Question

Picture the electron as a uniformly charged spherical shell, with charge e and radius R, spinning at angular velocity $\omega$.

(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:$L_{em}=\frac{ħ}{2}$ On these two assumptions, determine the radius and angular velocity of the electron. What is their product, $\omega R$? Does this classical model make sense?

Short answer

(a)The total energy contained in the electromagnetic fields is$W=\frac{1}{8\pi {\epsilon }_0}\frac{e^2}{R}+\frac{{\mu }_0{e}^2{\omega }^{2}R}{36\pi }$ .

(b)The total angular momentum contained in the electromagnetic fields is $\stackrel{\rightharpoonup }{L}=\frac{{\mu }_0\omega e^{2}R}{18\pi }\hat{z}$.

(c) The radius and angular velocity of an electron is $2.95\times {10}^{-11}$ and $3.12\times {10}^{-21}rad/s$ respectively.

Step-by-step solution

Expression for the total energy stored, total electrical energy stored and the total magnetic energy stored:

Write the expression for the total energy stored.

$w=w_{\mathrm{E}}+W_{\mathrm{B}}$…… (1)

Here, is the energy due to an electrical field and is the total magnetic energy stored.

Write the expression for the energy due to an electrical field.

$W_{\mathrm{E}}=\frac{{\mathrm{\epsilon }}_0}{2}\int E^{2}d\tau$…… (2)

Write the expression for the total magnetic stored.

$W_{\mathrm{B}}=W_{{\mathrm{B}}_{\mathrm{in}}}+W_{{\mathrm{B}}_{\mathrm{out}}}$…… (3)

Determine the total energy contained in the electromagnetic fields:

(a)

Using equation 5.70, write the expression for electric and magnetic field.

$$\begin{gathered} \left\{\begin{array}{l}r<R:E=0,B=\frac{2}{3}{\mu }_0\sigma R\omega \hat{z}with\sigma =\frac{e}{4\pi R^2} \\ r>R:E=\frac{1}{4\pi {\epsilon }_0{r}^2}\frac{e}{r^2}\hat{r},\frac{{\mu }_0}{4\pi }\frac{m}{r^3}(2\cos \theta \hat{r}+\sin \theta \hat{\theta })withm=\frac{4}{3}\pi \sigma \omega R^4\\ \end{array}\right. \end{gathered}$$

Substitute$\frac{1}{4\pi {\epsilon }_0}\frac{e}{r^2}\hat{r}$ for E and $\left(r^2\sin \theta drd\theta d\varphi \right)$ for $\hat{r}$ in equation (2).

$$\begin{gathered} W_E=\frac{{\epsilon }_0}{2}{\int }_R^{\infty }\left(\frac{1}{4\pi {\epsilon }_0}\frac{e}{r^2}\right)r^2\sin \theta drd\theta d\varphi \\ {W}_E=\frac{{\epsilon }_0}{2}\left(\frac{1}{4\pi {\epsilon }_0}\right)e^{2}4\pi {\int }_R^{\infty }\frac{1}{r^2}dr \\ {W}_E=\frac{1}{8\pi {\epsilon }_0}\frac{e^2}{R} \end{gathered}$$

Write the formula for the energy density due to an internal magnetic field.

$u_{B_{in}}=\frac{1}{2{\mu }_0}{B}^2$

Substitute $\frac{2}{3}{\mu }_0\sigma R\omega \hat{z}$ for B and $\frac{e}{4\pi R^2}$ for $\sigma$ in the above expression.

$$\begin{gathered} {\mu }_{B_{in}}\frac{1}{2{\mu }_0}{\left[\frac{2}{3}{\mu }_0\sigma R\omega \right]}^2 \\ {\mu }_{B_{in}}\frac{1}{2{\mu }_0}{\left[\frac{2}{3}{\mu }_{0}R\omega \left(\frac{e}{4\pi R^2}\right)\right]}^2 \\ {\mu }_{B_{in}}\frac{1}{2{\mu }_0}{\left[\frac{4}{9}{\mu }_0^2{R}^2{\omega }^2\frac{e^2}{16{\pi }^2{R}^4}\right]}^2 \\ {\mu }_{B_{in}}=\frac{{\mu }_0{\omega }^2{e}^2}{72{\pi }^2{R}^2} \end{gathered}$$

