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Article #56

Re: Pi day >> Pi for a Black Hole / Singularity

#56
From: Greffindel
Date: Wed, 09 Apr 1997 00:00
111 lines
5872 bytes
The Silicon Surfer (surfbaud@mail.zynet.co.uk) wrote:
: Martin Ward wrote:
: > 
: > In article <5hhge0$48q@raven.cc.ukans.edu>, Greffindel
: > <ipsifend@eagle.cc.ukans.edu> writes
: > >Jonathan Cunningham (jlc@sofluc.demon.co.uk) wrote:
: > -- snip --
: > >: Be careful if you are determining pi by measuring the ratio of the
: > >(thinks) (thinks) (thinks)  You know, that makes sense.  What could

: > Jonathan obviously understands the physics of this better than I do, but
: > putting in a weight must *reduce* the local "pi", surely?
: > 
: > Since there are algebraic and rational numbers arbitrarily close to any
: > transcendental number like pi the same would be true of the set of
: > values of our central mass.  If we can assign it a random real value
: > then most such would leave "pi" transcendental, but we could find
: > infinitely many values within any range >0 which make "pi" rational.
: > 
: > Of course quantum mechanics says, does it not, that masses cannot be
: > given an arbitrary real value but only one chosen from a set of discrete
: > values....

: > Martin Ward

: Hmmmm, hadn't thought of Pi as dependent on local conditions, but of
: course it's true.

: Suppose the one ton weight was a singularity with an mass x and an
: effective diameter of 1 millimetre. This could be easily drawn and
: visualised but how would you work out the radius/circumfrence.

Hmm.  Well, Suppose you draw the circle on "flat" (in a 3-dimensional
sense) space.  All nice and Euclidean, c = pi*d.  Now, you drop your
neutron star or whatever smack in the centre (Oh, what an ugly
calculation if the mass is off-centre!), and because of the mass,
space stretches, sort of like this:

-------------_     _-------------
              \   / 
               \ /
                *

Now, if you do this with a rubber sheet (the most common model that I
know of), you notice that all of the rubber stretches *some*-- but the
outer edges don't stretch very much.  The part right under the weight,
of course, stretches the most, but the variance is going to be
continuous-- there isn't a sharp line between stretched and
not-so-stretched rubber.  So there's some continuous function that
tells you how much space has been stretched at any given point based
on the "distance" from the weight.  I'd have to ask a physicist what
it was;  it would make an interesting calculus problem. (OK, maybe a
boring calculus problem)

So our "circle" will get pulled inward a little bit (because the space
outside our circle stretches too-- just not as much as the space
inside).  Therefore, the circumference is reduced a little bit [*].  But
the diameter changes far more drastically, since the diameter line
passes right through the centre of the circle (where the weight was
placed) and hence through the most stretched-out bits of space in the
whole experiment.  That line, being stretched, gets longer.  So,
taking pi as the ratio of circumference to diameter, you have a
smaller numerator, bigger denominator, and hence a smaller "pi".

-----------
[*] 'But how can the circle's circumference get smaller (squashed) if
space is getting stretched out?'  Because space is getting stretched
along any axis which intersects the centre of the circle (where the
weight is).  Along a perpendicular axis (the 'circle'-- or rather, any
tangent to the circle-- is always perpendicular to a radius of that
circle intersecting the point of tangency), space can and does get
squashed.  Complicated!

-----------


: Also, if my memory serves, object travelling at very near "c" will be
: "stretched" along the axis of travel, so the orientation of a circle to
: said axis would presumably affect pi, or maybe not as it would no longer
: be a circle.

Squashed, actually, in the direction of travel-- it's called, or at
least was once called "Fitzgerald Contraction" (named for
mathematician Albert Contraction).  But the result is an ellipse, and
the circumference is pi times the average of its major and minor axes.
This also reduces the circumference and therefore reduces "pi".




: OK, my head hurts.

: JB

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From: ipsifend@eagle.cc.ukans.edu (Greffindel)
Subject: Re: Pi day >> Pi for a Black Hole / Singularity
Date: 1997/04/09
Message-ID: <5iemjs$97s@raven.cc.ukans.edu>
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