# Hydrodynamics of the absorbing state model
###### tags: `Dissipative self-assembly`

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### Owners (the only one with the permission to edit the main test)

EF, AP, FS, GF

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# I'm creating a new page, the other one is approaching its max size and is becoming laggy...

### Hydrodynamics of the simple delta model with damping

#### Definition of our model
Our model is the $\Delta$ model with damping. We chose the convention defined [here](https://arxiv.org/pdf/1501.07190.pdf) and [here](https://arxiv.org/pdf/1803.03429.pdf)
The collision rule is (with $g = v_1-v_2$):

$$v_1=v_1 - \dfrac{1+\alpha}{2}(\hat\sigma\cdot g)\hat\sigma + \Delta\alpha^{-1}\hat\sigma$$
$$v_2=v_2 + \dfrac{1+\alpha}{2}(\hat\sigma\cdot g)\hat\sigma - \Delta\alpha^{-1}\hat\sigma$$

Our system is thus described by a Boltzmann equation (consistent with the Klein Framers equation):

$$\dfrac{\partial f(r, v, t)}{\partial t} + v\cdot\nabla_r f-\gamma \nabla_v \cdot(vf)=J(r, v|f, f)$$
with:

$$J(r, v_2|f, f) = g(\sigma^+)\int dv_1\int d\hat\sigma\left[\Theta(\hat \sigma \cdot g - 2\Delta)(\hat \sigma \cdot g - 2\Delta)\dfrac{f(v_2'')f(v_1'')}{\alpha^2}- \Theta(\hat \sigma \cdot g)(\hat \sigma \cdot g)f(v_2)f(v_1)\right]$$

where $v''_{1, 2}$ are the precolisionnal velocities leading to $v_{1, 2}$ after the collision:
$$v_1''=v_1 - \dfrac{1+\alpha^{-1}}{2}(\hat\sigma\cdot g)\hat\sigma - \Delta\alpha^{-1}\hat\sigma$$
$$v_2''=v_2 + \dfrac{1+\alpha^{-1}}{2}(\hat\sigma\cdot g)\hat\sigma + \Delta\alpha^{-1}\hat\sigma$$

__________
#### Collisional operator/Pseudo liouvillian
We can define the operator $\bar T_0(v_1, v_2)$ defined as:

$$\bar T_0(v_1, v_2) = \int d\hat\sigma\Theta(\hat \sigma \cdot g - 2\Delta)(\hat \sigma \cdot g - 2\Delta)\alpha^{-2}b^{-1}(1, 2)- \Theta(\hat \sigma \cdot g)(\hat \sigma \cdot g)$$

with $b^{-1}(1, 2)$ the operator transforming, all velocities ($v_{1, 2}$) in front of it into their precolisional values ($v''_{1, 2}$). It is also useful to define $b(1, 2)$ the operator that transform all velocities ($v_{1, 2}$) into their postcolisional ones ($v'_{1, 2}$).

Notably, we can rewrite $J$ as follow:

$$J(r, v_2|f, f) = g(\sigma^+)\int dv_1\bar T_0(v_1, v_2)f(v_1, r, t)f(v_2, r, t)$$

Moreover, if we define the formal conjuguate of $\bar T_0$ as $T_0$, we have the following relation:
$$\int dv_1\int dv_2 a(v_1, v_2)\bar T_0(v_1, v_2)b(v_1, v_2)\equiv \int dv_1\int dv_2 b(v_1, v_2) T_0(v_1, v_2)a(v_1, v_2)$$

This gives:

$$ T_0(v_1, v_2) = \int d\hat\sigma\Theta(-\hat \sigma \cdot g)|\hat \sigma \cdot g|(b(1, 2)-1)$$

Note that the conjuguate of $\bar T_0$: $T_0$ is considerably simpler than $\bar T_0$. Notably because it only involves post collisional velocities and does not involve $\Delta$ directly in the heaviside and cross section part of the collision operator.
___________
#### Derivation of the hydrodynamic equation
The hydrodynamic fields are the density, momentum and energy field:

$$\rho(r, t) = \int dv f(r,v, t)$$
$$\rho(r, t)u(r, t) = \int dv vf(r,v, t)$$
$$\dfrac{d}{2}\rho(r, t)T(r, t) = \int dv \dfrac{m}{2}(v - u)^2 f(r,v, t)$$

______

##### Continuity equation

\begin{split}
\dfrac{\partial \rho}{\partial t} &= \int dv \dfrac{\partial f}{\partial t}\\
&=\int dv [-v\cdot\nabla_r f+\gamma \nabla_v\cdot (v f) + J(r, v|f, f)]\\
&=-\nabla_r\cdot(\rho u) + 0 + 0
\end{split}

The first 0 comes from the nullity of $f$ at the boundaries and the last 0 comes from the fact that, during a collision, the number of particle is conserved.

$$\boxed{\dfrac{\partial \rho}{\partial t}=-\nabla_r\cdot(\rho u)}$$

or equivalently

$$\boxed{\dfrac{\partial \rho}{\partial t} + u \cdot \nabla_r \rho = - \rho \nabla_r \cdot u}$$

##### macroscopic velocity equation

\begin{split}
\dfrac{\partial \rho u}{\partial t} &= \int dv v\dfrac{\partial f}{\partial t}\\
&=\int dv [-v(v\cdot\nabla_r f)+\gamma v\nabla_v\cdot (v f) + vJ(r, v|f, f)]\\
&=-\nabla_r\cdot(\rho\langle v v\rangle) -\gamma\rho u + 0
\end{split}

where we defined $\displaystyle{\langle A\rangle=\dfrac{1}{\rho}\int dv A f}$. **Note the difference between $u$ the average velocity and $v$ the microscopic velocity. We can look at the average term and rewrite it as to extract the microscopic and macroscopic velocities:

$$\rho \langle v_iv_j\rangle=\rho(\langle (v_i - u_i)(v_j - u_j)\rangle + u_i\langle v_j\rangle + u_j\langle v_i\rangle - u_iu_j)=\rho\langle (v_i - u_i)(v_j - u_j)\rangle + \rho u_iu_j $$

We define the stress tensor $\Pi_{ij} = m\rho\langle (v_i - u_i)(v_j - u_j)\rangle$.
Finaly, our equation reads:
\begin{split}
\boxed{\dfrac{\partial \rho u}{\partial t}+ \nabla_r\cdot (\rho u u) =-\nabla_r\cdot\Pi/m -\gamma\rho u}
\end{split}

or equivalently using the conservation of particles:
\begin{split}
\boxed{\dfrac{\partial u}{\partial t}+ u\cdot\nabla_r u =-(m\rho)^{-1}\nabla_r\cdot\Pi -\gamma u}
\end{split}

##### macroscopic temperature equation

The temperature is defined as the microscopic velocity squared without taking into account global motion:

\begin{split}
\dfrac{\partial(\rho T) }{\partial t} &= \dfrac{m}{d}\int dv (v - u)^2 \dfrac{\partial f}{\partial t}\\
&=\frac m d\int dv [-(v-u)^2(v\cdot\nabla_r f)+\gamma (v-u)^2\nabla_v\cdot (v f) + (v-u)^2J(r, v|f, f)]\\
\end{split}

A new difficulty arises, the last term is not 0 because the energy is not conserved in a collision. It is the rate of change of energy during due to collision:

\begin{split}
G &= \int dv \dfrac{m}{2}(v-u)^2 J= \int dv_1dv_2 f(v_1) f(v_2) \dfrac{m}{2}T_0(1, 2) (v - u)^2\\
&= g(\sigma^+)\dfrac{m\pi^{(d - 1)/2}\sigma^{d-1}}{2}\int\int dv_1dv_2f(r, v_1, t)f(r, v_2, t)\left(\dfrac{\Delta^2 v_{12}}{\Gamma((d + 1)/2)} + \dfrac{\pi^{1/2}\alpha\Delta v_{12}^2}{2\Gamma((d+2)/2)} - \dfrac{(1-\alpha^2)v_{12}^3}{4\Gamma((d + 3)/2)}\right)
\end{split}

Back at the equation:

\begin{split}
\dfrac{\partial(\rho T) }{\partial t} &=\frac 2 d G + \frac m d\int dv [-(v-u)^2(v\cdot\nabla_r f)+\gamma (v-u)^2\nabla_v\cdot (v f) ]\\
&=\frac 2 dG - \frac m d[\nabla_r \cdot (\rho \langle v(v - u)^2 \rangle) + 2\rho \langle v \cdot \nabla_r u \cdot (v - u)\rangle]-2\rho\gamma T
\end{split}

