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Il Reconciliatio

Ian Beardsley

2022

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It would seem to reconcile quantum mechanics with relativity we need perhaps

more to find an equation that takes us between formal systems than to have a
synthesis.

Contents

1.0 Overview………………………………………………..3

2.0 The Conundrum……………………………………….5

3.0 Quantum Mechanics………………………………….7

4.0 Analysis………………………………………………….10

5.0 Formulating Relativity A La HG Wells………………11

6.0 Conclusion………………………………………………13

7.0 Outlining The Problem…………………………………14

8.0 Proton-Seconds And Proton Radius………………..15

9.0 Product Calculus……………………………………….22

10.0 Difference Between Classical and Quantum…….24

11.0 Natural Abundances………………………………….26

12.0 Formulation of Proton-Seconds………………………….34


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1.0 Overview

In summation mathematics, upon which calculus is founded, we can express all of the
means, like arithmetic, harmonic, root mean square,…by a summation formula

1/p
1 n p
(n ∑ )
Equation 1.1 xi

i=1

This is the arithmetic mean (x1+x2)/2 if p =1 and n=2, the harmonic mean if p=2,…and
so on…

But, it is the geometric mean if you take the limit as p goes to zero which means since
you have to take the limit it is no longer an equation in standard calculus (does not use
the 𝜮 operator) and only becomes defined in the product calculus because the
equation must now be notated with the product operator 𝚷. While calculus is based on
the sigma operator 𝜮, product calculus is based on the product operator. Where the
definite integral of a function over an interval is its arithmetic mean, the integral over an
interval in product calculus is its geometric mean. The geometric mean is defined by
the product operator 𝜫, and is given when its p is equal to zero is

∏
Equation 1.2 M0(x1, …xn) = n xi

i=1

Where in arithmetic calculus derivatives and integrals of sums of functions are easy, for
products of functions can be become complex, but in product calculus theorems for
products of function become simple. However functions involving sums are not so
easily express, but apparently product calculus for a special case scenario becomes
arithmetic calculus, thus the claim is that calculus is a subset of product calculus. This
is easy to see, what is multiplication but a form of addition:

x + x + x = 3x

I have two theorems, one involving arithmetic and harmonic means, the others the
geometric mean. So you can see why I have become interested in product calculus,
which was first discovered in Italy in 1887 by Vito Voltera, and textbooks have been
created for it today as a formal system by some American mathematicians who are
trying to get it included in university curriculum for mathematics majors. I have become
interested in it because I have the desire to bring arithmetic and harmonic means
together with the geometric mean as one expression. So I need a synthesis of
arithmetic calculus and product calculus. Indeed is this not the human story seeing this
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in mathematics because the same conundrum exists for physics in the disparity
between quantum mechanics and relativity theory.

We can of course already make one equation that includes the arithmetic, harmonic,
and geometric means. It is called the f-mean, but it requires a limit operation, and I
want an equation that is purely algebraic. The f-mean is:

1 n
(n ∑ )
−1
Equation 1.3 Mf (x1, …xn) = f f (xi )

i=0

f (x) = x p and i=0,1,2,3..

Then it is the arithmetic mean, harmonic mean,…

When the function f is log (x ) it it becomes the geometric mean.

We can see how arithmetic calculus becomes hampered if we have to use the f-mean
for problems with the geometric mean. because instead of using x p it uses log(x). This
is precisely where the natural log function was created because in general the integral
of

x (n+1)
∫
n
x dx = + C

n+1

When we apply it to 1/x we have:

x0
∫
−1
x dx = + C

0
Which in not defined. So we had to find a function for which the integral holds which
turned out to be ln(x). But the natural log of x for most values are decimal expressions,
usually irrational.

Further the natural log is log to the base e, Euler's number. It seems to me whenever
we run into a conundrum in a formal system we find the solution is something that has
very interesting properties and thus implications. Indeed Euler's number is that. The
derivative of e x is e x itself which means its rate of change is equal to itself, hence its
applications in biology, like exponential growth rates of reproducing organisms, or the
properties of Euler's extraordinary discovery often called the most beautiful equation in
mathematics e iπ = − 1 relating the imaginary unit to negative one and π .

