\chapter{STR solutions}
\begin{abox}
	Practice set 1 solutions
	\end{abox}
\begin{enumerate}
	\item Consider the decay process $\tau^{-} \rightarrow \pi^{-}+v_{\tau}$ in the rest frame of the $\tau^{-} .$The masses of the $\tau^{-}, \pi^{-}$and $v_{\tau}$ are $M_{\tau}, M_{\pi}$ and zero respectively.\\
	\textbf{A} The energy of $\pi^{-}$is
	{\exyear{NET JUNE 2011}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\frac{\left(M_{\tau}^{2}-M_{\pi}^{2}\right) c^{2}}{2 M_{\tau}}$
	\task[\textbf{B.}]$\frac{\left(M_{\tau}^{2}+M_{\pi}^{2}\right) c^{2}}{2 M_{\tau}}$
	\task[\textbf{C.}]$\left(M_{\tau}-M_{\pi}\right) c^{2}$
	\task[\textbf{D.}]$\sqrt{M_{\tau} M_{\pi}} c^{2}$
\end{tasks}
\begin{answer}
	\begin{align*}
	\intertext { From conservation of energy }
	 M_{\tau} c^{2}&=E_{\pi}+E_{v} \text {. }\\
	E_{\pi}^{2}&=p^{2} c^{2}+M_{\pi}^{2} c^{4} \text { and } E_{v}^{2}=p^{2} c^{2} \text { since momentum of } \pi^{-} \text {and } v_{\tau} \text { is same. }\\
	M_{\tau} c^{2}&=E_{\pi}+E_{v}, M_{\pi}^{2} c^{4}=E_{\pi}^{2}-E_{v}^{2} \Rightarrow E_{\pi}-E_{v}=\frac{M_{\pi}^{2} c^{4}}{M_{\tau} c^{2}}\\
	E_{\pi}-E_{v}&=\frac{M_{\pi}^{2} c^{2}}{M_{\tau}} \text { and } E_{\pi}+E_{v}=M_{\tau} c^{2} \Rightarrow E_{\pi}=\frac{\left(M_{\tau}^{2}+M_{\pi}^{2}\right) c^{2}}{2 M_{\tau}}
	\end{align*}
	The correct option is \textbf{(b)}
\end{answer}
\textbf{B} The velocity of  $\pi^{-} \text {is }$
\begin{tasks}(1)
	\task[\textbf{A.}] $\frac{\left(M_{\tau}^{2}-M_{\pi}^{2}\right) c}{M_{\tau}^{2}+M_{\pi}^{2}}$
	\task[\textbf{B.}]$\frac{\left(M_{\tau}^{2}+M_{\pi}^{2}\right) c}{M_{\tau}^{2}-M_{\pi}^{2}}$
	\task[\textbf{C.}] $\frac{M_{\pi} c}{M_{\tau}}$
	\task[\textbf{D.}]$\frac{M_{\tau} c}{M_{\pi}}$
\end{tasks}
\begin{answer}
\begin{align*}
\text { Velocity of } \pi^{-} \quad E_{\pi}&=\frac{\left(M_{\tau}^{2}+M_{\pi}^{2}\right) c^{2}}{2 M_{\tau}}=\frac{M_{\pi} c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \Rightarrow\left(1-\frac{v^{2}}{c^{2}}\right)=\frac{4 M_{\pi}^{2} M_{\tau}^{2}}{\left(M_{\tau}^{2}+M_{\pi}^{2}\right)^{2}}\\
\Rightarrow \frac{v^{2}}{c^{2}}&=1-\frac{4 M_{\pi}^{2} M_{\tau}^{2}}{\left(M_{\tau}^{2}+M_{\pi}^{2}\right)^{2}}\\
\frac{v^{2}}{c^{2}}&=\frac{M_{\tau}^{4}+M_{\pi}^{4}+2 M_{\tau}^{2} M_{\pi}^{2}-4 M_{\pi}^{2} M_{\tau}^{2}}{\left(M_{\tau}^{2}+M_{\pi}^{2}\right)^{2}}\\
v&=\left(\frac{M_{\tau}^{2}-M_{\pi}^{2}}{M_{\tau}^{2}+M_{\pi}^{2}}\right) c
\end{align*}
The correct option is \textbf{(a)}
\end{answer}

	\item A constant force $F$ is applied to a relativistic particle of rest mass $m$. If the particle starts from rest at $t=0$, its speed after a time $t$ is
	{\exyear{NET DEC 2011}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\mathrm{Ft} / \mathrm{m}$
	\task[\textbf{B.}]$c \tanh \left(\frac{F t}{m c}\right)$
	\task[\textbf{C.}]$c\left(1-e^{-F t / m c}\right)$
	\task[\textbf{D.}]$\frac{F c t}{\sqrt{F^{2} t^{2}+m^{2} c^{2}}}$
\end{tasks}
\begin{answer}
	\begin{align*}
	\frac{d p}{d t}&=F \Rightarrow p=F t+c . \text { At } t=0, p=0 \text { so, } c=0\\
	p&=F t \Rightarrow \frac{m u}{\sqrt{1-\frac{u^{2}}{c^{2}}}}\\
	u&=\frac{\left(\frac{F}{m}\right) t}{\sqrt{1+\left(\frac{F t}{m c}\right)^{2}}}=\frac{F c t}{\sqrt{F^{2} t^{2}+m^{2} c^{2}}}\\
	\end{align*}
	The correct option is \textbf{(d)}
\end{answer}

	\item Two events separated by a (spatial) distance $9 \times 10^{9} \mathrm{~m}$, are simultaneous in one inertial frame. The time interval between these two events in a frame moving with a constant speed $0.8 c$ (where the speed of light $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ) is
	{\exyear{NET JUNE 2012}}
\begin{tasks}(2)
	\task[\textbf{A.}] $60 \mathrm{~s}$
	\task[\textbf{B.}]$40 \mathrm{~s}$
	\task[\textbf{C.}]$20 s$
	\task[\textbf{D.}] $0 s$
\end{tasks}
\begin{answer}
	\begin{align*}
	x_{2}^{\prime}-x_{1}^{\prime}&=9 \times 10^{9} m \text { and } t_{2}^{\prime}-t_{1}^{\prime}=0 . \text { Then }\\
	t_{2}-t_{1}&=\left(\frac{t_{2}^{\prime}+\frac{v}{c^{2}} x_{2}^{\prime}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)-\left(\frac{t_{1}^{1}+\frac{v}{c^{2}} x_{1}^{\prime}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)\\
	t_{2}-t_{1}&=\frac{t_{2}^{\prime}-t_{1}^{\prime}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}+\frac{v}{c^{2}} \frac{\left(x_{2}^{\prime}-x_{1}^{\prime}\right)}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{v}{c^{2}} \frac{\left(x_{2}^{\prime}-x_{1}^{\prime}\right)}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\
	\text { Put } v&=0.8 c \quad \Rightarrow t_{2}-t_{1} \cong 40 \mathrm{sec}
	\end{align*}
	The correct option is \textbf{(b)}
\end{answer}

	\item What is proper time interval between the occurrence of two events if in one inertial frame events are separated by $7.5 \times 10^{8} \mathrm{~m}$ and occur $6.5 \mathrm{~s}$ apart?
	{\exyear{NET JUNE 2012}}
\begin{tasks}(2)
	\task[\textbf{A.}] $6.50 \mathrm{~s}$
	\task[\textbf{B.}]$6.00 \mathrm{~s}$
	\task[\textbf{C.}]$5.75 \mathrm{~s}$
	\task[\textbf{D.}]$5.00 \mathrm{~s}$
\end{tasks}
\begin{answer}
	\begin{align*}
	\text{Proper time interval}
	\Delta t=\sqrt{\left(\Delta t^{\prime}\right)^{2}-\frac{r^{2}}{c^{2}}}=\sqrt{(6.5)^{2}-\left(\frac{7.5}{3}\right)^{2}}=6 \text { sec. }
	\end{align*}
	The correct option is \textbf{(b)}
\end{answer}
	\item The muon has mass $105 \mathrm{MeV} / \mathrm{c}^{2}$ and mean life time $2.2 \mu \mathrm{s}$ in its rest frame. The mean distance traversed by a muon of energy $315 \mathrm{MeV}$ before decaying is approximately,
	{\exyear{NET DEC 2012}}

\begin{tasks}(2)
	\task[\textbf{A.}] $3 \times 10^{5} \mathrm{~km}$
	\task[\textbf{B.}]$2.2 \mathrm{~cm}$
	\task[\textbf{C.}]$6.6 \mu \mathrm{m}$
	\task[\textbf{D.}]$1.98 \mathrm{~km}$
\end{tasks}
\begin{answer}
	\begin{align*}
	\text { Since } E&=315 \mathrm{MeV} \text { and } m_{0}=105 \frac{\mathrm{MeV}}{c^{2}} \text {. }\\
	E=m c^{2} \Rightarrow E&=\frac{m_{0} c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \Rightarrow 315	\\
	315&=\frac{105}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \Rightarrow v=0.94 c\\
	\text{Now}, t&=\frac{t_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\
	t_{0}&=2.2 \mu s \Rightarrow t=\frac{2.2 \times 10^{-6}}{\sqrt{1-\frac{8}{9}}} \Rightarrow t=6.6 \mu s\\
	\text { Now the distance traversed by muon is } v t&=0.94 c \times 6.6 \times 10^{-6}=1.86 \mathrm{~km} \text {. }
	\end{align*}
	The correct option is \textbf{(d)}
\end{answer}
	\item The area of a disc in its rest frame $S$ is equal to 1 (in some units). The disc will appear distorted to an observer $O$ moving with a speed $u$ with respect to $S$ along the plane of the disc. The area of the disc measured in the rest frame of the observer $O$ is $(c$ is the speed of light in vacuum)
	{\exyear{NET JUNE 2013}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\left(1-\frac{u^{2}}{c^{2}}\right)^{1 / 2}$
	\task[\textbf{B.}]$\left(1-\frac{u^{2}}{c^{2}}\right)^{-1 / 2}$
	\task[\textbf{C.}]$\left(1-\frac{u^{2}}{c^{2}}\right)$
	\task[\textbf{D.}]$\left(1-\frac{u^{2}}{c^{2}}\right)^{-1}$
\end{tasks}
\begin{answer}
\begin{align*}
&\text { Area of disc from } \mathrm{S} \text { frame is } 1 \text { i.e. } \pi a^{2}=1 \text { or } \pi a \cdot a=1\\
&\text { Area of disc from } S^{\prime} \text { frame is } \pi a \cdot b=\pi a \cdot a \sqrt{1-\frac{u^{2}}{c^{2}}}=1 \cdot \sqrt{1-\frac{u^{2}}{c^{2}}}=\sqrt{1-\frac{u^{2}}{c^{2}}}\\
&\text { where } b=a \sqrt{1-\frac{u^{2}}{c^{2}}}
\end{align*}
The correct option is \textbf{(a)}
\end{answer}

	\item The recently-discovered Higgs boson at the LHC experiment has a decay mode into a photon and a $Z$ boson. If the rest masses of the Higgs and $Z$ boson are $125 \mathrm{GeV} / \mathrm{c}^{2}$ and $90 \mathrm{GeV} / \mathrm{c}^{2}$ respectively, and the decaying Higgs particle is at rest, the energy of the photon will approximately be
	{\exyear{NET JUNE 2014}}

\begin{tasks}(2)
	\task[\textbf{A.}] $35 \sqrt{3} \mathrm{GeV}$
	\task[\textbf{B.}]$35 \mathrm{GeV}$
	\task[\textbf{C.}]$30 \mathrm{GeV}$
	\task[\textbf{D.}]$15 \mathrm{GeV}$
\end{tasks}
\begin{answer}
	\begin{align*}
	H_{B} &\rightarrow P_{H}+Z_{B}\\
	\intertext { From conservation of momentum }
	 0&=\vec{P}_{1}+\vec{P}_{2} \Rightarrow \vec{P}_{1}=-\vec{P}_{2} \Rightarrow\left|P_{1}\right|=\left|P_{2}\right|\\
	\text { Now } E_{H_{B}}&=E_{P_{H}}+E_{Z_{B}} \Rightarrow E_{P_{H}}+E_{Z_{B}}=M_{H_{B}} c^{2}\\
	E_{P_{H}}^{2}&=P_{1}^{2} c^{2}+0 \text { and } E_{Z_{B}}^{2}=P_{2}^{2} c^{2}+M_{Z_{B}}^{2} c^{4}\\
	&\Rightarrow\left(E_{Z_{B}}-E_{P_{H}}\right)\left(E_{Z_{B}}+E_{P_{H}}\right)=M_{Z_{B}}^{2} c^{4} \quad \because\left|P_{1}\right|=\left|P_{2}\right|\\
	&\Rightarrow E_{Z_{B}}-E_{P_{H}}=\frac{M_{Z_{B}}^{2} c^{4}}{M_{H_{B}} c^{2}}=\frac{M_{Z_{B}}^{2} c^{2}}{M_{H_{B}}} \quad \because E_{Z_{B}}+E_{P_{H}}=M_{H_{B}} c^{2}\\
	&\Rightarrow 2 E_{P_{H}}=M_{H_{B}} c^{2}-\frac{M_{z_{B}}^{2} c^{2}}{M_{H_{B}}} \Rightarrow E_{P_{H}}=\frac{\left(M_{H_{B}}^{2}-M_{z_{B}}^{2}\right) c^{2}}{M_{H_{B}}}\\
	&\Rightarrow E_{P_{H}}=\left(\frac{125 \times 125-90 \times 90}{2 \times 125}\right) \times \frac{c^{4}}{c^{4}}=30.1 \mathrm{GeV}
	\end{align*}
	The correct option is \textbf{(c)}
\end{answer}
	\item Consider three inertial frames of reference $A, B$ and $C$. the frame $B$ moves with a velocity $\frac{c}{2}$ with respect to $A$, and $C$ moves with a velocity $\frac{c}{10}$ with respect to $B$ in the same direction. The velocity of $C$ as measured in $A$ is
	{\exyear{NET JUNE 2015}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\frac{3 c}{7}$
	\task[\textbf{B.}]$\frac{4 c}{7}$
	\task[\textbf{C.}]$\frac{c}{7}$
	\task[\textbf{D.}] $\frac{\sqrt{3} c}{7}$
\end{tasks}
\begin{answer}$\left. \right. $	\\
\begin{minipage}{0.5\textwidth}
		\begin{align*}
	v&=\frac{c}{2}, \quad u_{x}^{\prime}=\frac{c}{10}\\
	u_{x}&=\frac{u_{x}^{\prime}+v}{1+\frac{u^{\prime} v_{x}}{c^{2}}}=\frac{4 c}{7}
	\end{align*}
\end{minipage}
	\begin{minipage}{0.5\textwidth}
	\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=5cm]{NET 1}
	\end{figure}
	\end{minipage}
The correct option is \textbf{(b)}
\end{answer}

