## Unjustified Poisson's ratio

Rui,

you claimed that a zero sum of forces can nevertheless result in a non-zero pressure. If I misunderstood you, you have a highly misleading way of choosing your words. As of this point you are by no means on the safe side. And I do not accept calculations involving fudge factors, ok? I gave you my reasons why I think Poisson's ratio is not a physical term; you insisted, but without rationale other than that it is commonly accepted. I still think the energetic implications of this ratio need to be explored.

Plus, neither you nor anyone else has addressed the obvious contradiction between common understanding and Gibbs: common understanding has it that a bar may attenuate, and no work is done in this; Gibbs claimed that no work is done in a direction in which nothing happens. So either the thing attenuates with work (or relaxation) being done, or withhout, so which one is it? Just wrapping this in silence is not an answer.

Falk

## Re: Unjustified Poisson's ratio

Falk,

What I said is that the sum of external forces acting on a system is zero while the pressure in the system is not necessarily zero. I will leave this for the public to judge.

I used Poisson's ratio because we were discussing how the conventional theory calculates the strain and work during isochoric deformation. You refused Poisson's ratio but claimed that the conventional theory is wrong because it leads to zero work. Do you still think you have applied the conventional theory correctly (without using Poisson's ratio) to reach the wrong conclusion? I thought I have convinced you what is the right way to apply the conventional theory. You agreed in your email.

Gibbs' claim has no contradiction with the conventional theory. In a direction in which nothing happens, no displacement and thus no work done. Remember here we consider work to be force times displacement in the conventional theory. On the other hand, in a direction in which there is no force, no work is done either even when there is displacement (contraction or attenuation). The two cases are not equivalent however. The first case is uniaxial strain, and the second is uniaxial stress. If you do not understand the difference, you do not understand any mechanics, conventional or not.

Indeed there are many contradictions that you have claimed and we have not addressed. The reason however is not that we intentionally remain silent on any of those. Instead, we could not make any progress with you on any single point that we wish to get started. You simply refuse to correctly apply the conventional theory and yet claim the conventional theory is wrong. I hope you understand the simple logic: (1) the theory is right; (2) you applied it incorrectly (or simply did not follow the theory); (3) you reached an obviously wrong conclusion; (4) you think the theory is wrong. You may apply this to any theory.

RH

## Just do a real physical test

Falk,

See Chad's challenge here:

http://imechanica.org/node/5321

## See there. Falk

See there. Falk

## Poisson's ratio

Rui,

No, I have not given in. I was disappointed that you insisted that I accept a term "because it is part of the conventional theory" which I question because of its purely phenomenological nature. I was disappointed because I thought I had given you enough insight in my arguments, which you would not reply to. All you got from me was the admission that if you blindly follow the cookbook you get what you want; but I don't like the recipe at all, I consider it thoughtless, and I didn't eat it.

My critique of Poisson's ratio is this:

(1) It can be observed only if free surfaces are nearby; it is therefore dependent on the shape and the dimensions of the sample. There is no basis for a Poisson's ratio in an infinite continuum. Such an infinite continuum may be hard to find in engineering, but it is the standard situation when considering deformation in the Earth's crust & mantle, that is: for a geologist the constraints are tighter.

(2) Step 1: if a sample is stretched in X, and no attenuation is permitted in Y & Z, the work done per stretch in X is substantial because the sample is in effect expanded. Step 2: if then the boundary conditions are released in Y & Z, the sample will contract in Y & Z, and the energetic state will relax. Let's assume that now the deformation in the center of the bar is isochoric. Then it is obvious that in step 1 work was done in Y & Z _although_nothing_happened_ in these directions: without the constraints provided by the surrounding the sample would have contracted; and in step 2 the sample did work on the surrounding by pulling it inward. (I did not imply free surfaces, I implied external conditions in an infinite continuum that resemble the situation observed in a bar. If there is an energetic relaxation in step 2, it is wrong to assume that for a bar in freespace the attenuation comes free of work. )

(3) Consider an infinite continuum, and a subvolume in it (the system; 2D conditions – XZ – for simplicity, it works in 3D too). Boundary conditions are such that the sample is compressed in Z, and may expand in X. What we observe is: the sample is shortened in Z. The sample will also expand by itself in X. But since the system is part of a bonded continuum, such that system and surrounding are solidly bonded, the surface point on the X axis is under tension by the surrounding. Thus there are two independent stretch components in X: one provided by the system itself, the other one caused by the surrounding on the system. Without the second stretch component the surface point of the system on X would be in equilibrium after the system-caused stretch in X.

