\documentclass[10pt]{article} \usepackage{a4} \usepackage{amsmath} \oddsidemargin=0pt % No extra space wasted after first inch. \evensidemargin=0pt % Ditto (for two-sided output). \topmargin=0pt % Same for top of page. \headheight=0pt % Don't waste any space on unused headers. \headsep=0pt \textwidth=16cm % Effectively controls the right margin. \textheight=24cm \begin{document} \centerline{\Large \bf Notes on TPG} \bigskip \centerline{\large May 2016} \section{Optimisation of layer weights} Each event $e$ gives some values $d_{e,i}$ of the deposited energy in layer $i$; these can be in any units, e.g. MIPs. Assume these are to be multiplied by some constant coefficients $a_i$ (which are approximately the integrated dE/dx values if the $d_{e,i}$ are in MIPs) to give the estimate of the incoming EM photon or electron energy. Hence, the energy estimation for event $e$ is $$E_e = \sum_i a_i d_{e,i}$$ To find the optimal coefficients, then we need to know the truth energy per event $T_e$. For a given set of coefficients, the RMS$^2$ of the energy around the truth value is given by $$\mathrm{RMS}^2 = \frac{1}{N} \sum_e (E_e - T_e)^2 = \frac{1}{N} \sum_e \left(\sum_i a_i d_{e,i} - T_e\right)^2$$ 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 $$\frac{\partial \mathrm{RMS}^2}{\partial a_j} = \frac{1}{N} \sum_e 2d_{e,j} \left(\sum_i a_i d_{e,i} - T_e\right) = \frac{2}{N} \sum_i a_i \left(\sum_e d_{e,j} d_{e,i} \right) - \frac{2}{N} \sum_e d_{e,j} T_e$$ Hence, for the minimum, we require $$\sum_i \frac{\sum_e d_{e,j} d_{e,i}}{N} a_i = \frac{\sum_e d_{e,j} T_e}{N}$$ Writing in matrix notation with $M$ and $v$ defined as $$M_{ji} = \frac{\sum_e d_{e,j} d_{e,i}}{N},\qquad v_j = \frac{\sum_e d_{e,j} T_e}{N}$$ then the requirement is $$M a = v\qquad\mathrm{so}\qquad a = M^{-1}v$$ Inverting the large matrix $M$ is required to give the solution for the optimal $a_i$. Note, $M$ is similar (but not identical) to the error matrix of the $d_i$. The resulting RMS using the best fit values is given by \begin{eqnarray*} \mathrm{RMS}^2_\mathrm{min} &=& \frac{1}{N} \sum_e \left[ \left(\sum_i a_i d_{e,i} \right)^2 - 2 T_e \sum_i a_i d_{e,i} + T_e^2 \right] \\ &=& \frac{1}{N} \sum_j \sum_i a_j a_i \sum_e d_{e,j} d_{e,i} - \frac{2}{N} \sum_i a_i \sum_e T_e d_{e,i} + \frac{1}{N} \sum_e T_e^2 \\ &=& \sum_j \sum_i a_j a_i M_{ji} - 2 \sum_i a_i v_i + \frac{1}{N} \sum_e T_e^2 = a^T M a - 2 a^T v + \frac{1}{N} \sum_e T_e^2 \end{eqnarray*} But since the solution is defined by $Ma=v$, then $a^T M a = a^T v$. Hence $$\mathrm{RMS}^2_\mathrm{min} = \frac{1}{N} \left(\sum_e T_e^2\right) - a^T M a = \frac{1}{N} \left(\sum_e T_e^2\right) - v^T M^{-1} v = \frac{1}{N} \left(\sum_e T_e^2\right) - a^T v$$ The above can be extended slightly, which may improve the energy response linearity as well as the RMS. The energy estimation for the event (i.e. the first equation in this section) can be generally considered to be a polynomial in the $d_{e,i}$, but with the quadratic and higher