Commit 58a3450b by dauncey

parent 392a2e92
 \documentclass[10pt]{article} \usepackage{a4} \usepackage{amsmath} \usepackage{graphicx} \oddsidemargin=0pt % No extra space wasted after first inch. \evensidemargin=0pt % Ditto (for two-sided output). ... ... @@ -216,6 +217,15 @@ around the truth value is given by \mathrm{RMS}^2 = \frac{1}{N} \sum_e (E_e - T_e)^2 = \frac{1}{N} \sum_e \left(\sum_i a_i d_{e,i} - T_e\right)^2 Alternatively, we could optimise for $\Delta E/E$, i.e. \mathrm{RMS}^2 = \frac{1}{N} \sum_e \left(\frac{E_e - T_e}{T_e}\right)^2 = \frac{1}{N} \sum_e \left(\frac{\sum_i a_i d_{e,i} - T_e}{T_e}\right)^2 = \frac{1}{N} \sum_e \left(\sum_i a_i d_{e,i}^\prime - 1\right)^2 where $d_{e,i}^\prime = d_{e,i}/T_e$. Hence, this method will be identical in form in the following, but replacing $T_e$ by 1 and $d_{e,i}$ by $d_{e,i}^\prime$. This can be thought of as similar to a chi-squared; we want to minimise this expression. If all the $a_i$ are considered as independent parameters (so 28 for the EE only), then explicitly ... ... @@ -466,6 +476,29 @@ For the example of $M=2$, $E=2$ in the table above, this gives $2-1+4=5$ bits, i.e. numbers up to $2^5=32$ as shown and the reduction is 1 bit. The maximum bit lengths of the value, i.e. $M+2^E-1$, for various values of $M$ and $E$ are shown in the table below. \bigskip \begin{center} \begin{tabular}{r||c|c|c|c|c|c|c|c|c} \hline $M=$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \cr\hline $E=0$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \cr 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \cr 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \cr 3 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \cr 4 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 \cr 5 & 31 & 32 & 33 & 34 & 35 & 36 & 37 & 38 & 39 \cr 6 & 63 & 64 & 65 & 66 & 67 & 68 & 69 & 70 & 71 \cr 7 & 127 & 128 & 129 & 130 & 131 & 132 & 133 & 134 & 135 \cr 8 & 255 & 256 & 257 & 258 & 259 & 260 & 261 & 262 & 263 \cr \hline \end{tabular} \end{center} %\newpage For the extreme values of $E$, then $E=0$ and $E=1$ both give an exact representation as they only use the lowest range or two lowest ranges, respectively. The reduction is $2^0-0-1=0$ and $2^1-1-1=0$ bits in both cases. ... ... @@ -504,29 +537,6 @@ Representation & Exponent & Mantissa & Value \cr\hline \end{tabular} \end{center} The maximum bit lengths of the value, i.e. $M+2^E-1$, for various values of $M$ and $E$ are shown in the table below. \bigskip \begin{center} \begin{tabular}{r||c|c|c|c|c|c|c|c|c} \hline $M=$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \cr\hline $E=0$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \cr 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \cr 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \cr 3 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \cr 4 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 \cr 5 & 31 & 32 & 33 & 34 & 35 & 36 & 37 & 38 & 39 \cr 6 & 63 & 64 & 65 & 66 & 67 & 68 & 69 & 70 & 71 \cr 7 & 127 & 128 & 129 & 130 & 131 & 132 & 133 & 134 & 135 \cr 8 & 255 & 256 & 257 & 258 & 259 & 260 & 261 & 262 & 263 \cr \hline \end{tabular} \end{center} \newpage \section{Template fit of energy in depth} Assume a template shape for a photon of $P_l$ per photon energy GeV for layer $l$. ... ... @@ -708,6 +718,138 @@ Again, for small $\Delta\phi$, then $\overline{\rho}$ clearly approximates to the geometric mean of the two radii, as before. \section{Cartesian coordinates and pseudorapidity} As above, take $\rho = \sqrt{x^2+y^2}$ and define the arc length $s=\rho\phi$. In terms of scaled coordinates $x_s=x/z$ and $y_s=y/z$, then equivalently $\rho_s = \rho/z$ and $s_s = (\rho/z)\phi = \rho_s\phi$. Note, $\rho_s = \tan\theta$ and so $x_s = \tan\theta\cos\phi$ and $y_s = \tan\theta\sin\phi$. Generally $\sinh\eta = z/\rho = 1/\rho_s$, so $\rho_s = 1/\sinh\eta$ and $s_s = \phi/\sinh\eta$. To convert from $E$ to $E_T$ requires a factor of $\sin\theta$, which is given by \sin\theta = \sin\left(\tan^{-1}\rho_s\right) = \sin\left[\tan^{-1}\left(\frac{1}{\sinh\eta}\right)\right] % \begin{figure}[ht!] \begin{center} \includegraphics[width=9cm]{SinThetaVsEta.pdf} %\caption{ % Left: Active rotation by $30^\circ$ % of an object in a fixed coordinate system. % Right: Passive rotation by $30^\circ$ % of the coordinate system around a fixed object. %} \label{fig:rotations} \end{center} \end{figure} Comparing areas in $\Delta \rho_s \Delta s_s$ and $\Delta \eta \Delta \phi$ requires the Jacobian. From above \frac{\partial \rho_s}{\partial \eta} = - \frac{\cosh\eta}{\sinh^2\eta},\qquad \frac{\partial \rho_s}{\partial \phi} = 0 while \frac{\partial s_s}{\partial \eta} = - \frac{\phi\cosh\eta}{\sinh^2\eta},\qquad \frac{\partial s_s}{\partial \phi} = \frac{1}{\sinh\eta} Hence the Jacobian is J = \left|\frac{\partial \rho_s}{\partial \eta}\frac{\partial s_s}{\partial \phi} - \frac{\partial \rho_s}{\partial \phi}\frac{\partial s_s}{\partial \eta}\right| = \left|\frac{\cosh\eta}{\sinh^2\eta}\times\frac{1}{\sinh\eta}\right| = \left|\frac{\cosh\eta}{\sinh^3\eta}\right| Hence \Delta \rho_s \Delta s_s = \left|\frac{\cosh\eta}{\sinh^3\eta}\right| \Delta \eta \Delta \phi Cross-check the other way around, using \eta = \sinh^{-1}\left(\frac{1}{\rho_s}\right),\qquad \phi = \frac{s_s}{\rho_s} then \frac{\partial \eta}{\partial \rho_s} = - \frac{\sinh^2\eta}{\cosh\eta} = -\frac{1}{\rho_s^2\sqrt{1+1/\rho_s^2}} = -\frac{1}{\rho_s\sqrt{1+\rho_s^2}},\qquad \frac{\partial \eta}{\partial s_s} = 0 and \frac{\partial \phi}{\partial \rho_s} = - \frac{s_s}{\rho_s^2},\qquad \frac{\partial \phi}{\partial s_s} = \frac{1}{\rho_s} Hence the inverse Jacobian to the above is \frac{1}{J} = \left|\frac{\partial \eta}{\partial \rho_s}\frac{\partial \phi}{\partial s_s} -\frac{\partial \phi}{\partial \rho_s}\frac{\partial \eta}{\partial s_s}\right| = \left|\frac{1}{\rho_s\sqrt{1+\rho_s^2}}\times\frac{1}{\rho_s}\right| = \left|\frac{1}{\rho_s^2\sqrt{1+\rho_s^2}}\right| In terms of $\eta$, this is \frac{1}{J} = \left|\frac{1/\rho_s^3}{\sqrt{1+1/\rho_s^2}}\right| = \left|\frac{\sinh^3\eta}{\cosh\eta}\right| and so is consistent. The derivatives are shown below, showing that they are very similar in size, so that a circle in $\rho_s,s_s$ coordinates is also circular to a good approximation in $\eta,\phi$. At the innermost edge of the HGCAL, i.e. $\eta=3$, the derivatives are around 10, so for example $\Delta\rho_s=\Delta s_s=0.02$ is approximately equivalent to $\Delta\eta=\Delta\phi = 0.2$. % \begin{figure}[ht!] \begin{center} \includegraphics[width=7cm]{DerivativesVsEtaA.pdf} \hspace{1cm} \includegraphics[width=7cm]{DerivativesVsEtaB.pdf} %\caption{ % Left: Active rotation by $30^\circ$ % of an object in a fixed coordinate system. % Right: Passive rotation by $30^\circ$ % of the coordinate system around a fixed object. %} \label{fig:derivatives} \end{center} \end{figure} The size of the inverse Jacobian is shown below, which gives the factor for converting areas. Specifically, with a pileup transverse energy density at PU 200 of around 140\,GeV per unit area in $\Delta\eta\Delta\phi$, then the average transverse energy in GeV in a circle of $\Delta\rho_s=\Delta s_s=0.02$ is approximately equivalent to E_T = \frac{140}{J} 0.02^2 \pi At $\eta=1.4$, this is around 0.5\,GeV, while at $\eta=3.0$, this is around 18\,GeV. % \begin{figure}[ht!] \begin{center} \includegraphics[width=9cm]{JacobianVsEta.pdf} %\caption{ % Left: Active rotation by $30^\circ$ % of an object in a fixed coordinate system. % Right: Passive rotation by $30^\circ$ % of the coordinate system around a fixed object. %} \label{fig:jacobian} \end{center} \end{figure} \section{Shower position and direction fit} \section{Motion in a magnetic field} ... ...