Question

Classical Radius of the Electron Tutorial
Creator of Tutorial: Michele_Laino

Answer 1

\(\mathbf{CLASSICAL \; RADIUS \; OF \; THE \; ELECTRON}\)

by Michele Laino

\(\mathbf{abstract}\)
In this brief tutorial, I will show how to get the commonly used formula, for classical radius of electron. Before that, I will write a simple derivation in order to get the energy density formula, for an electrostatic field. A note on the units of measure: in all my computations, I will use the \({CGS}\) system of units of measure.


\(\mathbf{1) \;The\; Energy\; Density\; of\; the\; Electrostatic\; Field}\)
The energy \(W\), of an isolated distribution of electrical charge, is given by the subsequent formula:
\[W = \frac{1}{2}\int\limits_\tau {\rho \left( {x} \right)\;\varphi \left( {x} \right){d^3}x}\]
where \(\tau\) is the region of the euclidean space wherein such distribution of charge is located, \(\rho\) and \(\phi\) are the electrical charge density, and the electrical potential respectively, inside such region.

Recalling the Maxwell's equation for an electrostatic field:
\[\nabla \cdot {\mathbf{E}} = 4\pi \rho \left( x \right)\]
we can rewrite the energy \(W\), as below:
\[W = \frac{1}{{8\pi }}\int\limits_\tau {\varphi \left( {x} \right)\;\left( {\nabla \cdot {\mathbf{E}}} \right){d^3}x}\]
and, since outside such distribution of charge, the charge density is zero, we can extend the limits of such integration to all Euclidean space:

\[\begin{equation}
W = \frac{1}{{8\pi }}\int\limits_{all\;space} {\varphi \left( {x} \right)\;\left( {\nabla \cdot {\mathbf{E}}} \right){d^3}x}
\end{equation}\]
Next, I use this identity:
\[\nabla \cdot \left( {\varphi {\mathbf{E}}} \right) = {\mathbf{E}} \cdot \left( {\nabla \varphi } \right) + \varphi \;\left( {\nabla \cdot {\mathbf{E}}} \right)\]
which can be easily proved, using the tensor calculus, the Leibniz's rule for derivative of a product of two functions, and the Einstein's notation for repeated indices:

\[\begin{gathered}
\nabla \cdot \left( {\varphi {\mathbf{E}}} \right) = {\partial _i}\left( {\varphi {E_i}} \right) = \left( {{\partial _i}\varphi } \right){E_i} + \varphi \left( {{\partial _i}{E_i}} \right) = \hfill \\
\hfill \\
= {\mathbf{E}} \cdot \left( {\nabla \varphi } \right) + \varphi \;\left( {\nabla \cdot {\mathbf{E}}} \right) \hfill \\
\end{gathered}\]
wherein \(\partial_i\) stands for the derivative \(\partial /\partial {x_i}\) , and \(x_i=x,\;y,\;z,\;\;i=1,\;2,\;3\).
So, we can write this:
\[\varphi \;\left( {\nabla \cdot {\mathbf{E}}} \right) = \nabla \cdot \left( {\varphi {\mathbf{E}}} \right) - {\mathbf{E}} \cdot \left( {\nabla \varphi } \right) = \nabla \cdot \left( {\varphi {\mathbf{E}}} \right) + {E^2}\]
where, for last step, I have used the Maxwell's equation:
\[{\mathbf{E}} = - \nabla \varphi\]
After a substitution into equation \((1)\), we get:
\[\begin{gathered}
W = \frac{1}{{8\pi }}\int\limits_{all\;space} {\left\{ {\nabla \cdot \left( {\varphi {\mathbf{E}}} \right) + {E^2}} \right\}{d^3}x} = \hfill \\
\hfill \\
= \frac{1}{{8\pi }}\int\limits_{all\;space} {\nabla \cdot \left( {\varphi {\mathbf{E}}} \right){d^3}x} + \int\limits_{all\;space} {\frac{{{E^2}}}{{8\pi }}{d^3}x} \hfill \\
\end{gathered}\]
Now, I apply, to the first term at the right side, the Gauss Theorem of divergence:

\[W = \frac{1}{{8\pi }}{\left. {\Phi \left( {\varphi {\mathbf{E}}} \right)} \right|_S} + \int\limits_{all\;space} {\frac{{{E^2}}}{{8\pi }}{d^3}x}\]
where $S$ is the integration surface, located at infinite distance from the distribution of charge. Since at infinite distance the electrostatic field is zero, then that first term, at the right side, is also zero. Therefore, we get:

\[\begin{equation}
\boxed{W = \int\limits_{all\;space} {\frac{{{E^2}}}{{8\pi }}{d^3}x} }
\end{equation}\]
As we can see, we can speak about an energy density (energy over volume) of the electrostatic field, whose value is:
\[{\frac{{{E^2}}}{{8\pi }}}\]

Answer 2

\(2)\; \mathbf{Classical\; Radius\; of\; the\; Electron}\)

Let's consider an electron, whose radius is \(r_0\), inside the empty space. If we apply the Theorem of Gauss for electrostatic field, generated by a sphere of radius \(r_0\), we get:
\[E = \left\{ {\begin{array}{*{20}{c}}
{\frac{e}{{r_0^3}}r,\quad 0 \leqslant r \leqslant {r_0}\;\left( {region\;1} \right)} \\
\begin{gathered}
\hfill \\
\frac{e}{{{r^2}}},\quad r \geqslant {r_0}\;\;\left( {region\;2} \right) \hfill \\
\end{gathered}
\end{array}} \right.\]

Answer 3

where, as usually, \(e\) is the charge of the electron: \(e=4.8 \cdot 10^{-10}\) esu.
Now I substitute such electric field into equation \((2)\), and I use polar coordinates in order to evaluate the corresponding integral, so I get:
\[\begin{gathered}
W = \int\limits_{all\;space} {\frac{{{E^2}}}{{8\pi }}{d^3}x} \hfill \\
\hfill \\
= \int_0^{{r_0}} {\frac{{{E^2}}}{{8\pi }}{d^3}x} + \int_{{r_0}}^{ + \infty } {\frac{{{E^2}}}{{8\pi }}{d^3}x} = \hfill \\
\hfill \\
= \int_0^{{r_0}} {\frac{1}{{8\pi }}\frac{{{e^2}}}{{r_0^6}}{r^2}4\pi {r^2}} + \int_{{r_0}}^{ + \infty } {\frac{1}{{8\pi }}\frac{{{e^2}}}{{{r^4}}}} 4\pi {r^2} = \frac{3}{5}\frac{{{e^2}}}{{{r_0}}} \sim \frac{{{e^2}}}{{{r_0}}} \hfill \\
\end{gathered}\]
According to the Einstein's formula for the rest energy of a particle of mass \(m\), we can write:
\[{m_e}{c^2} = \frac{{{e^2}}}{{{r_0}}}\]
where \(m_e=0.911 \cdot 10^{-27}\) grams, is the electron mass.
Therefore:
\[{r_0} = \frac{{{e^2}}}{{{m_e}{c^2}}} = \frac{{\left( {{e^2}/\hbar c} \right)}}{{{m_e}{c^2}}} \cdot \hbar c = \frac{{1/137}}{{0.51}} \cdot 197 = 2.82\;{\text{fm}}\]
where \(1\; {\text{fm}} = 10^{-13}\) cm.

Answer 4

For a better comprehension of this tutorial, please refer to this tutorial: http://questioncove.com/study#/updates/5844866a4f71d41d88a8c800