The smaller spiral is caused by an electron The bigger spiral is caused by a positron

However, they have the same mass and magnitude of charge. So, during this pair production, why does the positron follow a spiral that has a larger radius?

enter image description here

  • $\begingroup$ Strictly speaking from this Beautiful picture ALONE you can not claim that these are Electron and Positron. All you can say is that particles have different charge signs and the fraction $\frac{Mass \times Speed}{Charge}$ looks same. Because they have almost same number of rotations - left particle has 4 turns and right one about 6. You need MORE data, this picture alone is not enough. And what is this third track coming from spirals origin? :) $\endgroup$
    – Asphir Dom
    Jan 30 '14 at 10:31
  • $\begingroup$ @AsphirDom Generally the experimentor also know some things (density and composition) about the fluid in the chamber, the strength of the magnetic field and the magnification of the image which mean that they can generally do very good PID from the image. $\endgroup$ Jan 31 '14 at 1:17
  • $\begingroup$ It seems to me that they start out with the same curvature. After that the rate of energy loss is different. $\endgroup$
    – user137289
    Mar 13 '20 at 11:35

In a magnetic field $\boldsymbol B$, a particle with charge $q$ moves in circles of radius $$r=\frac{m\,v}{|q|\,\|\boldsymbol B\|},\tag{1}$$ where $v$ is its speed. The orientation (clockwise or anticlockwise) depends on sign of $q$. Since electrons and positrons have the same masses and opposite charges, an electron and a positron in a magnetic field move in circles with opposite orientations. If they have the same speed they will move in circles of the same sizes because formula (1) gives the same radius. But if they have not the same speed, the circles will have different radii, accordingly.

As @annav said, when the electron and the positron are created, their momenta $p=mv$ have no reason to be equal, and one can observe either an electron with a larger circle than the positron or the opposite as in you picture. The case where they have exactly the same speed is highly improbable.

  • $\begingroup$ I see, so the difference in radius is just due to the difference in $p$ and thus a difference in $v$ of the electron and positron. Just out of interest, could it be that during such a pair production where a gamma ray hits a nucleus, the positron may have a higher $v$ due to the repulsion of the positively charged nuclei...and thus having a larger $r$ than the electron? $\endgroup$
    – Eliza
    Jan 30 '14 at 9:57
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    $\begingroup$ @Eliza. I'm not a specialist of nuclear physics and my answer might be uncomplete. What you say seems correct, the positron should be accelerated by repulsion of the protons in the nucleus, as long at it is under the atom's electrons cloud. After exiting the atom, as the atom is neutral, it will keep its momentum. The electron will be slowed down by the nucleus in the same way. As a conclusion, there should be a higher probability to see a positron with larger momentum, but I have no idea if this effect is significant or measurable. The picture is by no means evidence of such an effect. $\endgroup$
    – Tom-Tom
    Jan 30 '14 at 10:05
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    $\begingroup$ @V.Rossetto Your claim about the nuclear effect on the positron and electron is incorrect. The energy gain/loss approaching the nucleus is symmetric with the loss/gain leaving the nucleus. The biggest difference between electrons and positrons is their eventual fate: the positron will eventually be annihilate with an electron. $\endgroup$ Jan 31 '14 at 1:21

The electron and positron are the materialization of the energy of a gamma/photon hitting (let us assume) a proton in a hydrogen bubble chamber. The pair cannot be created without an interaction in the field of another particle, due to four momentum conservation in the center of mass system. The photon has zero mass, the electron positron pair are limited by the mass of each particle : the center of mass system of the e+ e- the four vector invariant mass is at least m_e+ + m_e-.

Energy and momentum balance will give the original energy of the photon. As a three body problem the three particles involved can share the momentum within the constraints of the conservation laws. Nature needs no calculator.

  • $\begingroup$ "share the momentum"... but why does the positron get a bigger share of the momentum? I mean because it gets a bigger share, it gets a larger kinetic energy and thus a larger radius... but why can't this be the case of an electron. Also, does it have to do with the repulsion of between the proton and positron? (taking the case of a gamma ray hitting a nucleus and pair production occuring) $\endgroup$
    – Eliza
    Jan 30 '14 at 7:14
  • $\begingroup$ If you accumulate a large statistical sample of the same interaction ( in energy of gamma) you might be right. What one event says is similar to one hit on the screen of a double slit experiment. It materializes the probability of the interaction, which has a freedom within the energy and momentum conservation limits and angular momentum of the interaction. One would have to solve the problem to see if the charge of the proton plays a role. $\endgroup$
    – anna v
    Jan 30 '14 at 7:18

This is a so-called 'trident' event: a gamma converts to an electron-positron pair in the field of an electron in the target liquid. The electron gets a big enough kick to produce a track of its own, which is why there are two tracks curving right and one (the positron) curving left.

The way the photon energy is shared between the 3 particles is random. It is not shared equally. It just happens that in this case the positron got more energy (hence more momentum, hence larger radius) than the low-energy electron.

Likewise lower down there is a separate gamma conversion, this time the more usual case of a conversion in the field of a nucleus, which does not leave a track, and the electron and positron have different energies/momenta/curvature.


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