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Taken from a bubble chamber track image,

enter image description here

(http://hst-archive.web.cern.ch/archiv/HST2004/bubble_chambers/BCwebsite/media/lesson071.gif)

I have marked several single (unpaired) creation of what I suppose are electrons and positrons:

enter image description here

Are my markings correct indicating the electrons and positrons being created in this image?

When a new unpaired particle is created seemingly out of nothing in bubble chamber images, is the cause always attributed to photon-photon collision, which do not show up in bubble chamber track images?

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Can you give a link for the picture so as to know the incoming beam? It is a very noisy beam, I have scanned thousand of the Cern two meter bubble chamber pictures and do not remember such noise.

Assuming that soft electrons and positrons are observed, the electrons may come from neutral particles in the beam interacting with the atoms of the medium of the bubble chamber.

When a new unpaired particle is created seemingly out of nothing in bubble chamber images, is the cause always attributed to photon-photon collision, which do not show up in bubble chamber track images?

No. Photon-photon interaction have very small probability . If it is photons coming with the beam the interaction is with the fields of the atoms of the medium.

Creation of positrons by real or virtual photons means a pair creation, for conservation of charge. At the center of mass the two are of equal and opposite momentum, but depending on the momentum in the lab of the creating photon the Lorentz transformation to the lab may give a very small momentum in the lab for one of them, so it would not have any length of track to be recorded.

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    $\begingroup$ thank you. the picture is from hst-archive.web.cern.ch/archiv/HST2004/bubble_chambers/… $\endgroup$
    – James
    Commented Nov 28, 2022 at 5:35
  • $\begingroup$ @James And the text gives the answer to your question? $\endgroup$
    – Farcher
    Commented Nov 28, 2022 at 8:40
  • $\begingroup$ Farcher yes, it is mentioned in the text, the question is primarily to test if I understood it right... so left-deflection means positive charge, right deflection means negative charge. More curvature of the deflection either means lower speed, or smaller mass. Less curvature means higher speed, or bigger mass. There is no simple way to determine which is the case (lower speed or smaller mass) given a steeper curvature, just from looking at the track? $\endgroup$
    – James
    Commented Nov 28, 2022 at 9:26
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    $\begingroup$ @James The link you gave contains a detailed explanation and description of the tracks in the picture. Bremsstrahlung and the Compton effect are also involved, not just pair production. $\endgroup$
    – Vladimir F Героям слава
    Commented Nov 28, 2022 at 13:13

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