# Hyperuniform interface
###### tags: `Dissipative self-assembly`

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## Interface and hyperuniformity

### Surface tension

#### Equilibrium

Let's call $\sigma$ the surface tension.

Active matter people have problem with surface tension because their mean field equations do not have a free energy structure. They have additional term that complicates the identification of a unique pressure, unique surface tension, etc... In our case, I believe that the equations are more or less given by a free energy at the mean field level for the A-B model (A chapman-Enskog procedure on the Boltzmann equation at least gives only free energy like term). And the non-equilibriumness comes from the noise (while for active matter people, their field theory have a white noise which respect a FDT in some way). Therefore, all mean field results from equilibrium should still work for our non-equilibrium system (Laplace pressure, surface tension as the excess free energy, ...) . The only point at which it will not work is for quantities that need average such as capillary wave theory results (sructure factor, W², coarsening, etc...). However, these quantities will still be linked to a mean field surface tension: $$\sigma_{cw} = \int_{-\infty}^{\infty}\kappa \left(\dfrac{d \phi}{dx}\right)^2dx$$

| Definition | Equilibrium | MIPS through active brownian particles | Active model B+ (F + $\lambda$ and $\zeta$ TRS breaking terms) | Ran ni + AB mixture | COM conserving model B |
| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
| Surface tension is the energy cost for doing work against it by expanding the interface’s area https://journals.aps.org/pre/abstract/10.1103/PhysRevE.107.015105 | $$\begin{split}E &= \sigma \left( \int_{0}^{L_y} \sqrt{1 + \left( \frac{\partial h}{\partial y} \right)^2} \, dx - L_y \right)\\&\simeq\sigma \int_0^{L_y}\dfrac{1}{2}\left(\dfrac{\partial h}{\partial y}\right)^2dy\end{split}$$ | Same | Same | Same | Same |
| The surface tension is the excess free energy stored in the interface region per unit area https://www.abebooks.com/9780521039055/Phase-Transition-Dynamics-Onuki-Akira-0521039053/plp and https://arxiv.org/abs/1806.01239 | $$\sigma(y)=\int_{-\infty}^{\infty} F(x, y)-F_{bulk}(x, y)dx$$ At mean field: $$\mathcal F(\phi) = -\dfrac{a}{2}\phi + \dfrac{u}{4}\phi^4+\dfrac{\kappa}{2} \dfrac{d^2\phi}{dy^2}$$ We define $F$ as $\mathcal{F}(\phi_{MF})$. $$\dfrac{\delta \mathcal F}{\delta \phi} = -a\phi + u\phi^3-\kappa \dfrac{d^2\phi}{dy^2}=0\Rightarrow \phi_{MF}(x)=\tanh(x/(2\xi))$$ with $\xi=\sqrt{\kappa/(2a)}$. It leads to: $$\sigma = \int_{-\infty}^{\infty}\kappa \left(\dfrac{d \phi}{dx}\right)^2dx$$ Here: $\sigma=\dfrac{\sqrt{\kappa (2a)^2}}{3 u}$ | See structure factor line. We cannot measure directly $\phi_{MF}=tanh(...)$ because this is really a result that makes sense for a field theory but not for a particle system. The measured of the interface will depend on the binning. Indeed, the intrinsic width $\xi$ defined in the $\tanh(x/(2\xi))$ is somehow impossible to measure "directly". See however: https://arxiv.org/abs/1904.11726 | Using the typical Ansatz to map active field theory to regular field theory: $$\sigma_{\text{cw}}(q) = \sigma_{\lambda} + \frac{3 \zeta}{4} \int dx_1 dx_2 \frac{(x_1 - x_2)}{x_1 - x_2} \frac{\psi'(x_1) \phi'^2(x_2)}{e^{q(x_1 - x_2)}}$$ $$\sigma_{\lambda} = \kappa \int dx \phi'(x) \psi'(x)$$ $$\psi=\kappa \dfrac{e^{(\zeta - 2\lambda)\phi/K}-1}{\zeta - 2\lambda}$$ $\sigma_{cw}$ is the surface tension ruling capillary wave dynamics. It can be negative, in which case the flat interface is unstable. In the limit $\zeta,\lambda\to 0$, the equilibrium result is recovered. https://link.aps.org/accepted/10.1103/PhysRevLett.127.068001 | See structure factor line. We cannot measure directly $\phi_{MF}=tanh(...)$ because this is really a result that makes sense for a field theory but not for a particle system. The measured of the interface will depend on the binning. Indeed, the intrinsic width $w$ defined in the $\tanh(x/(2\xi))$ is somehow impossible to measure "directly". See however: https://arxiv.org/abs/1904.11726 | The capillary surface tension of this model is the same as the corresponding equilibrium system (same free energy) because it is a mean field picture and we do not add any non-free energy term to the equation. |
| Laplace pressure: The surface tension is the restoring force in spherical bubble. | For a spherical nucleus of radius $R$, the difference of pressure $P$ between inside and outside is given by: $P_{liquid}- P_{gas}=(d-1)\sigma/R$ | Painful to measure but probably was looked at somewhere (perhaps https://arxiv.org/pdf/1803.06159) | $$P(\phi_+) = P(\phi_-) + \frac{(d-1)\sigma_{laplace}}{R} + \mathcal{O}\left(\frac{1}{R^2}\right)$$ $$\sigma_{laplace} = \frac{K}{\zeta - 2\lambda} \left[ \zeta S_0 - 2\lambda S_1 \right]$$ In general $\sigma_{laplace}\neq \sigma_{cw}$. $\sigma_{laplace}<0$ leads to reversed Ostwald ripening/microphase separation. Note that $\sigma_{cw}>0$ but $\sigma_{laplace}<0$ is possible and leads to a stable interface.In the limit $\zeta,\lambda\to 0$, the equilibrium result is recovered. Pressure is a slippery concept, using a slighlty different pressure definition, it is defined as equal to the Kirkwood Buff: https://arxiv.org/pdf/2407.06462 | Hell no, not doing that | Again, our equation has a free energy structure at the mean field level. Therefore, we obtain an equilibrium like surface tension arising from laplace pressure.|
| Kirkpatrick-Buff: The surface tension is the additional stress along the transverse direction required such that the stress is constant in the box (mechanical equilibrium)| $\sigma=\int_{\infty}^{\infty}P^{\perp}(x)-P^\parallel(x)dx$ | The first article finding a negative surface tension measuring this: https://arxiv.org/pdf/1412.4601 Pressure is still a slippery concept in active matter. These guys find a different expression in mips which is always positive: https://arxiv.org/abs/2308.04917 . In any case, the negativity comes from the non-zero magnetization field at the interfaces for mips systems I think | In active model B (not AMB+ just AMB): $$\sigma_{extra~stress}=\int_{\infty}^{\infty}P^{\perp}(x)-P^\parallel(x)dx= \kappa \int dx (\phi'(x))^2 R'(x)$$. Very similar to the laplace one. Can be negative. $R$ is defined here and is almost like $\psi$: https://arxiv.org/pdf/1803.06159. | Comparison equilibrium vs hyperuniform (same temperature same density). The surface tensions are really similar: | Same as equilibrium again... It's a mean field result. HOWEVER, when phases are at two different temperatures, things can get messy I think. Because all the mechanical derivations of the extra stress assume that the temperature is the same. Probably underdamped active matter people know something about that |
| Structure factor|$P[h]\sim e^{-\beta E[h]}\to S_h = T/(\sigma k^2)$|Always measured a scaling in $k^{-2}$ https://arxiv.org/pdf/2308.08531. Field theory result->|The new universality class of cesare put aside. $S_h=T_{eff}/(\sigma_{cw}k^2)$ https://link.aps.org/accepted/10.1103/PhysRevLett.127.068001 | $S_{h}\sim k^ {-1}$ for $k<\sqrt{\gamma/\eta}$ ($\eta=$viscosity), otherwise effective equilibrium $S_h\sim k^{-2}$ because not feeling the damping | $S_{h}= T/(2\sigma_{cw}k \sqrt{\gamma/\eta})$ for $k<\sqrt{\gamma/\eta}$ otherwise: $S_h\sim T/(\sigma_{cw}k^2)$|

Last note, the decay of $S_h(k, t)\sim e^{-k^3 t\sigma_{cw}/\gamma}$ for the hyperuniform system.

### Quick recap of what is below

$$S\sim \langle h_q h_{-q}\rangle~\sim q^{-d - 2\chi}$$
$$W^2\sim\langle h(x)h(x')\rangle\sim |x-x'|^{2\chi}$$
$$\langle h(t)h(t') \rangle\sim t^{2\chi/z}$$

#### THEORY:

| phase coexistence model | $S(k)$ | Dynamical critical exponent $z$ |
| ------------------------------------------- | -------- | --------------------------- |
| **Equilibrium** diffusive | $k^{-2}$ | $3$ |
| **Equilibrium** hydrodynamical (no drag)| $k^{-2}$ | $1$ |
| **Non-Equilibrium** random organization (hyperuniform bulk) | $k^{-1}$ | $3$ (diffusive because drag) |
| **Non-Equilibrium** qKPZ (perhaps mips/perhaps non equilibrium granular with noise: relevant term in RG)| $k^{-1.78<a<-1.61}$ | $2.18<z<2.79$ |
| **Non-Equilibrium** early time qKPZ | $k^{-3}$ | $1$ |

#### Measured Liquid-Solid :
| phase coexistence model | $S(k)$ | Dynamical critical exponent $z$ | Note |
| ----------------------- | -------- | ------------------------------- | --- |
| Ran ni | $k^{-2}$ | ? | We lose Hyperuniformity in the solid... Somehow |

#### Monodisperse Liquid-Gas thanks to a mips like instability :
| phase coexistence model | $S(k)$ | Dynamical critical exponent $z$ | Note |
| --------------------------- | -------- | ------------------------------- | ---- |
| With damping (hyperuniform) | $k^{-1}$ | ? | Surprising that with hyperuniormity it is not cst. However, the theory below shows how |
| Without damping | $k^{-3}$ | ? | Momentum is conserved, looks like early time qKPZ but Nardini says that when the momentum is conserved, the non linearity relevant in the RG cannot be created |
| With noise | $k^{\sim 1.55}$ | ? | Could definitively be qKPZ |

#### Liquid Liquid:
| phase coexistence model | $S(k)$ | Dynamical critical exponent $z$ | Note |
| --------------------------- | -------- | ------------------------------- | ---- |
| Ran ni dynamics With damping (hyperuniform) | $k^{-1}$ | ? | nice |
|Ran ni dynamics Without damping | ?| ? | |
|Equilibrium with noise|$k^{-2}$|$3$|Expected diffusive behavior|

### Derivation of an hyperuniform interface: explaining $S\sim q^{-1}$.
_______
#### Signs and prefactor are most likely wrong
_______

We observed below that, when the bulk of the phases in a phase separating system are hyperuniform, the structure factor of the interface $S(k)=\langle h_k h_{-k} \rangle$ behaves as $k^{-1}$ instead of $k^{-2}$ at equilibrium. Below we will show that for a model with hyperuniformity induced by momentum/COM conserving noise and global damping, it seems to be what is predicted theoretically.

