### Model B + conserved noise RG

What are the critical exponents of:

$$\partial_t\rho=\nabla^2\left(\dfrac{\delta F}{\delta \rho}+\eta\right)$$
with:
$$F = \int \left[\dfrac{\kappa}{2}(\nabla \rho)^2-\dfrac{a}{2}\rho^2+\dfrac{u}{4}\rho^4\right] d\boldsymbol r$$
and:
$$\langle\eta(r, t)\eta(r', t')\rangle=2D\delta(r-r')\delta(t-t')$$
The dynamics therefore reads:
$$\partial_t\rho=\nabla^2\left(-a\rho + u\rho^3 -\kappa \nabla^2\rho+\eta\right)$$

A rescaling of the parameters and of the field of the critica linear theory gives:

$$b^{-z+\Delta_\rho}\partial_t\rho=b^{-2}\nabla^2\left( -\kappa b^{-2+\Delta_\rho} \nabla^2\rho+b^{-\frac{d+z}{2}}\eta\right)$$

This gives:
$$-z+\Delta_\rho = -4+\Delta_\rho\Rightarrow z = 4$$
$$-4+\Delta_\rho=-\frac{d+z}{2}-2\Rightarrow \Delta_\rho = -d/2$$

Then: $\langle\rho(r)\rho(r')\rangle=|r-r'|^{-d}\Rightarrow S(|k|)\sim \log(k)$

Which does not really agree with a direct computation from the equation of motion:
$$S(k)=\int dw \frac{k^4}{w^2+k^4}\propto cst$$

More generally, at the linear level we expect the structure factor to behave as:

$$
S(k)=\dfrac{k^2}{\xi^{-2}+k^2}~~~\text{with}~~~ \xi^{-2}\sim a
$$

_____________

Now, if one considers the non linearity, then the picture is different, because a anomalous dimension $\eta$ will arise at two loop, changing the scaling to:

$$S(k)\sim k^{-\eta}$$

the non linearity will have scaling: $\Delta_u=z-\Delta_\rho-2+3\Delta \rho=2-d$

Thus the only non-trivial behavior is expected at $d=1$.

___________

We define the bare propagator:

$$G_0 = \dfrac{1}{-iw+k^2(a+\kappa k^2)}$$

The bare correlator:

$$C_0=\dfrac{2Dk^4}{w^2+k^4(a+\kappa k^2)^2}$$

The bare noise:

$$\langle \zeta^2\rangle\sim 2Dq^4$$

The bare 3 vertex: $$g_0(q)=-u q^2$$

_______

$$
\begin{split}
G(k)&=G_0(k)\left(1-3uk^2G_0(k)\int C_0(q,w)\dfrac{dw}{2\pi}\dfrac{dq}{(2\pi)^d}\right)\\
&=G_0(k)\left(1-3uk^2G_0(k)\int S_0(q) \dfrac{dq}{(2\pi)^d}\right)\\
\end{split}
$$
with $S_0(k)=\displaystyle{\int} \dfrac{dw}{2\pi}\dfrac{2Dk^4}{w^2+k^4(a+\kappa k^2)^2}=\dfrac{D k^2}{a+k^2\kappa}$. Therefore:

$$
\begin{split}
G(k)&=G_0(k)\left(1-3uk^2G_0(k) K_d\Lambda^d S_0(\Lambda)\delta l\right)\\
\end{split}
$$

And so:

$$
\begin{split}
k^2(a_I +\kappa_I k^2)&\simeq k^2(a +\kappa k^2)\left(1+3u \dfrac{1}{a +\kappa k^2}K_d\Lambda^d \dfrac{D\Lambda^2}{a+\Lambda^2\kappa}\delta l\right)\\
&\simeq k^2(a +\kappa k^2)+3uk^2K_d\Lambda^d \dfrac{D\Lambda^2}{a+\Lambda^2\kappa}\delta l
\end{split}
$$

therefore $\kappa$ is not renormalized but $a$ is as:

$$a_I \simeq a\left(1 + 3\dfrac{uD}{a}\dfrac{K_d\Lambda^{2+d}}{a+\Lambda^2\kappa}\delta l\right)$$
_____