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2.0 The Conundrum

If we refer back to the foundations of calculus, while the integral of simple functions
can be considered

x n+1
∫
n
x dx = + C

n+1
We have a conundrum for

1
f (x) = = x −1

x
That the power rule gives:

1 x −1+1
∫ x
dx =

0
Thus to get around this, we searched for a function such that the integral holds, and as
such we discovered the natural logarithm (ln) and Euler’s number e. And we have

1
∫ x
d x = ln(x) + C

d x
e = e x

dx
Where

ln(x) = loge(x)

And, the derivative of e x is itself and e is the transcendental and irrational number given
by

e=2.718…

That is, while

1
f −1ln(x) ≠

ln(x)

f −1ln(x) = e x

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We can approximate any function with a polynomial, the simplest example being the
linear approximation formed by writing the change in f(x) due to a change in x:

f (x) = f (a) + f′(a)(x − a)

This results in Taylor’s formula

f′′(a)
f (x) = f (a) + f′(a)(x − a) + (x − a)2 + …

2!
From which we derive the Taylor series

∞
f n(a) f′′(a)
(x − a)n = f (a) + f′(a)(x − a) + (x − a)2 + …

∑ n! 2!
n=0

We know the kth derivative of e to the x is e to the x itself. Thus,

f (k)(x) = e x

x x2 x3
e =1+x+ + + …

2! 3!
xn
lim = 0

n→∞ n!

∞
x xn x2 x3
∑ n!
e = =1+x+ + + …

n=0
2! 3!
∞
1 1 1 1
∑ n!
e= =1+ + + + … = 2.718

n=0
1! 2! 3!

I would like to now suggest that the schism between Quantum Mechanics and
Relativity theory, and indeed Classical Mechanics in general is the same as that
between arithmetic calculus and product calculus.

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3.0 Quantum Mechanics

Since the derivative of

d x
e = e x

dx

If we put in the imaginary number i = −1 , we have

e iθ = cosθ + isinθ

This becomes important because we can use it to write wave equations and quantum
mechanics is wave mechanics.

We start with the premise that light can behave like a particle and its energy, E, is given
by

hc
E=

λ
And that particles, like electrons, can behave like a wave given by

h
λ=

mv
Where h is a constant called Planck’s constant and is given by

h = 2πℏ = 6.626 × 10−34 J ⋅ s

We consider a wave

U(x, t) = A0 sin(k x − ωt)

But, since

e ix = cos(x) + isin(x)

We write it as

U(x, t) = A0e i(k x−ωt)

But we need to put it in terms of energy E, and momentum, p, which since

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p = ℏk

E = ℏω

mv 2
k=

2
mv 2 p2
=

2 2m
We have the wave for our particle is

i
ψ (x, t) = Ce ℏ ( px−Et)

We now make the guess that the equation which it satisfies is:

( ∂x c 2 ∂t )
∂2 1 ∂2
− ψ (x, t) = 0

If we put in ψ (x, t) we have

∂ i ( px−Et) ∂ i ( px) i
⋅ −e ℏ (Et) = p (e ℏ ( px−Et))

i i
eℏ = eℏ
∂x ∂x ℏ
Or quite simply:

∂ψ i
= pψ

∂x ℏ
Similarly, for the time derivative

∂ i ( px−Et) ∂ i i
= e ℏ ( px) ⋅ −e ℏ (Et) = − E (e ℏ ( px−Et))

i i
eℏ
∂t ∂t ℏ
Or quite simply

∂ψ i
= − E ψ

∂t ℏ
We now have the two first derivatives of the wave as taken by the wave equation:

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∂ψ i ∂ψ i
= pψ and = − E ψ

∂x ℏ ∂t ℏ
Just as when a ball roles down a parabolic surface, as its potential energy increase and
the potential at the bottom is zero when the kinetic energy, and hence velocity is
greatest, electrons in an atom occupy different energy levels. But in this case they are
discrete, that is to say they only exist at certain levels and not in between. The inner
levels are lower energies than the outer levels. Quantum mechanics can predict these
by solving the wave equation.

Schrodingers wave equation:

+ V( r) ⃗ ψ ( r,⃗ t)

( 2m )
∂ψ ℏ2 ∇2
iℏ =
∂t

Schrodinger had his equation published in The Physical Review in 1926. He was asked
by his mentor to make sense of the strange claims being made by leading researchers,
that particles needed to be described as waves, and waves needed to be described as
particles, and give a presentation. It is said he did not give credence to the idea, and
didn’t realized the importance of his equation at the time of publishing it, but he later
received a Nobel Prize in physics for it.

The next crucial development came from Max Born, who offered an interpretation for
the meaning of the equation, which though we use it today, and it works, it is still an
interpretation not a derivation, and thus professionals often debate its validity still.

The Born Interpretation was that the square of the absolute value of the wave function
times the volume element dxdydz:

2
ψ (x, y, z, t) d xdydz

Is the probability of finding a particle described by ψ ( r,⃗ t) in the volume element

dV = d xdydz at time t. Since ψ is a complex function, the square of its absolute
value is the product of it with its complex conjugate. Written:

2
ψ = [ψ (x, y, z, t)] [ψ*(x, y, z, t)]

For this Max Born received a Nobel Prize in 1954.

