EM energies in nucleons ?

[Widdekind]

total energy in Magnetic field

 

external field

[math]\vec{B}_{out} = \frac{\mu_0}{4 \pi r^3} \left( \frac{3 \vec{r} \left( \vec{m} \circ \vec{r} \right) }{r^2} - \vec{m} \right)[/math]

 

[math]\mathcal{E}_{out} = \frac{B^2}{2 \mu_0} = \frac{\mu_0 m^2}{2 \left( 4 \pi r^3 \right)^2} \left( 6 cos(\theta)^2 + 1\right)[/math]

 

[math]E_{out} = \int \mathcal{E}_{out} \; r^2 dr sin(\theta) d\theta d\phi[/math]

[math] = \frac{\mu_0 m^2}{2 \left( 4 \pi \right)^2} \left( 2 \pi \right) \left( 6 \right) \left( \frac{1}{3 R^3} \right)[/math]

[math] = \frac{\mu_0 m^2}{8 \pi R^3}[/math]

 

internal field

[math]\vec{B}_{in} = \frac{\mu_0 \vec{m}}{\left( \frac{4 \pi}{3} R^3\right) }[/math]

 

[math]\mathcal{E}_{in} = \frac{B^2}{2 \mu_0} = \frac{\mu_0 m^2}{2 \left( \frac{4 \pi}{3} R^3\right)^2}[/math]

 

[math]E_{in} = \mathcal{E}_{out} \; \left( \frac{4 \pi}{3} R^3 \right)[/math]

[math] = \frac{3 \mu_0 m^2}{8 \pi R^3}[/math]

[math] = 3 \times E_{out}[/math]

 

total field energy

[math]\boxed{E_{tot} = \frac{\mu_0 m^2}{2 \pi R^3}}[/math]

[math] = B_{pole} \times m[/math]

For nucleons, having nuclear magnetons

 

[math]\mu_q \equiv \frac{\frac{e}{3} \frac{\hbar}{2}}{2 \frac{m_P}{3}} = \frac{1}{2} \mu_N[/math]

 

[math]\mu_P \approx 6 \mu_q[/math]

[math]\mu_N \approx -4 \mu_q[/math]

then

 

[math]E_{tot} = \frac{\mu_0 g^2 \mu_q^2}{2 \pi R^3}[/math]

 

[math]\boxed{ \begin{array}{c} E_P \approx +700 KeV \\ E_N \approx +300 KeV \end{array} }[/math]

EDIT: however much energy magnetic moments infuse into their mutual magnetic field, that much energy is reduce, by their alignments with the overall field. Is magnetic self energy, of a group of magnetic moments, always zero?

 

 

 

total energy in Electric field

 

Simplistically, within nucleons, the two quarks having like electric charge would repel out into thin shells near the surface of the nucleon, whereas the lone oppositely-charged quark would reside near the center. Cp. neutrons are observed to have a positively-charged core, and negatively-charged periphery. Then:

 

[math]E \approx + \frac{q_1^2}{4 \pi \epsilon_0 R} - 2 \frac{q_1 q_2}{4 \pi \epsilon_0 R}[/math]

 

[math]\boxed{ \begin{array}{c} E_P \approx 0 \\ E_N \approx -500 KeV \end{array} }[/math]

Thus, the total electro-magnetic (EM) energy of a neutron is about 1MeV less than that of a proton; in the absence of EM interactions, protons would be even lighter, and neutrons slightly heavier, so that neutrons would be over 2MeV more massive than protons (nearly half of which is offset, by the neutron's negative EM energy). Are these Classical equations correct ?

Edited October 18, 2012 by Widdekind