Classical model of spin

Task number: 4384

We will assume the classical point of view of the electron as a small ball rotating around its axis with radius

\[ r_\mathrm{c} = \frac{e^\mathrm{2}}{4 \pi \varepsilon_\mathrm{0} mc^\mathrm{2}} \dot = 2.81·10^{\mathrm{-15}} \mathrm{m} \, . \]

This is considered to be the so-called classical radius of the electron. This radius can be obtained with assumption of the mass of the electron to be connected with the energy of its electrostatic field by the formula \(E=m{c^\mathrm{2}}\). We also assume that it spins with angular momentum \(\frac{\hbar}{2}\). How fast (in m s^-1) would the point on the electron’s „equator“ be moving? Does this model make sense?

Note: The real value of electron’s radius was experimentally measured as way smaller than \(r_\mathrm{c}\) but this makes the situation even worse.

Necessary constants

\(m\) = 9.11·10^-34 kg	electron’s mass
\(\varepsilon_\mathrm{0}\) = 8.85·10^-12 F m⁻¹	vacuum permittivity
\(\hbar\) = 1.055·10^-34 J s^-1	reduced Planck’s constant
\(e\) = 1.6·10^-19 C	magnitude of the elementary charge
\(c\) = 3·10⁸ m s^-1	speed of light in vacuum

Hint 1
Find out how angular momentum is connected with a angular velocity, respectively with tangential speed of the point on the sphere’s equator.
Solution of hint 1
The magnitude of angular momentum with respect to the rotation axis is given by

\[ \hspace{40px}L=J\omega \, , \]

where \(J\) is the moment of inertia with respect to the rotation axis and \(\omega\) is angular velocity.

The magnitude of the angular velocity is \(\omega = \frac{v}{r}\) for the sphere with radius \(r_\mathrm{c}\) where \(v\) is tangential speed of the point on equator.

Finally

\[ L=J \frac{v}{r_\mathrm{c}} \, . \]
Hint 2
Remind yourself or find out how to determine the moment of inertia \(J\) for a sphere.
Solution of hint 2
We will start with the general formula for moment of inertia of a rigid body
\[ J = \int_{\mathrm{V}} \rho R^\mathrm{2} \, \mathrm{d}V \, , \]
where \(R\) is the distance of the infinitesimal volume element \(\mathrm{d}V\) from the rotation axis.

The easiest way to describe a sphere is through the spherical coordinates. For conversion from Cartesian coordinates to spherical coordinates we will use following formulas
\[x=r \sin \theta \cos \varphi \, ,\] \[y=r \sin \theta \sin \varphi \, ,\] \[z=r \cos \theta \, .\]
Since the moment of inertia of the sphere is the same with respect to all axes that go through its centre, we will pick the rotation axis in the way so that the distance \(R\) from the axis will be the easiest. We will choose the axis \(z\) and therefore end up with \(R=r\sin \theta\).

Next we want to determine the volume element \(\mathrm{d}V\) which has a following form in spherical coordinates
\[ \mathrm{d}V = r^{\mathrm{2}} \sin \theta \, \mathrm{d}r \, \mathrm{d}\theta \, \mathrm{d}\varphi \, . \]
Therefore, we transferred volume integral into the triple integral. The last step before computing is to determine the limits of integration. Since we are integrating over the whole sphere the \(r\) must be from zero to its radius \(r_\mathrm{c}\), angle \(\theta\) has values from zero to \(π\) and angle \(\varphi\) goes from zero to \(2π\).

Now we substitute these values to general formula for the moment of inertia and compute it
\[ J=\int_{0}^{2π}\int_{0}^{r_\mathrm{c}}\int_{0}^{\pi} \rho\, (r \sin\theta)^2r^2 \sin\theta \, \mathrm{d}\theta \, \mathrm{d}r \, \mathrm{d}\varphi = [\varphi]_0^{2\pi} \rho \int_{0}^{r_\mathrm{c}}\int_{0}^{\pi} r^4 \sin ^3 \theta \, \mathrm{d}\theta \, \mathrm{d}r = \] \[ = 2 \pi \rho \left [\frac{r^5}{5} \right] _0^{r_c}\int_{0}^{\pi} \sin ^3 \theta \, \mathrm{d}\theta = \frac{2}{5} \pi \rho r_c^5 \int_{0}^{\pi} (1-\cos ^2 \theta) \sin \theta \, \mathrm{d}\theta \, . \]
Now we will apply the substitution \(\cos\theta = u\) therefore \(\sin\theta = - \mathrm{d}u\).