Thus, the object of quantum mechanics is to solve Schrodinger’s equation

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+ V( r) ⃗ ψ ( r,⃗ t)

( 2m )
∂ψ ℏ2 ∇2
iℏ =
∂t
For ψ ( r,⃗ t)

And then, apply the Born interpretation

2
ψ = [ψ (x, y, z, t)] [ψ*(x, y, z, t)]

To find the probability of finding a particle in a given region at a time t.

Finally we have:

4.0 Analysis

The f-mean becomes the geometric mean when f (xi ) = ln(x) because f −1(x) = e x.
That is,…

1 n
(n ∑ )
−1
Mf (x1, …xn) = f f (xi )

i=1

Becomes:

1/n
1 2
(2 ∑ )
Mf (x1, x2) = exp ln(xi )

i=1
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(2 )
1/2
1 1
=exp ln(x1) + ln(x2)

= e ln(a1⋅b2 ) = (a1)(b2) = ab

Is the geometric mean between Si and Ge. We have the operator that transforms the
arithmetic mean is when p=1:

1 2 p
(2 ∑ )
Mf (x1, x2) = x

i=1

(2 2 )
1 1 x + x2 a+b
= x1 + x2 = 1 =

2 2

Which means ln(x) which maps a and b into their geometric mean is where
ln(x) = loge(x) reconciles with the arithmetic mean.

5.0 Formulating Relativity A La HG Wells

In The Time Machine, a science fiction story by HG Wells, the time traveller describes
time as physical distance, the direction through which the universe is falling at the
speed of light, c. Thus, not only when we move through space do we travel through a
distance at a velocity v, but we travel through a distance t at a velocity c. If we draw the
picture and account for that distance and velocity as well, we arrive at time dilation as
given by relativity theory (see illustrations and computation on the next page). He
writes:

I think that at the time none of us quite believed in the Time Machine. The fact is, the
Time Traveller was one of those men who are too clever to be believed: you never felt
that you saw all around him; you always suspected some subtle reserve, some
ingenuity in ambush, behind his lucid frankness.
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6.0 Conclusion

In quantum mechanics we have

E = ℏω

Energy is energy over time, times a frequency. The wave for our particle is:

i
ψ (x, t) = Ce ℏ ( px−Et)

The equation which satisfies it is the wave equation

( ∂x c 2 ∂t )
∂2 1 ∂2
− ψ (x, t) = 0

This means that momentum is associated with frequency:

∂ψ i
= pψ

∂x ℏ
And momentum is measured by time. But in relativity time is

t
t0 =

v2
1−
c2

It would seem we have two formal systems that contradict one another. But, in
mathematics we can write the arithmetic and harmonic means as one equation:

1/p
1 n p
(n ∑ )
xi

i=1

We can only include the geometric mean in the f-mean:

1 n
(n ∑ )
−1
Mf (x1, …xn) = f f (xi )

i=0

But f(x) in this is different than f(x) for the geometric mean which is ln(x). We then have
to write the geometric mean in a separate formal system, the product calculus:

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∏
M0(x1, …xn) = n xi

i=1

But are the two different formal systems here mathematical parallels of the two
different formal systems Quantum Theory and Relativity? Can the f-mean handle both,
and are the physical manifestations of the periodic table of the elements Nature’s way
of handling this dichotomy?

7.0 Outlining The Problem

a
1 a+b
b − a ∫b
f (x) = dx =

2
1/p
1 n p
(n ∑ )
xi

i=1
b−a 2ab
f (x) = a dx
=

∫b a+b
x

But,

1 n
(n ∑ )
−1
Mf (x1, …xn) = f f (xi )

i=1

a+b
: p = 1

2
f (x) = x p ⟼

2ab
: p = 2

a+b
n

∏
f (x) = ln(x) ⟼ M0(x1, …xn) = n = ab

i=1

a+b 2ab
Together: and Separate Formal System: ab

2 a+b
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So perhaps it becomes more a thing of switching between formal systems by changing

the function in the f-mean than actually achieving a synthesis.

8.0 Proton-Seconds and Proton Radius

We need to have one equation that brings in constants of quantum mechanics and of
relativity. Like h of QM and G of gravity. I find I can do this with inertia.

In order to present the elements as mathematical structures we need to explain the matter from
which they are made as mathematical constructs. We need a theory for Inertia. I had found
(Beardsley Essays In Cosmic Archaeology. 2021) where I suggested the idea of proton
seconds, that is six proton-seconds, which is carbon the core element of biological life if we
can figure out a reason to divide out the seconds. I found

1 h 4π rp2
Equation 8.1 = 6proton s

t1α 2 mp Gc

Where h is Planck’s constant 6.62607E—34 Js, rp is the radius of a proton 0.833E-15m, G is

the universal constant of gravitation 6.67408E-11 (Nm2)/(kg2), and c is the speed of light
299,792,459 m/s. And t1 is t=1 second. α is the Sommerfeld constant (or fine structure
constant) is 1/137. The mass of a proton is mp = 1.67262E − 27kg.

