The theory of stellar energy generation in the Reciprocal System is stated qualitatively in various works by Mr. Larson, such as Quasars and Pulsars. For the benefit of new readers of Reciprocity, I quote Mr. Larson in full:

Inasmuch as a charge is a modification of the basic rotation, the number of charges that an atom can acquire, the degree of ionization, as it is called, is limited by the number of rotational units of the appropriate space-time direction that exist in the atomic structure; the number of units available for modification. Negative ionization is confined to low levels, as the effective negative rotation is never more than a few units. The limit of positive ionization is the atomic number, which represents the net total nurnber of units of rotational time displacement in the atom.

Electric ionization may be produced by any one of a number of agencies, inasmuch as the requirement for this process is essentially nothing more than the availability of sufficient energy under appropriate conditions. In the universe at large the predominant process is thermal ionization. Thermal or heat energy is linear motion of material particles, and it is therefore space displacement. In the ionization process this linear space displacement is transformed into rotational space displacement: positive charge. As the temperature increases, more and more space displacement becomes available for ionization, and the degree of ionization rises until the atom finally reaches the point where it is fully ionized; that is, each of its units of time displacement has acquired a positive charge.

If the temperature of the fully ionized atom continues to rise, a destructive limit is ultimately reached at the point where the total space displacement, the sum of the ionization and the thermal energy, is equal to the time displacement of one of the magnetic rotational units. Here the oppositely directed rotational displacements neutralize each other, and both revert to the linear basis, destroying this portion of the atomic structure. Since the maximum ionization increases with the atomic number, the amount of thermal energy required to bring the total space displacement of a fully ionized atom up to the destructive limit is less for heavier atoms, and the effect is to establish a temperature limit for each element that is inversely related to atomic number. As the temperature of an aggregate rises the heaviest elements are therefore the first to disintegrate.1

To sum up, when the destructive thermat limit is reached, the following word equation holds true:

ionization energy of atom + thermal energy of atom = energy  
          equivalent of one unit of magnetic time displacement

Let EI be the ionization energy, ET be the thermal energy, and EM be the oppositely directed magnetic rotational energy. Then in symbols,

EI + ET = EM

Each of the terms in the equation will now be discussed.

Equivalent energy of one unit of magnetic time displacement

Before we can find the energy equivalent of one unit of magnetic time displacement, we must find the mass equivalent. According to deductions previously made from the postulates of the Reciprocal System the electric equivalent of a magnetic displacement n is 2n²; this does not refer to the total from zero to n—it is the equivalent of the nth term alone. Each electrical unit is equal to two atomic mass units, and each atomic mass unit is equivalent to 931.48 MeV. For n equal to 3 and 4, the following table results:


Thus, the third magnetic time displacement is equivatent to 33533.28 MeV, and the fourth unit to 59614.72 MeV.

Ionization energy

At the present stage of development of the Reciprocal System we do not have a theoretical equation giving the energy needed to completely ionize at atom—but then neither does quantum mechanics. An empirical equation will have to do for now.

Reference three has the most comprehensive table of ionization values available, giving the complete ionization energy for the first twenty elements. From the values, I have derived the following empirical equation:

EI - 13.595 + 52.148(Z-1)² 10-6 MeV

where Z is the atomic number. Of course, other equations are possible.4 Extrapolating any empirical equation to high values of Z is risky, but this will have to do. For thorium, at no. 90, eq. 2 gives

EITh = .413 MeV

Thermal energy

Let k be Boltzmann’s constant in MeV/°K and T be the temperature of an atom in °K Then the standard equation for the thermal energy (based on the ideal gas taw) is

ET = 3/2 kt (mv²/2) = 1.292 x 10-10

Calculation of critical temperature and velocity

From eq. lb,

ET = EM - EI


TCRIT = EM - EI / 1.292 x 10-10 °K

For thorium, EM is 59614.72 MeV and EI is .413 MeV, so

TCRITTh = 4.614110449 x 1014 °K

This is fantastically high from our view as spectators on the earth, but in terms of natural units, the temperature is “only” 127.44.

With k in J/K, equation 3 can be solved for the velocity at the critical temperature.

vCRIT = (3kTCRIT/m)½

For thorium, this amounts to

vCRITTh = 2.5289 x 108 m/sec

This is 84% of the speed of light:

For iron, the critical temperature is

TCRITFe = 4.61413989 x 1014 oK

and the critical velocity is

vCRITFe = 2.7165 x 108 m/sec

This is 91% of the speed of light! No wonder atoms are accelerated to velocities above the speed of light during a supernova explosion:

Most likely the motion of the atoms in the core of a star is circular. The greater the temperature, the higher the velocity—thus as theoretically expected O and B type stars have much greater rotational velocities than G and K type stars.

