This paper will present in the most concrete, explicit manner the mathematics of space–time, radiation, and matter of the Reciprocal System. Readers without special knowledge of the Reciprocal System are first urged to study Larson’s books, especially Nothing But Motion¹ before undertaking the study of this paper.
I. Mathematics of Space–Time
A. Rectangular Coordinates
Starting from any reference point x_{0}, y_{0}, z_{0}, t_{0} in the 0–system, the space–time progression is a spherical expansion. In rectangular coordinates the equation is
(x–x_{0})^{²} + (y–y_{0})^{²} + (z–z_{0})^{²} = c^{²} t^{²}

(1)

where c is the speed of light. If we choose the reference point to be x_{0} = _{0}, y_{0} = _{0}, z_{0} = _{0}, t_{0} = 0, then the equation is simply
x^{²} + y^{²} + z^{²} = c^{²} t^{²} 
Now consider a second system, the 0´–system, moving translationally with respect to the 0´–system in the x–direction. What is the equation for the progression in the 0´–system? From the inverse Lortentz transformations,
[x´+vt´]² 
1


x²=

—————  = ————  (x´²+v²t´²+2v x´t´)  (2) 
[n1(v²/c²)]²  1(v²/c²) 
[t´+(v/c²)x´]² 
1


t²=

—————  = ————  (v²/c4)x´²+t´²+(2v/c²)x´t´)  (3) 
[n1(v²/c²)]²  1(v²/c²) 
y² = y´² 
(4)

z² = z´² 
(5)

Upon substitution, we obtain
x^{²} + y^{²} + z^{²} – c^{²} ^{t²} = x´^{²} + y´² + z´^{²} – c^{²}t´² 
(6)

But since the left side of the equation equals zero, so must the right side:
x´^{²} + y´^{²} + z´^{²} = c^{²}t´^{²} 
(7)

Thus the progression as determined by 0' is also spherical. And so the equation for the progression is invariant under a Lorentz transformation.
B. Polar Coordinates
In polar coordinates the equation is simply
r – r_{0} = c(t – t_{0}) 
(8)

Or, letting r_{0} = 0, t_{0} = 0,  (9) 
r/t = c 
In the Reciprocal System the speed of light is the natural unit of velocity and so r and t must take equal natural values. The space–time progression is thus ^{¹}/1, ^{²}/2, ^{3}/3, etc. Thus one unit of space is equivalent to one unit of time. If there are an infinite number of space units, there must be an infinite number of time units; if there are a finite number of space units, there must be a finite number of time units.
II. Mathematics of Radiation
In the Reciprocal System radiation is the combined motion of a simple harmonic oscillation in one dimension and a uniform translation in a perpendicular direction.
A. Simple Harmonic Oscillation
The equation for a simple harmonic oscillation in one dimension (say the y direction) is
y = A*SIN(–2pf_{os}t)  (10) 
where A is the amplitude and f_{os} is the frequency. Since the oscillation takes place over one natural space unit, the amplitude must be one–half a natural space unit and thus is
A = .5*S_{nat} = .5 x 4.558816 x 10^{–8}m = 2.279408 x 10^{–8}m  (11) 
for all photons. In observation from the time–space region this value is reduced by the interregional ratio142.222 to 1.6027 x 10^{–10}m = 1.6027 Å
The other variable to be determined in eq.(10) is the frequency, f_{os}. In one cycle the oscillation travels one space unit up and one space unit down, for a total of two units. The average velocity of the oscillation is th
v_{os} = (2*s_{nat}/cycle) * f_{os} (cycles/sec)  (12) 
The natural unit of frequency must occur when the average velocity is c.
v_{os} = c = 2*s_{nat} * f_{os nat}  (13) 
But c = s_{nat}/t_{nat´}  
so s_{nat}/t_{nat} = _{2*s}n_{at} * f_{os nat}  
Solving for f_{os nat} we have 
f_{os nat} = 1/(2*t_{nat}) = 1/(2*1.520655 * 10^{–16}sec) = 3.2880575 * 10^{15} cycles/sec  (14) 
the Rydberg frequency. (Actually, Larson derived the natural unit of time from the Rydberg frequency, but I think it was instructive to do the reverse, and this method will be used to calculate rotational and rotational vibration frequencies as well. Of course, this method assumes that the natural unit of time can be found by some other means.) Because of the discrete nature of the Reciprocal System, it is only possible to have integer multiples or reciprocal integer multiples of the Rydberg frequency.
Putting the values of A and f_{os} in eq.(10) we have
y = 1.6027*SIN(–2pn * 3.2880575 * 10^{15} * t)Å 
(15)