The third term can be written as $2\Pi:\nabla_r u/m$ while the second can b using the following definition for the heat flux $J$:
\begin{split}
\dfrac{1}{2}m\rho \langle v(v - u)^2 \rangle&=\dfrac{1}{2}m\rho \langle (v-u)(v - u)^2\rangle +\dfrac{1}{2}m\rho u\langle (v - u)^2\rangle\\
& = J + \frac d 2\rho u T
\end{split}

which lets us rewrite finaly:

$$\boxed{\dfrac{\partial(\rho T) }{\partial t}=\frac 2 dG - \frac 2 d[\nabla_r \cdot J +\Pi : \nabla_r u] - \nabla_r\cdot(\rho u T) -2\rho\gamma T}$$

Or equivalently:

$$\boxed{\dfrac{\partial T }{\partial t} + u\cdot \nabla_r T=\frac 2 {d\rho}G - \frac 2 {d\rho}[\nabla_r \cdot J +\Pi : \nabla_r u] -2\gamma T}$$

____________

### Chapman-Enskog theory and normal solution

#### Form of the fluxes

Now that the full hydrodynamic equations have been written, we have to find some constitutive relations for the stress tensor and heat flux in order to close the equations.

The Chapman-Enskog theory allows to do just that. We first suppose that the particle probability distribution dependence on the position and (explicit) time can be expressed trough the macroscopic quantities:

$$f(x, v, t) = f(v| n, u, T)$$

Then, it is expanded as a formal power series: $$f = f^{(0)} + f^{(1)}\epsilon +f^{(2)}\epsilon^2+\dots$$

$\epsilon$ is a parameter caracterising the homogeneity of the system. It is often taking to be the Knudsen number.

Perhaps later I'll try to derive it myself, but for now, I will just accept that:

$$\Pi = \rho T \mathbb{I} - \eta\left(\nabla u + (\nabla u)^T - \dfrac{2\mathbb{I}}{d}\nabla \cdot u\right)+\xi (\nabla \cdot u)\mathbb I$$
$$J = -\kappa \nabla T - \mu \nabla \rho$$

where the equilibrium ideal gas equation of state was assumed for the pressure. $\mu$ is the shear viscosity and $\xi$ the bulk viscosity (this quantity won't change anything at the linear order for the interesting fields, only the transversed field would be affected). $\kappa$ is the thermal conductivity and $\mu$ is the thermal diffusion (I think it was found by Soto?). It is not present in in neq liquid because the heat is exchanged only though conduction for conserving dynamics.

At the lowest order $G^{(0)}$ will simply be the constant computed from the gaussian and is the object found in the absorbing paper. According to Soto: $G^{(1)}\sim \bar G^{(1)}\nabla \cdot u$.

Anyway, we will see that later.

### Linearized hydrodynamics equation

The full hydrodynamics equations are (assuming back a generalized equation of state for the pressure):

\begin{split}
\dfrac{\partial \rho}{\partial t} + u \cdot \nabla \rho &= - \rho \nabla \cdot u\\
\dfrac{\partial u}{\partial t}+ u\cdot\nabla u &=-(m\rho)^{-1}\nabla\cdot\left(p \mathbb{I} - \eta\left(\nabla u + (\nabla u)^\dagger - \dfrac{2\mathbb{I}}{d}\nabla \cdot u\right)\right) -\gamma u\\
\dfrac{\partial T }{\partial t} + u\cdot \nabla T&=\frac 2 {d\rho}G - \frac 2 {d\rho}[\nabla \cdot (-\kappa \nabla T - \mu \nabla \rho) +\left(p \mathbb{I} - \eta\left(\nabla u + (\nabla u)^\dagger - \dfrac{2\mathbb{I}}{d}\nabla \cdot u\right)\right) : \nabla u] -2\gamma T
\end{split}

We linearize these equations through:

$\rho = \rho_0 + \delta \rho$
$u = 0 + \delta u$
$T = T_0 + \delta T$

#### Linearized density equation:

$$\dfrac{\partial \delta \rho}{\partial t}=-\rho_0\nabla \cdot \delta u$$
or
$$\dfrac{\partial \delta \rho_k}{\partial t}=i\rho_0 k u_k^{\parallel}$$

#### Linearized velocity equation:

Beware! The transport coefficient depends on $T$, $u$, $\rho$, ... But we are saved by the fact that they are in front of gradient. For which the homogeneous state is equal to 0. Hence, their $\delta$ will vanish.

$$m\rho_0\dfrac{\partial \delta u}{\partial t}=-\gamma m\rho_0\delta u-\nabla(p_\rho\delta \rho + p_T\delta T) + \eta_0 \nabla^2\delta u + (\xi_0 + \eta_0 (1-2/d))\nabla (\nabla \cdot \delta u) $$
or

\begin{split}
m\rho_0\dfrac{\partial \delta u_k^{\parallel}}{\partial t}&=-\gamma m\rho_0\delta u_k^{\parallel}+ik(p_\rho\delta \rho + p_T\delta T) - (\xi_0 + \eta_0 (2-2/d))k^2\delta u_k^{\parallel}\\
m\rho_0\dfrac{\partial \delta u_k^{\perp}}{\partial t}&=-\gamma m\rho_0\delta u_k^{\perp} - \eta_0 k^2\delta u_k^{\perp}
\end{split}

If we stop here, we have Ran Ni Model
#### Linearized energy equation:
The pressure term is easy to linearize:

\begin{split}
\delta\left[\left(p \mathbb{I} - \eta\left(\nabla u + (\nabla u)^\dagger - \dfrac{2\mathbb{I}}{d}\nabla \cdot u\right)\right) : \nabla u\right] = p_0(\nabla \cdot \delta u)
\end{split}

It matches with a choice by Trizac but not Hansen or Soto, strange. It matches for an ideal law equation of state. But otherwise, no. On the other side, Soto talks about the traceless part of the tensor. And say that it is the diagonal part that is different from mine... Which, in our case, is not a problem. Anyway, we continue.

$$\dfrac{\rho_0d}{2}\dfrac{\partial \delta T }{\partial t} = G_T \delta T + G_\rho \delta\rho + \kappa_0\nabla^2 \delta T + \mu_0 \nabla^2 \delta\rho -p_0(\nabla \cdot \delta u) -d\gamma (\rho_0 \delta T + T_0 \delta \rho)$$
or
$$\dfrac{\rho_0d}{2}\dfrac{\partial \delta T_k }{\partial t} = (G_T)_k \delta T_k + (G_\rho)_k \delta\rho - \kappa_0 k^2 \delta T_k - \mu_0 k^2 \delta\rho - ip_0k \delta u_k -d\gamma (\rho_0 \delta T_k + T_0 \delta \rho_k)$$

#### Hydrodynamic Matrix

The transverse velocity fields are uncoupled, we don't care about them:

We define the hydrodynamic matrix as follow:

$$
M(k) = \begin{pmatrix}
0 & -i\rho_0 k & 0\\
-\dfrac{ikp_\rho}{m\rho_0} & -\gamma - \dfrac{\xi_0 + \eta_0 (2-2/d)}{m\rho_0}k^2 & -\dfrac{ikp_T}{m\rho_0} \\
\dfrac{2}{\rho_0d}G_\rho^{(0)}-\dfrac{2\gamma T_0}{\rho_0}-\dfrac{2\mu_0}{\rho_0 d}k^2 & -2\dfrac{ik}{\rho_0 d}(p_0 + G^{(1)}) & \dfrac{2}{\rho_0d}G_T^{(0)}-2\gamma -\dfrac{2\kappa_0}{\rho_0 d}k^2
\end{pmatrix}
$$
Where we took into account the transport coefficient arising from the delta model $G^{(1)}$

For which we have the dynamics equation $\psi_k(t) = (\rho_k, u_k^{\parallel}, T_k)$:

$$\dfrac{\partial \psi}{\partial t} = M\psi$$

The eigenvalues of this matrix equation are the hydrodynamic modes of the system.

