Question

Is intrinsic angular momentum "real" angular momentum? The famous Einstein-de Haas effect demonstrates it. Although it actually requires rather involved techniques and high precision, consider a simplified case. Suppose you have a cylinder $\mathbf{2}\mathbf{}\mathbf{\text{cm}}$in diameter hanging motionless from a thread connected at the very center of its circular top. A representative atom in the cylinder has atomic mass 60 and one electron free to respond to an external field. Initially, spin orientations are as likely to be up as down, but a strong magnetic field in the upward direction is suddenly applied, causing the magnetic moments of all free electrons to align with the field.

(a) Viewed from above, which way would the cylinder rotate?

(b) What would be the initial rotation rate?

Short answer

(a) Counter clockwise.

(b)$1.06\times {10}^{-5}\text{rad}/\text{s}$

Step-by-step solution

Given information

A cylinder $2\text{cm}$in diameter, atomic mass$60$ amu

Step 2:Concept of  conservation of angular momentum

The law of conservation of angular momentum requires that the total angular momentum before applying the magnetic field equal the total angular momentum after. Before the magnetic field is applied, the total angular momentum of the whole cylinder ${\mathbf{J}}_{\mathrm{t}}={\mathbf{L}}_{\mathrm{t}}+{\mathbf{S}}_{\mathrm{t}}=\mathbf{0}$where${\mathbf{L}}_{\mathbf{t}}$is the total orbital angular momentum. As discussed before, here${\mathbf{S}}_{\mathrm{t}}=\mathbf{0}$. Therefore${\mathbf{L}}_{\mathrm{t}}=\mathbf{0}$as well. After applying the magnetic field, spins (and magnetic moments) get aligned and${\mathbf{S}}_{\mathrm{t}}\ne \mathbf{0}$, but${\mathbf{J}}_{\mathbf{t}}\mathbf{=}{\mathbf{L}}_{\mathbf{t}}\mathbf{+}{\mathbf{S}}_{\mathbf{t}}$must remain$\mathbf{0}$, so${\mathbf{L}}_{\mathbf{t}}\mathbf{=}\mathbf{-}{\mathbf{S}}_{\mathbf{t}}$

Illustration of Einstein-de Haas effect

The Einstein-de Haas effect is illustrated in the following figure.

Fig. 1 illustrates the Einstein-de Haas effect due to applying a magnetic field B upwards

Therefore, cylinder rotate counter clock wise.

Equate magnitudes

(b) To get the rotation rate, we begin with Eq. land equate the magnitudes (take the absolute value)

${\mathrm{L}}_{\mathrm{t}}={\mathrm{S}}_{\mathrm{t}}$

The total spin angular momentum equals the spin angular momentum of just one atom$\mathrm{S}$ multiplied by the number of atoms in the cylinder $\left(\mathrm{N}\right)$, so${\mathrm{L}}_{\mathrm{t}}=\mathrm{NS}$

We also know that,${\mathrm{L}}_{\mathrm{t}}=\mathrm{I\omega }$ where$I$ is the moment of inertia of the cylinder and equals ${\mathrm{mr}}^2/2$in this case, and$\mathrm{\omega }$ is the rotation rate/angular speed/angular frequency. So, we now have

$\frac{{\mathrm{mr}}^2\mathrm{\omega }}{2}=\mathrm{NS}$

Solving for the rotation rate from Eq. 2 gives us

$\mathrm{\omega }=\frac{2\mathrm{NS}}{{\mathrm{mr}}^2}$Where${\mathrm{m}}_{\mathrm{s}}$ for the electron is$\pm \frac{1}{2}$ . We are only interested in the magnitude so$\mathrm{\omega }=\frac{2\mathrm{N}\mathrm{\hslash }}{2{\mathrm{mr}}^2}=\frac{\mathrm{N}\mathrm{\hslash }}{{\mathrm{mr}}^2}$

In Eq. 3,$\mathrm{m}$ is the mass of the whole cylinder which we can rewrite using$\mathrm{N}$ as $\mathrm{m}={\mathrm{Nm}}_{\mathrm{a}}$, where${\mathrm{m}}_{\mathrm{a}}$ is the mass of just one atom. Thus, we get$\mathrm{\omega }=\frac{\mathrm{N}\mathrm{\hslash }}{{\mathrm{Nm}}_{\mathrm{a}}{\mathrm{r}}^2}=\frac{\mathrm{\hslash }}{{\mathrm{m}}_{\mathrm{a}}{\mathrm{r}}^2}$

We are given both the values of ${\mathrm{m}}_{\mathrm{a}}\left(60\text{amu}\right)$and the diameter $(2\text{cm})$, so the radius$\mathrm{r}=1\text{cm}$ .

Substitute values of diameter and atomic mass

Substituting these values and the known value of $\mathrm{\hslash }$, we get

$\mathrm{\omega }=\frac{1.055\times {10}^{-34}\text{J}\cdot \text{s}}{(60\times 1.66\times {10}^{-27}\text{kg}){(0.01\text{m})}^2}=1.06\times {10}^{-5}\text{rad}/\text{s}$

Therefore, initial rotation rate is $1.06\times {10}^{-5}\text{rad}/\text{s}$.