Re: Positrons

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)

    BTW:
    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?
There is no way you can measure anything about this system,
and the photons are detected individually.

    The total momentum of the system "γ + γ" is 0 when we assume that we
   had chosen coordinates in which the total momentum of the system
   "e⁺ + e⁻" was 0. So all the energy of "γ + γ" still is mass energy
   (zero momentum, non-zero mass).

The only sensible frame of reference is obviously the rest frame
of the gamma detectors, (and the patient's body).
There is no reason to believe that the speeds of the leptons
are equal, so they will have different kinetic energy.
However, the kinetic energy will generally be negligible compared
to then energy equivalent of the lepton mass.
That means that the gamma photons will have a negligible difference
in the energy. And a slight difference would be irrelevant for
the purpose.

    Only when one (mentally) changes the definition of the system and
   looks at one of those "γ" in isolation has one a system with zero
   mass and non-zero momentum!
 

The whole point is to observe the gamma photons individually.
There are two detectors which make it possible to "see" the point
in the body where the annihilation occurred.
--
Paul
https://paulba.no/