• # Langevins Equations for Magnetization in Classical Fluid

Consider a fluid at a temperature T, with dipole moments. As the molecules collide with each other, the magnetic field exerts a mean restoring force. That is, there is a resultant magnetic moment M.

In the absence of a field, the number of randomly oriented molecules satisfy the equation

$\LARGE \frac{dN}{N} = \frac{d\Omega }{\Omega }$

where N is the total number of dipoles &  $\inline \Omega$ is the solid angle (the whole of space).

Consider a sphere, $\inline \Omega$ = 4$\inline \pi$, then the solid angle at $\inline \theta$ is $\inline 2\pi\sin(\theta)d\theta$

$\LARGE \frac {dN}{N} = \frac {d\Omega }{\Omega } = \frac{2\pi\sin(\theta)d\theta}{4\pi} = \frac{\sin(\theta)d\theta}{2}$

This can also be shown by the fractional surface area of a surrounding sphere. That is, the area obtained on the surface of the sphere when we are situated at an angle $\inline \theta$ and we moved by an angle d$\inline \theta$, over the total surface area  of the sphere.

$\LARGE \frac {dN}{N} = \frac {dA }{A } = \frac{2\pi r^{2}\sin(\theta)d\theta}{4\pi r^{2}} = \frac{\sin(\theta)d\theta}{2}$

where dN lies between $\inline \theta$ and $\inline \theta$+d$\inline \theta$

One assumes that each of the atom has a magnetic moment $\inline \mu$, which always has the same magnitude but can point in any direction.

The orientation of atomic magnetic dipoles with respect to external magnetic field B give rise to a magnetic potential energy given by $-\overrightarrow{\mu} \cdot \overrightarrow{B} = -\mu B \cos(\theta)$, where $\inline \theta$ is the angle between the moment and the field. From statistical mechanics, the relative probability of having any angle is exp(-energy/kT), so angles near zero are more likely than angles near $\inline \pi$.

Therefore dN is proportional to $\large e^{\frac{\overrightarrow{\mu }\cdot \overrightarrow{B}}{kT}} \sin(\theta) d\theta$

$\LARGE dN = Ce^{\frac{\overrightarrow{\mu }\cdot \overrightarrow{B}}{kT}} \sin(\theta) d\theta$

where C is the Normalization constant.

$\LARGE N = C\int_{0}^{\pi}e^{ucos(\theta)} \sin(\theta) d\theta ;\: \; u = \frac{\mu B}{kT}$

Each moment contributes an amount $\inline \mu\cos(\theta)$ to the magnetization parallel to the field. This is because the components $\inline \sum\mu\sin(\theta)$ will give zero value because of the symmetry of the problem.  The magnetization resulting from all dipoles within $\inline (\theta,\theta+d\theta)$ is given by

$\LARGE dM = \mu dN\cos(\theta) = \frac{\mu NdN\cos(\theta)}{N} \\ \\ dM = \frac{N\mu e^{ucos(\theta)} \sin(\theta) \cos(\theta) d\theta}{\int_{0}^{\pi}e^{ucos(\theta)} \sin(\theta) d\theta} \\ \\ \\ M = \frac{N\mu \int_{0}^{\pi}e^{ucos(\theta)} \sin(\theta) \cos(\theta) d\theta}{\int_{0}^{\pi}e^{ucos(\theta)} \sin(\theta) d\theta}$

let $\inline u = \cos(\theta)$ so $\inline dx = -u\sin(\theta)d\theta$

$\LARGE M = \frac{\frac{N\mu}{u}\int_{-u}^{+u}xe^xdx}{\int_{-u}^{+u}e^xdx}$

Performing the integration and re-arranging gives

$\LARGE M = \frac{N\mu}{u}\left ( \frac{u(e^u + e^{-u})}{e^u - e^{-u}} -1\right )$
$\LARGE M = N\mu\left ( \cosh (u) - \frac{1}{u}\right )$
$\LARGE M = N\mu\left ( \cosh (\frac{\mu B}{kT}) - \frac{kT}{\mu B}\right )$

At normal temperature, cosh(u) can be expanded as

$\LARGE \cosh(u) = \frac{1}{u} + \frac{u}{3} - \frac{u^2}{45} + ... \approx \frac{1}{u} + \frac{u}{3}$

So, at normal temperature,

$\LARGE M \approx \frac{N\mu u}{3} \approx \frac{N\mu^2B}{3kT}$

So, magnetic effects will be noticeable small at lower temperature for the case of classical fluid.

A typical graph of M against u would be as follows: