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This question is about momentum in a system when an electron and positron annihilate each other.
Consider the following, please:
SITUATION
Consider a system of 3 particles. They are in motion (which I will explain with Earth as the geometric frame).
The electron (#1) is moving, coming from the north, moving towards the center.
The positron (#2) is moving, coming from the south, moving towards the center.
The neutron (#3) is in the equatorial plane, rotating in orbit around the system's center.
(#1 and #2 are collinear in their trajectory (sort of), while #3 is perpendicular).
As the electron and positron move towards each other, they actually have a minor spiral trajectory because of the neutron's circular rotation. Their mutual precession is opposite to the orbital orientation of the neutron. This minor feature should be immaterial to my later point. The neutron is able to orbit around the system's center because of the presence of the electron and positron (which offer a gravitational field). While the neutron has non-trivial rotational inertia (from the reference point of the first two particles), the system as a whole can have zero rotational inertia.
OUTCOME
Upon eventual impact, the electron and positron annihilate, with only photons emerging.
e− + e+ → γ + γ
The energy is converted as, E=m*c^2 => E=h*v. The mass of the two particles is converted to energy. But mass is now gone. Since the former pair of particles can no longer offer a gravitation field, then the remaining neutron can no longer orbit - there is no mass particles for it to interact with. So, mathematically the neutron's rotational inertia must go to zero. That is OK, since the system as a whole originally had zero rotational inertia, and now the photons also have zero rotational inertia. So far, so good regarding rotational inertia.
But, how does the neutron itself transition from having some angular momentum (on its associated orbital movement) to having no rotational inertia?
QUANDARY
1) Does this mean the neutron flies off on a tangential linear path? I hope not, because this would introduce a linear momentum that the original system never had.
2) Does this mean the resulting two photons have an equivalent virtual (or maybe real) linear momentum which together is opposite to the ensuing linear momentum of the neutron?
3) Does the annihilation reaction impart some of its mutual energy (via work) onto the neutron in reducing the neutron's angular momentum to zero? If so, then the gamma radiation would have reduced energy, where some of it was given to the neutron.
Force, F = delta(r X p)/delta(t)
Energy, E = F*r
Consider the following, please:
SITUATION
Consider a system of 3 particles. They are in motion (which I will explain with Earth as the geometric frame).
The electron (#1) is moving, coming from the north, moving towards the center.
The positron (#2) is moving, coming from the south, moving towards the center.
The neutron (#3) is in the equatorial plane, rotating in orbit around the system's center.
(#1 and #2 are collinear in their trajectory (sort of), while #3 is perpendicular).
As the electron and positron move towards each other, they actually have a minor spiral trajectory because of the neutron's circular rotation. Their mutual precession is opposite to the orbital orientation of the neutron. This minor feature should be immaterial to my later point. The neutron is able to orbit around the system's center because of the presence of the electron and positron (which offer a gravitational field). While the neutron has non-trivial rotational inertia (from the reference point of the first two particles), the system as a whole can have zero rotational inertia.
OUTCOME
Upon eventual impact, the electron and positron annihilate, with only photons emerging.
e− + e+ → γ + γ
The energy is converted as, E=m*c^2 => E=h*v. The mass of the two particles is converted to energy. But mass is now gone. Since the former pair of particles can no longer offer a gravitation field, then the remaining neutron can no longer orbit - there is no mass particles for it to interact with. So, mathematically the neutron's rotational inertia must go to zero. That is OK, since the system as a whole originally had zero rotational inertia, and now the photons also have zero rotational inertia. So far, so good regarding rotational inertia.
But, how does the neutron itself transition from having some angular momentum (on its associated orbital movement) to having no rotational inertia?
QUANDARY
1) Does this mean the neutron flies off on a tangential linear path? I hope not, because this would introduce a linear momentum that the original system never had.
2) Does this mean the resulting two photons have an equivalent virtual (or maybe real) linear momentum which together is opposite to the ensuing linear momentum of the neutron?
3) Does the annihilation reaction impart some of its mutual energy (via work) onto the neutron in reducing the neutron's angular momentum to zero? If so, then the gamma radiation would have reduced energy, where some of it was given to the neutron.
Force, F = delta(r X p)/delta(t)
Energy, E = F*r
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