where
n = 1, 2, 3, ...or  (16) 
n = ½, ^{¹}/_{3}, ^{¼}, .... 
B. Perpendicular Translation
Perpendicular to the oscillation is a translation at unit velocity (the speed of light). Let x be perpendicular to y. Then
x = c * t 
(17)

C. Combined Motion
From eq.(17) t can be found in terms of x and c and put in eq.(15). The result is
y = 1.6027 * SIN(–2pn * 3.2880575 * 10^{15} * x/2.997930 * 10^{8})  (18) 
or y = 1.6027 * SIN (–6.8912465 * 10^{7} * nx)Å 
if x is given in meters. This is the equation for a monochromatic wave of radiation in the Reciprocal System.
III. Mathematics of Matter
Particles of matter consist of rotating photons. Subatoms have one rotating photon; atoms have two rotating photons (both photons rotate about the same central point). The rotational motion has a translational effect, which will be discussed after the mathematics of the rotation has been worked out.
A. Rotation
1. single systems–particles
A photon can rotate around either of two horizontal axes passing through its midpoint, and also around itself. In the Reciprocal System the true physical zero is motion at unit speed. Anything physical must have a motion either greater or less than unit speed. This deviation is called a speed displacement by Larson. The first particle has 1 speed displacement around one horizontal axis of the photon and is called the rotational base. Actually there are two rotational bases: one with one speed displacement above unity, the other with one speed displacement below unity. As will be discussed later, the one displacement unit neutralizes the translational motion of the photon in the original dimension, but the progression now continues in the remaining dimension, so the effective displacement is zero. In the ground state condition, the photon that is being rotated is one vibrational displacement away from unity (either 2R or (^{1}/2)R). Here is a table giving the photon frequency, the rotational displacement, the effective rotational displacement, and rotational speed of the cosmic rotational base and the material rotational base:
Photon frequency  Rotational displacement 
Effective rotational displacement 
Rotational speed 

———— 
—————

————————

——————


Cosmic rotational base 
½R

(1)00

000

211

Material rotational base 
2R

100

000

½11

In the above table the speeds are calculated from the displacements as follows. For displace
ments of n_{p}, n_{s}, and n_{E}, the speeds are (n_{p}+1), (n_{s} + 1), and (n_{E} + 1) for a cosmic particle, and 1/(n_{p} + 1), 1/(n_{s} + 1), and 1/(n_{E} + 1) for a material particle. (Of course material particles could have high speed electric displacement, and cosmic particles could have low speed electric displacement).
These speeds can be converted to conventional units, such as revolutions per second, as follows. In one rotation of a photon about a horizontal axis the tip of the photon coverse a distance of p* s_{nat´} a circumference. The speed of the rotation is then
V_{rot} = [(p * s_{nat}) /rev] * f_{rot} (rev/sec)  (19) 
The natural frequency of rotation must occur when the speed is c.
c = ¹ * s_{nat} * f_{rot nat}  
But c = s_{nat}/t_{nat´} 

so solving for f_{rot nat}, we get 

f_{rot nat } = ^{1}/(¹t_{nat}) 