Since plank’s constant h is a measure of energy over time where space and time are
concerned it must play a role. Of course the radius of a proton plays a role since squared and
multiplied by 4π it is the surface area of our proton embedded in space. The gravitational
constant is force produced per kilogram over a distance, thus it is a measure of how the
surrounding space has an effect on the proton giving it inertia. The speed of light c has to play
a role because it is the velocity at which event are separated through time. The mass of a
proton has to play a role because it is a measurement of inertia itself. And alas the fine
structure constant described the degree to which these factors have an effect. We see the
inertia then in equation 3.1 is six protons over 1 second, by dimensional analysis.

The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed
of light squared:

Ue
α2 =

mec 2
Matter is that which has inertia. This means it resists change in position with a force applied to
it. The more of it, the more it resists a force. We understand this from experience, but what is
matter that it has inertia?

I would like to answer this by considering matter in one of its simplest manifestations, the
proton, a small sphere with a mass of 1.6726E-27 kg. This is a measure of its inertia.

I would like to suggest that matter, often a collection of these protons, is the three dimensional
cross-section of a four dimensional hypersphere.

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The way to visualize this is to take space as a two-dimensional plane and the proton as a two
dimensional cross-section of a sphere, which would be a circle.

In this analogy we are suggesting a proton is a three dimensional bubble embedded in a two
dimensional plane. As such there has to be a normal vector holding the higher dimensional
sphere in a lower dimensional space. Thus if we apply a force to to the cross-section of the
sphere in the plane there should be a force countering it proportional to the normal holding it in
a lower dimensional universe. This counter force would be experienced as inertia. It may even
induce in it an electric field, and we can see how it may do the same equal but opposite for the
electron.

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 18 of 37

Since we have

1 h 4π rp2 1 t2

Gc ∫t1 t 2
dt = ℕ

α 2 mp

Is a number of protons

1 h 4π rp2

α 2 mp Gc

Is proton-seconds. Divide by time we have a number of protons because it is a mass divided

by the mass of a proton. But these masses can be considered to cancel and leave pure
number.

Generating A Table

We make a program that looks for close to whole number solutions so we can create a table of
values for problem solving. I set it at decimal part equal to 0.25. You can choose how may
values for t you want to try, and by what to increment them. Here are the results for
incrementing by 0.25 seconds then 0.05 seconds. Constant to all of this is hydrogen and
carbon. The smaller integer value of seconds gives carbon (6 protons at 1 second) and the
largest integer value of seconds gives hydrogen (1 proton at six seconds) and outside of that
for the other integer values of protons you get are at t>0 and t<1. Equation 1 really has some
interesting properties. Here are two runs of the program( decpart is just me verifying that my
boolean test was working right to sort out whole number solutions):

By what value would you like to increment?: 0.25

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

24.1199 protons 0.250000 seconds 0.119904 decpart

12.0600 protons 0.500000 seconds 0.059952 decpart

8.0400 protons 0.750000 seconds 0.039968 decpart

6.0300 protons 1.000000 seconds 0.029976 decpart

4.0200 protons 1.500000 seconds 0.019984 decpart

3.0150 protons 2.000000 seconds 0.014988 decpart

2.1927 protons 2.750000 seconds 0.192718 decpart

2.0100 protons 3.000000 seconds 0.009992 decpart

1.2060 protons 5.000000 seconds 0.205995 decpart

1.1486 protons 5.250000 seconds 0.148567 decpart

1.0964 protons 5.500000 seconds 0.096359 decpart

1.0487 protons 5.750000 seconds 0.048691 decpart

1.0050 protons 6.000000 seconds 0.004996 decpart

0.2487 protons 24.250000 seconds 0.248659 decpart

0.2461 protons 24.500000 seconds 0.246121 decpart

0.2436 protons 24.750000 seconds 0.243635 decpart

By what value would you like to increment?: 0.05

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

40.1998 protons 0.150000 seconds 0.199837 decpart

30.1499 protons 0.200000 seconds 0.149879 decpart

24.1199 protons 0.250000 seconds 0.119904 decpart

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20.0999 protons 0.300000 seconds 0.099918 decpart

