Using the AD equations to calculate Pion decay is a redundant exercise, because AD explains every type of decay. Nevertheless, this paper illustrates the state-of-the-art in momentum conservation for particle decay, as accepted by the scientific mainstream.
The classical observation that a Pion decays “on the fly” to a Neutrino and to a Muon at 90 degrees is irrelevant, because this phenomenon was never observed in an absolute vacuum. Nevertheless, in the second part of the present calculation, the Pion decays on the fly to a Muon, which decays to Electrons in the Pion’s direction of motion, conserving energy and momentum when the AD concept of particle propellant is applied.
In the classical arrangement, a Pion beam is stopped by a target. The Pion is supposed to be at rest, but this is a euphemism. For a “target” to stop Pions in a beam simply means that other particles acquire kinetic energy equivalent to the momentum transferred from the Pion. Rigorously speaking, the Pion mass needs to be added to the particles’ mass, because both travel together after the collision. This will be shown by a counter-argument.
1.- Pion Decay to Muon
We start by supposing that the Pion is at rest, even though a residual energy appears at the end of the calculation.
In AD, the Muon kinetic energy simply equals the mass difference between the Pion and Muon, 33.94 MeV, as shown in the qBasic program PION1.BAS at the end of this paper.
We suppose that the Muon loses 3.94 MeV by ionization along its path. Using the concepts in “The Third Particle”, A17, the Muon’s velocity is given by
BM1 = (1 – (Mm / (Mm + Kem))2)1/2 (1)
= 0.627199159 c
The Muon’s momentum is
pM1 = Bm1 * Mm (2)
= 66.269865 MeV/c
We suppose that the Muon is stopped by another particle; in this case, a water molecule of 18 mass units.
The water molecule’s velocity is
B1H20 = 66.269864/ (18 * 931.5) (3)
= 0.003952398 c
The water molecule’s kinetic energy is then
KE1H20 = 1/2 * MH20 * B1H20 (4)
= 0.130962452 MeV
Rigorously speaking, this energy is a little smaller, since the water molecule is heavier due to the added Muon mass.
Adding all the kinetic energies plus the Muon’s rest mass energy, we have
TE = 105.66 + 3.94 + 30.- + 0.130962452
= 139.730962452. MeV
This confirms what we said before: The total energy (TE) is greater than the initial Pion energy of the Pion “at rest.” This excess energy is equal to
Ex = 139.730962452 – 139.6
= 0.130962452 MeV
What does this mean?
This energy is equivalent to the energy transferred by the Muon when it is stopped by the water molecule, through momentum conservation. This energy also must be vectori ally equal to the Pion absorption when stopped by the target, plus the Muon momentum at the time of its creation, in order to conserve momentum and energy.
Of course, to obtain a perfect mathemati cal solution, the calculation should introduce a general equation for energy and momentum conservation from the beginning.
AD conserves momentum and energy, confirming the experimental results. What matters is that all the experimental results contradict the SR kinetic energy, momentum and mass variation equations.
The Neutrino is always associated with the SR equations mentioned above.
When decay can be calculated with E = h v and p = h v / c (photonic emission), the Neutrino is totally irrelevant and unnecessary.
All modern experimental evidence leads to a clear conclusion: Those experiments contradictSR and support AD. We give examples extracted from the PIBETA experimental device.
In Fig.1, it is possible to see that the Electron, as a product of Pion decay, has an energy equal to 63.25 MeV plus 3,80 MeV loss in the target plus 0.9 MeV in the plastic veto. The total energy equals 67.95 MeV. Adding to this energy the Electron rest mass we get 68.461 MeV. The theoretical energy expected is equal to 139.6/2 = 69.8 MeV. The experimental result differs by only 69.800 – 68.461 = 1.1339 MeV. This small difference supposes that the Electron didn’t lose any energy and ignores the light emission that was also registered during the experiment. These values are tabulated as 68.87 or 68.90, and 68.88 or 68.92 in Table 0.3 of PIBETA-Calorimetric Lineshapes, Predicted Using Tomography Results.
