ex-4: went on writing the theory
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ex-4/main.c
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117
ex-4/main.c
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#include <math.h>
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#include <stdlib.h>
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#include <string.h>
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#include <gsl/gsl_rng.h>
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int main(int argc, char **argv)
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{
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// Set default options.
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//
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size_t N = 50000; // number of events.
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size_t n = 50; // number of bins.
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double p_max = 10; // maximum value of momentum module.
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// Process CLI arguments.
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//
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for (size_t i = 1; i < argc; i++)
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{
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if (!strcmp(argv[i], "-N")) N = atol(argv[++i]);
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else if (!strcmp(argv[i], "-n")) n = atol(argv[++i]);
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else if (!strcmp(argv[i], "-p")) p_max = atof(argv[++i]);
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else
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{
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fprintf(stderr, "Usage: %s -[hiIntp]\n", argv[0]);
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fprintf(stderr, "\t-h\tShow this message.\n");
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fprintf(stderr, "\t-N integer\tThe number of events to generate.\n");
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fprintf(stderr, "\t-n integer\tThe number of bins of the histogram.\n");
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fprintf(stderr, "\t-p float\tThe maximum value of momentum.\n");
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return EXIT_FAILURE;
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}
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}
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// Initialize an RNG.
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//
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gsl_rng_env_setup();
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gsl_rng *r = gsl_rng_alloc(gsl_rng_default);
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// Generate the angle θ uniformly distributed on a sphere using the
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// inverse transform:
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//
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// θ = acos(1 - 2X)
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//
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// where X is a random uniform variable in [0,1), and the module p of
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// the vector:
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//
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// p² = p_v² + p_h²
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//
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// uniformly distributed between 0 and p_max. The two components are
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// then computed as:
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//
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// p_v = p⋅cos(θ)
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// p_h = p⋅sin(θ)
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//
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// The histogram will be updated this way.
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// The j-th bin where p_h goes in is given by:
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//
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// step = p_max / n
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// j = floor(p_h / step)
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//
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// Thus an histogram was created and a structure containing the number of
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// entries in each bin and the sum of |p_v| in each of them is created and
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// filled while generating the events.
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//
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struct bin
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{
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size_t amo; // Amount of events in the bin.
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double sum; // Sum of |p_v|s of all the events in the bin.
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};
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struct bin *histo = calloc(n, sizeof(struct bin));
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// Some useful variables.
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//
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double step = p_max / n;
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struct bin *b;
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double theta;
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double p;
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double p_v;
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double p_h;
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size_t j;
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for (size_t i = 0; i < N; i++)
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{
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// Generate the event.
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//
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theta = acos(1 - 2*gsl_rng_uniform(r));
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p = p_max * gsl_rng_uniform(r);
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// Compute the components.
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//
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p_v = p * cos(theta);
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p_h = p * sin(theta);
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// Update the histogram.
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//
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j = floor(p_h / step);
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b = &histo[j];
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b->amo++;
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b->sum += fabs(p_v);
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}
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// Compute the mean value of each bin and print it to stodut
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// together with other useful things to make the histogram.
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//
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printf("bins: \t%ld\n", n);
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printf("step: \t%.5f\n", step);
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for (size_t i = 0; i < n; i++)
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{
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histo[i].sum = histo[i].sum / histo[i].amo;
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printf("\n%.5f", histo[i].sum);
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};
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// free memory
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gsl_rng_free(r);
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free(histo);
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return EXIT_SUCCESS;
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}
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Binary file not shown.
@ -6,16 +6,16 @@ Consider a great number of non-interacting particles, each of which with a
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random momentum $\vec{p}$ with module between 0 and $p_{\text{max}}$ randomly
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angled with respect to a coordinate system {$\hat{x}$, $\hat{y}$, $\hat{z}$}.
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Once the polar angle $\theta$ is defined, the momentum vertical and horizontal
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components of a particle, which will be referred as $p_v$ and $p_h$, are the
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ones shown in @fig:components.
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components of a particle, which will be referred as $\vec{p_v}$ and $\vec{p_h}$
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respectively, are the ones shown in @fig:components.
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If $\theta$ is evenly distributed on the sphere and the same holds for the
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module $p$, which distribution will the average value of the absolute value of
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$p_v$ follow as a function of $p_h$?
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module $p$, which distribution will the average value of $|p_v|$ follow as a
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function of $p_h$?
