# How does the extra $\frac{N_1}{2}k_\text{b}T$ term end up in the effective Hamiltonian after inverse temperature integration?

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:

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?

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?
• 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 $H_{\text{eff}}$. Oct 15, 2022 at 7:46
• 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.