Exact solution of the Percus Yevick integral equation for hard spheres: Difference between revisions

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m (I replaced x by \eta in all the eqs. refered to Thiele's work. This is correct because the "dimensionless number density" (eq.6 of Thiele) divided by 4 is the packing fraction. I also modify the cross ref to Wertheim1 ref. to solve a problem with it.)
 
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The exact solution for the [[Percus Yevick]] integral equation for [[hard sphere]]s
The exact solution for the [[Percus Yevick]] [[Integral equations |integral equation]] for the [[hard sphere model]]
was derived by M. S. Wertheim in 1963 Ref. 1 (See also Ref. 2)
was derived by M. S. Wertheim in 1963 <ref name="wertheim1" >[http://dx.doi.org/10.1103/PhysRevLett.10.321  M. S. Wertheim "Exact Solution of the Percus-Yevick Integral Equation for Hard Spheres", Physical Review Letters '''10''' 321 - 323 (1963)]</ref> (see also <ref>[http://dx.doi.org/10.1063/1.1704158  M. S. Wertheim "Analytic Solution of the Percus-Yevick Equation", Journal of Mathematical Physics, '''5''' pp. 643-651 (1964)]</ref>), and for [[mixtures]] by Joel Lebowitz in 1964 <ref>[http://dx.doi.org/10.1103/PhysRev.133.A895  J. L. Lebowitz, "Exact Solution of Generalized Percus-Yevick Equation for a Mixture of Hard Spheres", Physical Review '''133''' pp. A895 - A899 (1964)]</ref>.
(and for mixtures by in Lebowitz 1964 Ref. 3).
 
The direct correlation function is given by (Ref. 1 Eq. 6)
The [[direct correlation function]] is given by (Eq. 6 of <ref name="wertheim1" /> )


:<math>C(r/R) = - \frac{(1+2\eta)^2 - 6\eta(1+ \frac{1}{2} \eta)^2(r/R) + \eta(1+2\eta)^2\frac{(r/R)^3}{2}}{(1-\eta)^4}</math>
:<math>C(r/R) = - \frac{(1+2\eta)^2 - 6\eta(1+ \frac{1}{2} \eta)^2(r/R) + \eta(1+2\eta)^2\frac{(r/R)^3}{2}}{(1-\eta)^4}</math>
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:<math>\eta = \frac{1}{6} \pi R^3 \rho</math>
:<math>\eta = \frac{1}{6} \pi R^3 \rho</math>


and ''R'' is the hard sphere diameter.
and <math>R</math> is the hard sphere diameter.
The equation of state is (Ref. 1 Eq. 7)
The [[Equations of state | equation of state]] is given by (Eq. 7 of <ref name="wertheim1" />)


:<math>\beta P \rho = \frac{(1+\eta+\eta^2)}{(1-\eta)^3}</math>  
:<math>\frac{\beta P}{\rho} = \frac{(1+\eta+\eta^2)}{(1-\eta)^3}</math>  


Everett Thiele (1963  Ref. 4}) also studied this system,
where <math>\beta</math> is the [[inverse temperature]]. Everett Thiele also studied this system <ref>[http://dx.doi.org/10.1063/1.1734272  Everett Thiele "Equation of State for Hard Spheres", Journal of Chemical Physics, '''39'''  pp. 474-479 (1963)]</ref>,
resulting in (Eq. 23)
resulting in (Eq. 23)


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where (Eq. 24)
where (Eq. 24)


:<math>a = \frac{(2x+1)^2}{(x-1)^4}</math>
:<math>a = \frac{(2\eta+1)^2}{(\eta-1)^4}</math>


and
and


:<math>b= - \frac{12x + 12x^2 + 3x^3}{2(x-1)^4}</math>
:<math>b= - \frac{12\eta + 12\eta^2 + 3\eta^3}{2(\eta-1)^4}</math>


and
and


:<math>c= \frac{x(2x+1)^2}{2(x-1)^4}</math>
:<math>c= \frac{\eta(2\eta+1)^2}{2(\eta-1)^4}</math>


and where <math>x=\rho/4</math>.
The [[pressure]] via the pressure route (Eq.s 32 and 33) is
The pressure via the pressure route (Eq.s 32 and 33) is


