# Chapter 2: Could the beautiful, sexy Milady de Winter seduce all the men?

This post is based on my joint work with Toan Nguyen.

Milady de Winter is a spy for Cardinal Richelieu and the most beautiful female fictional character in the classic ”The Three Musketeers”

by Alexandre Dumas (father). Being twenty-two, uncommonly beautiful, with brilliant blue eyes and a bewitching voice, Milady can seduce any men she wants.

Our question in this chapter is if the greedy Milady-condensate can absorb all of the men-excited atoms around her. If the answer is yes, there would be a moment, when all of excited atoms fell into the condensate i.e. there exists a point $(t,p)\in \mathbb{R}_+\times\mathbb{R}^3$ such that the density function $f(t,p)$ of the excited atoms becomes $0$. In order to see this, let us recall the equation of $f(t,p)$ from Chapter 1:

$\frac{df}{dt}=C_{12}[f], f(0,\cdot)=f_0, (1)$

where the interaction operator is defined as

$C_{12}[f]:=\int _{ \mathbb{R}^3}\int _{ \mathbb{R}^3}dp_1dp_2 [R(p, p_1, p_2)-R(p_1, p, p_2)-R(p_2, p_1, p) ],$

$R(p, p_1, p_2):= |\mathcal{M}(p, p_1, p_2)|^2 [\delta (\mathcal{E}_p-\mathcal{E}_{p_1}-\mathcal{E}_{p_2}) \delta (p-p_1-p_2)][ f(p_1)f(p_2)(1+f(p))-(1+f(p_1)(1+f(p_2))f(p)].$

Here, we use the Bogoliubov dispersion law form of the particle energy $\mathcal{E}_p=\sqrt{\kappa_1 |p|^2 + \kappa_2 |p|^4} (2),$ for $\kappa_1$, $\kappa_2$ are positive constants.

The Dirac delta function in (1) ensures the conservation of momentum and energy after collision: $p = p_1 + p_2, \mathcal{E}_p = \mathcal{E}_{p_1} + \mathcal{E}_{p_2}.$

The above conservation laws dictate the structure of the energy level sets in $\mathbb{R}^3$, for each $p$, defined by $S_p: = \{ p_2\in \mathbb{R}^3~:~\mathcal{E}_{p-p_2} + \mathcal{E}_{p_2}= \mathcal{E}_{p} \},$ $S'_p : = \{p_2\in \mathbb{R}^3~:~ \mathcal{E}_{p+p_2} = \mathcal{E}_p+\mathcal{E}_{p_2} \},$ and $S''_p : = \{p_*\in \mathbb{R}^3~:~ \mathcal{E}_{p_*}= \mathcal{E}_{p}+\mathcal{E}_{p_*-p}\}.$

The Bogoliubov dispersion law form of energy functions (2) significantly complicates the analysis in treating the collision integral operator $C_{12}[f]$, which is now reduced to the surface integral on the energy surfaces $S_p$, $S'_p$ and $S''_p$. For instance, it is not clear whether the second moment of $f$ on these surfaces is bounded, even the second moment of $f$ in $\mathbb{R}^3$ is bounded.

In this work with Toan Nguyen (Penn State), we proved that under some conditions on the initial data, for any time $T>0$, there exist positive constants $\theta_0, \theta_1$ such that positive radial solutions $f(t,p)$ to the quantum Boltzmann equations (1), satisfy $f(t,p)\geq \theta_1\exp(-\theta_0|p|^2),$  $\forall t\ge T.$

By the H-Theorem, the equation has a family of equilibria: $n_0(p )=\frac {1} {e^{\frac{\mathcal{E}_p}{k_B T}}-1}, \beta >0,$ where $k_B$ is the Boltzmann’s constant and $T$ the temperature of the quasiparticles whose distribution is $n_0$. It is clear that $n_0(p)$ is above a Gaussian, as a consequence, our result is quite reasonable, mathematically speaking.

Physically speaking, our result asserts that given a condensate and its thermal cloud, we can prove that there will be some portion of excited atoms which remain outside of the condensate and the density of such atoms will be greater than a Gaussian, uniformly in time $t\geq T$ for any time $T>0$.

In other words, the greedy Milady-condensate cannot seduce all of the men-excited atoms around her!

Up to now, we have been discussing about the condensate growth term. A natural question would be: how does the full system look like? This is the topic of Chapter 3:

What else besides the condensate growth term?

Reference for the picture:

[1]   https://www.pinterest.com/pin/408631366161518678/

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