The average is 68.89, which is subtracted from the theoretical value of 69.8 – 68.98 = 0.9075 MeV, value that equals the Photon emission.
2.- Muon Decay to Electron
If the Muon is “at rest”, this repeats the situation where the Pion is “at rest.” The Electron momentum will be absorbed by the Muon “container” and the Electron will be stopped later by another water molecule. There is momentum conservation at the starting and at the end point.
3.- Pion and Muon decay in motion
We suppose here that the Pion has 100 MeV of kinetic energy.
KEPM = 100 MeV
Bp = 0.812 781 Beta’s Pion.
pP = 113.4318 Pion’s momentum in MeV/c.
KEMM = 239.66 – 105.66 = 133.90 in MeV.
Bm = 0.897 478 29 Beta’s Muon.
pM = 94.827 559 Beta’s momentum in MeV/c.
It is possible to see here that the Pion momentum on the fly is 113.4318 MeV/c and when it decays to a Muon, the Muon momentum has a different value equal to 94.8275 MeV/c. We cannot talk , apparently, about “momentum conservation” because the Pion momentum was not literally transfered to the Muon. As we saw before, the momentum conservation is always related to “momentum transfer” as a real action.
The momentum transformation without momentum transfer is not, apparently, conserved because “any action relates a particle with another.” This, of course, is only apparent, because the momentum will be conserved by each particle through Particle Propellant action, as is explained in A11.
Similar arguments hold for Muon decaying to Electron.
4.- Excess of Momentum
Now we will show the situation in the Conventional Wisdom that accepts the SR equation as applied to decay phenomena.
We illustrate the state-of-the-art regarding a particle’s momentum conservation as follows:
D. P. is an Associate Professor of Physics at the University of Virginia.
****Studying PIBETA experiments in the WWW, I found your name and this is the reason I am contacting you.
There is something that I cannot understand. All experimental evidence, PIBETA, etc., etc., shows that the Electrons or Photons from the Pion decay, directly or through Muon decay, conserve energy. Why are we talking about Neutrinos, if the energy of two Photons gives the Pion rest mass energy, “or the energy of two Electrons or an Electron and a Photon , gives the Muon rest mass energy?
.Why in all theoretical papers do the authors (including PIBETA) use expressions like pi+ —> e+ ve or m+—> e+ ve v-m, if the experiment shows that there is energy conservation?
Please, could you explain this to me?****
His answer was :
****Thank you for your inquiry. The reactions (decays) you were referring to in fact would not conserve energy (nor momentum) if we omitted the neutrino, i.e., they are definitely there. Consider, for example, the decay at rest:
pi+ —-> e+ nu_e.
Before the decay, the energy of the system equals 139.6 MeV (i.e., the mass of the pion times c^2, the speed of light squared) and the momentum is zero (the pion at rest).
Experiments show clearly that after the decay, the positron has the energy of about 69.5 MeV and momentum of 69.5 MeV/c. Obviously, we are missing about 1/2 of the initial energy and have an excess of almost 70 MeV/c of momentum. In order to conserve both energy and momentum there must be another particle that flies in the direction opposite of the positron and carries the missing energy and equal momentum to that of the positron. This particle must furthermore have negligible mass since the mass of the positron is about 0.5 MeV/c^2.
Similar arguments hold in the other reactions you brought up. I hope I have answered your question.****
It is interesting to point out here how he “repeats the lesson” mechanically. His answer is the completely stereotyped answer coined by all Physicists in the Conventional Wisdom. They believe in that as a religion and repeat the mantra automatically.
My answer was the following:
****To be stopped, that is, at rest, the Pion in the beam needs to be “trapped” in the Target by an atom, a group of atoms or molecules, which “absorb” the Pion momentum and serve as its “container.”
