---
# System prepended metadata

title: Comparison Ran Ni and our system/
tags: [Dissipative self-assembly]

---

# Comparison Ran Ni activated barrier and our system
###### tags: `Dissipative self-assembly`

<style>.markdown-body { max-width: 1950px; }</style>
---

### System description:

Both systems:

$$v(t) = v_0e^{-\gamma t}$$

##### Ran Ni collision rule:

$E'= E + \Delta E$ if relative energy of the colliding particle: $0.5m(\Delta v_\perp)^2> E_b$.

##### Our collision rule:

\begin{equation}
    \begin{split}
        \boldsymbol{v}_i'&= \boldsymbol v_i - \dfrac{1+\alpha}{2}(\boldsymbol v_{ij}\cdot \hat{\boldsymbol\sigma}_{ij})\hat{\boldsymbol\sigma}_{ij} - \Delta_{ij} \hat{\boldsymbol\sigma}_{ij} \\
        \boldsymbol v_j'&= \boldsymbol v_j + \dfrac{1+\alpha}{2}(\boldsymbol v_{ij}\cdot \hat{\boldsymbol\sigma}_{ij})\hat{\boldsymbol\sigma}_{ij} + \Delta_{ij} \hat{\boldsymbol\sigma}_{ij},
    \end{split}
\end{equation}

With $\Delta_{ij}=\Delta$ if $\min(\tau_i, \tau_j)<\tau_s$ with $\tau_i$ the time before the last collision of particle $i$. And $\Delta_{ij}=0$ otherwise.

##### Other collision rule (not tested):



\begin{equation}
    \begin{split}
        \boldsymbol v_i'&= \boldsymbol v_i - \dfrac{1+\alpha_0^{-|\boldsymbol{v}_{ij}|/v_0}}{2}(\boldsymbol v_{ij}\cdot \hat{\boldsymbol\sigma}_{ij})\hat{\boldsymbol\sigma}_{ij}  \\
        \boldsymbol v_j'&= \boldsymbol v_j + \dfrac{1+\alpha_0^{-|\boldsymbol{v}_{ij}|/v_0}}{2}(\boldsymbol v_{ij}\cdot \hat{\boldsymbol\sigma}_{ij})\hat{\boldsymbol\sigma}_{ij} ,
    \end{split}
\end{equation}

with $\alpha_0>0$. It also leads to a discontinuous transition (and probably a tricritical point). Did not try to look at the behavior of the metastable fluid.



### Comparison




|     | Ran Ni | Our collision rule $\alpha\neq 1$ |   Our collision rule $\alpha= 1$   |
| --- | ------ | --------------------------------- | --- |
|  Nucleation time   |  ------------------------------ ![image](https://hackmd.io/_uploads/SJsLbxvRA.png)     |             ![image](https://hackmd.io/_uploads/BkctGgD0R.png)           |   ![image](https://hackmd.io/_uploads/SkhmGgwRR.png)![image](https://hackmd.io/_uploads/rk3vGlPAC.png) The last points are not equilibrated, I just wanted to check if the nucleation time would drop for large $N$ but it is not the case. |
| Structure factor reaching the absorbing state   |   ![image](https://hackmd.io/_uploads/Hyt3GlDR0.png)    |          $N = 100k$ ![image](https://hackmd.io/_uploads/r1WitlP0A.png) (clustering/inelastic collapse affects large $k$) ![image](https://hackmd.io/_uploads/Sy3z9RPCC.png) |Not so useful $N = 10$k I'll need a system with $N>100$k that reaches an absorbing state to see something interesting (maybe later). At least we see that the pertubation comes from small $k$. $t = 594711$ we reached the absorbing state ![image](https://hackmd.io/_uploads/r1LBsbwCA.png) ![image](https://hackmd.io/_uploads/By6TsRD0A.png)|
| Structure factor active in the metastable |   ![image](https://hackmd.io/_uploads/Hyt3GlDR0.png)    |                $N = 600$k (I believe, it was not in a metastable zone but very close in the active zone, because such a high $N$ should lead to instant nucleation, which was not the case... I don't remember and I don't have the parameters I used) ![image](https://hackmd.io/_uploads/SkJbPgwCR.png)   | ($\tau_s= 7.692$ $\Delta = 0.08$ $N= 400000$ $\phi= 0.22$ )![image](https://hackmd.io/_uploads/SycJG7PAC.png) shitty ![image](https://hackmd.io/_uploads/ryR_THOC0.png) |
  | Velocity Velocity longitudinal structure factor  |  ?   | (active region) ![image](https://hackmd.io/_uploads/HJqV2xPCA.png)   | ![image](https://hackmd.io/_uploads/SyZHzmDC0.png)     |
 |Temperature temperature structure factor (**might be wrong, seems strange...**) |?| (active region) ![image](https://hackmd.io/_uploads/rkem0sePRC.png) Note the huge temperature fluctuations on a snapshot: ![image](https://hackmd.io/_uploads/S17lLWvAR.png)(color = velocity magnitude) Zoom: ![image](https://hackmd.io/_uploads/BJRB8-wR0.png)|![image](https://hackmd.io/_uploads/BJeoGmwRA.png)|

