Re: Positrons

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Sujet : Re: Positrons
De : ram (at) *nospam* zedat.fu-berlin.de (Stefan Ram)
Groupes : sci.physics.relativity
Date : 09. Jul 2025, 19:32:06
Autres entêtes
Organisation : Stefan Ram
Message-ID : <mass-20250709190843@ram.dialup.fu-berlin.de>
References : 1 2 3 4 5 6 7 8 9
"Paul.B.Andersen" <relativity@paulba.no> wrote or quoted:
The mass of the two leptons is 2m and the mass of the two
gamma photons is zero.
How can this be "to vague" and "not even wrong"?  :-D

  Well, when I said some wording upthread was "too vague", I also
  criticized myself, because I also was not specific enough.

  The mass of pair of particles depends on whether you

  A: measure the mass of each particle in isolation
     and then add up the results, or

  B: measure the mass of the combined system in a frame
     where the momentum of the combined system is 0.

  To get into this mindset just think about how the mass of an atom
  of hydrogen is smaller than the mass of the proton and electron
  taken in isolation. This would be another example where the
  mass of a combined system differs from the sum of the masses of its
  components.

So mass isn't invariant?

  For each system, mass does not depend on the frame of reference.
  In other words: A so-called boost does not change mass. In this
  sense mass is invariant.

  However, the mass of one system might differ from the mass of another
  system - or from the sum of the masses of two other systems.

So the mass of two photons isn't two times the mass of a photon?

  Yes. Some people call this effect, the "non-additivity of mass".

  A pair of photons can have a non-zero mass. Some people call
  this mass it "invariant mass" or "effective mass", but it is
  nothing else than mass, so I prefer just "mass" for it.

  I start with the general relation (let me use c=1 to simplify the
  notation):

E^2 = m^2 + p^2

  . When you have two photons "0" and "1" and the momentum p1 is -p0,
  then the momentum of the pair is 0 (=p0+p1=p0-p0). So we have:

E^2 = m^2 + 0^2

  for that pair, which means: All its energy is mass (mass energy).

  When you change the definition of the system and take one photon in
  isolation, then half of this mass energy becomes kinetic energy.

  (Strictly, one cannot separate one photon from the pair of
  photons generated by a pair production before it was measured,
  because it is entangled with the other photon - so that would
  be one reason to see them as one system. The moment one single
  photon is detected at a detector this entanglement is lost,
  and we now may speak of "a [single, separated] photon".)



Date Sujet#  Auteur
6 Jul05:27 * Re: Positrons23Bertietaylor
7 Jul19:40 +* Re: Positrons16Paul.B.Andersen
7 Jul19:53 i+- Re: Positrons1Athel Cornish-Bowden
7 Jul22:20 i+- Re: Positrons1Maciej Woźniak
7 Jul23:02 i+* Re: Positrons11Stefan Ram
8 Jul13:57 ii`* Re: Positrons10Paul.B.Andersen
8 Jul15:17 ii `* Re: Positrons9Stefan Ram
8 Jul20:38 ii  `* Re: Positrons8Paul.B.Andersen
8 Jul20:57 ii   `* Re: Positrons7Stefan Ram
8 Jul21:16 ii    +* Re: Positrons2Stefan Ram
8 Jul22:19 ii    i`- Re: Positrons1Stefan Ram
9 Jul18:43 ii    `* Re: Positrons4Paul.B.Andersen
9 Jul19:32 ii     `* Re: Positrons3Stefan Ram
9 Jul20:53 ii      `* Re: Positrons2Stefan Ram
10 Jul20:38 ii       `- Re: Positrons1Paul.B.Andersen
8 Jul20:50 i`* Re: Positrons2Aether Regained
9 Jul20:29 i `- Re: Positrons1Aether Regained
8 Jul21:00 `* Re: Positrons6Aether Regained
8 Jul23:45  +- Re: Positrons1William Hyde
9 Jul06:27  +* Re: Positrons3Thomas Heger
9 Jul20:24  i`* Re: Positrons2Aether Regained
10 Jul08:34  i `- Re: Positrons1Thomas Heger
11 Jul00:29  `- Re: Positrons1Bertietaylor

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