I think, the term $2\gamma T_0/\rho_0$ might be 0 since it is proportioanl to $\delta \rho/\rho_0$. Moreover, we define $\nu_l = \dfrac{\xi + \eta (2-2/d)}{m\rho}$ the kinematic longitudinal viscosity. We set $m = 1$ and $d = 2$ and $\tilde G=G-2\gamma T$

$$
M(k) = \begin{pmatrix}
0 & -i\rho_0 k & 0\\
-\dfrac{ikp_\rho}{\rho_0} & -\gamma - \nu_lk^2 & -\dfrac{ikp_T}{\rho_0} \\
\dfrac{\tilde G_\rho^{(0)}}{\rho_0}-\dfrac{\mu_0}{\rho_0}k^2 & -\dfrac{ik}{\rho_0- }(p_0 + G^{(1)}) & \dfrac{\tilde G_T^{(0)}}{\rho_0} -\dfrac{\kappa_0}{\rho_0 }k^2
\end{pmatrix}
$$

This is more or less the equation found by soto if we set $\gamma = 0$

### Expression for the hydrodynamic transport coefficient

From the Chapman-Enskog expansion, we can find theoretical expression for the transport coefficient in 2d.

$$\xi = \dfrac{1.022}{g^+}\sqrt{\dfrac{T}{2\pi}}(1 + (2\rho g^+)+0.8729 (2\rho g^+)^2)$$

$$\eta = \dfrac{1.022}{g^+}\sqrt{\dfrac{T}{2\pi}}(1.246 (2\rho g^+)^2)$$

$$\kappa = \dfrac{1.029}{g^+}\sqrt{\dfrac{2T}{\pi}}(1 + \dfrac{3}{2}(2\rho g^+)+0.8718(2\rho g^+)^2)$$

We take a pair correlation:

$$g^+=\dfrac{1-\dfrac{7}{16}\dfrac{\pi \rho}{4}}{(1-\rho\pi/4)^2}$$

and a pressure:
$$p = \rho T \dfrac{1+\dfrac{\rho^2\pi^2}{128}}{(1-\rho\pi/4)^2}$$

I don't have the values of $\mu$ and $G^{(1)}$.

### note
The ayre wrong so i used:

https://arxiv.org/pdf/1107.1446.pdf
### Numerical results

For the absorbing state model, if one neglect $\mu$ and $G^{(1)}$, the homogeneous state is always stable and looks very much like Ran ni's

In the active region (orange is imaginary and blue is real part of the eigenvalus of $M$):

Close to the absorbing region:

### Toward an hydrodynamic of a population model.

with the model that was keeping sync particle sync at collisions we saw waves. Can we observe them in the hydrodynamic theory?

$$\dfrac{dT}{dt} = (2P_A - P_A^2)G_A(T) + (1 - P_A)^2G_S(T)$$
$$\dfrac{dP_A}{dt} = \omega(T)P_A(1-P_A) -\omega e^{-\omega \tau_s}P_A$$

This type of model give a discontinuous transition:


(here $T$ as a function of $\gamma$), Mathematica is messing up the solving due to hysteresis.

But here is the energy landscape:

| active | transition|locally stable/hysteresis | absorbing |
| -------- | --------| ----- | -------- |
|  |  ||  |

Here are the evolution of the eigenvalues of the Jacobian (times -1):

We see that one of them goes to 0 telling us that the system slows down a lot at the transition. Note that one of them is staying finite. I have the impression that it tells us that we don't need two variable but only one since one is slaved to the other?
Nonetheless, the eigenvalues are always stables. Hence the need of a hydrodynamical theory to see the waves.

### Simple try:

We use back our old equation:

\begin{split}
\dfrac{\partial \rho}{\partial t} + u \cdot \nabla \rho &= - \rho \nabla \cdot u\\
\dfrac{\partial u}{\partial t}+ u\cdot\nabla u &=-(m\rho)^{-1}\nabla\cdot\left(p \mathbb{I} - \eta\left(\nabla u + (\nabla u)^\dagger - \dfrac{2\mathbb{I}}{d}\nabla \cdot u\right)\right) -\gamma u\\
\dfrac{\partial T }{\partial t} + u\cdot \nabla T&=\frac 2 {d\rho}G - \frac 2 {d\rho}[\nabla \cdot (-\kappa \nabla T - \mu \nabla \rho) +\left(p \mathbb{I} - \eta\left(\nabla u + (\nabla u)^\dagger - \dfrac{2\mathbb{I}}{d}\nabla \cdot u\right)\right) : \nabla u] -2\gamma T
\end{split}