### Notes:

*  It seems that even when $\alpha\neq 1$, the system is hyperuniform.

* If the hyperuniform structure factor is also observed in our case due to metastability, it seems that it's not making the waiting time/nucleation time shoots to infinity with $N$ increasing. Perhaps because the cluster scale sets a lenght scale at which the critical nucleus can easily form.

* The metastable hyperuniform scaling seems to be close to $k^{0.45} for $\alpha\neq 1$ which is strange because the system is far from the critical point in the parameter space, the configurations are not critical at all (cannot see avalanches, the velocity distribution is really gaussian, ...)

* The metstable fluid at $\alpha = 1$ seems to indeed develop a strong hyperuniformity at low $k$ but it requires huge systems

* Interesting that the velocity velocity structure factor also displays some kind of hyperunoformity. The hyperuniformity induced by the conservation of the center of mass by collision also induces a hyperuniform velocity velocity structure factor: $S_{u^\parallel u^\parallel}\propto k^2$. But the hyperuniformity induced by critical point is not visible in the velocity velocity structur factor. We also know from the conservation of the center of mass and from the damping that $S_{u^\parallel u^\parallel}(k = 0) = 0$, so it might not be too suprising


### Numerically solving the CDP equation

We find hyperuniformity if com conserving noise without first order phase transition (easy):

(Lx = 150.0
Nx = 300   
dx = Lx / (Nx - 1)    
dt = 0.01 
T = 100   
timesteps = int(T / dt)
lam = 1
a = 0.5)
![image](https://hackmd.io/_uploads/S1EbuGuAA.png)

We do find critical induced hyperuniformity (for strange parameter, moreover, here the multiplicative noise can be problematic so...):

![image](https://hackmd.io/_uploads/BkmDyXu0R.png)

And also the $k^1.2$:
(  D = 0.1   
    Lx = 300.0
    Nx = 600   
    dx = Lx / (Nx - 1)    
    dt = 0.01 
    T = 100  
    timesteps = int(T / dt)
    b = -1
    lam = 1
    a = 2)
![image](https://hackmd.io/_uploads/r1zgV7_CC.png)

### Nucleation time.

$\alpha = 1$:
(blue still running)
![image](https://hackmd.io/_uploads/rkglnrRykl.png)

Initial condition important!
- In my system, if we start from an hexagonal perfect crystal, we reach the absorbing state quicker.
- The system cool down at $\sim e^{-2\gamma t}$ at high energy. Meaning that we should really start from a system with very high energy to see the points around $N=10^7$ particles reach a steady state instead of immediately dying. 
- Maybe, the system dies quickly because $S(k_{small})$ does not have time to relax to an absorbing state. Moreover, the initial condition is not ideal since it has a large $S(k_{small})$:
![image](https://hackmd.io/_uploads/SkLpQI0kye.png)

### Renormalized mean field parameter through noise

Here is the field theory of the conserved directed percolation with an additional $A^3$ term:

\begin{align}
    \partial_t \rho &= D_\rho \nabla^2 A, \\
    \partial_t A &= - \mathcal F(A) + D_A \nabla^2 A +  \sqrt{\sigma A} \eta\\
    \mathcal F(A)&= -(-a + w\rho)A -b A^2 + cA^3
\end{align}

With $b>0$, $c>0$. The transition, at the mean field level takes place for some values of $\rho$ small enough.
At the mean field level, the two solutions for $A$ (slaved to $\rho$) are:
\begin{equation}
    A_0^0 = 0~~~~\text{ or } A_0^1 =  \frac{b + \sqrt{b^2 - 4ac+4cw\rho_o}}{2c}
\end{equation}




When $b^2 - 4ac+4cw\rho_o<0$, $A_0^0$ is the only stable solution. When $(-a + w\rho)>0$, $A_0^1$ is the only stable solution. In between the two the mean field theory predicts an hysteresis loop.

