\chapter{Central force motions}
It was Newton's fascination with planetary motion that led him to formulate his laws of motion and the law of universal gravitation. His success in explaining Kepler's empirical laws of planetary motion was an overwhelming argument in favor of the new mechanics and marked the beginning of modern mathematical physics. Planetary motion and the more general problem of motion under a central force continue to play an important role in most branches of physics and turn up in such topics as particle scattering, atomic structure, and space navigation.
\section{Central force}
A central force  acting on a particle only depends upon magnitude of a distance from a fixed center. If $r$ is the instanteneous position vector of the particle relative to the fixed center. Then the central force is represented by the relation,
\begin{equation}
F= f(r) \hat{r}
\end{equation}
Where, $f(r)$ is a scalar function of distance, $r$  and $\hat{r}=\frac{\vec{r}}{r}$. Then the torque acting on the particle is,
\begin{equation}
\tau =r\times F
\end{equation}
This type of motion is particularly relevant when studying the orbital movement of planets and satellites. The laws which govern this motion were first postulated by Kepler and deduced from observation. In this lecture, we will see that these laws are a consequence of Newton's second law. An understanding of central force motion is necessary for the design of satellites and space vehicles.
\subsection{General Properties of Central Force Motion}
\begin{enumerate}

	\item The central force $f(r) \hat{{r}}$ is along ${r}$ and can exert no torque on the reduced mass $\mu $ , therefore angular momentum about centre of force is conserved. This implies that central force motion is a planar motion.
	\begin{align*}
	\tau &=r\times F\\
	&=r  \hat{r} \times f(r) \hat{{r}}\\
	&=0 \quad (\text{Since,  } \hat{r} \times \hat{r} =0)
	\end{align*}
	\item Central forces are conservative, therefore total energy of a particle moving under central force is conserved.
	\item The magnitude of the angular momentum $|{L}| \equiv l$, and the total energy $E $ , of central force motion is constant.
	\item Areal velocity is constant, that is area swept by line joining the particle to the centre of force per unit time is constant.
	\begin{equation}
	\frac{\Delta A}{\Delta t}=\text { constant }
	\end{equation}
\end{enumerate}
\section{The two body central force problems}
\subsection{Reduction of two-body problems to one body problem}
A two-body system can be effectively reduced to one-body system by introducing the concept of reduced mass. Suppose a system is composed of two masses $m_{1}$ and $m_{2}$, then for an inertial observer the relative motion of these masses may be expressed by a fictitious particle of reduced mass $\mu$.\\
Let the instantaneous position of masses $m_{1}$ and $m_{2}$ be represented by position vectors $r_{1}$ and $r_{2}$ relative to an arbitrary origin 0\\
\begin{figure}[H]
	\centering
	\includegraphics[height=5cm,width=5cm]{diagram-20220217(5)-20220217171814-crop}
	\caption{}
	\label{}
\end{figure}
Let us assume that no external force acts on the system and the internal force is simply due to mutual interaction of the particles, so the potential energy function is dependent only on internal forces, so the potential energy function may be assumed to be the function of vector between two particles $\mathbf{r}_{2}-\mathbf{r}_{1}$ or their relative velocity $\left(\dot{\mathbf{r}}_{2}-\dot{\mathbf{r}}_{1}\right) .$ This system has six degrees of freedom and hence requires six independent generalised coordinates. We choose these to be three components of radius vector $\mathbf{R}$ of centre of mass and three components of difference vector $\mathbf{r}=\mathbf{r}_{2}-\mathbf{r}_{1}$.\\
