## Archive for August, 2009

### The Born and Debye-Huckel treatment of electrostatistics

Tuesday, August 11th, 2009

In solution, the total charges of all charges species is zero, charge neutral. The polymers or small molecules which are immersed in the solvent are far smaller than the solvent itself. So they can viewed as a spherical with all charges on the surface. And this approximation is the base of what is coming next.

The the distribution for all the charged species is described by the following
$\vec{n_{i}} = \frac{N_{i}z_{i}e}{V} e^{\frac{z_{i}e\varphi(\vec{r})} {kT}}$

while Ni is the number of species i, zi is the number of charges at species i, $\varphi(\vec{r})$ is the electric potential on the species i at position $\vec{r}$.

The whole system is charge neutral so
$\sum_{i=1}^{N}n_{i}e=0$

The electric field at position $\vec{r}$ while r>a, suppose a is the radius of the protein spherial. Remember we assume the protein is a charged spherial shape with all charges on the surface. The Poisson equation tells us just that, the field is ruled by the following form:

$\nabla^{2} \varphi(\vec{r}) = \sum_{i=1}^{N}\frac{\vec{\rho_{i}}}{4\pi\,\epsilon_{0}\epsilon_{r}} = \frac{1}{4\pi\,\epsilon_{0}\epsilon_{r}} \sum_{i=1}^{N}\frac{N_{i}z_{i}e}{V} e^{\frac{z_{i}e\varphi(\vec{r})} {kT}}$

The above equation might be solved exactly, but for the moment, I pretend that I can’t find the solution and have to resort to any kind of approximation. One of the simplest one is that the exponetial part vanishes:
$e^{x} \simeq 1+x for 0
when x is smaller and positive, it is a reasonable approximation.

$\nabla^{2} \varphi(\vec{r}) = \frac{1}{\epsilon_{0}\epsilon_{r}} \sum_{i=1}^{N}\frac{N_{i}z_{i}e}{V} [1+\frac{z_{i}e\varphi(\vec{r})} {kT}] = \frac{1}{\epsilon_{0}\epsilon_{r}} \sum_{i=1}^{N}\frac{N_{i}z_{i}e}{V} \frac{z_{i}e\varphi(\vec{r})} {kT}$

The Poisson eqn is then simplied as following:
$\nabla^{2} \varphi(\vec{r}) = \kappa^{2} \varphi(\vec{r})$

while $n_{i}^{0}=\frac{N_{i}}{V}$
and $\kappa^{2}= \sum_{i=1}^{N}\frac{N_{i}}{V} \frac{z_{i}^{2}e^{2}} {\epsilon_{0}\epsilon_{r}kT}.$

The general solution for the above equation is
$\varphi = A\frac{e^{-r}}{r}+B\frac{e^{r}}{r}$

Now comes to the boundary conditions to determine the coefficients A and B:
$\lim_{r \to +\infty} \varphi(r) = 0 \Rightarrow B=0$

For the charged sphere, the field at where its distance to the center is larger than a, we simply have
$\varphi_{sphere}=\frac{z_{i}e}{4\pi \epsilon_{0}\epsilon_{r}r}+C$
D is the constant field inside the sphere.

At the radius a, the field should be equal and also continuous, so we have the following two conditions:

$\left. \varphi(r) \right|_{r=a} = \varphi(r)_{sphere}.$
$\left. \varphi(r)^{'} \right|_{r=a} = \left. \varphi(r)_{sphere}^{'}\right|_{r=a}$

The only two coefficients in the feild formula, A and C are then solved. The final field is
$\varphi(a)=\frac{z_{i}e}{4\pi\epsilon_{0}\epsilon_{r}} \frac{1} {a(1+\kappa\,a))}$

Born said that the energy of the charged sphere is to charge the sphere, so
$E=\int_{0}^{z_{i}e}\frac{q}{4\pi\epsilon_{0}\epsilon_{r}}\frac{dq}{a(1+\kappa\,a)} = \frac{z_{i}^{2}e^{2}}{4\pi\epsilon_{0}\epsilon_{r}}\frac{1}{2a(1+\kappa\,a)}$

This is what Ken Dill uses in his PNAS paper to describe the free energy contribution. It seems all simple things have been introduced.

### Twenty-First Century Rocket Science – O’Reilly Media

Thursday, August 6th, 2009