The latter situation must be expected on free surfaces. Thus there must be a strain gradient near free surfaces: far inside a solid there will be two stretch components in X, but one of them must reach zero on an unloaded surface.

Thus the question: if you talk about strain in a sample, do you mean a situation deep inside, on the surface, or someplace in between?

All I want to show at this point is this: point (1) shows that a very careful analysis of the boundary conditions is required. In order to write a general approch one must take care not to imply specific boundary conditions. (2) The attenuation is indeed work saved. The real significance here is not Poisson's ratio; but it is evident that the attenuation is caused by the principle of least work. Thus we must talk about work as a function of the boundary conditions. (3) The fact that there is only one shortening component in Z, but two possible stretch components in X, one of which must be zero on free surfaces, such that inevitably there must be a strain gradient from the interior towards a free surface, is probably new to most readers. It is nonetheless a fact.

Summary. If you consider Poisson's ratio as a material property (which is the standard way) you cannot consider the implications I outlined. Rather, we must very carefully analyze the possible material reactions as a function of the boundary conditions. A material property called "Poisson's ratio" does not exist. The principle of least work certainly does exist.

Falk

## Framework of classical physics

Classical physics (pre-1900) began to find its first reliable foundations in the 17th C. Newton's three laws are indispensable for the mechanics of discrete bodies in freespace. Originally one used a force conservation law; but it soon became apparent that the entity to be conserved is some thing else for which the term 'energy' was proposed by Thomas Young in the 1790s, but the term did not come into common use before the late 19th C. It was certainly considered, under the name 'vis mortua'. The first energy conservation law was found by J.G. Leibnitz in the 1670s and Johann Bernoulli 1735 who concluded that E_kin + E_pot = const. Bernoulli had a student who would gain the stature of a giant: Leonhard Euler.

In 1776 two entirely different steps of importance for the discussion of deformation were made: Euler proposed the stress tensor; and James Watt invented the steam engine. Clearly the connection was not seen for another 70 years.

Watt's step culminated in the First Law of Thermodynamics in 1847 by Hermann Helmholtz. Since then the framework of classical physics has been completed:

1.There is the field ofconservativeprocesses under the rule of E_kin + E_pot = const. The 'const' is called the entire energy H of a kinetic system. This is Newton's mechanics.2.If a process isnon-conservativeit means that H is 'not conserved' i.e. it is a variable. The energy conservation law for such processes is the First Law dU = dw + dq.Two different types of non-conservative processes were soon distinguished:

2a.The process may bereversible.2b.The process may beirreversible, and entropy is produced.If a process is conservative, the energy of the system is invariant. The system is a kinetic system consisting of n bodies in freespace, and the system does not exchange energy with its surrounding. It is isolated. A conservative theory may be started with an equation of motion; it does not need an equation of state.

If a process is non-conservative, it must start with an equation of state.

The connection from Bernoulli's energy conservation law of conservative mechanics and the First Law of non-conservative mechanics was provided by Rudolf Clausius 1870. He considered the free vibrations of atoms in a solid or fluid and their kinetic energy E_kin and correlated them with the heat. He correlated the potential energy term with the bonds in a solid, or – my liberty – other, less permanent forms of atomic interactions. Thus he arrived at the

virial law,E_kin + E_pot = PV

Clausius interpreted E_kin as the heat. Thus if the internal energy PV = U of a system is a variable, we can consider it by virtue of the equation of state

PV = nRT,

and the energy conservation law for non-conservative physics, i.e. the physics of changes of the energetic state, is, of course,

dU = dw + dq.