We start from the model B of levine with a noise that conserves the center of mass:
$$
\partial_t \rho = \nabla^2 \frac{\delta F}{\delta \rho} + \nabla^2\eta
$$

with $F = \displaystyle{\int d \boldsymbol{ r}\left(\dfrac{1}{2}\left(\nabla \rho\right)^2 + V(\rho)\right)}$ the first term is a surface tension/diffusion (at some point I should include back $\sigma$ in it and not normalize). $V$ is such that we can obtain a phase coexistence (two minima).

The density follows (sign issue somewhere i think):

$$\partial_t \rho = \nabla^2\left(\nabla^2\rho - V'(\rho) - \eta\right)$$

with $\langle \xi(\boldsymbol r, t) \xi(\boldsymbol r', t')\rangle=2D\delta(\boldsymbol r-\boldsymbol r')\delta(t-t')$

Assuming a roughly flat interface in the plane perpendicular to $\hat{ r}_\perp$, we note the height of the interace $h(\boldsymbol r_\parallel)$ and use the following Ansatz:

$$\rho(\boldsymbol r, t) = f(r_\perp- h(\boldsymbol r_\parallel, t))$$

It should represent well the order parameter. Moreover, it will allow us to easily neglect terms in high power of $(\nabla h)^2$ (altough note that they could be relevant at the gaussian point in the RG sense). The interface being almost flat, we note that the function describing the change of order paramerer is almost a heaviside function interpolating between the gas and the liquid densities: $f(x)\sim \rho_g + (\rho_l - \rho_g )\Theta_{dx}(x)$. We regularizes it with a width $dx$. We note that the derivative is rougly a delta function with width $dx$ and height $\rho_l-\rho_g$: $f'(x)\sim (\rho_l - \rho_g )\delta_{dx}(x)$.

Puting the Ansatz into the evolution equation leads to:

\begin{split}
\left[\partial_th(\boldsymbol r_\parallel, t)\right]f'(r_\perp- h(\boldsymbol r_\parallel, t))=&-\nabla^2\left[\nabla_\parallel^2f(r_\perp- h(\boldsymbol r_\parallel, t))+\nabla_\perp^2f(r_\perp- h(\boldsymbol r_\parallel, t))\right.\\&\left.-V'(f(r_\perp- h(\boldsymbol r_\parallel, t)))+\eta(\boldsymbol r_\parallel, r_\perp, t)\right]\\
=&-\nabla^2\left[- h'' f'(r_\perp- h(\boldsymbol r_\parallel, t))+ (1+(h')^2)f''(r_\perp- h(\boldsymbol r_\parallel, t))\right. \\
&\left.-V'(f(r_\perp- h(\boldsymbol r_\parallel, t))) +\eta(\boldsymbol r_\parallel, r_\perp, t)\right]
\end{split}

We could also expand the other $\nabla^2$ but it seems to make everyhting very complicated (and also i don't obtain the result i want :) Neither do i obtain the results by bray If if do that. As Bray vs me in raw intellect, there is no question, I'll just follow him :) :) :) :) ) .
For simplicity, we change variable: $u(r_\perp, \boldsymbol r_\parallel, t)=r_\perp - h(\boldsymbol r_\parallel, t)$. We then multiply by $f'u (-\nabla^2)^{-1}$ and integrate over $u$ (we could directly integrate without multiplying anything, but it makes everything a mess). Note that such procedure effectively project our dynamics to the one of the interface since $f'(u)$ is more or less a dirac delta centered at the interface. Note also that $(-\nabla^2)^{-1}$ has to be understood as the functional inverse of the laplacian, i.e the Green function of the laplacian.

\begin{split}
\int du f'(u)(-\nabla^2)^{-1}\left(\left[\partial_th\right]f'(u)\right)=\int du\left(- h'' f'(u)^2+ (1+(h')^2)f'(u)f''(u) -V'(f(u))f'(u) +f'(u)\eta(\boldsymbol r_\parallel, r_\perp, t)\right)
\end{split}

Let's proceed term by term, from the simplest to the most complicated using the simple rules $f'(u)G(u)\simeq f'(u)G(0)$:

- $\int du f'(u)f''(u) \propto \int du (f(u)^2)'=0$ because $f'(\pm \infty) = 0$.
- $\int du V'(f(u))f'(u)\propto \int du (V(f(u)))'=V(\rho_g)-V(\rho_l)$. My stat mech is a bit rusty :) But I guess that both are at the same value of $V$ (altough I remember having in minde the common tangent argument.)
- $\xi=\int du f'(u)\eta(\boldsymbol r_\parallel, u+h, t)$. This creates a new noise. Approximating $u+h\simeq u$ we obtain that this new noise has correlation $\langle \xi(\boldsymbol r_\parallel, t)\xi(\boldsymbol r'_\parallel, t') \rangle \simeq 2D\sigma \delta(\boldsymbol r_\parallel - \boldsymbol r'_\parallel)\delta( t-t')$ (somehow, Bray says that $\int du (f'(u))^2 = \sigma$. See also [this](https://journals.aps.org/pre/pdf/10.1103/PhysRevE.81.031602?casa_token=90SRms0lZkIAAAAA%3A11elHewPwmv0XguDAg4uIvGL4-fMe5GmF6bQ9bLL3cXc8u4bD2xZTlSe0k45WVJOQJbeNnX8HkkBsQ) and [this](https://books.google.fr/books/about/Molecular_Theory_of_Capillarity.html?id=_ydSF_XUVeEC&redir_esc=y) ). I guess we could make the approximation: $\eta(u+h)\simeq \eta(u) + h'\eta(u)$. This would not change the final result at large lenghtscale since it would only add a noise with correlation in $k$ somehting. (with somehow assuming that everything is well defined and that the god of stochastic calculus is with me).
- $\int du f'(u)(-\nabla^2)^{-1}\left(\left[\partial_th(\boldsymbol r_\parallel)\right]f'(u)\right)$. We define $G(r_\perp, \boldsymbol r_\parallel)$ as the green function of $-\nabla^2$. Such that $-\nabla^2 f(x) = g(x)\Rightarrow f(x)=(-\nabla^2)^{-1}g(x)=\int G(x, x') g(x') dx'$:
$$-\left(\nabla^2_{\parallel}+\nabla^2_{\perp}\right)G(r_\perp-r'_\perp, \boldsymbol r_\parallel-\boldsymbol r'_\parallel)=\delta(r_\perp-r'_\perp)\delta( \boldsymbol r_\parallel-\boldsymbol r'_\parallel)$$
We take the partial fourier transform:
$$\left(\boldsymbol k_\parallel^2+\nabla^2_{\perp}\right)G(r_\perp-r'_\perp, \boldsymbol k_\parallel)=-\delta(r_\perp-r'_\perp)$$
This leads to: $G(r_\perp-r'_\perp, \boldsymbol k_\parallel)=\dfrac{1}{2|\boldsymbol k_\parallel|}e^{-|\boldsymbol k_\parallel||r_\perp-r'_\perp|}$
We proceed by taking the fourier transform along the parallel direction of the term we started with:
\begin{split}
\int du f'(u)(-\nabla^2)^{-1}\left(\left[\partial_th(\boldsymbol k_\parallel)\right]f'(u)\right)&=\int du\int dv f'(u)G(u-v, \boldsymbol k_\parallel) f'(v)\left[\partial_th(\boldsymbol k_\parallel)\right]\\
&\simeq(\rho_l - \rho_g)^2G(0, \boldsymbol k_\parallel)=\dfrac{(\rho_l - \rho_g)^2}{2|\boldsymbol k_\parallel|}
\end{split}
Where we used the fact that $f'(u)\simeq \Delta\rho\delta(u)$

Putting all of that together:

$$\partial_t h_{\boldsymbol k_\parallel}=\dfrac{2\sigma}{\Delta \rho^2}|\boldsymbol k_\parallel|^3h_{\boldsymbol k_\parallel}+\dfrac{2}{\Delta \rho^2}\sqrt{2\sigma D|\boldsymbol k_\parallel|^2}\zeta_{\boldsymbol k_\parallel}$$

With $\zeta$ white correlated. This leads to $S\sim k^{-1}$

### Model H (needed for binary mixture demixing)

#### Introduction
Note that we did not take into account hydrodynamic interactions before. We will do this now (we do not expect it to change anything, we simply want to flex a little, and ensure that model B with center of mass conservation was right). We will also discuss about liquid-liquid phase separation and binary mixture.

Model $H$ reads:

$$
\begin{split}
\partial_t \phi + \boldsymbol v \cdot \boldsymbol \nabla \phi &= M\nabla^2\mu + \sqrt{2MD}\boldsymbol \nabla\cdot \boldsymbol \eta\\
\rho\left(\partial_t\boldsymbol v + (\boldsymbol v \cdot \boldsymbol \nabla)\boldsymbol v \right)&=\eta \nabla^2\boldsymbol v-\boldsymbol\nabla p - \phi \boldsymbol \nabla \mu +\boldsymbol\nabla\cdot \boldsymbol\sigma\\
\boldsymbol \nabla \cdot \boldsymbol v &= 0
\end{split}
$$

Where $\boldsymbol \sigma$ is a random white stress such that:

$$\langle \sigma_{ij}(\boldsymbol r, t)\sigma_{kl}(\boldsymbol r', t')\rangle=2D\eta\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right)\delta(\boldsymbol r-\boldsymbol r')\delta(t-t').$$

$\eta$ is also a white noise (with unit variance). $\phi$ is an order parameter for a binary mixture. For example $(\rho_A-\rho_B)/(\rho_A+\rho_B)$.

$\mu$ derives from a free energy:

$$\begin{split}
\mu &= \frac{\delta F}{\delta \phi}\\
F&=\int d\boldsymbol r \left(V(\phi) + \frac{K}{2}\left(\nabla \phi\right)^2\right)\\
f(\phi)&=\dfrac{(\phi-1)^2(\phi+1)^2}{4}\\
\mu&=\phi(\phi^2-1)-K\nabla^2\phi\\
\nabla^2\mu &=(3\phi^2 - 1)\nabla^2\phi+6\phi(\boldsymbol \nabla \phi)^2 + K\nabla^4\phi
\end{split}$$

A number of remarks have to be made:

- The fluid is assumed to be incompressible $\boldsymbol \nabla\cdot\boldsymbol v=0$. Therefore the pressure fails to be a dynamical/thermodynamical variable and simply enforces the incompressibility condition. Which also implies that the fluid is driven by "chemical" gradient current and not by pressure. Indeed, spinodal decompisition of a monodisperse fluid should be triggered (hydrodynamically) by a negative compressibility ($\chi=\frac{1}{\rho}\frac{\partial \rho}{\partial p}$) which in turn would create an hydrodynamical instability. Note also that we did not include the isotropic contribution to the autocorrelation of the random noise tensor because the pressure is not a dynamical variable anymore. But nothing prevented us to add it.

- We have diffusion of the order parameter $\phi$. This does not happen only when the fluid is subject to random noise (and hence that there is an external thermal bath that does not conserve the momentum). To understand where this noise comes from, we hae to clearly see that all hydrodynamical interactions are mediated through $\boldsymbol v$ the **global** velocity of the fluid. Indeed, if in the equation we kept $\boldsymbol v_A$ and $\boldsymbol v_B$. Then, the field equations for $\rho_A$ and $\rho_B$ would both follow advection equation without diffusion. But here $\rho_A$ and $\rho_B$ are advected by the total velocity $\boldsymbol v$ hence, there is some velocity that is not associated to each of the specie and that enters into the diffusive part. Note altough that here (without external bath), the prefactor of the noise and of the diffusion should be dependent on $\phi$. Because of course, if we take the limit $\rho_A\to 0$ and $\rho_B\to \rho$ we should recover the Navier Stokes equation. We will do a quick detour to fluctuating hydrodynamic of binary mixture that might be instructive (I don't know yet if it is instructive or not but I want to make one).

#### Fluctuating hydrodynamics of binary mixture

We define $c=\rho_A/(\rho_A+\rho_B)$,[ we can show that](https://www.sciencedirect.com/book/9780444515155/hydrodynamic-fluctuations-in-fluids-and-fluid-mixtures#:~:text=Fluctuating%20hydrodynamics%20is%20a%20stochastic,by%20a%20fluctuation%2Ddissipation%20theorem.):

$$\frac{\partial c}{\partial t} + \boldsymbol{v} \cdot \boldsymbol\nabla c = D \nabla^2 c + D\frac{k_T}{T} \nabla^2 T +\boldsymbol \nabla \cdot J,
$$

The velocity equation is left unchanged. The temperature equation altough changes quite a lot. Let's not care since it is a highly out of equilibrium field that relaxes quickly. We also did not consider any [baro-diffusion](https://www.esaim-m2an.org/articles/m2an/pdf/2010/05/m2an0896.pdf) (I guess in the limit of an incompressible fluid it is alright). Again, we can show that $J$ has correlation:

$$\langle J_i(\boldsymbol r, t) J_j(\boldsymbol r', t')\rangle = 2D (\rho_A + \rho_B)\left(\dfrac{\partial \mu}{\partial c}\right)^{-1}\delta(t-t')\delta(\boldsymbol r - \boldsymbol r')\delta_{ij}$$

The equation for $\phi$ has the same form since, $1-c=\rho_B/(\rho_A+\rho_B)$ which means that $\phi = c - (1-c)=2c-1$.

It is unclear if the $\mu$ here is the same as the one coming from the Landau free energy (I suppose evaluated at its minimum). From the fluctuation dissipation they should be the same, somehow. In this case since $\mu=\phi(\phi^2 - 1)$, $\partial_\phi\mu = (\phi^2-1)+2\phi^2$ the noise does not vanish even if we are in a region in which $\phi=\pm 1$. Which is strange... [Taking a realistic chemical potential](https://www.esaim-m2an.org/articles/m2an/pdf/2010/05/m2an0896.pdf), we obtain $(\partial_c\mu)^{-1}\propto (c-1)c$. That is, it is 0 if $c = 0$ or $c=1$.

Note that [Cates says](https://www.cambridge.org/core/journals/journal-of-fluid-mechanics/article/theories-of-binary-fluid-mixtures-from-phaseseparation-kinetics-to-active-emulsions/5BD133CB20D89F47E724D77C296FEF80) that a more realistic free energy for liquid liquid/binary phase separation is:

$$f(\phi)=-a\dfrac{\phi^2}{2}-T\left((1-\phi)\log(1-\phi) + \phi\log(\phi)\right)$$

Which looks like a flory hugging free energy, but I don't know a single thing about these... The two phases seems to correspond to $\phi=0$ or $\phi = 1$. In this case we find $(\partial_\phi \mu)^{-1}(\phi=0\text{ or }\phi = 1)=0$ since:

$$\partial_\phi \mu\equiv \partial_{\phi\phi}f(\phi)=-a-T\left(\dfrac{1}{1-\phi}+\dfrac{1}{\phi}\right)$$

Therefore, I think that the messed up correlations of the noise comes from the $\phi^4$ theory we are using.

#### Derivation of the interface equation for model H

Anyway, we start from model $H$ and we neglect diffusion. I think that, it can really really messes up the analysis exactly as adding a white noise on random organization model destroys hyperuniformity. However, as a first approxmiation we will assume that it can bi ignored because it plays a role only at the interface while the momentum conserving noise act on the bulk which is coupled to the niterface through hydrodynamic forces via the Oseen tensor. We will also add a global damping such that we recover the model B with center of mass conservation in the limit of strong damping and such that the fluctuation dissiation theorem is broken. We again assume incompressibility to easily deal with our problem. The model is therefore:
$$
\begin{split}
\partial_t \phi + \boldsymbol v \cdot \boldsymbol \nabla \phi &= 0\\
\rho\left(\partial_t\boldsymbol v + (\boldsymbol v \cdot \boldsymbol \nabla)\boldsymbol v \right)&=-\gamma \boldsymbol v + \eta \nabla^2\boldsymbol v-\boldsymbol\nabla p - \phi \boldsymbol \nabla \mu +\boldsymbol\nabla\cdot \boldsymbol\sigma\\
\boldsymbol \nabla \cdot \boldsymbol v &= 0
\end{split}
$$

First, we assume the convective term to be negligible such that we end up with Stokes equation which is linear $(\boldsymbol v \cdot \boldsymbol \nabla)\boldsymbol v=0$. This should hold when the amplitudes of waves are small. We will also get rid of the time derivative even if it should play some role at the very end In any case, it is easy to incorporate it back in the equation. In $k$ space we obtain:

$$
\begin{split}
(\gamma +\eta\boldsymbol k^2) \boldsymbol v_k-i\boldsymbol k p_k &= \boldsymbol f_k\\
\boldsymbol k\cdot \boldsymbol v_k=0
\end{split}
$$

with $\boldsymbol f=\mathcal{F}(- \phi \boldsymbol \nabla \mu +\boldsymbol\nabla\cdot \boldsymbol\sigma)_k$. We obtain from the incompressibility:

$$p_k=i(\boldsymbol f_k\cdot \boldsymbol k)/\boldsymbol k^2.$$

This leads to:

$$
(\gamma +\eta\boldsymbol k^2) \boldsymbol v_k+\boldsymbol k \dfrac{\boldsymbol f_k\cdot \boldsymbol k}{\boldsymbol k^2} = \boldsymbol f_k
\Rightarrow \boldsymbol v_k=\dfrac{1}{\gamma + \eta \boldsymbol k^2}\left(\mathbb{I}-\frac{\boldsymbol k \otimes\boldsymbol k}{\boldsymbol k^2} \right)\boldsymbol f\equiv \boldsymbol{T}_k\boldsymbol f
$$

with $\boldsymbol T_k$ the (modified) Oseen tensor. Note that including the time dependence is fairly easy in $w$ space. It suffices to $\gamma\to \gamma -iw$.
##### Quick note on Oseen Tensor. Can skip.

