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- TL;DR Summary
- Calculating the potential energy of spin anti-alignment

Hello everybody, I consider two electrons that have enough kinetic energy to reach their respective classical electron radius. This would be:

2.0514016772310431402e-13 J

The corresponding speed is v = 287336682 m/s.

The electric field is

E = \frac{k_{e}}{R_e^2} = 1.8133774657059088443 × 10^{20}

The magnetic field is:

B = \frac{v × E}{c^{2}}

Multiplying that by the relativistic gamma factor, which in this case is 3.5056494831959322035, we have:

B=2.0323868283603503304e12 T

The magnetic moment of an electron is:

μ_{S} = - 9.2847647043 × 10^{−24} J⋅T^{-1}

The potential energy of spin anti-orthogonality is then:*

∆U = 2 μB Or: ∆U = γ_{e} B ℏ

That yields:

3.7740466978888805979e-11 J

Obviously too much. It exceeds the total mass energy of the electron, and even that of the muon. That means the resulting mass would be negative. It can't be, so there must be something terribly wrong with my calculation.

Kind regards.

*http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magpot.html7

2.0514016772310431402e-13 J

The corresponding speed is v = 287336682 m/s.

The electric field is

E = \frac{k_{e}}{R_e^2} = 1.8133774657059088443 × 10^{20}

The magnetic field is:

B = \frac{v × E}{c^{2}}

Multiplying that by the relativistic gamma factor, which in this case is 3.5056494831959322035, we have:

B=2.0323868283603503304e12 T

The magnetic moment of an electron is:

μ_{S} = - 9.2847647043 × 10^{−24} J⋅T^{-1}

The potential energy of spin anti-orthogonality is then:*

∆U = 2 μB Or: ∆U = γ_{e} B ℏ

That yields:

3.7740466978888805979e-11 J

Obviously too much. It exceeds the total mass energy of the electron, and even that of the muon. That means the resulting mass would be negative. It can't be, so there must be something terribly wrong with my calculation.

Kind regards.

*http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magpot.html7