Dugald Carmichael schrieb:
> >1. ... If f/A -> infinity if V -> 0 it shows that the Newtonian
> >definition of pressure cannot be used because the transition does not
> >exist. It should not bother anyone because the thermodynamic definition
> >of pressure P = dU/dV holds, and is _really_ and without doubt
> >scale-independent.
>
> If "scale-independent" implies that P = dU/dV would remain valid as V -> 0,
> I am not persuaded. Not only P but all the thermodynamic properties of
> matter presuppose a substantial volume of matter in the "system".
But that is precisely my point. In terms of potential theory, a thermodynamic
system is a _distributed source_ of fluxes; all its properties need to be scaled
per unit mass, e.g. one mol (implying a standard molar volume in the reference
state). P = dU/dV observes this relation, P = f/A does not. And it matters
tremendously whether the Newtonian equilibrium condition f_left + f_right = 0 is
used which refers to all the external forces acting on a system, or if the
thermodynamic equilibrium condition f_system + f_surrounding = 0 is used; only
this eq.condition can treat elastic deformation as a change of state, the
Newtonian eq.condition cannot.
If external forces of whichever configuration act upon a body of solid, they may
or may not balance, or they may balance in part. Externally unbalanced forces
will accelerate the system externally, so we can ignore them, they cannot cause
a deformation. It is important to note: only externally balanced forces can do
work on a system to the effect that a change of state occurs, i.e. only they can
interact with forces exerted by the system on the surrounding. This point is
consistently ignored by classical continuum mechanics.
> >2. ...Stress need not be "symmetric" by definition (the mathematically
> >correct term is "orthogonal").
>
> In my opinion "orthorhombic" is preferable. "Orthogonal" does not
> distinguish the symmetry of an orthorhombic prism (or triaxial ellipsoid)
> from that of a tetragonal prism or a cube. There is plenty of precedent in
> the literature of "petrofabrics".
I agree 100%!
> >The conclusion that stress is orthogonal, is due to a misunderstood
> >equilibrium condition: forces acting in a kinetic system of particles which
> >is (a) not interacting with a surrounding and (b) does not spin
> >about itelf must be balanced, hence the orthogonality condition. In essence,
> >such a system is in equilibrium with _itself_. But the "unit cube"
> >you mention was never understood as a thermodynamic system
> >(which should be the case). A thd.system is in equilibrium with its
> >_surrounding_, so the forces exerted by the system on the surrounding and
> > vice versa must balance. ...
>
> It seems to me that a system in equilibrium with its surroundings must also
> be in equilibrium with itself.
No. In the simplest case, isotropic loading of an isotropic material, a system
of solid is in equilibrium with itself if its bond lengths are determined only
by the energetics of its bonds. This is the unloaded state, or the zero
potential state. If the volume deviates from the volume which it would have in
the unloaded state, it is the surrounding doing work on the system to the effect
that its bond lengths are shorter (or longer) than in the unloaded state. The
system is thus in a non-zero potential state, i.e. it would spontaneously expand
(or contract) if left to itself. Therefore the system in the loaded state is not
in equilibrium with itself, but it is in equilibrium with its surrounding.
> Let us envisage the "cube" as a
> thermodynamic system in contact with its surroundings across a flexible
> diathermal boundary impervious to chemical transport. [snip]
> ...it must be possible to remove the surroundings, leaving in place
> the exact same set of surface forces that was exerted by
> the surroundings on the boundary, with no effect on the state.
I think that's a contradiction in terms. If you remove the surrounding, you
necessarily remove the forces exerted by it as well. Or you may replace the
surrounding, keeping the forces in place; but this is probably not what you
meant. If you expand a rubber band, it surely will not stay that way if you let
go on the left side, but it will snap with a vengeance on your fingers on the
right.
> So let us now remove the surroundings. If the system begins
> to spin with ever-increasing angular velocity, centripetal forces will be
> induced and its state of stress will not remain constant and uniform. If
> the system remains stationary, the set of surface forces on its boundary
> must be such as to generate within the system a stress field with
> orthorhombic (or higher) symmetry. Accordingly, I am not persuaded that the
> stress in a homogeneous system in thermodynamic equilibrium can have lower
> than orthorhombic symmetry.
You have slightly changed the subject here - from the nature of the equilibrium
in general to its geometric properties in particular. The first one is dealt
with above. - The second part I wish to answer by giving an example. Simple
shear is caused by a force field like this
---->
--->
-->
->
.
<-
<--
<---
<----
which surely is not orthorhombic. You can let it act on a thermodynamic system
to cause an elastic deformation. A ball in water would spin, but a sphere of
solid in a larger piece of solid will not because of the bonds across the
system-surrounding boundary. Thus you know that equilibrium exists. For the
following I just indicate the outline: calculate the average force exerted by
the above force field (lower and upper half separately), assume that its
rotational momentum is balanced by bonding forces, subtract that momentum from
the field above, and you get non-orthogonal eigendirections.
> But strain is a different beast. Natural consequences of irreversible
> strain often have lower than orthorhombic symmetry. I am really intrigued
> by the figure on Falk's website that illustrates non-orthogonal directions
> of principal compression and extension in a "shear zone". Presumably this
> results in "distorted" strain ellipsoids with monoclinic symmetry?
The strain ellipsoid is always at least orthorhombic, but strain is a
path-independent term and may have been reached by a zillion different ways. (Be
careful not to equate strain and deformation which many people do.) We are
talking displacements here, and yes, surely they are monoclinic.
_______________________________________________________________
Falk H. Koenemann Aachen, Germany [log in to unmask]
http://home.t-online.de/home/peregrine/hp-fkoe.htm
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