The diagram renormalizing the vertex is:

$$g(k)=g_0(k)\left(1+18\int \dfrac{dw}{2\pi}C_0(q)G_0(q-k)g_0(q-k)\dfrac{dq}{(2\pi)^d}\right)$$

The integral over $w$ reads:

at lowest order in $k$, it reads:

We also put $k=0$ in the vertex. We obtain:

$$g(k)=g_0(k)\left(1-9Du\int \dfrac{q^2}{(a+q^2)^2}\dfrac{dq}{(2\pi)^d}\right)$$

That is, $u_I\simeq u\left(1-9Du \dfrac{K^d\Lambda^{d+2}}{(a+\kappa \Lambda^2)^2}\delta l\right)$

____________________

We define $\tilde u=\dfrac{uD}{\kappa^2}K_d\Lambda^{d-4}$ and $\tilde a=a/(\kappa \Lambda^2)$

$$
\begin{split}
\partial_l \tilde u &= \left(2-d+9\tilde u\dfrac{1}{(\tilde a+1)^2}\right)\tilde u \\
\partial_l\tilde a &=2\tilde a+ \dfrac{3\tilde u}{\tilde a+1}
\end{split}
$$

Which leads to a Wilson Fischer fixed point but with 2 as the upper critical dimension/$\epsilon = 2-d$.

### Two loop computation

$D$ does not get renoramlized as at equilibrium due to conservation law.

However, we can use the 2 point vertex function $\Gamma^2$:

$$
\begin{split}
\Gamma^2(q, w=0, a)&\propto \iint \dfrac{1}{(2\pi)^{2d}}dk_1dk_2\dfrac{D^2k_1^4k_2^4}{F(k_1)F(k_2)\left(F(k_1)+F(k_2)+F(q-k_1-k_2)\right)}\\
&=\iiint \dfrac{1}{(2\pi)^{3d}}dk_1dk_2dk_3\dfrac{D^2k_1^4k_2^4\delta(k_3-(q-k_1-k_2))}{F(k_1)F(k_2)\left(F(k_1)+F(k_2)+F(k_3)\right)}\\
\end{split}
$$

with $F(k)=k^2(a+\kappa k^2)$. This form is useful at equilibrium because, at equilibrium, the numerator is in $k_1^2k_2^2$, and the static perturbation serie is recovered ([using permutations](https://www-thphys.physics.ox.ac.uk/people/JohnCardy/qft/dynamicRG.pdf) ). Here however, it is not useful. We set $a=0$ because we are interested at the divergence at the critical point, and we obtain:
$$
\begin{split}
\Gamma^2(q)&\propto \iint \dfrac{1}{(2\pi)^{2d}}dk_1dk_2\dfrac{D^2/\kappa^3}{k_1^4+k_2^4+(q-k_1-k_2)^4}\\
&=\dfrac{D^2K_d^2}{\kappa^3}\iint \dfrac{1}{(2\pi)^{2}}\dfrac{k_1^{d-1}k_2^{d-1}dk_1dk_2}{k_1^4+k_2^4+(q-k_1-k_2)^4}
\end{split}
$$

## Scaling relations

We do not have any free energy, thus we are somewhat restricted as to what we can do, however, we still have scaling law/scale invariance and a generating function should exist. This is what is done here for example: "https://arxiv.org/abs/2010.08387".

I think it's the most rigorous approach, but let's do something slightly different.