	\item Consider a particle of mass $m$ moving with a speed $v$. If $T_{R}$ denotes the relativistic kinetic energy and $T_{N}$ its non-relativistic approximation, then the value of $\frac{\left(T_{R}-T_{N}\right)}{T_{R}}$ for $v=0.01 c$, is
	{\exyear{NET DEC 2015}}
\begin{tasks}(2)
	\task[\textbf{A.}] $1.25 \times 10^{-5}$
	\task[\textbf{B.}]$5.0 \times 10^{-5}$
	\task[\textbf{C.}]$7.5 \times 10^{-5}$
	\task[\textbf{D.}]$1.0 \times 10^{-4}$
\end{tasks}
\begin{answer}
	\begin{align*}
	T_{N}&=\frac{1}{2} m_{0} v^{2}, T_{R}=m c^{2}-m_{0} c^{2}=\frac{m_{0} c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}-m_{0} c^{2} \quad(\because v=0.01 c)\\
	\text { Now, } \frac{\left(T_{R}-T_{N}\right)}{T_{R}}&=1-\frac{T_{N}}{T_{R}}=1-\frac{\frac{1}{2} m_{0} v^{2}}{\frac{m_{0} c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}-m_{0} c^{2}}\\
	&=1-\frac{\frac{\frac{v^{2}}{2}}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}-c^{2}}=1-\frac{\frac{(0.01)^{2}}{2}}{\frac{1}{\sqrt{1-(0.01)^{2}}}}-1\\
	\frac{T_{R}-T_{N}}{T_{R}}&=0.75
	\end{align*}
	None of the option is correct.
\end{answer}
	\item Let $(x, t)$ and $\left(x^{\prime}, t^{\prime}\right)$ be the coordinate systems used by the observers $O$ and $O^{\prime}$, respectively. Observer $O^{\prime}$ moves with a velocity $v=\beta c$ along their common positive $x$ axis. If $x_{+}=x+c t$ and $x_{-}=x-c t$ are the linear combinations of the coordinates, the Lorentz transformation relating $O$ and $O^{\prime}$ takes the form
	{\exyear{NET JUNE 2016}}
\begin{tasks}(2)
	\task[\textbf{A.}] $x_{+}^{\prime}=\frac{x_{-}-\beta x_{+}}{\sqrt{1-\beta^{2}}}$ and $x_{-}^{\prime}=\frac{x_{+}-\beta x_{-}}{\sqrt{1-\beta^{2}}}$
	\task[\textbf{B.}]$x_{+}^{\prime}=\sqrt{\frac{1+\beta}{1-\beta}} x_{+}$and $x_{-}^{\prime}=\sqrt{\frac{1-\beta}{1+\beta}} x_{-}$
	\task[\textbf{C.}]$x_{+}^{\prime}=\frac{x_{+}-\beta x_{-}}{\sqrt{1-\beta^{2}}}$ and $x_{-}^{\prime}=\frac{x_{-}-\beta x_{+}}{\sqrt{1-\beta^{2}}}$
	\task[\textbf{D.}]$x_{+}^{\prime}=\sqrt{\frac{1-\beta}{1+\beta}} x_{+}$and $x_{-}^{\prime}=\sqrt{\frac{1+\beta}{1-\beta}} x_{-}$
\end{tasks}
\begin{answer}
\begin{align*}
x_{+}^{\prime}&=x^{\prime}+c t^{\prime}\\
&=\frac{x-v t}{\sqrt{1-\frac{v^{2}}{c^{2}}}}+\frac{c\left(t-\frac{v x}{c^{2}}\right)}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{x\left(1-\frac{v}{c}\right)}{\sqrt{1-\frac{v^{2}}{c^{2}}}}+\frac{c t\left(1-\frac{v}{c}\right)}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\
&=x \sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}+c t \sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}=\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}(x+c t)\\
x_{+}^{\prime}&=\sqrt{\frac{1-\beta}{1+\beta}} x_{+}\\
x_{-}^{\prime}&=x^{\prime}-c t^{\prime}=\frac{x-v t}{\sqrt{1-\frac{v^{2}}{c^{2}}}}-\frac{c\left(t-\frac{v x}{c^{2}}\right)}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{x\left(1+\frac{v}{c}\right)}{\sqrt{1-\frac{v^{2}}{c^{2}}}}-\frac{c t\left(1+\frac{v}{c}\right)}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\
x_{-}^{\prime}&=x \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}-c t \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}} \Rightarrow x_{-}^{\prime}=\sqrt{\frac{1+\beta}{1-\beta}}(x-c t) \Rightarrow x_{-}^{\prime}=\sqrt{\frac{1+\beta}{1-\beta} x_{-}}
\end{align*}
The correct option is \textbf{(d)}
\end{answer}

	\item For a particle of energy $E$ and momentum $p$ (in a frame $F$ ), the rapidity $y$ is defined as $y=\frac{1}{2} \ln \left(\frac{E+p_{3} c}{E-p_{3} c}\right) .$ In a frame $F^{\prime}$ moving with velocity $v=(0,0, \beta c)$ with respect to $F$, the rapidity $y^{\prime}$ will be
	{\exyear{NET JUNE 2016}}
\begin{tasks}(2)
	\task[\textbf{A.}] $y^{\prime}=y+\frac{1}{2} \ln \left(1-\beta^{2}\right)$
	\task[\textbf{B.}]$y^{\prime}=y-\frac{1}{2} \ln \left(\frac{1+\beta}{1-\beta}\right)$
	\task[\textbf{C.}]$y^{\prime}=y+\ln \left(\frac{1+\beta}{1-\beta}\right)$
	\task[\textbf{D.}]$y^{\prime}=y+2 \ln \left(\frac{1+\beta}{1-\beta}\right)$
\end{tasks}
\begin{answer}
\begin{align*}
y&=\frac{1}{2} \ln \left(\frac{E+p_{3} c}{E-p_{3} c}\right)\\
\text { Then } y^{\prime}&=\frac{1}{2} \ln \left(\frac{E^{\prime}+p_{3}^{\prime} c}{E^{\prime}-p_{3}^{\prime} c}\right)\\
\text { Where } p_{3}^{\prime}&=\gamma\left(p_{3}-v\left(\frac{E}{c^{2}}\right)\right) \quad E^{\prime}=\gamma\left(E-v p_{3}\right)\\
\text { Put the value of } p_{3}^{\prime} \text { and } E^{\prime} \text { one will get } y^{\prime}&=\frac{1}{2} \ln \left(\frac{\left(E+p_{3} c\right)-\frac{v}{c}\left(E+p_{3} c\right)}{\left(E-p_{3} c\right)+\frac{v}{c}\left(E-p_{3} c\right)}\right)\\
\frac{1}{2} \ln \left(\frac{\left(E+p_{3} c\right)(1-\beta)}{\left(E-p_{3} c\right)(1+\beta)}\right) &\Rightarrow \frac{1}{2} \ln \left(\frac{\left(E+p_{3} c\right)}{\left(E-p_{3} c\right)}\right)+\frac{1}{2} \ln \left(\frac{1-\beta}{1+\beta}\right)\\
y+\frac{1}{2} \ln \left(\frac{1-\beta}{1+\beta}\right)& \Rightarrow y-\frac{1}{2} \ln \left(\frac{1+\beta}{1-\beta}\right)
\end{align*}
The correct option is \textbf{(b)}
\end{answer}
	\item A relativistic particle moves with a constant velocity $v$ with respect to the laboratory frame. In time $\tau$, measured in the rest frame of the particle, the distance that it travels in the laboratory frame is
	{\exyear{NET DEC 2016}}
\begin{tasks}(2)
	\task[\textbf{A.}] $v \tau$
	\task[\textbf{B.}]$\frac{c \tau}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$
	\task[\textbf{C.}]$v \tau \sqrt{1-\frac{v^{2}}{c^{2}}}$
	\task[\textbf{D.}]$\frac{v \tau}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$
\end{tasks}
\begin{answer}$\left. \right. $\\
	\begin{minipage}{0.5\textwidth}
	\begin{align*}
	\text { From Particle } x_{1}^{\prime}&=0 x_{2}^{\prime}=0\\
	t_{\text {initial }}&=t_{1}^{\prime} \quad t_{\text {final }}=t_{2}^{\prime}\\
	x_{1}&=\frac{x_{1}^{\prime}+v t_{1}^{\prime}}{\sqrt{1-v^{2} / c^{2}}}, x_{2}=\frac{x_{2}^{\prime}+v t_{2}^{\prime}}{\sqrt{1-v^{2} / c^{2}}}\\
	x_{2}-x_{1}&=\frac{x_{2}^{\prime}-x_{1}^{\prime}}{\sqrt{1-v^{2} / c^{2}}}+\frac{v\left(t_{2}^{\prime}-t_{1}^{\prime}\right)}{\sqrt{1-v^{2} / c^{2}}}\\
	\Delta x&=\frac{v\left(t_{2}^{\prime}-t_{1}^{\prime}\right)}{\sqrt{1-v^{2} / c^{2}}}=\frac{v \tau}{\sqrt{1-v^{2} / c^{2}}}
	\end{align*}
	\end{minipage}
\begin{minipage}{0.5\textwidth}
	\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=5cm]{problem 2}
	\end{figure}
\end{minipage}
The correct option is \textbf{(d)}
\end{answer}
	\item Consider a radioactive nucleus that is travelling at a speed $\frac{c}{2}$ with respect to the lab frame. It emits $\gamma$-rays of frequency $v_{0}$ in its rest frame. There is a stationary detector, (which is not on the path of the nucleus) in the lab. If a $\gamma$-ray photon is emitted when the nucleus is closest to the detector, its observed frequency at the detector is
	{\exyear{NET DEC 2016}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\frac{\sqrt{3}}{2} v_{0}$
	\task[\textbf{B.}]$\frac{1}{\sqrt{3}} v_{0}$
	\task[\textbf{C.}]$\frac{1}{\sqrt{2}} v_{0}$
	\task[\textbf{D.}]$\sqrt{\frac{2}{3}} v_{0}$
\end{tasks}
\begin{answer}
	\begin{align*}
	v&=v_{0} \sqrt{1-\frac{v^{2}}{c^{2}}} \quad \text { (If detector is not in the path at nucleus) }\\
	v&=v_{0} \sqrt{1-\frac{1}{4}}=v_{0} \frac{\sqrt{3}}{2}
	\end{align*}
	The correct option is \textbf{(a)}
\end{answer}
	\item An inertial observer sees two events $E_{1}$ and $E_{2}$ happening at the same location but $6 \mu s$ apart in time. Another observer moving with a constant velocity $v$ (with respect to the first one) sees the same events to be $9 \mu s$ apart. The spatial distance between the events, as measured by the second observer, is approximately
	{\exyear{NET JUNE 2017}}

\begin{tasks}(2)
	\task[\textbf{A.}] $300 m$
	\task[\textbf{B.}]$1000 m$
	\task[\textbf{C.}]$2000 m$
	\task[\textbf{D.}]$2700 m$
\end{tasks}
\begin{answer}
\begin{align*}
	x_{2}^{1}-x_{1}^{1}&=0, t_{2}^{1}-t_{1}^{1}=6 \times 10^{-6}, t_{2}-t_{1}=9 \times 10^{-6}, x_{2}-x_{1}=?\\
t_{2}-t_{2}&=9 \times 10^{-6}\\
\left(\frac{t_{2}^{1}+\frac{v}{c^{2}} x_{2}^{1}}{\sqrt{1-v^{2} / c^{2}}}\right)-\frac{\left(t_{1}^{\prime}+\frac{v}{c^{2}} x_{1}^{\prime}\right)}{\sqrt{1-v^{2} / c^{2}}}&=9 \times 10^{-6}\\
\frac{t_{2}^{\prime}-t_{1}^{\prime}}{\sqrt{1-v^{2} / c^{2}}}&=9 \times 10^{-6} \Rightarrow \frac{6 \times 10^{-6}}{\sqrt{1-v^{2} / c^{2}}}=9 \times 10^{-6}\\
v&=\sqrt{\frac{5}{9}} c \Rightarrow \sqrt{1-\frac{v^{2}}{c^{2}}}=2 / 3\\
\left(x_{2}-x_{1}\right)&=\left(\frac{x_{2}^{\prime}+v t_{2}^{\prime}}{\sqrt{1-v^{2} / c^{2}}}\right)-\left(\frac{x_{1}^{\prime}+v t_{1}^{\prime}}{\sqrt{1-v^{2} / c^{2}}}\right)\\
\frac{v}{\sqrt{1-v^{2} / c^{2}}}\left(t_{2}^{\prime}-t_{1}^{\prime}\right)\\
\left(x_{2}-x_{1}\right)&=\frac{\sqrt{5}}{3} c \times \frac{9}{6} \times\left(6 \times 10^{-6}\right)=\frac{\sqrt{5}}{3} \times 3 \times 10^{8} \times \frac{9}{6} \times 6 \times 10^{-6}\\
&=9 \times \sqrt{5} \times 10^{2}=20.12 \times 10^{2} \simeq 2000 m
\end{align*}
The correct option is \textbf{(c)}
\end{answer}
	\item A light signal travels from a point $A$ to a point $B$, both within a glass slab that is moving with uniform velocity (in the same direction as the light) with speed $0.3 c$ with respect to an external observer. If the refractive index of the slab is $1.5$, then the observer will measure the speed of the signal as
	{\exyear{NET DEC 2017}}
\begin{tasks}(2)
	\task[\textbf{A.}] $0.67 c$
	\task[\textbf{B.}]$0.81 c$
	\task[\textbf{C.}]$0.97 c$
	\task[\textbf{D.}] $c$
\end{tasks}
\begin{answer}$\left. \right. $\\
	\begin{minipage}{0.5\textwidth}
\begin{align*}
v&=0.3 c\\
u_{x}^{\prime}&=\frac{c}{n} n=1.5\\
u_{x}&=\frac{u_{x}^{\prime}+v}{1+\frac{u_{x}^{\prime} v}{c^{2}}}=\frac{0.3 c+\frac{c}{n}}{1+\frac{c}{n} \cdot \frac{0.3 c}{c^{2}}}\\
u_{x}&=0.81 c
\end{align*}
	\end{minipage}
\begin{minipage}{0.5\textwidth}
\begin{figure}[H]
	\centering
	\includegraphics[height=3cm,width=5cm]{problem 4}
\end{figure}
\end{minipage}
The correct option is \textbf{(b)}
\end{answer}
	\item Two particles $A$ and $B$ move with relativistic velocities of equal magnitude $v$, but in opposite directions, along the $x$-axis of an inertial frame of reference. The magnitude of the velocity of $A$, as seen from the rest frame of $B$, is
	{\exyear{NET JUNE 2018}}

\begin{tasks}(2)
	\task[\textbf{A.}] $\frac{2 v}{\left(1-\frac{v^{2}}{c^{2}}\right)}$
	\task[\textbf{B.}]$\frac{2 v}{\left(1+\frac{v^{2}}{c^{2}}\right)}$
	\task[\textbf{C.}] $2 v \sqrt{\frac{c-v}{c+v}}$
	\task[\textbf{D.}]$\frac{2 v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$
\end{tasks}
\begin{answer}
	\begin{align*}
	u_{x}^{\prime}&=v \quad V=v\\
	u_{x}&=\frac{u_{x}^{\prime}+V}{1+\frac{u_{x} V}{c^{2}}}\\
	u_{x}&=\frac{v+v}{1+\frac{v^{2}}{c^{2}}}=\frac{2 v}{1+\frac{v^{2}}{c^{2}}}
	\end{align*}
	The correct option is \textbf{(b)}
\end{answer}

