### 2D chiral BIO

#### Models

Tangential velocity gain at collision:
\begin{split}
\boldsymbol v_i'&= \boldsymbol v_i + \dfrac{1+\alpha}{2}(\boldsymbol v_{ij}\cdot \hat{\boldsymbol\sigma}_{ij})\hat{\boldsymbol\sigma}_{ij} - (\varepsilon\cdot\hat{\boldsymbol\sigma}_{ij} )\Delta \\
\boldsymbol v_j'&= \boldsymbol v_j - \dfrac{1+\alpha}{2}(\boldsymbol v_{ij}\cdot \hat{\boldsymbol\sigma}_{ij})\hat{\boldsymbol\sigma}_{ij} + (\varepsilon\cdot\hat{\boldsymbol\sigma}_{ij} )\Delta,
\end{split}

--------------------------------------------

Angular momentum gain at collisions:


#### Theory
Model:

\begin{aligned}
\partial_t n + \boldsymbol \nabla\cdot(n\boldsymbol{u}) &= 0, \\
m n (\partial_t \boldsymbol{u} + \boldsymbol{u}\cdot\boldsymbol \nabla \boldsymbol{u})
&= -\gamma\boldsymbol u-\boldsymbol \nabla p + \boldsymbol \nabla\cdot\boldsymbol{\sigma} + \zeta_0\varepsilon \cdot \boldsymbol \nabla n, \\
\frac{d}{2} n (\partial_t T + \boldsymbol{u}\cdot\boldsymbol \nabla T)
&= -p\boldsymbol \nabla\cdot\boldsymbol{u} + \boldsymbol{\sigma}:\boldsymbol \nabla \boldsymbol{u}
- \boldsymbol \nabla\cdot\boldsymbol{q} + \dot Q,
\end{aligned}

We assume $p=nT$ and $\boldsymbol{\sigma} = \eta_s \left( \nabla \boldsymbol{u} + (\nabla \boldsymbol{u})^T - \frac{2}{d} (\nabla\cdot \boldsymbol{u}) \boldsymbol{I} \right) + \eta_H \left( \varepsilon\cdot \nabla \boldsymbol{u} + (\varepsilon\cdot \nabla \boldsymbol{u})^T \right)$.

$T$ is a fast field, we do not care. $\boldsymbol u$ too but let's keep it at first.

We linearize around $\boldsymbol u=0$ and $n=n_0$:

\begin{aligned}
\partial_t \delta n &=- n_0\boldsymbol \nabla\cdot(\delta\boldsymbol{u}), \\
m n_0 \partial_t \delta\boldsymbol{u}
&= -\gamma\delta\boldsymbol u-mc_s^2\boldsymbol \nabla \delta n + \eta_s\boldsymbol \nabla^2\boldsymbol u+\eta_H\boldsymbol \nabla^2\varepsilon\cdot\boldsymbol u + \zeta_0\varepsilon \cdot \boldsymbol \nabla\delta n, \\
\end{aligned}

We decompose into transverse and longitudinal:

$$\partial_t
\begin{pmatrix}
\delta n_{\mathbf{k}} \\[4pt]
u_L \\[4pt]
u_T
\end{pmatrix}=
\begin{pmatrix}
0 & -i n_0 k & 0 \\[4pt]
-\dfrac{i c_s^2 k}{m} & -\dfrac{\gamma + \eta_s k^2}{m n_0} & \dfrac{\eta_H k^2}{m n_0} \\
\dfrac{i \zeta_0 k}{m n_0} & -\dfrac{\eta_H k^2}{m n_0} & -\dfrac{\gamma + \eta_s k^2}{m n_0}
\end{pmatrix}
\begin{pmatrix}
\delta n_{\mathbf{k}} \\
u_L \\
u_T
\end{pmatrix}.$$

## $\gamma = 0$

Assuming in the first time $\gamma= 0$ , a pertubative expansion leads the following eigenvalues:

\begin{aligned}
\lambda_{\pm} &= \pm\sqrt{\dfrac{n_0}{m}}c_s \boldsymbol i|k| - \boldsymbol k^2\dfrac{\zeta_0\eta_H + c_s^2n_0\eta_s}{2c_s^2mn_0^2}\\
\lambda_{\perp}&=-\boldsymbol k^2\dfrac{c_s^2n_0\eta_s-\zeta_0\eta_H}{2c_s^2mn_0^2}
\end{aligned}

The sound modes are always stable, however, the shear mode is not. At strong chirality vs pressure $c_s^2n_0\eta_s<\zeta_0\eta_H$. Indeed, the system is unstable to shear pertubation (circular current). Interestingly, both the chiral viscosity and the chiral pressure part are needed.

