5. Appendix 1: Proof of $\rho_{eq}(\vec{r})=- \delta \Omega [u]/\delta u(\vec{r})$

Grand Partition Function:

$$\Xi= \sum_{N=0}^{\infty} \frac{z^{N}}{N!} \int d\vec{1}..d\v... ...\vec{j})} e^{\textstyle -\beta \sum_{j=1}^{N}\sum_{k>j}u(jk)}$$ (5.1)

Let us write the external and internal Boltzmann factors as

$$B^{ex}(1..N) \equiv e^{\textstyle -\beta \sum_{j=1}^{N}\phi(... ....N) \equiv e^{\textstyle -\beta \sum_{j=1}^{N}\sum_{k>j}u(jk)}$$ (5.2)

Then

$$\Xi= \sum_{N=0}^{\infty} \frac{z^{N}}{N!} \int d\vec{1}..d\vec{N} \; B^{ex}(1..N) \; B^{in}(1..N)$$ (5.3)

Grand Potential:

$$\Omega [\phi] \equiv -kT \; \ln \Xi$$ (5.4)

Varying the external potential according to $\phi$ $\Longrightarrow$ $\phi+\delta \phi$ we find

$$\Omega [\phi+\delta \phi] =$$ (5.5)

$$= -kT \;\ln \sum_{N=0}^{\infty} \frac{z^{N}}{N!} \int d\vec{1}..d\v... ...extstyle -\beta \sum_{j=1}^{N}\delta\phi(\vec{j})} B^{ex}(1..N) \; B^{in}(1..N)$$ (5.6)

$$\approx -kT \;\ln \sum_{N=0}^{\infty} \frac{z^{N}}{N!} \int d\vec{1}..d\v... ...[1-\beta \sum_{j=1}^{N}\delta \phi(\vec{j})\right] B^{ex}(1..N) \; B^{in}(1..N)$$ (5.7)

$$= -kT \;\ln \Xi -kT \; \ln \left\{ 1- \frac{\beta}{\Xi} \sum_{N=0... ...N}\; \sum_{j=1}^{N}\delta \phi(\vec{j}) B^{ex}(1..N) \; B^{in}(1..N) \right\}$$ (5.8)

$$\equiv \Omega + \delta \Omega$$ (5.9)

It follows that

$$\delta \Omega = \frac{z}{\Xi} \sum_{N=1}^{\infty} \frac{z^{N... ...le -\beta \sum_{k>1}u(1k)}\right) B^{ex}(2..N) \; B^{in}(2..N)$$ (5.10)

On the other hand,

$$\rho(\vec{r}) \equiv \langle \hat{\rho}(\vec{r})\rangle_{ens} = \langle \sum_{i} \delta (\vec{r}-\vec{i}) \rangle_{ens}$$

$$= \sum_{N=0}^{\infty} \frac{z^{N}}{N!} \; \frac{1}{\Xi} \int d\vec{... ...ec{N} \; \sum_{i=1}^{N} \; \delta(\vec{r}-\vec{i}) B^{ex}(1..N) \; B^{in}(1..N)$$

$$= \frac{z}{\Xi} \sum_{N=1}^{\infty} \frac{z^{N-1}}{(N-1)!} \; \int ... ...style -\beta \sum_{k>1}u(\vec{r},\vec{k})} \right) B^{ex}(2..N) \; B^{in}(2..N)$$ (5.11)

Therefore

$$\int d\vec{r} \; \rho(\vec{r}) \delta \phi(\vec{r}) = \delta \Omega$$ (5.12)

Since $u(\vec{r}) \equiv \mu - \phi(\vec{r})$ we have

$$\rho_{eq}(\vec{r})= - \frac{\delta \Omega [u]}{\delta u(\vec{r})}$$ (5.13)

q.e.d.

Back to text $\Longrightarrow$1.2