17.2285 protons 0.350000 seconds 0.228500 decpart

15.0749 protons 0.400000 seconds 0.074938 decpart

12.0599 protons 0.500000 seconds 0.059950 decpart

10.0500 protons 0.600000 seconds 0.049958 decpart

8.0400 protons 0.750000 seconds 0.039966 decpart

7.0941 protons 0.850000 seconds 0.094088 decpart

6.0300 protons 1.000000 seconds 0.029975 decpart

5.2435 protons 1.150000 seconds 0.243457 decpart

5.0250 protons 1.200000 seconds 0.024980 decpart

4.1586 protons 1.450000 seconds 0.158605 decpart

4.0200 protons 1.500000 seconds 0.019985 decpart

3.1737 protons 1.899999 seconds 0.173673 decpart

3.0923 protons 1.949999 seconds 0.092296 decpart

3.0150 protons 1.999999 seconds 0.014989 decpart

2.2333 protons 2.699999 seconds 0.233325 decpart

2.1927 protons 2.749999 seconds 0.192719 decpart

2.1536 protons 2.799999 seconds 0.153564 decpart

2.1158 protons 2.849998 seconds 0.115782 decpart

2.0793 protons 2.899998 seconds 0.079303 decpart

2.0441 protons 2.949998 seconds 0.044061 decpart

2.0100 protons 2.999998 seconds 0.009993 decpart

1.2433 protons 4.850000 seconds 0.243294 decpart

1.2306 protons 4.900001 seconds 0.230607 decpart

1.2182 protons 4.950001 seconds 0.218177 decpart

Here is the code for the program:

#include <stdio.h>

#include <math.h>

int main(int argc, const char * argv[]) {

int n;

float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,

c=299792459,protons[100],r=0.833E-15;

do

printf("By what value would you like to increment?: ");

scanf("%f", &increment);

printf("How many values would you like to calculate for t in equation 1 (no more than 100?):
");

scanf("%i", &n);

while (n>=101);

for (int i=0; i<n;i++)

protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));

int intpart=(int)protons[i];

float decpart=protons[i]-intpart;

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t=t+increment;

if (decpart<0.25)

{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);

}}}}

A very interesting thing here is looking at the values generated by the program, the smallest
integer value, 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest
of the elements heavier than carbon are formed by fractional seconds. These are the
hydrocarbons the backbones biological chemistry.

1 h 4π rp2
Equation 8.2 = 1.004996352secon d s

6α 2 mp Gc

1 h 4π rp2
= (6proton s)(1secon d )

α 2 mp Gc

1 h 4π rp2
= (1proton)(6secon d s)

α 2 mp Gc
Ue
α2 =

mec 2
Thus we have six protons (carbon) is one second. We see 1 second predicts the radius of a
proton,…

In that we have

1 h 4π rp2 1 t2

Gc ∫t1 t 2
dt = ℕ

α 2 mp

And the periodic table of the elements is cyclical with 18 groups and

1 h 4π rp2
6= 2

α mp Gc

Then perhaps we are supposed to write

3 h 4π rp2 1 t2

Gc ∫t1 t 2
dt = 18

α 2 mp

In fact, what if the 3 is supposed to be pi, then

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π h 4π rp2 1 t2

Gc ∫t1 t 2
dt = 18

α 2 mp

Then we would say that

k=18/pi=5.7229577951

The parameter in our constant with the most uncertainty is the radius of a proton rp. If the 3 is
supposed to be pi, then the radius of a proton becomes:

Gc
Equation 8.3 rp = k α 2 mp

4πh
Which gives

rp = 8.790587E − 16m

About 95% raw most recent value measured. But, if

1 h 4π rp2

α 2 mp Gc

Is supposed to be 6 and it is supposed to be multiplied by three to give 18 even which we

need for chemistry so we have 18 protons in the last group of the periodic table which is
important because we need argon with 18 protons for predicting valence numbers of elements
in terms of their need to attain noble gas electron configuration. Then we get

rp = 8.288587E − 16m = 0.829f m

This is in very close agreement with the most recent value measured which is

rp = 0.833 + / − 0.014

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9.0 Product Calculus

The arithmetic mean of a function f(x) is the inverse operation to integration. The arithmetic
mean derivative is the one we are all familiar with, the Fermat-Newton-Leibnitz derivative:

dF(x)
D (1) F(x) =

dx
It is an additive operator:

Equation 9.1. D(F1(x) + F2(x)) = DF1(x) + DF2(x)

It was suggested by Michael Spivey that there be a generalized calculus for the product
integral which is related to the geometric mean over an interval, which is multiplicative. The
arithmetic mean is:

b
1
b − a ∫a
Equation 9.2. f (x)d x

It is an additive derivative and thus applies effectively to infinite series. The geometric mean

1 b
e b − a ∫a ln( f (x))d x

Is multiplicative and applies effectively to products:

Equation 9.3 D(F1(x)F2(x)) = DF1(x)DF2(x)

b
1
b − a ∫a
f (x)d x

Is the inverse operation of additive integration in the sense that

1
x = 1

x
b
1 1
b − a ∫a
dx = x ∣ ba = 1

b−a
The multiplicative integral is useful for growth:

