The Third Particle

The "Third Partcle"

The RaE (83BI210) decay has a double historic meaning. It pointed out the first failure of SR equations to explain a decay phenomenon and led to the invention of a second(1) powerful fantasy of the twenty century: The Neutrino.(See A18, E3)

RaE decay is a fundamental issue, contrary to what many professionals think. Event though Pauli⁽²⁾ paid no too much attention to momentum conservation, but only focused on energy conservation, today momentum conservation is a critical issue related to the Neutrino, to the extent that we speak about the Neutrino Spectrum.

The Neutrino was postulated by Pauli to save SR's failure to explain energy conservation in RaE decay, but was later used to carry away momentum, and this application created another ad hoc hypothesis: A (parent) particle decaying to two daughters cannot conserve energy momentum and simultaneously produce a spectrum without resorting to the Neutrino, as happens with Neutron decay in the RaE case.

The problem of momentum conservation is really irrelevant, because the Electron momentum carried away could be absorbed by the Nucleus⁽³⁾ (or Atom). Carezani, in his paper about a detailed RaE experiment follows this line of action, that he took from Pauli. AD postulates that the "third particle" is the original (parent) decaying particle: The Neutron inside the Nucleus, in the RaE case.

It is absolutely false that from the theoretical point of view a "third particle," the Neutrino, is needed to conserve energy and momentum, and simultaneously, given the Electron Spectrum. The only requirement is to have the correct equations and the parent particle in MOTION. The first condition is fulfilled by AD. The last condition is provided by Nature, because it is known, or postulated, that all the Neutrons and Protons that form the Nuclei are in motion. Taking into account the Strong Force, this motion is in a straight line⁽⁴⁾.

In the SR realm, the “third particle” is postulated due to the SR failure to explain energy and momentum conservation, no because a theoretical requirement is needed.

Of course, Autodynamics can explain the energy and momentum conservation generating simultaneously the electron spectrum, because it is the correct theory with the right equations, that can be applied to decay phenomena.

We said many times⁽⁵⁾ in the SAA list and in private answers to many questions about decay, especially on the RaE decay, that for AD the "third particle," currently taken as a Neutrino, is the same particle before decay, but in motion. That is to say, if a particle in motion (parent) decays to two particles (daughters), to conserve energy and momentum a "third particle" (Neutrino) is not needed, if we have the right equations for kinetic energy and momentum, as happens with AD. We also said that the Electron and Proton escape in opposite directions to conserve momentum, and that the Electron can take all the possible angles with respect to the Neutron motion line, that is, between 0 and 180 degrees.

We will demonstrate this with a simple example with real values: An AD application to RaE decay through a moving Neutron decaying to Proton and Electron, given energy and momentum conservation, and simultaneously, generating the Electron Spectrum.

The following notation will be used:

Nm = Neutron rest mass.

Pm = Proton rest mass.

em = Electron rest mass.

B = Beta = velocity in fractions of c. (c = light velocity). Applied to each particle.

KE = Kinetic energy. Applied to each particle.

Be = Beta Electron, variable to give the spectrum.

BN = Beta Neutron, variable to find momentum conservation.

Energy in MeV and Momentum in MeV/c.

The Neutron is moving from left to right on the horizontal axis (x), and the Electron is as explained in the text.

Starting from the classic representation

(1)

Where A, B and C are the total energy, that is, particle rest mass energy plus its kinetic energy of particles A, B and C respectively.

In the case of RaE⁽⁶⁾ applied to the Neutron, we have:

(2)

(3)

But

(4)

And

(5)

Three examples are given. The Electron flying at 90 degrees (program spec1.bas) from the Neutron line of motion (Axis (x) or horizontal (h)), and at 0 degree or 180 degrees (program spec2.bas).

In AD, the moving mass plus its kinetic energy is constant in each decay case. This is the original "particle rest mass energy." For the Electron, it will be taken as Me.⁽⁷⁾ Remembering the AD mass variation equation

(6)

The “particle(Electron) rest mass energy” is:

(7)

The Electron kinetic energy is:

(8)

The Electron momentum is:

(9)

Being

(10)

(11)

Or its vectorial sum in the 90 degrees case.

As we show later, the momentum sum will depend if the Electron velocity (Be) is larger or smaller than the Neutron velocity (BN), when the Electron escapes at 180 degrees, that is the example we are following now.

Applying equation (7) for Proton and remembering equation (5), the “particle rest mass energy” (Pmm) is:

(12)

As Pmm and Pm are related by

(13)

Working out the equation

(14)

(15)

(16)

The Proton kinetic energy is:

(17)

The Proton momentum(pPh) in the horizontal (x) axis is:

(18)

And the Proton-Electron (pPeh) momentum sum is:

(19)

The double sign, as we said before, is related to the value of Be and BN in equation (10) or (11).

