ex-4: went on writing the theory

ex-4/main.c

@@ -0,0 +1,117 @@
+#include <math.h>
+#include <stdlib.h>
+#include <string.h>
+#include <gsl/gsl_rng.h>
+
+int main(int argc, char **argv)
+  {
+
+  // Set default options.
+  //
+  size_t N = 50000;          // number of events.
+  size_t n = 50;             // number of bins.
+  double p_max = 10;         // maximum value of momentum module.
+
+  // Process CLI arguments.
+  //
+  for (size_t i = 1; i < argc; i++)
+    {
+         if (!strcmp(argv[i], "-N")) N     = atol(argv[++i]);
+    else if (!strcmp(argv[i], "-n")) n     = atol(argv[++i]);
+    else if (!strcmp(argv[i], "-p")) p_max = atof(argv[++i]);
+    else
+      {
+      fprintf(stderr, "Usage: %s -[hiIntp]\n", argv[0]);
+      fprintf(stderr, "\t-h\tShow this message.\n");
+      fprintf(stderr, "\t-N integer\tThe number of events to generate.\n");
+      fprintf(stderr, "\t-n integer\tThe number of bins of the histogram.\n");
+      fprintf(stderr, "\t-p float\tThe maximum value of momentum.\n");
+      return EXIT_FAILURE;
+      }
+    }
+
+  // Initialize an RNG.
+  //
+  gsl_rng_env_setup();
+  gsl_rng *r = gsl_rng_alloc(gsl_rng_default);
+
+  // Generate the angle θ uniformly distributed on a sphere using the
+  // inverse transform:
+  //
+  //   θ = acos(1 - 2X)
+  //
+  // where X is a random uniform variable in [0,1), and the module p of
+  // the vector:
+  //
+  //   p² = p_v² + p_h²
+  // 
+  // uniformly distributed between 0 and p_max. The two components are
+  // then computed as:
+  //
+  //   p_v = p⋅cos(θ)
+  //   p_h = p⋅sin(θ)
+  //
+  // The histogram will be updated this way.
+  // The j-th bin where p_h goes in is given by:
+  //
+  //   step = p_max / n
+  //   j = floor(p_h / step)
+  //
+  // Thus an histogram was created and a structure containing the number of
+  // entries in each bin and the sum of |p_v| in each of them is created and
+  // filled while generating the events.
+  //
+  struct bin
+    {
+    size_t amo;  // Amount of events in the bin.
+    double sum;  // Sum of |p_v|s of all the events in the bin.
+    };
+  struct bin *histo = calloc(n, sizeof(struct bin));
+
+  // Some useful variables.
+  //
+  double step = p_max / n;
+  struct bin *b;
+  double theta;
+  double p;
+  double p_v;
+  double p_h;
+  size_t j;
+
+  for (size_t i = 0; i < N; i++)
+    {
+    // Generate the event.
+    // 
+    theta = acos(1 - 2*gsl_rng_uniform(r));
+    p = p_max * gsl_rng_uniform(r);
+
+    // Compute the components.
+    //
+    p_v = p * cos(theta);
+    p_h = p * sin(theta);
+
+    // Update the histogram.
+    //
+    j = floor(p_h / step);
+    b = &histo[j];
+    b->amo++;
+    b->sum += fabs(p_v);
+    }
+
+  // Compute the mean value of each bin and print it to stodut
+  // together with other useful things to make the histogram.
+  //
+  printf("bins: \t%ld\n", n);
+  printf("step: \t%.5f\n", step);
+  for (size_t i = 0; i < n; i++)
+    {
+    histo[i].sum = histo[i].sum / histo[i].amo;
+    printf("\n%.5f", histo[i].sum);
+    };
+
+  // free memory
+  gsl_rng_free(r);
+  free(histo);
+
+  return EXIT_SUCCESS;
+  }

notes/sections/4.md

@@ -6,16 +6,16 @@ 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 $p_v$ and $p_h$, are the
+components of a particle, which will be referred as $\vec{p_v}$ and $\vec{p_h}$
-ones shown in @fig:components.  
+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 the absolute value of
+module $p$, which distribution will the average value of $|p_v|$ follow as a
-$p_v$ follow as a function of $p_h$?
+function of $p_h$?
 
 \begin{figure}
 \hypertarget{fig:components}{%
 \centering
-\begin{tikzpicture}
+\begin{tikzpicture}[font=\scriptsize]
   % Axes
   \draw [thick, ->] (5,2) -- (5,8);
   \draw [thick, ->] (5,2) -- (2,1);
@@ -41,178 +41,128 @@ $p_v$ follow as a function of $p_h$?
 }
 \end{figure}
 
-The aim is to compute $\langle |p_v| \rangle (p_h) dp_h$.
+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 $x$:
 
-Consider all the points with $p_h \in [p_h ; p_h - dp_h]$: the values of
+$$
-$p_v$ that these points can assume depend on $\theta$ and the total momentum
+  f (P_v | P_h = x) = \frac{f_{P_h , P_v} (x, P_v)}
-length $p$.
+  {\int_{\{ P_v \}} d P_v f_{P_h , P_v} (x, P_v)}
+  = \frac{f_{P_h , P_v} (x, P_v)}{I}
+$$
+
+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:
 
 $$
   \begin{cases}
-  p_h = p \sin{\theta} \\ p_v = p \cos{\theta}
+    P_v = P \cos(\theta) \\
+    P_h = P \sin(\theta)
   \end{cases}
-  \thus |p_v| = p |\cos{\theta}| = p_h \frac{|\cos{\theta}|}{\sin{\theta}}
-  = |p_v| (\theta)
 $$
 
-It looks like the dependence on $p$ has disappeared, but obviously it has
+with $\theta \in [0, \pi]$, the previous ones are written as:
-not. In fact, it lies beneath the limits that one has to put on the possible
-values of $\theta$. For the sake of clarity, take a look at @fig:sphere (the
-system is rotation-invariant, hence it can be drown at a fixed azimuthal angle).
-
-\begin{figure}
-\hypertarget{fig:sphere}{%
-\centering
-\begin{tikzpicture}
-  % p_h slice
-  \definecolor{cyclamen}{RGB}{146, 24, 43}
-  \filldraw [cyclamen!15!white] (1.5,-3.15) -- (1.5,3.15) -- (1.75,3.05)
-                             -- (2,2.85) -- (2,-2.85) -- (1.75,-3.05)
-                             -- (1.5,-3.15);
-  \draw [cyclamen] (1.5,-3.15) -- (1.5,3.15);
-  \draw [cyclamen] (2,-2.9)    -- (2,2.9);
-  \node [cyclamen, left] at (1.5,-0.3) {$p_h$};
-  \node [cyclamen, right] at (2,-0.3) {$p_h + dp_h$};
-  % Axes
-  \draw [thick, ->] (0,-4) -- (0,4);
-  \draw [thick, ->] (0,0)  -- (4,0);
-  \node at (0.3,3.8) {$z$};
-  \node at (4,0.3) {$hd$};
-  % p_max semicircle
-  \draw [thick, cyclamen] (0,-3.5) to [out=0, in=-90] (3.5,0);
-  \draw [thick, cyclamen] (0,3.5)  to [out=0, in=90]  (3.5,0);
-  \node [cyclamen, left] at (-0.2,3.5)  {$p_{\text{max}}$};
-  \node [cyclamen, left] at (-0.2,-3.5) {$-p_{\text{max}}$};
-  % Angles
-  \draw [thick, cyclamen, ->] (0,1.5) to [out=0, in=140] (0.55,1.2);
-  \node [cyclamen] at (0.4,2) {$\theta$};
-  \draw [thick, cyclamen] (0,0) -- (1.5,3.15);
-  \node [cyclamen, above right] at (1.5,3.15) {$\theta_x$};
-  \draw [thick, cyclamen] (0,0) -- (1.5,-3.15);
-  \node [cyclamen, below right] at (1.5,-3.15) {$\theta_y$};
-  % Vectors
-  \draw [ultra thick, cyclamen, ->] (0,0) -- (1.7,2.2);
-  \draw [ultra thick, cyclamen, ->] (0,0) -- (1.9,0.6);
-  \draw [ultra thick, cyclamen, ->] (0,0) -- (1.6,-2);
-\end{tikzpicture}
-\caption{Momentum space at fixed azimuthal angle ("$hd$" stands for
-"horizontal direction"). Some vectors with $p_h \in [p_h, p_h +dp_h]$
-are evidenced.}\label{fig:sphere}
-}
-\end{figure}
-
-As can be seen, $\theta_x$ and $\theta_y$ are the minimum and maximum tilts
-angles of these vectors respectively, because if a point had $p_h \in [p_h; p_h
-+ dp_h]$ and $\theta < \theta_x$ or $\theta > \theta_y$, it would have $p >
-p_{\text{max}}$. Therefore their values are easily computed as follow:
 