	\item The energy of a free relativistic particle is $E=\sqrt{|\vec{p}|^{2} c^{2}+m^{2} c^{4}}$, where $m$ is its rest mass, $\vec{p}$ is its momentum and $c$ is the speed of light in vacuum. The ratio $v_{g} / v_{p}$ of the group velocity $v_{g}$ of a quantum mechanical wave packet (describing this particle) to the phase velocity $v_{p}$ is
	{\exyear{NET JUNE 2018}}
\begin{tasks}(2)
	\task[\textbf{A.}] $|\vec{p}| c / E$
	\task[\textbf{B.}]$|\vec{p}| m c^{3} / E^{2}$
	\task[\textbf{C.}] $|\vec{p}|^{2} c^{3} / E^{2}$
	\task[\textbf{D.}]$|\vec{p}| c / 2 E$
\end{tasks}
	\begin{answer}
		\begin{align*}
		E^{2}&=p^{2} c^{2}+m^{2} c^{4} \text { and } v_{g}=\frac{d E}{d p}, v_{p}=\frac{E}{p}\\
		2 E \frac{d E}{d p}&=2 p c^{2} \Rightarrow \frac{E}{p} \frac{d E}{d p}=c^{2}\\
		\frac{v_{g}}{v_{p}}&=\frac{c^{2}}{v_{p}^{2}} \quad \frac{v_{g}}{v_{p}}=\frac{c^{3} p^{2}}{E^{2}}
		\end{align*}
		The correct option is \textbf{(c)}
\end{answer}
	\item An inertial frame $K^{\prime}$ moves with a constant speed $v$ with respect to another inertial frame $K$ along their common $x$ - direction. Let $(x, c t)$ and $\left(x^{\prime}, c t^{\prime}\right)$ denote the spacetime coordinates in the frames $K$ and $K^{\prime}$, respectively. Which of the following spacetime diagrams correctly describes the $t^{\prime}$ - axis $\left(x^{\prime}=0\right.$ line $)$ and the $x^{\prime}$ - axis $\left(t^{\prime}=0\right.$ line $)$ in the $x$-ct plane? (In the following figures $\tan \phi=v / c$ )
{	\exyear{NET JUNE 2018}}
\begin{tasks}(2)
	\task[\textbf{A.}]\begin{figure}[H]
		\centering
		\includegraphics[height=4cm,width=5cm]{PROBLEM 5}
	\end{figure}
	\task[\textbf{B.}]\begin{figure}[H]
		\centering
		\includegraphics[height=4cm,width=5cm]{PROBLEM 6}
	\end{figure}
	\task[\textbf{C.}]\begin{figure}[H]
		\centering
		\includegraphics[height=4cm,width=5cm]{PROBLEM 7}
	\end{figure}
	\task[\textbf{D.}]\begin{figure}[H]
		\centering
		\includegraphics[height=4cm,width=5cm]{PROBLEM8}
	\end{figure}
\end{tasks}
\begin{answer}
$	\left(\begin{array}{c}
		c t^{\prime} \\
		x^{\prime} \\
		y^{\prime} \\
		z^{\prime}
	\end{array}\right)=\left(\begin{array}{cccc}
		\cosh \phi & -\sinh \phi & 0 & 0 \\
		-\sinh \phi & \cos \phi & 0 & 0 \\
		0 & 0 & 1 & 0 \\
		0 & 0 & 0 & 1
	\end{array}\right)\left(\begin{array}{l}
		c t \\
		x \\
		y \\
		z
	\end{array}\right)$\\
	$\text { Where } v=\cosh \phi, \beta v=\sinh \phi \beta=\tanh \phi$\\
	The correct option is \textbf{(a)}
\end{answer}

	\item A relativistic particle of mass $m$ and charge $e$ is moving in a uniform electric field of strength $\varepsilon$. Starting from rest at $t=0$, how much time will it take to reach the speed $\frac{c}{2}$ ?
	{\exyear{NET DEC 2018}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\frac{1}{\sqrt{3}} \frac{m c}{e \varepsilon}$
	\task[\textbf{B.}]$\frac{m c}{e \varepsilon}$
	\task[\textbf{C.}]$\sqrt{2} \frac{m c}{e \varepsilon}$
	\task[\textbf{D.}]$\sqrt{\frac{3}{2}} \frac{m c}{e \varepsilon}$
\end{tasks}
\begin{answer}
\begin{align*}
\frac{d p}{d t}&=e \varepsilon\\
p&=e \varepsilon t+c\\
\text { At } t&=0, p=0, c=0\\
\frac{m v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}&=e \varepsilon t\\
t&=\frac{m}{e \varepsilon} \frac{v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\
\text { Put } v&=\frac{c}{2}, \quad t=\frac{m}{e \varepsilon} \frac{c / 2}{\sqrt{1-\frac{1}{4}}}=\frac{m c}{\sqrt{3} e \varepsilon}\\
t&=\frac{m c}{\sqrt{3} e E}
\end{align*}
The correct option is \textbf{(a)}
\end{answer}
\end{enumerate}
\newpage
\begin{abox}
	Practice set 2 solutions
	\end{abox}
\begin{enumerate}

	\item For the set of all Lorentz transformations with velocities along the $x$-axis consider the two statements given below:
	P: If $L$ is a Lorentz transformation, then, $L^{-1}$ is also a Lorentz transformation.
	Q: If $L_{1}$ and $L_{2}$ are Lorentz transformations, then $L_{1} L_{2}$ is necessarily a Lorentz transformation.
	Choose the correct option
	{\exyear{GATE 2010}}

\begin{tasks}(2)
	\task[\textbf{A.}] $P$ is true and $Q$ is false
	\task[\textbf{B.}]Both $P$ and $Q$ are true
	\task[\textbf{C.}]Both $P$ and $Q$ are false
	\task[\textbf{D.}]$P$ is false and $Q$ is true
\end{tasks}
\begin{answer}
The correct option is \textbf{(b)}
\end{answer}
	\item A $\pi^{0}$ meson at rest decays into two photons, which moves along the $x$-axis. They are both detected simultaneously after a time, $t=10 \mathrm{~s} .$ In an inertial frame moving with a velocity $v=0.6 c$ in the direction of one of the photons, the time interval between the two detections is
	{\exyear{GATE 2010}}

\begin{tasks}(2)
	\task[\textbf{A.}] $15 c$
	\task[\textbf{B.}]$0 \mathrm{~s}$
	\task[\textbf{C.}] $10 \mathrm{~s}$
	\task[\textbf{D.}]$20 s$
\end{tasks}
\begin{answer}
\begin{align*}
t_{1}&=t_{0} \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}=10 \sqrt{\frac{1+0.6}{1-0.6}}=10 \times 2=20 \mathrm{sec}\\
t_{2}&=t_{0} \sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}}=10 \sqrt{\frac{1-0.6}{1+0.6}}=10 \times \frac{1}{2}=5 \mathrm{sec}\\
\mathrm{t}_{1}-\mathrm{t}_{2}&=15 \mathrm{sec}
\end{align*}
The correct option is \textbf{(a)}
\end{answer}
	\item Two particles each of rest mass $m$ collide head-on and stick together. Before collision, the speed of each mass was $0.6$ times the speed of light in free space. The mass of the final entity is
{	\exyear{GATE 2011}}
\begin{tasks}(2)
	\task[\textbf{A.}] $5 m / 4$
	\task[\textbf{B.}]$2 m$
	\task[\textbf{C.}]$5 \mathrm{~m} / 2$
	\task[\textbf{D.}]$25 \mathrm{~m} / \mathrm{s}$
\end{tasks}
\begin{answer}
From conservation of energy\\
$$
\frac{m c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}+\frac{m c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=m_{1} c^{2} \Rightarrow \frac{2 m c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=m_{1} c^{2}
$$
Since $v=0.6 c \Rightarrow m_{1}=5 \mathrm{~m} / 2$	\\
The correct option is \textbf{(c)}
\end{answer}
	\item A rod of proper length $l_{0}$ oriented parallel to the $x$-axis moves with speed $2 c / 3$ along the $x$-axis in the $S$-frame, where $c$ is the speed of light in free space. The observer is also moving along the $x$-axis with speed $c / 2$ with respect to the $S$-frame. The length of the rod as measured by the observer is
	{\exyear{GATE 2012}}
\begin{tasks}(2)
	\task[\textbf{A.}] $0.35 l_{0}$
	\task[\textbf{B.}]$0.48 l_{0}$
	\task[\textbf{C.}]$0.87 l_{0}$
	\task[\textbf{D.}]$0.97 l_{0}$
\end{tasks}
\begin{answer}
$l=l_{0} \sqrt{1-\frac{u_{x}^{2}}{c^{2}}}=0.97 l_{0}$\\
The correct option is \textbf{(d)}
\end{answer}
	\item An electron is moving with a velocity of $0.85 c$ in the same direction as that of a moving photon. The relative velocity of the electron with respect to photon is
{	\exyear{GATE 2013}}
\begin{tasks}(2)
	\task[\textbf{A.}] $c$
	\task[\textbf{B.}]$-c$
	\task[\textbf{C.}] $0.15 c$
	\task[\textbf{D.}]$-0.15 c$
\end{tasks}
\begin{answer}
The correct option is \textbf{(b)}
\end{answer}

	\item  The relativistic form of Newton's second law of motion is
{	\exyear{GATE 2013}}
\begin{tasks}(2)
	\task[\textbf{A.}] $F=\frac{m c}{\sqrt{c^{2}-v^{2}}} \frac{d v}{d t}$
	\task[\textbf{B.}]$F=\frac{m \sqrt{c^{2}-v^{2}}}{c} \frac{d v}{d t}$
	\task[\textbf{C.}]$F=\frac{m c^{2}}{c^{2}-v^{2}} \frac{d v}{d t}$
	\task[\textbf{D.}]$F=m \frac{c^{2}-v^{2}}{c^{2}} \frac{d v}{d t}$
\end{tasks}
\begin{answer}
	\begin{align*}
	P&=\frac{m v}{\sqrt{1-\frac{v^{2}}{c^{2}}}}	\\
	F&=\frac{d P}{d t}=m \frac{d v}{d t} \cdot \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}+m v\left(-\frac{1}{2}\right) \cdot \frac{1}{\left(1-\frac{v^{2}}{c^{2}}\right)^{3 / 2}} \cdot \frac{-2 v}{c^{2}} \frac{d v}{d t}\\
	F&=m \frac{d v}{d t} \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\left(1+\frac{1}{2} \frac{v^{2} / c^{2}}{\left(1-\frac{v^{2}}{c^{2}}\right)}\right)\\
	&=m \frac{d v}{d t}\left(\frac{1-v^{2} / 2 c^{2}}{\left(1-\frac{v^{2}}{c^{2}}\right)^{3 / 2}}\right)\\
	&=m \frac{d v}{d t}\left[\frac{\left(1-v^{2} / c^{2}\right)^{1 / 2}}{\left(1-v^{2} / c^{2}\right)\left(1-v^{2} / c^{2}\right)^{1 / 2}}\right]=\frac{m c^{2}}{\left(c^{2}-v^{2}\right)} \frac{d v}{d t}
	\end{align*}
The correctoption is \textbf{(c)}
\end{answer}
	\item If the half-life of an elementary particle moving with speed $0.9$ c in the laboratory frame is $5 \times 10^{-8} s$, then the proper half-life is $\times 10^{-8}$ s. $\left(c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
	{\exyear{GATE 2014}}
\begin{answer}
\begin{align*}
	t&=\frac{t_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\
	t_{0}&=t \times \sqrt{1-\frac{v^{2}}{c^{2}}}=5 \times 10^{-8} \times \sqrt{0.19}=2.18 \times 10^{-8} \mathrm{~s}
\end{align*}
\end{answer}
	\item In an inertial frame $S$, two events $A$ and $B$ take place at $\left(c t_{A}=0, \vec{r}_{A}=0\right)$ and $\left(c t_{B}=0, \vec{r}_{B}=2 \hat{y}\right)$, respectively. The times at which these events take place in a frame $S^{\prime}$ moving with a velocity $0.6 c \hat{y}$ with respect to $S$ are given by
	{\exyear{GATE 2015}}
\begin{tasks}(2)
	\task[\textbf{A.}] $c t_{A}^{\prime}=0 ; c t_{B}^{\prime}=-\frac{3}{2}$
	\task[\textbf{B.}]$c t_{A}^{\prime}=0 ; c t_{B}^{\prime}=0$
	\task[\textbf{C.}]$c t_{A}^{\prime}=0 ; c t_{B}^{\prime}=\frac{3}{2}$
	\task[\textbf{D.}] $c t_{A}^{\prime}=0 ; c t_{B}^{\prime}=\frac{1}{2}$
\end{tasks}
\begin{answer}
\begin{align*}
&\text { Velocity of } S^{\prime} \text { with respect to } S \text { is } v=0.6 c\\
t_{A}^{\prime}&=\frac{t_{A}-\frac{v}{c^{2}} y}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\
&\text { For event } \mathrm{A}, t_{A}=0, y=0 \text {. So } \mathrm{ct}_{A}^{\prime}=0\\
t_{B}^{\prime}&=\frac{t_{B}-\frac{v}{c^{2}} y}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\
&\text { For event } \mathrm{B}, t_{B}=0, y=2 \text {. So } c t_{B}^{\prime}=-\frac{3}{2}
\end{align*}
The correct option is \textbf{(a)}
\end{answer}
	\item A particle with rest mass $M$ is at rest and decays into two particles of equal rest masses $\frac{3}{10} M$ which move along the $z$ axis. Their velocities are given by
	{\exyear{GATE 2015}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\vec{v}_{1}=\vec{v}_{2}=(0.8 c) \hat{z}$
	\task[\textbf{B.}]$\vec{v}_{1}=-\vec{v}_{2}=(0.8 c) \hat{z}$
	\task[\textbf{C.}]$\vec{v}_{1}=-\vec{v}_{2}=(0.6 c) \hat{z}$
	\task[\textbf{D.}]$\vec{v}_{1}=(0.6 c) \hat{z} ; \vec{v}_{2}=(-0.8 c) \hat{z}$
\end{tasks}
\begin{answer}
	\begin{align*}
	M &\rightarrow \frac{3}{10} M+\frac{3}{10} M\\
	\intertext { From momentum conservation }
	0&=\vec{P}_{1}+\vec{P}_{2} \Rightarrow \vec{P}_{1}=-\vec{P}_{2} \Rightarrow\left|P_{1}\right|=\left|P_{2}\right|\\
	\intertext { From energy conservation }
	E&=E_{1}+E_{2}\\
	\Rightarrow M c^{2}&=\frac{3}{10} \frac{M c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}+\frac{3}{10} \frac{M c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \Rightarrow M c^{2}=\frac{3}{5} \frac{M c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\
	\left(1-\frac{v^{2}}{c^{2}}\right)&=\frac{9}{25} \Rightarrow \frac{v^{2}}{c^{2}}=\frac{16}{25} \Rightarrow v=0.8 c
	\end{align*}
	The correct option is \textbf{(b)}
\end{answer}
	\item The kinetic energy of a particle of rest mass $m_{0}$ is equal to its rest mass energy. Its momentum in units of $m_{0} c$, where $c$ is the speed of light in vacuum, is
	(Give your answer upto two decimal places)
	{\exyear{GATE 2016}}
\begin{answer}
\begin{align*}
m_{0} c^{2}&=E-m_{0} c^{2} \Rightarrow E=2 m_{0} c^{2}\\
&\Rightarrow \frac{m_{0} c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=2 m_{0} c^{2} \Rightarrow v=\frac{\sqrt{3}}{2} c\\
\cdot E^{2}&=p^{2} c^{2}+m_{0}^{2} c^{4} \Rightarrow 4 m_{0}^{2} c^{4}-m_{0}^{2} c^{4}=p^{2} c^{2} \Rightarrow p=\sqrt{3} m_{0} c=1.732 m_{0} c
\end{align*}
\end{answer}
	\item In an inertial frame of reference $S$, an observer finds two events occurring at the same time at coordinates $x_{1}=0$ and $x_{2}=d$. A different inertial frame $S^{\prime}$ moves with velocity $v$ with respect to $S$ along the positive $x$-axis. An observer in $S^{\prime}$ also notices these two events and finds them to occur at times $t_{1}^{\prime}$ and $t_{2}^{\prime}$ and at positions $x_{1}^{\prime}$ and $x_{2}^{\prime}$ respectively.
	If $\Delta t^{\prime}=t_{2}^{\prime}-t_{1}^{\prime}, \Delta x^{\prime}=x_{2}^{\prime}-x_{1}^{\prime}$ and $\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$, which of the following statements is true?
	{\exyear{GATE 2016}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\Delta t^{\prime}=0, \Delta x^{\prime}=\gamma d$
	\task[\textbf{B.}] $\Delta t^{\prime}=0, \Delta x^{\prime}=\frac{d}{\gamma}$
	\task[\textbf{C.}]$\Delta t^{\prime}=\frac{-\gamma v d}{c^{2}}, \Delta x^{\prime}=\gamma d$
	\task[\textbf{D.}] $\Delta t^{\prime}=\frac{-\gamma v d}{c^{2}}, \Delta x^{\prime}=\frac{d}{\gamma}$
\end{tasks}
\begin{answer}
	\begin{align*}
	t_{2}^{\prime}-t_{1}^{\prime}&=\left(\frac{t_{2}-\frac{v x_{2}}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)-\left(\frac{t_{1}-\frac{v x_{1}}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right) \Rightarrow \Delta t^{\prime}=\gamma \Delta t-\frac{\gamma v \Delta x}{c^{2}}\\
	\text { It is given, } \Delta t&=0, \Delta x=d\\
	\Rightarrow \Delta t^{\prime}&=-\frac{\gamma v \Delta x}{c^{2}}=-\frac{\gamma v d}{c^{2}}\\
	x_{2}^{\prime}-x_{1}^{\prime}&=\left(\frac{x_{2}-v t_{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)-\left(\frac{x_{1}-v t_{1}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right) \Rightarrow \Delta x^{\prime}=\gamma(\Delta x-v \Delta t)\\
	\Rightarrow \Delta x^{\prime}&=\gamma d
	\end{align*}
	The correct option is \textbf{(c)}
\end{answer}
	\item A particle of rest mass $M$ is moving along the positive $x$-direction. It decays into two photons $\gamma_{1}$ and $\gamma_{2}$ as shown in the figure. The energy of $\gamma_{1}$ is $1 \mathrm{GeV}$ and the energy of $\gamma_{2}$ is $0.82 \mathrm{GeV}$. The value of $M$ (in units of $\frac{\mathrm{GeV}}{c^{2}}$ ) is upto two decimal places)
	{\exyear{GATE }}
	\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=5cm]{problem 17}
	\end{figure}
\begin{answer}
\begin{align*}
\sqrt{p^{2} c^{2}+M^{2} c^{4}}&=E_{1}+E_{2}=1.82 \mathrm{GeV}\\
p&=\frac{E_{1}}{c} \cos \theta_{1}+\frac{E_{2}}{c} \cos \theta_{2}=\frac{1 G e V}{c} \frac{1}{\sqrt{2}}+\frac{0.82 G e V}{c} \frac{1}{2}=\frac{1.11 G e V}{c}\\
&\Rightarrow p^{2} c^{2}+m^{2} c^{4}=3.312 \Rightarrow m^{2} c^{4}=3.312-1.23=2.08\\
&\Rightarrow m=\sqrt{2.076}=1.44
\end{align*}
\end{answer}
	\item An object travels along the $x$-direction with velocity $\frac{C}{2}$ in a frame $O$. An observer in a frame $O^{\prime}$ sees the same object travelling with velocity $\frac{C}{4}$. The relative velocity of $O^{\prime}$ with respect to $O$ in units of $c$ is.............. (up to two decimal places).
	{\exyear{GATE 2017}}
\begin{answer}
	\begin{align*}
	u_{x}^{\prime}&=\frac{C}{2}, v=\frac{C}{4}\\
	u_{x}&=\frac{u_{x}^{\prime}-v}{1-\frac{u_{x}^{\prime} v}{c^{2}}}=\frac{\frac{c}{2}-\frac{c}{4}}{1-\frac{c}{2} \cdot \frac{c}{4} \cdot \frac{1}{c^{2}}}=\frac{2 c}{7}=0.28 c
	\end{align*}
\end{answer}
	\item A spaceship is travelling with a velocity of $0.7 c$ away from a space station. The spaceship ejects a probe with a velocity $0.59$ c opposite to its own velocity. A person in the space station would see the probe moving at a speed $X_{C}$, where the value of $X$ is (up to three decimal places).
	{\exyear{GATE 2018}}
\begin{answer}$\left. \right. $\\
	\begin{minipage}{0.5\textwidth}
	\begin{align*}
	v&=0 \cdot 7 c, u_{x}^{\prime}=-0 \cdot 59 c\\
	u_{x}&=\frac{u_{x}^{\prime}+v}{1+\frac{u_{x}^{\prime} v}{c_{2}}}\\
	u_{x}&=\frac{-0.59 c+0.7 c}{1-0.7 \times 0.59}=\frac{0.11 c}{1-0.413}=\frac{0.11 c}{0.587}=0.187 c
	\end{align*}
	\end{minipage}
\begin{minipage}{0.5\textwidth}
\begin{figure}[H]
	\centering
	\includegraphics[height=3cm,width=5cm]{PROBLEM 18}
\end{figure}
\end{minipage}
\end{answer}
	\item Two spaceships $A$ and $B$, each of the same rest length $L$, are moving in the same direction with speeds $\frac{4 c}{5}$ and $\frac{3 c}{5}$, respectively, where $c$ is the speed of light. As measured by $B$, the time taken by $A$ to completely overtake $B$ [see figure below] in units of $L / c$ (to the nearest integer) is
	{\exyear{GATE 2019}}$\left. \right. $\\
	\begin{minipage}{0.5\textwidth}
	\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=5cm]{PROBLEM 19}
	\end{figure}
	\end{minipage}
\begin{minipage}{0.5\textwidth}
	\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=5cm]{PROBLEM 20}
	\end{figure}
\end{minipage}
\begin{answer}
	\begin{align*}
	u_{A, B}&=\frac{\frac{4}{5} c-\frac{3}{5} c}{1-\frac{4}{5} c \cdot \frac{3}{5} c \cdot \frac{1}{c^{2}}}=\frac{\frac{c}{5}}{\frac{13}{25}}=\frac{5}{13} c	\\
	&\text { Kinematic equation is given by }\\
	\frac{5}{13} c \times t&=L \sqrt{1-\frac{25}{169}}+L \Rightarrow t=\frac{5 L}{c} \Rightarrow \alpha=5
	\end{align*}
\end{answer}
	\item Two events, one on the earth and the other one on the Sun, occur simultaneously in the earth's frame. The time difference between the two events as seen by an observer in a spaceship moving with velocity $0.5 c$ in the earth's frame along the line joining the earth to the Sun is $\Delta t$, where $c$ is the speed of light. Given that light travels from the Sun to the earth in $8.3$ minutes in the earth's frame, the value of $|\Delta t|$ in minutes (rounded off to two decimal places) is
	(Take the earth's frame to be inertial and neglect the relative motion between the earth and the sun)
	{\exyear{GATE 2019}}
\begin{answer}
\begin{align*}
t_{2}^{\prime}-t_{1}^{\prime}&=0 \quad x_{2}^{\prime}-x_{1}^{\prime}=8.3 \times 3 \times 10^{8} \times 60 \quad v=0.5 c\\
\Delta t&=t_{2}-t_{1}=\left(\frac{t_{2}^{\prime}+\frac{v x_{2}^{\prime}}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)-\left(\frac{t_{1}^{\prime}+\frac{v x_{1}^{\prime}}{c^{2}}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)=\left(\frac{t_{2}-t_{1}^{\prime}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)+\frac{v}{c^{2}} \frac{\left(x_{2}^{\prime}-x_{1}^{\prime}\right)}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=4.77 \min
\end{align*}
\end{answer}
\end{enumerate}

Classical Mechanics  -CSIR/chapter/str solutions.tex


% !TEX root = ../main.tex
\section{Permutation-invariant Recommender Models}
\label{sec:models}
\Cref{prop:universal-approximation} shows that \gls{rfs} can approximate permutation-invariant recommendation models. We describe several common recommendation models and show that they are permutation-invariant, before comparing their performance to \gls{rfs} in \Cref{sec:rfs-experiments}.

\citet{gopalan2014content-based} develop a probabilistic matrix factorization model of user consumption data. \Gls{ctpf} models user preferences using a generative process,
\input{ch-rfs/ctpf_generative_process}
To show that \gls{ctpf} is permutation-invariant, consider the Poisson likelihood function over words $w_{dv}$. Conditional on the latent item representation $\theta_d$ and latent word representation $\beta_v$, every word in the document $w_{dv}$ is independent; the joint probability of words in a document factorizes:
\begin{align}
  p(w_{d} \mid \theta_d, \beta_v) &= \prod_{w_{dv} \in w_d} p(w_{dv} \mid \theta_d, \beta_v)\, .
\end{align}
\gls{ctpf} makes predictions using expectations under the posterior. The posterior is proportional to the the log joint of the model, and the attributes of items (words in documents) enter into the model only via the above product. The product of the probability of words in a document is invariant to a reordering of the words in the document, and therefore \gls{ctpf} is permutation-invariant.

Word embedding models~\citep{mikolov2013distributed} can be used as recommendation models if the embeddings are learned using a modified context window. For an item with attributes $x_m$, let the context window for attribute ${j\in x_m}$ be the set of other attributes of the same item ${j' \in x_m: j'\neq j}$. To recommend items using this model of attributes $\beta_j$ for $j \in V$, item embeddings are computed as the average of their attribute embeddings. Users are represented as the average of the embeddings of the items they consume, and recommendation is performed using the cosine similarity of user and item embeddings. This is a permutation-invariant model, as the output of the model depends on the sum of attribute embeddings (summation is invariant to permutation).

StarSpace is also an embedding model and represents users as a sum over a user's consumed items' attribute embeddings (there is no explicit user embedding). In contrast to the word embedding model, StarSpace is trained on a classification objective with negative samples drawn from the the set of items~\citep{wu2018starspace:}. As model predictions depend on sums of attributes, StarSpace is a permutation-invariant recommendation model.

We next consider LightFM~\citep{kula2015metadata}, a permutation-invariant recommendation model. We show that if the \gls{bpr} objective~\citep{rendle2009bpr:} is used, LightFM is an instance of \acrshort{rfs}. \footnote{The LightFM paper~\citep{kula2015metadata} uses a logistic objective to which \Cref{prop:maximizing-recall} applies. LightFM with the \gls{bpr} objective is unpublished but implemented in code released by the author. For completeness, we studied LightFM with both objectives to ensure its performance is equivalent to \gls{rfs} when the \gls{bpr} objective is used.} Although the \gls{bpr} objective is designed for ranking, models trained with it can be used to construct classifiers. The \gls{bpr} objective is
$${\log\sigma\left(f(u, x_m; \mbgamma) - f(u, x_k; \mbgamma)\right)}\, ,$$
where $m$ corresponds to a positive label $\yum = 1$, $k$ corresponds to a negative label $\yuk = 0$, and $f$ is parameterized as in \gls{rfs}~\citep{kula2015metadata}. A ranking function $f$ optimizes the \gls{bpr} objective if~${f\rightarrow \infty}$ for the positive example and $f$ is constant for the negative example; or, if $f$ is constant for the positive example and $f\rightarrow -\infty$ for the negative example. In either case, a constant can be added to yield a perfect classifier from the ranking function $f$ (positive examples are ranked higher than negative examples in the optimal ranking, so there exists such a constant). That we can construct a classifier from the \gls{bpr} objective means that \Cref{prop:maximizing-recall} applies: permutation-invariant models such as LightFM, trained with the \gls{bpr} objective, are instances of the \gls{rfs} class of recommendation models.

The regression function $f$ in \gls{rfs} can also be parameterized using a recurrent network, as in \citet{bansal2016ask-the-gru:}. Such a recommendation model can be made permutation-invariant if averaged over permutations of attributes fed to the network. Attributes are treated as a sequence and the marginalization is over these permutations,
\begin{equation}
p(\yum = 1 \mid x_m) =
  \frac{1}{\lvert \pi(x_m) \rvert}\sum_{\pi \in \pi(x_m)}\sigma\left(\phi(\theta_u,\{\beta_{\pi(1)}, \ldots,
  \beta_{\pi(J)}\})\right) \, .
\label{eq:rnn}
\end{equation}
Here $\beta_j$ are attribute embeddings, $\pi(x_m)$ denotes the set of all permutations of the attributes $x_m$, and $\phi$ is the output of a recurrent neural network architecture~\citep{bansal2016ask-the-gru:} projected to a scalar.


\section{Conclusion}
\subsection{Summary}
The goal of this paper was to find a model that would represent the neural network as defined by the Bio-M MC as closely as possible through the use of topological statistics. Given the array of constraints that we could apply to the model, we chose to incrementally add these constraints to the models to see if there is improvement in each one. We found that each implementation had their own unique improvement.

\subsection{Future Implementations}
A future implementation that could see an improvement in connectivity, especially on the local level, would be to consider a cloud connected model. The idea being that we take axon and dendrite lengths into account in the way where we add a volume around each neuron that has a certain axon and dendrite cloud covering, apply a probability of connection, then randomly select a connection until we have a matching amount of connections in the MC as the Bio-M MC and compare topological statistics as we have done above.

Further to this, there are multiple instantiations that we did not consider and if we had included them, this may give rise to models that could be better suited to these instantiations as compared to how they affected this instantiation (Bio-M).

\section{Acknowledgements}

This research was partially supported by the Wallenberg AI, Autonomous Systems and Software Program funded by Knut and Alice Wallenberg Foundation.
The distributed computing infrastructure for this pjoject was supported by Databricks University Alliance with AWS credits.

Many thanks to Kathryn Hess, Svante Janson, Wojciech Chachólski, Martina Scolamiero, and Michael Reimann.


% =========================================
% COMMAND: _RAND
% =========================================

\newpage
\section{\_RAND}
\label{cmd:_RAND}

\paragraph{Syntax:}
\subparagraph{}
\texttt{\_RAND <start> <end> <variable>}

\paragraph{Purpose:}
\subparagraph{}
Generates a random number between $<$start$>$ and $<$end$>$
and stores the result in $<$variable$>$.