#### Above the instability treshold:

#### Below the instability treshold:

### $\gamma \neq 0$

As expected, $\gamma\neq0$ stabilises the small $k$ modes and therefore selects the size of the bubble.

#### Above the instability treshold:
#### Below the instability treshold:

The pertubation theory is bound to be shitty to due the bifurcation. First we can look at the $k \to\infty$ behavior. It is as if $\gamma\to0$ and therefore, from this alone, we conclude that the instability is the same as when $\gamma = 0$, we don't have any new information. What we need to know is the wavector $k^*(\gamma)>0$, such that $Re(\lambda(k^*))=0$. It is easily found to be:

$$k^*=\sqrt{\dfrac{c_s^2\gamma n_0}{\eta_H\zeta_0-c_s^2 n_0\eta_s}}$$

We find this zone of stability:


which compares favorably (qualitatively) to :

### Adiabatic slaving

It should be possible to describe the system with only the density field. Whether it's possible to find back the (inertia induced) instability, is another question. In the limit $\gamma\to\infty$, for sure we can't have it.

$$\partial_t \begin{pmatrix}u_L \\u_T\end{pmatrix} = \underbrace{\boldsymbol A}_{\text{damping block}}\begin{pmatrix}u_L \\ u_T\end{pmatrix}+\underbrace{\boldsymbol B}_{\text{coupling to } \delta n}\delta n_{\mathbf{k}},$$
with
$$\boldsymbol A =\begin{pmatrix}-\dfrac{\gamma + \eta_s k^2}{m n_0} &\dfrac{\eta_H k^2}{mn_0} \\-\dfrac{\eta_H k^2}{m n_0} & -\dfrac{\gamma + \eta_s k^2}{m n_0}\end{pmatrix},\quad\boldsymbol B =\begin{pmatrix}-\dfrac{i c_s^2 k}{m} \\\dfrac{i \zeta_0 k}{m n_0}\end{pmatrix}.$$

Formally:

$$\begin{pmatrix}u_L \\ u_T\end{pmatrix}(t)= \int_{-\infty}^t e^{-\boldsymbol A (t'-t)} \boldsymbol B\delta n_{\boldsymbol{k}}$$

For large eigenvalues of $\boldsymbol A$:

$$\begin{pmatrix}
u_L \\ u_T \end{pmatrix}\simeq-\boldsymbol A^{-1} \boldsymbol B \delta n_{\mathbf{k}}- \boldsymbol A^{-2} \boldsymbol B \partial_t \delta n_{\mathbf{k}}+ \mathcal{O}(\boldsymbol A^{-3}).$$

Since $\partial_t \delta n_{\mathbf{k}} = -i n_0 k u_L$, we obtain:

$$u_L \simeq -(\boldsymbol A^{-1} \boldsymbol B)_L \delta n_{\mathbf{k}}+in_0k \left(\boldsymbol A^{-2} \boldsymbol B \right)_L(\boldsymbol A^{-1} \boldsymbol B)_L\delta n_k+ \mathcal{O}(\boldsymbol A^{-3}).$$

Keeping only the first term, we find:
$$u_L = -\frac{i c_s^2 k {n_0}}{\gamma }-\frac{i k^3 \left(\gamma \zeta_0\eta_H-c_s^4 m {n_0}^4+\gamma c_s^2 \eta_s n_0\right)}{\gamma ^3}+\mathcal O(k^3)$$

Using $\partial_t \delta n_{\mathbf{k}} = -i n_0 k u_L$, we find:

$$
\partial_t \delta n_{\mathbf{k}} = -\dfrac{n_0k^2}{\gamma}\left(c_s^2 n_0+\dfrac{k^2}{\gamma}\left(\zeta_0\eta_H + c_s^2\eta_sn_0-\dfrac{c_s^4mn_0^4}{\gamma}\right)\right)\delta n_k
$$

We again see the large $k$ instability. $\gamma\to 0$ is singular (anyway, we assumed large $\gamma$). The instability happens when $\zeta_0\eta_H + c_s^2\eta_sn_0-\dfrac{c_s^4mn_0^4}{\gamma}<0\Rightarrow \gamma\leq \dfrac{c_s^4mn_0^4}{\zeta_0\eta_H + c_s^2\eta_sn_0}$. Completely broken.