#### Placing ourselves at the critical density $\phi_c$!

We are placing ourselves at the critical packing fraction, see https://www.lptmc.jussieu.fr/files/chap_rg.pdf for when we are not exactly at $\phi_c$.
For example, let's start from the correlation function (we called it $C$ let's call it $S$ as it is the structure factor), at equilibrium, scale invariance implies (on large lenght scales/ small $\boldsymbol k$):

$$S(\boldsymbol k)=\dfrac{1}{|\boldsymbol k|^{2-\eta}}\mathcal{G}(|\boldsymbol k|\xi)$$

with $\xi$ the correlation lenght.
In our non-equilibrium case, we can propose:

$$S(\boldsymbol k)=|\boldsymbol k|^{\eta}\mathcal{G}(|\boldsymbol k|\xi)$$

as we know that the MF predicts a constant $S$ at small $\boldsymbol k$ or that the scaling dimension of $\rho$ is $-d/2$ and not $-d/2+1$ as in equilibrium. In both cases, probably: $\mathcal{G}\sim a^\eta$ as $\xi\to\infty$ with $a$ some lattice spacing.

We then recall that the correlation lenght diverges as one approaches the critical point: $\xi\sim|t|^{-\nu}$ where $t$ denotes the closeness to the critical point, for example: $t=(T-T_c)/T_c$.

Then, we can rewrite the structure factor to:

$$S(\boldsymbol k)=|\boldsymbol k|^{\eta}\mathcal{G}(|\boldsymbol k||t|^{-\nu})$$ or in real space (still have to understand this one):

$$S(\boldsymbol r)=\dfrac{1}{|\boldsymbol r|^{d+\eta}}\mathcal{G}(|\boldsymbol r||t|^{-\nu})$$.

#### Our scaling laws.

##### Fischer scaling law does not working.

Indeed, it depends on te fact that $\boldsymbol k\to 0$ does not depend on $\boldsymbol k$/is finite. Which sould leads to something like this: $\mathcal{G}( x\to 0)\sim x^{-\eta}$, from this, we obtain the exponent $\gamma$: $\chi\sim |t|^{-\gamma}$:

$$\chi\sim S(\boldsymbol k\to 0, t)\sim|\boldsymbol k|^{\eta}(|\boldsymbol k||t|^{-\nu})^{-\eta}\Rightarrow \gamma = \nu\eta$$

Of course as $\chi$ does not diverge it does not work.
##### Order parameter scaling law

From the correlation function, the order parameter has dimension: $\Delta_\rho=-d/2-\eta/2$. Therefore, $\langle \rho\rangle\sim\xi^{\Delta_\rho}\sim (-t)^{-\nu\Delta_\rho}\sim(-t)^{\beta}$ which implies: $$\boxed{\beta=\dfrac{\nu}{2}(d+\eta)}$$

Note that, as in equilibrium, at the upper critical dimension, we obtain $\beta = \nu$. At equilibrium, the relation reads: $\beta=\dfrac{\nu}{2}(d-2+\eta)$.

#### Fluctuation dissipation theorem

We show below that $P_S[\rho]\sim e^{-\displaystyle{\dfrac{1}{T}\int (f(\rho(k)) +\kappa (k \rho)^2/2+h(k)\rho(k))/k^2dk}}$

We define $Z=\int \mathcal D\rho P_S[\rho]$
Let's define the generalized susceptbility (up to some $T$):

$$\chi(k)=\dfrac{\delta \langle \rho\rangle}{\delta h(k)}=k^2\dfrac{\delta^2 \log Z}{\delta h(k)\delta h(-k)}=\left(\langle \rho(k)\rho(-k)\rangle-\langle\rho(k)\rangle^2\right)/k^2$$

Which means that $\chi(k)=S(k)/k^2\Rightarrow \chi(k)=\dfrac{1}{k^2+a^2}$
#### Finding $\phi_c$


$\phi_c\simeq 0.3162$

$N =10k$

$N = 25k$


$N = 25k$ close up:
If we look at the density profiles close to the critical poit $\Delta E = 35.7$:

| $N = 10k$| $N=20k$ |
| -------- | -------- |
|  |  |

#### Binder cumulant:

we see that we have interfacial contribution

##### Without damping:






##### With damping and $\alpha = 1$

Averaging probably also the transient:

Better average but less point

Zoom:

##### With damping and $\alpha < 1$
shitty statistics,, cannot trust

#### Elongated box (1/10):

Does not seem better but we are closer to the critical point than above ($\Delta E=34.7$)
| $N = 10k$| $N=20k$ |
| -------- | -------- |
|  |  |


## Nucleation

Let's do nucleation.