    \documentclass{article}
    \usepackage{amsfonts}
    \usepackage{amsmath,multicol,eso-pic}
    \begin{document}
    \title{Algebra 101 worksheet 1 Solutions}
 \date{\vspace{-5ex}}
 \maketitle

        \section{Linear equations}

        \begin{enumerate}
        \item$$a =\frac{g - 5}{H - 14}$$
\item$$N =\frac{16}{- S + w}$$
\item$$A =\frac{h + 4}{E - 6}$$
\item$$w =\frac{z}{15} - \frac{23}{15}$$
\item$$b =\frac{K - 5}{Y - 23}$$
\item$$M =\frac{10}{S - 4}$$
\item$$y =\frac{- V + z}{T - X}$$
\item$$p =\frac{6}{Q + 2}$$
\item$$p =\frac{- V + g}{t + 5}$$
\item$$H =\frac{y - 20}{E - u}$$
\item$$M =\frac{R + 10}{R - k}$$
\item$$M =\frac{- b + 9}{j + 26}$$
\item$$V =\frac{P - X}{Q - S}$$
\item$$P =- \frac{e}{13} + \frac{1}{13}$$
\item$$g =- \frac{37}{7}$$
\item$$K =\frac{S + 6}{T + 23}$$
\item$$d =\frac{Q + 24}{N + 10}$$
\item$$D =- \frac{27}{m - 8}$$
\item$$E =\frac{- y + 2}{q + 5}$$
\item$$r =\frac{c - 10}{x + 5}$$
        \end{enumerate}

        \section{Quadratic equations}

        \begin{enumerate}
        \item$$y = - \frac{15}{8} - \frac{\sqrt{65}}{8}, y = - \frac{15}{8} + \frac{\sqrt{65}}{8}$$
\item$$x = \frac{19}{18} + \frac{\sqrt{757}}{18}, x = - \frac{\sqrt{757}}{18} + \frac{19}{18}$$
\item$$x = -10, x = 23$$
\item$$y = \frac{1}{4} + \frac{\sqrt{177}}{4}, y = - \frac{\sqrt{177}}{4} + \frac{1}{4}$$
\item$$x = 0, x = 22$$
\item$$x = \frac{8}{13} - \frac{3 i}{13} \sqrt{29}, x = \frac{8}{13} + \frac{3 i}{13} \sqrt{29}$$
\item$$x = -2, x = 10$$
\item$$x = - \frac{1}{22} + \frac{\sqrt{309}}{22}, x = - \frac{\sqrt{309}}{22} - \frac{1}{22}$$
\item$$x = 0, x = \frac{24}{5}$$
\item$$y = - \frac{11}{20} - \frac{\sqrt{799} i}{20}, y = - \frac{11}{20} + \frac{\sqrt{799} i}{20}$$
\item$$x = 0, x = \frac{4}{15}$$
\item$$x = 12, x = 15$$
\item$$y = 2, y = 6$$
\item$$x = -17, x = 23$$
\item$$x = -7, x = -1$$
\item$$x = -13, x = -5$$
\item$$x = -3, x = \frac{7}{4}$$
\item$$x = \frac{5}{21} + \frac{2 \sqrt{85}}{21}, x = - \frac{2 \sqrt{85}}{21} + \frac{5}{21}$$
\item$$y = \frac{25}{26} + \frac{\sqrt{1041}}{26}, y = - \frac{\sqrt{1041}}{26} + \frac{25}{26}$$
\item$$x = 0, x = 1$$
        \end{enumerate}

        \section{Compute the derivative}

        \begin{enumerate}
        \item$$\frac{1}{x} \left(14 x + 12\right) - \frac{1}{x^{2}} \left(7 x^{2} + 12 x - 24\right)$$
\item$$\frac{2 \sqrt{x} \left(27 x^{2} - 18\right)}{\left(- 9 x^{3} + 18 x + 7\right)^{2}} + \frac{1}{\sqrt{x} \left(- 9 x^{3} + 18 x + 7\right)}$$
\item$$\frac{1}{\left(16 x^{3} - 23 x^{2} + 5 x\right)^{2}} \left(\log{\left (x \right )} + \tan{\left (x \right )}\right) \left(- 48 x^{2} + 46 x - 5\right) + \frac{\tan^{2}{\left (x \right )} + 1 + \frac{1}{x}}{16 x^{3} - 23 x^{2} + 5 x}$$
\item$$- \left(e^{x} + \tan{\left (x \right )}\right) e^{- x} + \left(e^{x} + \tan^{2}{\left (x \right )} + 1\right) e^{- x}$$
\item$$- \left(19 x + e^{x}\right) e^{- x} + \left(e^{x} + 19\right) e^{- x}$$
\item$$\frac{1}{x} \left(48 x - \sin{\left (x \right )} + 7\right) - \frac{1}{x^{2}} \left(24 x^{2} + 7 x + \cos{\left (x \right )}\right)$$
\item$$\frac{1}{\tan^{2}{\left (x \right )}} \left(\log{\left (x \right )} + \sin{\left (x \right )}\right) \left(- \tan^{2}{\left (x \right )} - 1\right) + \frac{\cos{\left (x \right )} + \frac{1}{x}}{\tan{\left (x \right )}}$$
\item$$\frac{1}{x} \left(12 x^{2} + \frac{1}{x}\right) - \frac{1}{x^{2}} \left(4 x^{3} + \log{\left (x \right )} - 17\right)$$
\item$$\frac{1}{\sin{\left (x \right )}} \left(- 51 x^{2} + 48 x + 14 + \frac{1}{x}\right) - \frac{\cos{\left (x \right )}}{\sin^{2}{\left (x \right )}} \left(- 17 x^{3} + 24 x^{2} + 14 x + \log{\left (x \right )} - 14\right)$$
\item$$\left(\frac{1}{x} + \frac{1}{2 \sqrt{x}}\right) e^{- x} - \left(\sqrt{x} + \log{\left (x \right )}\right) e^{- x}$$
        \end{enumerate}

    \end{document}


\section*{Standard Extensions}

\subsection*{RV32M Multiply Extension}
\begin{center}
\begin{tabular}
{T | l | c | c | c | c | C } \hline
\rm Inst & Name              & FMT & \rm Opcode & \rm funct3 & \rm funct7 & \rm Description (C)     \\ \hline
mul      & MUL               & R   & 0110011    & 000    & 0000001   & rd = (rs1 * rs2)[31:0]  \\
mulh     & MUL High          & R   & 0110011    & 001    & 0000001   & rd = (rs1 * rs2)[63:32] \\
mulsu    & MUL High (S) (U)  & R   & 0110011    & 010    & 0000001   & rd = (rs1 * rs2)[63:32] \\
mulu     & MUL High (U)      & R   & 0110011    & 011    & 0000001   & rd = (rs1 * rs2)[63:32] \\
div      & DIV               & R   & 0110011    & 100    & 0000001   & rd = rs1 / rs2          \\
divu     & DIV (U)           & R   & 0110011    & 101    & 0000001   & rd = rs1 / rs2          \\
rem      & Remainder         & R   & 0110011    & 110    & 0000001   & rd = rs1 \% rs2         \\
remu     & Remainder (U)     & R   & 0110011    & 111    & 0000001   & rd = rs1 \% rs2         \\
\hline
\end{tabular}
\end{center}

\subsection*{RV32A Atomic Extension}

%\begin{itemize}
%    \item \texttt{aq}: acquire access bit -- this operation must occur before later memory ops
%    \item \texttt{rl}: release access bit -- this operation must occur after earlier memory ops
%\end{itemize}

\begin{center}
\begin{tabular}{R@{}W@{}W@{}R@{}R@{}F@{}R@{}O}
\instbitrange{31}{27} &
\instbit{26} &
\instbit{25} &
\instbitrange{24}{20} &
\instbitrange{19}{15} &
\instbitrange{14}{12} &
\instbitrange{11}{7} &
\instbitrange{6}{0} \\
\hline
\multicolumn{1}{|c|}{funct5} &
\multicolumn{1}{c|}{aq} &
\multicolumn{1}{c|}{rl} &
\multicolumn{1}{c|}{rs2} &
\multicolumn{1}{c|}{rs1} &
\multicolumn{1}{c|}{funct3} &
\multicolumn{1}{c|}{rd} &
\multicolumn{1}{c|}{opcode} \\
\hline
5 & 1 & 1 & 5 & 5 & 3 & 5 & 7 \\
\end{tabular}

\begin{tabular}
{T | l | c | c | c | c | T } \hline
\rm Inst  & Name              & FMT   & \rm Opcode & \rm funct3 & \rm funct5 & \rm Description (C)         \\ \hline
lr.w      & Load Reserved     & R     & 0101111    & 010    & 00010   & rd = M[rs1], reserve M[rs1] \\
sc.w      & Store Conditional & R     & 0101111    & 010    & 00011   & \parbox[t]{2.5in}{ if (reserved) \{ M[rs1] = rs2; rd = 0 \} \\
                                                                        else \{ rd = 1 \}}   \\
amoswap.w & Atomic Swap       & R     & 0101111    & 010    & 00001   & rd = M[rs1]; swap(rd, rs2); M[rs1] = rd \\
amoadd.w  & Atomic ADD        & R     & 0101111    & 010    & 00000   & rd = M[rs1] + rs2; M[rs1] = rd \\
amoand.w  & Atomic AND        & R     & 0101111    & 010    & 01100  & rd = M[rs1] \& rs2; M[rs1] = rd \\
amoor.w   & Atomic OR         & R     & 0101111    & 010    & 01010   & rd = M[rs1] | rs2; M[rs1] = rd \\
amoxor.w  & Atomix XOR        & R     & 0101111    & 010    & 00100   & rd = M[rs1] \^{} rs2; M[rs1] = rd \\
amomax.w  & Atomic MAX        & R     & 0101111    & 010    & 10100   & rd = max(M[rs1], rs2); M[rs1] = rd \\
amomin.w  & Atomic MIN        & R     & 0101111    & 010    & 10000   & rd = min(M[rs1], rs2); M[rs1] = rd \\
\hline
\end{tabular}
\end{center}

\newpage

\subsection*{RV32F Single-Precision Floating-Point Extension}
\begin{center}
\textbf{R4-type instructions}

\begin{tabular}{R@{}W@{}W@{}R@{}R@{}F@{}O}
\instbitrange{31}{27} &
\instbitrange{26}{25} &
\instbitrange{24}{20} &
\instbitrange{19}{15} &
\instbitrange{14}{12} &
\instbitrange{11}{7} &
\instbitrange{6}{0} \\
\hline
\multicolumn{1}{|c|}{rs3} &
\multicolumn{1}{c|}{fmt} &
\multicolumn{1}{c|}{rs2} &
\multicolumn{1}{c|}{rs1} &
\multicolumn{1}{c|}{funct3} &
\multicolumn{1}{c|}{rd} &
\multicolumn{1}{c|}{opcode} \\
\hline
5 & 2 & 5 & 5 & 3 & 5 & 7 \\
\end{tabular}

\begin{tabular}
{T | l | c | c | c | c | c | c | T } \hline
\rm Inst  & Name                  & FMT   & \rm Opcode & \rm funct3/rm & \rm funct7 & \rm rs2 [20...24] & \rm fmt & \rm Description (C)         \\ \hline
flw       & Flt Load Word         & I     & 0000111    & 010    &        &        & & rd = M[rs1 + imm]    \\
fsw       & Flt Store Word        & I     & 0100111    & 010    &        &        & & M[rs1 + imm] = rs2   \\
\hline
fadd.s    & Flt Add               & R     & 1010011    &        & 0000000 &        & & rd = rs1 + rs2       \\
fsub.s    & Flt Sub               & R     & 1010011    &        & 0000100 &        & & rd = rs1 - rs2       \\
fmul.s    & Flt Mul               & R     & 1010011    &        & 0001000 &        & & rd = rs1 * rs2       \\
fdiv.s    & Flt Div               & R     & 1010011    &        & 0001100 &        & & rd = rs1 / rs2       \\
fsgnj.s   & Flt Sign Injection    & R     & 1010011    & 000    & 0010000 &        & & rd = abs(rs1) * sgn(rs2) \\
fsgnjn.s  & Flt Sign Neg Injection& R     & 1010011    & 001    & 0010000 &        & & rd = abs(rs1) * -sgn(rs2) \\
fsgnjx.s  & Flt Sign Xor Injection& R     & 1010011    & 010    & 0010000 &        & & rd = rs1 * sgn(rs2)  \\
fmin.s    & Flt Minimum           & R     & 1010011    & 000    & 0010100 &        & & rd = min(rs1, rs2)   \\
fmax.s    & Flt Maximum           & R     & 1010011    & 001    & 0010100 &        & & rd = max(rs1, rs2)   \\
fsqrt.s   & Flt Square Root       & R     & 1010011    &        & 0101100 & 00000  & & rd = sqrt(rs1)       \\
\hline
fle.s     & Float Less / Equal    & R     & 1010011    & 000    & 1010000 &        & & rd = (rs1 <= rs2) ? 1 : 0\\
flt.s     & Float Less Than       & R     & 1010011    & 001    & 1010000 &        & & rd = (rs1 <  rs2) ? 1 : 0 \\
feq.s     & Float Equality        & R     & 1010011    & 010    & 1010000 &        & & rd = (rs1 == rs2) ? 1 : 0 \\
\hline
fcvt.w.s  & Flt Convert to Int    & R     & 1010011    &        & 1100000 & 00000  & & rd = (int32\_t) rs1 \\
fcvt.wu.s & Flt Convert to Int    & R     & 1010011    &        & 1100000 & 00001  & & rd = (uint32\_t) rs1 \\
fmv.x.w   & Move Float to Int     & R     & 1010011    & 000    & 1110000 & 00000  & & rd = *((int*) \&rs1) \\
fclass.s  & Float Classify        & R     & 1010011    & 001    & 1110000 & 00000  & & rd = 0..9 \\
\hline
fcvt.s.w  & Flt Conv from Sign Int& R     & 1010011    &        & 1101000 & 00000  & & rd = (float) rs1 \\
fcvt.s.wu & Flt Conv from Uns Int & R     & 1010011    &        & 1101000 & 00001  & & rd = (float) rs1 \\
fmv.w.x   & Move Int to Float     & R     & 1010011    & 000    & 1111000 & 00000  & & rd = *((float*) \&rs1) \\
\hline
fmadd.s   & Flt Fused Mul-Add     & R4    & 1000011    &        &        &        & 00 & rd = rs1 * rs2 + rs3 \\
fmsub.s   & Flt Fused Mul-Sub     & R4    & 1000111    &        &        &        & 00 & rd = rs1 * rs2 - rs3 \\
fnmadd.s  & Flt Neg Fused Mul-Add & R4    & 1001011    &        &        &        & 00 & rd = -rs1 * rs2 + rs3 \\
fnmsub.s  & Flt Neg Fused Mul-Sub & R4    & 1001111    &        &        &        & 00 & rd = -rs1 * rs2 - rs3 \\
\hline
\end{tabular}
\end{center}

\newpage
\subsection*{RV32D Double-Precision Floating-Point Extension}

\begin{center}
\begin{tabular}
{T | l | c | c | c | c | c | c | T } \hline
\rm Inst  & Name                  & FMT   & \rm Opcode & \rm funct3/rm & \rm funct7 & \rm rs2 [20...24] & \rm fmt & \rm Description (C)         \\ \hline
fld       & Flt Load Word         & I     & 0000111    & 011    &        &        & & rd = M[rs1 + imm]    \\
fsd       & Flt Store Word        & I     & 0100111    & 011    &        &        & & M[rs1 + imm] = rs2   \\
\hline
fadd.d    & Flt Add               & R     & 1010011    &        & 0000001 &        & & rd = rs1 + rs2       \\
fsub.d    & Flt Sub               & R     & 1010011    &        & 0000101 &        & & rd = rs1 - rs2       \\
fmul.d    & Flt Mul               & R     & 1010011    &        & 0001001 &        & & rd = rs1 * rs2       \\
fdiv.d    & Flt Div               & R     & 1010011    &        & 0001101 &        & & rd = rs1 / rs2       \\
fsgnj.d   & Flt Sign Injection    & R     & 1010011    & 000    & 0010001 &        & & rd = abs(rs1) * sgn(rs2) \\
fsgnjn.d  & Flt Sign Neg Injection& R     & 1010011    & 001    & 0010001 &        & & rd = abs(rs1) * -sgn(rs2) \\
fsgnjx.d  & Flt Sign Xor Injection& R     & 1010011    & 010    & 0010001 &        & & rd = rs1 * sgn(rs2)  \\
fmin.d    & Flt Minimum           & R     & 1010011    & 000    & 0010101 &        & & rd = min(rs1, rs2)   \\
fmax.d    & Flt Maximum           & R     & 1010011    & 001    & 0010101 &        & & rd = max(rs1, rs2)   \\
fsqrt.d   & Flt Square Root       & R     & 1010011    &        & 0101101 & 00000  & & rd = sqrt(rs1)       \\
\hline
fle.d     & Float Less / Equal    & R     & 1010011    & 000    & 1010001 &        & & rd = (rs1 <= rs2) ? 1 : 0\\
flt.d     & Float Less Than       & R     & 1010011    & 001    & 1010001 &        & & rd = (rs1 <  rs2) ? 1 : 0 \\
feq.d     & Float Equality        & R     & 1010011    & 010    & 1010001 &        & & rd = (rs1 == rs2) ? 1 : 0 \\
\hline
fcvt.w.d  & Flt Convert to Int    & R     & 1010011    &        & 1100001 & 00000  & & rd = (int32\_t) rs1 \\
fcvt.wu.d & Flt Convert to Int    & R     & 1010011    &        & 1100001 & 00001  & & rd = (uint32\_t) rs1 \\
fcvt.s.d   & Double Flt to Single & R     & 1010011    & 000    & 0100000 & 00001  & & rd = *((float*) \&rs1) \\
fmv.x.d   & Move Int to Float     & R     & 1010011    & 000    & 1111001 & 00000  & & rd = *((float*) \&rs1) \\
fclass.d  & Float Classify        & R     & 1010011    & 001    & 1110001 & 00000  & & rd = 0..9 \\
\hline
fcvt.d.w  & Flt Conv from Sign Int& R     & 1010011    &        & 1101001 & 00000  & & rd = (float) rs1 \\
fcvt.d.wu & Flt Conv from Uns Int & R     & 1010011    &        & 1101001 & 00001  & & rd = (float) rs1 \\
fcvt.d.s   & Single Flt to Double & R     & 1010011    & 000    & 0100001 & 00001  & & rd = *((float*) \&rs1) \\
\hline
fmadd.d   & Flt Fused Mul-Add     & R4    & 1000011    &        &        &        & 01 & rd = rs1 * rs2 + rs3 \\
fmsub.d   & Flt Fused Mul-Sub     & R4    & 1000111    &        &        &        & 01 & rd = rs1 * rs2 - rs3 \\
fnmadd.d  & Flt Neg Fused Mul-Add & R4    & 1001011    &        &        &        & 01 & rd = -rs1 * rs2 + rs3 \\
fnmsub.d  & Flt Neg Fused Mul-Sub & R4    & 1001111    &        &        &        & 01 & rd = -rs1 * rs2 - rs3 \\
\hline
\end{tabular}
\end{center}

