258 lines
8.0 KiB
Markdown
258 lines
8.0 KiB
Markdown
# 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:
|
|
|
|
{#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:
|
|
|
|
|
|
|
|
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 = 0.295
|
|
- p = 0.768
|
|
|
|
which allows to assert that the sampled points actually follow the predicted
|
|
distribution. In @fig:fit, the fit function superimposed on the histogram is
|
|
shown.
|
|
|
|
{#fig:fit}
|