2
$\begingroup$

In Elena Pulvirenti's slides she introduced a $\textbf{static Widom-Rowlinson model of one species}$. Consider $\Lambda\subset R^2$ with periodic boundary conditions, $\Lambda$ set of particle configurations with $$\Gamma=\{\gamma\subset \Lambda: N(\gamma)\in \mathbb{N}\}, \, $$ where $N(\gamma)$ is the cardinality of $\gamma$. The halo of a configuration is $h(\gamma)=\cup_{x\in \gamma}B_2(x)$ with radius $2$. Let $H(\gamma)=|h(\gamma)|-N(\gamma)|B_2(0)|$ be the Hamiltonian. Define the grand-canonical Gibbs measure, $$(1) \mu(d\gamma)=\frac{z^{N(\gamma)}}{\Xi}e^{-\beta H(\gamma)}\mathbb{Q}(d\gamma) $$ where $\mathbb{Q}$ is Poisson point process with intensity 1 and $\Xi$ is the partition function.

Her result is that the 2-species Widom-Rowlinson model is equivalent to 1-species. $\textbf{The 2-species WR model}$ is two types of particles(blue and red) with configurations $\gamma^B, \gamma^R$. The grand-canonical Gibbs measure: $$(2) \hat{\mu}(d\gamma^R, d\gamma^B)=\frac{1}{\hat{\Xi}}1_{\{\text{red-blue hard-core}\}}z_R^{N(\gamma^R)}z_B^{N(\gamma^B)}\mathbb{Q}(d\gamma^R)\mathbb{Q}(d\gamma^B)$$ where $1_{\{\text{red-blue hard-core}\}}$ means it is $1$ if $d(\gamma^R, \gamma^B)\geq 1$, otherwise is 0, and $z_R=e^{\beta\lambda_R}$ and $z_B=e^{\beta\lambda_B}$.

$\textbf{My question is why 1-species and 2-species are equivalence?}$ I am confused about that fix the centers of the red discs and integrate over the centers of the blue disc, then: $$\frac{1}{\hat{\Xi}}\int_{\Gamma} 1_{\{\text{red-blue hard-core}\}}z_R^{N(\gamma^R)}z_B^{N(\gamma^B)}\mathbb{Q}(d\gamma^B)=C \frac{z^{N(\gamma^R)}}{\Xi}e^{-\beta H(\gamma^R)}$$ where $(z_B, z_R)\to (\beta, ze^{\beta V_0})$ and $V_0:=|B_2(0)|$.

enter image description here

$\endgroup$
0
$\begingroup$

So let's think about it this way: if $A$ is an event in the two-type model depending only on red then \begin{align*} \mathbb{P}(A) &= \frac{1}{\tilde{\Xi}} \sum_{j,k\geq 0} \frac{z_R^k}{k!} \frac{z_B^j}{j!} \int_{S^k} \int_{S^j} 1_{A} \cdot 1_{RBHC} \,dy\, dx \\ &=\frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \left(\sum_{j \geq 0} \frac{z_B^j}{j!}\int_{S^j} 1_{RBHC} \,dy \right)\,dx \\ &= \frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \left(\sum_{j \geq 0} \frac{z_B^j}{j!}\int_{(S \setminus h(\gamma_R))^j} \,dy \right)\,dx \\ &= \frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \left(\sum_{j \geq 0} \frac{z_B^j}{j!}(|S| - |h(\gamma_R)|)^j \right)\,dx \\ &= \frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \exp(z_B |S| - z_B |h(\gamma_R)|)\,dx \\ &= \frac{C}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A e^{- z_B |h(\gamma_R)|)}\,dx \\ &= \frac{C}{\tilde{\Xi}} \sum_{k\geq0}\frac{(z_R e^{-z_BV_0})^k}{k!} \int_{S^k} 1_A e^{- z_B (|h(\gamma_R)| - k V_0))}\,dx \,. \end{align*}

Doing the change of variables listed completes it.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So actually neither of those matters in the above computation, they're both just the normalizing constant that makes it a probability measure. My $S$ is the underlying space on which this model is defined (so some finite volume subset of R^d). $\endgroup$ – Marcus M Jun 28 at 20:10
  • $\begingroup$ If you'd like, you can use the identity $1/\Xi = P(\gamma_R = \emptyset)$, and then the above shows that $\Xi = \tilde{\Xi}/C$. This is equivalent to what I described in my previous comment. $\endgroup$ – Marcus M Jun 28 at 20:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.