Hence, the energy stored due to an internal magnetic field will be,

localid="1657612638384" $$\begin{gathered} W_{B_{in}}=u_{B_{in}}{v}_{sphere}\left(\because v_{sphere}=\frac{4}{3}{R}^3\right) \\ {W}_{B_{in}}=\frac{{\mu }_0{\omega }^2{e}^2}{72{\pi }^2{R}^2}\frac{4}{3}\pi R^3 \\ {W}_{B_{In}}=\frac{{\mu }_0{\omega }^2{e}^{2}R}{54\pi } \end{gathered}$$

Write the formula for the energy density due to an internal magnetic field.

${\mu }_{B_{out}}=\frac{1}{2{\mu }_0}{B}^2$

Substitute localid="1657601609018" $\frac{{\mu }_0}{4\pi }\frac{m}{r^3}(2\cos \theta \hat{r}+\sin \theta \hat{\theta })$ for B in the above expression.

localid="1657602324446" $$\begin{gathered} {\mu }_{B_{out}}=\frac{1}{2{\mu }_0}{\left[\frac{{\mu }_0}{4\pi }\frac{m}{r^3}(2\cos \theta \hat{r}+\sin \theta \hat{\theta })\right]}^2 \\ {\mu }_{B_{out}}=\frac{1}{2{\mu }_0}\frac{{\mu }_0}{16{\pi }^2}\frac{m^2}{r^6}(4{\cos }^2\theta +{\sin }^2\theta ) \end{gathered}$$…… (4)

Using trigonometric identity, it is known that:

${\sin }^2\theta =1-{\cos }^2\theta$

Hence, the equation (4) becomes,

$$\begin{gathered} {\mu }_{B_{out}}=\frac{e^2{\omega }^2{R}^4{M}_0}{18\left(16{\pi }^2\right)}\frac{1}{r^2}(4{\cos }^2\theta +1-{\cos }^2\theta ) \\ {\mu }_{B_{out}}=\frac{e^2{\omega }^2{R}^4{M}_0}{18\left(16{\pi }^2\right)}\frac{1}{r^2}(3{\cos }^2\theta +1) \end{gathered}$$

Hence, the energy stored due to an external magnetic field will be,

$$\begin{gathered} w_{B_{out}}=\frac{{\mu }_0{e}^2{\omega }^2{R}^4}{\left(18\right)\left(16{\pi }^2\right)}{\int }_R^a\frac{1}{r^6}{r}^{2}dr{\int }_0^{\pi }{(3{\cos }^2\theta +1)}^{\sin \theta }d\theta {\int }_0^{2\pi }d\varphi \\ {w_B}_{out}=\frac{{\mu }_0{e}^2{\omega }^{2}R}{108\pi } \end{gathered}$$

Substitute $\frac{{\mu }_0{\omega }^2{e}^{2}R}{54\pi }$ for $W_{B_{in}}$ and $\frac{{\mu }_0{e}^2{\omega }^{2}R}{108\pi }$ for$W_{B_{out}}$ in equation (3).

$$\begin{gathered} w_B=\left(\frac{{\mu }_0{\omega }^2{e}^{2}R}{54\pi }\right)+\left(\frac{{\mu }_0{e}^2{\omega }^{2}R}{108}\right) \\ {w}_B=\frac{{\mu }_0{\omega }^2{e}^{2}R}{36\pi } \end{gathered}$$

Substitute $\frac{1}{8\pi {\epsilon }_0}\frac{e^2}{R}$fo and $\frac{{\mu }_0{e}^2{\omega }^{2}R}{36\pi }$ for in equation (1).

localid="1657603883832" $$\begin{gathered} W=\left(\frac{1}{8\pi {\epsilon }_0}\frac{e^2}{R}\right)+\left(\frac{{\mu }_0{e}^2{\omega }^{2}R}{36\pi }\right) \\ W=\frac{1}{8\pi {\epsilon }_0}\frac{e^2}{R}+\frac{{\mu }_0{e}^2{\omega }^{2}R}{36\pi } \end{gathered}$$

Therefore, the total energy contained in the electromagnetic fields is $W=\frac{1}{8\pi {\epsilon }_0}\frac{e^2}{R}+\frac{{\mu }_0{e}^2{\omega }^{2}R}{36\pi }$.

