ex-4: revised and typo-fixed
This commit is contained in:
parent
4aa8eefaf9
commit
12f15e1998
@ -41,6 +41,9 @@ int main(int argc, char **argv)
|
||||
size_t n = 50;
|
||||
double p_max = 10;
|
||||
size_t go = 0;
|
||||
|
||||
// Get eventual CLI arguments.
|
||||
//
|
||||
int res = parser(&N, &n, &p_max, argc, argv, &go);
|
||||
if (go == 0 && res == 1)
|
||||
{
|
||||
|
||||
@ -15,7 +15,7 @@ edges = np.linspace(0, bins*step, bins+1)
|
||||
plt.rcParams['font.size'] = 17
|
||||
|
||||
plt.hist(edges[:-1], edges, weights=counts,
|
||||
color='#eacf00', edgecolor='#3d3d3c')
|
||||
histtype='stepfilled', color='#e3c5ca', edgecolor='#92182b')
|
||||
plt.title('Vertical component distribution', loc='right')
|
||||
plt.xlabel(r'$p_h$')
|
||||
plt.ylabel(r'$\left<|p_v|\right>$')
|
||||
|
||||
Binary file not shown.
Binary file not shown.
@ -3,14 +3,14 @@
|
||||
## 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
|
||||
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 $\vec{p_h}$
|
||||
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 $p_h$?
|
||||
module $P$, which distribution will the average value of $|P_v|$ follow as a
|
||||
function of $P_h$?
|
||||
|
||||
\begin{figure}
|
||||
\hypertarget{fig:components}{%
|
||||
@ -30,9 +30,9 @@ function of $p_h$?
|
||||
\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) {$\vec{p}$};
|
||||
\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) {$\theta$};
|
||||
@ -41,11 +41,11 @@ function of $p_h$?
|
||||
}
|
||||
\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 $x$:
|
||||
|
||||
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 $x$ over the total probability of
|
||||
getting that $x$:
|
||||
$$
|
||||
f (P_v | P_h = x) = \frac{f_{P_h , P_v} (x, P_v)}
|
||||
{\int_{\{ P_v \}} d P_v f_{P_h , P_v} (x, P_v)}
|
||||
@ -53,35 +53,31 @@ $$
|
||||
$$
|
||||
|
||||
where $f_{P_h , P_v}$ is the joint pdf of the two variables $P_v$ and $P_h$ and
|
||||
the integral runs over all the possible values of $P_v$ given $P_h$.
|
||||
|
||||
This joint pdf can simly be computed from the joint pdf of $\theta$ and $p$ with
|
||||
a change of variables. For the pdf of $\theta$ $f_{\theta} (\theta)$, the same
|
||||
considerations done in @sec:3 lead to:
|
||||
|
||||
the integral $I$ runs over all the possible values of $P_v$ given a certain
|
||||
$P_h$.