Note that the usual Oseen tensor is not well defined in 2D (physically, mathematically as a green tensor its fine) because its isotropic part reads $\boldsymbol{T}^{\text{iso}}(\boldsymbol r)\sim \log(|\boldsymbol r|)$ which diverges on large lenght scale (in 3D it decays as $r^{-1}$). Here, the $\gamma$ imposes a cutoff. The isotropic is immediatelly found because it is, as usual, the Hankel transform of the isotropic part of the tensor. We obtain (asymptotics by chat GPT, pretty sure bheavior is alright if I remember correctly but check later prefactors)
$$
\boldsymbol{T}^{\text{iso}}(\boldsymbol r)\sim \dfrac{K_0(|\boldsymbol r|\sqrt{\gamma/\eta})}{\eta}\sim\begin{cases}
-\dfrac{\ln(|\boldsymbol{r}| \sqrt{\gamma/\eta})}{\eta}, & |\boldsymbol{r}| \to 0, \\
\dfrac{1}{\eta} \sqrt{\dfrac{\pi}{2 |\boldsymbol{r}| \sqrt{\gamma/\eta}}} e^{-|\boldsymbol{r}| \sqrt{\gamma/\eta}}, & |\boldsymbol{r}| \to \infty.
\end{cases}
$$

with $K_0$ the 0th bessel function of second kind. Everything is well defined in the large system size limit. As in the Mermin-Wagner case, $\gamma$ acts as a mass that regularizes the IR sector.
##### End of note on Oseen Tensor

Using the Oseen tensor, we have a closed expression for $\phi$:

$$\partial_t \phi + \boldsymbol v \cdot \boldsymbol \nabla \phi= 0$$

since we know $\boldsymbol v(\boldsymbol r)=\int\boldsymbol T(\boldsymbol r'-\boldsymbol r)\boldsymbol f(\boldsymbol r')d\boldsymbol r'$.

$$
\begin{split}
\partial_t \phi(\boldsymbol r) &= - \int \partial_i \phi(\boldsymbol r) T_{ij}(\boldsymbol r'-\boldsymbol r) f_j(\boldsymbol r')d\boldsymbol r' \\
&= \int \partial_i \phi(\boldsymbol r) T_{ij}(\boldsymbol r'-\boldsymbol r) \phi(\boldsymbol r') \partial_i \mu(\boldsymbol r') d\boldsymbol r'+\zeta(\boldsymbol r)\\
\end{split}
$$

We defined the noise:
$$\zeta(\boldsymbol r) = \int \partial_i \phi(\boldsymbol r) T_{ij}(\boldsymbol r-\boldsymbol r')\partial_k\sigma_{kj}(\boldsymbol r')d\boldsymbol r'$$

we recall that:
$$\langle \sigma_{ij}(\boldsymbol r, t)\sigma_{kl}(\boldsymbol r', t')\rangle=2D\eta\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}\right)\delta(\boldsymbol r-\boldsymbol r')\delta(t-t').$$

I now realize that my notations with underscores for fourier transform are stupids. So i wont use them anymore :). To find the autocorrelation of the noise we go to Fourier Space:

$$
\begin{split}
\zeta(\boldsymbol k) &= -k_i\phi(\boldsymbol k) \star (T_{ij}(\boldsymbol k)k_k\sigma_{kj}(\boldsymbol k))\\
&=\int (\boldsymbol k_i'-\boldsymbol k_i)\phi(\boldsymbol k'-\boldsymbol k)T_{ij}(\boldsymbol k')k_k'\sigma_{kj}(\boldsymbol k')d\boldsymbol k'\\
\end{split}
$$

By direct computation we can show that (check fourier coeff) $\langle k_i\sigma_{ij}(\boldsymbol k, t)q_l\sigma_{lp}(\boldsymbol q, t')\rangle= 4\eta D\boldsymbol k^2\delta_{lp}\delta(\boldsymbol k + \boldsymbol q)\delta(t-t')$ which leads to:

$$
\begin{split}
\langle\zeta(\boldsymbol q, t) \zeta(\boldsymbol k, s)\rangle &=4\eta D\int\int (\boldsymbol k_i'-\boldsymbol k_i)(\boldsymbol q_l'-\boldsymbol q_l)\phi(\boldsymbol k'-\boldsymbol k)\phi(\boldsymbol q'-\boldsymbol q)T_{ij}(\boldsymbol k')T_{lp}(\boldsymbol q')\boldsymbol{k}'^2\delta(t-s)\delta(\boldsymbol k'+\boldsymbol q')\delta_{pj}d\boldsymbol k'd\boldsymbol q'\\
&=4\eta D\int\int (\boldsymbol k_i'-\boldsymbol k_i)(\boldsymbol q_l'-\boldsymbol q_l)\phi(\boldsymbol k'-\boldsymbol k)\phi(\boldsymbol q'-\boldsymbol q)T_{ij}(\boldsymbol k')T_{lj}(\boldsymbol q')\boldsymbol{k}'^2\delta(t-s)\delta(\boldsymbol k'+\boldsymbol q')d\boldsymbol k'd\boldsymbol q'
\end{split}
$$

We evaluate the product of Oseen tensor (using the $\delta(\boldsymbol q'+ \boldsymbol k')$):

$$
\begin{split}
T_{ij}(\boldsymbol k)T_{lj}(-\boldsymbol k)\boldsymbol{k}^2&=\dfrac{\boldsymbol k^2}{(\gamma + \eta \boldsymbol k^2)^2}\left(\delta_{ij}-\dfrac{k_ik_j}{\boldsymbol k^2}\right)\left(\delta_{lj}-\dfrac{k_lk_j}{\boldsymbol k^2}\right)\\
&=\dfrac{\boldsymbol k^2}{(\gamma + \eta \boldsymbol k^2)^2}\left(\delta_{il}-\dfrac{k_ik_l}{\boldsymbol k^2}-\dfrac{k_ik_l}{\boldsymbol k^2}+\dfrac{k_ik_jk_lk_j}{\boldsymbol k^2}\right)\\
&=\dfrac{\boldsymbol k^2}{(\gamma + \eta \boldsymbol k^2)^2}\left(\delta_{il}-\dfrac{k_ik_l}{\boldsymbol k^2}\right)
\end{split}
$$

We call this new tensor $\tilde {\boldsymbol T}$:

$$\tilde {\boldsymbol T}(\boldsymbol k)=\dfrac{\boldsymbol k^2}{(\gamma + \eta \boldsymbol k^2)^2}\left(\mathbb{1}-\dfrac{\boldsymbol k\otimes \boldsymbol k}{\boldsymbol k^2}\right) $$

Note that when $\gamma = 0$ or more generally when the noise respect the FDT, $\tilde {\boldsymbol T}={\boldsymbol T}$.

We continue:
$$
\begin{split}
\langle\zeta(\boldsymbol q, t) \zeta(\boldsymbol k, s)\rangle &=4\eta D\int\int (\boldsymbol k_i'-\boldsymbol k_i)(\boldsymbol q_l'-\boldsymbol q_l)\phi(\boldsymbol k'-\boldsymbol k)\phi(\boldsymbol q'-\boldsymbol q)\tilde T_{ij}(\boldsymbol q')\delta(t-s)\delta(\boldsymbol k'+\boldsymbol q')d\boldsymbol k'd\boldsymbol q'
\end{split}
$$

Therefore, in real space, it reads (the therefore is a bit too much, it's complicated and easier to show in real space):
$$\langle\zeta(\boldsymbol r, t) \zeta(\boldsymbol r', s)\rangle=4\eta D \partial_i\phi(\boldsymbol r) \tilde T_{ij}(\boldsymbol r - \boldsymbol r')\partial_j\phi(\boldsymbol r')\delta(t-s)$$

The final thing we use is that the Oseen Tensor is divergence free. It follows from the incompressibility condition. Therefore we can integrate by part the equation for $\phi$.

$$
\begin{split}
\partial_t \phi(\boldsymbol r) &= \int \partial_i \phi(\boldsymbol r) T_{ij}(\boldsymbol r'-\boldsymbol r) \phi(\boldsymbol r') \partial_i \mu(\boldsymbol r') d\boldsymbol r'+\zeta(\boldsymbol r)\\
&= -\int \left[\partial_i \phi(\boldsymbol r) T_{ij}(\boldsymbol r'-\boldsymbol r) \partial_i\phi(\boldsymbol r') \right] \mu(\boldsymbol r') d\boldsymbol r'+\zeta(\boldsymbol r)
\end{split}
$$

We use again the Ansatz: $\phi(\boldsymbol r, t)=f(r_\perp- h(\boldsymbol r_\parallel, t))$ and proceed mostly as before.

$$
\begin{split}
(\partial_t h(\boldsymbol r_\parallel))f'(r_\perp- h(\boldsymbol r_\parallel, t))
&= -\int \left[\partial_i f(r_\perp- h(\boldsymbol r_\parallel, t)) T_{ij}(\boldsymbol r'_\parallel-\boldsymbol r_\parallel, r_\perp - r'_\perp) \partial_if(r_\perp'- h(\boldsymbol r_\parallel', t)) \right]\\
&~ ~ ~ ~\times\left[- h'' f'(r_\perp'- h(\boldsymbol r_\parallel', t))+ (1+(h')^2)f''(r_\perp'- h(\boldsymbol r_\parallel', t))-V'(f(r_\perp'- h(\boldsymbol r_\parallel', t))) \right] d\boldsymbol r'+\zeta(\boldsymbol r_\parallel, r_\perp)
\end{split}
$$

As before, we will multiply with $f'$ and integrate. Before that, since $\mu$ is already of order $\mathcal{O}(\nabla^2 h)$ at best, we don't have to worry to much about a strickly well made expansion. We assume the interface to be flat in the convolution. This means that we effectively take:

$$\partial_i\phi(\boldsymbol r)=f'(r_\perp- h(\boldsymbol r_\parallel))e^\perp$$

Moreover, for simplicity we perform the change of variable $u = r_\perp - h(r_\parallel)$ (Note to me, find better notation to not confuse the prime for derivative with the prime for dummy variable):

$$
\begin{split}
(\partial_t h(\boldsymbol r_\parallel))f'(u, t)
&=- \int \left[ f'(u, t) T_{\perp \perp}(\boldsymbol r'_\parallel-\boldsymbol r_\parallel, u -u' -(\overbrace{h(r_\parallel) - h(r_\parallel')}^{0}) f'(u', t)) \right]\\
&~ ~ ~ ~\times\left[- h'' f'(u', t)+ (1+(h')^2)f''(u', t)-V'(f(u', t)) \right] d\boldsymbol r'+\zeta(\boldsymbol r_\parallel, u+h(r_\perp))
\end{split}
$$

We multiply and integrate with respect to $f'(u)$ (**not caring about prefactor at all**):

$$
\begin{split}
\sigma\partial_t h(\boldsymbol r_\parallel)
&= -\int \left[ \left(\int f'(u, t)^2 T_{\perp \perp}(\boldsymbol r'_\parallel-\boldsymbol r_\parallel, u -u') du\right) f'(u', t)) \right]\\
&~ ~ ~ ~\times\left[- h'' f'(u', t)+ (1+(h')^2)f''(u', t)-V'(f(u', t)) \right] d\boldsymbol r'+\int f'(u)\zeta(\boldsymbol r_\parallel, u+h(r_\perp))du\\
&\simeq\sigma^2\int T_{\perp\perp}(\boldsymbol r'_\parallel-\boldsymbol r_\parallel, r_\perp=0)\nabla^2_\parallel h(\boldsymbol r'_\parallel)dr'+\sigma \eta
\end{split}
$$
Where $$\langle\eta(\boldsymbol q_\parallel, t) \eta(\boldsymbol k_\parallel, s)\rangle=4D\sigma^2\tilde T_{\perp\perp}(\boldsymbol q_\parallel, r_\perp=0)\delta(\boldsymbol q_\parallel+\boldsymbol k_\parallel)\delta(t-s)$$

We finally obtain the linear equation for the Interface:

$$\partial_th(\boldsymbol k_\parallel, t)=\sigma T_{\perp\perp}(\boldsymbol k_\parallel, r_\perp=0)\boldsymbol k_\parallel^2h(\boldsymbol k_\parallel, t)+\eta(\boldsymbol k_\parallel, t)$$

#### What are $T_{\perp\perp}(\boldsymbol q_\parallel, r_\perp=0)$ and $\tilde T_{\perp\perp}(\boldsymbol q_\parallel, r_\perp=0)$?