The Neutron momentum is:

(20)

The Proton and Electron total energy after decay is:

(21)

The Neutron kinetic energy before decay is:

(22)

Knowing that

(23)

Of course, the Proton plus Electron total energy is always equal to the Neutron total energy. Simultaneously the Neutron momentum must equal the Proton-Electron vectorial sum momentum, given the Electron spectrum. This is fulfilled with the values show in Table I.

The simple program spec2.bas shows the factors in correct order. The calculation is for successive approximations, because it is very easy (no more than 20 runs on the average) to find the right value of BN (Neutron velocity) to achieve equality for the momentum.

Table I.- The Be column is the "internal" Electron velocity after decay, and Be1 is the Electron vector sum velocity that escapes from the nucleus to form the Electron spectrum, with values between 0.046 and 0.93. BN is the Neutron velocity, with values between 0.0198 and 0.7050. In the 90 degrees example, the Electron angle of flight is between 81.93 degrees for Be = 0.1 and 11.34 degrees for Be = .9

CONCLUSION.
It is absolutely false that from the theoretical point of view, a third particle is needed when a particle decays to two other particles, in order to conserve energy and momentum and explain the given the spectrum.

The Neutrino has, in consequence, no theoretical support for explaining energy and momentum conservation and the particle spectrum.

AD opens the gate to help in the study of nuclear component velocities, such as Neutrons and Protons in Nuclei, starting a possible connection between disintegration(decay) rate and other physical properties.

The gate to the Future is open.

REFERENCE.

(1).- The first is the Big Bang.

(2).- The details about the particle postulated by Pauli are unknown. He didn't publish a paper about the issue, and all that is known is through letters to other physicists with only general ideas.

(3).- The Electron momentum will be absorbed by the Proton, and this momentum will be absorbed by the Nucleus or Atom. The energy that this will generate, very small, could be converted into heat, or simply to change the excitation level in the Atom.

(4).- The Feynman Lectures on Physics, Volume III, 10-3, 12-2. R. P. Feynman, R. B. Leighton and M. Sands. Addison-Wesley Publishing Co., California, London, etc., 1966.

(5).- We know this because Carezani calculated, many years ago, a few examples with values selected arbitrarily.

(6).- Of course, if the total energy is taken as the difference between Bi (Bismuth) and Po (Polonium), the energy available as kinetic energy will not be 0.78232 MeV, but 0.649272 MeV. This is only interesting from the point of view that AD opens the gate to study the Nuclei and the energies and velocities of its components: Neutrons and Protons. This difference will not change what we are demonstrating here: energy and momentum conservation and Electron Spectrum generation without a "third particle."

(7).- Of course, since the Electron rest mass energy, em, is constant, Me only represents the Electron increasing kinetic energy.

10 CLS : PRINT "Disk 118-BIN -C:\Qc2\BIN qbasic - specf1.bas"

20 Nm = 939.5656320000001# 'Neutron rest mass.

30 Pm = 938.2723120000001# 'Proton rest mass.

40 em = .511 'Electron rest mass.

50 Be = .9 ' .9186 'Electron velocity.

60 BN = .019801 ' Neutron velocity.

70 PRINT "Be="; Be; "Electron velocity."

80 PRINT "BN="; BN; "Neutron velocity."

90 Be1 = Be + BN 'Electron vectorial sum.

100 PRINT "Be1="; Be1; "Electron vectorial SUM."

110 Me = em / (SQR(1 - (Be1 ^ 2)))'Rest mass energy available for Electron.

120 PRINT "Me="; Me; "Rest mass energy available for Electron.(em + KEe)"

130 KEe1 = Me * (1 - SQR(1 - (Be1 ^ 2)))'Electron kinetic energy.

140 PRINT "KEe1="; KEe1; "Electron kinetic energy."

150 Nmm = Nm / (SQR(1 - (BN ^ 2)))'Rest mass energy available for Neutron.

160 PRINT "Nmm="; Nmm; "Rest mass energy available for Neutron."

170 KEN = Nmm * (1 - SQR(1 - (BN ^ 2)))' Neutron kinetic energy.

180 PRINT "KEN="; KEN; " Neutron kinetic energy."

190 S = .78232 + KEN - KEe1 'Proton KE available.(Nm -(em + Pm)) = 0.78232)

200 PRINT "S="; S; "Proton KE available."

210 Pmm = Pm + S 'Proton Rest Mass energy available.

220 L = Nmm 'Left member in the equation of energy conservation.

230 pe1 = em * Be1 'Electron momentum.

240 PRINT "pe1="; pe1; "Electron momentum."

250 BP = SQR(1 - ((Pm / Pmm) ^ 2))'Beta Proton.

260 PRINT "BP="; BP; "Beta Proton."

270 KEP = Pmm * (1 - SQR(1 - (BP ^ 2)))'Proton kinetic energy.

280 pPh = Pm * BP 'Proton momentum in x axis.

