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 p_v and p_h, 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 p_v follow as a function of p_h?

\begin{figure} \hypertarget{fig:components}{% \centering \begin{tikzpicture} % 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}

The aim is to compute \langle |p_v| \rangle (p_h) dp_h.

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 length p.

  \begin{cases}
  p_h = p \sin{\theta} \\ p_v = p \cos{\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 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}
  \thus \sin{\theta_x} = \sin{\theta_y} = \frac{p_h}{p_{\text{max}}}

Since the average value of a quantity is computed by integrating it over all the possible quantities it depends on weighted on their probability, one gets:

  \langle |p_v| \rangle (p_h) dp_h = \int_{\theta_x}^{\theta_y} 
  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 $\theta \in [\theta; \theta + d\theta]$ 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

with:

  dp = p(p_h + dp_h) - p(p_h)
     = \frac{p_h + dp_h}{\sin{\theta}} - \frac{p_h}{\sin{\theta}}
     = \frac{dp_h}{\sin{\theta}}

hence

  P(p) \, dp = \frac{1}{p_{\text{max}}} \cdot \frac{1}{\sin{\theta}} \, dp_h

For d\theta P(\theta), instead, one has to do the same considerations done in @sec:3, from which:

  P(\theta) d\theta = \frac{1}{2} \sin{\theta} d\theta

Ultimately, having found all the pieces, the integral must be computed:

\begin{align*} \langle |p_v| \rangle (p_h) dp_h &= \int_{\theta_x}^{\theta_y} d\theta \frac{1}{2} \sin{\theta} \cdot \frac{1}{p_{\text{max}}} \frac{1}{\sin{\theta}} , dp_h \cdot 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:

\begin{align*} \mathscr{O} &= \int_{\theta_x}^{\theta_y} d\theta \frac{|\cos{\theta}|}{\sin{\theta}} \ &= \int_{\theta_x}^{\frac{\pi}{2}} d\theta \frac{\cos{\theta}}{\sin{\theta}}

• \int_{\frac{\pi}{2}}^{\theta_y} d\theta \frac{\cos{\theta}}{\sin{\theta}} \ &= \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} \hypertarget{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},{12ln(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

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: