Richard A. Ketcham schrieb:
Hi Richard,
> >1. Equivalence of f/A and dU/dV. 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.
> >
> >Hence U/V is scale-dependent. This is physically impossible.
> >Furthermore, as V vanishes, so does f, but f vanishes faster than r.
>
> I'm not sure what's been demonstrated here -- all you seem to have shown is
> that area doesn't equal volume.
It goes a little deeper. The question is: what is the relevant term in this
problem: force or work. Mechanical forces can do work only across the surface of
a system, but work is done per mass, or per volume. Thus if work is proportional
to volume - and this ratio must be scale-independent - but since V/A cannot be
scale-independent, f must be a function of scale.
> Please enlighten me on the following points:
> 1) Pressure is dU/dV, not U/V, so I don't see why U/V is of any relevance
> to this argument.
dU/dV says that the ratio of change of U is equivalent to the ratio of a volume
change (at constant conditions). In this case mass is a variable, and V is a
measure of the scale considered. In integrated form it becomes U/V which says
the same: U and V are proportional as the scale of consideration is changed.
I cannot see a difference. All it says is that the internal energy is an
extensive parameter.
> 2) Your column calculating U/V assumes that U is constant. You define an
> exact initial value for U, but none of your equations constrain subsequent
> values (unless somehow you're assuming that all of your states are related
> by compression?). So, it seems that you're stuffing more and more energy
> into a smaller and smaller volume, while simultaneously saying that the
> pressure is constant. On the face of it, this does not make much sense.
I did not make the assumption that U is constant. I made the assumption that f/A
is constant, AND that f/A and U/V are equivalent at all scales. Since I have V
and U/V, I can get U which then blows up as V -> 0.
In symbolic form, say f/A = U/V.
Replace A by r^2 pi and V by 4/3 r^3 pi and cancel redundant terms.
Result is: f = 3U/r, or U = fr/3.
Make the assumption that f is constant, hence U is proportional to r; thus
if V -> 0, U/V -> infinity.
This cannot be, hence f/A and U/V are not equivalent at varying scale of
consideration.
> 3) A simplifying feature of using flat surfaces is that it's easy to
> resolve force vectors into normal and parallel components. When you're
> using the surface of a sphere, this would seem to make balance-of-force
> computations much more difficult to do correctly.
I think that this is due to the definition of a shear force as a force parallel
to a surface, according to Euler. I cannot see why this should be correct, and I
believe that this is simply a faulty definition, for the following reason:
Newton defined the rotational momentum as f x r where r is the distance from the
center of mass of a discrete body to the point of action of f on the surface of
the body. The rotational momentum is thus independent of the orientation of the
surface, what matters is the shape of the body. Euler defined the shear force as
f x n where n is the surface-normal vector. The stress tensor is assumed to be
orthogonal; this assumption is based on the physical significance of f x n.
I cannot see how Newton's r can be replaced by Euler's n, and the transformation
must be made if f x n is to have a physical significance. If it is assumed that
f and n are physically of similar relevance it is like saying that since n is by
definition a unit vector, so is r, hence the shape of the body is a sphere! Then
what do you do with an elliptical body? Say, its principal axes have length
x = 5 and y = 2; then radius-normal forces with magnitude 2 and -5,
respectively, would balance the rotational momentum. But a vector field with
maximum radius-normal components in the directions x and y with the magnitudes
given above is not orthogonal. Hence it is possible to balance the momentum of a
non-orthogonal force field with an elliptical body.
The question may be permitted what we do with the concept of shape (eg. of a
thermodynamic system) in the case of a perfect continuum within a solid - after
all, there are no discrete surfaces, and any shape could be the right one, thus
shape may be arbitrary. This is an argument from the Euler-Cauchy theory. But I
am just taking Newton serious: if f x r is the only term we can use, we cannot
get rid of the shape when we consider equilibrium, hence we can only make
minimum assumptions. - In my theory I made the assumption that the system has
spherical shape (because a sphere minimizes the A/V ratio at constant V) unless
there is reason to believe that it is not so; in that case I have one more
degree of freedom to balance the rotational momentum because I can choose the
shape SINCE I know that in the elastic case, equilibrium is maintained.
Falk Koenemann
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