|
||||
$f_{P_h , P_v}$ can simply be computed from the joint pdf of $\theta$ and $P$
|
||||
with a change of variables. For the pdf of $\theta$ $f_{\theta} (\theta)$, the
|
||||
same considerations done in @sec:3 lead to:
|
||||
$$
|
||||
f_{\theta} (\theta) = \frac{1}{2} \sin{\theta} \chi_{[0, \pi]} (\theta)
|
||||
$$
|
||||
|
||||
whereas, being $p$ evenly distributed:
|
||||
|
||||
whereas, being $P$ evenly distributed:
|
||||
$$
|
||||
f_p (p) = \chi_{[0, p_{\text{max}}]} (p)
|
||||
f_P (P) = \chi_{[0, P_{\text{max}}]} (P)
|
||||
$$
|
||||
|
||||
where $\chi_{[a, b]} (y)$ is the normalized characteristic function which value
|
||||
is $1/N$ between $a$ and $b$, where $N$ is the normalization term, and 0
|
||||
is $1/N$ between $a$ and $b$ (where $N$ is the normalization term) and 0
|
||||
elsewhere. Being a couple of independent variables, their joint pdf is simply
|
||||
given by the product of their pdfs:
|
||||
|
||||
$$
|
||||
f_{\theta , p} (\theta, p) = f_{\theta} (\theta) f_p (p)
|
||||
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)
|
||||
\chi_{[0, P_{\text{max}}]} (P)
|
||||
$$
|
||||
|
||||
Given the new variables:
|
||||
|
||||
$$
|
||||
\begin{cases}
|
||||
P_v = P \cos(\theta) \\
|
||||
@ -89,12 +85,11 @@ $$
|
||||
\end{cases}
|
||||
$$
|
||||
|
||||
with $\theta \in [0, \pi]$, the previous ones are written as:
|
||||
|
||||
with $\theta \in [0, \pi]$, the previous ones can be written as:
|
||||
$$
|
||||
\begin{cases}
|
||||
P = \sqrt{P_v^2 + P_h^2} \\
|
||||
\theta = \text{atan}^{\star} ( P_h/P_v ) :=
|
||||
\theta = \text{atan}_2 ( P_h/P_v ) :=
|
||||
\begin{cases}
|
||||
\text{atan} ( P_h/P_v ) &\incase P_v > 0 \\
|
||||
\text{atan} ( P_h/P_v ) + \pi &\incase P_v < 0
|
||||
@ -103,24 +98,21 @@ $$
|
||||
$$
|
||||
|
||||
which can be shown having Jacobian:
|
||||
|
||||
$$
|
||||
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{1}{2} \sin[ \text{atan}_2 ( P_h/P_v )]
|
||||
\chi_{[0, \pi]} (\text{atan}_2 ( 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}}
|
||||
$$
|
||||
|
||||
from which, the integral $I$ can now be computed. The edges of the integral
|
||||
are fixed by the fact that the total momentum can not exceed $P_{\text{max}}$:
|
||||
|
||||
$$
|
||||
I = \int
|
||||
\limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
|
||||
@ -128,39 +120,32 @@ $$
|
||||
$$
|
||||
|
||||
after a bit of maths, using the identity:
|
||||
|
||||
$$
|
||||
\sin[ \text{atan}^{\star} ( P_h/P_v )] = \frac{P_h}{\sqrt{P_h^2 + P_v^2}}
|
||||
\sin[ \text{atan}_2 ( P_h/P_v )] = \frac{P_h}{\sqrt{P_h^2 + P_v^2}}
|
||||
$$
|
||||
|
||||
and the fact that both the characteristic functions are automatically satisfied
|
||||
within the integral limits, the following result arises:
|
||||
|
||||
and the fact that both the characteristic functions play no role 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
|
||||
Finally, putting all the pieces together, the average value of $|P_v|$ can be
|
||||
computed:
|
||||
$$
|
||||
\langle |P_v| \rangle = \int
|
||||
\limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
|
||||
f (P_v | P_h = x)
|
||||
\\
|
||||
&= [ \dots ]
|
||||
\\
|
||||
&= x \, \frac{\ln \left( \frac{P_{\text{max}}}{x} \right)}
|
||||
f (P_v | P_h = x) = [ \dots ]
|
||||
= x \, \frac{\ln \left( \frac{P_{\text{max}}}{x} \right)}
|
||||
{\text{atan} \left( \sqrt{ \frac{P^2_{\text{max}}}{x^2} - 1} \right)}
|
||||
\end{align*}
|
||||
$$ {#eq:dip}
|
||||
|
||||
Namely:
|
||||
|
||||
@ -171,80 +156,71 @@ Namely:
|
||||
## 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
|
||||
proper way. A number of $N = 50'000$ points were generated as a couple of values
|
||||
($P$, $\theta$), with $P$ evenly distributed 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:
|
||||
|
||||
with $x$ uniformly distributed between 0 and 1.