We start from:

$$\partial_t\rho=\nabla^2\left(\mu+\eta\right)\tag{1}$$
with:
$$\mu = f'(\rho)-\kappa \nabla^2(\phi)$$

and [assume an ansatz](https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.130.098203?casa_token=YdDwZRVlu4UAAAAA%3A6pjEeJ-287BNSpYcuajg2r7K8mTzpvmJIBRBM5urIrFuAdnmSwTDV75QkjUeVkyoVEiSvO90EUpq1g): $\rho(\boldsymbol r, t) = g(r - R(\theta, t))$ (in 2D). This assumes that the bubble dynamics is controlled on large time scale by growth and shrinkage and not by effects along the tangential direction.
We fix notation as follow:

- $\phi_{1, 2}$ are the coexistence densities for flat interfaces.
- $\phi_s=\phi_1+\epsilon$ is the the supersaturated value of the majority phase. It is then the value of the density at infinity. I don't really know what is $\phi_-$. But it does not enter in our computation
- $\phi_+=\phi_2+\varepsilon$ is the density of the minority phase.

Using the Ansatz in Eq.(1), we obtain:

$$-\dot R g'(r-R)=\nabla^2\left(\mu+\chi\right)$$

Now, assuming that $R(\theta, t)\equiv R(t)$ (spherical symetry) and averaging over the angles:

$$-\dot R g'(r-R)=\nabla^2\left(\mu+\tilde\chi\right)$$

with $\langle\tilde\chi(r_1, t_1)\tilde\chi(r_2, t_2)\rangle=2D\delta(r_1-r_2)\delta(t_1-t_2)/(S_dr_1^{d-1})$ with $S_d$ the area of the unit d-sphere.

We now have to invert the radially symmetric Laplacian: $\nabla^2 f(r)=f''+\dfrac{d-1}{r}f'$:

$$-\nabla^{-2}(\dot Rg'(r-R))=\mu(r)+\tilde\chi(r)-\mu(\infty)$$

Where we defined $\nabla^{-2}s$ as the solution $l$ to $\nabla^2l = s$ which vanishes at infinity and is regular at $r=0$: $l(r)=-\int_r^\infty dr_2\int_0^{r_2}dr_1 (r_1/r_2)^{d-1}s(r_1)$. We also required that the determinstic part of the chemical potential follows $\mu(r\to\infty)=\mu(\infty)$ by introducing constant.

Now: $$\mu(r)= f'(g(r-R))-\kappa\left(\dfrac{d-1}{r}g'(r-R)+g''(r-R)\right)$$

Next, we multiply the dynamics equation by $g'(r-R)$ and integrate across the interface:

$$
\begin{split}
\int_0^\infty dr g'(r-R)\nabla^{-2}(\dot Rg'(r-R))=&-\int_0^\infty drg'f'(g)-\kappa\left(\dfrac{d-1}{r}(g'(r-R))^2+g'g''\right)\\
&-g'f'(\phi_s)+g'\tilde\chi(r)\\
=&-\left[f(\phi_s)-f(\phi_+)-\dfrac{d-1}{R}\sigma - (\phi_s-\phi_+)f'(\phi_s)\right]\\& - \int_0^\infty dr \tilde\chi(r)g'(r-R)
\end{split}
$$

Where we have partially integrated most terms and used the fact that $g'$ is peaked at $R$:

$$\kappa\int_0^\infty dr g'^2(r-R)/r=\int_0^\infty dr g'^2(r-R)/R+\mathcal{O}(R^{-2})=\sigma/R+\mathcal{O}(R^{-2})$$