Determine the total angular momentum contained in the fields:

(b)

Write the expression for the angular momentum.

$$\begin{gathered} \stackrel{\rightharpoonup }{l}={\epsilon }_0\left(\stackrel{\rightharpoonup }{r}\times \left({\stackrel{\rightharpoonup }{E}}_{out}\times \stackrel{\rightharpoonup }{B_{out}}\right)\right) \\ \stackrel{\rightharpoonup }{l}={\epsilon }_0\left(\stackrel{\rightharpoonup }{r\times }\left(\frac{e}{4\pi {\epsilon }_0{r}^2}\hat{r}\times \frac{{\mu }_0\omega e{R}^2}{12\pi r^3}(2\cos \theta \hat{r}+\sin \theta \widehat{\theta )}\right)\right) \end{gathered}$$

Solve the above expression as:

$$\begin{gathered} \stackrel{\rightharpoonup }{l}={\epsilon }_0\left(\overrightarrow{r}\times \left(\frac{e}{4\pi {\epsilon }_0{r}^2}\right)\frac{{\mu }_0\omega e{R}^2}{12\pi r^3}\sin \theta \left(\hat{\varphi }\right)\right) \\ \stackrel{\rightharpoonup }{l}=\frac{{\mu }_0\omega e^2{R}^2}{48{\pi }^2{r}^3}\sin \theta \left(-r\hat{\varphi }\right) \\ \stackrel{\rightharpoonup }{l}=\frac{{\mu }_0\omega e^2{R}^2}{48{\pi }^2{r}^3}\sin \theta \hat{\theta } \end{gathered}$$

Integrate the above equation by using spherical coordinates for the total angular momentum.

$$\begin{gathered} \stackrel{\rightharpoonup }{\stackrel{\rightharpoonup }{L}={\int }_R^{\infty }{\int }_0^{\pi }}{\int }_0^{2\pi }\left(-\frac{{\mu }_0\omega E^2{R}^2}{48{\pi }^2{r}^4}\sin \theta \hat{\theta }\right)\left(r^2\sin \theta drd\theta d\varphi \right) \\ \stackrel{\rightharpoonup }{L}=-\frac{{\mu }_0\omega E^2{R}^2}{48{\pi }^2{r}^4}\left({\int }_0^{2\pi }d\varphi \right)\left({\int }_0^{\infty }{\int }_0^{\pi }\frac{{\sin }^2}{r^2}drd\theta \hat{\theta }\right) \\ \stackrel{\rightharpoonup }{L}=-\frac{{\mu }_0\omega E^2{R}^2}{48{\pi }^2{r}^4}\left(2\pi \right){\left(-\frac{1}{r}\right)}_R^{\infty }{\int }_0^{\pi }{\sin }^2\theta d\theta \hat{\theta } \\ \stackrel{\rightharpoonup }{L}=-\frac{{\mu }_0\omega E^2{R}^2}{48{\pi }^2{r}^4}{\int }_0^{\pi }{\sin }^2\theta d\theta \hat{\theta } \end{gathered}$$

On further solving,

$$\begin{gathered} \stackrel{\rightharpoonup }{L}=-\frac{{\mu }_0\omega e^2{R}^2}{24\pi }{\int }_0^{\pi }{\sin }^2\theta d\theta \left(-\sin \theta \hat{z}\right) \\ \stackrel{\rightharpoonup }{L}=\frac{{\mu }_0\omega e^2{R}^2}{24\pi }{\int }_0^{\pi }{\sin }^3\theta d\theta \hat{z} \\ \stackrel{\rightharpoonup }{L}=\frac{{\mu }_0\omega e^2{R}^2}{24\pi }\left(\frac{4}{3}\right)\hat{z} \\ \stackrel{\rightharpoonup }{L}=\frac{{\mu }_0\omega e^2{R}^2}{18\pi } \end{gathered}$$

Therefore, the total angular momentum contained in the electromagnetic fields is .