But t_{nat} = ^{1}/_{(2R)} 

so f_{rot nat} = ^{2R}/_{¹} 
(20) 
where R is the Rydberg frequency, as before. In these terms, then, the cosmic rotational base is a photon that has a vibrational oscillation of 1.6440288 x 10^{15} cycles/sec and is rotating at 4.1864848 x 10^{15} revolutions/sec around one axis, and 2.0932424 x 10^{15 }revolutions/sec around the other two axes. Likewise, the material rotational base is a photon that has a vibrational oscillation of 6.576115 x 10^{15} cycles/sec and is rotating at 1.0466212 x 10 ^{15} revolutions/sec around one axis, and 2.093242 x 10^{15} revolutions/sec around the other two axes.
All the other particles have photon vibrational frequencies, rotational displacements, effective rotational displacements, rotational speeds, and rotational frequencies. Here is a complete tabulation:
Photon frequency

Rotational
displacement 
Effective rot.
displacement 
Rotational
speed 
Rotational
frequency 

————

—————

—————

————————

——————


M–positron

2R

1–0–1

0–0–1

^{½}1–^{½}

^{R}/_{p}^{2R}/_{p}^{R}/_{p}

C–positron 
½R

(1)–0–(1)

0–0–(1)

2–1–2

^{4R}/_{p}^{2R}/_{p}^{4R}/_{p}

M–electron

2R

1–0–(1)

0–0–(1)

^{½}1–2

^{R}/_{p}^{2R}/_{p}^{4R}/_{p}

C–electron

½R

(1)–0–1

0–0–1

2–1–^{½}

^{4R}/_{p}^{2R}/_{p}^{R}/_{p}

Photon frequency

Rotational
displacement 
Effective rot.
displacement 
Rotational
speed 
Rotational
frequency 

————

—————

—————

————————

————————


Mmassless neutron

2R

1–1–0

½ –½–0

^{½}–^{½1}

^{R}/_{p}^{R}/_{p}^{²R}/_{p}

Cmassless neutron 
½R

(1)–(1)–0

(½)–(½)–0

2–2–1

^{4R}/_{p}^{4R}/_{p}^{²R}/_{p}

Mneutrino

2R

1–1–(1)

½–½–(1)

^{½}½–2

^{R}/_{p}^{R}/_{p}^{4R}/_{p}

C–neutrino

½R

(1)–(1)–1

(½)–(½)–1

2–2–^{½}

^{4R}/_{p}^{4R}/_{p}^{R}/_{p}

Many more permutations appear to be possible, but the probability principles keep eccentricity to a minimum. Since none of the above particles has an effective displacement of 1 or more, they are all massless (aside from the mass contribution of an electric charge). The diameter of all the particles is one natural space unit, reduced by the (one–photon) interregional ratio, or 3.2054 Å. However, because these particles do not exert any force in the uncharged state, a particle–measuring probe would not be able to detect any size of these particles at all.
2. intermediate systems
Intermediate particles have two rotating photons, but one of the two sets has no effective displacement and thus contributes no primary mass. The two intermediate particles are the neutron and the mass one hydrogen isotope (and their cosmic analogs). There are only two kinds of rotations that can combine to form this kind of particle, the proton type and the neutrino type We identify the combination of the material proton rotation and the material neutrino rotation as the mass one hydrogen atom; the combination of the material proton rotation and the cosmic neutrino rotation as the neutron; the combination of the cosmic proton rotation and the cosmic neutrino rotation as the mass one atom of cosmic hydrogen; and the combination of the cosmic proton rotation and the material neutrino rotation as the cosmic neutron. The proton is a single system with displacements 2–1–(1), effective displacements 1–1–(1), speeds ^{½}^{½}2, and rotational frequencies 2R/3p–R/p–4R/p. Then we would have the following table for the neutron and mass one hydrogen.
Photon frequency

Rotational
displacement 
Effective rot.
displacement 
Rotational
speed 
Rotational
frequency 

————

—————

—————

————————

————————


Neutron

{

2R

2–1–(1)

1–1–(1)