We can now write:
$$
\begin{split}
\int_0^\infty dr g'(r-R)\nabla^{-2}(\dot Rg'(r-R))=&f(\phi_+)-f(\phi_s) - (\phi_+-\phi_s)f'(\phi_s)+\dfrac{d-1}{R}\sigma\\& - \int_0^\infty dr \tilde\chi(r)g'(r-R)\\
=&\sigma(d-1)\left(\dfrac{1}{R}-\dfrac{1}{R_c}\right) - \int_0^\infty dr \tilde\chi(r)g'(r-R)\\
\end{split}
$$
with $R_c^{-1}=\dfrac{(\phi_+-\phi_s)f'(\phi_s)- (f(\phi_+)-f(\phi_s))}{\sigma(d-1)}$.

______

Now let's obtain the time dependence. Recall that $g'(r-R)\sim \delta(r-R')(\phi_+-\phi_s)$.

##### In 3D:
$$
\begin{split}
\nabla^{-2}(g'(r-R))&=-\int_r^\infty \dfrac{dr_2}{r_2^2}\int_0^{r_2}dr_1 r_1^{2}g'(r_1-R)\\
&\simeq-(\phi_s-\phi_-)\int_r^\infty \dfrac{R^2}{r_2^2}\Theta(r_2-R)dr_2\\
&=-(\phi_s-\phi_-)\left(R-\dfrac{R}{r}(r-R)\Theta(r-R)\right)
\end{split}
$$

Therefore:

$$
\begin{split}
\int_0^\infty dr g'(r-R)\nabla^{-2}(\dot Rg'(r-R))&\simeq-\dot R(\phi_s-\phi_-)\int_0^\infty dr g'(r-R) \left(R-\dfrac{R}{r}(r-R)\Theta(r-R)\right)\\
&\simeq -\dot R(\phi_s-\phi_-)^2R
\end{split}
$$

##### In D>2:
$$
\begin{split}
\int_0^\infty dr g'(r-R)\nabla^{-2}(\dot Rg'(r-R))&\simeq -\dot R(\phi_s-\phi_-)^2\dfrac{R}{d-2}
\end{split}
$$

##### In 2D:

$$
\begin{split}
\int_0^\infty dr g'(r-R)\nabla^{-2}(\dot Rg'(r-R))&\simeq -\dot R(\phi_s-\phi_-)^2R\log(R_+/R)
\end{split}
$$

with $R_+$ the upper limit of integration of the $r_2$ integral.
_____

The noise (let's call it $\eta$) statistics reads:

$$
\begin{split}
\left\langle\eta(t_1)\eta(t_2) \right\rangle&=\left\langle\int_0^\infty dr_1 \tilde\chi(r_1)g'(r_1-R)\int_0^\infty dr_2 \tilde\chi(r_2)g'(r_2-R) \right\rangle\\
&=\int_0^\infty dr_2 \int_0^\infty dr_1 g'(r_1-R)g'(r_2-R) 2D\delta(r_1-r_2)\delta(t_1-t_2)/(S_dr_1^{d-1})\\
&=2D\int_0^\infty dr_1 g'(r_1-R)^2 \delta(t_1-t_2)/(S_dr_1^{d-1})\\
&\simeq\dfrac{2D\sigma}{\kappa S_d R^{d-1}}\delta(t_1-t_2)
\end{split}
$$
________________________

