analistica/notes/sections/4.md

# Exercise 4

## Kinematic dip PDF derivation

Consider a great number of non-interacting particles, each of which with a
random momentum $\vec{p}$ with module between 0 and $p_{\text{max}}$ randomly
angled with respect to a coordinate system {$\hat{x}$, $\hat{y}$, $\hat{z}$}.
Once the polar angle $\theta$ is defined, the momentum vertical and horizontal
components of a particle, which will be referred as $\vec{p_v}$ and $\vec{p_h}$
respectively, are the ones shown in @fig:components.  
If $\theta$ is evenly distributed on the sphere and the same holds for the
module $p$, which distribution will the average value of $|p_v|$ follow as a
function of $p_h$?

\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 probablity of
getting a fixed value of $P_v$ given $x$ over the total probability of $x$:

$$
  f (P_v | P_h = x) = \frac{f_{P_h , P_v} (x, P_v)}
  {\int_{\{ P_v \}} d P_v f_{P_h , P_v} (x, P_v)}
  = \frac{f_{P_h , P_v} (x, P_v)}{I}
$$

where $f_{P_h , P_v}$ is the joint pdf of the two variables $P_v$ and $P_h$ and
the integral runs over all the possible values of $P_v$ given $P_h$.

This joint pdf can simly 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 $p$ evenly distributed:

$$
  f_p (p) = \chi_{[0, p_{\text{max}}]} (p)
$$

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. Being a couple of independent variables, their joint pdf is simply
given by the product of their pdfs:

$$
  f_{\theta , p} (\theta, p) = f_{\theta} (\theta) f_p (p)
  = \frac{1}{2} \sin{\theta} \chi_{[0, \pi]} (\theta)
    \chi_{[0, p_{\text{max}}]} (p)
$$

Given the new variables:

$$
  \begin{cases}
    P_v = P \cos(\theta) \\
    P_h = P \sin(\theta)
  \end{cases}
$$

with $\theta \in [0, \pi]$, the previous ones are written as:

$$
  \begin{cases}
    P = \sqrt{P_v^2 + P_h^2} \\
    \theta = \text{atan}^{\star} ( P_h/P_v ) :=
      \begin{cases}
        \text{atan} ( P_h/P_v )       &\incase P_v > 0 \\
        \text{atan} ( P_h/P_v ) + \pi &\incase P_v < 0
      \end{cases} 
  \end{cases}
$$

which can be shown having Jacobian:

$$
  J = \frac{1}{\sqrt{P_v^2 + P_h^2}}
$$

Hence:

$$  
  f_{P_h , P_v} (P_h, P_v) =
    \frac{1}{2} \sin[ \text{atan}^{\star} ( P_h/P_v )]
    \chi_{[0, \pi]} (\text{atan}^{\star} ( P_h/P_v )) \cdot \\
    \frac{\chi_{[0, p_{\text{max}}]} \left( \sqrt{P_v^2 + P_h^2} \right)}
    {\sqrt{P_v^2 + P_h^2}}
$$

from which, the integral $I$ can now be computed. The edges of the integral
are fixed bt the fact that the total momentum can not exceed $P_{\text{max}}$:

$$
  I = \int
  \limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
  dP_v \, f_{P_h , P_v} (x, P_v)
$$

after a bit of maths, using the identity:

$$
  \sin[ \text{atan}^{\star} ( P_h/P_v )] = \frac{P_h}{\sqrt{P_h^2 + P_v^2}}
$$

and the fact that both the characteristic functions are automatically satisfied
within the integral limits, the following result arises:

$$
  I = 2 \, \text{atan} \left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)
$$

from which:

$$
  f (P_v | P_h = x) = \frac{x}{P_v^2 + x^2} \cdot
  \frac{1}{2 \, \text{atan}
           \left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)}
$$

Finally, putting all the pieces together, the average value of $|P_v|$ can now
be computed:

\begin{align*}
  \langle |P_v| \rangle &= \int
  \limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
  f (P_v | P_h = x)
  \\
  &= [ \dots ]
  \\
  &= x \, \frac{\ln \left( \frac{P_{\text{max}}}{x} \right)}
  {\text{atan} \left( \sqrt{ \frac{P^2_{\text{max}}}{x^2} - 1} \right)}
\end{align*}

Namely:

![Plot of the expected distribution with
  $P_{\text{max}} = 10$.](images/expected.pdf){#fig:plot}


## Monte Carlo simulation

The same distribution should be found by generating and binning points in a
proper way.  
A number of $N = 50'000$ points were generated as a couple of values ($p$,
$\theta$), with $p$ evenly distrinuted between 0 and $p_{\text{max}}$ and
$\theta$ given by the same procedure described in @sec:3, namely:

$$
  \theta = \text{acos}(1 - 2x)
$$

with $x$ uniformely distributed between 0 and 1.  
The data binning turned out to be a bit challenging. Once $p$ was sorted and
$p_h$ was computed, the bin in which the latter goes in must be found. If $n$ is
the number of bins in which the range [0, $p_{\text{max}}$] is willing to be
divided into, then the width $w$ of each bin is given by:

$$
  w = \frac{p_{\text{max}}}{n}
$$

and the $j^{th}$ bin in which $p_h$ goes in is:

$$
  j = \text{floor} \left( \frac{p_h}{w} \right)
$$

where 'floor' is the function which gives the upper integer lesser than its
argument and the bins are counted starting from zero.  
Then, a vector in which the $j^{\text{th}}$ entry contains the sum $S_j$ of all
the $|p_v|$s relative to each $p_h$ fallen into the $j^{\text{th}}$ bin and the
number num$_j$ of entries in the $j^{\text{th}}$ bin was reiteratively updated.
At the end, the average value of $|p_v|_j$ was computed as $S_j /
\text{num}_j$.  
For the sake of clarity, for each sorted couple, it works like this:

  - the couple $[p; \theta]$ is generated;
  - $p_h$ and $p_v$ are computed;
  - the $j^{\text{th}}$ bin in which $p_h$ goes in is computed;
  - num$_j$ is increased by 1;
  - $S_j$ (which is zero at the beginning of everything) is increased by a factor
    $|p_v|$.

At the end, $\langle |p_h| \rangle_j$ = $\langle |p_h| \rangle (p_h)$ where:

$$
  p_h = j \cdot w + \frac{w}{2} = w \left( 1 + \frac{1}{2} \right)
$$

The following result was obtained:

![Histogram of the obtained distribution.](images/dip.pdf)

In order to check wheter the expected distribution properly metches the
produced histogram, a chi-squared minimization was applied. Being a simple
one-parameter fit, the $\chi^2$ was computed without a suitable GSL function
and the error of the so obtained estimation of $p_{\text{max}}$ was given as
the inverese of the $\chi^3$ second derivative in its minimum, according to the
Cramér-Rao bound.  
The following results were obtained:

$$
  p^{\text{oss}}_{\text{max}} = 10.005 \pm 0.018 \with \chi^2 = 0.071
$$

In order to compare $p^{\text{oss}}_{\text{max}}$ with the expected value
$p_{\text{max}} = 10$, the following compatibility t-test was applied:

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

where $\Delta p_{\text{max}}$ is the absolute error of $p_{\text{max}}$. At 95%
confidence level, the values are compatible if $p > 0.05$.
In this case:

  - t = 0.278
  - p = 0.781

which allows to assert that the sampled points actually follow the predicted
distribution.