Let's see what happens with noise. We follow https://arxiv.org/pdf/2402.05078 and https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.126.148001 .

We expand our equation pertubatively in $\sqrt{\sigma}$.

\begin{split}
\rho &= \rho_0+\sqrt{\sigma}\delta\rho_1 + \sigma\delta\rho_2 + \dots\\
A &= A_0+\sqrt{\sigma}\delta A_1 + \sigma \delta A_2 + \dots
\end{split}

The zero-th order is the mean field computation.

The first order terms are:
\begin{align}
    \partial_t \delta \rho_1 &= D \nabla^2 \delta A_1 , \\
    \partial_t \delta m_1 &= D \nabla^2 \delta A_1 
    - \frac{\partial \mathcal{F}}{\partial \rho} \delta \rho_1 
    - \frac{\partial \mathcal{F}}{\partial A} \delta A_1 + \sqrt{2 \rho_0} \eta,
\end{align}

where every derivative are evaluated at the mean field values of the field $A_0$ and $\rho_0$.

Same thing for the second order term
\begin{align}
    \partial_t \delta \rho_2 &= D \nabla^2 \delta A_2 , \\
    \partial_t \delta m_2 &= D \nabla^2 \delta A_2
    - \frac{\partial \mathcal{F}}{\partial \rho} \delta \rho_2 
    - \frac{\partial \mathcal{F}}{\partial A} \delta A_2 \\
   & - \frac{\partial^2 \mathcal{F}}{\partial A^2} \frac{\delta A_1^2}{2} 
    - \frac{\partial^2 \mathcal{F}}{\partial \rho^2} \frac{\delta \rho_1^2}{2}
    - \frac{\partial^2 \mathcal{F}}{\partial A \partial \rho} \delta A_1 \delta \rho_1
    + \frac{\delta \rho_1}{\sqrt{2 \rho_0}} \eta .
\end{align}

Using Ito convention for the noise, we can average these things and obtain an equation for the average values of $\rho$ (we call $\tilde \rho=\langle \rho \rangle$) and $A$ (we call $\tilde A=\langle A\rangle$) by summing the contributions at every order:

\begin{split}
\partial_t \delta \tilde \rho &= D \Delta \delta \tilde A ~~~\text{linear hence exact at all order (fluctuations cannot renormalize conservation law)} ,\\
\partial_t \delta \tilde A&= -\mathcal{F}(\rho_0, A_0) + D\Delta \tilde A - \sqrt{\sigma}\left(  \frac{\partial \mathcal{F}}{\partial \rho} \langle\delta  \rho_1\rangle 
    + \frac{\partial \mathcal{F}}{\partial A} \langle \delta A_1\rangle\right) -\sigma\left(\frac{\partial \mathcal{F}}{\partial \rho} \langle\delta  \rho_2\rangle 
    + \frac{\partial \mathcal{F}}{\partial A} \langle \delta A_2\rangle\right)\\
   & -\sigma\left(\frac{\partial^2 \mathcal{F}}{\partial \rho^2} \dfrac{\langle\delta  \rho_1^2\rangle }{2}
    + \frac{\partial^2 \mathcal{F}}{\partial A^2} \dfrac{\langle \delta A_1^2\rangle}{2}+ \frac{\partial^2 \mathcal{F}}{\partial A\partial\rho} \langle \delta A_1\delta \rho_1\rangle\right) + \mathcal{O}(\sigma^{3/2})
\end{split}

We can do everyhting from this, but it is useful to use the expression of $\mathcal{F}$ evaluated at $\tilde \rho$ and $\tilde A$ and give its expansion around mean field values **using the averaged expansion above** ($\tilde \rho = \rho_0 + \sqrt{\sigma}\langle \delta \rho_1\rangle+\sigma\langle \delta\rho_2\rangle+\dots)$:

\begin{split}
\mathcal{F}(\tilde \rho, \tilde A)&=\mathcal{F}(\rho_0, A_0) + \sqrt{\sigma}\left(  \frac{\partial \mathcal{F}}{\partial \rho} \langle\delta  \rho_1\rangle 
    + \frac{\partial \mathcal{F}}{\partial A} \langle \delta A_1\rangle\right) +\sigma\left(\frac{\partial \mathcal{F}}{\partial \rho} \langle\delta  \rho_2\rangle 
    + \frac{\partial \mathcal{F}}{\partial A} \langle \delta A_2\rangle\right)\\
   & +\sigma\left(\frac{\partial^2 \mathcal{F}}{\partial \rho^2} \dfrac{\langle\delta  \rho_1\rangle^\color{red}{2} }{2}
    + \frac{\partial^2 \mathcal{F}}{\partial A^2} \dfrac{\langle \delta A_1\rangle^\color{red}{2}}{2}+ \frac{\partial^2 \mathcal{F}}{\partial A\partial\rho} \langle \delta A_1{\color{red}{\rangle}\color{red}{\langle}}\delta \rho_1\rangle\right) + \mathcal{O}(\sigma^{3/2})
\end{split}