But now we have two type of particles $\rho_S$ and $\rho_A$. Let's do it the stupid way. We simply add an additional field $\rho_A$ and change the temperature change accordingly:

\begin{split}
\dfrac{\partial \rho}{\partial t} + u \cdot \nabla \rho &= - \rho \nabla \cdot u\\
\dfrac{\partial \rho_A}{\partial t} + u \cdot \nabla \rho_A &= - \rho_A \nabla \cdot u +\underbrace{\omega(T)\rho_A\dfrac{\rho - \rho_A}{\rho}-\omega(T) e^{-\omega(T) \tau_s}\rho_A}_{L}\\
\dfrac{\partial u}{\partial t}+ u\cdot\nabla u &=-(m\rho)^{-1}\nabla\cdot\left(p \mathbb{I} - \eta\left(\nabla u + (\nabla u)^\dagger - \dfrac{2\mathbb{I}}{d}\nabla \cdot u\right)\right) -\gamma u\\
\dfrac{\partial T }{\partial t} + u\cdot \nabla T&=\frac 2 {d\rho}G - \frac 2 {d\rho}[\nabla \cdot (-\kappa \nabla T - \mu \nabla \rho) +\left(p \mathbb{I} - \eta\left(\nabla u + (\nabla u)^\dagger - \dfrac{2\mathbb{I}}{d}\nabla \cdot u\right)\right) : \nabla u] -2\gamma T
\end{split}

with $G=\dfrac{\rho_A^2 + 2\rho_A(\rho - \rho_A)}{\rho^2}G_A(T) + \dfrac{(\rho-\rho_A)^2}{\rho^2}G_S(T)$

This gives the hydrodynamical matrix (with simplified transport coefficient):
### to change sign!
$$M(k) = \begin{pmatrix}
0 & 0& i\rho_0 k & 0\\
L_\rho & L_{\rho_A}&i\rho_0^A k&L_T\\
\dfrac{ikp_\rho}{\rho_0}&0 & -\gamma - \nu_lk^2 & \dfrac{ikp_T}{\rho_0} \\
\dfrac{\tilde G_\rho^{(0)}}{\rho_0}-\dfrac{\mu_0}{\rho_0}k^2&\dfrac{\tilde G_{\rho_A}^{(0)}}{\rho_0} & -\dfrac{ik}{\rho_0 }(p_0 + G^{(1)}) & \dfrac{\tilde G_T^{(0)}}{\rho_0} -\dfrac{\kappa_0}{\rho_0 }k^2
\end{pmatrix}$$

And we obtain the same thing aas above:


## Hydrodynamics of the $\Delta$ model.

We saw that the $\Delta$ model has some very weird structure factor (red):

Let's see how we can rationalize that.

\begin{equation}
\begin{split}
w\delta &\rho= \rho_0 k u_{\parallel}\\
-iwu_\parallel &= -k^2(\Gamma(w) + \nu^\parallel)u_{\parallel} - ikc_s^2\delta \rho/\rho_0 + \sqrt{-2\nu^\parallel k^2 T/\rho_0}\sigma^r_\parallel + \sqrt{-2\Gamma(w) k^2 T/\rho_0}\tilde\sigma^r_\parallel
\end{split}
\end{equation}

We obtain

$$ -iwu_\parallel = -k^2(\Gamma(w) + \nu^\parallel)u_{\parallel} - ik^2c_s^2u_\parallel/w + \eta$$

or:

$$-iw^2\delta\rho=-k^2(\Gamma(w) +\nu^\parallel)w\delta\rho -ik^2c_s^2\delta\rho + k\eta$$

It gives:

\begin{split}
S(k, w) &= \langle \delta\rho(k, w)\delta\rho(-k, -w)\rangle\\
&=\left|\dfrac{k}{i(-w^2+k^2c_s^2)+k^2w(\Gamma(w) + \nu^\parallel)}\right|^2\langle \eta(k, w)\eta(-k, -w)\rangle
\end{split}

Let's use an exponentially correlated noise.

If we plot the dynamic structure factor as a function of $k$, here is what we got:


the blue curve is with exponential noise and delta noise, the orange one is without the exponential noise.

Same here, as a function of $w$:



The exponential noise increases the sound peak height.. But nothing that we cannot find with a delta noise.. Unfortunately.

#### Comparison eq/Delta model

_____________________________________

We can also find the structure factor from:

$$
M(k) = \begin{pmatrix}
0 & -i\rho_0 k & 0\\
-\dfrac{ikp_\rho}{\rho_0} & - \nu_lk^2 &-\dfrac{ikp_T}{\rho_0} \\
\dfrac{\tilde G_\rho^{(0)}}{\rho_0}-\dfrac{\mu_0}{\rho_0}k^2 & -\dfrac{ik}{\rho_0 }(p_0 + G^{(1)}) & \dfrac{\tilde G_T^{(0)}}{\rho_0} -\dfrac{\kappa_0}{\rho_0 }k^2
\end{pmatrix}
$$

Which gives $S(k)=\frac{T \left(\rho -\frac{\text{Gt} \text{Pt}^2 \rho T \left(\text{Gt}+k^2 (\kappa +\rho \text{vl})\right)}{\left(\text{Gt}+\kappa k^2\right) \left(\text{Pt}^2 T \left(\text{Gt}+\kappa k^2\right)+\rho \text{vl} \left(\left(\text{Gt}+\kappa k^2\right)^2+k^2 \text{Pt}^2 T\right)+k^2 \rho ^2 \text{vl}^2 \left(\text{Gt}+\kappa k^2\right)+k^2 \text{Pp} \rho ^3 \text{vl}\right)}\right)}{\text{Pp} \rho }$


More explicitely:

$$S_{\rho\rho}(0)/\rho = \dfrac{T}{P_\rho}\left(1-\dfrac{1}{1+\dfrac{G_T \nu_l \rho^2}{P_T^2 T}}\right)=\dfrac{T}{P_\rho}\left(\dfrac{1}{1+\dfrac{P_T^2T}{G_Tv_l\rho^2}}\right)$$

and:
$$S_{\rho\rho}(\infty)/\rho = \dfrac{T}{P_\rho}$$

Note that the limit $G_T \to 0$ is singular.

It also gives:

$$\rho S_{u u}(0) = P_\rho S_{\rho\rho}(0)/\rho$$

and

$$\rho S_{u u}(\infty) = T$$

And interestingly:

$$S_{TT}(0) = 0$$ and $$\rho S_{TT}(\infty) = T^2$$

_________



(highly dependents on the transport coefficient used)
________

### Marginalization

$$
{\begin{pmatrix}
\delta \dot\rho \\
\delta\dot u_\parallel\\
\delta \dot T
\end{pmatrix}}= \begin{pmatrix}
0 & -i\rho_0 k & 0\\
-\dfrac{ikp_\rho}{\rho_0} & - \nu_lk^2 & -\dfrac{ikp_T}{\rho_0} \\
0 & -\dfrac{ik}{\rho_0 }p_0 & \tilde G_T^{(0)} -\dfrac{\kappa_0}{\rho_0 }k^2
\end{pmatrix}{\begin{pmatrix}
\delta \rho \\
\delta u_\parallel\\
\delta T
\end{pmatrix}} +
{\begin{pmatrix}
0 \\
\eta_u\\
\eta_T
\end{pmatrix}}
$$

We can rewrite this as:

$$
\begin{split}\delta \ddot u_{\parallel} &= -k^2p_\rho \delta u_{\parallel}-\nu_lk^2\delta \dot u_\parallel-ik\dfrac{p_T}{\rho_0}\delta \dot T + \dot\eta_u\\
\delta T(t) &= e^{\underbrace{(\tilde G_T^0-\kappa_0k^2/\rho_0)}_{a_1}t }\left[\delta T(0) + \int_0^te^{-(\tilde G_T^0-\kappa_0k^2/\rho_0)t' }\left(\eta_T(t') -ik\dfrac{p_0}{\rho_0}\delta u_{\parallel}(t')\right)\right]
\end{split}
$$

Thus we find the expression for $\ddot u$:

$$
\delta \ddot u_{\parallel} = -k^2p_\rho \delta u_{\parallel}-\nu_lk^2\delta \dot u_\parallel-ik\dfrac{p_T}{\rho_0}\left(a_1\delta T_0 e^{a_1t}-ik\dfrac{p_0}{\rho_0}\delta u_{\parallel}+ \eta_T+a_1\int_0^te^{-a_1(t'-t) }\left(\eta_T(t') -ik\dfrac{p_0}{\rho_0}\delta u_{\parallel}(t')\right)dt'\right) + \dot\eta_u
$$