290 pPeh = pPh + pe1 ' Proton-Electron vector momentum.

300 pN = Nm * BN 'Neutron momentum.

310 R = em + KEe1 + Pm + KEP 'Right member in equation of energy conservation.

320 PRINT "L="; L; "Left member in the equation.Total Energy."

330 PRINT "R="; R; "Right member in the equation.Total energy."

340 PRINT "pPeh="; pPeh; "Proton-Electron vector momentum."

350 PRINT "pN="; pN; "Neutron vector momentum."

360 X = L - R' Difference for calculation."

370 PRINT "X="; X; " (L - R)"; " Difference for calculation."

380 Pm1 = Pmm * (SQR(1 - BP ^ 2))' Test Proton mass.

390 PRINT "Pm1="; Pm1; " Test Proton mass."

10 CLS : PRINT "Disk M -C:\QC2\BIN qbasic - specf2.bas"

20 Nm = 939.5656320000001# 'Neutron rest mass.

30 Pm = 938.2723120000001# 'Proton rest mass.

40 em = .511 'Electron rest mass

45 pi = 3.141592

50 Be = .4' .9186 'Electron velocity.

60 BN = .68245 'Neutron velocity.

70 PRINT "Be="; Be; "Electron velocity."

80 PRINT "BN="; BN; "Neutron velocity."

90 Be1 = SQR((Be ^ 2) + (BN ^ 2))'Electron vectorial SUM.

95 PRINT "Be1="; Be1; "Electron vectorial SUM."

100 Me = em / (SQR(1 - (Be1 ^ 2)))'Rest mass energy available for Electron.

110 PRINT "Me="; Me; "Rest mass energy available for Electron.(em + KEe)"

120 KEe1 = Me * (1 - SQR(1 - (Be1 ^ 2)))'Electron kinetic energy.

130 PRINT "KEe1="; KEe1; "Electron kinetic energy."

140 Nmm = Nm / (SQR(1 - (BN ^ 2)))'Rest mass energy available for Neutron."

150 PRINT "Nmm="; Nmm; "Rest mass energy availavle for Neutron."

160 KEN = Nmm * (1 - SQR(1 - (BN ^ 2)))' Neutron kinetic energy.

170 PRINT "KEN="; KEN; "Neutron kinetic energy."

180 S = .78232 + KEN - KEe1 'Proton KE available.(Nm -(em+Pm)) = 0.78232)

190 PRINT "S="; S; "Proton KE available."

200 Pmm = Pm + S 'Proton Rest Mass energy available.

210 L = Nmm 'Left member in the equation of energy conservation.

220 pe1 = em * Be1'Electron momentum.

224 PRINT "pe1="; pe1; "Electron momentum."

230 tgalpha = BN / Be 'Tangent of alpha angle.

240 alpha = ATN(tgalpha) * (180 / pi) 'Alpha angle.

245 PRINT "alpha="; alpha; "Alpha angle in degree."

250 gamma = 90 - alpha 'Gamma angle.

260 pe1v = pe1 * SIN(gamma * (pi / 180))'Electron momentum on y axis.

262 PRINT "pe1v="; pe1v; "Electron momentum on y axis."

270 pe1h = pe1 * COS(gamma * (pi / 180))'Electron momentum on x axis.

280 BPv = pe1v / Pm ' Beta Proton on y axis.

290 BP = SQR(1 - ((Pm / Pmm) ^ 2)) 'Beta Proton.

300 PRINT "BP="; BP; "Beta Proton."

310 BPd = SQR((BP ^ 2) + (BPv ^ 2))'Beta Proton on the diagonal.

320 KEPd = Pmm * (1 - SQR(1 - (BPd ^ 2)))'Proton kinetic energy.

330 pPh = Pm * BP'Proton momentum on x axis.

340 pPeh = pPh + pe1h'Proton-Electron momentum sum on the x axis.

350 pN = Nm * BN'Neutron momentum.

360 R = em + KEe1 + Pm + KEPd 'Right member on the equation. Total energy..

370 PRINT "L="; L; "Left member in the equation. Total energy."

380 PRINT "R="; R; "Right member in the equation. Total energy."

390 PRINT "pPeh="; pPeh; "Proton-Electron momentum sum on the x axis."

400 PRINT "pN="; pN; "Neutron momentum."

410 X = L - R'Difference for calculation.

420 PRINT "X="; X; " (L - R)"; " Difference for calculation."

430 Pm1 = Pmm * (SQR(1 - BP ^ 2))' Test Proton mass.

440 PRINT "Pm1="; Pm1; " Test Proton mass."