The same atom, particle or molecule, the “container,” that stopped the Pion, will “absorb” (reaction equals action) the momentum of any Particle or Photon when the Pion decays. Consequently, the argument that a Neutrino needs to be emitted in the opposite direction of the Positron is irrelevant and disconnected from the experimental setup****
In the original email there was a long detailed explanation with references directly connected with this topic, but they are non relevant here.
His answer was the Silence, the most powerful weapon that the Conventional Wisdom uses when it is trapped by a flagrant proof that they are totally wrong. He, probably, is now teaching the truth to his students, because miracles sometimes occur! More likely, he is deceiving his students and they will repeat the false lesson as their Professor is doing. Of course, what he is doing is not too important. More important is that hundreds of “Professors” and “Associate Professors” are spreading false concepts to sustain SR through a false invention like the Neutrino.
Is there no Authority in the Physics Department at the University of Virginia to correct this flagrant suppression of the truth? Is there no serious, intelligent and honest Physicist in the PIBETA group to correct this absurd conclusion obtained by the “Scientific Community?”
Of course, correcting this absurd “excess of momentum” leads us to show that the Neutrino doesn’t exist. The “Scientific Community” prefers to disown those who accept the truth!
This is the state-of-the-art of momen-tum conservation using the Ghostly Neutrino.
10 CLS : PRINT “A31. PION1.BAS, Disk M.”: PRINT
20 PRINT “PION DECAY TO MUON”: PRINT
30 U = 931.5 ‘ Mass unit in MeV.
40 MP = 139.56 ‘ Pion rest mass in MeV.
50 MM = 105.66 ‘ Muon rest mass in MeV.
60 ME = .511 ‘ Electron rest mass in MeV.
70 MH2O = 18 * U ‘ Mass of water molecule in MeV.
80 ECM = 30 ‘ Residual Muon Kinetic Energy in MeV.
90 DPM = MM + ECM ‘ Muon Total Energy in MeV.
100 BM1# = SQR(1 – (MM / DPM) ^ 2) ‘ Beta Muon.
110 PRINT “BM1=”; BM1#; “Beta Muon.”
120 PM1# = MM * BM1# ‘ Muon momentum in MeV/c.
130 PRINT “pM1=”; PM1#; ” Muon momentum in Mev/c.”
140 ‘ H2O (water molecule) momentum equal to Muon momentum.