Ae r(t)dt

Where we have the sequence of infinite products

Ae r(t1)Δt e r(t2 )Δt…e r(tn )Δt = Ae [r(t1)+r(t2 )+…+r(xn )]Δt

We say the product integral of e r(t) is

 23 of 37

b
b
e r(t)dt = e ∫a r(t)dt

∏
Equation 9.4
a

And, of f(x) over an interval [a,b] is

b
b
f (x)d x = e ∫a (lnf (x))d x

∏
Equation 9.5.
a

The harmonic mean in terms of discrete product notation can be written as the the
ratio between between the geometric and arithmetic means:

n
n ⋅ ∏i=1 xi
H=

∑i=0 { x1 ∏i=1 xi}

n n
i

n n

∏ ∑
is a product analog to and

i=0 i=0

b a

∏ ∫b
is a product analog to

That is, the harmonic mean has in the denominator the arithmetic mean of the product
of numbers n times, and the numerator is the nth power of the geometric mean. This is
written

H(x1, …, xn) =

(G(x1, …, xn))n
=

A(x2 x3…xn, x1x3…xn, x1x2…xn−1)

(G(x1, …, xn))n

A ( x1 ∏i=1 xi)
n n n
∏i=1 xi, x1 ∏i=1 xi, …, x1
1 2 n
 24 of 37

10.0. Difference Between Classical And Quantum

In the sense in classical deterministic physics, which indeed include relativity as far as
determinism goes, it is a formulation of arithmetic calculus:

D(F1(x) + F2(x)) = DF1(x) + DF2(x)

Whereas we might say quantum mechanics is a natural formulation of product

calculus:

D(F1(x)F2(x)) = DF1(x)DF2(x)

In the sense that classical mechanics has that energy is:

∫b
W= F ⋅ d x

Whereas quantum mechanics has the wave for a particle is

i
ψ (x, t) = Ae ℏ ( px−Et)

And since we have the probability of its position at a time t is

2
ψ = [ψ (x, y, z, t)] [ψ*(x, y, z, t)]

It is more characteristic of

Ae r(t1)Δt e r(t2 )Δt…e r(tn )Δt = Ae [r(t1)+r(t2 )+…+r(xn )]Δt

Lending itself to

b
b
e r(t)dt = e ∫a r(t)dt

∏
a

But what expression is capable of both? It is the f-mean:

1 n
(n ∑ )
−1
Mf (x1, …xn) = f f (xi )

i=1
 25 of 37

It is the arithmetic mean character is to of 9.3 when

f (x) = x p

And is the geometric mean characteristic of 9.4

f (x) = ln(x)

 26 of 37

11.0 Natural Abundances

Indeed I find the size of the solar system is connected to Silicon through its counterpart as a
semiconductor material germanium. Let’s look at that:

While we have

1 h 4π rp2 1 t2

Gc ∫t1 t 2
dt = ℕ

α 2 mp

Is a number of protons

1 h 4π rp2

α mp
2 Gc

Is proton-seconds. Divide by time we have a number of protons because it is a mass divided

by the mass of a proton. But these masses can be considered to cancel and leave pure
number. We have that

( 2 3)
1 1 1
∫
6 dt = 6 − = 0.78

2 t2

3
1 π −4
∫ 6( )
cos −1(x /2)d x = 3π − 6 − = 0.21

2 2 2
Interestingly 78% is the percent of N2 in the atmosphere and 21% is the percent of O2 in the
atmosphere (by volume). These are the primary constituents that make it up. The rest is
primarily argon and CO2. This gives the molar mass of air as a mixture is:

0.78N 2 + 0.21O2 = 29.0g/m ol

Now interestingly, I have found this connected to the solar system by considering a mixture of
silicon, the primary constituent of the Earth crust, and germanium just below it in the periodic
table, in the same proportions of 78% and 21%. Silicon is also the primary second generation
semiconductor material (what we use today) and germanium is the primary first generation
semiconductor material (what we used first). The silicon is directly below our carbon of one
proton-second, silicon directly below that, and germanium directly below that, in the periodic
table. So we are moving directly down the periodic table in group 14. The density of silicon is
2.33 g/cm3 and that of germanium is 5.323 g/cm3. Let us weight these densities with our 0.21
and 0.78:

0.21Si + 0.78Ge = 4.64124g/cm 3

Now consider this the starting point for density of a thin disc decreasing linearly from the Sun
to Pluto (49.5AU=7.4E14cm). Thus,…

