# Exercise 4

## Kinematic dip derivation

Consider a great number of non-interacting particles having random momenta
$\vec{P}$, with magnitude between 0 and $P_{\text{max}}$, at an angle $\theta$
wrt to some coordinate system ($\hat{x}$, $\hat{y}$, $\hat{z}$).
The vertical and horizontal components of a particle momentum, which will be
referred as $\vec{P_v}$ and $\vec{P_h}$ respectively, are shown in
@fig:components.  
If $\theta$ is uniformly distributed on the unit sphere and $P$ is uniformly
distributed in $[0, P^\text{max}]$, what will be the average value
$|P_v|$ of the particles with a given $P_h$?

\begin{figure}
\hypertarget{fig:components}{%
\centering
\begin{tikzpicture}[font=\scriptsize]
  % Axes
  \draw [thick, ->] (5,2) -- (5,8);
  \draw [thick, ->] (5,2) -- (2,1);
  \draw [thick, ->] (5,2) -- (8,1);
  \node at (1.5,0.9) {$x$};
  \node at (8.5,0.9) {$y$};
  \node at (5,8.4)   {$z$};
  % Momentum
  \definecolor{cyclamen}{RGB}{146, 24, 43}
  \draw [ultra thick, ->, cyclamen] (5,2)     -- (3.8,6);
  \draw [thick, dashed, cyclamen]   (3.8,0.8) -- (3.8,6);
  \draw [thick, dashed, cyclamen]   (5,7.2)   -- (3.8,6);
  \draw [ultra thick, ->, pink]     (5,2)     -- (5,7.2);
  \draw [ultra thick, ->, pink]     (5,2)     -- (3.8,0.8);
  \node at (4.8,1.1) {$\vec{P_h}$};
  \node at (5.5,6.6) {$\vec{P_v}$};
  \node at (3.3,5.5) {$\vec{P}$};
  % Angle
  \draw [thick, cyclamen] (4.4,4) to [out=80,in=210] (5,5);
  \node at (4.7,4.2) {$\theta$};
\end{tikzpicture}
\caption{Momentum components.}\label{fig:components}
}
\end{figure}


Since the aim is to compute $\langle |P_v| \rangle (P_h)$, the conditional
distribution probability of $P_v$ given a fixed value of $P_h = x$ must first
be determined. It can be computed as the ratio between the probability of
getting a fixed value of $P_v$ given $x$ to the total probability of
getting that $x$:
$$
  f (P_v | P_h = x) = \frac{f_{P_h , P_v} (x, P_v)}
  {\int_{\{ P_v \}} d P_v f_{P_h , P_v} (x, P_v)}
  = \frac{f_{P_h , P_v} (x, P_v)}{I}
$$

where $f_{P_h , P_v}$ is the joint PDF of the two variables $P_v$ and $P_h$ and
the integral $I$ runs over all the possible values of $P_v$ given a certain
$P_h$.  
$f_{P_h , P_v}$ can simply be computed from the joint PDF of $\theta$ and $P$
with a change of variables. For the PDF of $\theta$ $f_{\theta} (\theta)$, the
same considerations done in @sec:3 lead to:
$$
  f_{\theta} (\theta) = \frac{1}{2} \sin{\theta} \chi_{[0, \pi]} (\theta)
$$

whereas, being $P$ uniform:
$$
  f_P (P) = \chi_{[0, P_{\text{max}}]} (P)
$$

where $\chi_{[a, b]} (y)$ is the normalized characteristic function which value
is $1/N$ between $a$ and $b$ (where $N$ is the normalization term) and 0
elsewhere. Since $P,\theta$ are independent variables, their joint PDF is
simply given by the product:
$$
  f_{\theta , P} (\theta, P) = f_{\theta} (\theta) f_P (P)
  = \frac{1}{2} \sin{\theta} \chi_{[0, \pi]} (\theta)
    \chi_{[0, P_{\text{max}}]} (P)
$$
and they are related to the vertical and horizontal components
by a standard polar coordinate transformation:

\begin{align*}
  \begin{cases}
    P_v = P \cos(\theta) \\
    P_h = P \sin(\theta)
  \end{cases}
  &&
  \begin{cases}
    P = \sqrt{P_v^2 + P_h^2} \\
    \theta = \text{atan2}(P_h, P_v)
  \end{cases}
\end{align*}

where:

- $\theta \in [0, \pi]$,

- and atan2 is defined by:
$$
\begin{cases}
  \arctan(P_h/P_v) &\incase P_v > 0 \\
  \pi/2 &\incase P_v = 0            \\
  \arctan(P_h/P_v) + \pi &\incase P_v < 0
\end{cases}
$$

The Jacobian of the inverse transformation is easily found to be:
$$
  |J^{-1}| = \frac{1}{\sqrt{P_v^2 + P_h^2}}
$$
Hence, the PDF written in the new coordinates is:
$$
  f_{P_h , P_v} (P_h, P_v) =
    \frac{1}{2} \sin\left[ \text{atan2}(P_h, P_v) \right]
    \chi_{[0, \pi]} \left[\text{atan2}(P_h, P_v)\right] \cdot \\
    \frac{\chi_{[0, p_{\text{max}}]} \left(\sqrt{P_v^2 + P_h^2} \right)}
    {\sqrt{P_v^2 + P_h^2}}
$$