150 B1H2O# = PM1# / MH2O ‘ H2O’s velocity.
160 PRINT “B1H2O=”; B1H2O#; ” Velocity of water molecule.”
170 KE1H2O# = ((MH2O * (B1H2O# ^ 2)) / 2) ‘ Kinetic Energy of water molecule in MeV.
180 PRINT “KE1H2O=”; KE1H2O#; ” Kinetic Energy of water molecule in MeV.”
190 KEM# = DPM * (1 – SQR(1 – BM1# ^ 2)) ‘Muon KE test.
200 PRINT “KEM=”; KEM#; ” Muon Kinetic Energy test.”
210 TEST# = ECM + KE1H2O# ‘ Muon plus H2O KE.
220 PRINT “Test=”; TEST#; ” Muon plus H2O KE”
240 PRINT “MUON DECAY TO ELECTRON”
250 ECE = 105.149 ‘ Electron Kinetic Energy.(105.66 – 0.511)
260 BETAE# = SQR(1 – (1 – (ECE / MM)) ^ 2) ‘ Beta electron.
270 PRINT “Betae=”; BETAE#; ” Beta Electron.”
280 PE# = ME * BETAE# ‘ Electron momentum.
290 PRINT “pe=”; PE#; ” Electron momentum in MeV/c.”
300 VW# = PE# / MH2O ‘ Water molecule velocity.
310 PRINT “Vw=”; VW#; ” Velocity of water molecule.”
320 KEW# = (MH2O * (VW# ^ 2)) / 2 ‘ KE of water molecule.
330 PRINT “KEw=”; KEW#; ” KE of water molecule in MeV.”
350 PRINT ” PION AND MUON DECAY IN MOTION”
360 PRINT “PION DECAY TO MUON”
370 REM We suppose that the residual Pion KE is KEPM=100 MeV.
380 KEPM = 100 ‘ Pion KE in MeV when the Muon decay.
390 PRINT “KEPM=”; KEPM; ” Pion KE in MeV when Muon decay.”
400 TKEP = MP + KEPM
410 BP# = SQR(1 – (1 – (KEPM / TKEP)) ^ 2) ‘ Beta Pion.
420 PRINT “Bp=”; BP#; ” Beta Pion.”
430 PP# = MP * BP# ‘ Pion momentum in MeV/c.
440 PRINT “pP=”; PP#; ” Pion momentum in MeV/c.”
450 TKEMM = KEPM + MP ‘ Total energy of Pion rest mass plus its KE in MeV.
460 PRINT “TKEMM=”; TKEMM; ” Total Energy: Pion rest mass plus its KE in MeV.”
470 KEMM = TKEMM – MM ‘ Muon KE in MeV.
480 PRINT “KEMM=”; KEMM; ” Muon KE in MeV.”
490 BETAMM# = SQR(1 – (1 – (KEMM / TKEMM)) ^ 2) ‘ Beta Muon.
500 PRINT “BetaMM=”; BETAMM#; ” Beta Muon.”
510 PMM# = MM * BETAMM# ‘ Muon momentum.
520 PRINT “pM=”; PMM#; ” Muon momentum in MeV/c.”
530 KEPMM# = TKEMM * (1 – SQR(1 – (BETAMM# ^ 2))) ‘ Muon KE in MeV. (TEST).
540 PRINT “KEPMM=”; KEPMM#; ” Muon KE in MeV. (TEST).”
560 PRINT “MUON DECAY TO ELECTRON”: PRINT
570 ‘ We suppose that the Muon loses 114 MeV in the ionization path.
580 ML = 114 ‘ Energy lost by Muon.
590 PRINT “ML=”; ML; ” Energy lost by Muon.”
600 MRE = KEMM – ML ‘ Muon residual energy in MeV.
610 PRINT “MRE=”; MRE; ” Muon residual energy.”
620 S = MM + MRE ‘ Muon rest mass plus KE in MeV.
630 PRINT “S=”; S; ” Muon total energy.”
640 E = S – .511 ‘ Electron KE in MeV.
650 PRINT “E=”; E; ” Electron KE in MeV.”
660 BE# = SQR(1 – (1 – (E / S)) ^ 2) ‘ Beta electron.
670 PRINT “Be=”; BE#; ” Beta electron.”
Pion decay to Muon
BM1 = 0.627199159 Beta Muon.
pM1 = 66.269865 Muon momentum in MeV.
B1H20 = 3.952398E-3 Velocity of water molecule.
KE1H20 = .130962458 Kinetic Energy of water molecule in MeV.
KEM = 30 Muon Kinetic Energy test.
Test = 30.130964 Muon plus H20 KE.
Muon Decay to Electron
Betae = .9999883 Beta Electron.
pe = 0.510994 Electron momentum in MeV/c.
Vw = 3.047617E-5 Velocity of water molecule in MeV.
Pion and Muon Decay in Motion.
Pion Decay to Muon
KEPM = 100 Pion KE in MeV when Muon decay.
Bp = .812781940 Beta Pion.
pP = 113.431855 Pion momentum in MeV/c
TKEMM = 239.56 Total Energy: Pion rest mass plus its KE in MeV.
KEMM = 133.9 Muon KE in MeV.
BetaMM = .89747829 Beta Muon.
pM = 94.82755978 Muon Momentum in MeV/c.
KEPMM = 133.899993 Muon KE in MeV (TEST).
Muon decay to Electron.
ML 114 Energy lose by Muon.
MRE = 19.89999 Muon residual energy.
S = 125.56 Electron KE in MeV.
Be = .999991718 Beta electron.