:<math>P=nkT\frac{(1+2x+3x^2)}{(1-x)^2}</math>
:<math>P=nk_BT\frac{(1+2\eta+3\eta^2)}{(1-\eta)^2}</math>


and the compressibility route is
and the [[Compressibility equation |compressibility]] route is


:<math>P=nkT\frac{(1+x+x^2)}{(1-x)^3}</math>
:<math>P=nk_BT\frac{(1+\eta+\eta^2)}{(1-\eta)^3}</math>
 
==A derivation of the Carnahan-Starling equation of state==
It is interesting to note (Ref <ref> [http://dx.doi.org/10.1063/1.1675048    G. A. Mansoori, N. F. Carnahan, K. E. Starling, and T. W. Leland, Jr. "Equilibrium Thermodynamic Properties of the Mixture of Hard Spheres", Journal of Chemical Physics  '''54''' pp. 1523-1525 (1971)] </ref>  Eq. 6) that one can arrive at the [[Carnahan-Starling equation of state]] by adding two thirds of the exact solution via the compressibility route, to one third via the pressure  route, i.e.
 
:<math>Z = \frac{ p V}{N k_B T} =  \frac{2}{3} \left[  \frac{(1+\eta+\eta^2)}{(1-\eta)^3}  \right] +  \frac{1}{3} \left[    \frac{(1+2\eta+3\eta^2)}{(1-\eta)^2}  \right] = \frac{ 1 + \eta + \eta^2 - \eta^3 }{(1-\eta)^3 }</math>
 
The reason for this seems to be a slight mystery (see discussion in Ref. <ref>[http://dx.doi.org/10.1021/j100356a008 Yuhua Song, E. A. Mason, and Richard M. Stratt "Why does the Carnahan-Starling equation work so well?", Journal of Physical Chemistry '''93''' pp. 6916-6919 (1989)]</ref> ).


==References==
==References==
#[http://dx.doi.org/10.1103/PhysRevLett.10.321  M. S. Wertheim "Exact Solution of the Percus-Yevick Integral Equation for Hard Spheres", Physical Review Letters '''10''' 321 - 323 (1963)]
<references/>
#[http://dx.doi.org/10.1063/1.1704158  M. S. Wertheim "Analytic Solution of the Percus-Yevick Equation", Journal of Mathematical Physics, '''5''' pp. 643-651 (1964)]
 
#[http://dx.doi.org/10.1103/PhysRev.133.A895  J. L. Lebowitz, "Exact Solution of Generalized Percus-Yevick Equation for a Mixture of Hard Spheres", Physical Review '''133''' pp. A895 - A899 (1964)]
#[http://dx.doi.org/10.1063/1.1734272  Everett Thiele "Equation of State for Hard Spheres", Journal of Chemical Physics, '''39'''  pp. 474-479 (1963)]


[[Category: Integral equations]]
[[Category: Integral equations]]

Latest revision as of 16:25, 22 January 2018

The exact solution for the Percus Yevick integral equation for the hard sphere model was derived by M. S. Wertheim in 1963 [1] (see also [2]), and for mixtures by Joel Lebowitz in 1964 [3].

The direct correlation function is given by (Eq. 6 of [1] )

where

and is the hard sphere diameter. The equation of state is given by (Eq. 7 of [1])

where is the inverse temperature. Everett Thiele also studied this system [4], resulting in (Eq. 23)

where (Eq. 24)

and

and

The pressure via the pressure route (Eq.s 32 and 33) is

and the compressibility route is

A derivation of the Carnahan-Starling equation of state[edit]

It is interesting to note (Ref [5] Eq. 6) that one can arrive at the Carnahan-Starling equation of state by adding two thirds of the exact solution via the compressibility route, to one third via the pressure route, i.e.

The reason for this seems to be a slight mystery (see discussion in Ref. [6] ).

References[edit]