 27 of 37

( R)
r
ρ(r) = ρ0 1 −

Thus,…

∫0 ( R)
2π R
r
∫0
M= ρ0 1 − rdrdθ

Or,…

πρ0 R 2
M=

3
Thus,…

π (4.64124)(7.4E14)2
M= = 2.661E 30

3
If we add up the masses of the planets in our solar system they are 2.668E30 grams.

Since

2.661
(100) = 99.736

2.668
Meaning mixing germanium and silicon in the same proportion that occurs with N2 and O2 in
the atmosphere and with

( 2 )
1 1 1
∫
6 dt = 6 − = 0.78

2 t2 3

3
1 π −4
∫
cos −1(x /2)d x =
6 ( 3π − 6) − = 0.21

2 2 2
Where

1 h 4π rp2
6= 2

α mp Gc

In the first integral. See the following pages to see the computation of the mass of the planets
in the solar system…

 28 of 37

As we can see Jupiter carries the majority of the mass, Saturn a pretty large piece, and
somewhat large Uranus and Neptune. We don’t even list Pluto’s mass. When we consider

( 2 )
1 1 1
∫
6 dt = 6 − = 0.78

2 t2 3
3
1 π −4
∫
cos −1(x /2)d x =
6 ( 3π − 6) − = 0.21

2 2 2

We are considering 2 and 3 . These come from

2cos(45∘) = 2

2cos(30∘) = 3

From 30 degrees to 45 degrees is 15 degrees. The Earth rotates through 360/24 is 15 degrees
per hour. The hour is divided into 60 minutes and minutes into 60 seconds…

We have said

πρ0 R 2
M=

3
 29 of 37

For a thin disc. Thus we have a definition for the radius of the solar system, Rs:

3Mp
Equation 11.1 Rs =

π (0.78Ge + 0.21Si )

Where

1 h 4π rp2 3
1
Gc ∫
Equation 11.2 dt = 0.78

α 2 mp 2 t2

∫
Equation 11.3 cos −1(x /2)d x = 0.21

Equation 11.4 air = 0.78N2 + 0.21O2

air
Equation 11.5 ≈ Φ

H2O

Mp is the mass of all the planets. We also have that the molar mass of air to the molar mass of
water is approximately the golden ratio. The interesting thing is we determine a definition for
the radius of the solar system and predict the hydrocarbons (backbones of life) all in one fell
swoop. Thus since if the universe is mostly hydrogen and helium if we can predict their relative
abundances with proton-seconds then we can feel quite certain we are formulating things
correctly, consider the following gaussian distribution illustrated…

 30 of 37



 31 of 37

We consider a Gaussian wave-packet at t=0:

x2 −
ψ (x,0) = Ae

2d 2
We say that d which in quantum mechanics would be the delocalization length when squared is
Si − C Ga − P
. A is the amplitude and we might say it is . We write the wave packet as a
C C
Fourier transform which is:

x2 dp
∫ 2π ℏ
i
ψ (x,0) = Ae = −
ϕpe h px

2d 2

We use the identity that gives the integral of a quadratic:

∞
π β2
∫−∞
2 x+βx
e −α dx = e 4α

α
Solve the equation

p̂
iℏ∂t ψ (x, t) = ψ (x, t)

2m
With the initial condition

p2 d2

∫
i
ψ (x,0) = dp ⋅ e 2ℏ2 ⋅ e − h px

A plane wave is the solution:

i
e ℏ ( px−ϵ( p)t)

p2
Where, ϵ( p) =

2m
The wave-packet evolves with time as

p2 d2 p2

∫
− hi ( px− 2m t)
ψ (x, t) = dp ⋅ e 2ℏ2 ⋅e

Calculate the Gaussian integral of dp

d2 it ix
α= + and β =

2ℏ2 2mℏ ℏ
The solution is:

 32 of 37

[ d 1 + t 2 /τ 2 ]
2 x2 1 m d2
ψ = exp − 2 ⋅ where τ =

We notice here one of the things you can do with our equation for proton-seconds is integrate
from 0.5 sec to 1 sec and you get one which multiplied by the constant which is six yields six.
Now look up 0.5 seconds from the data output from the program and it is magnesium, then go
to one second and it is carbon, thus the integral from magnesium in time to carbon in time is
carbon in protons. Now consider life as we know it is based on carbon because it has four
valence electrons, but it is not based on silicon, which has four valence electrons as well,
because in the presence of oxygen it readily forms SiO2 (sand or glass) leaving it unavailable to
nitrogen, phosphorus, and hydrogen to make make amino acids the building blocks of life. But
silicon can be doped with phosphorus, boron, gallium, and arsenic to make semiconductors --
transistor technology from which we can build artificial life (artificial intelligence, AI). We can
integrate over many time ranges to explore millions of more facets to the equation:

1 h 4π rp2 tC
1
Gc ∫tMg
dt = 6 = carbon(C )

α 2 mp t2

tMg = 0.5secon d s

tC = 1secon d

We should say

Si − C 4 16
d= = or d 2 =

C 3 9
The way I am using equation 1 is τ = d 2. We have:

Equation 3.14

2 C2 x2 1
ψ = exp − ⋅

[ m(Si − C)2 ]
(Si − C )2 ℏC 2
2
1+ t2
 33 of 37

Thus for hydrogen:

2 9 2 1
ψ = exp − x ⋅

16 2
1 + ℏ2 81 t 2
m 256

2 9 1
ψ = (1)exp − (1proton)2 ⋅ =74%.

