At 03:45 PM 2001/06/20 +0200, Falk Koenemann wrote:
>...
>... Assume shape of V is spherical.
>Assume: f = 5, f/A = P is scale-independent.
>Assume that f/A and U/V are equivalent for the case r = 1.
>Calculate  U/V and f as V approaches zero at constant P.
>...[table of numbers omitted]
>Hence U/V is scale-dependent. This is physically impossible.

Agreed - there must be some mistake. The calculation seems to ignore two
necessary conditions. In order for dU/dV to be equal to P (actually to
negative P if we assume compressive P is positive), the change in volume of
the system (dV) must be carried out (1) at constant mass and (2) at
constant entropy, so that no other variation in U (except mechanical work
PdV) is simultaneously permitted.

>Furthermore, as V vanishes, so does f, but f vanishes faster than r.

Certainly "stress at a point" has no meaning in a real solid. A continuum
differs from a real solid in being strictly homogeneous even at the
subatomic scale. By the same token, "pressure at a point" and "temperature
at a point" would be meaningless; P and T are properties of macroscopic
systems. P or stress = f/A where A refers to the external surface of a
system containing a  finite quantity of "statistically homogeneous" matter.

>Assume: U/V = 5 is scale-independent. Calculate f and f/A.
>From the divergence theorem it follows that f and r are proportional if V is
>allowed to approach zero at constant P.
>...[table of numbers omitted]
>Hence f/A approaches infinity as V approaches zero.
>It follows that f/A and U/V are not equivalent if A is a closed surface.

But P could not remain constant during compression of a real system at
constant mass and entropy.

>2. Measurability. Some standard unit volume must be assumed to which the
>measurements refer. This may be the volume of some unit mass, or a unit
volume
>in the standard state. Then V is no longer a variable, and then f/A works.

I would contend that P = f/A works for any thermodynamic system at
equilibrium or during any reversible process, but -P = dU/dV works only for
systems at constant mass and entropy.

>Potential theory says: the shape can be arbitrary if the system contains n
>discrete bodies such that the surface of the system does not pass through
>mass. ...

Thus potential theory joins continuum theory in making stipulations that
cut down on complexity. :-)

>The shape of the system is not arbitrary if rotational/shear forces can do
>mechanical work. ...

Agreed. The forces applied to the outer surface of the system, whatever its
shape, must be mechanically balanced so as not to cause either linear or
angular acceleration of the system. This is the condition that restricts
the symmetry of stress to be not less than orthorhombic.

>If we consider a thermodynamic system within a solid, ie. within a larger
>volume that is continuously filled with mass, we get one more problem: ...
>system and surrounding are permanently bonded to one another. This
>point is of utmost importance for the equilibrium conditions; however it can
>only be considered if the existence of bonds is indeed taken into account,
>which is not done in continuum mechanics.

Thermodynamics deals routinely with this type of system, taking account
neither of bonds nor of strength of materials. The prevailing theory of
metasomatic zoning assumes constant P during reactions that predicate
finite changes of volume; hence it treats rocks as if they have zero
strength. But a continuum is resistant to deformation, and hence continuum
mechanics generalizes P (an isotropic scalar) into stress (a second-rank
tensor with orthorhombic symmetry). Seems to me this is an important step
towards taking account of the existence of bonds. Further steps are clearly
desireable.

>...
>Thus I concede that the shape of the system may be a cube, but it is
certainly
>not arbitrary. And whether it is permitted to be a cube is a question that
>touches the rotational equilibrium condition f x r.

Let's face it - forces are just things we invent to explain what we
observe; they are strictly empirical. The whole trick is to keep the forces
as simple as possible - hence the unit cube.

Dugald M Carmichael                    Phone/V-mail: 613-533-6182
Dept of Geological Sciences and Geological Engineering
Queen's University                              FAX: 613-533-6592
Kingston  ON  K7L3N6             E-mail: [log in to unmask]