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\begin{figure}
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\hypertarget{fig:components}{%
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\centering
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\begin{tikzpicture}
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\begin{tikzpicture}[font=\scriptsize]
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% Axes
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\draw [thick, ->] (5,2) -- (5,8);
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\draw [thick, ->] (5,2) -- (2,1);
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@ -41,178 +41,128 @@ $p_v$ follow as a function of $p_h$?
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}
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\end{figure}
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The aim is to compute $\langle |p_v| \rangle (p_h) dp_h$.
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Since the aim is to compute $\langle |p_v| \rangle (p_h)$, the conditional
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distribution probability of $p_v$ given a fixed value of $p_h = x$ must first
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be determined. It can be computed as the ratio between the probablity of
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getting a fixed value of $P_v$ given $x$ over the total probability of $x$:
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Consider all the points with $p_h \in [p_h ; p_h - dp_h]$: the values of
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$p_v$ that these points can assume depend on $\theta$ and the total momentum
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length $p$.
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$$
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f (P_v | P_h = x) = \frac{f_{P_h , P_v} (x, P_v)}
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{\int_{\{ P_v \}} d P_v f_{P_h , P_v} (x, P_v)}
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= \frac{f_{P_h , P_v} (x, P_v)}{I}
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$$
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where $f_{P_h , P_v}$ is the joint pdf of the two variables $P_v$ and $P_h$ and
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the integral runs over all the possible values of $P_v$ given $P_h$.
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This joint pdf can simly be computed from the joint pdf of $\theta$ and $p$ with
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a change of variables. For the pdf of $\theta$ $f_{\theta} (\theta)$, the same
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considerations done in @sec:3 lead to:
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$$
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f_{\theta} (\theta) = \frac{1}{2} \sin{\theta} \chi_{[0, \pi]} (\theta)
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$$
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whereas, being $p$ evenly distributed:
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$$
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f_p (p) = \chi_{[0, p_{\text{max}}]} (p)
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$$
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where $\chi_{[a, b]} (y)$ is the normalized characteristic function which value
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is $1/N$ between $a$ and $b$, where $N$ is the normalization term, and 0
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elsewhere. Being a couple of independent variables, their joint pdf is simply
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given by the product of their pdfs:
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$$
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f_{\theta , p} (\theta, p) = f_{\theta} (\theta) f_p (p)
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= \frac{1}{2} \sin{\theta} \chi_{[0, \pi]} (\theta)
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\chi_{[0, p_{\text{max}}]} (p)
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$$
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Given the new variables:
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$$
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\begin{cases}
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p_h = p \sin{\theta} \\ p_v = p \cos{\theta}
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P_v = P \cos(\theta) \\
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P_h = P \sin(\theta)
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\end{cases}
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\thus |p_v| = p |\cos{\theta}| = p_h \frac{|\cos{\theta}|}{\sin{\theta}}
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= |p_v| (\theta)
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$$
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It looks like the dependence on $p$ has disappeared, but obviously it has
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not. In fact, it lies beneath the limits that one has to put on the possible
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values of $\theta$. For the sake of clarity, take a look at @fig:sphere (the
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system is rotation-invariant, hence it can be drown at a fixed azimuthal angle).
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\begin{figure}
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\hypertarget{fig:sphere}{%
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\centering
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\begin{tikzpicture}
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% p_h slice
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\definecolor{cyclamen}{RGB}{146, 24, 43}
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\filldraw [cyclamen!15!white] (1.5,-3.15) -- (1.5,3.15) -- (1.75,3.05)
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-- (2,2.85) -- (2,-2.85) -- (1.75,-3.05)
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-- (1.5,-3.15);
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\draw [cyclamen] (1.5,-3.15) -- (1.5,3.15);
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\draw [cyclamen] (2,-2.9) -- (2,2.9);
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\node [cyclamen, left] at (1.5,-0.3) {$p_h$};
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\node [cyclamen, right] at (2,-0.3) {$p_h + dp_h$};
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% Axes
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\draw [thick, ->] (0,-4) -- (0,4);
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\draw [thick, ->] (0,0) -- (4,0);
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\node at (0.3,3.8) {$z$};
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\node at (4,0.3) {$hd$};
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% p_max semicircle
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\draw [thick, cyclamen] (0,-3.5) to [out=0, in=-90] (3.5,0);