 $$
-  p_h = p_{\text{max}} \sin{\theta_x} = p_{\text{max}} \sin{\theta_y}
+  \begin{cases}
-  \thus \sin{\theta_x} = \sin{\theta_y} = \frac{p_h}{p_{\text{max}}}
+    P = \sqrt{P_v^2 + P_h^2} \\
+    \theta = \text{atan}^{\star} ( P_h/P_v ) :=
+      \begin{cases}
+        \text{atan} ( P_h/P_v )       &\incase P_v > 0 \\
+        \text{atan} ( P_h/P_v ) + \pi &\incase P_v < 0
+      \end{cases} 
+  \end{cases}
 $$
 
-Since the average value of a quantity is computed by integrating it over all the
+which can be shown having Jacobian:
-possible quantities it depends on weighted on their probability, one gets:
 
 $$
-  \langle |p_v| \rangle (p_h) dp_h = \int_{\theta_x}^{\theta_y} 
+  J = \frac{1}{\sqrt{P_v^2 + P_h^2}}
-  d\theta P(\theta) \cdot  P(p) \, dp \cdot |p_v| (\theta)
 $$
 
-where $d\theta P(\theta)$ is the probability of generating a point with $\theta
+Hence:
-\in [\theta; \theta + d\theta]$ and $P(p) \, dp$ is the probability of
-generating a point with $\vec{p}$ in the pink region in @fig:sphere, given a
-fixed $\theta$.  
-The easiest to deduce is $P(p) \, dp$: since $p$ is evenly distributed, it
-follows that:
 
 $$  
-  P(p) \, dp = \frac{1}{p_{\text{max}}} dp
+  f_{P_h , P_v} (P_h, P_v) =
+    \frac{1}{2} \sin[ \text{atan}^{\star} ( P_h/P_v )]
+    \chi_{[0, \pi]} (\text{atan}^{\star} ( P_h/P_v )) \cdot \\
+    \frac{\chi_{[0, p_{\text{max}}]} \left( \sqrt{P_v^2 + P_h^2} \right)}
+    {\sqrt{P_v^2 + P_h^2}}
 $$
 
-with:
+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 $P_{\text{max}}$:
 
 $$
-  dp = p(p_h + dp_h) - p(p_h)
+  I = \int
-     = \frac{p_h + dp_h}{\sin{\theta}} - \frac{p_h}{\sin{\theta}}
+  \limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
-     = \frac{dp_h}{\sin{\theta}}
+  dP_v \, f_{P_h , P_v} (x, P_v)
 $$
 
-hence
+after a bit of maths, using the identity:
 
 $$
-  P(p) \, dp = \frac{1}{p_{\text{max}}} \cdot \frac{1}{\sin{\theta}} \, dp_h
+  \sin[ \text{atan}^{\star} ( P_h/P_v )] = \frac{P_h}{\sqrt{P_h^2 + P_v^2}}
 $$
 
-For $d\theta P(\theta)$, instead, one has to do the same considerations done
+and the fact that both the characteristic functions are automatically satisfied
-in @sec:3, from which:
+within the integral limits, the following result arises:
 
 $$
-  P(\theta) d\theta = \frac{1}{2} \sin{\theta} d\theta
+  I = 2 \, \text{atan} \left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)
 $$
 
-Ultimately, having found all the pieces, the integral must be computed:
+from which:
 