### Odd transverse pressure

We start with this collision rule:

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

The force is:

\begin{equation}
\boldsymbol F_{ij} = -m\left(\dfrac{1 + \alpha}{2}|\boldsymbol v_{ij} \cdot \hat{\boldsymbol \sigma}_{ij} |\hat{\boldsymbol \sigma}_{ij} + (\varepsilon\cdot\hat{\boldsymbol\sigma}_{ij} )\Delta\right)\delta(t - t^{\text{coll}}_{ij}),
\end{equation}

The stress tensor is:

\begin{equation}\sigma_{\alpha\beta} = \frac{1}{V} \left\langle m\sum_{i=1}^{N} (v_{i\alpha}-\langle v_{i\alpha}\rangle) (v_{i\beta}-\langle v_{i\beta}\rangle)- \sum_{i<j}^{N}r_{ij,\alpha} F_{ij,\beta}\right\rangle\end{equation}

By isotropy, $v_x$ and $v_y$ are uncorrelated and the first term is the regular ideal gas.

\begin{equation}
\begin{split}
\sigma_{\alpha\beta} &= n\left[T\delta_{\alpha\beta} + \frac{m\sigma \omega(T, \phi g^+)}{4}\left\langle \dfrac{1 + \alpha}{2}|\boldsymbol v_{ij} \cdot \hat{\boldsymbol \sigma}_{ij} |\hat{\boldsymbol \sigma}_{ij\alpha}\hat{\boldsymbol \sigma}_{ij\beta} + \Delta(\hat{\boldsymbol \sigma}_{ij\alpha}\hat{\boldsymbol \sigma}_{ij\beta}^\perp)\right\rangle_\text{coll}\right]\\
&=n \left[T\delta_{\alpha\beta} + \phi g^+\sqrt{T}\left((1+\alpha)\sqrt{T}\delta_{\alpha\beta}+ 2\sqrt{\pi m}\Delta\varepsilon_{\alpha\beta}\right)\right],
\end{split}
\end{equation}

with $\omega$ the frequency of collision and $\phi$ the packing fraction. Therefore, there is a tranverse contribution arising from the collisions, to the stress tensor.

### Odd Viscosity

The collision rule:

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

can be inverted to give the precollisional velocities $v{''}$ that lead to the postcollisional velocities $v$:

\begin{split}
\boldsymbol v_i''&= \boldsymbol v_i + \dfrac{1+\alpha^{-1}}{2}(\boldsymbol v_{ij}\cdot \hat{\boldsymbol\sigma}_{ij})\hat{\boldsymbol\sigma}_{ij} - (\varepsilon\cdot\hat{\boldsymbol\sigma}_{ij} )\Delta \\
\boldsymbol v_j''&= \boldsymbol v_j - \dfrac{1+\alpha^{-1}}{2}(\boldsymbol v_{ij}\cdot \hat{\boldsymbol\sigma}_{ij})\hat{\boldsymbol\sigma}_{ij} + (\varepsilon\cdot\hat{\boldsymbol\sigma}_{ij} )\Delta,
\end{split}

Therefore, the Boltzmann equation is:

$$\partial_t f(\boldsymbol r, \boldsymbol v, t) + \boldsymbol v\cdot \partial_{\boldsymbol r} f(\boldsymbol r, \boldsymbol v, t)=\sigma\chi\tilde J(\boldsymbol r, \boldsymbol v|f, f),$$
with
\begin{split}
\tilde J(\boldsymbol r_i, \boldsymbol {v}_i|f, f) = \iint &\left[\Theta(-\boldsymbol {v}_{ij}''\cdot \hat{ \boldsymbol\sigma}_{ij} )(-\boldsymbol {v}_{ij}''\cdot \hat{ \boldsymbol\sigma}_{ij} )\dfrac{f(\boldsymbol r_j, \boldsymbol v_j'')f(\boldsymbol r_i+\boldsymbol \sigma_{ij}, \boldsymbol v_i'')}{\alpha}- \right.\\&~\left.\vphantom{\frac 1 2 } \Theta(-\boldsymbol {v}_{ij}\cdot \hat{ \boldsymbol\sigma}_{ij})(-\boldsymbol {v}_{ij}\cdot \hat{ \boldsymbol\sigma}_{ij})f(\boldsymbol r_j, \boldsymbol v_j)f(\boldsymbol r_i+\boldsymbol \sigma_{ij}, \boldsymbol v_i)\right]d \boldsymbol v_jd\hat{\boldsymbol\sigma}_{ij}.
\end{split}

We can simplify things by taking the dilute limit in which we neglect collisional contributions. Explicityl, we just evaluta all functions at the same point in space. We can also expand the $v''$:

\begin{split}
\tilde J(\boldsymbol r, \boldsymbol {v}_i|f, f) = \iint &\Theta(-\boldsymbol {v}_{ij}\cdot \hat{ \boldsymbol\sigma}_{ij} )(-\boldsymbol {v}_{ij}\cdot \hat{ \boldsymbol\sigma}_{ij} )\left[\dfrac{f(\boldsymbol r, \boldsymbol v_j'')f(\boldsymbol r, \boldsymbol v_i'')}{\alpha^2}- f(\boldsymbol r, \boldsymbol v_j)f(\boldsymbol r, \boldsymbol v_i)\right]d \boldsymbol v_jd\hat{\boldsymbol\sigma}_{ij}.
\end{split}

### Including inertia:

$$m n (\partial_t \boldsymbol{u} + \boldsymbol{u}\cdot\boldsymbol \nabla \boldsymbol{u})
= -\gamma\boldsymbol u-\boldsymbol \nabla p + \boldsymbol \nabla\cdot\boldsymbol{\sigma} + \boldsymbol \nabla\cdot (\tau\boldsymbol\varepsilon)$$

Where we replaced the last part by a torque.


The convective term is

$$(\boldsymbol u\cdot\nabla)\boldsymbol u =
\Big[
u_r \partial_r u_r + \frac{u_\theta}{r}\partial_\theta u_r - \frac{u_\theta^2}{r}
\Big]\boldsymbol e_r
+
\Big[
u_r \partial_r u_\theta + \frac{u_\theta}{r}\partial_\theta u_\theta + \frac{u_r u_\theta}{r}
\Big]\boldsymbol e_\theta.$$

In a first step, we can begin by setting $u_r\to 0$. By symmetry, $\partial_\theta = 0$. This will make us look at the steady state bubble: $(\boldsymbol u\cdot\nabla)\boldsymbol u=-u_\theta^2/r \boldsymbol e_r$.

We now use:

$$\begin{split}
(\nabla\!\cdot\!\boldsymbol{\sigma})_r = \frac{1}{r}\partial_r(r\sigma_{rr}) - \frac{\sigma_{\theta\theta}}{r} + \frac{1}{r}\partial_\theta\sigma_{\theta r}, \\
(\nabla\!\cdot\!\boldsymbol{\sigma})_\theta = \frac{1}{r}\partial_r(r\sigma_{r\theta}) + \frac{1}{r}\partial_\theta\sigma_{\theta\theta} + \frac{\sigma_{r\theta}}{r}.
\end{split}$$

### $e_\theta$ part!
Therefore, for the azimuthal part, we have in the steady state:

$$0=-\gamma u_\theta +\frac{1}{r}\partial_r(r\sigma_{r\theta}) + \frac{\sigma_{r\theta}}{r}-\partial_r\tau$$

Which leads to:

$$0=-\gamma u_\theta + \eta\partial_r\left(\dfrac{1}{r}\partial_r(ru_\theta)\right) - \partial_r\tau$$

The odd viscosity term does not enter if we assume $u_r = 0$. It seems that there is a bessel solution to this stupidity, i'll instead just assume $\gamma=0$ now. Assuming $\tau = \tau_0\Theta(r-R)$ We find:

$$u_\theta(r) =
\begin{cases}
-\dfrac{\tau_0}{2\eta} r, & r<R \\
-\dfrac{\tau_0 R^2}{2\eta r}, & r > R
\end{cases}$$

And therefore, $V_R = u_\theta(R)=-\tau_0R/2\eta$.

