Why is this to the power of $\frac{3}{2}$

ordinary differential equationsphysics

I was reading Paul Nylander's website and I read about this thing called the "magnetic pendulum fractal". Essentially, imagine a pendulum whose bob is made of a magnetic metal and placed above some magnets. See this video for the actual experiment. Paul Nylander's website gives this equation for the pendulum's position:

$\begin{align*} m\ddot{\mathbf{x}} = \sum_j \frac{\mathbf{r}_j}{(h^2 + \mathbf{r}_j \cdot \mathbf{r}_j)^{3/2}} – g\mathbf{x} – \mu\dot{\mathbf{x}} \end{align*}$

In this equation:

  • $\mathbf{x}$ is the position of the pendulum bob in the x-y coordinate plane. (The bob is assumed to be locked to a plane some fixed height above the plane the magnets lie on.)
  • $\mathbf{r}_j$ is the distance between the bob's position and that of the $j$th magnet, i.e. $\mathbf{r}_j := \mathbf{x}_j – \mathbf{x}$, where $\mathbf{x}_j$ is the center of the $j$th magnet.
  • $h$ is the height of the bob above the magnets.
  • $g$ is the kickback force of the pendulum (think Hooke's law).
  • $\mu$ is friction/air resistance.
  • $m$ is the bob's mass.
  • We assume the magnets are of equal strength.

I'm trying to understand this equation and I understand most of it. From what I understand, the $-g\mathbf{x}$ term comes from Hooke's law, and describes the force restoring the pendulum back to its state of rest. The $\mu\dot{\mathbf{x}}$ term just damps the velocity. We put it all together using Newton's second law.

However, I don't understand why the denominator in the summation is raised to the power of $\frac{3}{2}$. This source says it's to eliminate the vertical component from what is obviously a distance formula in the equation, but I don't see why that is. From what I understand, magnetic force obeys Coulomb's law, which is essentially a type of inverse square law. So why is there a 3 here?

Please try to keep in mind that I don't know too much about electromagnetics. Thanks for your help.

Best Answer

It is still the inverse square law, as the full expression for this term is $$ \frac{{\bf r}_j}{\|({\bf r_j},h)\|^3} =-\nabla_{{\bf r_j}}\frac1{\|({\bf r_j},h)\|} $$ so the power 1 in the numerator and the power 3 in denominator cancel to an overall power of $-2$. The dipole expression usually has the power 5 in the denominator

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