-\begin{align*}
+$$
-  \langle |p_v| \rangle (p_h) dp_h &= \int_{\theta_x}^{\theta_y} 
+  f (P_v | P_h = x) = \frac{x}{P_v^2 + x^2} \cdot
-  d\theta \frac{1}{2} \sin{\theta} \cdot
+  \frac{1}{2 \, \text{atan}
-  \frac{1}{p_{\text{max}}} \frac{1}{\sin{\theta}} \, dp_h \cdot
+           \left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)}
-  p_h \frac{|\cos{\theta}|}{\sin{\theta}}
+$$
-  \\
-  &= \frac{1}{2} \frac{p_h dp_h}{p_{\text{max}}} \int_{\theta_x}^{\theta_y}
-  d\theta \frac{|\cos{\theta}|}{\sin{\theta}} 
-  \\
-  &= \frac{1}{2} \frac{p_h dp_h}{p_{\text{max}}} \cdot \mathscr{O}
-\end{align*}
 
-Then, with a bit of math:
+Finally, putting all the pieces together, the average value of $|P_v|$ can now
+be computed:
 
-\begin{align*}
+$$
-  \mathscr{O} &= \int_{\theta_x}^{\theta_y} d\theta
+  \langle |P_v| \rangle = \int
-  \frac{|\cos{\theta}|}{\sin{\theta}}
+  \limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
-  \\
+  f (P_v | P_h = x) = [ \dots ] =
-  &= \int_{\theta_x}^{\frac{\pi}{2}} d\theta \frac{\cos{\theta}}{\sin{\theta}}
+  x \, \frac{\ln \left( \frac{P_{\text{max}}}{x} \right)}
-  - \int_{\frac{\pi}{2}}^{\theta_y} d\theta \frac{\cos{\theta}}{\sin{\theta}}
+  {\text{atan} \left( \sqrt{ \frac{P_{\text{max}}}{x^2} - 1} \right)}
-  \\
+$$
-  &= \left[ \ln{(\sin{\theta})} \right]_{\theta_x}^{\frac{\pi}{2}}
-  - \left[ \ln{(\sin{\theta})} \right]_{\frac{\pi}{2}}^{\theta_y}
-  \\
-  &= \ln{(1)} -\ln{ \left( \frac{p_h}{p_{\text{max}}} \right) }
-  - \ln{ \left( \frac{p_h}{p_{\text{max}}} \right) } + \ln{(1)}
-  \\
-  &= 2 \ln{ \left( \frac{p_{\text{max}}}{p_h} \right) }
-\end{align*}
-
-\newpage
-Hence, in conclusion:
-
-\begin{align*}
-  \langle |p_v| \rangle (p_h) dp_h &= \frac{1}{2} \frac{p_h dp_h}{p_{\text{max}}}
-  \cdot 2 \ln{ \left( \frac{p_{\text{max}}}{p_h} \right) }
-  \\
-  &= \ln{ \left( \frac{p_{\text{max}}}{p_h} \right) }
-  \frac{p_h}{p_{\text{max}}} dp_h
-\end{align*}
 
 Namely:
 
-\begin{figure}
+![Plot of the expected distribution with
-\hypertarget{fig:plot}{%
+  $P_{\text{max}} = 10$.](images/expected.pdf){#fig:plot}
-\centering
-\begin{tikzpicture}
-  \definecolor{cyclamen}{RGB}{146, 24, 43}
-  % Axis
-  \draw [thick, <->] (0,5) -- (0,0) -- (11,0);
-  \node [below right] at (11,0) {$p_h$};
-  \node [above left] at (0,5) {$\langle |p_v| \rangle$};
-  % Plot
-  \draw [domain=0.001:10, smooth, variable=\x,
-         cyclamen, ultra thick] plot ({\x},{12*ln(10/\x)*\x/10});
-  \node [cyclamen, below] at (10,0) {$p_{\text{max}}$};
-\end{tikzpicture}
-\caption{Plot of the expected distribution.}\label{fig:plot}
-}
-\end{figure}
 
 
 ## Monte Carlo simulation