text:
\section{Introduction}

$N$-particle systems have been a target of flourishing research for the better part of the last 50 years, in particular due to their connection to classical statistical physics PDE's via hydrodynamic limits. Thoroughly inspecting these collision processes allows us to better understand limiting cases such as the Boltzmann equation, while avoiding dealing with the equation explicitly, since it involves strong non-linearity.

In 1956, Marc Kac \cite{kac} introduced what we refer to as the \textit{Kac process}, a reversible Markov process for $N$ spatially homogeneous particles with one dimensional velocities, which describes the evolution of the system as particles randomly collide in a momentum and energy conserving fashion.

A behaviour exhibited in many naturally occurring particle systems, and a desirable property of these models, is that of exponential decay to equilibrium. The easiest way one can prove such property is by showing the existence of a spectral gap for the generator of the process, and a uniform bound lets us draw conclusions as the number of particles becomes infinite. Recall that the spectral gap of an operator is defined as the infimum of its positive spectrum, which may coincide with the lowest non-zero eigenvalue, if these exist.

In essence, under the Kac process, if $L$ is the infinitesimal generator and $f_0$ is an initial distribution in our state space, then it evolves as
$f_t = e^{tL}f_0$,
and the existence of a spectral gap $\Delta$ implies an exponential decay with rate at least equal to $\Delta$.

It is hard to relate relaxation in the $N$-particle system to relaxation in the limit equation, unless one has uniform bounds on the spectral gap, independent of $N$. Kac conjectured this to be true in his toy model, and this was successfuly proven later by Janvresse \cite{janvresse}, but with one dimensional velocities and uniform jump rates, which nevertheless allows us to efficiently generate uniform points on the $n$-sphere.

In a series of papers, Carlen, Carvalho and Loss \cite{spectralgap} managed to prove the existence of a spectral gap for a one parameter family of Kac processes in 3-dimensions, undergoing more general collisions with rates that are not bounded away from zero. We will explore next the background of these papers.

\subsection{The Kac process}

More precisely, consider an additional physical parameter $0\leq\alpha\leq2$ and an $N$-tuple of velocities $v_j$ in ${{\mathbb R}}^3$ such that
\begin{equation}
    \frac{1}{N}\sum_{j=1}^N v_j = 0, \frac{1}{N}\sum_{j=1}^N |v_j|^2 = 1,
\end{equation}
meaning that the total momentum is zero and the average kinetic energy per particle (with mass 2) is 1. These constraints describe our state space $\mathcal{S}_N$, isometric to a sphere with $3N-4$ dimensions and radius $\sqrt{N}$.

Before we move on, it is important to fix some notation. We will refer to elements of this state space $\mathcal S_N$ as a single letter $v$, $y$, and sometimes as a vector $\vec v$, $\vec y$, if it's not clear from context. Letters with subscripts such as $v_j$ shall always denote velocities in ${{\mathbb R}}^3$.

When the process begins, there is an associated exponential clock $T_{i,j}$ for each pair of velocities $(v_i,v_j)$, with parameter
\begin{equation}
    \lambda_{i,j} = N \binom{N}{2}^{-1} |v_i - v_j|^\alpha,
\end{equation}
and the minimum
\begin{equation}
    T := \min_{i,j} T_{i,j}
\end{equation}
yields our first collision time (and colliding pair). Given the restrictions above, these collisions can be parametrized in terms of a vector $\sigma \in \mathbb{S}^2$:
\begin{align}
    v_i^*(\sigma) & = \frac{v_i+v_j}{2} + \frac{|v_i-v_j|}{2}\sigma, \\
    v_j^*(\sigma) & = \frac{v_i+v_j}{2} - \frac{|v_i-v_j|}{2}\sigma.
\end{align}
Another source of randomness on the system is then the determination of this direction vector $\sigma$. The particular choice we follow is uniform sampling, as it turns out to be more physically relevant, even though other distributions or rules could be used.

The cases $0\leq \alpha \leq 1$, with 0 being the Maxwellian molecule case and 1 the hard sphere case, are the ones of physical interest, but it turns out, as shown by Villani (regarding entropy production inequalities) \cite{sometimes} and as seen in the work of Carlen, Carvalho and Loss, that the "super hard sphere" $\alpha=2$ case provides powerful insights for lower values of the parameter, and will later be our main focus.

Their proof is based on chaos estimates that quantify the "near independence" of a finite set of coordinates. If $\sigma_N$ denotes the uniform measure on the sphere $\mathbb{S}^N(\sqrt{N})$ with radius $\sqrt{N}$, it is known that
\begin{equation}
    \lim_{N\to+\infty} \int_{\mathbb{S}^N(\sqrt{N})} \phi(v_1,...,v_k) d\sigma_N
    = \int_{{{\mathbb R}}^k}  \phi(v_1,...,v_k) d\gamma^{\otimes k},
\end{equation}
with $\gamma$ being the Gaussian measure in ${{\mathbb R}}$, and we say that the measure $\sigma_N$ is $\gamma$-chaotic. In general, if the above equality holds for $\mu$ and a sequence of measures $\mu_N$, $\mu_N$ is said to be $\mu$-chaotic.

Another important step of the proof is the introduction of a conjugate process, which allows for a clever induction argument. With the right choice of coordinates, integrals over the state space can be factored as
\begin{equation}\label{measurefactor}
    \int_{\mathcal{S}_N} \phi(\vec v) d\sigma_N = \int_B \bigg[ \int_{S_{N-1}} \phi(T_j(\vec y, v)) d\sigma_{N-1} \bigg] d\nu_N(v),
\end{equation}
with $B$ the unit ball in ${{\mathbb R}}^3$. $T_1: \mathcal{S}_{N-1}\times B \to \mathcal{S}_N$ is given by
\begin{equation}
T_1(y,v) = \bigg( \sqrt{N-1}v, \beta(v)y_1 - \frac{1}{\sqrt{N-1}}v, ..., \beta(v)y_{N-1} - \frac{1}{\sqrt{N-1}}v \bigg),
\end{equation}
while $T_j$ is defined analogously but with $v$ going to the $j$-th coordinate, and
\begin{equation}
    \beta^2(v) := \frac{N}{N-1}(1-|v|^2).
\end{equation}
Finally, the measure $\nu_N$ on $B$ is defined as
\begin{equation} \label{equilibriumradial}
    d\nu_N(v) = \frac{|\mathbb S^{3N-7}|}{|\mathbb S^{3N-4}|} (1-|v|^2)^{(3N-8)/2}dv,
\end{equation}
which is what we refer to as the equilibrium radial measure, as it is the radial marginal of the velocities for the uniform measure in $\mathcal S_N$.

As we will verify later, this formula allows for an inductive inequality of the type
\begin{equation}
    \Delta_{N,\alpha} \geq \frac{N}{N-1}\Delta_{N-1,\alpha} \hat \Delta_{N,\alpha},
\end{equation}
where $\hat \Delta_{N,\alpha}$ is the spectral gap of a \textit{conjugate Kac process}, with vastly simplified dynamics. Furthermore, the authors make use of this fact to prove that there are $K_0 > 0$ and $N_0 \in {{\mathbb N}}$ such that
\begin{equation}
    \Delta_{N,\alpha} \geq K_0 \Delta_{N_0, \alpha} > 0,
\end{equation}
leaving only the base case to show. For $N=2$, the process is trivial, as the constraints completely determine one velocity from the other. Finally, making use of a compactness argument to show that $\hat \Delta_{3,\alpha}>0$, it follows that
\begin{equation}
    \Delta_{3,\alpha} \geq \frac32 \Delta_{2,\alpha} \hat \Delta_{3,\alpha} > 0.
\end{equation}

However, unlike previous bounds that could be explicitly computed, this does not give any quantitative information about $\hat \Delta_{3,\alpha}$, besides positivity, and it remains to be shown that this gap is physically meaningful when interpreted as an expected relaxation time. In this paper, we compute a lower bound for the spectral gap of the conjugate Kac process with 3 particles.

Even though the proof provided in \cite{spectralgap} makes use of asymptotic methods in the form of quantitative chaos estimates, the results are strong enough to hold already for all $N\geq 4$, and provide a positive gap. This fails for $N=3$.

Due to the fact that $N=2$ is already trivial, and most methods allow for good estimates for $N>3$, one would expect this case to be accessible. Unfortunately, this is not true, as most methods fail to provide a positive gap in this number of dimensions.

\subsection{Structure of the paper}
This paper is organized as follows.

In section \ref{conjkac}, we introduce our main object of study, the conjugate Kac process, building up the necessary tools and finally stating our main result. We also explore the possibility of obtaining spectral gap bounds via entropy production inequalities and, while not being able to prove it analytically, present some numerical evidence of such results.

The proof is developed over section \ref{mainproof}. We begin by inspecting the spectrum of our main tool, the correlation operator $K$, and proving suitable upper bounds on the eigenvalue sequence $\kappa_{n,\ell}$.

The proof itself is then broken down over the anti-symmetric and symmetric sectors, as they are orthogonal to each other. The first one is handled fairly quickly, while the symmetric sector, which is of bigger physical interest due to its relation to the limit equation, requires a more hands-on approach. We treat different angular momentum sectors, where different methods handle 'large' and 'small' values of $\ell$, leaving only a finite amount of eigenvalues to be computed.

\clearpage

\section{The conjugate Kac process}\label{conjkac}

We will now define the family of reversible Markov jump processes on $\mathcal S_N$, conjugate to the Kac process. Fixing a number of interacting particles $N$, $0<\alpha\leq2$, and given $\vec v \in S_N$, let $\{\hat T_1, ..., \hat T_N\}$ be a set of $N$ independent exponential variables with parameter $\lambda_k (\vec v)$ of $\hat T_k$ given by
\begin{equation}
    \lambda_k (\vec v) = \frac1N \bigg[ \frac{N^2-(1+|v_k|^2)N}{(N-1)^2} \bigg]^{\alpha/2}.
\end{equation}
Due to the restraints on energy and momentum, the maximum of $|v_k|^2$ is $N-1$, and thus the parameter is always positive.

Let $\hat T = \min \{\hat T_1, ..., \hat T_N\}$ be the first jump time, and let $k$ be the corresponding index. At this time, the process makes a conditional jump to uniform to a new point $\vec v \in S_N$, fixing $v_k$, and all other velocities are re-sampled uniformly, conditional on $v_k$. Even though this process is trivial for $N=2$, since the momentum equation completely determines one velocity from the other, our $N=3$ case is subtle.

The generator of this process can be obtained via projection (or conditional expectation) operators. Recall that, for any $\phi \in L^2(\sigma_N)$, there is a unique element $f(v_k)$ of $L^2(\sigma_N)$ such that
\begin{equation}
    \int_{S_N}\phi(\vec v) g(v_k) d\sigma_N = \int_{S_N} f(v_k) g(v_k) d\sigma_N,
\end{equation}
and we define this way the conditional expectation of $\phi$, given $v_k$:
\begin{equation}
    P_k \phi := {{\mathbb E}} [\phi | v_k].
\end{equation}
This can also be understood as an orthogonal projection onto the subspace of square integrable functions that depend only on the $k$-th coordinate. Furthermore, the factorization formula \eqref{measurefactor} gives us an expression for this operator:
\begin{equation}
    P_k \phi (v) = \int_{\mathcal S_{N-1}} \phi (T_k(y, v_k/\sqrt{N-1}) d\sigma_{N-1}.
\end{equation}

The generator of this process can then be written as
\begin{equation}
    \hat L_{N,\alpha} f = -\frac1N \sum_{k=1}^N \frac1N \bigg[ \frac{N^2-(1+|v_k|^2)N}{(N-1)^2} \bigg]^{\alpha/2} [f-P_k f],
\end{equation}
and it is also useful to define
\begin{equation}
    w_{N}(v_k) := \frac{N^2-(1+|v_k|^2)N}{(N-1)^2}.
\end{equation}

It is easy to prove self-adjointness, as we would expect from the reversibility of the process, and its null space is spanned by the constants. Similarly as before, the spectral gap of the conjugate Kac process is defined as
\begin{equation}\label{conjgap}
    \hat \Delta_{N,\alpha} := \inf \{ \mathcal{D}_{N,\alpha} (f,f) | \langle f,1 \rangle_{L^2(\sigma_N)}, \|f\|^2_{L^2(\sigma_N)} = 1 \},
\end{equation}
with
\begin{align}\label{conjugatedirichlet}
    \mathcal{D}_{N,\alpha} (f,f) & := - \langle f, \hat L_{N,\alpha} f \rangle_{L^2(\sigma_N)}\nonumber \\
    & = \frac1N \sum_{k=1}^N \frac1N \int_{S_N} \bigg[ \frac{N^2-(1+|v_k|^2)N}{(N-1)^2} \bigg]^{\alpha/2} [f^2-f P_k f] d\sigma_N.
\end{align}

We are at this moment in conditions of explaining how this process arises. Let $L_{N,\alpha}$ be the infinitesimal generator of the Kac process,
\begin{equation}
    \mathcal{E}_{N,\alpha}(f,f) = -\langle f, L_{N,\alpha} f \rangle
\end{equation}
the Dirichlet form associated with $L_{N,\alpha}$ and $\mathcal{E}_{N,\alpha}(f,f|v_k)$ the \textit{conditional Dirichlet form} obtained by factorizing, so that
\begin{equation}
     \mathcal{E}_{N,\alpha}(f,f) = \frac{N}{N-1} \bigg( \frac1N \sum_{k=1}^N \int_B w_N^{\alpha/2}(v_k) \mathcal{E}_{N,\alpha}(f,f|v_k) d\nu_N(v_k/\sqrt{N-1}) \bigg).
\end{equation}
Then, using the spectral gap for $N-1$ particles, one has
\begin{equation}
     \mathcal{E}_{N,\alpha}(f,f) \geq \frac{N}{N-1} \Delta_{N-1,\alpha} \bigg( \frac1N \sum_{k=1}^N \int_{\mathcal{S}_N} w_N^{\alpha/2} (v_k) [f-P_k f]^2 d\sigma_N\bigg),
\end{equation}
and the quantity in brackets corresponds exactly to \eqref{conjugatedirichlet}.

