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\exnumber{2173}
\heading{Polarization of classical polar molecule} 
\auname{Arka Prabha Banik}


{\bf The problem: }

\Dn

Find the polarization ${{\tilde{P}}\left(\xi\right)}$ and
the electric susceptibility ${\chi}$ for  gas of ${N}$ classical
molecules with dipole moment ${\mu}$, The system's temperature is
${T}$.

\bigskip
{\bf The solution:}

\Dn
This problem has been approached in different branches of Physics , let's discuss it in the way of  Statistical Mechanics .

\bigskip

Suppose, each polar molecule of the cluster having Mass and Moment
of Inertia $m$ and $I$ respectively is feeling an external Electric
field $E$.

\medskip
\noindent

 we can write down the Lagrangian for each molecule in $(r,\theta,\phi)$
co-ordinate 

\[
\quad L(x,y,z,\theta,\phi)=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2})+\frac{1}{2}I(\dot{\theta}^{2}+ \sin^{2}\theta\dot{\phi}^{2})
\]


next comes the Energy consideration :

\medskip


The conjugate momenta as we know are $p_{j}=\frac{\partial L}{\partial\dot{q}_{j}}$
where $j=r,\theta,\phi$ 

\medskip

furthar if we notice that we can write the Hamiltonian from Lagrangian
with those conjugate momenta as , 

\[
H_{0}=\Sigma p_{j}\dot{q}_{j}-L(r,\theta,\phi)
\]
 

With furthar simplification it yields 

\[
H_{0}=\frac{1}{2m}p_{r}^{2}+\frac{1}{2I}p_{\theta}^{2}+\frac{p_{\phi}^{2}}{2I\sin^{2}\theta}
\]
 

or more specifically 

\[
H_{0}=\frac{1}{2m}(p_{x}^{2}+p_{y}^{2}+p_{z}^{2})+\frac{1}{2I}p_{\theta}^{2}+\frac{p_{\phi}^{2}}{2I\sin^{2}\theta}
\]


perturbation on the molecule due to the presence of external electric
field

\[
H_{perturb}=-\mathcal{E}\mu \cos\theta
\]


henceforth the modified Hamiltonian reads 

\[
H_{1}=H_{0}+H_{perturb}=\frac{1}{2m}(p_{x}^{2}+p_{y}^{2}+p_{z}^{2})+\frac{1}{2I}p_{\theta}^{2}+\frac{p_{\phi}^{2}}{2Isin^{2}\theta}-\mathcal{E}\mu \cos\theta
\]


we are now ready to write the single molecule Partition Function :

\[
Z_{1}=\int..\int\frac{dxdydzdp_{x}dp_{y}dp_{z}}{(2\pi)^{3}}\int\int\frac{d\theta dp_{\theta}}{2\pi}\int\int\frac{d\phi dp_{\phi}}{2\pi}e^{-\beta H_{1}}
\]
 

now, Writing $H_{1}$ explicitly , we 've 

\[
Z_{1}=\int..\int\frac{dxdydzdp_{x}dp_{y}dp_{z}}{(2\pi)^{3}}e^{-^{\frac{\beta}{2m}(p_{x}^{2}+p_{y}^{2}+p_{z}^{2})}}\int\int\frac{d\theta dp_{\theta}}{2\pi}e^{-\frac{\beta}{2I}p_{\theta}^{2}}e^{\beta \mathcal{E}\mu \cos\theta}\int\int\frac{d\phi dp_{\phi}}{2\pi}e^{-\beta\frac{p_{\phi}^{2}}{2I \sin^{2}\theta}}=I_{1}.I_{2}.I_{3}
\]


the first Integral is as usual 

\[
I_{1}=\int..\int\frac{dxdydzdp_{x}dp_{y}dp_{z}}{(2\pi)^{3}}e^{-^{\frac{\beta}{2m}(p_{x}^{2}+p_{y}^{2}+p_{z}^{2})}}=V(\frac{m}{2\pi\beta})^{3/2}
\]


the second integral gives

\[
I_{2}=\int\int\frac{d\theta dp_{\theta}}{2\pi}e^{-\frac{\beta}{2I}p_{\theta}^{2}}e^{\beta \mathcal{E}\mu \cos\theta}=(\frac{2\pi I}{\beta})^{1/2}\int\frac{d\theta}{2\pi}e^{\beta \mathcal{E} \mu \cos\theta}
\]


and the third integral yields

\[
I_{3}=\int\int\frac{d\phi dp_{\phi}}{2\pi}e^{-\beta\frac{p_{\phi}^{2}}{2I \sin^{2}\theta}}=\int dp_{\phi}e^{-\beta\frac{p_{\phi}^{2}}{2I \sin^{2}\theta}}=(\frac{2\pi I \sin^{2}\theta}{\beta})^{1/2}
\]
 

hence , the single molecule partition function reads 

\[
Z_{1}=I_{1}.I_{2}.I_{3}=V(\frac{m}{2\pi\beta})^{3/2}\frac{I}{\beta}\int  \sin\theta d\theta e^{\beta \mathcal{E}\mu \cos\theta}=V(\frac{m}{2\pi\beta})^ {3/2} \frac{2I}{\beta}\frac{ \sinh\beta\mu \mathcal{E}}{\beta\mu \mathcal{E}}
\]


now Gibbs Prescription is to be followed , i.e, For N molecules the
Partition function is

\[
Z_{N}=\frac{1}{N!}Z_{1}^{N}
\]


So that 
\[
Z_{N}=\frac{1}{N!}V^{N}(\frac{m}{2\pi\beta})^{3N/2}(\frac{2I}{\beta})^{N}(\frac{\sinh\beta\mu \mathcal{E}}{\beta\mu E})^{N}
\]

since we've learned Polarisation and Electric fields are conjugate variables , so we can write :

\[
\widetilde{P}=\frac{1}{\beta}\frac{\partial lnZ_{N}}{\partial \mathcal{E}}
\]


hence in this particular problem , it becomes :

\[
\widetilde{P}=N\mu[\coth(\beta\mu \mathcal{E})-\frac{1}{\beta\mu\mathcal{E}}]
\]
 


Let's  define a function as , $ \mathcal{L}(u) =\coth(u)-\frac{1}{u}$

\medskip

If we expand the function $\mathcal{L}(u)$ in the following Taylor expansion series :

\[
\mathcal{L}(u)=\coth(u)-\frac{1}{u}=\frac{e^{u}+e^{-u}}{e^{u}-e^{-u}}-\frac{1}{u}
\]


\[
\quad\;=\frac{(1+u+\frac{1}{2}u^{2}+\frac{1}{6}u^{3}+....)+(1-u+\frac{1}{2}u^{2}-\frac{1}{6}u^{3}+....)}{(1+u+\frac{1}{2}u^{2}+\frac{1}{6}u^{3}+....)-(1-u+\frac{1}{2}u^{2}-\frac{1}{6}u^{3}+....)}-\frac{1}{u}
\]


\[
\quad\;=\frac{1+\frac{1}{2}u^{2}+\frac{1}{24}u^{4}+\frac{1}{720}u^{6}+....}{u+\frac{1}{6}u^{3}+\frac{1}{120}u^{5}+....}-\frac{1}{u}
\]
 

\[
\quad\;=\frac{\frac{1}{3}u^{3}+\frac{1}{30}u^{5}+\frac{1}{840}u^{7}+....}{u(u+\frac{1}{6}u^{3}+\frac{1}{120}u^{5}+....)}
\]
 
\medskip

since , $u=\beta\mu\mathcal{E}$ is usually very small , so in this limit $u\rightarrow0$

\[
\mathcal{L}(u)=u/3
\]

so that it yields , $\widetilde{P}\rightarrow\frac{1}{3}N \beta \mu^{2}\mathcal{E}$
and hence ...

\bigskip

Susceptibility,  $\chi=\frac{N}{3V} \beta \mu^{2}$  
    


\end{document} 