\newpage
\subsection*{RV32Q Quad-Precision Floating-Point Extension}

\begin{center}
\begin{tabular}
{T | l | c | c | c | c | c | c | T } \hline
\rm Inst  & Name                  & FMT   & \rm Opcode & \rm funct3/rm & \rm funct7 & \rm rs2 [20...24] & \rm fmt & \rm Description (C)         \\ \hline
flq       & Flt Load Word         & I     & 0000111    & 100    &        &        & & rd = M[rs1 + imm]    \\
fsq       & Flt Store Word        & I     & 0100111    & 100    &        &        & & M[rs1 + imm] = rs2   \\
\hline
fadd.q    & Flt Add               & R     & 1010011    &        & 0000011 &        & & rd = rs1 + rs2       \\
fsub.q    & Flt Sub               & R     & 1010011    &        & 0000111 &        & & rd = rs1 - rs2       \\
fmul.q    & Flt Mul               & R     & 1010011    &        & 0001011 &        & & rd = rs1 * rs2       \\
fdiv.q    & Flt Div               & R     & 1010011    &        & 0001111 &        & & rd = rs1 / rs2       \\
fsgnj.q   & Flt Sign Injection    & R     & 1010011    & 000    & 0010011 &        & & rd = abs(rs1) * sgn(rs2) \\
fsgnjn.q  & Flt Sign Neg Injection& R     & 1010011    & 001    & 0010011 &        & & rd = abs(rs1) * -sgn(rs2) \\
fsgnjx.q  & Flt Sign Xor Injection& R     & 1010011    & 010    & 0010011 &        & & rd = rs1 * sgn(rs2)  \\
fmin.q    & Flt Minimum           & R     & 1010011    & 000    & 0010111 &        & & rd = min(rs1, rs2)   \\
fmax.q    & Flt Maximum           & R     & 1010011    & 001    & 0010111 &        & & rd = max(rs1, rs2)   \\
fsqrt.q   & Flt Square Root       & R     & 1010011    &        & 0101111 & 00000  & & rd = sqrt(rs1)       \\
\hline
fle.q     & Float Less / Equal    & R     & 1010011    & 000    & 1010011 &        & & rd = (rs1 <= rs2) ? 1 : 0\\
flt.q     & Float Less Than       & R     & 1010011    & 001    & 1010011 &        & & rd = (rs1 <  rs2) ? 1 : 0 \\
feq.q     & Float Equality        & R     & 1010011    & 010    & 1010011 &        & & rd = (rs1 == rs2) ? 1 : 0 \\
\hline
fcvt.w.q  & Flt Convert to Int    & R     & 1010011    &        & 1100011 & 00000  & & rd = (int32\_t) rs1 \\
fcvt.wu.q & Flt Convert to Int    & R     & 1010011    &        & 1100011 & 00001  & & rd = (uint32\_t) rs1 \\
fcvt.s.q   & Quad Flt to Single   & R     & 1010011    & 000    & 0100000 & 00001  & & rd = *((float*) \&rs1) \\
fcvt.d.q   & Quad Flt to Double   & R     & 1010011    & 000    & 0100001 & 00001  & & rd = *((float*) \&rs1) \\
fmv.x.q   & Move Int to Float     & R     & 1010011    & 000    & 1111011 & 00000  & & rd = *((float*) \&rs1) \\
fclass.q  & Float Classify        & R     & 1010011    & 001    & 1110011 & 00000  & & rd = 0..9 \\
\hline
fcvt.q.w  & Flt Conv from Sign Int& R     & 1010011    &        & 1101011 & 00000  & & rd = (float) rs1 \\
fcvt.q.wu & Flt Conv from Uns Int & R     & 1010011    &        & 1101011 & 00001  & & rd = (float) rs1 \\
fcvt.q.s   & Single Flt to Quad   & R     & 1010011    & 000    & 0100011 & 00000  & & rd = *((float*) \&rs1) \\
fcvt.q.d   & Double Flt to Quad   & R     & 1010011    & 000    & 0100011 & 00001  & & rd = *((float*) \&rs1) \\
\hline
fmadd.q   & Flt Fused Mul-Add     & R4    & 1000011    &        &        &        & 11 & rd = rs1 * rs2 + rs3 \\
fmsub.q   & Flt Fused Mul-Sub     & R4    & 1000111    &        &        &        & 11 & rd = rs1 * rs2 - rs3 \\
fnmadd.q  & Flt Neg Fused Mul-Add & R4    & 1001011    &        &        &        & 11 & rd = -rs1 * rs2 + rs3 \\
fnmsub.q  & Flt Neg Fused Mul-Sub & R4    & 1001111    &        &        &        & 11 & rd = -rs1 * rs2 - rs3 \\
\hline
\end{tabular}
\end{center}

\newpage
\subsection*{RV32C Compressed Extension}

\small
\begin{center}
\begin{tabular}
{p{0in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}p{0.05in}l}
%&
%\multicolumn{1}{l}{\instbit{31}} &
%\multicolumn{1}{r}{\instbit{27}} &
%\instbit{26} &
%\instbit{25} &
%\multicolumn{1}{l}{\instbit{24}} &
%\multicolumn{1}{r}{\instbit{20}} &
%\instbitrange{19}{15} &
%\instbitrange{14}{12} &
%\instbitrange{11}{7} &
%\instbitrange{6}{0} \\
%\cline{2-11}

&
\instbit{15} &
\instbit{14} &
\instbit{13} &
\multicolumn{1}{c}{\instbit{12}} &
\instbit{11} &
\instbit{10} &
\instbit{9} &
\instbit{8} &
\instbit{7} &
\instbit{6} &
\instbit{5} &
\instbit{4} &
\instbit{3} &
\instbit{2} &
\instbit{1} &
\instbit{0} \\
\cline{2-17}

&
\multicolumn{4}{|c|}{funct4} &
\multicolumn{5}{c|}{rd/rs1} &
\multicolumn{5}{c|}{rs2} &
\multicolumn{2}{c|}{op} &
CR-type \\
\cline{2-17}

&
\multicolumn{3}{|c|}{funct3} &
\multicolumn{1}{c|}{imm} &
\multicolumn{5}{c|}{rd/rs1} &
\multicolumn{5}{c|}{imm} &
\multicolumn{2}{c|}{op} &
CI-type \\
\cline{2-17}

&
\multicolumn{3}{|c|}{funct3} &
\multicolumn{6}{c|}{imm} &
\multicolumn{5}{c|}{rs2} &
\multicolumn{2}{c|}{op} &
CSS-type \\
\cline{2-17}

&
\multicolumn{3}{|c|}{funct3} &
\multicolumn{8}{c|}{imm} &
\multicolumn{3}{c|}{rd'} &
\multicolumn{2}{c|}{op} &
CIW-type \\
\cline{2-17}

&
\multicolumn{3}{|c|}{funct3} &
\multicolumn{3}{c|}{imm} &
\multicolumn{3}{c|}{rs1'} &
\multicolumn{2}{c|}{imm} &
\multicolumn{3}{c|}{rd'} &
\multicolumn{2}{c|}{op} &
CL-type \\
\cline{2-17}

&
\multicolumn{3}{|c|}{funct3} &
\multicolumn{3}{c|}{imm} &
\multicolumn{3}{c|}{rd'/rs1'} &
\multicolumn{2}{c|}{imm} &
\multicolumn{3}{c|}{rs2'} &
\multicolumn{2}{c|}{op} &
CS-type \\
\cline{2-17}

&
\multicolumn{3}{|c|}{funct3} &
\multicolumn{3}{c|}{imm} &
\multicolumn{3}{c|}{rs1'} &
\multicolumn{5}{c|}{imm} &
\multicolumn{2}{c|}{op} &
CB-type \\
\cline{2-17}

&
\multicolumn{3}{|c|}{funct3} &
\multicolumn{11}{c|}{offset} &
\multicolumn{2}{c|}{op} &
CJ-type \\
\cline{2-17}

\end{tabular}
\end{center}

\begin{center}
\begin{tabular}
{T | l | c | c | c | c | c | c | c | T } \hline
\rm Inst   & Name                    & FMT & \rm OP & \rm funct4 & \rm funct3 & \rm [12] & \rm [11..10] & \rm [6..5] & \rm Description          \\ \hline
c.addi4spn & ADD Imm * 4 + SP        & CIW & 00     & & 000 & & &       & addi rd', sp, 4*imm      \\
c.fld      & Flt Load Double         & CL  & 00     & & 001 & & &       & flw rd', (8*imm)(rs1')   \\
c.lw       & Load Word               & CL  & 00     & & 010 & & &       & lw rd', (4*imm)(rs1')    \\
%c.ld       & Load Double             & CL  & 00     & & 011 & & &       & ld rd', (8*imm)(rs1')    \\
%c.lq       & Load Quad               & CL  & 00     & & 001 & & &       & lq rd', (16*imm)(rs1')   \\
c.flw      & Flt Load Word           & CL  & 00     & & 011 & & &       & flw rd', (4*imm)(rs1')   \\
c.fsd      & Flt Store Double        & CS  & 00     & & 101 & & &       & fsd rs1', (8*imm)(rs2')  \\
c.sw       & Store Word              & CS  & 00     & & 110 & & &       & sw rs1', (4*imm)(rs2')    \\
%c.sd       & Store Double            & CS  & 00     & & 111 & & &       & sd rs1', (8*imm)(rs2')   \\
%c.sq       & Store Quad              & CS  & 00     & & 101 & & &       & sq rs1', (16*imm)(rs2')  \\
c.fsw      & Flt Store Word          & CS  & 00     & & 111 & & &       & fsw rs1', (4*imm)(rs2')  \\
\hline
c.addi     & ADD Immediate           & CI  & 01     & & 000 & & &       & addi rd, rd, imm         \\
%c.addiw    & ADD Immediate Word      & CI  & 01     & & 001 & & &       & addiw rd, rd, imm        \\
c.jal      & Jump And Link           & CJ  & 01     & & 001 & & &       & jal ra, 2*offset         \\
c.li       & Load Immediate          & CI  & 01     & & 010 & & &       & addi rd, x0, imm         \\
c.lui      & Load Upper Imm          & CI  & 01     & & 011 & & &       & lui rd, imm              \\
c.addi16sp & ADD Imm * 16 to SP      & CI  & 01     & & 011 & & &       & addi sp, sp, 16*imm      \\
c.srli     & Shift Right Logical Imm & CB  & 01     & & 100 & & 00 & & srli rd', rd', imm       \\
c.srai     & Shift Right Arith Imm   & CB  & 01     & & 100 & & 01 & & srai rd', rd', imm       \\
c.andi     & AND Imm                 & CB  & 01     & & 100 & & 10 & & andi rd', rd', imm       \\
c.sub      & SUB                     & CS  & 01     & & 100 & 0 & 11 & 00 & sub rd', rd', rs2'       \\
c.xor      & XOR                     & CS  & 01     & & 100 & 0 & 11 & 01 & xor rd', rd', rs2'       \\
c.or       & OR                      & CS  & 01     & & 100 & 0 & 11 & 10 & or rd', rd', rs2'        \\
c.and      & AND                     & CS  & 01     & & 100 & 0 & 11 & 11 & and rd', rd', rs2'       \\
c.subw     & SUB Word                & CS  & 01     & & 100 & 1 & 11 & 00 & subw rd', rd', rs2'      \\
c.addw     & ADD Word                & CS  & 01     & & 100 & 1 & 11 & 01 & addw rd', rd', rs2'      \\
c.j        & Jump                    & CJ  & 01     & & 101 & & &       & jal x0, 2*offset         \\
c.beqz     & Branch == 0             & CB  & 01     & & 110 & & &       & beq rs', x0, 2*imm       \\
c.bnez     & Branch != 0             & CB  & 01     & & 111 & & &       & bne rs', x0, 2*imm       \\
%c.nop      & No OPeration            & CI  & 01     & & 000 & & &       & addi x0, x0, 0           \\
\hline
c.slli     & Shift Left Logical Imm  & CI  & 10     & & 000 & & &       & slli rd, rd, imm         \\
c.lwsp     & Load Word from SP       & CI  & 10     & & 010 & & &       & lw rd, (4*imm)(sp)       \\
c.fldsp    & Flt Load Double from SP & CI  & 10     & & 001 & & &       & flw rd, (8*imm)(sp)      \\
c.flwsp    & Flt Load Word from SP   & CI  & 10     & & 011 & & &       & flw rd, (4*imm)(sp)      \\
c.fswsp    & Flt Store Word to SP    & CSS & 10     & & 111 & & &       & fsw rs2, (4*imm)(sp)     \\
c.mv       & MoVe                    & CR  & 10     & 1000 & & & &      & add rd, x0, rs2          \\
c.add      & ADD                     & CR  & 10     & 1001 & & & &      & add rd, rd, rs2          \\
c.fsdsp    & Flt Store Double to SP  & CSS & 10     & & 101 & & &       & fsd rs2, (8*imm)(sp)     \\
c.swsp     & Store Word to SP        & CSS & 10     & & 110 & & &       & sw rs2, (4*imm)(sp)      \\
%c.ldsp     & Load Double from SP     & CI  & 10     & & 011 & & &       & ld rd, (8*imm)(sp)       \\
%c.lqsp     & Load Quad from SP       & CI  & 10     & & 001 & & &       & lq rd, (16*imm)(sp)      \\
%c.sdsp     & Store Double to SP      & CSS & 10     & & 111 & & &       & sd rs2, (8*imm)(sp)      \\
%c.sqsp     & Store Quad to SP        & CSS & 10     & & 101 & & &       & sq rs2, (16*imm)(sp)     \\
%c.jr       & Jump Reg                & CR  & 10     & 1000 & & & &      & jalr x0, rs1, 0          \\
%c.jalr     & Jump And Link Reg       & CR  & 10     & 1001 & & & &      & jalr ra, rs1, 0          \\
%c.ebreak   & Environment BREAK       & CR  & 10     & 1001 & & & &      & ebreak                   \\
\hline
\end{tabular}
\end{center}


\section{Technique}

This scheme is beneficial in cases with a high ratio of communication to computation and a non-negligible overhead associated with sending messages.

\begin{figure}
  \centering
  \input{fig/lightcone}
  \caption{Space-Time Diagram}
  \label{fig:lightcone}
\end{figure}

\begin{figure}
  \centering
  \input{fig/conventional_trans_1d.tex}
  \caption{Conventional 1D Boundary Exchange}
  \label{fig:exch1d}
\end{figure}

\begin{figure}
  \centering
  \input{fig/cascade_trans_1d.tex}
  \caption{Cascade 1D Boundary Exchange}
  \label{fig:casc1d}
\end{figure}

\subsection{Streaming Communications}
\begin{figure}
  \centering
  \input{fig/simcone}
  \caption{Simulation Dependencies}
  \label{fig:simcone}
\end{figure}

\begin{figure}
  \booltrue{commstl}
  \booltrue{commsbl}
  \centering
  \input{fig/comms}
  \caption{Streaming Communications Pattern}
  \label{fig:streamingcomms}
\end{figure}

\begin{figure}
  \centering
  \input{fig/ssa_concept}
  \caption{SSA Architecture}
  \label{fig:ssa_concept}
\end{figure}

\begin{table}
  \scriptsize
  \centering
  \caption{Stencil Code Architecture Comparrison}
  \label{tab:comparrison}
  \input{tab/comparrison.tex}
\end{table}

\subsection{Alternating Communications}

Requires half-duplex communications links between processing elements.
\begin{figure}
  \booltrue{commstr}
  \booltrue{commsbl}
  \centering
  \input{fig/comms}
  \caption{Alternating Communications Pattern}
  \label{fig:alternatingcomms}
\end{figure}

\subsection{Communication Overheads}
The cascade communication scheme introduces additional overheads relative to a conventional system. $2wrh$.
Firstly, the imposition of a staged pattern of communication (North then East in reference that figure)
That said, this pattern is common in conventional decompositions with $r>1$.

