Re: How can the Planck length be claimed to be the smallest length?

On May 9, 2:47 am, JohnMS <john_m_stan...@xxxxxxxxxxx> wrote:

If one takes 10^24 atoms of silicon in a single
crystal (around 1 kg)
it is undisputed that the whole object has a
measureable gravitational length.
The crystal bends space-time around it and attracts
other masses;
that is easy to measure. And there is a definite
metric in our environment.

It is also undisputed that all atom masses
in the crystal essentially add up
(the crystal binding energy can be neglected here).
Since the gravitational length of the silicon
crystal is defined as R=2GM/c^2,
it is very hard to avoid saying that every silicon atom
has a gravitational length given by the
same formula, this time using the atomic mass.

It is true that a kilogram of silicon has a measurable gravitational
field. This field corresponds to space-time curvature of order 1/L,
where L is some length. We know for a fact that these gravitational
effects are very weak, which in turn implies that L must be very
large. When we measure gravitational effects due to this hunk of
silicon, it is L that we measure, not the R that you've defined above.
R would be the size of the hunk of silicon if it were dense enough to
become a black hole. Since it is not a black hole, we have another
demonstration that R is irrelevant to the physical situation.

In short, your paradox is avoided because, no matter how small R is,
it never comes up as an experimental measurement; only L does. And L
is of regular macroscopic proportions.

Hope this helps.

Igor

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