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The Centripetal Accleration and the Larmor Effect for the Proton in a Deuteron |
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Joseph Larmor derived a formula for the radiation of electromagnetic waves from the acceleration of a point charge. That formula, in cgs units, is
where R is the rate of energy generation, α is the acceleration, q is the charge and c is the speed of light.
Acceleration is the rate of change of velocity and since velocity is a vector quantity acceleration arises from the change in directionof velocity as well as changes in its magnitude. A particle traveling in a circular orbit of radius r at a roatatiion rate of ω experiences a centripetal acceleration of rω².
The purpose of this material is to evaluate the Larmor effect for the proton in a deuteron.
The radius of the proton's orbit in a deuteron can be obtained from the separation distnce s of the ceners of the two nucleons. Let r_{p} and r_{n} be the radii of the proton and neutron, respectively and m_{p} and m_{n} their masses. Then
Therefore
The separation distance of the centers is obtained by subtracting the radii of the nucleons from the diameter of the deuteron. The measured radius of the deuteron is 2.1424×10^{-15} m and hence its diameter is 4.2824×10^{-15} m. The particle radius of the proton is 0.877551×10^{-15} m. The radius of the neutron is less certain but the figure of 1.0×10^{-15} m is a reasonable value. This gives the separation distance of the centers as 2.4072×10^{-15} m. This puts the value of r_{p} at approximately 1.2×10^{-15} meters.
One measured quantity that might depend directly upon the rate of rotation of a nucleus is its magnetic dipole moment. That of a deuteron is 0.8574 magnetons. One magneton is equal to 5.0508×10^{-27} joules per second. Thus the magnetic dipole moment of a deuteron is 4.3307346×10^{-27} joules per second. The net sum of the magnetic dipole moments of the proton and neutron is 0.86242 magnetons or 4.4437070×10^{-27} J/T.
This might mean that the moment due to the rotation of the deutron is 0.112972×10^{-27} J/T or rounded off 1.13×10^{-28} joules per second.
Since the magnetic dipole moment M is given by the formula
where q is charge. The magnetic dipole moment due to the rotation of the deuteron would depend only on the motion of the proton since the neutron is electrically neutral. This gives the rate of rotation ω as
The tangential velocity of the nucleons would then be
This is slightly less than 4 percent of the speed of light so non-relativistic analysis is appropriate.
The centripetal acceleration of the proton is then
The Larmor formula gives the rate of radiant energy generation. There is no provision for the storage of energy to go into a photon. Therefore the energy of the photon generated is the energy involved in one time-step or quantum of time. Let the quantum of time be denoted as δ and that of length as ε , where ε=cδ. Therefore the energy of the photon generated by the Larmor effect should be δR, which is given by
The largest value ε could take is the charge diameter of a proton which is 1.775×10^{-15} m. This makes the largest value for δ equal to 5.9×10^{-24} seconds.
The charge of a proton is 1.602×10^{-19} coulombs. This gives δR as
The rotational kinetic energy of the deuteron is
The ratio of δR to K is thus 2.14×10^{-19}
Thus the Larmor effect, if it exists, would be an insignificant amount of energy compared to the kinetic energy of the deuteron which itself is an insignificant amount of energy compared with that of the gamma ray created with its formation.
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