Re: Positrons

Liste des GroupesRevenir à sp relativity 
Sujet : Re: Positrons
De : relativity (at) *nospam* paulba.no (Paul.B.Andersen)
Groupes : sci.physics.relativity
Date : 08. Jul 2025, 20:38:08
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <104js26$3o3rc$1@dont-email.me>
References : 1 2 3 4 5 6
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Den 08.07.2025 16:17, skrev Stefan Ram:
"Paul.B.Andersen" <relativity@paulba.no> wrote or quoted:
Den 08.07.2025 00:02, skrev Stefan Ram:
"Paul.B.Andersen" <relativity@paulba.no> wrote or quoted:
e⁺ + e⁻  →  γ + γ
Energy and momentum are preserved.
The energy equivalence of an electron and a positron is 0.511 MeV,
so the energy of each of the gamma particles is 0.511 MeV. (0.024 Å)
(Mass is converted to pure kinetic energy of photons)
Mass is /not/ changed in this process when the mass of the system
"γ + γ" is being compared with the mass of the system "e⁺ + e⁻".
Why would you consider γ + γ as a closed system?
    Well, I was talking about a "system", not a "closed system".
    Many people go with
 E^2 = (mc^2)^2 + (pc)^2
    and emphasize that energy and momentum are conserved in a process
   "-->", while mass can be converted into "pure kinetic energy".
    Now, let's call the values before a process (left of the arrow "-->")
   "0" and those after the process (right of the arror "-->") "1".
    Then, what the people say would be:
 E_1 = E_0     "Energy is conserved"
p_1 = p_0     "momentum is conserved"
m_1 < m_0     "mass is converted into energy"
Please explain why these three assumptions are not compatible with:
   e⁺ + e⁻  →  γ + γ

    But these three assumptions are not compatible with
 E_0^2 = (m_0c^2)^2 + (p_0c)^2 and
E_1^2 = (m_1c^2)^2 + (p_1c)^2
    !
Let the mass of the leptons be m.
Let the speed of the leptons be equal v in opposite directions.
  γ = 1/√(1 − v²/c²)
For Each of the leptons we get: (E₀/2)² = (mc²)² + (γmvc)²
   E₀² = 4(mc²)² + 4(γmvc)²
The momentum for each of the gamma particles is p = E₀/2c
  E₁/2 = pc = E₀/2
  E₁² = E₀² = 4(mc²)² + 4(γmvc)²
  E₁ = E₀ "Energy is conserved"
           -------------------
Since both the leptons and the gamma particles are moving
with the same speed in opposite direction, the net momentum
is zero both before and after annihilation.
  p₁ = E₀/2c - E₀/2c = 0    p₀ = γmv - γmv = 0
p₁ = p₀ "momentum is conserved"
          ----------------------
  m₀ = 2m, m₁ = 0
   m₁ < m₀  "mass is converted into energy"
             ------------------------------
--
Paul
https://paulba.no/

Date Sujet#  Auteur
6 Jul05:27 * Re: Positrons14Bertietaylor
7 Jul19:40 +* Re: Positrons11Paul.B.Andersen
7 Jul19:53 i+- Re: Positrons1Athel Cornish-Bowden
7 Jul22:20 i+- Re: Positrons1Maciej Woźniak
7 Jul23:02 i+* Re: Positrons7Stefan Ram
8 Jul13:57 ii`* Re: Positrons6Paul.B.Andersen
8 Jul15:17 ii `* Re: Positrons5Stefan Ram
8 Jul20:38 ii  `* Re: Positrons4Paul.B.Andersen
8 Jul20:57 ii   `* Re: Positrons3Stefan Ram
8 Jul21:16 ii    `* Re: Positrons2Stefan Ram
8 Jul22:19 ii     `- Re: Positrons1Stefan Ram
8 Jul20:50 i`- Re: Positrons1Aether Regained
8 Jul21:00 `* Re: Positrons2Aether Regained
8 Jul23:45  `- Re: Positrons1William Hyde

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