We therefore have the following equation:

$$
\begin{split}
\dot R&=\dfrac{\sigma(d-1)}{(d-2)\Delta \phi^2R}\left(\dfrac{1}{R_c}-\dfrac{1}{R}\right)+\sqrt{\dfrac{2D\sigma}{\Delta \phi^4(d-2)^2\kappa S_d R^{d+1}}}\xi~~~\text{ for }~~ d>2\\
\dot R&=\dfrac{\sigma}{\Delta \phi^2R\log(R_+/R)}\left(\dfrac{1}{R_c}-\dfrac{1}{R}\right)+\sqrt{\dfrac{2D\sigma}{\Delta \phi^4\kappa S_d R^{3}\log(R_+/R)^2}}\xi~~~\text{ for }~~ d=2
\end{split}
$$

or equivalently:
$$
\dot R=-\mathcal{M(R)}\frac{\partial U}{\partial R}+\sqrt{2D\mathcal{M}(R)}\xi
$$
with

| | $\mathcal{M}(r)$ | $U(r)$ |
| -------- | -------- | -------- |
|$d>2$| $\dfrac{\sigma}{(d-2)^2\Delta \phi^4\kappa S_d R^{d+1}}$|$-\dfrac{(d-1)(d-2)}{d+1}\kappa S_d\Delta \phi^2R^{d+1}\left(1-\dfrac{1+d}{d}\dfrac{R_c}{R }\right)$|
|$d=2$|$\dfrac{\sigma}{2\pi\Delta \phi^4\kappa R^{3}\log(R_+/R)^2}$|$\dfrac{\kappa \Delta \phi^2\pi}{18}R^2\left(9-4\dfrac{R}{R_c}+6\left(3-2\dfrac{R}{R_c}\right)\log(R_+/R)\right)$|

Interestingly, $U^{2D}(R)$ still has a local maximum at $R_c$, but it grows with $R_+$:

$$U^{2D}(R_c)=\dfrac{\Delta\phi^2\kappa\pi }{18}R_c^2\left(5+6\log(R_+/R_c)\right)$$


Cannot really do spherical boundary conditions


### First tests:

#### 2D hyperuniform


Nothing works! No $R^2$ dep (perhaps too small nucleus), no size dep.

## Fokker Planck equation

When doing it correctly, goldenfeld: https://students.iiserkol.ac.in/~mms15ms051/courses/PH4202/[Nigel%20Goldenfeld]%20Lectures%20on%20Phase%20Transitions%20and%20the%20Renormalization%20Group.pdf#page=239&zoom=100,0,0 page 250 appendix

The following equation:
$$\partial_t\rho=\nabla^2\left(\dfrac{\delta F}{\delta \rho}+\sqrt{2T}\eta\right)\Rightarrow \partial_t\rho=-k^2\left(\dfrac{\delta F}{\delta \rho}-\sqrt{2T}k^2\eta\right)$$

leads to the Fokker-Planck equation:

$$\partial_tP[\rho(k, t)]=\int d^dk \left[k^2\dfrac{\delta}{\delta\rho}\left(\dfrac{\delta F}{\delta \rho}P\right)+Tk^4\dfrac{\delta^2P}{\delta \rho^2}\right]$$

In the steady state, we assume no current:

$$\dfrac{\delta F}{\delta \rho}=-Tk^2\dfrac{\delta\log(P_S)}{\delta \rho}$$

Let's assume that $P_S$ can be written as $$P_S=Ce^{-\dfrac{1}{T}\int \tilde f(\rho(k), k)dk}$$

with $\tilde f$ a function like a lagrangian.

We recall that $F[\rho]=\int f(\rho(k))+\kappa/2k^2|\rho|^2 dk$ and that for this: $\delta F/\delta\rho = f'(\rho) +\kappa k^2\rho$.

With this ansatz:

$$f' +\kappa k^2\rho=k^2\tilde f'\Rightarrow \tilde f = \dfrac{f +\kappa k^2\rho^2/2}{k^2}+g(k)$$

Therefore, we define $\tilde F[\rho]=\int (f(\rho(k))+\kappa/2k^2|\rho|^2)/k^2 dk$ and find that:

$$P_S=Ce^{-\dfrac{\tilde F[\rho]}{T}}$$

This tells us that configurations for which $\rho(k\to 0)\neq 0$ are really suppressed.