I am working through an old research paper which derives an expression for the effective Hamiltonian in a classical system of point ions and colloidal particles (see https://doi.org/10.1016/0021-9797(85)90362-5). There is no excess salt in the system. By virtue of electroneutrality $N_0 z_0=-N_1 z_1$, where $z_0$ is the charge on the colloids, $N_0$ the number of colloids, $z_1$ the charge of the counterions and $N_1$ the number of counterions.

In this paper the authors derive an expression for the effective Hamiltonian by first taking the derivative of $\beta H_\text{eff}$ with respect to $\beta$, where $\beta=\frac{1}{k_\text{b}T}$, and $H_\text{eff}$ is defined as: $$\exp{(-\beta H_\text{eff})}=\frac{1}{V^{N_1}}\int \exp{\left(-\beta H_{pot}\right)}\text{d} \textbf{r}^{N_1},$$ where $H_{pot}$ is the potential energy part of hamiltonian of the complete system.

After some formal manipulations and using linearized Poisson-Boltzmann theory they arrive at the following expression (EQ. [3.37] from their article):

$$\left(\frac{\partial \beta H_\text{eff}}{\partial \beta}\right)=\langle U_\text{sr} \rangle+N_1 \frac{\kappa e^2 z_1^2}{2 \epsilon}-N_0 \frac{3\kappa z_0^2 e^2}{4\epsilon}+\frac{z_0^2 e^2}{2\epsilon}\sum_{i\neq j}\frac{e^{-\kappa R_{ij}}}{R_{ij}}-\frac{\kappa z_0^2 e^2}{4 \epsilon}\sum_{i \neq j}e^{-\kappa R_{ij}}$$ where $U_\text{sr}$ are short-ranged interactions of the original hamiltonian and $\kappa$ is the usual Debye length. The rest of the symbols have their usual meaning. I've rewritten it a little bit by getting rid of the $i=j$ terms (there are $N_0$ such terms) in the last sum before integration, this does not alter the results. Up to this point I can reproduce all steps and follow the derivation.

Next, the authors integrate the previous expression with respect to $\beta$ by recognizing the $\beta$ dependence of $\kappa^2=\frac{4\pi \beta e^2 n_1 z_1^2}{\epsilon}$. After performing this integration the authors obtain the following expression: $$H_\text{eff}=W_\text{sr}+N_0 \frac{\kappa z_0 z_1 e^2}{3\epsilon}-N_0 \frac{\kappa z_0^2 e^2}{2\epsilon}+\frac{z_0^2 e^2}{2\epsilon}\sum_{i \neq j}\frac{e^{-\kappa R_{ij}}}{R_{ij}}+N_0\frac{z_0}{2 z_1}k_\text{b}T.$$

This is where it goes wrong, I am able to reproduce the first four terms from the $\beta$ integration, however the last term $N_0\frac{z_0}{2 z_1}k_\text{b}T$ seems to come out of nowhere. Note that this term can also be rewritten as $-\frac{N_1}{2}k_\text{b}T$.

Does anyone have any idea where this term comes from?

$\textbf{edit:}$ I'm quite certain it is an integration constant. Working back with their expression:

$$\beta H_\text{eff}=\beta W_\text{sr}+\beta N_0 \frac{\kappa z_0 z_1 e^2}{3\epsilon}-\beta N_0 \frac{\kappa z_0^2 e^2}{2\epsilon}+\beta \frac{z_0^2 e^2}{2\epsilon}\sum_{i \neq j}\frac{e^{-\kappa R_{ij}}}{R_{ij}}+N_0\frac{z_0}{2 z_1}.$$

then taking the derivative of this expression with respect to $\beta$ we obtain EQ [3.37] from their article from the first four terms, the last term is a constant so disappears after taking the derivative.

The physical reasoning and the form of the constant is not clear to me.

$\textbf{edit2:}$ I've found the original PhD thesis of the main author of the paper. The term is indeed an integration constant. They use the following justification to derive its form: enter image description here

I understand their definition of the zero energy state. However, the full justification of their derivation is not clear at all. Maybe someone has some insight?


1 Answer 1


I believe this comes from the constant of integration. After integrating with respect to $\beta$ and then dividing both sides by $\beta$, you find $H_\mathrm{eff} = H_\mathrm{int}+ C k_b T$, where $C$ is some constant to be determined. Here, $H_\mathrm{int}$ lumps together all of the terms coming from the electrostatic interactions. Now, the question is how to obtain the constant $C$. Consider the limit where $\epsilon \rightarrow \infty$. In this case, all of the electrostatic interactions are screened, so we must have $H_\mathrm{int} = 0$. In this case, the only energy should be the kinetic energy of the particles which are free to move. I assume there are $N_1$ of these, so that the kinetic energy is $N_1 k_b T/2.$ This fixes the constant $C = N_1 k_b/2.$ Why isn't the kinetic energy $3N_1 k_b T/2$? I'm not sure, but I don't know anything about colloidal systems. Maybe the $N_1$ ions can only move along the direction of the colloidal particles?

  • $\begingroup$ I thought something along the same lines, however, kinetic energy terms are already included in the kinetic part of the hamiltonian, which is not a part of $H_{\text{eff}}$. $\endgroup$
    – M . M
    Oct 15, 2022 at 7:46
  • $\begingroup$ I'm confused too then. If you take $\epsilon \rightarrow \infty$ (or $e \rightarrow 0$), the electrostatic interactions should be zero. I'm not sure I understand why there should be any interaction between the particles in this case. $\endgroup$
    – dan
    Oct 15, 2022 at 14:06
  • $\begingroup$ I fully agree with you, very strange. $\endgroup$
    – M . M
    Oct 15, 2022 at 21:21
  • $\begingroup$ I've found the original PhD thesis of the main author of the paper. It is indeed an integration constant, however they define the zero energy state differently. See the main post. $\endgroup$
    – M . M
    Oct 19, 2022 at 18:51

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