¹/_{²} –^{½2}

^{2R}/_{³}_{p}^{R}/_{p4}^{R}/_{p}


½R

(1)–(1)–1

(½)–(½)–1

2–2–½

^{4R}/_{p}^{4R}/_{p}^{R}/_{p}


H¹

{

2R

2–1

>  (1) 
1–1

>  (1) 
¹/_{³}½

> 
2

^{2R}/_{³}_{p}^{R}/_{p}

>4R/p 
2R

1–1

½–½

½ –^{½}

^{R}/_{p}^{R}/_{p}

The new notation makes clear the two photons involved and the five rotations (to be discussed next).
3. Atomic cycles
Atoms have two rotating photons, but here both systems have effective displacements and both systems ordinarily have the same velocities. Let the first photon be called A and the second be called B. A and B are mutually perpendicular. We have the following five rotations:
(i) the rotation of A about B produces disk a;
(ii) the rotation of B about A produces disk b;
(iii) then disk a can be rotated about A;
(iv) then disk b can be rotated about B;
(v) finally the whole structure can be rotated in the electric dimension.
This last rotation is in the scalar direction opposite to that of the previous rotations. Cosmic atoms have speeds above unity for the first four types of rotations, whereas material atoms have speeds below unity for the first four types. The electric rotation may be above or below unity for both cosmic and material atoms.
The first particle with two effective rotating systems is deuterium, the second is helium, etc. A table similar to that for the intermediate particles can be made.
Photon frequency

Rotational
displacement 
Rotational
speed 
Rotational
frequency 

————

—————

———————

————————


Deuterium

{

2R

2–1

>  (1) 
¹/_{³} –^{½}

> 
2

^{2R}/_{³}_{p}^{R}/_{p}

>4R/p

2R

2–1

2_{³}–½

^{2R}/_{³}_{p}^{R}/_{p}


Helium

{

2R

2–1

> 
0

¹/_{³}½

> 
1

^{2R}/_{³}_{p}^{R}/_{p}

>2R/p 
2R

2–1

½_{³} –^{½}

^{R}/_{³}_{p}^{R}/_{p}

All other atoms can be given appropriate values in the same manner. In the solid state, however, the values that govern the physical properties are not the actual rotations, but the relative rotations, and the different values there are not due to inherent differences in the rotational speeds, but to differences in the orientations of the interacting atoms, and this will be discussed further later.
4. Electric charges and magnetic charges
According to the Reciprocal System an electric charge is a rotational vibration about the electric axis, and the magnetic charge is a rotational vibration about one of the magnetic axes. Both charges have the same natural frequency, calculated as follows. In one cycle the motion covers a distance of ¹ * s_{nat} one way and p * s_{nat} back, for a total of 2p * s_{nat}. So we have
v_{ch} = (2p * s_{nat}/cycle) * f_{ch} (cycles/sec)  (21) 
At the unit level, v_{ch} = c = s_{nat}/t_{nat}, so
s_{nat}/t_{nat} = (2p * s_{nat}/cycle) * f_{ch nat}  (22) 
Solving for f_{ch nat} and recalling that t_{nat} = ^{1}/_{(2R)},
f_{ch nat} = ^{R}/p 
(23)

This frequency is one–half that of a full rotation and can thus be considered to be effective in one direction only half the time. One negative electric charge is a rotational vibration of R/2p = 5.233106 x 10^{14} cycles/sec. One positive electric charge is a rotational vibration of 2R/p = 2.093242 x 10^{15} cycles/sec. Similarly one unit of magnetic charge is a rotational vibration of 2R/p = 5.233106 x 10^{14} cycles/sec, whereas one unit of isotopic charge is a rotational vibration of R/2p = 5.233106 * 10^{14} cycles/sec. The isotopes of atoms result from the addition of isotopic charges.
B. Translation
The rotational motion of particles has a translational effect. The maximum inward translation is two full units, giving one net inward unit. In terms of rotation we can have 2^{³} = 8 one–dimensional rotational electric displacements or 4 two–dimensional rotational magnetic displacements. Note that since 1^{³} = 1, the first magnetic rotational displacement, which is ^{½} unit rotational speed, produces one unit of inward translation and thus neutralizes the original translational motion of the photon, but the progression still continues in the third dimension. Thus the rotational base and all the single system massless particles previously discussed move at the speed of light. Additional magnetic and electric displacements produce a net inward motion, and the inward motion of a group of atoms is termed gravitation.
For atoms with magnetic displacements of less than 4 and electric displacements of less than 8, the frequency of the rotating photons is normally one displacement above unity, or 2R (the frequency of photons in cosmic atoms is (^{½})R). When the magnetic displacement reaches 4 or the electric displacement reaches 8, the rotation must be extended to a second vibrational displacement unit–which means that the frequencies of the photons are now 3R (or (^{¹}/_{3})R for cosmic atoms). As Larson points out, though, it is possible to have these higher frequency photons even when the rotational displacements are less than 4 or 8, in which case we can say that the atom is “excited”.
After the change to vibration two, two units of vibrational displacement exist to be rotated, and so each added unit of rotational displacement corresponds to only one–half unit of added specific speed. Thus the speeds corresponding to magnetic displacements can be listed as follows:
_{Magnetic} _{Displacement}