We recall:
$$
\begin{split}
T_{\perp\perp}(\boldsymbol q_\parallel, q_\perp)&=\dfrac{1}{\gamma + \eta (\boldsymbol q_\parallel^2 + q_\perp^2)}\left(1-\dfrac{q_\perp ^2}{(\boldsymbol q_\parallel^2 + q_\perp^2)^2}\right)\\
\tilde T_{\perp\perp}(\boldsymbol q_\parallel, q_\perp)&=\dfrac{\boldsymbol q_\parallel^2 + q_\perp^2}{(\gamma + \eta (\boldsymbol q_\parallel^2 + q_\perp^2))^2}\left(1-\dfrac{q_\perp^2}{(\boldsymbol q_\parallel^2 + q_\perp^2)^2}\right)
\end{split}
$$

Mighty Mathematica gives:
$$
\begin{split}
T_{\perp\perp}(\boldsymbol q_\parallel, r_\perp)&=\dfrac{|\boldsymbol q_\parallel|}{2\gamma}\left(e^{-|\boldsymbol q_\parallel| |r_\perp|}-|\boldsymbol q_\parallel|\dfrac{1}{\sqrt{\gamma/\eta + \boldsymbol q_\parallel^2}}e^{-\sqrt{\gamma/\eta +\boldsymbol q_\parallel^2}|r_\perp|}\right)\\
\tilde T_{\perp\perp}(\boldsymbol q_\parallel, r_\perp)&=\dfrac{\boldsymbol q_\parallel^2e^{-\sqrt{\gamma/\eta}|r_\perp|}}{4\eta(\gamma + \boldsymbol q_\parallel^2\eta)}\left(|r_\perp|+1/\sqrt{\gamma/\eta+\boldsymbol q_\parallel^2}\right)
\end{split}
$$

Therefore:
$$
\begin{split}
T_{\perp\perp}(\boldsymbol q_\parallel, r_\perp=0)&=\dfrac{|\boldsymbol q_\parallel|}{2\gamma}\left(1-|\boldsymbol q_\parallel|\dfrac{1}{\sqrt{\gamma/\eta + \boldsymbol q_\parallel^2}}\right)=\beta |\boldsymbol q_\parallel|+\mathcal{O}(\boldsymbol q_\parallel^2)\\
\tilde T_{\perp\perp}(\boldsymbol q_\parallel, r_\perp=0)&=\dfrac{\boldsymbol q_\parallel^2}{4\eta(\gamma + \boldsymbol q_\parallel^2\eta)\sqrt{\gamma/\eta+\boldsymbol q_\parallel^2}}=\alpha \boldsymbol q_\parallel^2 + \mathcal{O}(\boldsymbol q_\parallel^4)
\end{split}
$$

This gives our final result for model H with damping:

$$\partial_t h(\boldsymbol q, t)=a |\boldsymbol q|^3h(\boldsymbol q, t) + \sqrt{b\boldsymbol q^2}\eta(\boldsymbol q, t)$$

which again shows $S(k)\sim k^{-1}$.

Note that in the limit $\gamma\to 0$, the usual result of model $H$ is recovered.

### Conservative model B with noise (Andrea's point) (can skip)

We will deal with this model:
$$\partial_t \rho = -\nabla^2\left(\nabla^2\rho - V'(\rho) + \eta\right)+\boldsymbol \nabla\cdot\boldsymbol \zeta$$

where both $\eta$ and $\zeta$ are white noise with correlation proportional to $D_{loc}$ and $D_{com}$ respectively.

We perform the same trick as before again not caring about prefactor too much:
$$
\begin{split}
\int du f'(u)(-\nabla^2)^{-1}\left(\left[\partial_th\right]f'(u)\right)=\sigma h''+\int du f'(u)\eta(\boldsymbol r_\parallel, r_\perp, t)+\int du f'(u)(-\nabla^2)^{-1}\boldsymbol \nabla\cdot\boldsymbol \zeta
\end{split}
$$

We will focus now on the divergence of $\zeta$:

$$
\begin{split}
(-\nabla^2)^{-1}\boldsymbol \nabla\cdot \boldsymbol\zeta(\boldsymbol r)=\int d\boldsymbol r' G(\boldsymbol r - \boldsymbol r')[\nabla\cdot \boldsymbol\zeta](\boldsymbol r')
\end{split}
$$

We now take the fourier transform along $\boldsymbol r_\parallel$:

$$
\begin{split}
\mathcal{F}_\parallel\left\{(-\nabla^2)^{-1}\boldsymbol \nabla\cdot \boldsymbol\zeta(\boldsymbol r)\right\}(\boldsymbol k_\parallel, r_\perp)&=\int d\boldsymbol r_\parallel e^{i\boldsymbol k_\parallel\cdot\boldsymbol r_\parallel}\int d\boldsymbol r'G(\boldsymbol r_\parallel - \boldsymbol r'_\parallel, r_\perp - r'_\perp)\left(\boldsymbol\nabla'_\parallel\cdot\boldsymbol \zeta_\parallel(\boldsymbol r')+\nabla'_\perp\zeta(\boldsymbol r')\right)\\
&=\int dr'_\perp G(\boldsymbol k_\parallel, r_\perp-r_\perp')(-i\boldsymbol k_\parallel\cdot\boldsymbol\zeta_\parallel(\boldsymbol k_\parallel, r_\perp')+\nabla'_\perp\zeta(\boldsymbol k_\parallel, r_\perp'))\\
&=\int dr'_\perp \dfrac{1}{2|\boldsymbol k_\parallel|}e^{-|\boldsymbol k_\parallel||r_\perp-r'_\perp|}(-i\boldsymbol k_\parallel\cdot\boldsymbol\zeta_\parallel(\boldsymbol k_\parallel, r_\perp')+\nabla'_\perp\zeta(\boldsymbol k_\parallel, r_\perp'))\\
\end{split}
$$

We call this noise $\xi$:
$$
\begin{split}
\langle \xi(\boldsymbol k_\parallel, r_\perp) \xi(\boldsymbol q_\parallel, x_\perp)\rangle&=\int dr'_\perp\int dx'_\perp \dfrac{1}{4|\boldsymbol k_\parallel||\boldsymbol q_\parallel|}e^{-|\boldsymbol k_\parallel||r_\perp-r'_\perp|-|\boldsymbol q_\parallel||x_\perp-x'_\perp|}\\
&\times\left\langle-k_\parallel^i\zeta_\parallel^i(\boldsymbol k_\parallel, r_\perp')q_\parallel^j\zeta_\parallel^j(\boldsymbol q_\parallel, x_\perp')+\nabla'_\perp\zeta(\boldsymbol k_\parallel, r_\perp')\nabla'_\perp\zeta(\boldsymbol q_\parallel, x_\perp')\right\rangle\\
&=2D_{com}\int dr'_\perp\int dx'_\perp \dfrac{1}{4|\boldsymbol k_\parallel|^2}e^{-|\boldsymbol k_\parallel||r_\perp-r'_\perp|-|\boldsymbol k_\parallel||x_\perp-x'_\perp|}\\
&\times\left(\boldsymbol k_\parallel^2+\nabla_\perp^2\right)\delta(\boldsymbol k_\parallel+\boldsymbol q_\parallel)\delta(x_\perp'-r_\perp')
\end{split}
$$

The fist term is easy to compute:

$$
\begin{split}
(1)&\propto\int dr'_\perp\int dx'_\perp e^{-|\boldsymbol k_\parallel||r_\perp-r'_\perp|-|\boldsymbol k_\parallel||x_\perp-x'_\perp|}\delta(\boldsymbol k_\parallel+\boldsymbol q_\parallel)\delta(x_\perp'-r_\perp')\\
&=\int dr'_\perp e^{-|\boldsymbol k_\parallel||r_\perp-r'_\perp|-|\boldsymbol k_\parallel||x_\perp-r'_\perp|}\delta(\boldsymbol k_\parallel+\boldsymbol q_\parallel)\\
&=\dfrac{1}{|\boldsymbol k_\parallel|}e^{-|\boldsymbol k_\parallel|r_\perp-x_\perp|}
\end{split}
$$

The second is not much complicated:
$$
\begin{split}
(2)&\propto\int dr'_\perp\int dx'_\perp \dfrac{1}{|\boldsymbol k_\parallel|^2} e^{-|\boldsymbol k_\parallel||r_\perp-r'_\perp|-|\boldsymbol k_\parallel||x_\perp-x'_\perp|}\delta(\boldsymbol k_\parallel+\boldsymbol q_\parallel)\partial_{x_\perp x_\perp}\delta(x_\perp'-r_\perp')\\
&\propto\int dr'_\perp e^{-|\boldsymbol k_\parallel||r_\perp-r'_\perp|-|\boldsymbol k_\parallel||x_\perp-r'_\perp|}\delta(\boldsymbol k_\parallel+\boldsymbol q_\parallel)\\
&=\dfrac{1}{|\boldsymbol k_\parallel|}e^{-|\boldsymbol k_\parallel|r_\perp-x_\perp|}
\end{split}
$$

We end up with our final expression for the interface dynamic (forgetting prefactor):

$$\partial_t h = |q|^3h+\sqrt{2D_{com}q}\eta+\sqrt{2D_{loc}q^2}\zeta$$

### Adding time dependence

With model H, we effectively have an underdamped dynamics while our interface is described by an overdamped dynamics. In order to take into account the underdamped nature of the system, we just have to generalize our Oseen tensor:

$$
\begin{split}
T(\boldsymbol k, w)&=\dfrac{1}{\gamma -iw + \eta \boldsymbol k^2}\left(\mathbb{I}-\frac{\boldsymbol k \otimes\boldsymbol k}{\boldsymbol k^2} \right)\\
&=\dfrac{1}{\gamma + \eta \boldsymbol k^2}\left(\mathbb{I}-\frac{\boldsymbol k \otimes\boldsymbol k}{\boldsymbol k^2} \right)-\dfrac{iw}{(\gamma + \eta \boldsymbol k^2)^2}\left(\mathbb{I}-\frac{\boldsymbol k \otimes\boldsymbol k}{\boldsymbol k^2} \right) + \mathcal{O}(w^2)
\end{split}
$$

We see that these terms are really really irrelevant because they make factors of k^4 enters at minimum.