Which leads to the simpler dynamics for $\tilde A$:
\begin{split}
\partial_t \delta \tilde A&= -\mathcal{F}(\tilde \rho,\tilde  A) + D\Delta \tilde A \\
&-\sigma\left(\frac{\partial^2 \mathcal{F}}{\partial \rho^2} \dfrac{\langle\delta  \rho_1^2\rangle -\langle\delta \rho_1\rangle^2}{2}
    + \frac{\partial^2 \mathcal{F}}{\partial A^2} \dfrac{\langle \delta A_1^2\rangle - \langle \delta A_1\rangle^2}{2}+ \frac{\partial^2 \mathcal{F}}{\partial A\partial\rho} \left(\langle \delta A_1\delta \rho_1\rangle-\langle A_1\rangle\langle \rho_1\rangle\right)\right) + \mathcal{O}(\sigma^{3/2})
\end{split}

Again, all derivative are evaluated at the mean field values:


\begin{split}
\dfrac{1}{2}\left.\frac{\partial^2 \mathcal{F}}{\partial \rho^2}\right|_{\rho_0, A_0}&=0\\
\dfrac{1}{2}\left.\frac{\partial^2 \mathcal{F}}{\partial A^2}\right|_{\rho_0, A_0}&=-2b + 6cA_0
\end{split}
______

In any case, we need to find the averaged correlators. In q space, the dynamic of the first order term is a multivariate ornstein uhlenbeck process:

$$\dot \psi_1(q, t) = M(q) \psi_1(q, t) + \zeta(q, t)$$
with $\psi_1(q, t) =(\delta \rho_1(q, t), \delta A_1(q, t))$, $\zeta=(0, \sigma\rho_0 \eta(q, t))$ and $M=$

\begin{bmatrix}
0 & -Dq^2 \\
wA_0 & -Dq^2 + (-a+\omega\rho_0)+2bA_0-3cA_0^2
\end{bmatrix}


We find:

\begin{split}
f(q)&=\dfrac{\rho_0\sigma/2}{Dq^2-(-a+\omega\rho_0)+2bA_0-3cA_0^2}\\
&=\left( -\frac{2 A_0 b}{(a + D q^2)^2} + \frac{1}{a + D q^2} + \frac{A_0^2 (4 b^2 + 3 a c + 3 c D q^2)}{(a + D q^2)^3}\right.\rho_0 \\
&-\left.\frac{4 A_0^3 (2 b^3 + 3 a b c + 3 b c D q^2)}{(a + D q^2)^4} \right) \rho_0+\mathcal{O}(\rho_0, A_0^3)\\
\langle \delta \rho_1(q_1)\delta \rho_1(q_2)\rangle&=\dfrac{D}{A_0}q^2f(q)\delta(q_1 + q_2)\\
\langle \delta A_1(q_1)\delta A_1(q_2)\rangle&=f(q)\delta(q_1 + q_2)\\
\langle \delta \rho_1(q_1)\delta A_1(q_2)\rangle&=0
\end{split}

____________

We compute the values of the correlator through integration:

\begin{equation}
    f(0) = 2\pi\int_0^{\infty} \frac{dq}{(2\pi)^2} q f(q)
\end{equation}
Which gives for $f(q)$:

\begin{split}
    \int dq q\frac{2 A_0 b}{(a + D q^2)^2} &= \frac{A_0b}{aD}\\
    \int_0^L dq q\frac{1}{a + D q^2} &=\dfrac{1}{2D}\log(1+DL^2/a)~~~~\text{diverges with L}\\
    \int dq q\frac{A_0^2 (4 b^2 + 3 a c + 3 c D q^2)}{(a + D q^2)^3}&=\dfrac{A_0^2(2b^2+3ac)}{2a^2D}\\
    \int dq q\frac{4 A_0^3 (2 b^3 + 3 a b c + 3 b c D q^2)}{(a + D q^2)^4}&=\dfrac{A_0^3(4b^3+9abc)}{3a^3D}
\end{split}

The second term diverges with system size. As with Wijland theory. But except that, nothing too suspicious that could make us get rid of the metastable region.