_{Magnetic} _{Speeds}


1

½

½

½ 
2

¹/_{3}

²/_{5}

¹/_{3} 
3

¹/_{4}

¹/_{3}

¹/_{4} 
4

¹/_{9}

²/_{7}

¹/_{5} 
5

²/_{5}

¹/_{4}

in one displacement 
axis only 
And the speeds corresponding to electric displacements can be listed as follows:
_{Electric Displacement}

_{Electric Speeds}_{ }

_{Electric Displacement}

_{Electric Speeds}


1

½

½

9

¹/_{9}

¹/_{6}

2

¹/_{3}

²/_{5}

10

²/_{19}

²/_{13}

3

¹/_{4}

¹/_{3}

11

¹/_{10}

¹/_{7}

4

¹/_{5}

²/_{7}

12

²/_{21}

²/_{15}

5

¹/_{6}

¹/_{4}

13

¹/_{11}

¹/_{8}

6

¹/_{7}

²/_{9}

14

²/_{23}

²/_{17}

7

¹/_{8}

¹/_{5}

15

¹/_{12}

¹/_{9}

8

²/_{17}

²/_{11}

16

²/_{25}

²/_{19}

In the solid state, the values for electric rotation can be further altered. Larson states that a combination of one atom of electric displacement x with another atom of electric displacement 8–x results in a neutral bond. This bond gives rise to an electric speed of ^{¹}/_{10} for vibration one, and ^{¹}/_{5} for vibration two. Also there can be a combination of two 8–x atoms, which Larson calls a secondary positive bond. In this case the rotational speed comes to ^{¹}/(18–2x).
One final set of complications involves the lower group elements. Here there is just one subordinate magnetic displacement unit and thus these elements have less rotational force and thus are closer together in the solid state. The force is proportional to ln t, where t is the inverse of the magnetic speed, and since ln 2 is less than 1, atoms that have magnetic speed greater than ^{¹}/3 in any dimension have no effective force in that dimension. The number of “active” dimensions is given in the following tabulation.
_{Lower Group Atomic Table in Solid Stat}
Atom.
No 
Elem.

Oscillation
Frequency 
Rotational
Displ. 
Bond

Rotational
Speed 
Rotational
Frequency 
Active
Dim. 





{

2R 
21

>

(1)