|
||||
The data binning turned out to be a bit challenging. Once $P$ was sampled and
|
||||
$P_h$ was computed, the bin containing the latter's value must be found. If $n$
|
||||
is the number of bins in which the range $[0, P_{\text{max}}]$ is divided into,
|
||||
then the width $w$ of each bin is given by:
|
||||
$$
|
||||
w = \frac{p_{\text{max}}}{n}
|
||||
w = \frac{P_{\text{max}}}{n}
|
||||
$$
|
||||
|
||||
and the $j^{th}$ bin in which $p_h$ goes in is:
|
||||
|
||||
and the $i^{th}$ bin in which $P_h$ goes in is:
|
||||
$$
|
||||
j = \text{floor} \left( \frac{p_h}{w} \right)
|
||||
i = \text{floor} \left( \frac{P_h}{w} \right)
|
||||
$$
|
||||
|
||||
where 'floor' is the function which gives the upper integer lesser than its
|
||||
where 'floor' is the function which gives the bigger integer smaller 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:
|
||||
Then, a vector in which the $j^{\text{th}}$ entry contains both the sum $S_j$
|
||||
of all the $|P_v|$s relative to each $P_h$ fallen into the $j^{\text{th}}$ bin
|
||||
itself and the number num$_j$ of the bin entries 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 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 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
|
||||
$|p_v|$.
|
||||
- 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 $|P_v|$.
|
||||
|
||||
At the end, $\langle |p_h| \rangle_j$ = $\langle |p_h| \rangle (p_h)$ where:
|
||||
For $P_{\text{max}} = 10$ and $n = 50$, the following result was obtained:
|
||||
|
||||
$$
|
||||
p_h = j \cdot w + \frac{w}{2} = w \left( 1 + \frac{1}{2} \right)
|
||||
$$
|
||||
|
||||
|
||||
For $p_{\text{max}} = 10$, 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
|
||||
In order to check whether the expected distribution (@eq:dip) properly matches
|
||||
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
|
||||
and the error of the so obtained estimation of $P_{\text{max}}$ was given as
|
||||
the inverse of the $\chi^2$ second derivative in its minimum, according to the
|
||||
Cramér-Rao bound.
|
||||
|
||||
The following results were obtained:
|
||||
|
||||
$$
|
||||
p^{\text{oss}}_{\text{max}} = 10.005 \pm 0.018 \with \chi^2 = 0.071
|
||||
P^{\text{oss}}_{\text{max}} = 10.005 \pm 0.018 \with \chi_r^2 = 0.071
|
||||
$$
|
||||
|
||||
In order to compare $p^{\text{oss}}_{\text{max}}$ with the expected value
|
||||
$p_{\text{max}} = 10$, the following compatibility t-test was applied:
|
||||
where $\chi_r^2$ is the reduced $\chi^2$, proving a good convergence.
|
||||
In order to compare $P^{\text{oss}}_{\text{max}}$ with the expected value
|
||||
$P_{\text{max}} = 10$, the following compatibility t-test was applied:
|
||||
|
||||
$$
|
||||
p = 1 - \text{erf}\left(\frac{t}{\sqrt{2}}\right)\ \with
|
||||
t = \frac{|p^{\text{oss}}_{\text{max}} - p_{\text{max}}|}
|
||||
{\Delta p_{\text{max}}}
|
||||
t = \frac{|P^{\text{oss}}_{\text{max}} - P_{\text{max}}|}
|
||||
{\Delta P^{\text{oss}}_{\text{max}}}
|
||||
$$
|
||||
|
||||
where $\Delta p_{\text{max}}$ is the absolute error of $p_{\text{max}}$. At 95%
|
||||
confidence level, the values are compatible if $p > 0.05$.
|
||||
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
|
||||
@ -254,4 +230,5 @@ 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}
|
||||
{#fig:fit}
|
||||
|
||||
Loading…
Reference in New Issue
Block a user