8.0 KiB
Exercise 4
Kinematic dip derivation
Consider a great number of non-interacting particles having random momenta
\vec{P}, with magnitude between 0 and P_{\text{max}}, at an angle $\theta$
wrt to some coordinate system (\hat{x}, \hat{y}, \hat{z}).
The vertical and horizontal components of a particle momentum, which will be
referred as \vec{P_v} and \vec{P_h} respectively, are shown in
@fig:components.
If \theta is uniformly distributed on the unit sphere and P is uniformly
distributed in [0, P^\text{max}], what will be the average value
|P_v| of the particles with a given 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 \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 probability of
getting a fixed value of P_v given x to 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)}
= \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 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 uniform:
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. Since P,\theta are independent variables, their joint PDF is
simply given by the product:
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)
and they are related to the vertical and horizontal components by a standard polar coordinate transformation: \begin{align*} \begin{cases} P_v = P \cos(\theta) \ P_h = P \sin(\theta) \end{cases} && \begin{cases} P = \sqrt{P_v^2 + P_h^2} \ \theta = \text{atan2}(P_h, P_v) \end{cases} \end{align*} where:
-
\theta \in [0, \pi], -
and atan2 is defined by:
\begin{cases}
\arctan(P_h/P_v) &\incase P_v > 0 \\
\pi/2 &\incase P_v = 0 \\
\arctan(P_h/P_v) + \pi &\incase P_v < 0
\end{cases}
The Jacobian of the inverse transformation is easily found to be:
|J^{-1}| = \frac{1}{\sqrt{P_v^2 + P_h^2}}
Hence, the PDF written in the new coordinates is:
f_{P_h , P_v} (P_h, P_v) =
\frac{1}{2} \sin\left[ \text{atan2}(P_h, P_v) \right]
\chi_{[0, \pi]} \left[\text{atan2}(P_h, P_v)\right] \cdot \\
\frac{\chi_{[0, p_{\text{max}}]} \left(\sqrt{P_v^2 + P_h^2} \right)}
{\sqrt{P_v^2 + P_h^2}}
The integral I can now be computed. Note that the domain is implicit in the
characteristic functions:
I(x) = \int\limits_{-\infty}^{+\infty} dP_v \, f_{P_h , P_v} (x, P_v)
= \hspace{-20pt} \int \limits_{- \sqrt{P_{\text{max}}^2 - P_h}}
^{\sqrt{P_{\text{max}}^2 - P_h}}
\hspace{-20pt} dP_v \, f_{P_h , P_v} (x, P_v)
With some basic calculus and the identity
\sin[ \text{atan2}(P_h, P_v)] = \frac{P_h}{\sqrt{P_h^2 + P_v^2}},
the integral can be evaluated to give
I = 2 \, \arctan \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{\chi_{[0, \pi]} \left[\text{atan2}(P_h, P_v)\right]
\chi_{[0, p_{\text{max}}]} \left(\sqrt{P_v^2 + P_h^2}\right)}{2 \, \arctan
\left( \sqrt{\frac{P_{\text{max}}^2}{x^2} - 1} \right)}
Finally, putting all the pieces together, the average value of |P_v| can be
computed:
\langle |P_v| \rangle(x)
= \int P_v f (P_v | P_h = x) dP_v
= \frac{x \ln \left( \frac{P_{\text{max}}}{x} \right)}
{\arctan \left( \sqrt{ \frac{P^2_{\text{max}}}{x^2} - 1} \right)}
$$ {#eq:dip}
The result is plotted in the figure below:
{#fig:plot}
## Monte Carlo simulation
This dependence should be found by running a Monte Carlo simulation and
computing a binned average of the vertical momentum. A number of $N = 50000$
particles was generated as pairs of values ($P$, $\theta$), with $P$ uniformly
distributed between 0 and $P_{\text{max}}$ and $\theta$ given by the same
procedure described in @sec:3, namely:
\theta = \arccos(1 - 2x)
where $x$ is uniformly distributed between 0 and 1.
The binning turned out to be quite a challenge: once a $P$ is sampled and
$P_h$ computed, the bin containing the latter has to be found. If
the range $[0, P_{\text{max}}]$ is divided into $n$ equal bins
of width:
w = \frac{P_{\text{max}}}{n}
then (counting from zero) $P_h$ goes into the $i$-th bin, where:
i = \left\lfloor \frac{P_h}{w} \right\rfloor
Then, the sum $S_j$ of all the $|P_v|$ values relative to the $P_h$ of the
$j$-th bin and the number num$_j$ of the bin counts are stored in an array
and iteratively updated. Once every point has been sampled, the average value
of $|P_v|_j$ is 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 containing $P_h$ is found,
- num$_j$ is increased by 1,
- $S_j$ is increased by $|P_v|$.
For $P_{\text{max}} = 10$ and $n = 50$, the following result was obtained:
In order to assert the compatibility of the expected function (@eq:dip)
with the histogram, a least squares minimization was applied. Being a simple
one-parameter fit, the $\chi^2$ was implemented manually and minimised
without using a general LSQ routine. The error of the estimation of
$P_{\text{max}}$ was computed as the inverse of the $\chi^2$ second derivative
at the minimum, according to the Cramér-Rao bound.
The following results were obtained:
\begin{align*}
P^{\text{oss}}_{\text{max}} = 10.005 \pm 0.018 && \chi^2 &= 0.071 \\
&& \text{P}(x > \chi^2) &= 0.79
\end{align*}
The $\chi^2$ and $p$-value show a very good agreement.
In order to compare $P^{\text{oss}}_{\text{max}}$ with the expected value
$P_{\text{max}} = 10$, the usual 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{oss}}_{\text{max}}}
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
- p = 0.768
which allows to assert that the sampled points actually follow the predicted
function. In @fig:fit, the fit function superimposed on the histogram is
shown.
{#fig:fit}