\subsection{Instruction Pipelining}

In the

Floating point operations take multiple clock cycles to complete.
To hide this latency, floating point hardware has been designed in a pipelined fashion, allowing multiple floating point operations to be executed concurrently.

\begin{figure}
  \centering
  \input{fig/cascade_placement}
  \caption{Data Movement in the Cascade architecture}
  \label{fig:cascade_placement}
\end{figure}

\subsection{Hybrid Schemes}
In one dimension, CASTLE can be thought of as a variant of a block-based version of SSA in which vectors rather than cells are exchanged.


\documentclass{article} %[11pt]{amsart}
\usepackage{geometry}
\geometry{letterpaper}
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{siunitx}
\usepackage{tikz}
\usepackage{lscape}
\usepackage{pgfplots}
\usetikzlibrary{
  arrows,
  calc,
  decorations.markings,
  decorations.pathreplacing,
  dsp,
  fit,
  positioning
}
\input{../util/airfoils.tex}
\input{../util/wing.tex}
\input{../util/coordinate_systems.tex}
\input{../util/control.tex}

\pgfplotsset{grid style={dashed,gray!25}}

\newcommand{\qhat}{\hat{q}}
\newcommand{\cbar}{\bar{c}}
\newcommand{\qbar}{\bar{q}}
\newcommand{\cmd}{\mathrm{cmd}}
\newcommand{\ff}{\mathrm{ff}}
\newcommand{\eff}{\mathrm{eff}}
\newcommand{\app}{\mathrm{app}}
\newcommand{\wind}{\mathrm{wind}}
\newcommand{\grav}{\mathrm{grav}}
\newcommand{\kite}{\mathrm{kite}}
\newcommand{\nom}{\mathrm{nom}}
\newcommand{\aero}{\mathrm{aero}}
\newcommand{\geom}{\mathrm{geom}}
\newcommand{\cw}{\mathrm{cw}}

\begin{document}

\section{Curvature from force balance}

\begin{figure}[h]
\begin{center}
  \begin{tikzpicture}[scale=0.75]
    \begin{scope}[shift={(-5, 0)}, scale=0.3, rotate=-10]
      \DrawWingFront[]
      \draw[line width=1.5pt, -latex] (4, 1) -- (9, 1) node[midway, above] {$-Y$};
      \draw[line width=1.5pt, -latex] (0, 5) -- (0, 12) node[midway, right] {$L$};
      \draw (-5.86, -0.13) -- (0, -4.92);
      \draw (5.86, -0.13) -- (0, -4.92);
      \draw[dashed] (0, -4.92) -- (0, -15);
    \end{scope}
    \node at (-5, -3) {$\phi_t$};
    \draw[line width=1.5pt, -latex] (-5.25, -1.45) --
    (-3.5, -4.3) node[midway, above right] {$t$};
    \draw[line width=1.5pt, -latex] (-3.5, -1) -- (-2, -2)
    node[midway, below left] {$W$};
    \draw (-5.25, -1.45) -- (0, -10) node[midway, above right] {$l_t$};
    %\draw[dashed] (-5.25, -1.45) -- (-0.975, 1.175);
    \node at (-0.5, -8.5) {$\gamma$};
    \draw[dashed] (0, -10) -- (0, -5);
    %\draw[dashed] (-5, 0) -- (-10, 0) node[near end, above] {$-\phi$};
    %\draw[line width=1.5pt, -latex] (-7, 0) -- (-9, 0) node[midway, below]
    %{$mv^2/R$};
    \DrawCoordinateSystem{shift={(0, -10)}, scale=1}
                         {$x_{\cw}$}{$y_{\cw}$}{$z_{\cw}$}
  \end{tikzpicture}
  \caption{Diagram of force balance in the stability axes' y-z plane.}
\end{center}
\end{figure}

By balancing forces in the stability axes' y-z plane, it is possible
to relate the curvature of a kite's flight path to a few known
parameters of the kite (e.g. lift coefficient, wing area, and mass)
and the easily measured roll angle between the kite and the tether.
This implies that it is possible to steer the kite simply by
controlling the tether roll angle.  Indeed, this is the approach taken
in the current flight controller.  Here an appoximate relationship
between tether roll angle and curvature is derived.  To simplify the
relationship, the tether is appoximated as a straight line from the
ground-station to the bridle point.  Other effects on the tether such
as the catenary from gravity are ignored.

It is easiest to conduct the force balance along axes that are
parallel and perpendicular to the tether.  Because the wing is forced
to fly on the surface of a sphere with radius equal to the tether
length, $l_t$, the acceleration parallel to the tether is
$a_{\parallel} = v_i^2/l_t$.  The lift, pylon side force, and weight
each have components along the direction perpendicular to the tether.
Together, these create an acceleration perpendicular to the tether
given by
\begin{equation}
m_{\eff} a_{\bot} = L \sin \phi_t - Y \cos \phi_t -
W_{\cw, x} \cos \gamma - W_{\cw, z} \sin \gamma
\end{equation}
Here $m_{\eff}$ is the effective mass of the kite, which is given by
$m_{\eff} = m_{\kite} + m_{\mathrm{tether}} / 3$.  The extra term from
the tether mass comes from calculating the acceleration of a point
mass at the end of a rigid rod due to a force applied at the point
mass.  Also, a few small approximations were made such as ignoring
forces from motor thrust, ignoring the component of drag along the
stability z-axis, and ignoring the cosine term from projecting the
lift vector onto the stability z-axis.

The curvature and thus acceleration that is relevant for control is
that in the crosswind flight plane.  The parallel and perpendicular
accelerations can be combined, projected onto the crosswind $x$-axis,
and converted to a curvature as follows:
\begin{equation}
\kappa_{\cw} = \frac{1}{l_t} \sin \gamma +
\frac{\rho A v_{\app}^2}{2 m_{\eff} v_i^2}
\left(C_L \sin \phi_t - C_Y \cos \phi_t \right) \cos \gamma -
\frac{m}{m_{\eff} v_i^2}
(g_{\cw,x} \cos \gamma + g_{\cw,z} \sin \gamma) \cos \gamma
\end{equation}
%
It is somtimes useful to decompose this total, in-plane curvature into
components caused by the tether sphere, aerodynamic angles, wind, and
gravity:
\begin{equation}
\kappa_{\cw} = \kappa_0 + \kappa_{\aero} + \kappa_{\wind} + \kappa_{\grav}
\end{equation}
%
where
%
\begin{eqnarray}
\kappa_0 &=& \frac{1}{l_t} \sin \gamma \\
\kappa_{\aero} &=&
\frac{\rho A}{2 m_{\eff}} \left(C_L \sin \phi_t - C_Y \cos \phi_t \right) \cos \gamma \\
\kappa_{\wind} &=&
\left( \frac{v_w^2 - 2 \vec{v}_i \cdot \vec{v}_w}{v_i^2} \right) \kappa_{\aero} \\
\kappa_{\grav} &=&
-\frac{m}{m_{\eff} v_i^2}
(g_{\cw,x} \cos \gamma + g_{\cw,z} \sin \gamma) \cos \gamma
\end{eqnarray}

\section{Curvature mixer}

The curvature mixer takes as input the curvature command,
$\kappa_{\cmd}$, and a component of the curvature command due to lift
and tether roll, $\kappa^{\aero}_{\cmd}$.  It returns the inner loop
incidence and tether angle commands and angular rate commands that
will generate these curvatures.

\begin{equation}
[\alpha_{\cmd},\; \Delta {C_L}_{\cmd},\; \beta_{\cmd},\; {\phi_t}_{\cmd}]^T
= F(\kappa_{\cmd}^{\aero})
\end{equation}

\begin{equation}
\kappa_{\cw} \approx v \cdot |\vec{\omega}|
\end{equation}

\begin{equation}
\kappa^{\cw} = \frac{|\vec{a}_b \times \vec{v}_b|}{|\vec{v}_b|^3}
\end{equation}

\begin{equation}
[p_{\cmd},\; r_{\cmd}]^T = G(\kappa_{\cmd}^{\geom})
\end{equation}

\begin{eqnarray}
\alpha_{\cmd} &=& \alpha_{\nom} \\
\Delta {C_L}_{\cmd} &=& 0 \\
\beta_{\cmd} &=& 0 \\
{\phi_t}_{\cmd} &=& \sin^{-1} \left(
\frac{2 m \kappa_{\cmd}^{\aero}}{\rho A {C_L}_{\nom}} +
\frac{C_{Y_0}}{{C_L}_{\nom}} \right)
\end{eqnarray}

\end{document}


\section{Optimization of Curry Programs}

After the invocation of the Curry front end,
which parses a Curry program and translates it into the intermediate FlatCurry
representation, \CYS applies a transformation
to optimize Boolean equalities occurring in the Curry program.
The ideas and details of this optimization are described
in \cite{AntoyHanus15LOPSTR}.
Therefore, we sketch only some basic ideas and options
to influence this optimization.

Consider the following definition of the operation \code{last}
to extract the last element in list:
%
\begin{curry}
last xs | xs == _++[x]
        = x
 where x free
\end{curry}
%
In order to evaluate the condition \ccode{xs == \us{}++[x]},
the Boolean equality is evaluated to \code{True} or \code{False}
by instantiating the free variables \code{\us} and \code{x}.
However, since we know that a condition must be evaluated to
\code{True} only and all evaluations to \code{False} can be ignored,
we can use the constrained equality to obtain a more efficient program:
%
\begin{curry}
last xs | xs =:= _++[x]
        = x
 where x free
\end{curry}
%
Since the selection of the appropriate equality operator
is not obvious and might be tedious, \CYS encourages
programmers to use only the Boolean equality operator \ccode{==}
in programs.
The constraint equality operator \ccode{=:=} can be considered
as an optimization of \ccode{==} if it is ensured that only
positive results are required, e.g., in conditions of program rules.

To support this programming style, \CYS has a built-in optimization phase
on FlatCurry files. For this purpose, the optimizer analyzes
the FlatCurry programs for occurrences of \ccode{==}
and replaces them by \ccode{=:=} whenever the result \code{False}
is not required.
The usage of the optimizer can be influenced by setting
the property flag \code{bindingoptimization} in the
configuration file \code{\curryrc}.
The following values are recognized for this flag:
\begin{description}
\item[\code{no}:] Do not apply this transformation.
\item[\code{fast}:] This is the default value.
The transformation is based on pre-computed values for
the prelude operations in order to decide whether the
value \code{False} is not required as a result of a Boolean equality.
Hence, the transformation can be efficiently performed
without any complex analysis.
\item[\code{full}:] Perform a complete ``required values'' analysis
of the program (see \cite{AntoyHanus15LOPSTR})
and use this information to optimize programs.
In most cases, this does not yield better results so that
the \code{fast} mode is sufficient.
\end{description}
%
Hence, to turn off this optimization, one can either modify
the flag \code{bindingoptimization} in the
configuration file \code{\curryrc} or dynamically pass this change
to the invocation of \CYS by
\begin{quote}
\ldots{} \code{-Dbindingoptimization=no} \ldots
\end{quote}


\documentclass[12pt]{cdblatex}
\usepackage{bssn-eqtns}

\begin{document}

\section*{PhysRevD.62.044034 equation (14)}

The advice given by Miguel Alcubierre, Bernd Brugmann etal (Phys Rev D (67) 084023,
2nd-3rd paragraph on pg. 084023-4)

\begin{quote}
... if one wants to achieve numerical stability. In the computer code we do not use the numerically
evolved ${\bar{\Gamma}}^i$ in all places, but we follow this rule:

Partial derivatives $\partial_j {\bar{\Gamma}}^i$ are computed as finite differences
of the independent variables ${\bar{\Gamma}}^i$ that are evolved using ...
\end{quote}

The Einstein Toolkit code uses the same rule -- the only place where the \emph{evolved} ${\bar{\Gamma}}^i$
are used is in computing the $\partial_j {\bar{\Gamma}}^i$ terms in the equation
for ${\bar{R}}_{ij}$, that is equation (18) of the Phys Rev D (62) 044034 paper.

\clearpage

\begin{cadabra}
   from shared import *
   import cdblib

   jsonfile = 'bssn-eqtns-14.json'
   cdblib.create (jsonfile)

   # --------------------------------------------------------------------------

   Rphi := -2 DBar_{a b}{\phi} - 2 gBar_{a b} gBar^{c d} DBar_{c d}{\phi}
           +4 DBar_{a}{\phi} DBar_{b}{\phi} - 4 gBar_{a b} gBar^{c d} DBar_{c}{\phi} DBar_{d}{\phi}.

                                                           # cdb(eq15.prd,Rphi)

   RBar := - (1/2) gBar^{l m} \partial_{l m}{gBar_{a b}}
           + (1/2) gBar_{k a} \partial_{b}{GammaBar^{k}}
           + (1/2) gBar_{k b} \partial_{a}{GammaBar^{k}}
           + (1/2) GammaBar^{k} GammaBar_{a b k}
           + (1/2) GammaBar^{k} GammaBar_{b a k}
           + gBar^{l m} gBar^{k e} (  GammaBar_{e l a} GammaBar_{b k m}
                                    + GammaBar_{e l b} GammaBar_{a k m}
                                    + GammaBar_{k a m} GammaBar_{e l b}).

                                                           # cdb(eq18.prd,RBar)

   defRab := R_{a b} -> @(Rphi) + @(RBar).