Now, define the self-adjoint operators
\begin{equation}
    W^{(\alpha)} = W^{(\alpha)}(v) := \frac1N \sum_{k=1}^N w_N^{\alpha/2}(v_k)
\end{equation}
and
\begin{equation}
    P^{(\alpha)} = \frac1N \sum_{k=1}^N w_N^{\alpha/2}(v_k) P_k,
\end{equation}
an average of weighted projections, so that we have $\hat L_{N,\alpha} = W^{(\alpha)} - P^{(\alpha)}$. For $\alpha=2$ and $N=3$, we see that $W^{(2)}=\frac34$, and \eqref{conjgap} becomes
\begin{equation}
    \hat \Delta_{3,2} = \frac34 - \mu_3,
\end{equation}
with
\begin{equation}
    \mu_3 := \sup \{ \langle f, P^{(2)} f \rangle_{ L^2(d\sigma_3) } | \|f\|_{ L^2(d\sigma_3) } = 1, \langle f, 1 \rangle_{ L^2(d\sigma_3) } = 0 \}.
\end{equation}
The spectrum of $P^{(\alpha)}$ is also studied on \cite{kspectrum}. More importantly, for $\alpha=2$, the following is true

\begin{Lem}
    The essential spectrum of $P^{(2)}$ is the interval $[0,\frac12]$, and the remaining spectrum can only consist of isolated eigenvalues in $\big(\frac12,\frac34\big)$.
\end{Lem}
Note that, in particular, $\frac34$ cannot be an accumulation point, and thus this proves that the spectral gap exists. Moreover, if $\mu_3\leq\frac12$, we immediately have that $\hat \Delta_{3,2}\geq0.25$, and if $\frac12 < \mu_3 < \frac34$, it is an eigenvalue of $P^{(2)}$.

We then turn to the eigenvalue equation $P^{(2)} f = \lambda f $ to obtain the only missing estimate for a fully quantitative result on the three particle spectral gap $\hat \Delta_{3,2}$. In \cite{spectralgap}, the authors break this down to the \textit{symmetric} and \textit{anti-symmetric} eigen-sectors, as they are mutually orthogonal. A computation for the anti-symmetric case is provided, and a suggestion is made on how to proceed for the remaining case. In section \ref{antisymsec} we explain and slightly improve the former, while the latter is treated in section \ref{symsec} - this last case proved to be much harder, due to the analytical difficulties in dealing with the spectrum of the conditional expectation operator (see section \ref{corroperator}).

Using this method, we were able to prove the following main result:
\begin{Thm}\label{maintheorem}
The spectral gap of the conjugate Kac process in 3 dimensions, with $\alpha = 2$, is at least $0.01984$.
\end{Thm}

The actual computations we provide show a result slightly less then $0.02$ in order to simplify and reduce the amount of cases, but could easily be adapted to achieve this quantity at the expense of explicitly determining a larger number of eigenvalues. Note that it is expected that the true value of the gap is larger, and we will illustrate this next.

\subsection{Spectral gap bounds via entropy production inequalities}

Cedric Villani \cite{sometimes} was the first to prove entropy production inequalities for variants of the Kac process, namely with one dimensional velocities.
Even though obtaining such bounds for the 3-dimensional Kac process is still an open problem, one can adopt a similar reasoning and attempt to derive entropy production inequalities for the simpler conjugate process. Recall that the entropy functional of a suitable density $f$ on $\mathcal S_N$ is defined as
\begin{equation}
    H(f) := \int_{\mathcal{S}_N} f\log f d\sigma_N,
\end{equation}
and the entropy production of a process with generator $\hat L_{N,\alpha}$ as
\begin{equation}
    D_{\hat L_{N,\alpha}}(f) := -\frac{d}{dt}\bigg ( H( e^{t\hat L_{N,\alpha}}f ) \bigg ) \geq 0,
\end{equation}
so that we can look for estimates of the type
\begin{equation}
    D_{\hat L_{N,\alpha}}(f) \geq C_{N,\alpha} H(f),
\end{equation}
with $C_{N,\alpha}$ a constant independent of $f$, and depending on $\hat L_{N,\alpha}$ only via the parameters describing the process.

Results of the above type are stronger than spectral gap estimates, as a classical linearization procedure implies the latter. More precisely, taking $f := 1 + \varepsilon h$, with $\varepsilon>0$ and $$\int_{\mathcal{S}_N} h d\sigma_N= 0,$$and expanding in first order leads us to
\begin{equation}
    \hat \Delta_{N,\alpha} \geq \frac{1}{2} C_{N,\alpha}.
\end{equation}
One advantage of the entropy production approach is that we can actually measure the empirical entropy in Monte Carlo simulations, and obtain an estimate of the true spectral gap by analyzing the log-entropy decay, which we will do next.

First, by symmetrizing one can compute the entropy production functional of the conjugate process:
\begin{equation}
    D_{N,\alpha}(f) := \frac1N \sum_{k=1}^N \int_{\mathcal S_N} w_N^{\alpha/2}(v_k) \log f [f - P_k f] d\sigma_N.
\end{equation}

Specializing to $\alpha=2$, and using the convexity of $-\log t$ and the fact that
\begin{equation}
    \frac{1}{N}\sum_{k=1}^N w_N(v) = 1,
\end{equation}
we obtain
\begin{align}
    D_{N,2} (f) \geq \bar D_{N,2} (f) := &
    \frac1N \sum_{k=1}^N \int_{\mathcal S_N} w_N(v_k) [f \log f  - P_k f \log P_k f] d\sigma_N \\
    = & H(f) - \frac{1}{N} \sum_{k=1}^N \int_{\mathcal S_N} w_N(v_k) [P_k f \log P_k f - P_k f + 1] d\sigma_N.
\end{align}

Now, as $t \log t - t + 1 \geq 0$ for all $t > 0$ and $w_N(v) \leq \frac{N}{N-1}$ for all $v$, we see that the $k$-th summand in the second term is less than $\frac{N}{N-1} H(P_k f)$, and thus we have

\begin{equation}
    \bar D_{N,2} (f) \geq H(f) - \frac{1}{N-1} \sum_{k=1}^N H(P_k f)
\end{equation}

To continue, we need to somehow relate the entropy of the marginals $P_k f$ with the entropy of the joint distribution $f$. Fortunately, this is  precisely developed in \cite{sphericalmarginals} and \cite{youngsn}, and we make use of the CCL inequality (in 3 dimensions),
\begin{equation}\label{CCL}
    \frac{N-1}{N} \sum_{k=1}^N H(P_k f) \leq 2 H(f),
\end{equation}
thus giving us
\begin{equation}
    \bar D_{N,2} (f) \geq \bigg(1-\frac{2N}{(N-1)^2}\bigg) H(f).
\end{equation}

As for the much simpler case of $\alpha=0$ with constant jump rates, one can easily verify by slightly adapting the computations above that this yields
\begin{equation}
    C_{N,0} = \frac{N-3}{N-1}.
\end{equation}

With this quick argument, we can then state the following:
\begin{Prop}
    The entropy production functional of the Kac process in 3 dimensions, $D_{N,\alpha}$, satisfies
    \begin{equation}
        D_{N,2} (f) \geq \bigg(1-\frac{2N}{(N-1)^2}\bigg) H(f),
    \end{equation}
    and
    \begin{equation}
        D_{N,0} (f) \geq \frac{N-3}{N-1} H(f).
    \end{equation}

    Thus, for $N \geq 4$, the spectral gap of the conjugate Kac process in 3 dimensions is at least $\frac{1}{18}$ for the super hard sphere case, and $\frac{1}{6}$ for the Maxwellian molecule case.
\end{Prop}

Sadly, this is not quite enough, and the bounds degenerate at $N=3$, and adapting Villani's approach seems to give degenerate bounds at $N=3$ as well, even though the technique is much stronger than the one we use above. The main culprit seems to be the CCL inequality \eqref{CCL} as the extra $N/(N-1)$ factor, which isn't present for the 1-dimensional case, loses us enough for the method to fail.

This makes it clear that we're dealing with an edge case, just outside the reach of most methods, even if we can simultaneously handle both the entropy production and the spectral gap problems for all $N$ above 3.

Thus, a careful analysis of the dynamics and geometry of the problem is needed to obtain such a bound, and we make use of the information we have about the spectrum of the conditional expectation operator $K$ on the sphere and properties of orthogonal polynomials for this purpose.

\subsection{Monte Carlo simulations}

In order to observe this relaxation to equilibrium, we perform Monte Carlo simulations of the Kac process with 3 interacting particles. Even if we can't prove entropy production inequalities for the conjugate Kac process when $N=3$, this method allows us to obtain an estimate of 0.3 for the exponential entropy decay rate, and thus an estimate of about 0.15 for the spectral gap, which is only about an order of magnitude larger than the theoretical result we achieve.

Based on the factorization formula \eqref{measurefactor}, we notice that the uniform measure in $S_3$, $\sigma_3$ factorizes as
$$ d\sigma_3 = d\sigma_2 \times d\nu_3, $$
with
$$ d\nu_3 := \frac{|\mathbb{S}^2|}{|\mathbb{S}^5|} (1-|v|^2)^\frac12 dv,$$
induced by the transformation $T_1: S_2 \times B \to S_3$,
$$T_1(y,v) := \bigg(\sqrt{2}v, \beta(v)y_1-\frac{v}{\sqrt{2}}, \beta(v)y_2-\frac{v}{\sqrt{2}}\bigg),$$
$$ \beta (v) := \sqrt{\frac32 (1-|v|^2)}.$$

As the system decays to equilibrium and the distribution of the velocities approaches uniform, the marginal distributions become closer do $d\nu_3$. In particular, by symmetry, we will only look at the radial component of this measure, which is proportional to $r^2(1-r^2)^\frac12 dr$. Additionally, the factorization formula and its symmetries reduce the conditional sampling problem to sampling from $S_2$, which is trivial. We refer to the radial distribution of the first particle as the sampled distribution, and the others as the implied distributions.

Thus, we can proceed as follows. Fixing the first coordinate, we sample $r = |v|$ from a given distribution using an acceptance/rejection mechanism, and sample $y_1$, $y_2$ uniformly, so that the point is in $S_3$. We then let the system evolve sampling first the jump time and the particle that stays fixed, identified with $\sqrt{2}v$, and sample $y_1$, $y_2$ uniformly on $S_2$.

We perform 1 million simulations of the process in Python, for $\alpha=0,2$, and use these simulations to build a histogram approximating the probability distribution over 100 bins and several timestamps. We then measure the relative entropy between the empirical measure and the equilibrium radial measure over time, and use a log-plot to showcase and estimate the exponential decay. This was done over a selection of initial distributions, and a "worst case scenario" is showcased here with the sampling distribution being $2(1-x)$, as most of the probability mass is concentrated close to zero.

In figures 1, 2 and 3, the rather quick relaxation is evident. By computing the relative entropy using a first order Euler method, we see that there is a log-linear relationship up to a certain point, where the distributions are close enough so that the numerical approximations and random noise prevent the entropy from decaying below $e^{-6}$. Meanwhile, figures 4, 5 and 6 show the evolution of the distributions over several timestamps.