Considering an elastic process, it is clearly non-conservative-yet-reversible. Thus it must start with an equation of state. I have not seen a textbook in which the attempt was made to derive the theory of stress from an equation of state. Rather, I have quoted enough textbooks in which an elastic deformation is understood as a variation. But a variation is a concept that was invented by Euler, it is strictly under the rule of E_kin + E_pot = const. Maybe the theory of variations has its benefits in other parts of physics, but this energy conservation law is the wrong one for elasticity and deformation of solids. Work in conventional CM is defined as the negative of the kinetic potential,

dw = d(E_pot – E_kin)

which makes it clear that this work is acceleration work, but not change-of-state work.

The difference between conservative and non-conservative physics is so profound that they are commonly explained in different textbooks. Mixing them up is arguably the worst mistake one can make in classical physics. But the difference between conservative and nonconservative physics is systematically blurred in continuum mechanics.

The form of the "First Law" given in the textbooks is

int (d E_kin + dU) dV = dw + dq

where U is the internal energy. (the d is a lower case greek delta, indicating a very small, but finite quantity.) I cannot see how E_kin and U can be reasonably summed. This "law" and the virial law above cannot be reconciled.

The fire I have drawn here on iMechanica so far did not address these points. I have some human understanding for those who have difficulties to accept this difference because an academic education, and possibly a decade or three of professional work and teaching, form the mind very strongly, and it is not easy to find out of that. However, the mixup must be cleared up. The form of the First Law as used in CM is invalid, it is not the First Law.

Falk H. Koenemann

## Trace of stress s_ii: physical significance

With one recent correspondent I had the following exchange:

Just answer: f is a flux.

(1) Consider heat flow. For a typical region the condition is div f = 0. I conclude that the temp of the region is constant. Agree or disagree?

(2) Consider mass flow, e.g. water. For a typical region the flow is characterized by div f = 0. I conclude that no water is stored in the region, and no stored water is released from the region. Agree or disagree?

(3) Consider energy. A volume of gas is deformed (equilibrium flow) such that its shape changes, but its volume is constant. The flow is characterized by div f = 0. Has work been done, has a change of state occurred?

To which he replied:

to (1): This is not even a relevant question. But I will answer briefly. If f is the heat flux tensor, which is typically defined as proportional to the gradient of temperature, div f = 0 implies no heat source in the region and leads to a Laplace's equation for temperature.

to (2) Again, this is an irrelevant question. If f is the mass flux vector, div f = 0 is the simple mass conservation rule. Water comes in and out, but the total mass of water in the region does not change.

to (3): What is f in this case? If you are thinking about ideal gas, only pressure does work. Then, if the volume of the gas does not change, no work is done and no change of state occurs (assuming isothermal condition). However, we cannot extend this to solids […].

I wrote:

I agree with your last sentence. Question now: why not? There is no mathematical reason. Of course I do not maintain that no work is done in deforming a solid. But if the condition s_ii = 0 is assigned the importance it has in conventional CM, it must have a meaning. It cannot have another meaning than "what goes in, goes out".

Or do you see an alternative? What does the condition mean physically in your view?

Why is the statement correct for a gas, but not correct for a solid, when the external boundary conditions are the same?

If you consider the situation that you have two pistons with only air in between, and you close the pistons slowly, the gas will deform. It will spread laterally, just as a solid. It will maintain its volume. Thus you can say that a gas has Poisson's ratio 0.5. As long as you are under the rule of dw = d(E_kin – E_pot), you deal with acceleration work only. For a gas with n particles it works: the work done on all particles sums to zero.

Since then he does not answer any more. That's just the easy way out.