terms neglected. However, it also neglects any constant term. A more general expression would then be to add another coefficient $b$ to give $$E_e = b + \sum_i a_i d_{e,i}$$ The easiest way to handle this is to allow the index $i$ to go one higher than previously, specifically change from $i=0,27$ to $i=0,28$ and then define $a_{28}=b$ and $d_{e,28}=1$. This means the expression simplifies to $$E_e = \sum_{i=0}^{28} a_i d_{e,i}$$ and so an identical calculation to previously holds, simply with the index running over a larger range. Explicitly, the matrix $M$ is now $29\times 29$ with the extra elements being $$M_{i,28} = M_{28,i} = \frac{1}{N} \sum_e d_{e,28}d_{e,i} = \frac{1}{N} \sum_e d_{e,i}$$ and $$M_{28,28} = \frac{1}{N} \sum_e d_{e,28}d_{e,28} = 1$$ while the extra element in $v$ is $$v_{28} = \frac{1}{N} \sum_e T_e d_{e,28} = \frac{1}{N} \sum_e T_e$$ \section{Units} Keeping quantities to 16-bit integers. The FE ASIC works in fC with an overall LSB of 0.1\,fC and upper range of 10\,pC $= 10^4$\,fC. This requires 17 bits total (although represented as a 10-bit and a 12-bit pair of values. Reconstructed energy (not deposited energy) with an LSB of 10\,MeV and 16-bit unsigned representation gives a maximum energy of 655\,GeV. These are initially MIPS $\times \int (dE/dx)\,dx$ for each layer until after forming the 3D clusters when the total energy is set more exactly. Position in $x$ and $y$ with an LSB of 100\,$\mu$m and a 16-bit signed representation gives a range of $\pm 328$\,cm (with $\pm 190$\,cm required). If needed, $z$ can be represented in a 16-bit unsigned representation with the same LSB, giving a range up to 655\,cm (with 408\,cm required). Note, the endcaps are handled separately so the negative $z$ endcap can be treated like the positive $z$ one. Sine and cosines can be represented in a 16-bit signed representation where they are multiplied by $2^{15}$. Hence, the result of a multiplication by this value needs to be stored in up to 31 bits and then bitshifted by 15. Note, this does not allow a representation of exactly $+1$, i.e. for angles of 0 or $\pi/2$. Neither of these should occur in the HGC. Similarly, if needed, $\tan(\theta)$ is in the appproximate range $\pm 0.1$ to $\pm 0.5$ and so can be represented in the same way (and hence is similar to $\sin\theta$ for small angles). Hence, the scaled variables $x/z = \tan\theta \cos\phi$ and $y/z = \tan\theta \sin\phi$ can also have the same representation. \section{FE ASIC TOT non-linearity} Modelled as a response $r$ for an input charge $q$ given by $$r = 0\quad\mathrm{for}\ q<100\,\mathrm{fC}, \qquad r = q - \frac{100(100-q_0)}{q-q_0} = q \left[1 - \frac{100(100-q_0)}{q(q-q_0)}\right] \quad \mathrm{otherwise}$$ where value of the parameter is chosen to be $q_0=90$\,fC. For $q(q-q_0) \gg 100(100-q_0)=1000$\,fC$^2$, the non-linear term becomes negligible. E.g. for $q=200$\,fC, then $q(q-q_0) = 22000$ and so is a 5\% correction, while for $q=400$\,fC, then $q(q-q_0) = 124000$ and