### $e_r$ part!

$$-mnu_\theta^2/r=-\partial_rp+\eta_H\partial_r\left(\dfrac{1}{r}\partial_r(r u_\theta)\right)$$

We have that $\partial_r(r u_\theta)/r=0$ outside the bubble and $\partial_r(r u_\theta)/r=-\tau_0/\eta$ inside. In both case, the derivative with respect to $r$ is 0 (except at the interface).

#### Integration for $r>R$

We can integrate the equation in the region $r>R$. We find:

$$p_\infty-p(R^+) = \int_R^{\infty}mn u_\theta^2/r dr=mnV_R^2/2$$

Which is Bernouilli's law.

#### (parenthesis) Integration for $r<R$

$$p(R^-)-p(r=0) = \left[-\tau_0\eta_H/\eta\right]_0^R + mn_{in}\int_0^{R^-}u_\theta^2/rdr=0$$

We assume the pressure at $0$ and the density inside to be vanishing. IT's fishy, and i think that integrating at the interface is better.

#### Integration for $R^-<r<R^+$

Since we forgot to include the stupid surface tension $\sigma$, let's add it now :)

$$p(R^-)-p(R^+)-\sigma/R=-\tau_0\eta_H/\eta$$

We recall that the odd stress is: $-\tau_0\eta_H/\eta\Theta(R-r)$ and we integrate in $R - \delta R$ and $R+\delta R$. We also have $P(R^-)=0$
#### Matching

Surface tension and pressure are trying to reduce the bubble size, while odd viscosity and inertial forces, are trying to expand it.
$$p_\infty+\sigma/R=mnV^2_R/2 + \tau_0\dfrac{\eta_H}{\eta}$$

If we get rid of surface tension and inertia, we cannot get the size of the bubble. That is not to say that the system cannot lead to a bubble (I believe it can from the hydrodynamics). When we add the surface tension:

$$R = \dfrac{\sigma}{\tau_0\eta_H/\eta-p_\infty}$$

There is a stable bubble, but it's rather strange, as its minimal size is taken at low pressure (altough, it becomes unstable when the pressure is too high). When everything is taken into account:

$$p_\infty - \tau_0\dfrac{\eta_H}{\eta} = \dfrac{\rho}{2}(\tau_0^2 R^2)/(4\eta^2)-\sigma/R$$

When we neglect surface tension, we obtain:

$$R = \dfrac{\alpha}{\tau_0}\sqrt{p_\infty-\tau_0\eta_H/\eta}$$

Which is strange, because we find the bubble size increases with $p_\infty$. This is not so unexpected, as in our case, if we impose a pressure and ask: "what is the size of the bubble $R$ such that the system can wistand this pressure", of course we will need a larger $R$. It seems broken still. Because it does not include the instability found before and the radius does not vanish at $\tau=0$ but instead diverges, such as to compensate for the low velocity of the bubble.

The problem is that, we did not find a stable fixed point, but an unstable one!

Indeed, if we define a 'force', it should be something like this:
$$F(R)\equiv \frac{\rho\tau_0^2}{8\eta^2}R^2 + \tau_0\frac{\eta_H}{\eta} - p_\infty - \frac{\sigma}{R}.$$

The bubble is pushed toward larger $R$ (positive F) by the chiral terms and toward small $R$ by the other terms. When $\tau_0\frac{\eta_H}{\eta} > p_\infty$ we find that the bubble is unstable, always, and reaches $R\to\infty$.


With surface tension:


we have a minimum size below which the bubble is unstable. This can also be modelized in the hydrodynamics theory by a Korteweg stress.

When the pressure is winning $\tau_0\frac{\eta_H}{\eta} < p_\infty$. Small bubble are unstables while big ones are growing to infinity (this is similar to the surface tension). It's similar to nucleation in a way, but it's non linear (cannot be seen from the linear stability analysis) and only mediated by inertia. It's also always there (whatever the parameters are):


Nucleation barrier: