I have read this (or something like this) in Wikipedia, and found some modelling of inhomogeneities in magnetic fields in clinical magnetic resonance imaging (MRI) through the Cauchy distribution, which turns the well behaved Gaussian into a fat tail distribution with no convergence to a mean or a stable variance.

However, the solution of differential equations in resonant states sounds like a closed-form equation that might bear some resemblance, or be identical, to the Cauchy distribution (cdf) or density (pdf), whereas modelling randomness in a magnetic field sounds like an empirical modelling problem.

I would like to ask for a sketch of this connection - nothing too intimidating, but an idea of where these to fields of knowledge (resonance in differential equations) and Cauchy distribution (probability) meet algebraically.


1 Answer 1


For nuclear resonance, the nuclear spin is set up in discretized state by applying an external static magnetic field, $\mathbf{B}$ to separate the spin up and spin down states:

$$ E_\pm = \mp \frac{\mu B}{2} $$

Then observe how is a microwave radiation with angular frequency $\omega$ is absorbed by the system. The peak of absorption will occur when $\omega$ is near the nature frequency of the system $\omega_o = \mu B / \hbar$, the energy separation of spin states. Practical, we will scan the static magnetic field $\mathbf{B}$ to find the resonance absorption. We can model this absorption by a driven damping oscillation.

$$ \frac{d^2x}{dt^2} + 2 \gamma \frac{dx}{dt} + \omega_o^2 x = A \sin \omega t. $$ Where $\gamma$ is the damping term determining the width at the resonace absorption. The steady solution of the driven oscillator is readily given:

$$ x_s(t) = \frac{A}{Z} \sin(\omega t + \phi). $$ where $$ Z = \sqrt{(2\omega\gamma)^2 + (\omega_o^2 - \omega^2)^2};\\ \tan^{-1}\phi = \frac{2\omega\gamma}{\omega^2 - \omega_o^2}. $$

The power absorped per cycle of oscillation:

$$ P_{abs} = \int F(t) v(t) dt = \int_0^{\frac{2\pi}{\omega}} A \sin\omega t \times \frac{d}{dt} \frac{A}{Z} \sin(\omega t + \phi) dt. \\ $$ $$ = \frac{\omega A^2}{Z} \int_0^{\frac{2\pi}{\omega}} \sin\omega t \cos(\omega t + \phi) dt.\\ $$ $$ = \frac{\pi A^2}{Z} (-\sin\phi) = \frac{2\pi \omega \gamma A^2}{Z^2} $$

The absoption coefficient as function of static $\mathbf{B}$ field:

$$ \eta = \frac{P_{abs}}{A^2} = \frac{2\pi\omega\gamma}{(2\omega\gamma)^2 + (\omega_o^2 - \omega^2)^2} = \frac{2\pi\omega\gamma}{(2\omega\gamma)^2 + ((\mu B / \hbar)^2 - \omega^2)^2} $$ This renders a Cauchy distribution containing the magnetic field.

  • $\begingroup$ Thank you! This is a great answer. I have a couple of questions: 1. Shouldn't we see $\pi \gamma$ as a factor in the denominator as in the PDF of the Cauchy, with $\gamma$ being the scale parameter (not clear to me in the last equation)? And 2. Is there a physical interpretation of $A$? $\endgroup$ Commented Mar 11, 2021 at 12:19
  • $\begingroup$ Also, is $\omega$ the support $x$? I'm trying to match your last equation with the pdf in here. $\endgroup$ Commented Mar 11, 2021 at 13:03
  • $\begingroup$ 2. A denotes the amplitude of the microwave, and $\omega$ the frequency of the microwave, the driing force. $\pi$ is the damping parameter, you can replace all $\gamma$ by $\gamma\pi$. $\endgroup$
    – ytlu
    Commented Mar 13, 2021 at 1:05

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