The integral $I$ can now be computed. Note that the domain
is implicit in the characteristic function:
$$
  I(x) = \int_{-\infty}^{+\infty} dP_v \, f_{P_h , P_v} (x, P_v)
       = \int \limits_{- \sqrt{P_{\text{max}}^2 - P_h}}
         ^{\sqrt{P_{\text{max}}^2 - P_h}}
         dP_v \, f_{P_h , P_v} (x, P_v)
$$

With some basic calculus and the identity
$$
  \sin[ \text{atan2}(P_h, P_v)] = \frac{P_h}{\sqrt{P_h^2 + P_v^2}},
$$
the integral can be evaluated to give
$$
  I = 2 \, \arctan \left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right),
$$
from which:
$$
  f (P_v | P_h = x) =
  \frac{x}{P_v^2 + x^2} \cdot
  \frac{\chi_{[0, \pi]} \left[\text{atan2}(P_h, P_v)\right]
    \chi_{[0, p_{\text{max}}]} \left(\sqrt{P_v^2 + P_h^2}\right)}{2 \, \arctan
           \left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)}
$$

Finally, putting all the pieces together, the average value of $|P_v|$ can be
computed:
$$
  \langle |P_v| \rangle(x)
  = \int P_v f (P_v | P_h = x) dP_v
  = \frac{x \ln \left( \frac{P_{\text{max}}}{x} \right)}
  {\arctan \left( \sqrt{ \frac{P^2_{\text{max}}}{x^2} - 1} \right)}
$$ {#eq:dip}

The result is plotted in the figure below:

![Plot of the expected dependence of $\langle |P_v| \rangle$ with
  $P_{\text{max}} = 10$.](images/4-expected.pdf){#fig:plot}


## Monte Carlo simulation

This dependence should be found by running a Monte Carlo simulation and
computing a binned average of the vertical momentum. A number of $N = 50000$
particles were generated as pair of values ($P$, $\theta$), with $P$
uniformly distributed between 0 and $P_{\text{max}}$ and $\theta$ given by the
same procedure described in @sec:3, namely:
$$
  \theta = \arccos(1 - 2x)
$$

where $x$ is uniformly distributed between 0 and 1.  
The binning turned out to be quite a challenge: once a $P$ is sampled and
$P_h$ computed, the bin containing the latter has to be found. If
the range $[0, P_{\text{max}}]$ is divided into $n$ equal bins
of the width
$$
  w = \frac{P_{\text{max}}}{n}
$$
then (counting from zero) $P_h$ goes into the $i$-th bin where
$$
  i = \left\lfloor \frac{P_h}{w} \right\rfloor
$$

Then, the sum $S_j$ of all the $|P_v|$ values relative to the $P_h$ of the
$j$-th bin itself and number num$_j$ of the bin counts are stored in an array
and iteratively updated.  Once every bin has been updated, the average value of
$|P_v|_j$ is computed as $S_j / \text{num}_j$.

For the sake of clarity, for each sampled couple the procedure is the
following. At first $S_j = 0 \wedge \text{num}_j = 0 \, \forall \, j$, then:

  - the couple $(P, \theta)$ is generated,
  - $P_h$ and $P_v$ are computed,
  - the $j^{\text{th}}$ bin containing $P_h$ is found,
  - num$_j$ is increased by 1,
  - $S_j$ is increased by $|P_v|$.

For $P_{\text{max}} = 10$ and $n = 50$, the following result was obtained:

![Sampled points histogram.](images/4-dip.pdf)

In order to assert the compatibility of the expected function (@eq:dip)
with the histogram, a least squares minimization was applied. Being a simple
one-parameter fit, the $\chi^2$ was implemented manually and minimised
without using a general LSQ routine. The error of the estimation of
$P_{\text{max}}$ was computed as the inverse of the $\chi^2$ second derivative
at the minimum, according to the Cramér-Rao bound.

The following results were obtained:

\begin{align*}
  P^{\text{oss}}_{\text{max}} = 10.005 \pm 0.018 && \chi^2 &= 0.071 \\
  && \text{P}(x > \chi^2) &= 0.79
\end{align*}

The $\chi^2$ and $p$-value show a very good agreement.
In order to compare $P^{\text{oss}}_{\text{max}}$ with the expected value
$P_{\text{max}} = 10$, the following compatibility $t$-test was applied:

$$
  p = 1 - \text{erf}\left(\frac{t}{\sqrt{2}}\right)\ \with
  t = \frac{|P^{\text{oss}}_{\text{max}} - P_{\text{max}}|}
           {\Delta P^{\text{oss}}_{\text{max}}}
$$

where $\Delta P^{\text{oss}}_{\text{max}}$ is the $P^{\text{oss}}_{\text{max}}$
uncertainty. At 95% confidence level, the values are compatible if $p > 0.05$.
In this case:

  - t = 0.295
  - p = 0.768

which allows to assert that the sampled points actually follow the predicted
function. In @fig:fit, the fit function superimposed on the histogram is
shown.

![Fitted sampled data. $P^{\text{oss}}_{\text{max}} = 10.005
  \pm 0.018$, $\chi^2 = 0.071$.](images/4-fit.pdf){#fig:fit}