\clearpage

\begin{figure}[!htb]\centering
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{21entropy.png}
  \caption{$\alpha=2$, sampled entropy}
  \label{fig:awesome_image1}
\endminipage\hfill
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{22entropy.png}
  \caption{$\alpha=2$, implied entropy}
\endminipage\hfill
\minipage{0.32\textwidth}%
\centering
  \includegraphics[width=\linewidth]{0entropy.png}
  \caption{$\alpha=0$, sampled entropy}
\endminipage
\end{figure}

\begin{figure}[!htb]\centering
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{0t0.png}\\
  t=0s
  \label{fig:awesome_image1}
\endminipage\hfill
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{0t05.png}\\
  t=0.5s
  \label{fig:awesome_image2}
\endminipage\hfill
\minipage{0.32\textwidth}%
\centering
  \includegraphics[width=\linewidth]{0t2.png}\\
  t=2s
\endminipage
\end{figure}
\begin{figure}[!htb]\centering
\minipage{0.32\textwidth}\centering
  \includegraphics[width=\linewidth]{0t35.png}\\
  t=3.5s
\endminipage\hfill
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{0t5.png}\\
  t=5s
\endminipage\hfill
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{0t10.png}\\
  t=10s
\endminipage
\caption{$\alpha=0$, sampled velocity}
\end{figure}

\begin{figure}[!htb]\centering
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{21t0.png}\\
  t=0s
  \label{fig:awesome_image1}
\endminipage\hfill
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{21t2.png}\\
  t=2s
  \label{fig:awesome_image2}
\endminipage\hfill
\minipage{0.32\textwidth}%
\centering
  \includegraphics[width=\linewidth]{21t6.png}\\
  t=6s
\endminipage
\end{figure}
\begin{figure}[!htb]\centering
\minipage{0.32\textwidth}\centering
  \includegraphics[width=\linewidth]{21t10.png}\\
  t=10s
\endminipage\hfill
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{21t16.png}\\
  t=16s
\endminipage\hfill
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{21t24.png}\\
  t=24s
\endminipage
\caption{$\alpha=2$, sampled velocity}
\end{figure}

\begin{figure}[!htbp]
\minipage{0.32\textwidth}\centering
  \includegraphics[width=\linewidth]{22t0.png}\\
  t=0s
  \label{fig:awesome_image1}
\endminipage\hfill
\minipage{0.32\textwidth}\centering
  \includegraphics[width=\linewidth]{22t2.png}\\
  t=2s
  \label{fig:awesome_image2}
\endminipage\hfill
\minipage{0.32\textwidth}%
\centering
  \includegraphics[width=\linewidth]{22t6.png}\\
  t=6s
  \label{fig:awesome_image3}
\endminipage
\end{figure}

\begin{figure}[!htbp]
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{22t10.png}\\
  t=10s
\endminipage\hfill
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{22t16.png}\\
  t=16s
\endminipage\hfill
\minipage{0.32\textwidth}\centering
  \includegraphics[width=\linewidth]{22t24.png}\\
  t=24s
\endminipage
\caption{$\alpha=2$, implied velocity}
\end{figure}

\clearpage

\section{Proof of the main result}\label{mainproof}

\subsection{The correlation operator on the sphere} \label{corroperator}

The main result is based on careful analysis of the eigenfunctions and the spectrum of $K$. We turn our attention to the study of this operator.

Let $B$ be the unit ball in ${{\mathbb R}}^3$, $N\geq3$, $\nu_N$ as defined in \eqref{equilibriumradial} and consider the projections $\pi_k (\vec v): S_N \to B$,
\begin{equation}
    \pi_k (\vec v) := \frac{1}{\sqrt{N-1}}v_k.
\end{equation}
By formula \eqref{measurefactor}, we then have that
\begin{equation*}
    \int_B \psi (v) d\nu_N = \int_{S_N}\psi(\pi_k(\vec v)) d\sigma_N,
\end{equation*}
for any integrable function $\psi$ on $B$. We then define the operator $K$ on $L^2(B,\nu_N)$ by

\begin{equation}
    \langle \psi_1, K\psi_2 \rangle_{L^2(B,\nu_N)} := \int_{S_N} \psi_1^*(\pi_1(\vec v)) \psi_2(\pi_2(\vec v)) d\sigma_N.
\end{equation}
Note that the symmetries of the variables make $K$ self-adjoint. Furthermore, a simple rescaling, to extend $K$ to $L^2(\sigma_N)$, allows us to interpret this operator simply as a conditional expectation on the uniform sphere
\begin{equation*}
    K\xi (v) = {{\mathbb E}} [ \xi (v_1) | v_2 = v],
\end{equation*}
but we will not consider this scaling as the former simplifies the treatment of eigenvalues.

Carlen, Carvalho and Loss make use of this operator to prove quantitative chaos estimates; that is, to quantify how far from independent the coordinates are, asymptotically in $N$, and most of their work is based on extensive study of the spectrum of $K$, built up over \cite{kspectrum}, \cite{markovsp}.

Before we continue, define the functions
\begin{equation}
    \eta_0 = 1, \eta_j(v) = v_j,\text{ for } 1\leq j \leq 3, \text{ and } \eta_4(v) = (|v|^2-1)/(N-1).
\end{equation}

We will now present the most important facts obtained in these papers for general $N$, and later turn to the our $N=3$ case. Following the results in \cite{kspectrum}, we have this next lemma:
\begin{Lem}
    Let $N\geq3$. The operator $K$ is compact. The function $\eta_0$ is an eigenfunction of $K$ with eigenvalue 1, and spans the corresponding eigenspace. The functions $\eta_j$, for $1\leq j \leq 4$, are eigenfunctions with eigenvalue $-\frac{1}{N-1}$, and they form an orthogonal basis for this subspace. No other eigenvalues of $K$ are larger in absolute value then $\frac{5N-3}{3(N-1)^3}$. Thus, for all $\psi \in L^2(B,\nu_N)$ orthogonal to these $\eta_j$,
    \begin{equation}
        \|K\psi\|_2^2 \leq \frac{5N-3}{3(N-1)^3}\|\psi\|^2_2.
    \end{equation}
    Finally, every eigenvalue $\kappa$ of $K$, other than 1, $-\frac{1}{N-1}$ and $\frac{5N-3}{3(N-1)^3}$, satisfies
    \begin{equation}
        -\frac{7N-3}{3(N-1)^4} \leq \kappa \leq \frac{5N-3}{3(N-1)^3}.
    \end{equation}
\end{Lem}

Unfortunately, these bounds are not enough when $N=3$ and are just slightly off to allow us to obtain the main result more easily. On the other hand, the authors also manage to establish that $K$ has a complete basis of eigenfunctions of the form
\begin{equation}\label{basis}
    g_{n,\ell,m} = h_{n,\ell}(|v|)|v|^\ell \mathcal{Y}_{\ell,m}(v/|v|),
\end{equation}
where $\mathcal{Y}_{\ell,m}$ denotes the $\ell$-th degree spherical harmonic and, with $t=|v|^2-1$, $\alpha=\frac{3N-8}{2}\bigg ( =\frac12 \bigg)$ and $\beta = \ell + \frac12$,
\begin{equation*}
    h_{n,\ell}(|v|) = P_n^{(\alpha,\beta)}(t),
\end{equation*}
the $n$-th degree Jacobi polynomial (recall that the Jacobi polynomials are the orthogonal polynomials on the interval $[-1, 1]$ for the weight $(1-x)^\alpha (1+x)^\beta$). For convenience, we let capital $P$ denote these polynomials under Rodrigues' normalization,
\begin{equation}
    P_n^{(\alpha,\beta)}(t) = \frac{(-1)^n}{2^n n} (1-x)^{-\alpha}(1+x)^{-\beta} \frac{d^n}{dx^n}\bigg( (1-x)^{\alpha+n}(1+x)^{\beta+n}\bigg),
\end{equation}
while $p_n^{(\alpha,\beta)}$ will correspond to the $L^2$ orthonormal basis. Equation \eqref{basis} also allows us to conclude that the spectrum of $K$ consists of eigenvalues $\kappa_{n,\ell}$, independent of $m$ and given by the explicit formula
\begin{equation}
    \kappa_{n,\ell} = \frac{P_n^{(\alpha, \beta)}\Big(-1+\frac{2}{(N-1)^2}\Big)}{P_n^{(\alpha, \beta)}(1)} \bigg(-\frac{1}{N-1} \bigg)^\ell.
\end{equation}

While similar formulas derived from the above were useful in the asymptotic $N$ case, a large amount of cancellation happens and makes the eigenvalues take values much smaller than implied by naive bounds. Furthermore, it appears that the eigenvalues display an explicit $\mod 3$ behavior. By this, we mean that for each $\ell$, plotting the eigenvalues for $n \mod 3$ separately yields 3 eventually monotonous sequences, as we can observe bellow:

\begin{figure}[!htb]\centering
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{l5.png}
  \caption{First 300 values of $\kappa_{n,5}$}
  \label{fig:awesome_image1}
\endminipage\hfill
\minipage{0.32\textwidth}
\centering
  \includegraphics[width=\linewidth]{l15.png}
  \caption{First 300 values of $\kappa_{n,15}$}
\endminipage\hfill
\minipage{0.32\textwidth}%
\centering
  \includegraphics[width=\linewidth]{l5to10.png}
  \caption{Overlap of the plots from $\ell=6$ to $10$}
\endminipage
\end{figure}

As these 3 different behaviours are included simultaneously in the sequences, it further makes difficult the tasks of obtaining good uniform bounds.

One can also make use of the recurrence relations for Jacobi polynomials \cite{orthogonalpolynomials} to obtain the same type of relations for the eigenvalues $\kappa_{n,\ell}$. More precisely, for each $\ell$, we have that
\begin{equation}\label{krecurrence}
    -\frac12 \kappa_{n,\ell} = \frac{n+\ell+\frac12}{2n+\ell+1}\kappa_{n,\ell-1} + \frac{n+\frac12}{2n+\ell+1} \kappa_{n+1,\ell-1}\ .
\end{equation}

Since we also know that
\begin{equation}\label{kzero}
    \kappa_{n,0} = \frac{1}{n+1} (-1)^n \sin\bigg(\frac{n\pi}{3}\bigg),
\end{equation}
this provides a fast and efficient way to at least numerically compute all the needed eigenvalues. We can, however, solve this recurrence relation in terms of the $\ell=0$ values, even if the obtained representation isn't very useful:

\begin{equation}
    \kappa_{n,\ell} = (-2)^\ell \sum_{j=0}^\ell \binom{\ell}{j} C_{n,\ell}^{(j)} \kappa_{n+j,0}\ ,
\end{equation}
with
\begin{equation}
    C_{n,\ell}^{(j)} = \frac{\prod_{i=1}^\ell \{n+i+1/2\}}{\prod_{i=0,i\neq j}^\ell \{2n+i+j+2\}}\ .
\end{equation}
One can see that the missing term in the denominator, when $i=j$, is exactly 2 times the numerator of $\kappa_{n+j,0}$. Unfortunately, this representation shows no indication of decay, meaning that there is a large amount of cancellation occurring (the terms are large, but very close in absolute value). Collecting the sine terms may, however, shed some light on the $\mod 3$ question.

Thus, while explicit values of Jacobi polynomials are not trivial to compute, the authors make use of uniform pointwise bounds obtained by Nevai, Erdelyi and Magnus \cite{unifjacobi}, a surprisingly deep result, to obtain monotonous bounds on these eigenvalues:

\begin{Lem}
    We have that
    \begin{equation}
        \kappa^2_{n,\ell} \leq \frac{2e}{\pi}g_1(n,\ell,N)g_2(N)g_3(n,N)g_4(n,\ell,N)\ ,
    \end{equation}
    with
    \begin{align}
        g_1(n,\ell,N) & = \bigg( \frac{4+\sqrt{9N^2-48N+65+4\ell^2+4\ell}}{3N+4n+2\ell-5} \bigg)\ ,\nonumber \\
        g_2(N) & = \bigg( \frac{(N-1)^2}{N(N-2)} \bigg)^{(3N-7)/2}\ ,\nonumber \\
        g_3(n,N) & = \frac{\Gamma(n+1)\Gamma(\frac32N-3)}{\Gamma(n+\ell+\frac32 N - \frac52)}\ ,\nonumber \\
        g_4(n,\ell,N) & =  \frac{(N-1)^2\Gamma(n+\ell+\frac32)\Gamma(\frac32N-3)}{\Gamma(n+\ell+\frac32N-\frac52)}\ .
    \end{align}
\end{Lem}

This bound can be enlarged to allow for the monotonous behaviour. This time, however, when $N=3$, we are actually in a better position to further simplify the achieved bound, yielding
\begin{equation}\label{kappahatbound}
    \kappa_{n,\ell}^2 \leq \frac{8e}{3}\frac{\Gamma(n+1)}{\Gamma(n+3/2)} \frac{\Gamma(n+\ell+3/2)}{\Gamma(n+\ell+2)} \leq \frac{8e}{3} \frac{1}{(n+1)^{1/2}} \frac{1}{(n+\ell+3/2)^{1/2}} =: \hat \kappa_{n,\ell}^2 \ .
\end{equation}

Now notice that, for all $\ell \geq 4$,
\begin{equation}
    (n+1)(n+\ell+3/2) \geq (n + \sqrt{\ell})^2\ ,
\end{equation}
so that if we define
\begin{equation}\label{kappatildebound}
\widetilde{\kappa}_{n,\ell} = \sqrt{\frac{8e}{3} }\frac{1}{\sqrt{n + \sqrt{\ell}}}\ ,
\end{equation}
we have $\hat{\kappa}_{n,\ell} \leq \widetilde{\kappa}_{n,\ell}$. Both upper bounds will be useful later on. Inequality \eqref{kappahatbound} is more precise, especially when we are looking for bounds independent of $\ell$. Nevertheless, inequality \eqref{kappatildebound} has a far simpler representation and will be considered when we need to take derivatives.