Now it is necessary to recalculate the limits of integration:

\(\theta = 0\) → \(u=1\), \(\theta = \pi\) → \(u=-1\).

After substitution and simplifying we obtain
\[ J=\frac{2}{5} \rho \pi r_c^5 \int_{-1}^{1} (1-u^2) \, \mathrm{d}u = \frac{2}{5} \rho \pi r_c^5 \left [u-\frac{u^3}{3} \right]_{-1}^1 = \frac{2}{5} \rho \frac{4}{3} \pi r_c^5 \, . \]
In the end we will use the formula for sphere’s volume \(V=\frac{4}{3} \pi r_\mathrm{c}^3\) and the formula for a mass \(m=\rho V\) and get
\[ J=\frac{2}{5}\rho V r_\mathrm{c}^2=\frac{2}{5}mr_\mathrm{c}^2 \, . \]
Solution
We will use the relation between the angular momentum \(L\) and the angular velocity \(\omega\) \[ L = J \omega \, , \] where \(J\) is the moment of inertia and we can rewrite tangential velocity \(v = r_\mathrm{c} \, \omega\) as
\[ v=r_\mathrm{c} \frac{L}{J} \, . \]
We will substitute the moment of inertia for rigid sphere \(J = \frac{2}{5}mr_\mathrm{c}^2\) (see solution of hint 2 above) to the formula for peripheral velocity and simplify
\[ v=r_\mathrm{c} \frac{\frac{\hbar}{2}}{\frac{2}{5}mr_\mathrm{c}^2}= \frac{5 \hbar}{4mr_\mathrm{c}} \, . \]
We will substitute for radius \(r_c\) to the obtained formula, simplify it and calculate the value of the velocity
\[ v= \frac{5 \hbar}{4m \frac{e^2}{4 \pi \varepsilon_\mathrm{0} m\mathrm{c}^2}} = \frac{5 \pi \varepsilon_\mathrm{0} \hbar c^2}{e^2} \dot= 5.15·10^{10} \mathrm{m s^{-1}} \, . \]
Answer
The fastest point on the surface of the electron would be moving with velocity 5.15\(\cdot\)10¹⁰ m s^-1.

This is approximately hundred times bigger than the speed of light. Since we didn’t consider relativity in our calculations we cannot make any assumption when it comes to deciding if it is possible or not. When applying relativity with the same angular momentum we would get different angular velocity (as with momentum we would obtain two velocities with classical and relativity calculations). Still it gives us a hint that this model doesn’t look realistic.
Note – the derivation of the classical electron’s radius
The classical electron’s radius can be derived with assumption that energy of its electrostatic field corresponds to its rest energy \(E = m {c}^2\).

We assume the electron to be a small rigid ball with elementary charge \(e\) equally spread on its surface. We will determine the energy \(E_\mathrm{p}\) of its electrostatic field with following formula
\[ E_\mathrm{p} = e \varphi \, , \]
where \(\varphi\) is electric potential.

Since we know where the charge is we will substitute phi with electric potential on the surface of homogenously charged sphere with radius \(r_\mathrm{c}\) that we know from the task Field of homogenously charged sphere. From there we know that it corresponds to the potential of point charge in the centre of the sphere. We get the energy of the electron’s electrostatic field
\[ E_\mathrm{p} = e \frac{e}{4 \pi \varepsilon_\mathrm{0}} \, \frac{1}{r_\mathrm{c}} = \frac{e^2}{4 \pi \varepsilon_\mathrm{0}} \, \frac{1}{r_\mathrm{c}} \, . \]
Now we will compare it with rest energy and derive the radius \(r_\mathrm{c}\)
\[ E_\mathrm{p} = E \, , \] \[ \frac{e^2}{4 \pi \varepsilon_\mathrm{0}} \, \frac{1}{r_\mathrm{c}} = m {c}^2 \, , \] \[ r_\mathrm{c} = \frac{e^2}{4 \pi \varepsilon_0 m{c}^2} \, . \]
We obtained the relation given in the task assignment.

Difficulty level: Level 3 – Advanced upper secondary level

Tasks to identify relationships between facts

×Original source: GRIFFITHS, David J. Introduction to Quantum Mechanics. 2nd ed. Upper Saddle River: Pearson Prentice Hall, 2005