¹/_{³}–^{½}

>

¹/_{10}

^{2R}/_{³}_{p}^{R}/_{p}

>

R/_{5}p



1

H


Neutral



1




2R 
21

¹/_{³}–^{½}

^{2R}/_{³}_{p}^{R}/_{p}






{

2R 
21

>

0

¹/_{³}–^{½}

>

1

^{2R}/_{³}_{p}^{R}/_{p}

>

2R/p



2

He


Zero



1




2R 
21

¹/_{³}–^{½}

^{2R}/_{³}_{p}^{R}/_{p}






{

3R 
21

>

1

²/_{5}–²/_{5}

>

½

^{4R}/_{5}_{p4}^{R}/_{5}_{p}

>

R/p



3

Li


Positive



2




3R 
21

²/_{5}–^{²/}_{5}

^{4R}/_{5}_{p4}^{R}/_{5p}






{

3R 
21

>

2

¹/_{³}–½

>

²/_{5}

^{2R}/_{³}_{p}^{R}/_{p}

>

4R/_{5}p



4

Be


Positive



2




3R 
21

¹/_{³}–^{½}

^{2R}/_{³}_{p}^{R}/_{p}






{

3R 
21

>

3

¹/_{³}–½

>

¹/_{5}

^{2R}/_{³}_{p}^{R}/_{p}

>

2R/_{5}p



5

B


Neutral



1




3R 
21

¹/_{³}–^{½}

^{2R}/_{³}_{p}^{R}/_{p}






{

2R 
22

>

(5)

¹/_{³}–¹/_{³}

>

¹/_{10}

^{2R}/_{³}_{p}^{2R}/_{³}_{p}

>

R/_{5}p



or

B


Neutral



1




2R 
22

¹/_{³}–¹/_{³}

^{2R}/_{³}_{p}^{2R}/_{³}_{p}






{

2R 
21

>

4

¹/_{³}–^{½}

>

¹/_{10}

^{2R}/_{³}_{pR/p}

>

R/_{5}p



6

C


Neutral



2




2R 
21

¹/_{³}–^{½}

^{2R}/_{³}_{p}^{R}/_{p}






{

2R 
22

>


¹/_{³}–¹/_{³}

>

1

^{2R}/_{³}_{p2R/³p}

>

2R/p



or

C


Zero



3




2R 
22

¹/_{³}–¹/_{³}

^{2R}/_{³}_{p}^{2R}/_{³}_{p}






{

2R 
22

>

(3)

¹/_{³}–¹/_{³}

>

¹/_{10}

^{2R}/_{³}_{p2R/³p}

>

R/_{5}p



7

N


Neutral



1½




2R 
22

¹/_{³}–¹/_{³}

^{2R}/_{³}_{p}^{2R}/_{³}_{p}






{

2R 
22

>

(3)

¹/_{³}–¹/_{³}

>

1

^{2R}/_{³}_{p2R/³p}

>

2R/p



or

N


Zero



3




2R 
22

¹/_{³}–¹/_{³}

^{2R}/_{³}_{p}^{2R}/_{³}_{p}






{

2R 
22

>

(2)

¹/_{³}–¹/_{³}

>

¹/_{10}

^{2R}/_{³}_{p2R/³p}

>

R/_{5}p



8

O


Neutral



1½




2R 
22

¹/_{³}–¹/_{³}

^{2R}/_{³}_{p}^{2R}/_{³}_{p}






{

2R 
22

>

(2)

¹/_{³}–¹/_{³}

>

1

^{2R}/_{³}_{p2R/³p}

>

2R/_{5}p



or

O


Zero



3




2R 
22

¹/_{³}–¹/_{³}

^{2R}/_{³}_{p}^{2R}/_{³}_{p}






{

2R 
22

>

(1)

¹/_{³}–¹/_{³}

>

¹/_{10}

^{2R}/_{³}_{p2R/³p}

>

R/_{5}p



9

F


Neutral



2




2R 
22

¹/_{³}–¹/_{³}

^{2R}/_{³}_{p}^{2R}/_{³}_{p}


The reader can continue the table all the way to element 118. Again, one must first determine the kind of bond involved before the electric rotational speed can be determined.
Since different atoms have different rotational speeds and thus different rotational forces, a particle probe of equal energy shot at atoms of different elements would “penetrate” to different depths. Thus experimenters have concluded that “nuclear” size is proportional to atomic weight. Actually what they are measuring is atomic size, and according to the Reciprocal System this is constant (2.914 Å diameter)–but the force is proportional to the atomic weight of the atom. Also, where interatomic distances are less than 2.914 Å, the atoms are partially merged; where distances are greater than 2.914Å, the atoms are separate.
Reference
1. Dewey B. Larson, Nothing But Motion (North Pacific Publishers: Portland, Oregon, 1979)