\subsection{Anti-symmetric sector}\label{antisymsec}

Now that we have built the necessary tools, we first treat the simpler anti-symmetric case, which can be done mostly through clever manipulation and simple bounds. We reproduce here the approach done in \cite{spectralgap}, with some minor numerical improvements.

\begin{Lem}
    For $N=3$, the largest eigenvalue of $P^{(2)}$ on the orthogonal complement of the symmetric sector is no larger than $0.729$. Thus, either we have $\hat{\Delta}_{3,2}\geq 0.021$ or else the gap eigenfunction is symmetric.
\end{Lem}

Suppose $f$ is anti-symmetric and we assume, without loss of generality, that $f(v)=\varphi(v_1)-\varphi(v_2)$, with $\varphi$ orthogonal to the constants. Then, $P^{(2)}f=\lambda f$ becomes
\begin{equation}
    \frac{1}{N}w_{N,2}(v_1)(1-K)\varphi(v_1) -\frac{1}{N}w_{N,2}(v_2)(1-K)\varphi(v_2)=\lambda(\varphi(v_1)-\varphi(v_2))
\end{equation}

Multiplying by $\varphi(v_1)$ and integrating over $\mathcal{S}_N$, we obtain

\begin{equation}\label{eigen2form}
    \frac{1}{N}\int_{\mathcal{S}_N} w_{N,2}(v_1)|(1-K)\varphi(v_1)|^2d\sigma_N = \lambda \langle \varphi, (1-K)\varphi\rangle.
\end{equation}
From
\begin{equation}
    w_{N,2}(v_k)=\frac{N}{N-1}-\frac{N}{(N-1)^2}|v_k|^2 ,
\end{equation}
we can write \eqref{eigen2form} as
\begin{equation}
    \frac{1}{N-1} \langle \varphi, (1-K)^2 \varphi \rangle - \frac{1}{(N-1)^2} \langle (1-K)\varphi,|v|^2 (1-K)\varphi\rangle = \lambda \langle \varphi, (1-K)\varphi\rangle
\end{equation}

After decomposing $\sqrt{1-K}\varphi = \psi + \zeta$, where $\psi$ is orthogonal to the constants, the components $v_j$ and $|v|^2$, we have that:
\begin{align}
\langle \varphi, (1-K)^2 \varphi \rangle
& = \langle \psi + \zeta, (1-K)(\psi + \zeta) \rangle \nonumber \\
& = \langle \psi, (1-K)\psi \rangle + (1+1/(N-1))\|\zeta\|^2_2 \nonumber \\
& = \|\sqrt{1-K}\psi\|^2_2+\frac{N}{N-1}\|\zeta\|^2_2
\end{align}

and

\begin{align}
\langle (1-K)\varphi,|v|^2(1-K)\varphi \rangle = & \langle \sqrt{1-K}(\psi+\zeta),|v|^2\sqrt{1-K}(\psi+\zeta) \rangle \nonumber \\
= & \langle \sqrt{1-K}\psi,|v|^2\sqrt{1-K}\psi \rangle + \langle \sqrt{1-K}\zeta,|v|^2\sqrt{1-K}\zeta\rangle \nonumber \\
+ &  2\langle |v| \sqrt{1-K}\psi,|v|\sqrt{1-K}\zeta \rangle,
\end{align}
since $\zeta$ is an eigenfunction of $K$ with eigenvalue $-1/(N-1)$, it is also an eigenfunction of $\sqrt{1-K}$, with eigenvalue $\sqrt{1+\frac{1}{N-1}}=\sqrt{\frac{N}{N-1}}$.

It then follows that
\begin{align}
    \langle \sqrt{1-K}\varphi,|v|^2 \sqrt{1-K}\varphi\rangle
    & \geq \langle \sqrt{1-K}\psi,|v|^2\sqrt{1-K}\psi \rangle + \langle \sqrt{1-K}\zeta,|v|^2\sqrt{1-K}\zeta\rangle \nonumber \\
    & - 2\||v|\sqrt{1-K}\psi\||_2\sqrt{\frac{N}{N-1}}\||v|\zeta\|_2 \nonumber \\
    & \geq \bigg(1-\frac{1}{t}\bigg)\langle \sqrt{1-K}\psi,|v|^2\sqrt{1-K}\psi \rangle + (1-t)\frac{N}{N-1}\||v|\zeta\|_2^2,
\end{align}
using the Cauchy-Schwarz and arithmetic-geometric mean inequalities.

Before we move on, it is useful to note that $\||v|\zeta\|_2^2 \geq \|\zeta\|^2_2$. To see this for $N=3$, write
\begin{equation}\label{zeta}
    \zeta = \sum_{j=1}^4 a_j \eta_j,
\end{equation}
with $\eta_j(v) = \boldsymbol{e}_j \cdot v$ for $j=1,2,3$ and $\eta_4 = |v|^2-1$ as discussed in section \ref{corroperator}.

From \eqref{zeta}, a simple computation yields
\begin{equation}
    \| |v|\zeta\|^2_2 = \sum_{j=1}^4 |a_j|^2 \||v|\eta_j\|^2_2\ ,
\end{equation}
and while a bit more cumbersome, one can also show that
\begin{equation}
    \int_{\mathcal S_3} |v_1|^4 d\sigma_3 = \frac54
\end{equation}
and
\begin{equation}
    \int_{\mathcal S_3} |v_1|^6 d\sigma_3 = \frac74.
\end{equation}
Together, the equations above give us
\begin{equation}
    \||v|\eta_j \|^2_2 = \frac{5}{4} \text{ for } j=1,2,3 \text{ and } \||v|\eta_4\|=1,
\end{equation}
which prove what we needed.

Finally, from the trivial bound $\langle \sqrt{1-K}\psi,|v|^2\sqrt{1-K}\psi \rangle \leq (N-1)\|\sqrt{1-K}\psi\|^2_2$, we have that, for all $0<t<1$,
\begin{equation}
    \langle \sqrt{1-K}\varphi,|v|^2 \sqrt{1-K}\varphi\rangle
    \geq \bigg(1-\frac{1}{t}\bigg)(N-1)\|\sqrt{1-K}\psi\|^2_2 + (1-t)\frac{N}{N-1}\|\zeta\|_2^2\ .
\end{equation}

Specializing to $N=3$, the second most negative eigenvalue of $K$ is $-\frac{3}{8}$, which yields $\|\sqrt{1-K}\psi\|^2_2\leq \frac{11}{8}\|\psi\|^2_2$. Combining all of the above,

\begin{align}
    \lambda (\|\psi\|^2_2+\|\zeta\|^2_2)  & \leq \frac{1}{2}\cdot\frac{11}{8} \|\psi\|^2_2+\frac{3}{4}\|\zeta\|^2_2-\frac{1}{2}\cdot\frac{11}{8}\bigg(1-\frac{1}{t}\bigg)\|\psi\|^2_2-\frac{1}{4}\frac{3}{2}(1-t)\|\zeta\|^2_2 \nonumber \\
    & = \frac{11}{16t}\|\psi\|_2^2+	\frac{3}{8}\bigg(1+t\bigg)\|\zeta\|^2_2 \ .
\end{align}
With $t=0.943$, this yields $\lambda\leq 0.729$.

\subsection{Symmetric sector} \label{symsec}

The remaining task is to bound the second largest eigenvalue of $P^{(2)}$
in the symmetric sector. Assume $f$ is of the form
\begin{equation}
    f(\Vec{v})=\sum_{j=1}^N \varphi(v_j).
\end{equation}

Then, we can write $P^{(2)}f = \lambda f$ as
\begin{equation}
    \frac{1}{N}\sum_{j=1}^N w_{N,2}(v_j)(\varphi(v_j) + (N-1)K\varphi(v_j))=\lambda\sum_{j=1}^N \varphi(v_j).
\end{equation}
Theorem 2.5 in \cite{spectralgap} is here extremely important, as it allows us to treat the above sum component wise:

\begin{equation}
    \frac{1}{N}w_{N,2}(v)(\varphi(v) + (N-1)K\varphi(v))=\lambda\varphi(v).
\end{equation}
By integrating against $\varphi(v)$, it follows that
\begin{equation}\label{inteq}
    \frac{1}{N}\int_{\mathcal{S}_N}\varphi(v_1)w_{N,2}(v_1)(\varphi(v_1)+(N-1)K\varphi(v_1))d\sigma_N = \lambda \|\varphi\|^2_2.
\end{equation}

As, once again,
\begin{equation}
    w_{N,2}(v)=\frac{N}{N-1}-\frac{N}{(N-1)^2}|v|^2,
\end{equation}
and focusing on $N=3$, \eqref{inteq} becomes
\begin{eqnarray}\label{inteq2}
 \lambda \|\varphi\|^2_2   &=& \frac{1}{2}\int_{\mathcal{S}_3}\varphi(v_1)(1- \tfrac12 |v|^2)(1+2K)\varphi(v_1)d\sigma_N\nonumber\\
& = & \frac12 \langle (1- \tfrac12 |v|^2)\varphi, (1+2K) \varphi \rangle .
\end{eqnarray}

Since
\begin{equation}
    -1 \leq \bigg(1 -\frac12 |v|^2\bigg) \leq 1,\text{ }\bigg\| \bigg(1- \frac12 |v|^2\bigg)\varphi\bigg\|_2 \leq \|\varphi\|_2,
\end{equation}
we obtain the first bound
\begin{equation}\label{crude}
    \lambda \leq \frac12 \|1+2K\|_2.
\end{equation}
Unfortunately, this bound is useless as is, since the largest eigenvalue of $K$, other than $1$, is $\frac12$, giving us $\lambda \leq 1$. Thus, we need to be more incisive on the spectrum of $K$ and its interaction with the multiplication operator $(1-\frac12 |v|^2)$.

Consider the operator
\begin{equation}
    M:=|v|^2(1+(N-1)K).
\end{equation}

Recalling the eigenbasis of $K$,
\begin{equation}
    g_{n,l,m}=h_{n,l}(|v|^2)|v|^l \mathcal{Y}_{l,m}(v/|v|),
\end{equation}
we see that $M$ commutes with rotations and thus different angular momentum sectors will be mutually orthogonal, so we can consider separately the dependence in $\ell$ and $m$ (in fact, we will ignore $m$ completely, as $g_{n,\ell,m}$, $g_{n,\ell,m'}$ belong to the same $\ell$-eigenspace). Thus, the quantities we are interested in estimating are
\begin{equation}
    \lambda_\ell := \sup_{\|\varphi\|=1} \frac14 \langle (1-t)\varphi, (1+2K) \varphi \rangle ,
\end{equation}
with the supremum being taken over the $\ell$ sector.
Moreover, as $h_{n,l}(|v|^2)=P_n^{(\alpha,\beta)}(t)$, with $t=|v|^2-1$, $\beta=l+\frac{1}{2}$, we see that $1-\frac12|v|^2 = \frac12 (1-t)$ in the Jacobi variables, and we can exploit the properties of Jacobi polynomials.

The slow simultaneous decay in $n$ and $\ell$ we obtain from expressions like \eqref{kappatildebound} pose an obstacle for computing meaningful bounds. Thus, we break down the problem over several ranges.

First, we look for a uniform bound given large enough $\ell>\ell^*$, but such that $\ell^*$ is as small as possible.

For $5<\ell<\ell^*$, the eigenvalues are already small enough for \eqref{crude} to work as well, and we make use of a uniform bound in $n$ to reduce this problem to explicitly computing a finite amount of these. Note that the recurrence formula (recurrence) makes this fairly trivial and efficient.

Finally, we provide an argument based on linear algebra for $\ell \leq5$, where closed form expressions are simple enough for this.

\subsubsection{Large $\ell$ case}

Let $f_{n,\ell,m}$ be a re-scaling of $g_{n,\ell,m}$ such that the basis is orthonormal, and fix $\ell$, $m$.