so is a 0.8\% correction. Inverting the above response function for $q \ge 100$\,fC, then $$r(q-q_0) = q(q-q_0) - 100(100-q_0) \qquad\mathrm{so}\qquad q^2-q(q_0+r)+rq_0 -100(100-q_0)$$ Hence $$q = \frac{1}{2}\left[q_0+r \pm \sqrt{(q_0+r)^2-4rq_0+400(100-q_0)}\right] =\frac{q_0+r}{2} \pm \sqrt{\left(\frac{q_0-r}{2}\right)^2 + 100(100-q_0)}$$ where the positive sign is required for $q>100$\,fC. \section{Link data representation} The selected trigger cell data are calculated to a large number of bits, typically 16-18. On the links, they need to be represented in a small number of bits $n$, typically $\sim 8$. This could be linear or logarithmic or floating. \subsection{Linear representation} For linear, then in general it can be linear betwen $x_\mathrm{min}$ and $x_\mathrm{max}$ and 0 or $2^n-1$ outside this range. This can be represented within the range by $$y = \frac{(2^n-1)(x-x_\mathrm{min})}{x_\mathrm{max}-x_\mathrm{min}}$$ which can be inverted to give $$x = x_\mathrm{min} + \frac{y(x_\mathrm{max}-x_\mathrm{min})}{2^n-1}$$ \subsection{Logarithmic representation} For logarithmic, the general case would be $y=a\log(x)+b$ but with $c=b/a$ and $x_\mathrm{min}=e^{-c}$, then $$y = a\log(x)+b = a\log(x)+ac = a(\log(x)+c) = a(\log(x)-\log(x_\mathrm{min})) = a \log(x/x_\mathrm{min})$$ Therefore $$y = (2^n-1) \frac{\log(x/x_\mathrm{min})}{\log(x_\mathrm{max}/x_\mathrm{min})}$$ This can be inverted to give $$x = x_\mathrm{min}\left(\frac{x_\mathrm{max}}{x_\mathrm{min}}\right)^{y/(2^n-1)}$$ \subsection{Float representation} Here, the $2^n$ values are split into an exponent of $E$ bits and a mantissa of $M$ bits. The naive approach is simply to take the actual value as the mantissa shifted up by $E$ bits. For example, for $E=2$ and $M=2$, then the 16 possible values would give the table below. \bigskip \begin{center} \begin{tabular}{c|c|c|c} \hline Representation & Exponent & Mantissa & Value \cr\hline 0 & 0b00 & 0b00 & 0b00000 = \phantom{2}0\cr 1 & 0b00 & 0b01 & 0b00001 = \phantom{2}1\cr 2 & 0b00 & 0b10 & 0b00010 = \phantom{2}2\cr 3 & 0b00 & 0b11 & 0b00011 = \phantom{2}3\cr 4 & 0b01 & 0b00 & 0b00000 = \phantom{2}0\cr 5 & 0b01 & 0b01 & 0b00100 = \phantom{2}2\cr 6 & 0b01 & 0b10 & 0b01000 = \phantom{2}4\cr 7 & 0b01 & 0b11 & 0b01100 = \phantom{2}6\cr 8 & 0b10 & 0b00 & 0b00000 = \phantom{2}0\cr 9 & 0b10 & 0b01 & 0b00100 = \phantom{2}4\cr 10 & 0b10 & 0b10 & 0b01000 = \phantom{2}8\cr 11 & 0b10 & 0b11 & 0b01100 = 12 \cr 12 & 0b11 & 0b00 & 0b00000 = \phantom{2}0\cr 13 & 0b11 & 0b01 & 0b01000 = \phantom{2}8\cr 14 & 0b11 & 0b10 & 0b10000 = 16 \cr 15 & 0b11 & 0b11 & 0b11000 = 24 \cr \hline \end{tabular} \end{center} It is clear this is neither monotonic nor efficient, as the same values appear for several representations. A better representation is made by realising that for all but the lowest exponent representations, there is always a leading bit. Hence, this does not have to be stored explicitly. This means this leading bit must be added to the mantissa before