16 1+
(0.075)81
(6secon d s)2
(1)256

For Helium:

2 9 1
ψ = (2)exp − (2proton)2 ⋅ =26%

16 1+
(0.075)81
(3secon d s)2
(4)256

This is interesting because the Universe is about 74% Hydrogen and 24% Helium, the rest of
the elements making up the other 2%. Thus we can say ℏ2 = 0.075 or ℏ = 0.27386. We have
multiplied the first by 1 for Hydrogen element 1 (1 proton), and the second by 2 for helium
element 2 (2 protons). In a sense then, the probabilities represent the probability of finding
hydrogen and helium in the Universe. Hydrogen much of the helium were made theoretically in
the Big Bang of the big bang theory. The other elements were synthesized from these by the
stars. The difference between hydrogen and helium is that some of the helium was made in
stars, that may be why it is 26% not 24% because it could include the remaining 2% of
elements in the universe. It must be kept in mind the data we have on the universe as a whole
for relative abundances can only currently be ballpark figures.

 34 of 37

12.0 Formulation of Proton-Seconds

We can actually formulate this differently than we have. We had

1 1 h 4π rp2
= 6proton s

t1 α 2 mp Gc

1 1 h 4π rp2
= 1proton

t6 α 2 mp Gc

But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2
are lower and upper limits in an integral, then we have:

1 h 4π rp2 1 t2

Gc ∫t1 t 2
Equation 12.1 dt = ℕ

α 2 mp

This Equation is the generalized equation we can use for solving problems.

Essentially we can rigorously formulate the notion of proton-seconds by considering

∫t ∬S
Equation 12.2 qdt = t 2 ρ(x, y, z)d x d y

Is protons-seconds squared where current density is J ⃗ = ρ v ⃗ and ρ = Q /m 3 (ρ can also be

Q /m 2). We say

∫V
Equation 12.3. Q= ρdV

Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a
number of protons. It is

1 h 4π rp2
Equation 12.4 ℕ= 2

α mp Gc

Dividing Equation 4.2 through by by t:

Equation 12.5

1 h 4π rp2dt
Gc ∫t t ∬S
=t ρ(x, y, z)d x d y

α 2 mp

Which is proton-seconds. Dividing through by t again:

 35 of 37

1 h 4π rp2dt
Gc ∫t t 2
Equation 12.6 = proton s

α 2 mp

We see that if J ⃗ = ρ v ⃗ where ρ = Q /m 3 and v = m /s then J is I/m2 (current per square

meter) is analogous to amperes per per square meter which are coulombs per second through
a surface. Thus we are looking at a number of protons per second through a surface. Thus we
write:

1 h 4π rp2 tC
dt 1.0

Gc ∫tMg ∫0.5
= 6 t −2 dt = − 6(1 − 2) = 6

α 2 mp t 2

Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to
protons in the output from our program and 1.0 seconds is carbon (C). We see we have the
following theorem:

h 4π rp2
J ⃗ ⋅ d S
⃗
1 dt
Gc ∫t t 3 ∬S
Equation 12.7 =
α 2 mp

So as an example,…

( 0.25 )
h 4π rp2 1.0
J ⃗⋅ d S ⃗ = − 3 1 −
1 dt 1 proton s
Gc ∫0.5 ∬S
= =9

α mp
2 t 3 secon d

Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second
which is carbon and we have protons per square meter. Charge conservation gives:

+ ∇ ⋅ J ⃗ = 0

∂ρ
∂t
Is 9-9=0. In general…

⃗ y, z) = (0,0,J ) = − J k
⃗
J (x,

d S ⃗ = d x d y k
⃗

J ⃗ ⋅ d S ⃗ = (0,0,J ) ⋅ (0,0,d x d y) = − Jd x d y

 36 of 37

We consider the surface through which J passes a circle of radius R where J decreases linearly
as r with J0 at the center we have:

Equation 12.8

∫0 ( R)
2π R
π J0 R 2
J ⃗ ⋅ d S ⃗ = J0
r
∬S ∫0
1− rdrdθ =



 37 of 37

The Author