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\exnumber{4016}
\heading{Polar adsorption of particles to a surface} 
\auname{Guy Zisling}


{\bf The problem:}

\Dn
Consider an $M$ site system in an equilibrium with gas of particles that have mass $\mass$. 
The chemical potential of the gas is~$\mu$ and its temperature is~$T$.  
A particle can bind to a site. Each site can absorb at most one atom. 
The binding energy is~$\varepsilon$, and the length of the bonds is~$a$. 
In such state it behaves as a rotor that has moment of inertia $I=ma^2$, and a dipole moment $qa$.         
The polarization can be in any direction away from the surface ($2\pi$ staradians). \\ 

{\bf Tip:} The kinetic part in a rotor Hamiltonian is 
%
\[ \frac{1}{2I} \left[ p_{\theta}^2 + \frac{p_{\varphi}^2}{\sin^2(\theta)} \right] \]

% In the presence of a perpendicular electric field $\mathcal{E}$ 
% the dipole has energy is ${-\mathcal{E} d \cos(\theta)}$, 
% where ${|\theta|<\pi/2}$ is the angle between $d$ and $\mathcal{E}$.

\Dn

(1) Calculate the partition function $Z_{\perp}(\beta,f)$ for an occupied site, 
assuming electric field $f$ perpendicular to the surface.

\Dn

(2) Calculate the partition function $Z_{\parallel}(\beta,f)$ for an occupied site, 
assuming electric field $f$ parallel to the surface.   

\Dn

(3) Express the $M$ site grand partition function $\mathcal{Z}(\beta,\mu,f)$ in terms of $Z$. 
Additionally, write an explicit expression for zero field.   

\Dn

(4) Express the average number $N$ of adsorbed particles in terms of $Z$.
Additionally, write an explicit expression for zero field.  

\Dn

(5) Find a leading order expression for the average polarization $D/N$ for weak perpendicular~$f$. 

\Dn

(6) Find a leading order expression for the average polarization $D/N$ for weak parallel~$f$. 

\Dn

(*) Tip: one can use a shortcut in the calculation of $Z$, bypassing the integration over the momentum variables.  


\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn
(1) The Hamiltonian is: \Dn


\[ 
\mathcal{H} = \frac{1}{2I} \left[ p_{\theta}^2 + \frac{p_{\varphi}^2}{\sin^2(\theta)} \right]  - \varepsilon a \cos{\theta} 
\\ \varepsilon \equiv fq\]

For the Dipole:
\[
Z_1=\intt{\frac{d\theta dP_\theta}{2\pi}}\intt{\frac{d\varphi dP_\varphi}{2\pi}e^{-\beta \matcal{H}}} =\intt{\sqrt{\frac{2\pi I}{\beta}}\sqrt{\frac{2\pi I \sin^2{\theta}}{\beta}}e^{\beta \varepsilon a \cos{\theta}} \frac{d\theta d\varphi}{(2\pi)^2}}=\frac{I}{2\pi\beta}\intt{\sin{\theta}e^{\beta \varepsilon a \cos{\theta}} d\theta d\varphi}
\]

\[
Z_1^{\perp} = \frac{I}{2\pi\beta}\int_0^{2\pi}{d\varphi}\int_0^{\frac{\pi}{2}}{e^{\beta \varepsilon a \cos{\theta}}\sin{\theta}d\theta } = \frac{I}{\beta}\cdot\frac{e^{\beta \varepsilon a} - 1}{\beta \varepsilon a}
\]

%%%%%%%%%%%%%%
(2) For the parallel electric field:
\Dn
\[
Z_1^{\parallel} = \frac{I}{2\pi\beta}\int_0^{\pi}{d\varphi}\int_0^{\pi}{e^{\beta \varepsilon a \cos{\theta}}\sin{\theta}d\theta } = \frac{I}{\beta}\cdot\frac{\sinh{\beta \varepsilon a}}{\beta \varepsilon a}
\]

%%%%%%%%%%
(3) Partition function of M sites in a grand canonical formalism \Dn
\[
\mathcal{Z}(\beta,\mu,\varepsilon) = [1 + Z_1(\beta,\varepsilon)e^{\beta \mu}]^M
\]
With zero field:
\[
\mathcal{Z}(\beta,\mu,\varepsilon) = [1 + \frac{I}{\beta}\cdot e^{\beta \mu}]^M
\]

%%%%%%
(4) Number of particles can be obtained with: \Dn
\[
N = \frac{1}{\beta}\cdot \frac{\partial \ln{\mathcal{{Z}}}}{\partial \mu} = \frac{M}{1 + \frac{1}{Z_1}e^{-\beta \mu}}
\]

With Zero Field:
\[
N = \frac{1}{\beta}\cdot \frac{\partial \ln{\mathcal{{Z}}}}{\partial \mu} = \frac{M}{1 + \frac{\beta}{I}e^{-\beta \mu}}
\]

%%%%
(5) Polarization can be obtained by\Dn
\[
D = \frac{1}{\beta}\cdot\frac{\partial \ln{\mathcal{Z}}}{\partial \varepsilon} = \frac{N}{\beta}\cdot\frac{\partial \ln{Z_1}}{\partial \varepsilon}
\]

For weak perpendicular field $\varepsilon^\perp << 1,\ \varepsilon^\parallel=0$:
\[
Z_1^\perp = \frac{I}{\beta}\cdot\frac{e^{\beta \varepsilon a} - 1}{\beta \varepsilon a} \propto 1 + \frac{1}{2} \beta \varepsilon a
\]
And the polarisation is:
\[
\frac{D(\varepsilon)}{N} = \frac{1}{2}a
\]

%%%%%%%%
(6) For weak parallel field $\varepsilon^\perp = 0,\ \varepsilon^\parallel << 1$: \Dn
\[
Z_1^\parallel = \frac{I}{\beta}\cdot\frac{\sinh{\beta \varepsilon a}}{\beta \varepsilon a} \propto 1 + \frac{1}{6}(\beta \varepsilon a)^2  
\]
And the polarisation is:
\[
\frac{D(\varepsilon)}{N} = \frac{1}{3} \cdot \frac{a^2}{T} \varepsilon
\]

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