The idea that the trace of a square matrix is meaningful comes from linear algebra and from the theory of potentials; and for heat, mass flow and for the volume-neutral 'deformation' of a gas the zero result is correct, showing the accuracy of the concept. If this is not applicable to elastic deformation of solids, the reason is that work is done which is stored in the system, and an elastic potential builds up. But this cannot be considered under the Laplace condition div f = 0: there are no excess forces left that can be used to define a non-zero work term.

Sharper: if a physical situation is described involving the Laplace condition div f = 0, it is the most stringent description of the fact that no work is done upon a system. If a theory containing this condition comes to a non-zero result anyway, by applying proportionality factors etc. (such as Poisson's ratio) it is not a sign that the theory is reliable nonetheless. Rather, it is a classic perpetuum mobile of the second kind: more energy comes out than goes in.

However, if the contention is dropped that the condition div f = 0 (or tr s = 0) is applicable to solids, if it is acknowledged that the divergence (or an equivalent expression) must be non-zero if work is done upon a system, it would be the proper step to understand elasticity better. In the light of potential theory this is the only way to go. But Euler-Cauchy will fall.

I could not get the correspondent to address my question what s_ii = 0 physically means in his opinion. This is a pity, this is the core of the entire dispute.

As long as it is maintained that the condition tr s = 0 is meaningful in deformation of solids, and correct to use, continuum mechanics remains out of touch with the rest of physics.

There are no sanctuaries for endangered theories.Falk H. Koenemann

## Please, enough of this

Dear Mr Koenemann,

I am certain I will regret re-engaging with this discussion, but I cannot let your comments pass unchallenged.

Several posts above you make a serious mistake in your interpretation of the virial theorem. Virial-type relations are a consequence of the vanishing of the time-average of bounded functions as time increases. An arbitrary deformation is not a bounded function. Furthermore, a second requirement of the virial theorem is that the time average of the external force resultants be zero, which is not the case on the time scale of a deformation. One of the consequences of your errors is that your theory is not invariant under translations. This means that a rigid-body motion will change the stress-state (or whatever you want to call it in your theory). Clearly this is absurd. This is also one of the foundations of your theory and, by itself, this mistake invalidates your claims.

I believe Schweitz did some work on this topic back in the 1970s, which may provide you with some greater insight. I expect that you will find that your theory is incorrect and you may wish to recant.

Sincerely,

Craig

## the virial law

Craig,

elastic behavior is time-independent. The situations you refer to are clearly relating to heat, not mechanics. Having said that, I do not question what you wrote. I just think there is more to it than is apparent so far. I would have to second-guess your words, and I'd rather not; but I am not talking about heat effects and/or diffusion. I am talking about the loaded state below the threshold temp at which diffusion becomes important, such that the elastically loaded state can be maintained, indefinitely if necessary. (A few billion years should suffice; I have seen recent cracks in rocks that developed due to an elastic potential that had waited 2,4by for its release.) My theory has nothing to do whatsoever with rigid-body motion, it has nothing to do with motion (in the Newtonian sense) at all, and I cannot see a reason why it should not translate. My theory is a generalization of thermodynamics; the theory you know to be written in P and V, I have rewritten in f and r, with the condition (always observed) that the surface integral of the f-r calculations must be compatible with the P-V calculations for identical situations. It does work.

I understand elastic deformation as a change of state. The simplest deformation conceivable is isotropic compression of an ideal gas. If you take that theory, transform it into vector form, and spend a few thougths on boundary conditions, you arrive at my theory. But you will not get into conventional CM because the Cauchy theory cannot consider the distinction of system and surrounding. It is simply wrong to start a theory of stress – implying a change of state – with an equation of motion (implying Newtonian mechanics, E_kin + E_pot = const, and therefore by definition excluding changes of state from consideration).

Falk

## ?

why there isn't more comments? it was very funny reading this. all the best to everyone:)

## As a good painkiller...

I would recommend this beautiful

article:

The Existence of the Flux Vector and.the Divergence Theorem for General Cauchy Fluxes

Silhavy, M., 1985. Archive for Rational

Mechanics and Analysis, 90, 3, 195-212.

I hope it will be a relief for much of

the torment, if not all...