Seems to me a simple dimensional analysis shows the fallacy of
Koenemann's argument: The dimensions of f/A are [M][L][T^(-2)] / [L^2]
and of U are [M][L^2][T^(-2)] / [L^3], so the dimensions of both
quantities are the same: [M]/[L][T^2]. Thus both quantities scale the
same way with the dimension of the system, and Koenemann's conclusion
must be incorrect.
Note that if the dimension of the system changes you cannot assume
that both f/A and f are constant, as Koenemann does in the argument
below (see underlined statements).
Rob Twiss
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.
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Robert J. Twiss email: [log in to unmask]
Geology Department telephone: (530) 752-1860
University of California at Davis FAX: (530) 752-0951
One Shields Ave.
Davis, CA 95616-8605, USA
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