### Tring to explain failure of solid-liquid (don't work)

We saw that at a liquid-solid phase transition, the solid fails to be hyperuniform. Maybe we could try to understand why. Take model C with $\rho$ the density and $z$ the crystallization.

The free energy describing this system should look like a double well potential (keep in mind that technically in liquid solid, we have two pressure branches that cannot technically be joined, but let's ignore that.) In this case, the free energy might be guessed to be:

$$\mathcal{F}(\rho, \phi)=\dfrac{\kappa_\rho}{2}\left(\nabla \rho\right)^2+\dfrac{\kappa_\phi}{2}\left(\nabla \phi\right)^2+\frac{\alpha}{2}\left((\rho+1)^2\phi^2+(\rho-1)^2(\phi-1)^2\right)$$

Where the well $\rho=-1$ and $\phi = 0$ corresponds to the liquid well (not so dense, and no crystallinity) and the well $\rho=1$ and $\phi = 1$ is the crystal.

In which case the evolution equation can be guessed to be:
$$
\begin{split}
\partial_t \rho &= \nabla^2 \left(\frac{\delta \mathcal F}{\delta \rho} + \eta\right)\\
\partial_t \phi&=-\frac{\delta \mathcal F}{\delta \phi}+\zeta
\end{split}
$$

Where both noises are white and the com is conserved thanks to the non equilibrium noise. This turns into:
$$
\begin{split}
\partial_t \rho &= \nabla^2 \left(-\kappa_\rho\nabla^2\rho + \partial_\rho f(\rho, \phi) + \eta\right)\\
\partial_t \phi&=\kappa_\phi\nabla^2\phi-\alpha\left((\rho-1)^2\phi+(\rho+1)^2(\phi - 1)\right)+\zeta
\end{split}
$$
Assuming $\phi$ to relax quickly (it's not conserved), we can get rid of the derivative (highly suspicious manipulation :)). We could also keep it by fourier transforming. However, we need to get rid of $\kappa_\phi$ to do anything. In which case we can isolate $\phi$:

$$\phi =\frac{\zeta+\alpha(1+\rho)^2}{2\alpha(1+\rho)(1-\rho)}$$

If we replace this thing in the equation for $\partial_t\rho$ we obtain:

$$
\partial_t\rho=\nabla^2\left(\kappa_\rho\nabla^2\rho + \partial_\rho\tilde f (\rho)+\tilde\eta\right)
$$

With:

$$\tilde f(\rho)=\dfrac{1}{2}\left(\frac{\rho^2}{2}+\frac{2}{1+\rho^2}\right)$$

which is a double well potential (no surprise):

and $$\tilde \eta = \eta+\frac{\zeta(2+(\zeta-2\rho)\rho)}{2(1+\rho^2)^2}$$
which is nastly multiplicative. Maybe this could be some role in the loss of hyperuniformity... I don't see how though..

### Renormalization group

We found our equation to be: $$\partial_t h(\boldsymbol q, t)=a |\boldsymbol q|^3h(\boldsymbol q, t) + \sqrt{b\boldsymbol q^2}\eta(\boldsymbol q, t)$$

With unstable Ran Ni model, I believe that the temperature difference betweren the phases breaks the symmetry: $h\to-h$ and allows for a non-equilibrium term. Let's try the qKPZ one:

$$\partial_t h(\boldsymbol q, t)=a |\boldsymbol q|^3h(\boldsymbol q, t) - \beta |\mathbf q|\int_0^{\Lambda}\mathbf k\cdot (\mathbf k - \mathbf q)h(\mathbf k, t)h(\mathbf k-\mathbf q, t)\dfrac{d\mathbf k}{(2\pi)^d}+\tilde\eta(\boldsymbol q, t)$$

Note that this term becomes irrelevant at $d>1$. Thus we do not expect it to play a big role. Except if there exists a strong coupling as for the KPZ or cKPZ+ equation. It's more a sanity check than anything. We could perform the renormalization procedure using the MSRJ path integral formalism. However, It's easier and equivalent to do so using a self-consistent approach based on the above equation. We rewrite it as:

$$h(\boldsymbol q, w)= G_0(\mathbf q, w)\tilde\eta(\boldsymbol q, t)+ G_0(\mathbf q, w)\int_{\mathbf q < \Lambda}\int_{-\infty}^{\infty} g_0(\mathbf q, \mathbf k,\mathbf q - \mathbf k)h(\mathbf k, \Omega)h(\mathbf k-\mathbf q, \Omega - w)\dfrac{d\mathbf k}{(2\pi)^d}\dfrac{d\Omega}{2\pi} \tag{1}$$

with the bare propagator:
$$G_0(q, w)=\dfrac{1}{-i\omega+a|\mathbf q|^3}$$

and the bare 2 vertex:

$$g_0(\mathbf q, \mathbf k, \mathbf p)=\alpha |\mathbf q|\mathbf k\cdot \mathbf p$$
Momentum conservation imposes that anytime a vertex arises in an equation: $\mathbf q = \mathbf k + \mathbf p$. $\Lambda$ is an ultraviolet cutoff. It is also useful to define the bare correlator $C_0(\mathbf q, w)$ (dynamic structure factor of the linear theory):

$$
\begin{split}
G_0(\mathbf q, w)\langle \tilde \eta(\mathbf q, w)\tilde\eta(\mathbf k, \omega)\rangle G_0(\mathbf k, \omega)&=(2\pi)^{d+1}\delta(w+w')\delta(\mathbf k+\mathbf k')b|\mathbf k|^2\dfrac{1}{w^2+a^2|\mathbf k|^6}\\
&=(2\pi)^{d+1}\delta(w+w')\delta(\mathbf k+\mathbf k')C_0(\mathbf q, w)
\end{split}
$$
Eq. (1) can be iteratively solved by adding the RHS solution to the non linear term. This will generate a serie in power of $\alpha$. With $G_0$, $g$ and $\eta$. This serie can be arranged through Feynman Diagrams. As usual, it diverges below the critical dimension which shows that the renormaliation group must be used in order to correctly resum the serie. The integration and possible average will be performed over $\mathbf q$'s such that $\Lambda (1-\delta l)\leq\mathbf q \leq\Lambda$. Then as usual, we rescale the fields and parameters to get the same form as the original equation and see how they changed under a partial integration of the fast degrees of freedom. We will ignore, in what follows, all renormalization of $w$.

We can show that the contribution to the renormalized correlator (noise)reads at first order (again, setting $w$ = 0 because we know it won"'t get renormalized):

$$
\begin{split}
C(\mathbf q, 0)&=C_0(\mathbf q, 0)+2G_0(\mathbf q, 0)G(-\mathbf q, 0)\int\frac{ d \mathbf k}{(2\pi)^d} \int\frac{ d \Omega}{2\pi} g_0(\mathbf q, \mathbf k, \mathbf q-\mathbf k)C_0(\mathbf q-\mathbf k, 0 - \Omega)C_0(\mathbf k, \Omega)g_0(-\mathbf q, -\mathbf k, \mathbf k -\mathbf q)
\end{split}
$$

The second term comes from this diagram:

We begin by computing the frequency integral exactly (note that we don't tkae the factor $2\pi$ for the correlator, they already have been taken care of):

$$
\begin{split}
I_w&=\int_{-\infty}^{\infty}\dfrac{d\Omega}{2\pi} C_0(\mathbf q-\mathbf k, \Omega)C_0(\mathbf k, \Omega)\\
&=b^2|\mathbf k|^2|\mathbf q - \mathbf k|^2\int_{-\infty}^{\infty}\dfrac{d\Omega}{2\pi} \dfrac{1}{w^2+a^2|\mathbf q -\mathbf k|^6}\dfrac{1}{w^2+a^2|\mathbf k|^6}\\
&=\dfrac{b^2}{2a^3}\dfrac{1}{|\mathbf q -\mathbf k||\mathbf k|\left(|\mathbf q -\mathbf k|^3 + |\mathbf k|^3\right)}
\end{split}$$

Therefore, the renormalized $C$ reads:

$$
\begin{split}
C(\mathbf q, 0)&=C_0(\mathbf q, 0)+ \dfrac{\alpha^2b^2}{a^3}G_0(\mathbf q, 0)G(-\mathbf q, 0)\int\frac{ d \mathbf k}{(2\pi)^d} \dfrac{\mathbf q^2(\mathbf k\cdot(\mathbf q - \mathbf k))^2}{|\mathbf q -\mathbf k||\mathbf k|\left(|\mathbf q -\mathbf k|^3 + |\mathbf k|^3\right)}
\end{split}
$$

We are interested in the lowest order approximation of $\mathbf q$ of this integral. We want to know if a new term is generated or if $b$ is renormalized.