Writing $P_n := P_n^{(\alpha,\beta)}$, the three-term relation for Jacobi polynomials \cite{orthogonalpolynomials} is
\begin{equation}
    P_{n+1}(t)=(A_n t + B_n)P_n(t) + C_n P_{n-1}(t)\ ,
\end{equation}
with
\begin{align}
    A_n & = \frac{(2n+\alpha + \beta +1)(2n+\alpha+\beta+2)}{2(n+1)(n+\alpha+\beta+1)}\ , \nonumber \\
    B_n & = \frac{(\alpha^2-\beta^2)(2n+\alpha+\beta+1)}{2(n+1)(n+\alpha+\beta+1)(2n+\alpha+\beta)}\ , \nonumber \\
    C_n & = \frac{(n+\alpha)(2n+\alpha+\beta+2)(n+\beta)}{(n+1)(n+\alpha+\beta+1)(2n+\alpha+\beta)}\ ,
\end{align}
which gives
\begin{equation}
  tP_n = (1/A_n) P_{n+1} - (B_n/A_n) P_n + (C_n/A_n) P_{n-1}\ .
\end{equation}

As we are working with orthonormal functions, we have to re-scale the equality. By the orthogonality relations for Jacobi polynomials,
\begin{equation}
    \langle P_n^{(\alpha, \beta)}, P_n^{(\alpha, \beta)}\rangle_{L^2(\mu^{(\alpha, \beta)})} = \frac{2^{\alpha+\beta+1}}{2n+\alpha+\beta+1}
    \frac{\Gamma(n+\alpha+1)\Gamma(n+\beta+1) }{\Gamma(n+\alpha+\beta+1)n!}\delta_{nm}\ ,
\end{equation}
we first compute
\begin{align}
    & \|P_n^{(\alpha,\beta)}\|_2^2 = \frac{2^{\alpha+\beta+1}}{2n+\alpha+\beta+1}
    \frac{\Gamma(n+\alpha+1)\Gamma(n+\beta+1) }{\Gamma(n+\alpha+\beta+1)n!}\ ,\nonumber \\
    & \frac{\|P_{n-1}^{(\alpha,\beta)}\|_2^2}{\|P_n^{(\alpha,\beta)}\|_2^2} = \frac{2n+\ell+2}{2n+\ell} \frac{n(n+\ell+1)}{(n+1/2)(n+\ell+1/2)}\ , \nonumber\\
    & \frac{\|P_{n+1}^{(\alpha,\beta)}\|_2^2}{\|P_{n}^{(\alpha,\beta)}\|_2^2} = \frac{2n+\ell+2}{2n+\ell+4}\frac{(n+3/2)(n+\ell+3/2)}{(n+1)(n+\ell+2)}\ .
\end{align}
Defining the scaling factors
\begin{align}
    F_n^-  & := \frac{\|P_{n-1}\|}{\|P_n\|} = \sqrt{\frac{n(2n+\alpha+\beta+1)(n+\alpha+\beta)}{(n+\alpha)(n+\beta)(2n+\alpha+\beta-1)}}\ , \nonumber \\
    F^+_n & := \frac{\|P_{n+1}\|}{\|P_n\|} = \sqrt{\frac{(n+\alpha+1)(n+\beta+1)(2n+\alpha+\beta+1)}{(n+1)(n+\alpha+\beta+1)(2n+\alpha+\beta+3)}}\ ,
\end{align}
the three-term relation for the multiplication operator in the orthonormal Jacobi polynomials is
\begin{equation}
    tp_n = c_n p_{n-1} + a_n p_n + b_n p_{n+1},
\end{equation}
with
\begin{equation}
    a_n := -\frac{B_n}{A_n} \text{, }b_n :=\frac{F^+_n}{A_n}\text{, }c_n := \frac{F^-_n C_n}{A_n}.
\end{equation}

Notice that, by the self-adjointness of multiplication by $t$, we must have $c_n = b_{n-1}$. After some algebraic manipulation, we verify this and also obtain
\begin{equation}
    b_{n,\ell} = 2\sqrt{\frac{(n+1)(n+3/2)(n+\ell+3/2)(n+\ell+2)}{(2n+\ell+2)(2n+\ell+3)^2(2n+\ell+4)}},
\end{equation}
\begin{equation}
    a_{n,\ell} = \frac{\ell(\ell+1)}{(2n+\ell+1)(2n+\ell+3)}.
\end{equation}
Evidently $a_{n,\ell} \geq \widetilde{a}_{n,\ell}$ where
\begin{equation}\label{atilde}
\widetilde{a}_{n,\ell} := \left(\frac{\ell}{2n+\ell +3}\right)^2 = \left(1 - \frac{2n+3}{2n+\ell + 3}\right)^2\ .
\end{equation}
It is clear that $\widetilde{a}_{n,\ell}$ is decreasing in $n$ and increasing in $\ell$.
It will also be useful to have a bound on $ b_{n,\ell}$ that is simpler to estimate. For this, notice that we can pair the 'close' terms in the numerator, and compare with those in the denominator. We then have

\begin{align}
    b_{n,\ell} \leq & 2\sqrt{\frac{(n+3/2)^2(n+\ell+2)^2}{(2n+\ell+2)^2(2n+\ell+3)^2}}
    =  \frac{2n+3}{2n+\ell+3} \frac{n+\ell+2}{2n+\ell+2} \nonumber \\
    =  &\bigg(1-\frac{\ell}{2n + \ell + 3}\bigg)\bigg(1 - \frac{n}{2n+\ell+2}\bigg)
    \leq  \bigg(1-\frac{\ell}{2n + \ell + 3}\bigg)\bigg(1 - \frac{n}{2n+\ell+3}\bigg)\ .
\end{align}
We therefore define
\begin{equation}\label{btil}
\widetilde{b}_{n,\ell} :=  \bigg(1-\frac{\ell}{2n + \ell + 3}\bigg)\bigg(1 - \frac{n}{2n+\ell+3}\bigg)\ .
\end{equation}

In order to better understand the behaviour of these sequences, we treat them as functions of continuous variables. Differentiating in $\ell$ we find
\begin{equation}
    \frac{{\partial}}{{\partial}\ell}\widetilde{b}_{n,\ell} = -\frac{(2\ell +6)n + 3\ell + 9}{(2n + \ell +3)^3}\ ,
\end{equation}
so that $\widetilde{b}_{n,\ell} $ is decreasing in $\ell$.  Differentiating in $n$, we see that
\begin{equation}
    \frac{{\partial}}{{\partial}n}\widetilde{b}_{n,\ell} = \frac{2\ell^2 + 3\ell - (6n+9)}{(2n + \ell +3)^3}\ ,
\end{equation}
so that $\widetilde{b}_{n,\ell}$ is increasing in $n$ for
$$
n \leq  \frac{2\ell^2 + 3\ell -9}{6}\ .
$$

Suppose now $\varphi$ belongs to the $(\ell,m)$ angular momentum sector, so that we can expand it as
\begin{equation}
    \varphi = \sum_{n=0}^{+\infty} \varphi_n f_{n,\ell,m} \ .
\end{equation}

Observing that
\begin{align}
    & (1+2K)\varphi = \sum_{n=1}^{+\infty}(1+2k_{n,\ell})\varphi_n f_{n,\ell,m}\ , \\
    & (1-t)\varphi = \sum_{n=1}^{+\infty}(b_{n-1} \varphi_{n-1} + (1-a_n)\varphi_n + c_{n+1} \varphi_{n+1})f_{n,\ell,m}\ ,
\end{align}
we finally derive
\begin{align}
    \langle (1-t)\varphi,(1+2K)\varphi\rangle =
    & - \sum_{n=1}^{+\infty}b_{n-1,\ell} (1+2k_{n,\ell})\varphi_{n-1}\varphi_n \nonumber\\
    & + \sum_{n=1}^{+\infty}(1-a_{n,\ell})(1+2k_{n,\ell})\varphi_n^2   \nonumber\\
    & - \sum_{n=1}^{+\infty}b_{n, \ell} (1+2k_{n,\ell})\varphi_{n+1}\varphi_n\ .
\end{align}

Taking the supremum on each term and using Cauchy-Schwarz on the $\varphi_n \varphi_{n\pm 1}$ terms yields, and recalling the upper bound \eqref{kappatildebound} $\kappa_{n,\ell} \leq \widetilde \kappa_{n,\ell}$,
\begin{align}
    \lambda_\ell & \leq \frac14 \bigg[2\sup_n \{ b_{n,\ell} (1+2\kappa_{n,\ell}) \} + \sup_n \{(1-a_{n,\ell})(1+2\kappa_{n,\ell}) \}\bigg]  \nonumber\\
    & \leq \frac14 \bigg[2\sup_n \{ \widetilde{b}_{n,\ell} (1+2 \widetilde \kappa_{n,\ell}) \} + \sup_n \{(1-\widetilde{a}_{n,\ell})(1+2\widetilde \kappa_{n,\ell}) \}\bigg].
\end{align}

We can then turn our attention to the sequences $\widetilde{b}_{n,\ell} (1+2 \widetilde \kappa_{n,\ell})$ and $(1-\widetilde{a}_{n,\ell})(1+2\widetilde \kappa_{n,\ell})$.

\begin{Prop}
For all $\ell \geq 70$, $\lambda_{\ell} \leq 0.73016$.
\end{Prop}

\begin{proof}
We first compute the derivative in $n$ of $(1-\widetilde{a}_{n,\ell})(1+2\widetilde{ \kappa}_{n,\ell})$,
 which we find
$$
\frac{4\ell^2}{3} \frac{ 4\sqrt{6e} + 3\sqrt{n+\sqrt{\ell}}}{(2n+\ell + 3)^3  \sqrt{n+\sqrt{\ell}}}  - \frac23 \frac{\left(1- \frac{k^2}{(2n+\ell+3)^2}\right)\sqrt{6e}}{(n + \sqrt{\ell})^{3/2}}\ .
$$
which has the same sign as
\begin{equation}\label{abound}
{2\ell^2} \frac{ 4\sqrt{6e} + 3\sqrt{n+\sqrt{\ell}}}{(2n+\ell + 3)^3 }  - \frac{\left(1- \frac{k^2}{(2n+\ell+3)^2}\right)\sqrt{6e}}{n + \sqrt{\ell}}\ .
\end{equation}
Now assume $n\geq \ell^{3/2}$.  Then, we have that ${2\ell^2} \leq (2n+\ell + 3)^{4/3}$, $\sqrt{n+\sqrt{\ell}} \leq (2n+\ell + 3)^{1/2}$,
and
$$
\frac{k^2}{(2n+\ell+3)^2} \leq \frac{1}{4\ell}\ .
$$
Thus, when $n\geq \ell^{3/2}$, the quantity in \eqref{abound} is bounded above by
$$
 \frac{ 4\sqrt{6e/\sqrt{\ell}} + 3}{(n + \sqrt{\ell})^{7/6} }  - \frac{\left(1- \frac{1}{4\ell}\right)\sqrt{6e}}{n + \sqrt{\ell}} \ .
$$
Hence our derivative is negative when $n \geq \ell^{3/2}$ and
$$
(n + \sqrt{\ell})^{1/6} \geq  \frac{ 4\sqrt{6e/\sqrt{\ell}} + 3}{\left(1- \frac{1}{4\ell}\right)\sqrt{6e}}\ .
$$
For $\ell = 70$, this is satisfied for $n\geq 86$, and $70^{3/2} \leq 586$. Hence, for $\ell =70$, $(1-\widetilde{a}_{n,\ell})(1+2\widetilde{ \kappa}_{n,\ell})$ is decreasing in $n$ for al $n\geq 586$, and these values of $n$ and $\ell$ is it less than $1.22$.
Checking the finitely many remaining cases, one finds that the maximum occurs at $n=66$, and is less than $1.4351$.

Proceeding similarly for $2 \widetilde{b}_{n,\ell} (1+2\tilde \kappa_{n,\ell})$ we find the for $\ell =70$, the maximum occurs at $n= 53$, and is no more than $1.4855$.
Altogether we find,
$$\lambda_{70} \leq \frac{1.4855+1.4351}{4} \leq 0.73016\ .$$
Since it is clear that $b_{n,\ell} (1+2\tilde \kappa_{,\ell})$ and $(1-a_{n,\ell})(1+2\tilde \kappa_{n,\ell})$ are decreasing in $\ell$, this proves that $\lambda_\ell \leq 0.73016$ for all $\ell \geq 70$.
\end{proof}

\subsubsection{Small $\ell$}

By inequality \eqref{kappatildebound}, and ignoring the dependence in $\ell$, we see that for $n\geq151$ we have $\kappa_{n,\ell} \leq 0.23$.
Numerically checking, this holds as well for all $6\leq \ell \leq 50$ and $0\leq n \leq 151$.  Thus, for $5 < \ell < 70$, the norm of $K_\ell$, the restriction of $K$ to the $\ell$ angular momentum sector satisfies
$$
\|K_\ell\| \leq  0.23\ .
$$
It follows from \eqref{crude} that for $5 < \ell < 70$,
$$
\lambda_\ell \leq \frac12(1+2\cdot0.23) = 0.73\ .
$$
Taking into consideration the $\ell$ dependency can also help reduce the amount of cases needed to check.

