The Foundations of the Second.
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The Foundations of the Second.         

Group: alt.philosophy · Group Profile
Author: Y
Date: Aug 16, 2008 23:43

The Rule : Every Structure requires a foundation.

The Equations.

Introduction

Let us assume then that we have limited our recording capaity using
quantum computers and three new units have come into being. These
units are almost already into being and at current our technology
supports down to approx 3 Atoms per bit, or .333 bits per atom.

These units are - Bits/Atom Digit/Atom and Symbol/Atom

This means Bits per Atom, Digits per Atom and Symbol Per Atom. You
see, just like our MP3 files which record music at 197Kilobits per
Second, we must not forget that the MP3 file also has a mass.
According to my Hard drive on a clunkier technology which weighs .6kg
ang contains 2781197205504 bits a bit has a mass of 2.15 X
10^(-13)kgs.

As an architect, I am accustomed to the fact that my ruler, i.e. the
thing I am measuring with has a mass. If we were to count seconds with
a machine, we would also require a certain mass in order to keep
count, just like our ancient abacus beads.

The Equation.

Eternity = Mass(universe) / Mass (Unit of record or the SI unit)

If quantum computers manages to utilise the atom, as a unitary
recording device, where 1 bit, digit or symbol can be stored using a
single atom. The total number of bits that can be used for any number.
Which also means the total number of seconds countable

Eternity = Mass(universe) / Mass(Unit of record or the SI unit)

Eternity = 1.59486 × 10^(55) kg / 1.660538782(83)×10(-27)kg

Even if in the most effient of number representation such as. . .

9^9^9^9^9^9^9^9^9^9^9^9^9^9 etc etc. . .

And after that, we could resort to other symbols which gave meaning to
numbers for example Where k=1000

K^K^K^K^K^K^K^K^K^K^K^K^K^K etc etc. . .

However there is a problem with this which I will provide in a
following Example. You see, once you run out of numbers, you run out
of numbers of
symbols. Beyond this, whatever it was that was recording will have
lost mathematical meaning. You will find examples of this after the
following questions.

First then I will ask some very easy High School difficulty questions,
and then I will provide examples. I have complied some short
Questions, which are not too difficult to answer. The questions regard
the concept of Eternity, where Eternity is not the stuff of the bible,
but rather the stuff defined as a limit of count. (it is important to
understand this definition before making any sense of the above.)

Question 1A. Difficulty Level (low)

What is the maximum recordable ammount of seconds in a 12 Atom
universe, where

Eternity = mass(universe) / mass(unit of record)

Mass of Atom = 1.6 x 10^(-27)kg

Recording unit = 1 digit / atom

Question 1B.
How many times more than the mass of the 12 Atom universe was your
answer to the last question, if one of your digits was measured to be
2.15 X 10^(-13)kg ?

(2.15 X 10^(-13)kg is the approx average mass of a digit on my
computer's hard disk where I aquired this mass by

1. Measuring the weight of my hard drive
2. Dividing this mass by number of available bits on the Hard drive.
3. Multiplying by number bits required of the digits.

---------

Question 2. Difficulty Level (low)

What is the maximum recordable ammount of seconds in a 12 atom
universe, where

Eternity = mass(universe) / mass(unit of record)

Mass of Atom = 1.6 x 10^(-27)kg

Recording unit = 1 bit / atom.

(just to clarify: a bit is considered as either 1 or 0. There will be
a maximum number of bits for example on your PC's hard disk.) In this
case, we would have already figured out that the mass of 1bit of
information = the mass of the atom.

Question 2B.
How many times quicker or slower did the 12 Atom universe run out of
time, if your answer took 35.5 years to complete. ?

---------

Question 3. Difficulty Level (medium)

What is the maximum recordable ammount of seconds in a 12 Atom
universe, where

Eternity = mass(universe) / mass(unit of record)

Mass of Atom = 1.6 x 10^(-27)

Recording unit = 1 symbol/atom.

Question 3B.

Providing they began at the same moment, How many seconds will have
been recorded in our our own universe, at the same moment that the
last integer was erased from the memory of the 12 atom universe ?

(this is a trick question, the answer will be the same as 3.)

EXAMPLES.

Now here is an example of a 10 atom universe.

Because this little universe is trying to survive counting seconds on
its parts, its life in seconds is as follows.

BINARY

Lets say this were binary bits per atom (bit/atom). . .i.e each second
needed to be 1 or 0

1,1,1,1,1,1,1,1,1,1 - it would now have run out of space to record
seconds. 10 seconds.

or

DIGIT

Lets say this were a Digit per atom universe. . .i.e each second
needed to be integer's 1 to 9

from begining

0000000001
to
9999999999 = maximum seconds

SYMBOL

Lets say this were Symbol per atom universe, (similar to our own) and
when it got to 10 seconds it was unable to place a two digit integer
on the storage medium. Therefore it had to invent a new symbol.

1 2 3 4 5 6 7 8 9 R (where to us R = 10)

R 1 2 3 4 5 6 7 8 G (where to us G = 19)

R G 1 2 3 4 5 6 7 Y (where to us Y = 27)

R G Y 1 2 3 4 5 6 B (where to us B = 34)

R G Y B 1 2 3 4 5 P (where to us P = 40)

R G Y B P 1 2 3 4 O (where to us O = 45)

R G Y B P O 1 2 3 C (where to us C = 49)

R G Y B P O C 1 2 M (where to us M = 52)

R G Y B P O C M 1 W (where to us W = 54)

R G Y B P O C M W N (where to us N = 55)

To us, we could continue counting at 56 seconds from here. We also
wouldn't
need numbers because we have the space to write things like

Example of a rule.
(where to us R = 10)(where to us G = 19)(where to us Y = 27)(where to
us B = 34)(where to us P = 40)(where to us O = 45)(where to us C = 49)
(where to us M = 52)(where to us W = 54)(where to us N = 55)

But as you can see, by the end of this graph the 10 atom universe @ 1
Symbol/atom has gradually erased the digits to keep up with the count.
Also, it did not maintain a correct record of its rules regarding the
quantity of 10 seconds when it got to G. When it got to G, there only
remained 8 digits to express the rule of R which required 9 other
digits as well as itself. The theory of inevitable lack of resource
for numbers would eventually be the same for a 10^80atoms universe
(such as our own).

There are ways that this 10 atom (10symbol) universe could become more
efficient. But only again by a finite amount before the rules fill up
the space and begin to write over eachother and numbers will begin to
lose meaning thereafter until altogether.
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