Question 8.17: ^18 Picture the electron as a uniformly charged spherical sh...

^{18} Picture the electron as a uniformly charged spherical shell, with charge e and radius R, spinning at angular velocity ω.

(a) Calculate the total energy contained in the electromagnetic fields.

(b) Calculate the total angular momentum contained in the fields .

(c) According to the Einstein formula \left(E=m c^{2}\right) , the energy in the fields should contribute to the mass of the electron. Lorentz and others speculated that the entire mass of the electron might be accounted for in this way: U_{ em }=m_{e} c^{2} . Suppose, moreover, that the electron’s spin angular momentum is entirely attributable to the electromagnetic fields: L_{ em }=\hbar / 2 . On these two assumptions, determine the radius and angular velocity of the electron. What is their product, ωR? Does this classical model make sense? 

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(a) From Eq. 5.70 and Prob. 5.37,

B = \nabla \times A =\frac{2 \mu_{0} R \omega \sigma}{3}(\cos \theta \hat{ r }-\sin \theta \hat{ \theta })=\frac{2}{3} \mu_{0} \sigma R \omega \hat{ z }=\frac{2}{3} \mu_{0} \sigma R \omega                      (5.70)

\left\{\begin{array}{l}r<R: E = 0 , B =\frac{2}{3} \mu_{0} \sigma R \omega \hat{ z }, \text { with } \sigma=\frac{e}{4 \pi R^{2}} \\r>R: E =\frac{1}{4 \pi \epsilon_{0}} \frac{e}{r^{2}} \hat{ r }, B=\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}(2 \cos \theta \hat{ r }+\sin \theta \hat{ \theta }), \text { with } m=\frac{4}{3} \pi \sigma \omega R^{4}\end{array}\right.

The energy stored in the electric field is (Ex. 2.9):

W_{E}=\frac{1}{8 \pi \epsilon_{0}} \frac{e^{2}}{R} .

The energy density of the internal magnetic field is:

u_{B}=\frac{1}{2 \mu_{0}} B^{2}=\frac{1}{2 \mu_{0}}\left(\frac{2}{3} \mu_{0} R \omega \frac{e}{4 \pi R^{2}}\right)^{2}=\frac{\mu_{0} \omega^{2} e^{2}}{72 \pi^{2} R^{2}}, \text { so } W_{B_{ in }}=\frac{\mu_{0} \omega^{2} e^{2}}{72 \pi^{2} R^{2}} \frac{4}{3} \pi R^{3}=\frac{\mu_{0} e^{2} \omega^{2} R}{54 \pi} .

The energy density in the external magnetic field is:

u_{B}=\frac{1}{2 \mu_{0}} \frac{\mu_{0}^{2}}{16 \pi^{2}} \frac{m^{2}}{r^{6}}\left(4 \cos ^{2} \theta+\sin ^{2} \theta\right)=\frac{e^{2} \omega^{2} R^{4} \mu_{0}}{18\left(16 \pi^{2}\right)} \frac{1}{r^{6}}\left(3 \cos ^{2} \theta+1\right) , so

W_{B_{\text {out }}}=\frac{\mu_{0} e^{2} \omega^{2} R^{4}}{(18)(16) \pi^{2}} \int_{R}^{\infty} \frac{1}{r^{6}} r^{2} d r \int_{0}^{\pi}\left(3 \cos ^{2} \theta+1\right) \sin \theta d \theta \int_{0}^{2 \pi} d \phi=\frac{\mu_{0} e^{2} \omega^{2} R^{4}}{(18)(16) \pi^{2}}\left(\frac{1}{3 R^{3}}\right)(4)(2 \pi)=\frac{\mu_{0} e^{2} \omega^{2} R}{108 \pi} .

W_{B}=W_{B_{ in }}+W_{b_{ out }}=\frac{\mu_{0} e^{2} \omega^{2} R}{108 \pi}(2+1)=\frac{\mu_{0} e^{2} \omega^{2} R}{36 \pi} ; W=W_{E}+W_{B}=\frac{1}{8 \pi \epsilon_{0}} \frac{e^{2}}{R}+\frac{\mu_{0} e^{2} \omega^{2} R}{36 \pi} .

(b) Same as Prob. 8.10(a), with Q \rightarrow e \text { and } m \rightarrow \frac{1}{3} e \omega R^{2}: L =\frac{\mu_{0} e^{2} \omega R}{18 \pi} \hat{ z } .

\text { (c) } \frac{\mu_{0} e^{2}}{18 \pi} \omega R=\frac{\hbar}{2} \Rightarrow \omega R=\frac{9 \pi \hbar}{\mu_{0} e^{2}}=\frac{(9)(\pi)\left(1.05 \times 10^{-34}\right)}{\left(4 \pi \times 10^{-7}\right)\left(1.60 \times 10^{-19}\right)^{2}}=9.23 \times 10^{10} m / s .

\frac{1}{8 \pi \epsilon_{0}} \frac{e^{2}}{R}\left[1+\frac{2}{9}\left(\frac{\omega R}{c}\right)^{2}\right]=m c^{2} ;\left[1+\frac{2}{9}\left(\frac{\omega R}{c}\right)^{2}\right]=1+\frac{2}{9}\left(\frac{9.23 \times 10^{10}}{3 \times 10^{8}}\right)^{2}=2.10 \times 10^{4};

R=\frac{\left(2.10 \times 10^{4}\right)\left(1.6 \times 10^{-19}\right)^{2}}{8 \pi\left(8.85 \times 10^{-12}\right)\left(9.11 \times 10^{-31}\right)\left(3 \times 10^{8}\right)^{2}}=2.95 \times 10^{-11} m ; \quad \omega=\frac{9.23 \times 10^{-10}}{2.95 \times 10^{-11}}=3.13 \times 10^{21} rad / s .

Since \omega R, the speed of a point on the equator, is 300 times the speed of light, this “classical” model is clearly unrealistic.

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