San José State University |
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Relativistic Angular Momentum |
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In 1922 the German physicists Otto Stern and Walther Gerlach ejected a beam of silver atoms into a sharply varying magnetic field. The beam separated into two parts. In 1926 Samuel A. Goudsmit and George E. Uhlenbeck showed that this separation could be explained by the valence electrons of the silver atoms having a spin that is oriented in either of two directions. It has been long asserted that this so-called spin is not literally particle spin. It is often referred to as intrinsic spin whatever that might mean. This was because a calculation of the tangential velocity at the equator of a spinning spherical particle might involve matter traveling faster than the speed of light in a vacuum. This was a conceptual mistake. If the analysis is done properly in relativistic terms there is no possibility that the maximum velocity on a spinning particle would exceed the speed of light. Thus here in the material that follows it is accepted that the magnet moment of any particle is due to its actual spinning and the spin rate can be computed from its measured magnetic moment.
Classically the Lagrangian of a physical system is the difference between its kinetic energy and its potential energy. For relativistic analysis the first term instead of just being kinetic energy could include the rest mass energy of the system. The momentum associated with a state variable z of the system is the partial derivative of the Lagrangian with respect to the time derivative of z. Rest mass energy is a constant so it does not matter in the determination of momentum whether it is included or not included in the Lagrangian.
Let θ be the angle of orientation of the with respect to a plane passing through the spin axis of the spherical particle and ω is its time derivative (dθ/dt). Then the Lagrangian L is of the form
where E(ω), as noted above, might include rest mass energy as well as kinetic energy. For a freely spinning spherical particle the potential energy is identically zero.
Because the tangential velocity on a rotating sphere varies from a maximum at the equator to minima of zero at the poles the computation of the energy E with relativistic adjustments would a very complicated affair. That won't be fully attempted here. Instead a general approximation will be used.
Let v_{max} and v_{m} stand for the maximum and mean tangential velocity, respectively, for the the sphere and β_{max} and β_{m} are the corresonding velocities relative to the speed of light. Then
where g is a coefficient to be determined.
Suppose the relativistic energy E of a rotating spherical surface of radius R and rest mass m_{0} is
When ω=0 the energy is m_{0}c², the rest mass energy of the particle.
The Maclaurin series for E is
The kinetic energy of a spinning sphere is
where J is the moment of inertia of the sphere. The moment of inertia of the sphere of mass m_{0} and radius R is given by
where k=2/3 for a spherical shell and k=2/5 for a spherical ball.
For the second term of the Maclaurin series for E to be equal to the kinetic energy of a spinning sphere it is necessary for g² to be equal to k. For a spinning spherical particle then
There is no potential energy for the free particle so E is the same as the Lagrangian function for the system. Therefore angular momentum L is given by
For the above energy function this evaluates to
The quantity m_{0}/(1 − k(ωR/c)²)^{½} could be denoted as the relativistic mass m, but that would not serve any purpose in the analysis.
The relativistic angular momentum L is
The relevant variable is β_{max}, the maximum velocity ωR relative to the speed of light c. Thus the above equation is
Consider solving for β_{max} as a function of L.
Now raise both sides of the equation to the 2/3 power to obtain
Let λ=k^{1/3}β_{max}^{2/3} so kβ_{max}^{2}=λ^{3} and hence the above equation can be expressed as
where σ=(k^{½}m_{0}cR/L)^{2/3}. This is a cubic equation in λ which can be solved for analytically or numerically by iteration.
A significant aspect of this equation is that the solution for λ does not go to infinity as m_{0} goes to zero. Instead the maximum relative tangential velocity β_{max} satisfies the condition
Of course this not is not exactly what is needed. What is needed that a spinning particle satisfies Special Relativity is β_{max} not exceeding 1. To get this the change in the shape of the particle at higher rates of rotation has to be taken into account. There is a contraction in length a higher speeds. This means the circumference of the sphere contracts more at the equator than those at higher latitudes. Consequently at higher rotation rates the shape of the particle becomes more like a cylinder.
For a cylinder β_{max}=β_{m}. ====================================================================================================================================================================================================
On the other hand, for large m_{0} the solution for β_{max} approaches the result of strictly classical analysis.
Thus the solution is alway compatible with the Special Theory of Relativity.
Here is a depiction of the situation.
For a small mass the relative velocity asymptotically approaches the relativistic limit of 1.0. For a larger mass the relative velocity asymptotically approaches the result that comes out of classical quantum analysis. What comes out of classical quantum analysis is that the momentum of a particle always has a positive minimum value. Momentum is in the nature of a product of mass and velocity. So if that positive minimum value for momentum is to be maintained for a different mass the velocity must change in the opposite direction.
There is no problem concerning Special Relativity in treating the spin of any small particle as being due to lts actual physical spinning.
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