   Rab := RBar_{a b} + Rphi_{a b}.                         # cdb(eq14.01,Rab)
   Rab := R_{a b}.                                         # cdb(eq14.00,Rab)

   substitute (Rab, defRab)                                # cdb(eq14.02,Rab)
   substitute (Rab, defDBar1)                              # cdb(eq14.03,Rab)
   substitute (Rab, defDBar2)                              # cdb(eq14.04,Rab)
   substitute (Rab, defGamma2GammaBar)                     # cdb(eq14.05,Rab)
   distribute (Rab)                                        # cdb(eq14.06,Rab)
   eliminate_kronecker (Rab)                               # cdb(eq14.07,Rab)

   Rab = product_sort (Rab)                                # cdb(eq14.08,Rab)

   rename_dummies (Rab)                                    # cdb(eq14.09,Rab)
   canonicalise   (Rab)                                    # cdb(eq14.10,Rab)

   foo := GammaBar^{a} GammaBar_{b c a} -> gBar^{d e} GammaBar^{a}_{d e} GammaBar_{b c a}.

   substitute (Rab, foo)                                   # cdb(eq14.11,Rab)
   substitute (Rab, defGBarSq)                             # cdb(eq14.12,Rab)
   substitute (Rab, defGammaBarD)                          # cdb(eq14.13,Rab)
   substitute (Rab, defGammaBarU)                          # cdb(eq14.14,Rab)
   distribute (Rab)                                        # cdb(eq14.15,Rab)

   foo := \partial_{a}{gBar_{b c}} gBar^{b c} -> 0.   # follows from det(g) = 1

   substitute   (Rab,foo)                                  # cdb(eq14.16,Rab)
   canonicalise (Rab)                                      # cdb(eq14.17,Rab)

   foo := gBar^{b e} gBar^{c f} \partial_{a}{gBar_{b c}}  -> - \partial_{a}{gBar^{e f}}.
   bah := gBar^{e b} gBar^{f c} \partial_{a}{gBar_{b c}}  -> - \partial_{a}{gBar^{e f}}.
   moo := gBar^{e b} gBar^{c f} \partial_{a}{gBar_{b c}}  -> - \partial_{a}{gBar^{e f}}.

   substitute (Rab,foo)                                    # cdb(eq14.18,Rab)
   substitute (Rab,bah)                                    # cdb(eq14.19,Rab)
   substitute (Rab,moo)                                    # cdb(eq14.20,Rab)

   Rab = product_sort (Rab)                                # cdb(eq14.21,Rab)
                                                           # cdb(eq14.99,Rab)

   defRab := R_{a b} -> @(Rab).   # used later in bssn-ricci-scalar.tex

   cdblib.put ('Rab',Rab,jsonfile)
   cdblib.put ('defRab',defRab,jsonfile)
\end{cadabra}

\clearpage

\begin{dgroup*}
   \begin{dmath*}
      \cdb{eq14.00} = \Cdb*{eq14.01}
                    = \Cdb*{eq14.02}
                    = \Cdb*{eq14.03}
                    = \Cdb*{eq14.04}
                    = \Cdb*{eq14.05}
                    = \Cdb*{eq14.06}
                    = \Cdb*[\hskip 2cm\hfill]{eq14.07}
                    = \Cdb*{eq14.08}
                    = \Cdb*{eq14.09}
                    = \Cdb*{eq14.10}
   \end{dmath*}
\end{dgroup*}

\clearpage

\begin{dgroup*}
   \begin{dmath*}
      \cdb{eq14.00} = \Cdb*{eq14.11}
                    = \Cdb*{eq14.12}
                    = \Cdb*{eq14.13}
                    = \Cdb*[\hskip 2cm\hfill]{eq14.14}
                    = \Cdb*{eq14.15}
   \end{dmath*}
\end{dgroup*}

\clearpage

\begin{dgroup*}
   \begin{dmath*}
      \cdb{eq14.00} = \Cdb*[\hskip 2cm\hfill]{eq14.16}
                    = \Cdb*[\hskip 2cm\hfill]{eq14.17}
                    = \Cdb*{eq14.18}
                    = \Cdb*{eq14.19}
                    = \Cdb*{eq14.20}
                    = \Cdb*{eq14.21}
   \end{dmath*}
\end{dgroup*}

\clearpage

\def\gBar{{\bar{g}}}
\def\dgBar#1{{\partial_{#1}}\gBar}

There is a single term in this final expression that appears to be neither symmetric in $ab$ nor part of a symmetric pair, namely
\begin{gather*}
   \dgBar{b}_{cd} \dgBar{a}^{cd}
\end{gather*}
It is, however, easy to show that this term is symmetric in $ab$. Start by noting that, for any $\gBar_{ab}$,
\begin{align*}
   \dgBar{a}^{cd} = -\gBar^{ce}\gBar^{df}\dgBar{a}_{ef}
\end{align*}
Now contract both sides with $\dgBar{b}_{cd}$ to obtain
\begin{align*}
   \dgBar{a}^{cd}\dgBar{b}_{cd} = -\gBar^{ce}\gBar^{df}\dgBar{a}_{ef}\dgBar{b}_{cd}
\end{align*}
The right hand side is clearly symmetric in $ab$ and thus the left hand must also be symmtric in $ab$.

\end{document}


\documentclass{article}

\usepackage[preprint]{./template/neurips_2020}

\usepackage[utf8]{inputenc} % allow utf-8 input
\usepackage[T1]{fontenc}    % use 8-bit T1 fonts
\usepackage{hyperref}       % hyperlinks
\usepackage{url}            % simple URL typesetting
\usepackage{booktabs}       % professional-quality tables
\usepackage{amsfonts}       % blackboard math symbols
\usepackage{nicefrac}       % compact symbols for 1/2, etc.
\usepackage{microtype}      % microtypography
\usepackage{physics}
\usepackage{amsmath}

\title{Chapter 1: Introduction and Overview}

\author{
  John Martinez \\
  \texttt{john.r.martinez14@gmail.com} \\
  % examples of more authors
  % \And
  % Coauthor \\
  % Affiliation \\
  % Address \\
  % \texttt{email} \\
  % \AND
  % Coauthor \\
  % Affiliation \\
  % Address \\
  % \texttt{email} \\
  % \And
  % Coauthor \\
  % Affiliation \\
  % Address \\
  % \texttt{email} \\
  % \And
  % Coauthor \\
  % Affiliation \\
  % Address \\
  % \texttt{email} \\
}

\begin{document}
\maketitle

%\begin{abstract}
%  The abstract paragraph should be indented \nicefrac{1}{2}~inch (3~picas) on
%  both the left- and right-hand margins. Use 10~point type, with a vertical
%  spacing (leading) of 11~points.  The word \textbf{Abstract} must be centered,
%  bold, and in point size 12. Two line spaces precede the abstract. The abstract
%  must be limited to one paragraph.
%\end{abstract}

\section{Introduction}

%%% Section 1.1
\subsection{Global Perspective}
n/a

%%% Section 1.2
\subsection{Quantum Bits}
What is a qubit? Just as a classical bit has a \emph{state} - either $0$ or $1$
- a qubit also has a state. Two possible states for a qubit are the states
$\ket{0}$ and $\ket{1}$, which as you might guess corresponds to the states
$0$ and $1$ for a classical bit. The difference between bits and qubits is
that a qubit can be in a state \emph{other} than $\ket{0}$ or $\ket{1}$. It
is also possible to form \emph{a linear combination} of states, often
called \emph{superpositions}:

\begin{center}
  $\ket{\psi} = \alpha\ket{0} + \beta\ket{1}$
\end{center}

Where $\alpha, \beta \in \mathbb{C}$.  Unfortunately, we cannot examine a qubit
to determine its quantum state, that is, the values of $\alpha$ and
$\beta$.  Instead, quantum mechanics tells us that we can only aquire
much more restricted information about the quantum state. When we measure a
qubit we get either the result $0$, with probability $\abs{\alpha}^{2}$, or the
result $1$, with probability $\abs{\beta}^{2}$ with
$\abs{\alpha}^{2} + \abs{\beta}^{2} = 1$, since probabilities must sum to $1$.

This equation can also be represented as

\begin{center}
  $\ket{\psi} = \cos(\frac{\theta}{2})\ket{0} + e^{i\phi}\sin(\frac{\theta}{2})\ket{1}$
\end{center}

%%% Section 1.2.1
\subsubsection{Multiple Qubits}
A two qubit system will have four \emph{computational basis states} denoted
$\ket{00}, \ket{01}, \ket{10}, \ket{11}$. The state vector describing two qubits is

\begin{center}
  $\ket{\psi} = \alpha_{00}\ket{00} + \alpha_{01}\ket{01} + \alpha_{10}\ket{10} + \alpha_{11}\ket{11}$
\end{center}
And again, the probabilities must sum to one, satisfying the condition
$\sum_{x \in \{0, 1\}^{2}}{\abs{\alpha_{x}^{2}} = 1}$, where the notation
'$x \in \{0, 1\}^{2}$' means 'the set of strings of length two with each letter
being either zero or one'.  Subsets of multi-qubit systems can be measured.
Measuring the first qubit of the above two qubit system gives $0$ with
probability $\abs{\alpha_{00}^{2}} + \abs{\alpha_{01}^{2}}$. The
post-measurement state would then be

\begin{center}
    $\ket{\psi'} = \frac{\alpha_{00}\ket{00} + \alpha_{01}\ket{01} }{\sqrt{\abs{\alpha_{00}^{2}} + \abs{\alpha_{01}^{2}}}}$
\end{center}

An important two qubit state is the \emph{Bell state} or \emph{EPR pair},

\begin{center}
    $\frac{\ket{00} + \ket{11}}{\sqrt{2}}$
\end{center}

Bell states are important because their measurement outcomes are
\emph{correlated}.

%%% Section 1.3
\subsection{Quantum Computation}

%%% Section 1.3.1
\subsubsection{Single Qubit Gates}
The quantum $\mathbf{NOT}$ gate acts \emph{linearly}.
\begin{center}
  $\mathit{X} \equiv \begin{bmatrix}
     1 & 0 \\
     0 & 1
   \end{bmatrix}$
\end{center}
If we write the quantum state $\ket{\psi} = \alpha\ket{0} + \beta\ket{1}$ in
vector notation
\begin{center}
    $\begin{bmatrix}
       \alpha \\
       \beta
     \end{bmatrix}$
\end{center}
then the $\mathbf{NOT}$ gate operation can be represented as
\begin{center}
  $\mathit{X} \begin{bmatrix}
     \alpha \\
     \beta
   \end{bmatrix} = \begin{bmatrix}
     \beta \\
     \alpha
   \end{bmatrix}$
\end{center}
There is a restriction on the the type of matrix that can be a quantum gate and
that is that the operation must preserve the normalization condition (i.e.
$\abs{\alpha}^{2} + \abs{\beta}^{2} = 1$). The appropriate condition on the
matrix representing the gate is that the matrix $\mathit{U}$ describing the
single qubit gate must be \emph{unitary}, that is
$\mathit{U}^{\dagger}\mathit{U} = \mathit{I}$ (where $\mathit{X}^{\dagger}$
is the transpose of the complex conjugate of $\mathit{X}$).

Other important single qubit quantum gates are the $\mathit{Z}$ gate
\begin{center}
  $\mathit{Z} \equiv \begin{bmatrix}
    1 & 0 \\
    0 & -1
  \end{bmatrix}$
\end{center}
which leaves $\ket{0}$ unchanged and flips the sign of $\ket{1}$, and the
\emph{Hadamard} gate
\begin{center}
  $\mathit{H} \equiv \frac{1}{\sqrt{2}}\begin{bmatrix}
    1 & 1 \\
    1 & -1
  \end{bmatrix}$
\end{center}
The Hadamard operation is just a rotation of the sphere about the $\hat{y}$
axis by $90^{\circ}$, followed by a rotation about the $\hat{x}$ axis by
$180^{\circ}$.

%%% Section 1.3.2
\subsubsection{Multiple Qubit Gates}
The prototypical multi-qubit logic gate is the \emph{controlled}-$\mathbf{NOT}$
or $\mathbf{CNOT}$ gate.

\begin{center}
  $ \mathit{U_{\mathit{CN}}} \equiv \begin{bmatrix}
    1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 0 & 1 \\
    0 & 0 & 1 & 0
  \end{bmatrix}$
\end{center}
The gate has two inputs, a \emph{control} qubit and a \emph{target} qubit. If
the control qubit is set to $0$, then the target qubit is left alone. If the
control qubit is set to $1$, then the target qubit is flipped.

\begin{center}
  $\ket{00} \to \ket{00}$;
  $\ket{01} \to \ket{01}$;
  $\ket{10} \to \ket{11}$;
  $\ket{11} \to \ket{10}$;
\end{center}

%%% Section 1.3.3
\subsubsection{Measurement in non-standard basis}
The states $\ket{0}$ represent just one of many possible choices of basis
states for a qubit. Another possible choice is

\begin{center}
  $\ket{+} \equiv \frac{(\ket{0} + \ket{1})}{\sqrt{2}} $ and
  $\ket{-} \equiv \frac{(\ket{0} - \ket{1})}{\sqrt{2}} $
\end{center}

%%% Section 1.3.4
\subsubsection{Quantum Circuits?}
%%% Need to learn how to draw circuits in LaTex

%%% Section 1.3.5
\subsubsection{Copying Circuit?}

%%% Section 1.3.6
\subsubsection{Example: Bell States}

\begin{center}
  $\ket{\beta_{00}} = \frac{\ket{00} + \ket{11}}{\sqrt 2}$;

  $\ket{\beta_{01}} = \frac{\ket{01} + \ket{10}}{\sqrt 2}$;

  $\ket{\beta_{10}} = \frac{\ket{00} - \ket{11}}{\sqrt 2}$;

  $\ket{\beta_{11}} = \frac{\ket{01} - \ket{10}}{\sqrt 2}$;
\end{center}
These are known as the \emph{Bell states}, or \emph{EPR states or pairs}.

%%% Section 1.4
\subsection{Quantum Algorithms}

%%% Section 1.4.1
\subsubsection{Classical Computations on a Quantum Computer?}

%%% Section 1.4.2
\subsubsection{Quantum Parallelism}
Suppose $f(x) : \{0, 1\} \to \{0, 1\}$ is a function with a one-bit domain
and range. A convenient way of computing this function on a quantum computer is
to consider a two qubit quantum computer which starts in the state $\ket{x, y}$.
It is possible with a sequence of gates to turn this into the state
$U_f : \ket{x, y} \to \ket{x, y \oplus f(x)}$.  In particular, if we apply
the Hadamard gate to
$\mathit{H} \ket{0}$ we obtain $\frac{\ket{0} + \ket{1}}{\sqrt 2}$. Now
applying
$U_f$ we get $\frac{\ket{y, y \oplus f(0)} + \ket{y, y \oplus f(1)}}{\sqrt 2}$.
And now letting $y = \ket{0}$, we get
$\frac{\ket{0, f(0)} + \ket{0, f(1)}}{\sqrt 2}$.  This is a remarkable state,
in that we have information about $f(0)$ and $f(1)$ simultaneously!

This procedure can me generalized to functions on an arbitrary number of bits,
by using the \emph{Walsh-Hadamard transform}. For $n = 2$ qubits we get

\begin{center}
  $\frac{\ket{0} + \ket{1}}{\sqrt 2}\frac{\ket{0} + \ket{1}}{\sqrt 2} =
   \frac{\ket{00} + \ket{01} + \ket{10} + \ket{11}}{2}$
\end{center}
We can write $\mathit{H}^{\otimes 2}$ to denote the parallel action of two
Hadamard gates. More generally, the result of performing the Walsh-Hadamard
transform on $n$ qubits initially in the $\ket{0}$ state is

\begin{center}
  $\frac{1}{\sqrt{2^{n}}}\displaystyle\sum_{x} \ket{x}$
\end{center}
And we write $\mathit{H}^{\otimes n}$ to denote this operation.

% \begin{center}
% \end{center}

\end{document}