At large time:

$$
\delta \ddot u_{\parallel} = -k^2\left(p_\rho\delta u_{\parallel}(t) + \dfrac{p_Tp_0}{\rho_0^2}\left(\delta u_{\parallel}(t)+a_1\int_0^te^{-a_1(t'-t) }\delta u_{\parallel}(t')dt'\right)\right) -\nu_lk^2\delta \dot u_\parallel+ \dot\eta_u - \dfrac{ik p_T}{\rho_0}\left(a_1\int_0^te^{-a_1(t'-t) }\eta_T(t')+ \eta_T\right)dt'
$$

Integrating by part the integralS:

$$
\delta \ddot u_{\parallel} = -k^2\left(p_\rho\delta u_{\parallel}(t) + \dfrac{p_Tp_0}{\rho_0^2}\left(\delta u_{\parallel}(t)+\int_0^te^{-a_1(t'-t) }\delta\dot u_{\parallel}(t')dt'-\delta u_\parallel\right)\right) -\nu_lk^2\delta \dot u_\parallel+ \dot\eta_u - \dfrac{ik p_T}{\rho_0}\int_0^te^{-a_1(t'-t) }\dot \eta_T(t')dt'
$$

Rearanging:

$$
\delta \ddot u_{\parallel}
= -k^2 p_\rho \delta u_\parallel-k^2\left(\nu_l\delta \dot u_\parallel + \dfrac{p_Tp_0}{\rho_0^2}\int_0^te^{-a_1(t'-t) }\delta\dot u_{\parallel}(t')dt'\right) + \dot\eta_u - \dfrac{ik p_T}{\rho_0}\int_0^te^{-a_1(t'-t) }\dot \eta_T(t')dt'
$$

_______

Even easier, adiabatically getting rid of T:

$$
\delta \ddot u_{\parallel}
= -k^2 p_\rho \delta u_\parallel-\delta\dot u_\parallel k^2\left(\nu_l - p_Tp_0/(a_1\rho_0^2)\right) + \dot\eta_u -\dfrac{ik p_T}{\rho_0}\dot \eta_T/a_1
$$


We can get the structure factor through integration of tyhe dynamic structure factor (maybe using a lyupanov expression we could fint it more easily, but the noise have a strange time correlation).

$$
\delta \ddot u_{\parallel}
= -k^2 p_\rho \delta u_\parallel-\delta\dot u_\parallel k^2\left(\nu_l - p_Tp_0/(a_1\rho_0^2)\right) + \tilde\eta_u +\tilde \eta_T
$$

$$\mu_1 = \langle \tilde\eta_T(w, k)\tilde\eta_T(w', k')\rangle = 2k^4p_T^2T^2w^2\kappa/a_1^2\rho^4\delta(k+k')\delta(w+w')$$

$$\mu2 = \langle \tilde\eta_u(w, k)\tilde\eta_u(w', k')\rangle = 2k^2w^2T\nu_l/\rho\delta(k+k')\delta(w+w')$$

This gives:

$$\langle {u_\parallel}_k(w){u_\parallel}_{-k}(-w)\rangle=\dfrac{\mu_1 + \mu_2}{(k^2p_\rho-w^2)^2+w^2k^4\left(p p_T/a1\rho^2 - \nu_l\right)^2}$$

From which we find:

$$S_{uu}(k) = \langle {u_\parallel}_k{u_\parallel}_{-k}\rangle=\dfrac{T}{\rho}\dfrac {k^2p_T^2T\kappa/a_1^2\rho^2 + \rho\nu_l}{-p p_T/a1\rho + \rho\nu_l}$$

We recall $$a_1 = \tilde G_T^0-\kappa k^2/\rho_0$$. Indeed, if $G_0=0$, we recover the equilibrium limit.

This gives:

$$S_{uu}(k) = \dfrac{T}{\rho}\left(\dfrac{\kappa k^2}{\kappa k^2+G_T\rho}+\dfrac{G_T\nu_l\rho^2}{P_T^2T+\nu_l\rho(\kappa k^2 + G_T\rho)}\right)$$

Which has the same limit as the one found above!