This leaves only the $\ell = 0,1,2,3,4,5$ cases. Below, we provide a computation for $\ell=0$, making use of the explicit expressions obtained via \eqref{krecurrence} and \eqref{kzero}. The same computation can be carried out analogously for the remaining cases, at the expense of explicitly computing the three term relation factors and the eigenvalues, and taking an appropriate choice of index for a break point.

For $\ell = 0$, the operator $\varphi(v) \mapsto  (1 - \tfrac12 |v|^2)\varphi(v)$  is given in the Jacobi basis by the matrix
$$
X := \frac14 \left[\begin{array} {cccccc}
\phantom{-}2 & -1 & \phantom{-}0 & \phantom{-}0 & \phantom{-}0 & \dots\\
-1 & \phantom{-}2 & -1 &\phantom{-}0 & \phantom{-}0  & \dots\\
\phantom{-}0 & -1 & \phantom{-}2 & -1 & \phantom{-}0  & \dots\\
\phantom{-}0 & \phantom{-}0 & -1 & \phantom{-} 2 & -1 & \dots \\
\phantom{-}0 & \phantom{-}0 &\phantom{-} 0 &-1& \phantom{-}2  & \dots\\
\phantom{-}\vdots & \phantom{-}\vdots &\phantom{-}\vdots & \phantom{-}\vdots & \vdots &\ddots
\end{array}\right] \ .
$$

Let $Y$ be the diagonal matrix whose $n$th diagonal entry is $1 + 2\kappa_{n+1,0}$.
The upper $5\times 5$ block of $Z := \frac12 (YX +XY)$ is
$$ \left[\begin{array}{ccccc}
\phantom{-}\frac12 & -\frac{5}{16} & \phantom{-}0 & \phantom{-}0  & \phantom{-}0 \\
-\frac{5}{16} & \phantom{-}\frac34 & -\frac{21}{80} &\phantom{-} 0 & \phantom{-}0 \\
\phantom{-}0& -\frac{21}{80} & \phantom{-}\frac{3}{10} & -\frac15 & \phantom{-}0 \\
\phantom{-}0 & \phantom{-}0 & -\frac15 & \phantom{-}\frac12  & -\frac27  \\
\phantom{-}0 & \phantom{-}0 & \phantom{-}0 & -\frac27 & \phantom{-}\frac{9}{14}
\end{array}\right]\ .
$$
According to Maple, the largest eigenvalue of  this matrix  is less than $1.0412$.   We also have that
$$
Z_{6,5} = Z_{5,6} = -\frac{57}{224}\ .
$$
Let $\varphi$ be any unit vector in $\ell^2$, and let $\eta$ be the vector with $\eta_j = \varphi_j$ for $j=1,2,3,4$, and $\eta_j =0$ for $j > 4$. Let $\zeta := \varphi - \eta$. Then
$$\langle \varphi, Z \varphi\rangle =  \langle \eta, \widehat{Z}\eta\rangle
 +    \langle \zeta, Z \zeta\rangle - \frac{57}{224} 2\varphi_4\varphi_5 \ .$$
 We have
 $$
 \langle \eta, \widehat{Z}\eta\rangle  \leq 1.0412 \|\eta\|^2\ ,
 $$
 and by the Schwarz argument, since the largest relevant eigenvalue of $K$ is $\tfrac{1}{10}$,
 $$
 \langle \zeta, Z \zeta\rangle  \leq \frac{6}{5}\|\zeta\|^2\ .
 $$
 Thus, $\langle \varphi, Z \varphi\rangle$ is less than the largest eigenvalue of
 $$
 \left[ \begin{array}{cc}
 1.0412 & -\frac{57}{224}\\  -\frac{57}{224} & \frac65\end{array}\right] \ ,
 $$
 which is no more than $1.388$. Dividing by $2$, we get our bound on the top eigenvalue in the $\ell = 0$ sector, $0.694$.

\section{Conclusion}
Comparing the bounds obtained in each sector, theorem \ref{maintheorem} follows. Obviously, the result is only as good as the worst bound we obtain. While a bound of $0.02$ is certainly achievable as we have seen, it is not clear if this method could yield a much better result.

Nevertheless, we appear to be off by one order of magnitude, which is decent enough, and we saw how the erratic behaviour of the eigenvalues $\kappa_{n,\ell}$ poses an obstacle towards this.

Thus, a future goal would be proving entropy production inequalities with non-degenerate 'constants', first for the conjugate process and possibly for the Kac process itself following Villani's work. This is out of reach at the moment, and careful geometrical analysis is needed due to the state space constraints.

It is possible, however, that these quantities can be related analogously to their respective spectral gaps, and new insights can be obtained this way.

\renewcommand{\abstractname}{Acknowledgements}
\begin{abstract}
The author would like to thank CMAF-CIO for the opportunity to develop this work and professor José Francisco Rodrigues for bringing this topic to my attention.

I also would like to extend my deepest gratitude to professor Eric Carlen, my supervisor, for all the patience, clear and intuitive explanations and everything that I have learned regarding functional analysis and mathematical physics for the duration of this project.
\end{abstract}

\clearpage

redpajama_set_name: RedPajamaArXiv


Winston Churchill famously said: "A pessimist sees the difficulty in every opportunity; an optimist sees the opportunity in every difficulty." We've all been through tough times at some point in our lives. We wouldn't be human if we hadn't. For the more resilient ones among us, it is our ability to see the positive in negative situations that helps us get through challenges and to overcome adversity.
I'm a strong believer that things happen because we're meant to learn from them. If we didn't have that negative hump to climb over, we wouldn't recognise the positive staring us right in the face. I like to think of it as the universe (or G-d, depending on your beliefs) throwing you a curveball to test your bounce-back ability and how you can grow from it. And if that setback didn't happen now, it would have to happen sooner or later, because it was placed in our path so that we could learn from it.
So the big question is: in tough times, instead of dwelling on the negative, shouldn't we concentrate on how our setbacks transform us and transform our lives for the better? Shouldn't we seek out the positive and what we're meant to learn from difficult situations?
Israel provides a great example that really drives this point home, and it's happening as we speak.
As Vocativ author Assaf Uni explains: "it's become a daily ritual here – like brushing your teeth before bed". When the shrill bomb sirens scream out, everybody instantly stops what they're doing and sprints to the nearest shelter. Families, friends and strangers crowd together in the first shelter or stairwell they can find. They wait anxiously for the boom of the Iron Dome missile defense system, the signal that it's safe to venture outside again. According to Israel's 1951 civil defense law, all homes, residential buildings and industrial buildings must be equipped with bomb shelters. The "protected space" is like a shelter but is found in individual apartment units, building floors, or other public areas. It is a reinforced concrete room that people can get into from the building's individual apartment. Some even have reinforced steel. Thanks to these protected spaces, more people can get to shelter fast. And they need to. In Israeli towns like Sderot, roughly one mile from Gaza, residents have just 15 seconds to find shelter from the time that the sirens blare to when the bomb hits. As this eye-opening video shows, not even the world's fastest man Usain Bolt, holding the world-record of 19.19 seconds for the 200m, would make it.
If the people in Israel can find the positive in their circumstances, why do many of us in South Africa battle to do this? Just listen to conversations at South African dinner tables, and the discussions shift from the impact of debilitating strikes on our economy, to crime, corruption, service inefficiencies and e-toll. All negative topics, despite everything we have going for us. We can definitely learn a thing or two from Israeli optimism in times of crisis.
In the business world, another fantastic example of positivity in tough conditions happens right here in our very own country.
South Africa tends to lag behind First-World countries in the adoption of certain trends and technologies. For example, daily deal sites like Groupon, online shopping platforms like eBay, and online music stores like Apple iTunes were well-established overseas before coming to South Africa. The iTunes Store launched in South Africa in 2012, some nine years after Apple first launched its iTunes Store in 2003 in the US. For South Africa's consumers, it is incredibly frustrating that it takes so long for American products to be launched here. To make matters worse, international courier costs and the weak rand make it prohibitive for the early adopters among us to buy these goods on overseas sites like Amazon or eBay.
But there is a positive side. This lag-time means that South Africa holds a crystal ball on global trends and the businesses that arise from them. We can see the trends happening in First-World countries before they happen in South Africa. In the eyes of an entrepreneur, there is huge opportunity in this. Based on these trends, we can start local businesses that copy successful overseas ventures, before these trends become established in South Africa. This means it is a matter of when, not if, they take off here. Me-too concepts that have worked overseas and are copied to South Africa are proven concepts. So your odds of building a successful business out of them are a lot higher.
Worldwide, Rocket Internet, the European incubator of Internet start-ups, has made this copycat approach famous (or infamous in some people's eyes). They clone the top Silicon Valley start-ups and implement them in other countries. This strategy has made Oliver Samwer, Rocket Internet's co-founder, an extremely wealthy man. His net worth is estimated at over $1 billion – not bad for a 40-year-old. If Rocket Internet can make a fortune out of the lag-time in US trends reaching other countries, so too can astute entrepreneurs in South Africa if they get the timing right.
So what is the key take-away from all of this? In every circumstance, there is something positive. Most of the time it's not noticeable. We need to stop and look. And sometimes we need to search really hard to find it. But being positive and seeing the positive in negative circumstances is a choice. A choice that only you can make, and nobody can take that away from you. So start making that choice today, no matter how tough your situation is.
Categories: Business, Entrepreneurship, Leadership | Tags: Groupon, optimism, pessimism, rocket internet, Usain Bolt | Permalink.


Vintage 1930s silk crepe kimono robe in cream and green striped floral print. Long sleeves and lined in white silk. This kimono is in very good to excellent condition, there are a few very faint marks, see last photo.


Sky Bet, Title Sponsor of The Football League, have pledged to offer the best price on promotion for all of the teams in the Sky Bet Play-Offs – and will give fans a £10 free bet on each Final.
The Play-Offs Finals at Wembley Stadium kick off with the League 2 clash between Southend United v Wycombe Wanderers on Saturday May 23, followed by the League 1 encounter between Swindon Town and Preston North End on Sunday, May 24.
The Sky Bet Championship Final, worth an estimated £120 million to the winners, sees Middlesbrough face Norwich City on Bank Holiday Monday May 25.
Sky Bet have guaranteed fans they will offer the best price on each team to be promoted from now until the games kick off.
And there will also be a £10 in-play free bet up for grabs for each of the Finals, when punters place a qualifying bet on that match.
Head of Sponsorship Edwin Martin said: "We want to make sure that Football League fans who bet with Sky Bet are given a great deal for the Play-Off Finals.