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\draw [thick, cyclamen] (0,3.5) to [out=0, in=90] (3.5,0);
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\node [cyclamen, left] at (-0.2,3.5) {$p_{\text{max}}$};
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\node [cyclamen, left] at (-0.2,-3.5) {$-p_{\text{max}}$};
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% Angles
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\draw [thick, cyclamen, ->] (0,1.5) to [out=0, in=140] (0.55,1.2);
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\node [cyclamen] at (0.4,2) {$\theta$};
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\draw [thick, cyclamen] (0,0) -- (1.5,3.15);
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\node [cyclamen, above right] at (1.5,3.15) {$\theta_x$};
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\draw [thick, cyclamen] (0,0) -- (1.5,-3.15);
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\node [cyclamen, below right] at (1.5,-3.15) {$\theta_y$};
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% Vectors
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\draw [ultra thick, cyclamen, ->] (0,0) -- (1.7,2.2);
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\draw [ultra thick, cyclamen, ->] (0,0) -- (1.9,0.6);
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\draw [ultra thick, cyclamen, ->] (0,0) -- (1.6,-2);
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\end{tikzpicture}
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\caption{Momentum space at fixed azimuthal angle ("$hd$" stands for
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"horizontal direction"). Some vectors with $p_h \in [p_h, p_h +dp_h]$
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are evidenced.}\label{fig:sphere}
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}
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\end{figure}
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As can be seen, $\theta_x$ and $\theta_y$ are the minimum and maximum tilts
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angles of these vectors respectively, because if a point had $p_h \in [p_h; p_h
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+ dp_h]$ and $\theta < \theta_x$ or $\theta > \theta_y$, it would have $p >
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p_{\text{max}}$. Therefore their values are easily computed as follow:
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with $\theta \in [0, \pi]$, the previous ones are written as:
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$$
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p_h = p_{\text{max}} \sin{\theta_x} = p_{\text{max}} \sin{\theta_y}
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\thus \sin{\theta_x} = \sin{\theta_y} = \frac{p_h}{p_{\text{max}}}
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\begin{cases}
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P = \sqrt{P_v^2 + P_h^2} \\
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\theta = \text{atan}^{\star} ( P_h/P_v ) :=
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\begin{cases}
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\text{atan} ( P_h/P_v ) &\incase P_v > 0 \\
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\text{atan} ( P_h/P_v ) + \pi &\incase P_v < 0
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\end{cases}
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\end{cases}
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$$
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Since the average value of a quantity is computed by integrating it over all the
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possible quantities it depends on weighted on their probability, one gets:
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which can be shown having Jacobian:
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$$
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\langle |p_v| \rangle (p_h) dp_h = \int_{\theta_x}^{\theta_y}
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d\theta P(\theta) \cdot P(p) \, dp \cdot |p_v| (\theta)
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J = \frac{1}{\sqrt{P_v^2 + P_h^2}}
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$$
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where $d\theta P(\theta)$ is the probability of generating a point with $\theta
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\in [\theta; \theta + d\theta]$ and $P(p) \, dp$ is the probability of
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generating a point with $\vec{p}$ in the pink region in @fig:sphere, given a
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fixed $\theta$.
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The easiest to deduce is $P(p) \, dp$: since $p$ is evenly distributed, it
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follows that:
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Hence:
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$$
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P(p) \, dp = \frac{1}{p_{\text{max}}} dp
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$$
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f_{P_h , P_v} (P_h, P_v) =
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\frac{1}{2} \sin[ \text{atan}^{\star} ( P_h/P_v )]
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\chi_{[0, \pi]} (\text{atan}^{\star} ( P_h/P_v )) \cdot \\
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\frac{\chi_{[0, p_{\text{max}}]} \left( \sqrt{P_v^2 + P_h^2} \right)}
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{\sqrt{P_v^2 + P_h^2}}
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$$
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with:
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from which, the integral $I$ can now be computed. The edges of the integral
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are fixed bt the fact that the total momentum can not exceed $P_{\text{max}}$:
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$$
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dp = p(p_h + dp_h) - p(p_h)
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= \frac{p_h + dp_h}{\sin{\theta}} - \frac{p_h}{\sin{\theta}}
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= \frac{dp_h}{\sin{\theta}}
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I = \int
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\limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
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dP_v \, f_{P_h , P_v} (x, P_v)
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$$
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hence
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after a bit of maths, using the identity:
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$$
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P(p) \, dp = \frac{1}{p_{\text{max}}} \cdot \frac{1}{\sin{\theta}} \, dp_h
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\sin[ \text{atan}^{\star} ( P_h/P_v )] = \frac{P_h}{\sqrt{P_h^2 + P_v^2}}
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$$
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For $d\theta P(\theta)$, instead, one has to do the same considerations done
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in @sec:3, from which:
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and the fact that both the characteristic functions are automatically satisfied
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within the integral limits, the following result arises:
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$$
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P(\theta) d\theta = \frac{1}{2} \sin{\theta} d\theta
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I = 2 \, \text{atan} \left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)
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$$
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Ultimately, having found all the pieces, the integral must be computed:
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from which:
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\begin{align*}
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\langle |p_v| \rangle (p_h) dp_h &= \int_{\theta_x}^{\theta_y}
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d\theta \frac{1}{2} \sin{\theta} \cdot
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\frac{1}{p_{\text{max}}} \frac{1}{\sin{\theta}} \, dp_h \cdot
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p_h \frac{|\cos{\theta}|}{\sin{\theta}}
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\\
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&= \frac{1}{2} \frac{p_h dp_h}{p_{\text{max}}} \int_{\theta_x}^{\theta_y}
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d\theta \frac{|\cos{\theta}|}{\sin{\theta}}
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\\
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&= \frac{1}{2} \frac{p_h dp_h}{p_{\text{max}}} \cdot \mathscr{O}
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\end{align*}
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$$
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f (P_v | P_h = x) = \frac{x}{P_v^2 + x^2} \cdot
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\frac{1}{2 \, \text{atan}
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\left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)}
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$$
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Then, with a bit of math:
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Finally, putting all the pieces together, the average value of $|P_v|$ can now
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be computed:
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\begin{align*}
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\mathscr{O} &= \int_{\theta_x}^{\theta_y} d\theta
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\frac{|\cos{\theta}|}{\sin{\theta}}
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\\
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&= \int_{\theta_x}^{\frac{\pi}{2}} d\theta \frac{\cos{\theta}}{\sin{\theta}}
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- \int_{\frac{\pi}{2}}^{\theta_y} d\theta \frac{\cos{\theta}}{\sin{\theta}}
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\\
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&= \left[ \ln{(\sin{\theta})} \right]_{\theta_x}^{\frac{\pi}{2}}
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- \left[ \ln{(\sin{\theta})} \right]_{\frac{\pi}{2}}^{\theta_y}
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\\
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&= \ln{(1)} -\ln{ \left( \frac{p_h}{p_{\text{max}}} \right) }
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- \ln{ \left( \frac{p_h}{p_{\text{max}}} \right) } + \ln{(1)}
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\\
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&= 2 \ln{ \left( \frac{p_{\text{max}}}{p_h} \right) }
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\end{align*}
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\newpage
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Hence, in conclusion:
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\begin{align*}
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\langle |p_v| \rangle (p_h) dp_h &= \frac{1}{2} \frac{p_h dp_h}{p_{\text{max}}}
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\cdot 2 \ln{ \left( \frac{p_{\text{max}}}{p_h} \right) }
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\\
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&= \ln{ \left( \frac{p_{\text{max}}}{p_h} \right) }
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\frac{p_h}{p_{\text{max}}} dp_h
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\end{align*}
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$$
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\langle |P_v| \rangle = \int
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\limits_{- \sqrt{P_{\text{max}}^2 - P_h}}^{\sqrt{P_{\text{max}}^2 - P_h}}
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f (P_v | P_h = x) = [ \dots ] =
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x \, \frac{\ln \left( \frac{P_{\text{max}}}{x} \right)}
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{\text{atan} \left( \sqrt{ \frac{P_{\text{max}}}{x^2} - 1} \right)}
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$$
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Namely:
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\begin{figure}
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\hypertarget{fig:plot}{%
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\centering
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\begin{tikzpicture}
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\definecolor{cyclamen}{RGB}{146, 24, 43}
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% Axis
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\draw [thick, <->] (0,5) -- (0,0) -- (11,0);
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\node [below right] at (11,0) {$p_h$};
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\node [above left] at (0,5) {$\langle |p_v| \rangle$};
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% Plot
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\draw [domain=0.001:10, smooth, variable=\x,
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cyclamen, ultra thick] plot ({\x},{12*ln(10/\x)*\x/10});
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\node [cyclamen, below] at (10,0) {$p_{\text{max}}$};
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\end{tikzpicture}
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\caption{Plot of the expected distribution.}\label{fig:plot}
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||||
}
|
||||
\end{figure}
|
||||
{#fig:plot}
|
||||
|
||||
|
||||
## Monte Carlo simulation
|
||||
|
||||
Loading…
Reference in New Issue
Block a user