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I saw this equation in a paper while ago. I remember that the paper was on Larmor precession; i.e., $$\boldsymbol{\mathrm{\omega}}_{\text{}} = \frac{q\boldsymbol{\mathrm{B}} }{2m}$$ How is the formula equating the torque $\boldsymbol{\mathrm{\tau}} $ with the cross product of Larmor frequency and angular momentum derived?

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The relation \begin{equation} \boldsymbol{\tau} = \boldsymbol{\omega} \times \boldsymbol{L} \end{equation} is a result of vector calculus independent of the phenomenon of Larmor precession.

It follows from the formula of the time derivative of a vector with constant modulus.

Assume that we have a vector $\boldsymbol{a}=a\,\boldsymbol{\hat{a}}(t)$, where $a$ is its constant modulus and $\boldsymbol{\hat{a}}(t)$ is the unit vector, whose direction depends on time. Let's take the time derivative \begin{equation} \frac{\mathrm{d}\boldsymbol{a}}{\mathrm{d}t} = \underbrace{\frac{\mathrm{d}a}{\mathrm{d}t}}_0\boldsymbol{\hat{a}} + a\frac{\mathrm{d}\boldsymbol{\hat{a}}}{\mathrm{d}t} \end{equation} The first term is zero because we assumed that the modulus of the vector is constant. The second term is slightly more difficult to treat. Consider the image below (interpret $\vec{a}$ as unit vectors).

enter image description here

As the derivative can be considered as \begin{equation} \frac{\mathrm{d}\boldsymbol{\hat{a}}}{\mathrm{d}t} = \lim_{\Delta t \to0}\frac{\Delta \boldsymbol{\hat{a}}}{\Delta t} \end{equation} we can see that if $\Delta t$ becomes very small, also the angle $\Delta\phi\to0$, and hence $\Delta \boldsymbol{\hat{a}}$ tends to be orthogonal to $\boldsymbol{\hat{a}}$. As the derivative must have the same direction as $\Delta \boldsymbol{\hat{a}}$, we conclude \begin{equation} \frac{\mathrm{d}\boldsymbol{\hat{a}}}{\mathrm{d}t} \parallel \Delta \boldsymbol{\hat{a}} \,\bot\, \boldsymbol{\hat{a}} \end{equation} Let's compute the modulus now. \begin{equation} \left| \frac{\mathrm{d}\boldsymbol{\hat{a}}}{\mathrm{d}t}\right| = 2\lim_{\Delta t\to0}\left|\frac{\Delta \boldsymbol{\hat{a}}}{\Delta t} \right|= \lim_{\Delta t\to0}\frac{\left|\Delta \boldsymbol{\hat{a}}\right|}{\Delta t} \end{equation} From the figure, we can easily see that $|\Delta \boldsymbol{\hat{a}}| = 2 |\boldsymbol{\hat{a}}| \sin(\Delta\phi / 2) =2 \sin(\Delta\phi / 2) $, because the unit vectors have modulus $1$. Hence the previous equation becomes \begin{equation} \begin{split} \left| \frac{\mathrm{d}\boldsymbol{\hat{a}}}{\mathrm{d}t}\right| &=\lim_{\Delta t\to0, \Delta \phi \to 0}\frac{2 \sin(\Delta\phi / 2)}{\Delta t} \\ &= \lim_{\Delta t\to0, \Delta \phi \to 0}2\frac{ \sin(\Delta\phi / 2)}{\Delta\phi / 2}\frac{\Delta\phi / 2}{\Delta t}\\ &=\lim_{\Delta \phi \to 0}\frac{\sin(\Delta\phi / 2)}{\Delta\phi / 2}\lim_{\Delta t \to 0}\frac{\Delta\phi}{\Delta t}\\ & = 1 \cdot \frac{\mathrm{d}\phi}{\mathrm{d}t}\\ &=\omega \end{split} \end{equation} where in the last line we introduced the angular velocity.

Putting all together we have that the derivative $\frac{\mathrm{d}\boldsymbol{\hat{a}}}{\mathrm{d}t}$ is orthogonal to $\boldsymbol{\hat{a}}$ and has modulus $\omega$. We can cleverly use the vector product to combine these 2 properties in one equation. \begin{equation} \frac{\mathrm{d}\boldsymbol{\hat{a}}}{\mathrm{d}t} = \boldsymbol{\omega}\times \boldsymbol{\hat{a}} \end{equation}

Finally, \begin{equation} \frac{\mathrm{d}\boldsymbol{a}}{\mathrm{d}t} = \underbrace{\frac{\mathrm{d}a}{\mathrm{d}t}}_0\boldsymbol{\hat{a}} + a\frac{\mathrm{d}\boldsymbol{\hat{a}}}{\mathrm{d}t} = a\boldsymbol{\omega}\times \boldsymbol{\hat{a}} = \boldsymbol{\omega}\times \boldsymbol{a}\qquad\mbox{if}\qquad|\boldsymbol a | = \mbox{const.} \end{equation} which is the Poisson formula.

In your case

As the torque is the time derivative of the angular momentum and the modulus of the angular momentum is constant, from the Poisson formula you have that

\begin{equation} \boldsymbol{\tau} =\frac{\mathrm{d}\boldsymbol{L}}{\mathrm{d}t}= \boldsymbol{\omega} \times \boldsymbol{L} \end{equation}

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  • $\begingroup$ Thank you this was a great answer to my question! $\endgroup$
    – innating
    Apr 30, 2021 at 4:50
  • $\begingroup$ @innating If my post helped you feel free to accept it as the right answer. I can provide more detail if you need it. $\endgroup$ Apr 30, 2021 at 9:18
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From Wikipedia:

the torque by a magnetic field induced on a magnetic moment is: $$\vec{\tau}=\gamma \vec{L}\times\vec{B}$$

where $\vec{L}$ is the angular momentum, $\gamma=-\frac{eg}{2m}$ is the gyromagnetic ratio and $\vec{B}$ is the magnetic field.

The Larmor Frequency can be written in terms of the gyromagnetic ratio as follows: $$\omega=-\gamma B$$

therefore: $$\vec{\tau}=\gamma \vec{L}\times\vec{B}=-\gamma\vec{B}\times\vec{L}=\vec{\omega}\times\vec{L}$$

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