text:
Add / Correct Share Experiences UFOs Extras
Aircraft Lighter than Air Gliders
Helicopters LSA / UL Power Current Production
Gyrocopter Hang-glider/Paraglider Military
Mikoyan-Gurevich MiG-15 / I-310
Shenyang F-2
Shenyang FT-2
Aero Vodochody S-103
Aero Vodochody CS-102
PZL-Mielec Lim-2
PZL-Mielec SBLim-1
The I-310(S) was designed to meet a March 1946 requirement for a high-altitude day interceptor. The Mikoyan/Gurevich, Lavochkin and Yakovlev design bureau competed for the production contract, striving to meet the specification of a Mach 0.9 top speed, high rate of climb to 10000 m (32800 ft), good manoeuvrability at this height and above, a minimum of one hour's duration and cannon armament, combined with simplicity of design and operation.
The need for a near-sonic speed demanded the adoption of wing sweep, and the design team was able to draw on the expertise of Gurevich himself, in addition to other work carried out by Russian and German engineers. A swept-forward wing layout was examined but discarded in favour of the swept back solution. The aircraft was designed around the RD-10A turbojet of 1000 kg (2205 lb) thrust, based on the German Jumo 004. A Russian-designed powerplant with twice the thrust was in the pipeline and was expected to become available within a year, but the Anglo-Soviet trade agreement of 1946 offered a more attractive possibility.
The trade pact included the supply of 25 Rolls-Royce Nene turbojets, most of which were allocated to the Klimov engine design bureau or to research establishments. One was supplied to the Mikoyan/Gurevich team and the MiG-15 was redesigned to accommodate the fatter Nene, which had a centrifugal compressor compared with the axial unit or the Russian engine under development, and which also produced greater thrust. The Rolls-Royce Nene immediately went into production at No 45 production factory in Moscow, designated RD-45 after the factory.
The layout involved air fed from a bifurcated nose intake via four ducts which passed either side of the cockpit and then over and under the unbroken wing centre section. The wing, of almost parallel chord, was swept back by 35 degrees at the leading edge and was built up from two main spars skinned with light alloy. The upper surfaces carried two full chord fences on each side and large Fowler flaps, set at 20 degrees for takeoff and 55 degrees for landing, were attached to the wing, just forward of the trailing edge. The ailerons were the only power-operated aerodynamic controls.
The circular-section fuselage was constructed in two halves which could be separated by means of quick-release bolts at the attachment point for the rear wing spar, exposing the complete engine for maintenance. Air brakes were fitted on either side of the rear fuselage. The tailplane was swept back by 40 degrees and its incidence could be adjusted manually before takeoff. Two fuel tanks in the rear fuselage carried 90 litres (19.8 Imp gal) each, but the majority was contained in a 1225-litre (269-Imp gal) tank fitted between the wing spars. The undercarriage had a wide track - 4 m (13 ft 1.5 in) to allow operation from rough fields; the levered-suspension mainwheels retracted inwards to lie within the fork of the front main spar.
Navigation, communication and fire-control equipment was extremely simple but this solution proved to be the correct solution when the type was blooded in combat. A gyro gunsight (copied from the British GGS Mk 2) with a maximum range of 800 m (875 yards) was used to aim the two 23-mm (0.90-in) NS-23 cannon mounted in a pack under the nose. This arrangement was later replaced by a single 37-mm (1.46-in) N-37 cannon with 40 rounds on the right-hand side and a pair of NS-23s with 80 rounds each on the left. The wing hardpoints were stressed to carry up to 500 kg (1102 lb) of bombs, although two weapons of 100 kg (220 lb) each were more usual, and rockets were also fitted. Alternatively, auxiliary fuel tanks could be carried to increase endurance.
It seems likely that the first prototype of the MiG-15, designated I-310, made its maiden flight in July 1947, but crashed during low-speed trials. Several design changes were introduced as a result, including the adoption of 2 degrees of wing anhedral in place of the dihedral layout, installation of wing fences and several changes to the back end. The rear fuselage was shortened and the jet pipe cut back to reduce the amount of engine power being lost, the tailplane was removed from the top of the fin and repositioned two-thirds of the way up, and the fin itself was swept back by 56 degrees. The first of two replacement prototypes, the S-01, was flown on 30 December 1947. As the I-350 prototype, it flew for the first time in September 1953.
The revised design was far from perfect - it tended to enter a spin from a tight turn, necessitating the fitting of recovery rockets - but it was apparent that the layout was basically sound. The Mikoyan/Gurevich team had six months' headway over their competitors, and this proved decisive, although both the La-168 and Yak-30 proceeded to the flight-test stage and the former entered limited production.
Reverse engineering of the Nene by Vladimir Klimov's bureau had paralleled design development of the I-310, and as the RD-45 of 4,850 lb st (2 200 kgp), this engine powered the initial series fighter, which, as the MiG-15, was cleared for production in March 1948 (NATO Fagot), the first pre-series aircraft being delivered to the NII for evaluation seven weeks later, on 10 May.
Early production MiG-15s powered by the RD-45, copied from the Nene, reached the squadrons before the end of 1948. The RD-45F (Forsirovanny, meaning boosted), uprated from 2200 kg (4850 lb) to 2270 kg (5000 lb) thrust for take-off, soon replaced the earlier powerplant at an early production stage. Armament comprising one 37-mm N-37 and two 23-mm NS-23KM cannon.
Production under licence was begun in Poland as the LlM-1 and in Czechoslovakia as the S-102.
Variants of the RD-45F-powered fighter included the MiG-15PB escort fighter with two 132 Imp gal (600 lt) underwing slipper tanks, the MiG-15P with a pre-series Izumrud (Emerald) radar and the MiG-15SV with faster-firing NR-23 cannon, none of these being built in quantity.
A structural reappraisal of the basic MiG-15 resulting in a 198 lb (90 kg) weight reduction, minor aerodynamic changes, upgraded equipment and a Klimov-developed VK-1 turbojet (origi-nally designated RD-45FA) rated at 5,952 lb st (2 700 kgp) for takeoff, or 3000 kg (6615 lb) with water injection produced the MiG-15SD flown in September 1949. With some further changes (which included modifications to the ailerons and air brakes) this was placed in production in the following year as the MiG-15bis. The engine's external dimensions remained the same, but the mass flow was increased and the larger diameter hot end resulted in dry weight rising from 870 kg (1918 lb) in the RD-45 to 875 kg (1930 lb) in the VK-1. Fuel capacity was increased by 160 litres (35 Imp gal) and improved equipment was fitted. Perforated flaps were therefore adopted to save airframe weight, offsetting increases in other areas. Late production models of the MiG-15bis, which was the variant built in the largest numbers, carried 23-mm (0.90-in) NR-23 revolver cannon in place of the slow-firing NS-23s. Polish and Czech versions were designated LIM-2 and S-103.
Lim-3
The MiG-15bis was followed by the two-seat MiG-15UTI, code-named Midget, with the instructor sitting behind and slightly above his pupil. Some fuel capacity was sacrificed, and the top speed dropped slightly. The principal production version of the basic design, with several thousand being built until late 1951, plus others licence-built in Czechoslovakia and Poland (as the LIM-3).
The MiG-15bis standardised on an armament of two 23-mm NR-23 cannon and one 37-mm N-37 cannon.
Limited all-weather versions with Izumrud radar included the MiG-15SP-1 and SP-2. The two-seater was also used for ejection-seat experiments and formed the basis of the SP-5, the first all-weather fighter variant. An Izumrud (Emerald) fire-control radar (NATO code-name Scan Fix) was fitted, the antenna being mounted in a bullet fairing in the centre of the intake splitter. A complementary ranging radar was installed in the top lip of the intake. The definitive all-weather fighter variant, the MiG-15P, was, however, a single-seater derived from the MiG-15bis.
Other experimental models were the MiG-15SU for ground attack and featuring a pack containing cannon angled to fire downward, the MiG-15bisSB (ISh) intended for the shturmovik role and the MiG-15bisS escort fighter with 132 Imp gal (600 1t) slipper tanks, reduced armament and other changes.
A reconnaissance version, the MiG-15bisR with full armament and fitted with a single nose-mounted vertical camera below the gun magazines was developed for high -altitude sorties over Korea and was additionally employed in Europe.
A further variant was the MiG-15SB, which had twin beams extending from the wing leading edges to carry two 100-kg (220-1b) bombs, eight 55-mm (2.16-in) rockets or auxiliary fuel tanks. Rockets were fitted to reduce the takeoff run and a braking parachute similarly shortened the landing roll.
Production in the Soviet Union ended in 1953.
One of the most widely used jet fighters of all time, the MiG-15bis was also manufactured in China (as the Shenyang F-2, and the MiG-15UTI as the Shenyang FT-2), Czechoslova-kia (Mig-15bis as the Aero Vodochody S-103 and the Mig-15UTI as the CS-102) and Poland Czechoslova-kia (Mig-15bis as the PZL-Mielec Lim-2 and the Mig-15UTI as the SBLim-1) in considerable num-bers and it is known to have served with close to 30 air arms throughout the world as a front-line fighter and as an advanced trainer.
Large numbers (over 1,000) of early versions were supplied to China and North Korea, entering combat in 1951. No Allied fighter could stay with it and even the technically superior F-86 had inferior climb, ceiling and high altitude turn radius. Many thousands (14,000+) of all versions were built, and the MiG-15 remained an invaluable aircraft for air forces with little jet experience, being supplied to at least 18.
The MiG-15 took part in the first ever all jet air combat when one was destroyed by a Lockheed F-80 Shooting Star of the US Air Force over Korea on November 7, 1950, some six days after the Russian type was deployed to that theatre. The Soviet aircraft was more manoeuvrable and had a higher ceiling than its US adversaries, and the slow -firing but hard-hitting cannon carried a heavier punch than the machine-guns of the opposing fighters. The USAF's response was to introduce the North American F-86A Sabre, which scored its first MiG kill on December 17, 1950. The North Korean and Chinese pilots could not hope to match the experience of their adversaries, and on May 20, 1951, Capt James Jabara bagged his fifth and sixth MiG-15s to become the first jet ace. The USAF alone claimed 792 MiG-15s destroyed over Korea, while the US Navy also notched up a creditable score, bringing the claimed kill-to-loss ratio to about 12:1.
Engine: 1 x Klimov RD-45F, 5,005 lb st (2270 kgp)
Max speed, 648 mph (1043 km/h) at 9,845 ft (3000 m)
Max initial climb, 8,268 ft/min (42 m/sec)
Range (clean), 882 mls (1 420 km) at 236 mph (380 km/h) at 39,370 ft (12 000 m)
Empty weight, 7,456 lb (3 382 kg)
Loaded weight (clean), 10,595 lb (4 806 kg)
Span, 33 ft 1 in (10,08 m)
Length, 32 ft 1¼ in (10,04 m)
Height, 12 ft 1 2/3 in (3,70 m)
Wing area, 221.75 sq ft (20,60 sq.m).
Engine: 1 x VK-1 turbo-jet, 26.5kN
Max take-off weight: 4960 kg / 10935 lb
Wingspan: 10.1 m / 33 ft 2 in
Length: 10.1 m / 33 ft 2 in
Height: 3.7 m / 12 ft 2 in
Wing area: 20.6 sq.m / 221.74 sq ft
Max. speed: 1076 km/h / 669 mph
Ceiling: 16000 m / 52500 ft
Range w/max.fuel: 2000 km / 1243 miles
Range w/max.payload: 1400 km / 870 miles
Armament: 1 x 37mm cannon, 2 x 23mm machine-guns, 400kg of bombs or missiles
MiG-15P
Equipment: Izumrud (Emerald) radar.
MiG-15PB
Type: escort fighter
Fuel cap: two 132 Imp gal (600 1t) underwing slipper tanks
MiG-15SV
Armament: faster-firing NR-23 cannon
MiG-l5bis
Powerplant: one 2700-kg (5,952-lb) thrust Klimov VK-1 turbojet / later models one 6990-lb thrust Klimov VK-1A turbojet.
Maximum speed 1100 km/h (684 mph) at 12000 m (39,370 ft)
Initial climb rate 3500 m (11,800 ft) per minute
Service ceiling 15550 m(51,015 ft)
Ferry range 2000 km (1,242 miles)
Empty weight 3400 kg (7,495 lb)
Maximum take-off 5785 kg(12,756 lb)
Wingspan 10.08 m (33 ft 0¾ in)
Length 11.05 m (36 ft 3¼ in)
Height 3.40 m(11 ft 1¾ in)
Wing area 20.60 sq.m (221.7 sq.ft)
Armament: one 37-mm NR-37 cannon and two 23-mm NR-23 cannon, plus up to 1000 kg (2,205 lb) of external stores.
MiG-15Bis (SD)
Max speed, 692 mph (1114 km/h) at 7,220 ft (2 200 m)
Time to 16,405 ft (5 000 m), 2.1 min.
Range (clean), 826 mls (1330 km)
Loaded weight (clean), 11,175 lb (5069 kg)
Length, 35 ft 7½ in (10,86 m)
Wing area, 221.75 sq ft (20,60 sq.m)
Armament: two 23-mm NR-23 cannon and one 37-mm N-37 cannon.
Engine one 7,452-lb Klimov VK- 1F turbojet
Gross wt. 12,000 lb
Empty wt. 8,115 lb
Max speed 668 mph
Range 450 nm
Ceiling 50,855 ft
Seats 1-2.
Engine: LIS-2, 5500 lb thrust.
Fuel cap: Internal: 1000 lt, external 2 x 400 lt.
ROC: 6000+ fpm.
Service ceiling: 48,000 ft.
Loading: Clean, +8g, With drop tanks full, +4g.
Max speed: 570 kts (M 0.92), with drop tanks, 430 kts.
Stall: 110-103 kts.
Type: Two seat fighter/trainer
Engine: One Klimov RD-45 turbojet, 5,005 lbs thrust
Empty weight: 7,900 lb
Loaded weight: 10,935 lb
Wing Span: 33 ft 1 in
Height: 12 ft 2 in
Ceiling: 50,580 ft
Speed: 668 mph
Range: 1,225 miles (ferry)
Armament: 2 x 23mm cannon, 1 x 37 mm cannon, 2 x 100 kg bombs or rockets
Mikoyan/Gurevich MiG-15
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text:
Energy, Empathy & Engagement
Here at Homewise, we've created our own Culture for Success - a focused set of skills, attributes and behaviour expectations that align with our core company values and drive our team to be successful in their roles.
You'll often hear us referring to the saying "Teamwork makes the dream work", and it really does at Homewise!
Creating Success
We established our Culture for Success in the summer of 2017, and have seen the team embrace and demonstrate the values, mindset and skills of our culture to make the team and the company what it is today.
"We are one team with one shared vision that values the contribution of every individual. We are tenacious in the pursuit of our goals and accountable to ourselves and each other. We work collaboratively to overcome challenges and create solutions."
Culture Champions
The Culture for Success was born from staff ideas and is regularly championed and celebrated. It is so important to the company that we have our own Culture Club!
The small team of volunteers champion the culture traits and aim to motivate and embed the culture into day-to-day life through regular activities and incentives, also providing a forum for everyone's ideas to be heard.
The Culture Club ensures we continue to recognise and celebrate success across all teams.
We have one of the best culture's I've seen or heard of - it's inspiring & fun!
Charlotte, Office Manager
The Culture for Success here at Homewise is brilliant! It was one of the many reasons I was so keen to be successful during my interview process.
Josh, Contact Centre Team Leader
The company really cares. It's not like any other business, they treat the team like a family and the office atmosphere is positive, friendly, supportive and fun!
Kathryn, Customer Service Consultant
It's refreshing to work in a business that cares about people, both employees and customers. The office has a family feel about it, people genuinely care about it each other and want to help each other to succeed.
Vicki, Head of Sales
I am a member of the Culture Club and in 2.5 years since I have worked here I have never not wanted to come to work - it's wonderful business to work for, especially after working in a corporate environment.
Karen, Learning and Development Consultant
The company cares about it's staff and there is a great support network in place. I enjoy the uniqueness of the business and our place in the local community.
Julian, Head of Investor Relations
We're keen to celebrate success and have fun at Homewise. But we're also quick to take ownership and responsibility, and are determined to reach solutions for ourselves, the company and importantly our customers.
Susan, Finance & Operations Director
I love that the company is really ambitious about what it wants to do, it means there's an opportunity for everyone to do exciting stuff. But it's also grounded, family oriented and determined to go about fulfilling those ambitions in the right way, never at all costs.
David, Data Analyst
Working at Homewise
Homewise is a values driven company and committed to providing a supportive and thriving working environment for all. Discover why we love working at Homewise and what you can expect when you join our team!
We pride ourselves on being able to give back to the local community here at Homewise. Read all about our previous volunteering projects, large scale team challenge events and fundraising days!