At lowest order in $\mathbf q$, we simply set: $\mathbf k - \mathbf q\to\mathbf k$

$$
\begin{split}
C(\mathbf q, 0)&=C_0(\mathbf q, 0)+ \dfrac{\alpha^2b^2}{2a^3}G_0(\mathbf q, 0)G(-\mathbf q, 0)\mathbf q^2\int\frac{ d \mathbf k}{(2\pi)^d} |\mathbf k|^{-1}+\mathcal{O}(|\mathbf q|^4)\\
&=C_0(\mathbf q, 0)+ \dfrac{\alpha^2b^2}{2a^3}G_0(\mathbf q, 0)G(-\mathbf q, 0)\mathbf q^2\dfrac{S_d}{(2\pi)^d}\Lambda^{d-1}+\mathcal{O}(|\mathbf q|^4)\
\end{split}
$$

With $S_d$ the hypersphere surface of dimension d. Note here that we computed the new term not in a shell but over all wavevector, so we are not renormalizing anything yet. We also note that $b$ is getting renormalized since the first term is proportional to $q^2$.

We perform the (first step) of renormalization by restricting the integral to: $\Lambda (1-\delta l)\leq\mathbf q \leq\Lambda$. In which case:

$$b_{I}\simeq b\left(1+\dfrac{\alpha^2b}{2a^3}K_d\Lambda^{d-1}\delta l\right)$$

with $K_d=S_d/(2\pi)^d$. In order to continue and get the RG flow, we will need to also rescale the fields and parameter, but we will do that later.
_____________________________

We can also show that the contribution to the renormalized propagator $G$ reads at first order:

$$
G(\mathbf q, w = 0) = G_0(\mathbf q, 0)+4G_0(\mathbf q, 0)^2\int \dfrac{d \mathbf k}{(2\pi)^d} \int \dfrac{d \Omega}{2\pi} g_0(\mathbf q, \mathbf k, \mathbf q-\mathbf k)G_0(\mathbf q-\mathbf k, 0 - \Omega)C_0(\mathbf k, \Omega)g_0(\mathbf q-\mathbf k, -\mathbf k, \mathbf q)$$

the second term comes from this diagram

We begin by computing the frequency integral exactly:

$$
\begin{split}
I_w&=\int_{-\infty}^{\infty}\dfrac{d\Omega}{2\pi} G_0(\mathbf q-\mathbf k, 0 - \Omega)C_0(\mathbf k, \Omega)\\
&=b|\mathbf k|^2\int_{-\infty}^{\infty}\dfrac{d\Omega}{2\pi} \dfrac{1}{i\Omega+a|\mathbf q-\mathbf k|^3}\dfrac{1}{w^2+a^2|\mathbf k|^6}\\
&=\dfrac{b}{2a^2}\dfrac{1}{|\mathbf k|(|\mathbf k|^3 + |\mathbf k - \mathbf q|^3)}
\end{split}$$

Therefore:
$$
\begin{split}
G(\mathbf q, w = 0) &= G_0(\mathbf q, 0)+4G_0(\mathbf q, 0)^2\int \dfrac{d \mathbf k}{(2\pi)^d} \int \dfrac{d \Omega}{2\pi} g_0(\mathbf q, \mathbf k, \mathbf q-\mathbf k)G_0(\mathbf q-\mathbf k, 0 - \Omega)C_0(\mathbf k, \Omega)g_0(\mathbf q-\mathbf k, -\mathbf k, \mathbf q)\\
&=G_0(\mathbf q, 0)-\dfrac{2\alpha^2 b}{a^3}\dfrac{G_0(\mathbf q, 0)}{|\mathbf q|^3}\int \dfrac{d \mathbf k}{(2\pi)^d} \dfrac{|\mathbf q||\mathbf q-\mathbf k| (\mathbf k\cdot (\mathbf q-\mathbf k))\mathbf k\cdot\mathbf q}{|\mathbf k|(|\mathbf k|^3 + |\mathbf k - \mathbf q|^3)}
\end{split}
$$

The expansion of the integrand is a bit more painful because the first order vanish upon integration. We will perform a Taylor expansion which is not so trivial due to the dot product:

$$|\mathbf k-\mathbf q|=\sqrt{\mathbf k^2 -2\mathbf k\cdot\mathbf q+ \mathbf q^2}= |\mathbf k|-\frac{\mathbf k\cdot \mathbf q}{|\mathbf k|}+\mathcal{O}(\mathbf q^2)$$

$$\dfrac{1}{|\mathbf k|^3+|\mathbf k-\mathbf q|^3}=\dfrac{1}{2|\mathbf k|^3}+\dfrac{3}{4}\dfrac{\mathbf k\cdot\mathbf q}{|\mathbf k|^5}+\mathcal{O}(\mathbf q^2)$$

We now obtain:

$$
\begin{split}
I&=\dfrac{|\mathbf q||\mathbf q-\mathbf k| (\mathbf k\cdot (\mathbf q-\mathbf k))\mathbf k\cdot\mathbf q}{|\mathbf k|(|\mathbf k|^3 + |\mathbf k - \mathbf q|^3)}=\dfrac{|\mathbf q||\mathbf q-\mathbf k| (\mathbf k\cdot\mathbf q)^2}{|\mathbf k|(|\mathbf k|^3 + |\mathbf k - \mathbf q|^3)}-\dfrac{|\mathbf q||\mathbf q-\mathbf k| |\mathbf k|\mathbf k\cdot\mathbf q}{|\mathbf k|^3 + |\mathbf k - \mathbf q|^3}\\
&=\dfrac{|\mathbf q| (\mathbf k\cdot\mathbf q)^2}{|\mathbf k|^3 + |\mathbf k - \mathbf q|^3}-\dfrac{|\mathbf q| (\mathbf k\cdot\mathbf q)^3}{|\mathbf k|^2(|\mathbf k|^3 + |\mathbf k - \mathbf q|^3)}-\dfrac{|\mathbf q| |\mathbf k|^2\mathbf k\cdot\mathbf q}{|\mathbf k|^3 + |\mathbf k - \mathbf q|^3}+\dfrac{|\mathbf q|(\mathbf k\cdot\mathbf q)^2}{|\mathbf k|^3 + |\mathbf k - \mathbf q|^3}\\
&=\dfrac{|\mathbf q| (\mathbf k\cdot\mathbf q)^2}{2|\mathbf k|^3}-\dfrac{|\mathbf q| (\mathbf k\cdot\mathbf q)^3}{2|\mathbf k|^5}-\dfrac{|\mathbf q| \mathbf k\cdot\mathbf q}{2|\mathbf k| }+\dfrac{|\mathbf q|(\mathbf k\cdot\mathbf q)^2}{2|\mathbf k|^3 }\\
&~+ \dfrac{3}{4}\dfrac{|\mathbf q| (\mathbf k\cdot\mathbf q)^3}{|\mathbf k|^5 }-\dfrac{3}{4}\dfrac{|\mathbf q| (\mathbf k\cdot\mathbf q)^4}{|\mathbf k|^7}-\dfrac{3}{4}\dfrac{|\mathbf q| (\mathbf k\cdot\mathbf q)^2}{|\mathbf k|^3}+\dfrac{3}{4}\dfrac{|\mathbf q|(\mathbf k\cdot\mathbf q)^3}{|\mathbf k|^5} + \mathcal{O}(\mathbf q^?)
\end{split}$$

All the terms odd in $\mathbf k$ vanishes at integration. The lowest non vanishing terms are in $\mathcal{O}(\mathbf q^4)$ which gives:
$$\int I=\int \left(\dfrac{|\mathbf q|(\mathbf k\cdot\mathbf q)^2}{2|\mathbf k|^3 }-\dfrac{3}{4}\dfrac{|\mathbf q| (\mathbf k\cdot\mathbf q)^2}{|\mathbf k|^3}\right)+\mathcal{O}(\mathbf q^4)=\int\left(\dfrac{|\mathbf q| (\mathbf k\cdot\mathbf q)^2}{4|\mathbf k|^3}\right)+\mathcal{O}(\mathbf q^4)$$

Therefore:

$$
\begin{split}
G(\mathbf q, w = 0) &\simeq G_0(\mathbf q, 0)-\dfrac{\alpha^2 b}{2a^3}G_0(\mathbf q, 0)\int \dfrac{d \mathbf k}{(2\pi)^d} \dfrac{\cos(\theta)^{2}}{|\mathbf k|^{-1}} \\
&\simeq G_0(\mathbf q, 0)-\dfrac{\alpha^2 b}{2a^3}G_0(\mathbf q, 0)\dfrac{K_d}{d}\Lambda^{d-1}\\
\end{split}
$$

This leads to the following intermediate renormalization for $a$:

$$
\begin{split}
\dfrac{1}{a_I}&\simeq\dfrac{1}{a}\left(1-\dfrac{\alpha^2b}{2a^3}\dfrac{\Lambda^{d-1}K_d}{d}\delta l\right)\\
a_I&\simeq a\left(1+\dfrac{\alpha^2b}{2a^3}\dfrac{\Lambda^{d-1}K_d}{d}\delta l\right)
\end{split}
$$

__________

We know make the big assumption that $\alpha$ is not renormalized at one loop as is usually the case. At some point I should check it. Note that nothing prevents it to be renormali

The RG flow therefore reads:

\begin{split}
\dfrac{d \alpha}{dl}&=(z-3+\chi)\alpha\\
\dfrac{d a}{dl}&= \left(z-3+\dfrac{\alpha^2b}{2da^3}\Lambda^{d-1}K_d\right)a \\
\dfrac{d b}{dl}&= \left(z-2\chi-d-2+\dfrac{\alpha^2b}{2a^3}K_d\Lambda^{d-1}\right)b
\end{split}

We define the reduce variable $g=b\alpha^2/a^3\Lambda^{d-1}K_d$ which controls the non-linearity strenght. Its RG flow reads:

$$
\begin{split}
\dfrac{d g}{dl}=&\dfrac{d g}{db}\dfrac{d b}{dl}+\dfrac{d g}{da}\dfrac{d a}{dl}+\dfrac{d g}{d\alpha}\dfrac{d \alpha}{dl}\\
=&\dfrac{g}{b}\dfrac{d b}{dl}+2\dfrac{g}{\alpha}\dfrac{d \alpha}{dl}-3\dfrac{g}{a}\dfrac{d a}{dl}\\
=&(z-2\chi-d-2+g/2\\
&+2z-6+2\chi\\
&-3z+9-\dfrac{3g}{2d})g\\
&=\left(1-d+\dfrac{g}{2d}(d-3)\right)g
\end{split}
$$

at $d>1$ (and probably at $d=1$ with log correction) the gaussian fixed point is stable. At $d<1$ (which does not mean much), a new fixed point appears. But anyway, we got what we wanted, there is no strong coupling indication at one loop, therefore the hyperuniform interface gaussian fixed point is probably stable!