Rate of energy generation

Since both the unit of magnetic time displacement and the opposing space displacement revert to linear motion, the total energy radiated per critical atom is

ERAD = 2xEM =119229.44 MeV

for n = 4.

The rate of energy generation depends on the number of atoms at the critical temperature, NCRIT. This in turn depends on the total mass of the star M, the average mass per critical atom, m, and on the fraction FCRIT of the mass M that is critical. Thus


Let PST be the total power output of a star. Then assuming no accretion whatever, the litetime of a star can be calculated as follows:


For the sun,

M = 2 x 1030 kg

PST = 2.43x1039 Mev/ sec

Taking thorium as representative of the critical elements,

m = 2.988 x 10-25 kg/atom

Assuming various values of FCRIT we can calculate the lifetime of a star with no accretion. The following table results.

LST (no accretion) years
1.0389 x 1011
1.0389 x 1010
1.0389 x 1009
1.0389 x 1008
1.0389 x 1007

According to the Reciprocal System, net accretion does occur over the life of a star, but there may be periods where there is a net loss. Since such a period may last as long as a billion years, I believe we are on good ground assuming that FCRIT is equal to .0001. At present we have no way of deducing theoretically the fraction of the mass of a star that is critical. Certainly, observation is no help; observation can only indicate the composition of the stellar atmosphere, not that of the central core.

Rate of accretion

The sun appears to be one-third along its way on the Herzsprung-Russell diagram. Since the sun has been estimated to be in existence for about 5 billion years, we can roughly assume that the average lifetime of a star is 15 billion years. According to the theory, a star slowly increases in temperature until the critical temperature of the iron group of elements is reached, at which point the life of the star is terminated in a supernova1 explosion. From the equations presented in this paper, the critical temperature of iron is 3,091,400,000 oK above that of uranium. Thus the rate of change of temperature with time can be roughly expressed as follows:

dT/dt / DT/Dt = 3,091,400,000/15 x 109 = .206 °K

(Even if L were only 7.5x109 years, the increase in T per year would be less than .5 °K).

Thus stars are for most of their lives very stable energy generators. From this we can conclude that the rate of accretion is just slightly greater than the rate of mass lost through burning. For calculating the rate of accretion we can assume that for the short term they are identical.

Let RACC be the rate of accretion in kg/sec. Then


Using previous values of PST, ERAD, mCRIT and FCRIT equal to .0001,

RACC = 6.099 x 1013 kg/ sec = 1.925 x 1021 kq/ yr

This amounts to .000000096% of the mass of the star per year. It would take over 3108 years for the accretion to amount to the mass of the earth:

Clearly we cannot observe this small rate of accretion. Observation cannot tell us whether the mass of the sun is remaining constant or slowly increasing, as we believe, or whether the mass of the sun is slowly decreasing, as present theory suggests.


The current theory of stellar energy generation has been criticized elsewhere, and a summary of that criticism is presented in reference five. The basic differences between the new theory and the current one are as follows:

  1. In the new theory, energy is generated by disintegration of heavy elements; in the current theory, energy is generated by fusion of light etements.
  2. In the new theory the temperature of the stellar core is of the order of 4.6 x 1014 °K in current theory, it is 3x107 °K for the first phase, and 109 for later phases.
  3. In the new theory, ordinary stellar energy generating processes do not give rise to neutrino emission, but in current theory they do. So far, no neutrinos have been found to emanate from the sun.
  4. In the new theory, one method for energy generation serves all types of stars; current theory proposes that various stars have different energy schemes: proton-proton reaction: the CNO bi-cycle; helium burning; (y, a) reaction of C12 , O16 , Ne20 nuclei; e-process; r-process.

Thus, though observation (other than neutrino counts) cannot at present decide in favor of one theory over the other, Occam’s Razer can: the new theory wins hands down.


  1. Dewey B. Larson, Quasars and Pulsars (Portland, Oregon: North Pacific Publishers, 1971), p. 60.
  2. Dewey B. Larson, New Light on Space and Time (Portland, Oregon: North Pacific Publishers 1965) p. 234
  3. Von W. Finkelnburg and W. Humbach, “Ionisierungsenergien von Atomen and Atomionen,” Die Naturwissenschaften, Heft 2, Jg. 42, 1955, pp. 35-37.
  4. For instance a Polynomial equation in Z has been worked out by computer by Frank V. Meyer: EI = 78.6411 - 72.8213 x Z + 33.675 x Z² + .801221 x Z³ .
  5. Ronald W. Satz, The Unmysterious Universe (Troy, NY: Troy Printers, 1971),
    p. 11.
  6. R. Davis, Jr., D. S. Harmer, K. C. Hoffman, “Search for Neutrinos from the Sun,” Phys. Rev. Let., 20, 1205 (1968).

Author’s Note: This paper is not meant to be the last word on subject of stellar energy. Rather it is meant only to be the second word. Constructive criticism would be welcome.