$\stackrel{\rightharpoonup }{L}=\frac{{\mu }_0\omega e^{2}R}{18\pi }\hat{z}$

Determine the radius and angular velocity of an electron:

(c)

From Einstein’s energy-mass relation, write the equation for the energy in the fields that contributes to the electron.

$W=m_e{c}^2$

Substitute $\frac{1}{8\pi {\epsilon }_0}\frac{e^2}{R}+\frac{{\mu }_0{e}^2{\omega }^{2}R}{36\pi }$ for w in the above equation.

$$\begin{gathered} \frac{1}{8\pi {\epsilon }_0}\frac{e^2}{R}+\frac{{\mu }_0{e}^2{w}^{2}R}{36\pi }=m_e{c}^2 \\ \frac{1}{4\pi {\epsilon }_0}\frac{e^2}{R}\left[1+\frac{2}{9}{\left(\frac{\omega R}{C}\right)}^2\right]=M_e{c}^2 \end{gathered}$$…… (5)

Write the expression for the angular momentum.

$L=\frac{\sout{h}}{2}$

Substitute $\stackrel{\rightharpoonup }{L}=\frac{{\mu }_0\omega e^{2}R}{18\pi }\hat{z}$ in the above expression.

$$\begin{gathered} \frac{{\mu }_{0\omega }{e}^{2}R}{18\pi }=\frac{\sout{h}}{2} \\ \omega R=\frac{9\pi \sout{h}}{{\mu }_0{e}^2} \end{gathered}$$…… (6)

Here, $\sout{h}$ is the Planck’s constant $\left(1.05\times {10}^{-34}\right)$, ${\mu }_0$ is the permeability of free space $\left(4\pi \times {10}^{7}h/m\right)$ , and e is the charge on an electron $\left(1.6\times {10}^{-19}C\right)$.

Substitute $1.05\times {10}^{-34}$J.s for $\sout{h}$, $4\pi \times {10}^{7}H/m$ for ${\mu }_0$and $1.6\times {10}^{-19}$ for e in equation (6).

$$\begin{gathered} \omega R=\frac{9\pi (1.05\times {10}^{-34}J.s)}{\left(4\pi \times {10}^{7}H/m\right){\left(1.6\times {10}^{-19}C\right)}^2} \\ \omega R=9.23\times {10}^{10}m/s \end{gathered}$$

Substitute$9\times {10}^{9}W{m}^2/C^2$ for $\frac{1}{4\pi {\epsilon }_0},1.6\times {10}^{-19}C$, for e , $9.11\times {10}^{-31}kg$for $m_e$ and $C=3\times {10}^{8}m/s$in equation (5).

$$\begin{gathered} 9\times {10}^{9}w{m}^2/C^2\frac{{\left(1.6\times {10}^{-19}C\right)}^2}{R} \\ \left[1+\frac{2}{9}{\left(\frac{9.23\times {10}^{10}m/s}{3\times {10}^{8}m/s}\right)}^2\right] \\ R=\left(\frac{48.384\times {10}^{-25}W.m^2}{\left(9.11\times {10}^{-31}kg\right)\left(3\times {10}^{8}m/s\right)2}\right) \\ R=2.95\times {10}^{11}m \end{gathered}$$

Write the expression for the angular velocity of an electron.

$\omega =\frac{v}{R}$

Substitute$9.23\times {10}^{10}m/s$ for v and $2.95\times {10}^{-11}$ for R in the above expression.

$$\begin{gathered} \omega =\frac{9.23\times {10}^{10}m/s}{2.95\times {10}^{-11}m} \\ \omega =3.12\times {10}^{21}rad/s \end{gathered}$$

As $\omega R$ is the speed of the point in an equator, it will be 300 times the speed of light. Therefore, the classical model is unrealistic.

Therefore, the radius and angular velocity of an electron is $2.95\times {10}^{11}$ and $3.12\times {10}^{21}rad/s$ respectively.