### Liquid-Solid dissipative hard disks

$N = 4\times 10^5$

$N = 2\times 10^5$

$q^{-1}$ or $q^{-2}$ hard to say. In any case, the interface is rough at least on very large lenght scales:


It seems that the coexisting solid lose hyperuniformity:

| solid at coexistence density ($\phi = 0.6475$ (strong energy kick)) | solid slightly above ($\phi = 0.66$)|
| -------- | -------- |
|  |  |

Note that hyperuniformity above the solid coexistence density is very easily obtained.

Note also that the relaxation times for the structure factor become completely crazy close to the coexisting densities...

### Gas- Liquid through unstable mips like dynamics

Here, the energy injection at collision depends on the time of free fly. If the particles have underwent a collision not so long ago, the energy injection is weak. It is strong otherwise. This creates an instability with a highly energetic gas and a cold liquid.

Dynamics:

$$\partial_t v_i=-\gamma v_i+\sqrt{2\gamma T}\eta$$

Collision:

\begin{split}
v_i'+ v_j'&= v_i+ v_j\\
\dfrac{1}{2}m\left( { v_i'}^2+{v_j'}^2\right)&=\dfrac{1}{2}m\left(v_i^2+ v_j^2\right)+\Delta E.
\end{split}

with:
\begin{equation}
\Delta E = \Delta E^+-m\dfrac{1-\alpha^2}{4}( v_{ij}\cdot \hat{\sigma}_{ij})^2.
\end{equation}

$$\Delta E^+ = 2\delta E_0 + \delta E\left(1 - e^{-\tau_i/\tau_r}\right)^\beta+\delta E\left(1 - e^{-\tau_j/\tau_r}\right)^\beta, $$

| model: | with damping ($T=0, \alpha = 1$) | without damping ($\gamma=0, \alpha<1$) | with noise (everything, $\alpha\leq1$) |
| ----------- | --------------------------------------------------- | --------------------------------------------------- | --------------------------------------------------- |
| gif: |  |  |  |
| $S(k)$: |  |  | ([might be qKPZ](https://arxiv.org/abs/2209.05096)) |
| $W^2(L_y):$ |  |  | ?! |
| Typical lenghtscale (starting from homogeneous) (bad stat) | [close to underdamped mips (probably due to temp diff, still no global noise)](https://arxiv.org/pdf/1902.06116)|  ~~~~~~~~~~~ |  might be underdamped mips |
|Coarsening of interface||||
|Evolution equation for the order parameter|Navier stokes + damping or $$\partial_t \rho = \nabla^2\frac{\delta F}{\delta \rho} + \sqrt{2D}\nabla^2\xi $$|See below|AMB+?|

Navier stokes + damping
\begin{split}
\partial_t \rho + \nabla \cdot ( \rho u ) &= 0\\
\rho\left(\partial_t u + ( u \cdot \nabla) v \right)&=-\gamma\rho u + \nabla\cdot\left( \Pi^d + \Pi^\rho + \Pi^r\right)\\
\Pi^d&= \eta(\nabla u + \nabla u^\intercal) + \mu ( \nabla\cdot u)\mathbb I\\
\langle\Pi_{ij}^r( r, t)\Pi_{kl}^r( r', t')\rangle &= 2 T \left[\eta\left(\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}-\dfrac{2}{d}\delta_{ij}\delta_{kl}\right) +\mu\delta_{ij}\delta_{kl}\right]\delta( r - r')\delta(t - t')\\
\nabla \cdot \Pi^\rho &= -\rho\nabla\dfrac{\delta F}{\delta \rho}\\
F &= \int V(\rho) + \frac{\kappa}{2}(\nabla\rho)^2 d\boldsymbol r,
\end{split}

without damping, we are not in an equilibrium setting since $S\sim k^{-3}$ and equivalently $W^2\sim L^2$. Note also that an equilibrium white noise does not restore the $k^{-2}$ scaling but instead we get something in between $k^{-2}$ and $k^{-1}$ as recently proposed by Nardini et al with the qKPZ equation. When we add damping, the scaling goes to $k^{-1}$ and $W^2\sim \log L$. The dynamical scaling are too shitty to say something interesting.

We also verify that the coexisting gas and liquid are hyperuniform (separately):


And even in the coexistence (by just cutting the snapshots):


Here is the corresponding structure factor of the interface:

#### Solid gas:

initial condition: 

Energy injection at collision with noise:


definitively not equilibrated

Energy injection at collision without damping

(might have not equilibrated $N = 580000)

### Ludovic's liquid liquid demixing

Here, in both cases the square well width $U$ is $|U|\sim 1.2T$.

| Model: | Equilibrium (with noise) | Ran ni model with damping | Each phase driven at different temperature |
| ---------- | --------------------------------------------------- | ---------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------ |
| gif: |  |  | wip |
| $S(k)$: |  |  |  |
| $W^2(L):$ |  |  (really really flat.. Is is really a log?) | wip |
| coarsening |  |  (model B) |  (pure hydro scaling) |
| Time evolution of $W^2$ |  | |  |

Strange to find $S\sim k^{-1\sim1.2}$ with damping but some systems lead to a pleateau at $S(k\to 0)$ (!!!) some don't. I'll have to do multiple independent run

We checked that the phases were hyperuniform even at coexistence (here left phase $N = 170k$):

### Attraction

Not working.

## More on theory
### Gaussian/Linear theory of interfaces

Without bothering too much with the derivations from microscopic dynamics. Let's assume that our interface is given by:

$$\partial_t h_q = \gamma q^a h_q + \sqrt{2D q^b}\xi$$

We require the equation to be invariant term by term under: $x\to b x$,$t\to b^{z}t$ and $h\to b^{\chi}h$. Note that $f_q = \int dx e^{ikx} f(x)$ which means that $f_q\sim b^{d}b^ff_q$ We obtain that $S\sim \langle h_q h_{-q}\rangle~\sim q^{-d - 2\chi}$ (or equivalently: $W^2\sim\langle h(x)h(x')\rangle\sim |x-x'|^{2\chi}$) and $\langle h(t)h(t') \rangle\sim t^{2\chi/z}$. At the linear level, since the fixed point is gaussian, the power counting is exact (it's dimensional analysis).

This leads to:

\begin{split}
-z + \chi &=-a+\chi\\
-z + \chi &= \dfrac{-b -d -z}{2}
\end{split}

From which we obtain:

\begin{split}
z &=a\\
\chi &= \dfrac{-b -d+a}{2}
\end{split}

Equilibrium requires that $-b=2-a$ which in turns gives: $\chi = \dfrac{2-d}{2}$ and $S\sim q^{-2}$
________

Above, for unstable mips like dynamics in 1D. We obtained $2\chi = 0$ (log increase of width) and $\alpha = 1/4$ where $\alpha$ is the dynamical exponent for the coarsening in spinodal decomposition. Which I believe is not the same as for the dynamical exponent coarsening of the interface width.

#### non linearity

The first non linearity consistent with the conservation of interface ($\int h(t)=cst$) [is termed qKPZ ](https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.130.187102):

$$\alpha q\mathcal{F}\left[(\nabla h)^2\right]$$

The equation is now:
$$\partial_t h_q = \gamma q^a h_q +\alpha q\mathcal{F}\left[(\nabla h)^2\right]+ \sqrt{2D q^b}\xi$$

We rescale and see when the gaussian fixed point renders irrelevant the non linearity:

$$b^{-z+\chi}\partial_t h_q = b^{-a+\chi}\gamma q^a h_q +b^{-1+2(\chi-1)}\alpha q\mathcal{F}\left[(\nabla h)^2\right]+ b^{\frac{-b-d-z}{2}}\sqrt{2D q^b}\xi$$

We put everything on the same side and replace the values of the exponent by their gaussian counterpart:
$$\partial_t h_q = b^{-a+z}\gamma q^a h_q +b^{-3+\chi+z}\alpha q\mathcal{F}\left[(\nabla h)^2\right]+ \sqrt{2D q^b}\xi$$

(In Fourier space, just add -d to every scaling. $\mathcal{F}a$ has a scaling $b^{\Delta_a-d}$)

With the gaussian scaling, the non linearity scales as:

$$\beta = -3 + \chi + a$$

| model | Equilibrium diffusive | Equilibrium momentum conserving | Non equilibrium noise conserving momentum and diffusive relaxation |
| ----- | --------------------- | ------------------------------- | --- |
|1D| $\beta =0.5$ | $\beta=-1.5$ | $\beta = 0$ |
|2D| $\beta =0$ | $\beta = -2$ | $\beta = -1$ |

Note that the non-linearirty is completely irrelevant in the case where the momentum is fully conserved where we managed to obtain a $q^{-3}$. Therefore, we must seek our answer elsewhere concerning why our interface displays such strong scaling. Probably in simple linear terms that are modified by the temperature gradient. We know that it changes [the coarsening scaling for mips due to heat flux](https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.123.228001?casa_token=wyUW-ovFB6IAAAAA%3AEp-mTXMLj2L9s7ukSMpmsS9p3dG_5EA6HuKwY_GyRwXsS2eFclES-1DraRat8Ce9A6DTB3MVmyublA). Note that [in this article](https://arxiv.org/abs/2409.02288) they have a term in $1/q(\nabla h)^2$ which is relevant and generate a $q^{-3}$ scaling. But I think it's not allowed as it violate conservation of $h$. I'm a bit confused..