the bit shift, and since this increments the length by one bit, then the shift up needed is only $E-1$. The table below shows this improved representation. It is monotonic, there are no duplicates, and the lowest two exponent ranges give an exact representation. \bigskip \begin{center} \begin{tabular}{c|c|c|c} \hline Representation & Exponent & Mantissa & Value \cr\hline 0 & 0b00 & 0b00 & 0b00000 = \phantom{2}0\cr 1 & 0b00 & 0b01 & 0b00001 = \phantom{2}1\cr 2 & 0b00 & 0b10 & 0b00010 = \phantom{2}2\cr 3 & 0b00 & 0b11 & 0b00011 = \phantom{2}3\cr 4 & 0b01 & 0b00 & 0b00100 = \phantom{2}4\cr 5 & 0b01 & 0b01 & 0b00101 = \phantom{2}5\cr 6 & 0b01 & 0b10 & 0b00110 = \phantom{2}6\cr 7 & 0b01 & 0b11 & 0b00111 = \phantom{2}7\cr 8 & 0b10 & 0b00 & 0b01000 = \phantom{2}8\cr 9 & 0b10 & 0b01 & 0b01010 = 10 \cr 10 & 0b10 & 0b10 & 0b01100 = 12 \cr 11 & 0b10 & 0b11 & 0b01110 = 14 \cr 12 & 0b11 & 0b00 & 0b10000 = 16 \cr 13 & 0b11 & 0b01 & 0b10100 = 20 \cr 14 & 0b11 & 0b10 & 0b11000 = 24 \cr 15 & 0b11 & 0b11 & 0b11100 = 28 \cr \hline \end{tabular} \end{center} In this improved representation, the mantissa has $M+1$ bits (except in the lowest exponent range). The exponent can represent numbers up to $2^E-1$ and hence will bit shift by a maximum of $2^E-2$ bits. Hence, the number of bits in the representation is $M+E$ bits, while the maximum number represented has $M+1+2^E-2 = M+2^E-1$ bits, i.e. is is less than $2^{M+2^E-1}$. The reduction is $2^E-E-1$ bits. 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. 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. Explicitly, for $E=0$, then the representation has $M$ bits while the value represented has $M$ bits also, i.e. the reduction is 0 bits. For $E=1$, the representation has $M+1$ bits, while the value represented has $M+1$ bits also, again with a reduction of 0 bits. For the extreme value of $M=0$, then the representation is just the number of bits in the input word. E.g. for $M=0$, $E=4$, then the table is given below. The value range is less thn $2^{15}=32768$. \bigskip \begin{center} \begin{tabular}{c|c|c|c} \hline Representation & Exponent & Mantissa & Value \cr\hline 0 & 0b0000 & 0 & 0b000000000000000 = \phantom{1222}0\cr 1 & 0b0001 & 0 & 0b000000000000001 = \phantom{1222}1\cr 2 & 0b0010 & 0 & 0b000000000000010 = \phantom{1222}2\cr 3 & 0b0011 & 0 & 0b000000000000100 = \phantom{1222}4\cr 4 & 0b0100 & 0 & 0b000000000001000 = \phantom{1222}8\cr 5 & 0b0101 & 0 & 0b000000000010000 = \phantom{122}16\cr 6 & 0b0110 & 0 & 0b000000000100000 = \phantom{122}32\cr 7 & 0b0111 & 0 & 0b000000001000000 = \phantom{122}64\cr 8 & 0b1000 & 0 & 0b000000010000000 = \phantom{12}128\cr 9 & 0b1001 & 0 & 0b000000100000000 = \phantom{12}256 \cr 10 & 0b1010 & 0 & 0b000001000000000 = \phantom{12}512 \cr 11 & 0b1011 & 0 & 0b000010000000000 = \phantom{1}1024 \cr 12 & 0b1100 & 0 & 0b000100000000000 = \phantom{1}2048 \cr 13 & 0b1101 & 0 & 0b001000000000000 = \phantom{1}4096 \cr 14 & 0b1110 & 0 & 0b010000000000000 = \phantom{1}8192 \cr 15 & 0b1111 & 0 & 0b100000000000000 = 16384 \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} \section{Template fit of energy in depth} Assume a template shape for a photon of $P_l$ per photon energy GeV for layer $l$. The minimum bias gives a template shape of $M_l$ for layer $l$ in some arbitrary units. The total expected per layer will then be $E_l = E_pP_l + E_m M_l$ for a photon energy $E_p$ and some scaling $E_m$ of the minimum bias template. Hence, the chi-squared compared to the observed energy $O_l$ will be $$\chi^2 = \sum_l \frac{(E_p P_l + E_m M_l - O_l)^2}{\sigma_l^2}$$ The $\sigma_l$ are given by the photon and minimum bias shower fluctuations around the average of the template. As such, the errors will depend on $E_p$ and $E_m$. However, expected'' values can be used initially to fix the $\sigma_l$ so that the problem remains linear. It could then be iterated several times with improved values to get a better fit. Minimising the chi-squared requires $$\frac{d\chi^2}{dE_p} = \sum_l \frac{2P_l(E_p P_l + E_m M_l - O_l)}{\sigma_l^2}=0,\qquad \frac{d\chi^2}{dE_m} = \sum_l \frac{2M_l(E_p P_l + E_m M_l - O_l)}{\sigma_l^2}=0$$ such that $$E_p \sum_l \frac{P_l^2}{\sigma_l^2} + E_m\sum_l \frac{P_lM_l}{\sigma_l^2} = \sum_l \frac{O_l P_l}{\sigma_l^2}, \qquad E_p \sum_l \frac{P_l M_l}{\sigma_l^2} + E_m\sum_l \frac{M_l^2}{\sigma_l^2} = \sum_l \frac{O_l M_l}{\sigma_l^2},$$ This can be written as a matrix equation $$\begin{pmatrix} \sum_l \frac{P_l^2}{\sigma_l^2} & \sum_l \frac{P_l M_l}{\sigma_l^2} \\ \sum_l \frac{P_l M_l}{\sigma_l^2} & \sum_l \frac{M_l^2}{\sigma_l^2} \end{pmatrix} \begin{pmatrix} E_p \\ E_m \end{pmatrix} = \begin{pmatrix} \sum_l \frac{O_l P_l}{\sigma_l^2} \\ \sum_l \frac{O_l M_l}{\sigma_l^2} \end{pmatrix}$$ As long as the $P_l$ and $M_l$ are not proportional to each other, the matrix on the left can be inverted to solve for $E_p$ (and $E_m$). This matrix is a constant for all events and so can be precalculated and inverted once, offline. The vector on the right must be calculated per event. However, explicitly the matrix determinant is $$\Delta = \left(\sum_l \frac{P_l^2}{\sigma_l^2}\right) \left(\sum_l \frac{M_l^2}{\sigma_l^2}\right) - \left(\sum_l \frac{P_l M_l}{\sigma_l^2} \right)^2$$ so the inverse is $$\frac{1}{\Delta} \begin{pmatrix} \sum_l \frac{M_l^2}{\sigma_l^2} & -\sum_l \frac{P_l M_l}{\sigma_l^2} \\ -\sum_l \frac{P_l M_l}{\sigma_l^2} & \sum_l \frac{P_l^2}{\sigma_l^2} \end{pmatrix}$$ and hence $$\begin{pmatrix} E_p \\ E_m \end{pmatrix} = \frac{1}{\Delta} \begin{pmatrix} \sum_{l^\prime} \frac{M_{l^\prime}^2}{\sigma_{l^\prime}^2} & -\sum_{l^\prime} \frac{P_{l^\prime} M_{l^\prime}}{\sigma_{l^\prime}^2} \\ -\sum_{l^\prime} \frac{P_{l^\prime} M_{l^\prime}}{\sigma_{l^\prime}^2} & \sum_{l^\prime} \frac{P_{l^\prime}^2}{\sigma_{l^\prime}^2} \end{pmatrix} \begin{pmatrix} \sum_l \frac{O_l P_l}{\sigma_l^2} \\ \sum_l \frac{O_l M_l}{\sigma_l^2} \end{pmatrix}$$ which means $$\begin{pmatrix} E_p \\ E_m \end{pmatrix} = \begin{pmatrix} \sum_l O_l \left[\frac{P_l}{\Delta \sigma_l^2} \left( \sum_{l^\prime} \frac{M_{l^\prime}^2}{\sigma_{l^\prime}^2}\right) - \frac{M_l}{\Delta \sigma_l^2} \left( \sum_{l^\prime} \frac{P_{l^\prime} M_{l^\prime}}{\sigma_{l^\prime}^2}\right) \right]\\ \sum_l O_l \left[\frac{M_l}{\Delta \sigma_l^2} \left(\sum_{l^\prime} \frac{P_{l^\prime}^2}{\sigma_{l^\prime}^2}\right) -\frac{P_l}{\Delta \sigma_l^2} \left(\sum_{l^\prime} \frac{P_{l^\prime} M_{l^\prime}}{\sigma_{l^\prime}^2}\right)\right] \end{pmatrix}$$ Hence $$E_p = \sum_l O_l A_l,\qquad E_m = \sum_l O_l B_l$$ where $A_l$ and $B_l$ correspond to the quantities in the square brackets and can be precalculated, except for any subtleties with the errors. ERROR MATRIX \section{Comparing coordinates in plane polars} On a given layer, then comparing e.g. a track extrapolation to a cluster position requires a difference of the two points in 2D; $x_1$, $y_1$ and $x_2$, $y_2$. This should be done in coordinates which preserve the cylindrical (i.e. plane polar for a layer) geometry. The obvious two are $$\Delta\rho = \rho_2-\rho_1 = \sqrt{x_2^2+y_2^2}-\sqrt{x_1^2+y_1^2}, \qquad \Delta\phi = \phi_2 - \phi_1 = \tan^{-1}(y_2/x_2) - \tan^{-1}(y_1/x_1)$$ However, $\Delta\phi$ has two issues; firstly is that it is not a length variable and so makes comparison with $\Delta\rho$ difficult, and secondly that there is a discontinuity in $\phi$ which needs to be handled. A length variable can be formed using some radius value $\overline{\rho}$ to give $\overline{\rho}\Delta\phi$ but there is an ambiguity about which radius to use; $\rho_1$, $\rho_2$ or some average of these. One desirable property is that the two variables should preserve the total length, i.e. $$\Delta\rho^2 + \overline{\rho}^2\Delta\phi^2 = \Delta x^2 + \Delta y^2 = (x_2-x_1)^2 + (y_2-y_1)^2 = x_2^2+y_2^2+x_1^2+y_1^2 - 2(x_1x_2+y_1y_2)$$ Using the expression for $\Delta\rho$ above, then $$\Delta\rho^2 = x_2^2+y_2^2+x_1^2+y_1^2 - 2\sqrt{x_2^2+y_2^2}\sqrt{x_1^2+y_1^2}$$ so that $$\overline{\rho}^2\Delta\phi^2 =2\sqrt{x_2^2+y_2^2}\sqrt{x_1^2+y_1^2}- 2(x_1x_2+y_1y_2)$$ Expressing the right hand side in plane polars gives $$\overline{\rho}^2\Delta\phi^2 =2\rho_1\rho_2 - 2\rho_1\rho_2 (\cos\phi_1\cos\phi_2 + \sin\phi_1\sin\phi_2) =2\rho_1\rho_2 [1-\cos(\phi_2-\phi_1)]=2\rho_1\rho_2 (1-\cos\Delta\phi)$$ This effectively defines $\overline{\rho}$ and hence the second variable directly. Note there is no issue with the $\phi$ discontinuity as this is handled automatically by the cosine. For small $\Delta\phi$, then the above expression is approximated by $$\overline{\rho}^2\Delta\phi^2 \approx 2\rho_1\rho_2 \frac{\Delta\phi^2}{2} \approx \rho_1\rho_2 \Delta\phi^2$$ so that $\overline{\rho} \approx \sqrt{\rho_1\rho_2}$, i.e. the geometric mean. Note that the sign of the second variable is not defined by the above; it should be the same as the sign of $\Delta\phi$. However, since $$1 - \cos\Delta\phi = 2\sin^2\left(\frac{\Delta\phi}{2}\right)$$ then $$\overline{\rho}^2\Delta\phi^2 =4\rho_1\rho_2 \sin^2\left(\frac{\Delta\phi}{2}\right)$$ and so $$\overline{\rho}\Delta\phi =2\sqrt{\rho_1\rho_2} \sin\left(\frac{\Delta\phi}{2}\right)$$ where the positive sign for the square-root is taken to agree with $\Delta\phi$. Again, for small $\Delta\phi$, then $\overline{\rho}$ clearly approximates to the geometric mean of the two radii, as before. \section{Shower position and direction fit} \section{Motion in a magnetic field} \section{Inverting matrices} A symmetric $3\times 3$ matrix can be written as $$M= \begin{pmatrix} M_{00} & M_{01} & M_{02} \\ M_{01} & M_{11} & M_{12} \\ M_{02} & M_{12} & M_{22} \end{pmatrix}$$ Its determinant is then \begin{eqnarray} \Delta &=& M_{00} \begin{vmatrix} M_{11} & M_{12} \\ M_{12} & M_{22} \end{vmatrix} -M_{01} \begin{vmatrix} M_{01} & M_{12} \\ M_{02} & M_{22} \end{vmatrix} +M_{02} \begin{vmatrix} M_{01} & M_{11} \\ M_{02} & M_{12} \end{vmatrix}\\ &=& M_{00}M_{11}M_{22}-M_{00}M_{12}M_{12} -M_{01}M_{01}M_{22}+M_{01}M_{12}M_{02} +M_{02}M_{01}M_{12}-M_{02}M_{11}M_{02}\\ &=& M_{00}M_{11}M_{22}+2M_{01}M_{12}M_{02} -M_{00}M_{12}^2 -M_{11}M_{02}^2 -M_{22}M_{01}^2 \end{eqnarray} This can be written is several ways \begin{eqnarray} \Delta &=& M_{00}(M_{11}M_{22}-M_{12}^2) +M_{01}(M_{12}M_{02}-M_{22}M_{01}) +M_{02}(M_{01}M_{12}-M_{11}M_{02})\\ &=& M_{01}(M_{12}M_{02}-M_{22}M_{01}) +M_{11}(M_{00}M_{22}-M_{02}^2) +M_{12}(M_{01}M_{02}-M_{00}M_{12})\\ &=&M_{02}(M_{01}M_{12}-M_{11}M_{02}) +M_{12}(M_{01}M_{02}-M_{00}M_{12}) +M_{22}(M_{00}M_{11}-M_{01}^2) \end{eqnarray} which means the inverse must be $$M^{-1}= \frac{1}{\Delta} \begin{pmatrix} M_{11}M_{22}-M_{12}^2 & M_{12}M_{02}-M_{22}M_{01} & M_{01}M_{12}-M_{11}M_{02} \\ M_{12}M_{02}-M_{22}M_{01} & M_{00}M_{22}-M_{02}^2 & M_{01}M_{02}-M_{00}M_{12} \\ M_{01}M_{12}-M_{11}M_{02} & M_{01}M_{02}-M_{00}M_{12} & M_{00}M_{11}-M_{01}^2 \end{pmatrix}$$ Note, if variable 2 becomes uncorrelated with variables 0 and 1, then $M_{02}=M_{12}=0$ so $$\Delta =M_{00}M_{11}M_{22}-M_{22}M_{01}^2= M_{22}(M_{00}M_{11}-M_{01}^2) = M_{22}\Delta_2$$ and $$M^{-1}= \frac{1}{\Delta} \begin{pmatrix} M_{11}M_{22} & -M_{22}M_{01} & 0 \\ -M_{22}M_{01} & M_{00}M_{22} & 0 \\ 0 & 0 & M_{00}M_{11}-M_{01}^2 \end{pmatrix} =\frac{1}{\Delta_2} \begin{pmatrix} M_{11} & -M_{01} & 0 \\ -M_{01} & M_{00} & 0 \\ 0 & 0 & \Delta_2/M_{22} \end{pmatrix} = \begin{pmatrix} M_2^{-1} & 0 \\ 0 & 1/M_{22} \end{pmatrix}$$ as expected. The inverse of the $2\times 2$ matrix is not the submatrix in the inverse of the $3\times 3$ matrix. E.g. for the first element in the inverse of the $2\times 2$ matrix, this is $$M_{00}^{-1} = \frac{M_{11}}{M_{00}M_{11}-M_{01}^2}$$ while in the $3 \times 3$ case, this is \begin{eqnarray} M_{00}^{-1} &=& \frac{M_{11}M_{22}-M_{12}^2}{M_{00}M_{11}M_{22}+2M_{01}M_{12}M_{02} -M_{00}M_{12}^2 -M_{11}M_{02}^2 -M_{22}M_{01}^2}\\ &=& \frac{M_{11}-(M_{12}^2/M_{22})}{M_{00}M_{11}-M_{01}^2 +(2M_{01}M_{12}M_{02} -M_{00}M_{12}^2 -M_{11}M_{02}^2)/M_{22}} \end{eqnarray} \end{document}