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Q: Export from Access to Excel, add worksheet to existing Workbook I would like to automate a process of exporting data from Access query to Excel workbook.  Each query relates to a batch.  Each worksheet in a workbook relates to a batch.  For example: Excel workbook called Jun2019.xlsx, might have batch1 relating to the worksheet named batch1.  Then save each workbook into a different file path.
I am looking for some guidance, whether it can be done via Access VBA or would need a separate step to export xlsx from access and run procedure in Excel?

A: Working from Access VBA your code should look something like this.
Please note where some actions are done by Apxl object, while others must be done by the object's Application property. I have found this issue vital in various Excel actions done from MS Access.
First create an Excel objet:
dim Apxl as Object, xlWorkbook as Object ,xlWorksheet as Object
Set Apxl = CreateObject("Excel.Application")

Next If the Workbook (the file) exists already,  open it:
ApXl.Workbooks.Open("C\:MyPath\MyFileName.xlsx")

Or create a new one:
Set xlWorkbook = Apxl.Application.WorkBooks.Add
xlWorkbook.SaveAs "C\:MyPath\MyFileName.xlsx"

Now you can approach the Worksheat Object by it's name, or ordinal number (startin by 1 ):
Set xlWorksheet = xlWorkbook.Worksheets("MySheetName")
Set xlWorksheet = xlWorkbook.Worksheets(1)

If sheet does not exist, you can create a new one:
xlWorkbook.Worksheets.Add

The new worksheet will be added to the beginning of the collection.
You can also rename the WorkSheet:
Set xlWorksheet = xlWorkbook.Worksheets(1)
xlWorksheet.Name = "MyNewName"

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For many of us, home is a sanctuary and living there for as long as possible is a priority. But during a person's life, especially a senior, certain circumstances and events can challenge their ability to perform even some of the most basic daily tasks or live at home without any assistance. It's not easy to admit that you or your Mom and Dad are in need of elderly care. But when that time comes, it doesn't mean that they have to give up their home. Also, you don't have to let other obligations go unfulfilled or end your career to take care of your elderly Mom, Dad, or other loved one who is at a point where he or she needs in-home care. There is no shame in reaching out to get the help you need and it doesn't mean that you are abandoning your loved one if you hire a trained, professional caretaker.
The decision to enlist in-home care services can be difficult and fraught with doubt and anxiety for many people. As the premier, non-medical home health care company in Beverly Hills, CA and surrounding neighborhoods, Mom's Home Care is there to help bear the burden associated with taking care of a loved one in the home. Our trained and highly skilled staff can provide you with exactly the type of home care services you need, whether temporarily, permanently, or on an on-call basis.
If you can't provide fully satisfactory care for your Mom or Dad on a continual basis, it means that there will be a time when you need temporary or additional help. And we at Mom's Home Care will be there to give your loved ones the best care possible and make them as comfortable as we can, while you maintain your job and make sure that all other obligations are met.
Who Can Benefit from Mom's Home Care Services?
We'll be there for you or your loved one to make sure that things are healthier, safer, and more convenient, be it temporarily, permanently, or on an on-call basis.
How Do You Get Started with Mom's Home Care?
Whether you're loved one needs someone urgently or you're trying to introduce the idea of a caregiver, Mom's Home Care is here to help. Simply call us at 323-244-4789 to learn more about us and the services we offer, or to set up a free, no-obligation consultation with one of our seasoned staff members. They'll come to your loved one's home, care facility, or hospital to assess the situation, tailor a care plan, and provide you with the pricing. They will also ask you about your preferences when it comes to the caregiver.
And once you start receiving our services, we don't keep you in the dark about what's going on. We're proud to offer an optional feature known as the Family Room. This feature provides family members and associate caregivers with information on the care team, medications, upcoming schedules, and even details of the tasks that were completed during each shift.
Also, we offer new clients a 10% discount on services for the first two weeks. This only applies if the services are scheduled for over 6 hours a day for 3 or more days per week and you continue services for a month or longer. Hospice care is not included.
When those who have been there for you are in their golden years and need assistance with daily living, it's your turn for you to be there for them as well. But that doesn't mean being there physically round the clock. To one degree or another, you'll have to enlist the services of a professional home care provider.
Mom's Home Care is a state-licensed home care agency and a trusted elderly care provider in Beverly Hills since 1997. Feel free to look us up by our Department of Social Services Home Care Services Bureau license number (194700009) to see our BBB rating. We also invite you to check our online reviews or ask any of our former and current clients.
We've been providing all manner of non-medical health care services to the elderly of Beverly Hills and we stand ready to provide the same great care to you. We have over three decades and running of experience, highly skilled and trained staff, respectful caregivers, and a name you can trust. We've built a reputation for skill, integrity, and conscientious care that is well recognized across Beverly Hills. All our services are geared to enrich lives and form strong bonds.
Because we look for people we would feel comfortable entrusting the care of our elderly parents too, we are very careful about how we hire our staff. We only go for the highly trained and the most reliable candidates. We make sure that all caregivers are fully qualified to complete their tasks and demonstrate a loving concern for all our clients. In fact, we run a background check on all candidates and put them through a thorough screening process. Those who will be responsible for running errands or transporting our clients must have a valid current California driver's license and a good driving record.
Another reason to choose Mom's Home Care over the competition is that we cover all our employees with workers comp and general liability insurance. As such, you're not left with any of the liability or risk-taking. We monitor our caregivers with cutting-edge technologies and we know if they're late. If your scheduled caretaker is so much as 10 minutes off schedule, we send out an urgent alert to get you another one. We always communicate with you if there are any significant developments.
We can provide part of the day, all-day, or on-call care as needed for any and all of our services. How long you use our services is 100% up to you. We are flexible to your needs and will work with you to set up a plan of care for you and your loved one based on your situation, needs, and preferences. And you're free to adjust your plan of care as needed going forward. We at Mom's Home Care also have a people standing by 24/7 ready to assist you with emergency care requests. We know that you can't schedule emergencies and we'll do everything possible to get you one of our caregivers quickly to help with areas of daily living and non-medical home health care services.
We provide a wide range of services for seniors across Beverly Hills and Southern California. And as we deliver all our services, we're committed to maintaining a high level of personal dignity and respect for our clients at all times. We're flexible to meet client needs in any way possible. You can visit our individual service pages on this website to delve deeper into more detailed information about each of our various offered services. However, here's a basic overview of the elderly care services that we commonly provide.
Simple companionship is a basic human need. And for senior citizens who live alone for extended periods of time, the risk of dementia or Alzheimer's, depression, and general cognitive decline conditions develops sooner due to the lack of social interaction. So it's not just a matter of aging and hereditary predisposition. Seniors who stay engaged and social have a better quality of life.
Our caregivers provide friendly companionship, conversation, day-to-day social interactions, and mealtime companionship to keep your loved one engaged and upbeat. This can mean spending time conversing, watching TV together, going shopping, playing board games, eating together, and more.
There are often errands that need to be run either on a day to day or weekly/monthly basis. However, as we age, getting around can be difficult, especially if unable to drive or even walk to/from the store. That's why we at Mom's Home Care include shopping and errand –running on our list of services. Be it stopping at the pharmacy, picking up food to-go at a restaurant, running a letter to the post office, or any other errand, we can take care of it. Your loved one is welcome is welcome to come along for the ride or stay home.
While it's an unpleasant subject that nobody likes to talk about, the fact is that most senior citizens often need special assistance in toileting. At Mom's Home Care, we understand the sensitive nature of toileting and incontinence care and also the embarrassment and frustration this situation can bring. Without proper care, the elderly can face many issues, including unpleasant odors, skin infections, and general hygiene problems.
We take a "matter of fact" approach and offer compassionate, dignified help with these personal aspects of daily living. We can help ensure the toileting environment is usable and safe for your loved one. If there's need to get the TP roll repositioned, grip rails installed, or the toilet height adjusted, we can assess it and get it taken care of. Also, we deal with all the various types of incontinence and can help access which toileting options are best.
Cleaning the house and handling other simple tasks becomes a major challenge for the elderly and those who have limited strength to expend each day. Mom's Home Care can provide housekeeping services for your elderly Mom or Dad, including, dusting, vacuuming, sweeping, mopping, cleaning bathrooms, cleaning kitchens, straightening all rooms, and organizing closets and drawers. We take care of all these household tasks and keep the home clean, livable, and healthy.
Healthy eating is of the utmost importance, especially for seniors. But as we age, our bodies lose lean mass and need more nutrient dense foods to prevent malnutrition and development of new or worsening of existing medical conditions. Unfortunately, many seniors are not eating balanced diets because they don't have an appetite, lack the strength to cook or are simply unable to go to buy all the groceries.
This is where we step in and help in the kitchen when needed. Our caregivers prepare wholesome and flavorful meals to meet all the nutritional and dietary needs. If your loved one has health conditions or restrictions, we can manage his/her menu plan and monitor his/her diet and health.
While Mom's Home Care only provides non-medical care, we also handle medication management and ensure that those under our care take their medication as prescribed by reminding them when it is time for the next dose. We can document the dispensing of the medications, maintain your loved one's pill box, and note any side effects that might occur, if necessary. We do not, however, administer invasive medications or treatment, change IVs, or give injections.
As we age, we reach a point where we can no longer continue doing simple tasks like bathing and keeping up with good hygiene, especially for those suffering from brain disorders that add further complications. Mechanical movements become more difficult and strenuous, making a simple hygienic task a complicated process that may require assistance. Also, brushing, flossing, shaving, combing, getting dressed in the morning and the like are all necessary tasks that can't be avoided.
Paying particular attention to grooming can help seniors can look and feel their best, boosting their self-esteem. Unfortunately, not all of us have the time to assist our favorite seniors with basic daily tasks. Mom's Home Care caregivers are intensely trained to handle all these and other personal care needs for our clients in a respectful yet adequate and reliable way. We'll provide you with a peaceful mind.
If you have any questions, call us today at 323-244-4789 for a free, in-home consultation.


At 11am on November 11th, the sun will shine through the windows of the Hall and onto the plaque.
At the kitchen end of the Great Hall, underneath the portrait of former Principal Dame Beryl Paston Brown, a copper plaque is inlaid in the wooden panelling. This is Homerton's memorial to the end of the First World War. It reads: Armistice Day – November 11th, 1918 – Recorded by Students 1917-1919. On either side of the text there is a conventional symbol combining a laurel wreath and a downturned sword, the first representing victory, the second peace.
At many other Cambridge Colleges, First World War memorials list the men who died as soldiers. As a training institution for female teachers, Homerton College's commitment was different. With so many men recruited to the war, the training of women teachers was extremely important and the College also took advantage of its large grounds by growing sorely needed fresh fruit and vegetables. The plaque commemorating the end of the war on November 11th is specifically placed so that at 11am on that day the sun will shine through the windows of the Hall and illuminate the plaque.