The Lagrangian of the system has the form $L=T(\dot{\mathbf{R}}, \dot{\mathbf{r}})-V(\mathbf{r}, \dot{\mathbf{r}}, \ldots)$
The kinetic energy of the system may be expressed as the sum of kinetic energy of the motion of centre of mass plus the kinetic energy of the motion about the centre of mass, i.e.,
$$
T=\frac{1}{2}\left(m_{1}+m_{2}\right) \dot{\mathbf{R}}^{2}+\left(\frac{1}{2} m_{1} \cdot{\mathbf{r}}_{C_{1}}^{2}+\frac{1}{2} m_{2} \cdot{\mathbf{r}}_{C_{2}}^{2}\right)
$$
where $\mathbf{r}_{C_{1}}$ and $\mathbf{r}_{C_{2}}$ are the radii vectors of the two particles relative to centre of mass. $\mathbf{r}_{C_{1}}$ and $\mathbf{r}_{C_{2}}$ are related to $\mathbf{r}$ by the equations\\
$$\left.\begin{array}{l}
	\mathbf{r}_{C_{1}}=-\frac{m_{2}}{m_{1}+m_{2}} \mathbf{r} \\
	\mathbf{r}_{C_{2}}=\frac{m_{1}}{m_{1}+m_{2}} \mathbf{r}
\end{array}\right\}$$
Substituting these values, we get
$$
T=\frac{1}{2}\left(m_{1}+m_{2}\right) \dot{\mathbf{R}}^{2}+\frac{1}{2}\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right) \dot{\mathbf{r}}^{2}
$$
so the Lagrangian of the system
$$
L=T-V=\frac{1}{2}\left(m_{1}+m_{2}\right) \dot{\mathbf{R}}^{2}+\frac{1}{2}\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right) \dot{\mathbf{r}}^{2}-V(\mathbf{r}, \dot{\mathbf{r}})
$$
$\mathbf{R}$ does not occurs in Lagrangian, so three components of $\mathbf{R}$ are cyclic, so centre of mass is either at rest or in uniform motion. No equation of motion for $\mathbf{r}$, will contain terms containing $\mathbf{R}$ or $\dot{\mathbf{R}}$, so assuming centre of mass at rest, me may drop the first term from the Lagrangian, i.e.,
$$
L=\frac{1}{2}\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right) \dot{\mathbf{r}}^{2}-V(\mathbf{r}, \dot{\mathbf{r}}, \ldots)
$$
$\text { This is the Lagrangian of a single particle of a fictitious mass } \mu=\frac{m_{1} m_{2}}{m_{1}+m_{2}} \text { placed at the }$ at the location of $m_2$ with the original potential function.Thus two body problem is equivalent to one body problem.
\subsection{The equation of motion and first integral}
Let us consider only the central force, where the potential $V$ is the function of $r$ only, so that the force is always directed along $\mathbf{r}$. Let a single particle move about a fixed centre of force which we assume to be the origin of co-ordinate system. Using polar co-ordinates $(r, \theta)$, the kinetic energy of particle is given by\\
$T=\frac{1}{2} \mu\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}\right), \mu$ being reduced mass.\\
The potential energy $V=V(r)$.\\
The Lagrangian of the system is given by

\begin{align}
&L=T-V \\
&=\frac{1}{2} \mu\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}\right)-V(r) .\\
\intertext { Lagrange's equation for } \theta \text { is } &\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{\theta}}\right)-\frac{\partial L}{\partial \theta}=0
\end{align}
$\therefore \quad$ From (2.5)
$$
\frac{\partial L}{\partial \theta}=0 \text { and } \frac{\partial L}{\partial \dot{\theta}}=\mu r^{2} \dot{\theta}
$$
$\therefore \quad$ From (2.6)
$$
\frac{d}{d t}\left(\mu r^{2} \dot{\theta}\right)=0
$$
Integrating, we get\\
\begin{align}
\mu r^{2} \dot{\theta}= constant =J (say),
\end{align}
where $J$ is first integral (constant of motion) and represents the magnitude of angular momentum.\\
As $\mu$ is constant, equation (2.6) gives\\
\begin{align}
	&\frac{d}{d t}\left(r^{2} \dot{\theta}\right)=0 \notag\\
	&\frac{d}{d t}\left(\frac{1}{2} r^{2} \dot{\theta}\right)=0 \notag\\
	&\frac{1}{2} r^{2} \dot{\theta}=\text { constant }\label{eq6}
\end{align}
The term $\frac{1}{2} r^{2} \dot{\theta}$ represents the areal velocity, i.e., the area swept out by the radius vector per unit time.
\begin{figure}[H]
	\centering
	\includegraphics[height=4cm,width=5cm]{diagram-20220217(4)-20220217171419-crop}
	\caption{}
	\label{}
\end{figure}
If vector $\mathbf{r}$ rotates by an angle $d \theta$ in time $d t$, the area swept out by $r$ in time $d t$ is $\frac{1}{2} r .(r d \theta)=d A$ (say), so that
$$
\frac{d A}{d t}=\frac{1}{2} r^{2} \frac{d \theta}{d t}=\frac{1}{2} r^{2} \dot{\theta}
$$
From equation (\ref*{eq6});
\begin{align}
\frac{d A}{d t}=\frac{1}{2} r^{2} \dot{\theta}=\text { constant }
\end{align}
Equation (2.4) gives
\begin{align*}
\frac{\partial L}{\partial \dot{r}}=\mu \dot{r}\\
\frac{\partial L}{\partial r}=\mu \dot{\theta}^{2}-\frac{\partial V}{\partial r}
\end{align*}
The Lagrangian equation in terms of $r$ is given by
\begin{align}
\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{r}}\right)-\frac{\partial L}{\partial r}=0 \notag\\
\frac{d}{d t}(\mu \dot{r})-\mu r \dot{\theta}^{2}+\frac{\partial V}{\partial r}=0\label{eq7}
\end{align}
If we represent the force along $\mathbf{r}$ by $F(r)$, then we have
$$
F(r)=-\frac{\partial V}{\partial r},
$$
so that equation \ref{eq7} can be written as
\begin{align}
\mu \ddot{r}-\mu r \dot{\theta}^{2}=F(r) \label{eq9}
\end{align}
This is the general equation of motion.\\
 But from equation (2.7),
 \begin{align*}
  \dot{\theta}=\frac{J}{\mu r^{2}}\\
 \dot{\theta}^{2}=\frac{J^{2}}{\mu^{2} r^{4}},
 \end{align*}
so that equation (\ref{eq9}) gives
\begin{align}
\mu \ddot{r}-\frac{J^{2}}{\mu r^{3}}=F(r) \label{eq10}
\end{align}
This is second order differential equation in $r$ only\\
 Equation (\ref{eq10}) gives\\
$
\mu \ddot{r}=\frac{J^{2}}{\mu r^{3}}+F(r),
$
\begin{align}
	&\mu \ddot{r}=\frac{j^{2}}{\mu r^{3}}-\frac{\partial V}{\partial r}=-\frac{1}{2} \frac{\partial}{\partial r}\left(\frac{J^{2}}{\mu r^{2}}\right)-\frac{\partial V}{\partial r} \notag \\
	&\mu \ddot{r}=-\frac{\partial}{\partial r}\left(\frac{1}{2} \frac{J^{2}}{\mu r^{2}}+V\right) \label{eq11}
\end{align}
Multiplying both sides of this equation by $\dot{r}$, we get
\begin{align*}
&\mu \dot{r} \ddot{r}=-\frac{\partial}{\partial r}\left(\frac{1}{2} \cdot \frac{J^{2}}{\mu r^{2}}+V\right) \dot{r} \\
&\frac{d}{d t}\left(\frac{1}{2} \mu \dot{r}^{2}\right)=-\frac{d}{d t}\left(\frac{1}{2} \frac{J^{2}}{\mu r^{2}}+V\right) \\
&\frac{d}{d t}\left(\frac{1}{2} \mu \dot{r}^{2}+\frac{1}{2} \frac{J^{2}}{\mu r^{2}}+V\right)=0 \\
&\frac{1}{2} \mu \dot{r}^{2}+\frac{1}{2} \frac{j^{2}}{\mu r^{2}}+V=\mathrm{constant}
\end{align*}
But
\begin{align*}
&\text { K.E. }=T=\frac{1}{2} \mu\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}\right) \\
&=\frac{1}{2} \mu\left(\dot{r}^{2}+\frac{J^{2}}{\mu^{2} r^{2}}\right) \\
&=\frac{1}{2} \mu \dot{r}^{2}+\frac{1}{2} \frac{J^{2}}{\mu r^{2}} \\
&\text { potential energy } =V\\
&\text { total energy } \quad E
=T+V=\frac{1}{2} \mu \dot{r}^{2}+\frac{1}{2} \frac{j^{2}}{\mu r^{2}}+V .
\end{align*}
From these equations we get
\begin{align}
\frac{1}{2} \mu \dot{r}^{2}+\frac{1}{2} \mu^{2}+V=E=\text { constant } \label{eq12}
\end{align}
ie total energy of the system is constant ie the total energy E , constant of motion.This is the another first integral of motion .This equation (\ref{eq10}) represents equation of motion while angular momentum and total energy are constant of motion.\\
From equation (\ref{eq12}) we have
\begin{align}
	&\frac{1}{2} \mu \dot{r}^{2}=E-\frac{J^{2}}{2 \mu r^{2}}-V \notag\\
	&\dot{r}=\sqrt{\left[\frac{2}{\mu}\left\{E-\frac{J^{2}}{2 \mu r^{2}}-V\right\}\right]} \notag\\
	&\frac{d r}{d t}=\sqrt{\left[\frac{2}{\mu}\left(E-\frac{J^{2}}{2 \mu r^{2}}-V\right)\right]} \notag\\
	&\sqrt{\left[\frac{2}{\mu}\left(E-\frac{j^{2}}{2 \mu r^{2}}-V\right)\right]} \label{eq15}
\end{align}
$\text { Let the initial value of } r \text { be } r_{0} \text {; then integrating equation (\ref{eq15}), we get }$
\begin{align}
	&\int_{r_{\theta}}^{r} \frac{d r}{\sqrt{\left\{\frac{2}{\mu}\left(E-V-\frac{J^{2}}{2 \mu r^{2}}\right)\right\}}}=\int_{\theta}^{t} d t \notag \\
	&\left.t=\int_{r_{\theta}}^{r} \frac{d r}{\sqrt{\left\{\frac{2}{\mu}\left(E-V-\frac{J^{2}}{2 \mu r^{2}}\right)\right.}}\right\} \label{eq16}
\end{align}
This equation gives $t$ as a function of $r$. However, from this equation we can find $r$ as function of $t$ and the constants.
From equation (2.7), we have
$$
d \theta=\frac{J}{\mu r^{2}} d t
$$
If initially $\theta=\theta_{\theta}$, then integration of above equation yields
\begin{align}
&\int_{\theta_{0}}^{\theta} d \theta=\int_{0}^{t} \frac{J}{\mu r^{2}} d t \notag \\
&\theta-\theta_{0}=\int_{0}^{t} \frac{J}{\mu r^{2}} d t \notag \\
&\theta=\int_{0}^{1} \frac{J}{\mu r^{2}} d t+\theta_{0}
\end{align}
This equation gives $\theta$ as a function of $t$.\\
Equation(2.17 and 2.16) are the only integration to be solved .Therefore the problems have been reduced to quadratures with four constants $E,J,r_0,\theta_0$
\subsection{The equivalent one-dimensional problem, and classification of orbits}
Although we have solved the one dimensional problem formally ,practically speaking the integrals (1.16) and (1.17) are usually quite unmanageable and in specific case it is often more convenient to perform the integration in some other fashion.But before obtaining the solution for any specific force laws,let us see what can be learned about the motion in the general case using only the equation of motion and conservation theorems,without requiring explicit solutions.
\par The equation of motion in r,with $\dot{\theta}$ expressed in terms of l,quation     $m\ddot{r}-\frac{l^2}{mr^3}=f(r)$ involves only r and its derivatives.It is the same equation as would be obtained  for a fictitious one-dimensional problem in which a particle of mass $m$ is subject to a force
$$f^{\prime}=f+\frac{l^{2}}{m r^{3}}$$
The significance of the additional term is clear if it is written as $m r \dot{\theta}^{2}=m v_{\theta}^{2} / r$, which is the familiar centrifugal force. An equivalent statement can be obtained from the conservation theorem for energy. By Eq. $\frac{1}{2}m\dot{r}^2+\frac{1}{2}\frac{l^2}{mr^2}+V=constant$ the motion of the particle in $r$ is that of a one-dimensional problem with a fictitious potential energy:\\
$$V^{\prime}=V+\frac{1}{2} \frac{l^{2}}{m r^{2}}$$
As a check, note that
$$
f^{\prime}=-\frac{\partial V^{\prime}}{\partial r}=f(r)+\frac{l^{2}}{m r^{3}}
$$
 The energy conservation theorem can thus also be written as
$$
E=V^{\prime}+\frac{1}{2} m \dot{r}^{2}
$$
As an illustration of this method of examining the motion, consider a plot of $V^{\prime}$ against $r$ for the specific case of an attractive inverse-square law of force:
$$
f=-\frac{k}{r^{2}} .
$$
(For positive $k$, the minus sign ensures that the force is toward the center of force.) The potential energy for this force is
$$
V=-\frac{k}{r}
$$
and the corresponding fictitious potential is
$$
V^{\prime}=-\frac{k}{r}+\frac{l^{2}}{2 m r^{2}}
$$
Such a plot is shown in Figure the two dashed lines represent the separate components
$$
-\frac{k}{r} \quad \text { and } \quad \frac{l^{2}}{2 m r^{2}}
$$
and the solid line is the sum $V^{\prime}$.
\begin{figure}[H]
	\centering
	\includegraphics[height=9cm,width=9cm]{diagram-20220219(10)}
	\caption{The equivalent one dimensional potential for attractive inverse square law of force}
	\label{}
\end{figure}
Let us consider now the motion of a particle having the energy $E_{1}$, as shown in Figures. Clearly this particle can never come closer than $r_{1}$  Otherwise with $r<r_{1}, V^{\prime}$ exceeds $E_{1}$ and by conservation of energy the kinetic energy would have to be negative, corresponding to an imaginary velocity! on the other hand, there is no upper limit to the possible value of $r$, so the orbit is not bounded. A particle will come in from infinity, strike the "repulsive centrifugal barrier," be repelled, and travel back out to infinity. The distance between $E$ and $V^{\prime}$ is $\frac{1}{2} m \dot{r}^{2}$, i.e., proportional to the square of the radial velocity, and becomes zero, naturally, at the turning point $r_{1}$. At the same time. the distance between $E$ and $V$ on the plot is the kinetic energy $\frac{1}{2} m v^{2}$ at the given value of $r$. Hence, the distance between the $V$ and $V^{\prime}$ curves is $\frac{1}{2} m r^{2} \theta^{2}$. These curves therefore supply the magnitude of the particle velocity and its components for any distance r,at the given energy and angular momentum .This information is sufficient to produce an approximate picture of the form of the orbit.\\
\begin{minipage}{0.5\textwidth}
\begin{figure}[H]
	\centering
	\includegraphics[height=7cm,width=10cm]{diagram-20220219(11)}
	\caption{Unbounded motion at positive energies for inverse square law of force}
	\label{}
\end{figure}
\end{minipage}
\begin{minipage}{0.5\textwidth}
\begin{figure}[H]
	\centering
	\includegraphics[height=5cm,width=7cm]{diagram-20220219(6)}
	\caption{The orbit for $E_1$ corresponding to unbounded motion}
	\label{}
\end{figure}
\end{minipage}
\par For the energy $E_{2}=0$, a roughly similar picture of the orbit behavior is obtained. But for any lower energy, such as $E_{3}$ indicated in fig (1.6) we have a different story. In addition to a lower bound $r_{1}$, there is also a max. imum value $r_{2}$ that cannot be exceeded by $r$ with positive kinetic energy. The motion is then "bounded," and there are two turning points, $r_{1}$ and $r_{2}$, also known as apsidal distances. This does not necessarily mean that the orbits are closed. All that can be said is that they are bounded, contained between two circles of radius $r_{1}$ and $r_{2}$ with turning points always lying on the circles\\
\begin{minipage}{0.5\textwidth}
	\begin{figure}[H]
		\centering
		\includegraphics[height=8cm,width=9cm]{diagram-20220219(8)}
		\caption{The equivalent one dimensional potential for inverse square law of force illustrating bounded motion at negative energies.}
		\label{}
	\end{figure}
\end{minipage}
\begin{minipage}{0.5\textwidth}
\begin{figure}[H]
	\centering
	\includegraphics[height=5cm,width=7cm]{diagram-20220219(7)}
	\caption{The nature of the orbit for bounded motion}
	\label{}
\end{figure}
\end{minipage}
\par If the energy is $E_{4}$ at the minimum of the fictitious potential as shown in Fig. (1.8), then the two bounds coincide. In such case, motion is possible at only one radius; $\dot{r}=0$, and the orbit is a circle.
\begin{figure}[H]
	\centering
	\includegraphics[height=7cm,width=7cm]{diagram-20220219(9)}
	\caption{The equivalent one dimensional potential of inverse square law of force,illustrating the condition for circular orbits}
	\label{}
\end{figure}
\subsection{Condition for circular orbit}
Remembering that the effective "force" is the negative of the slope of the $V^{\prime}$ curve, the requirement for circular orbits is simply that $f^{\prime}$ be zero, or
$$
f(r)=-\frac{l^{2}}{m r^{3}}=-m r \dot{\theta}^{2}
$$
which can be also derived by $\left.\frac{\partial V_{\text {effective }}}{\partial r}\right|_{r=r_{0}}=0$ and $\dot{\theta}=\omega_{0}$ is identified as angular frequency circular orbit.
where $$V_{e f f}=\frac{J^{2}}{2 m r^{2}}-\frac{k}{r^{n}}$$
Radius $r=r_0$ of circular orbit is also identified as stable equilibrium point so $\frac{\partial^2V_{effective}}{\partial r}|_{r=r_0}\geq 0$
Somehow particle of mass m changes its orbit without changing its angular momentum and orbit is bounded then new orbit is identified as elliptical orbit .The angular frequency in new elliptical orbit is
$$\omega=\sqrt{\frac{\frac{\partial^2 V_{effective}}{\partial r}|_{r-r_0}}{m}}$$
\subsection{The differential equation for the orbit}
Under central force
\begin{align}
\text{We have }J=\mu r^{2} \dot{\theta}= \text{constant}\\
and E=\frac{1}{2} \mu \dot{r}^{2}+\frac{J^{2}}{2 \mu r^{2}}+V= \text{constant}\\
\text{Differential equation in r is} \notag\\
\mu \ddot{r}-\frac{J^{2}}{\mu r^{3}}=F(r)
\end{align}
From equation (2.18)
\begin{align}
	&J=\mu r^{2} \dot{\theta}\notag  \\
	&J=\mu r^{2} \frac{d \theta}{d t}\notag \\
	&J d t=\mu r^{2} d \theta
\end{align}
The corresponding relation between the derivative relative to $t$ and $\theta$ can be written as
\begin{align}
\frac{d}{d t}=\frac{J}{\mu r^{2}} \frac{d}{d \theta}
\end{align}
second derivative w.r.t. t can be written as \\
\begin{equation}
\frac{d^{2}}{d t^{2}}=\frac{J}{\mu r^{2}} \frac{d}{d \theta}\left[\frac{J}{\mu r^{2}} \frac{d}{d \theta}\right]
\end{equation}
from equation (2.20)\\
\begin{align}
	&\mu \frac{J}{\mu r^{2}} \frac{d}{d \theta}\left\{\frac{J}{\mu r^{2}} \frac{d r}{d \theta}\right\}-\frac{J^{2}}{\mu r^{3}}=F(r) . \notag \\
	&\frac{J}{r^{2}} \frac{d}{d \theta}\left\{\frac{J}{\mu r^{2}} \frac{d r}{d \theta}\right\}-\frac{J^{2}}{\mu r^{3}}=F(r)
\end{align}
To simplify above equation we must remember that
\begin{equation}
\frac{1}{r^{2}} \frac{d r}{d \theta}=-\frac{d(1 / r)}{d \theta}
\end{equation}
$\text { Using }(2.25) \text {, equation (2.24) gives }$\\
\begin{equation}
\frac{J^{2}}{\mu r^{2}} \frac{d}{d \theta}\left[-\frac{d(1 / r)}{d \theta}\right]-\frac{J^{2}}{\mu r^{3}}=F(r)
\end{equation}
Substituting
$$
u=\frac{1}{r}
$$
equation (2.26) gives \\
\begin{align}
	&-\frac{J^{2} u^{2} d^{2} u}{\mu} \frac{J^{2}}{\mu^{2}}-\frac{J^{2}}{\mu}=F\left(\frac{1}{u}\right) \notag \\
	&\frac{J^{2} u^{2}}{\mu}\left[\frac{d^{2} u}{d \theta^{2}}+u\right]=-F\left(\frac{1}{u}\right)
\end{align}
This is the differential equation for the orbit if the force F is known.
\subsection{The kepler problem:Inverse square law of force}
The inverse square law is most important of all central force laws.It results in the deduction of Kepler's laws of planetary motion.\\
\textbf{The kepler's laws of planetary motion are:}\\
(i) All planets move in elliptical orbits having the sun as one focus.\\
(ii)The area swepts out by the radius vector of planet relative to the sun in equal times are equal.\\
(iii)The square of the period of revolution of any planet about the sun is proportional to the cube of the semi major axis\\
\textbf{Deduction of Kepler's laws}\\
\textbf{Kepler's first law}\\
Under central force the constant of motion are angular momentum and energy
\begin{align}
J&=\mu r^{2} \dot{\theta} \\
 E&=\frac{1}{2} \mu \dot{r}^{2}+\frac{J^{2}}{2 \mu r^{2}}+V
\end{align}
From equation (2.28 and 2.29)\\
\begin{align}
	&\frac{d \theta}{d t}=\frac{J}{\mu r^{2}} \\
	&\frac{d r}{d t}=\sqrt{\frac{2}{\mu}\left(E-\frac{J^{2}}{2 \mu r^{2}}-V\right)}
\end{align}
Dividing equation (2.31) by (2.30)
\begin{align}
	\frac{dr}{d \theta} &=\frac{\mu r^{2}}{J} \times \sqrt{\left\{\frac{2}{\mu}\left(E-V-\frac{j^{2}}{2 \mu r^{2}}\right)\right\}} \\
	d \theta &=\frac{J d r}{\mu r^{2} \sqrt{\left\{\frac{2}{\mu}\left(E-V-\frac{J^{2}}{2 \mu r^{2}}\right)\right\}}}
\end{align}
Under inverse square law of force, we have

\begin{align*}
&F(r)=-\frac{k}{r^{2}} \\
&F(r)=-\frac{\partial V}{\partial r} \\
&-\frac{\partial V}{\partial r}=-\frac{k}{r^{2}} \Rightarrow d V=\frac{k}{r^{2}} d r
\end{align*}
Integrating $\quad V=\int^{r} \frac{k}{r^{2}} d r$\\
or the potential energy $V=-\frac{k}{r}$.\\
Substituting this value of $V$ in equation (2.33), we get
$$
d \theta=\frac{J d r}{\sqrt[\mu r^{2}]{\left\{\frac{2}{\mu}\left(E+\frac{k}{r}-\frac{J^{2}}{2 \mu r^{2}}\right)\right\}}}
$$
Integrating, we get
$$
\theta=\int \frac{J d r}{\mu r^{2} \sqrt{\left\{\frac{2}{\mu}\left(E+\frac{k}{r}-\frac{J^{2}}{2 \mu r^{2}}\right)\right\}}}+\theta^{\prime}
$$
where $\theta^{\prime}$ is constant of integration.\\
Substituting $r=1 / u$, we get\\
\begin{align}
\theta=-\int \frac{J d u}{\mu \sqrt{\left\{\frac{2}{\mu}\left(E+k u-\frac{J^{2} u^{2}}{2 \mu}\right)\right\}}}+\theta^{\prime} \notag \\
=\theta^{\prime}-\int \frac{d u}{\sqrt{\left(\frac{2 \mu E}{J^{2}}+\frac{2 \mu k u}{J^{2}}-u^{2}\right)}} \notag \\
=\theta^{\prime}-\int \frac{d u}{\sqrt{\left[\left(\frac{2 \mu E}{J^{2}}+\frac{\mu^{2} k^{2}}{J^{4}}\right)-\left(u-\frac{\mu k}{J^{2}}\right)^{2}\right]}} \notag\\
=\theta^{\prime}-\cos ^{-1} \frac{u-\frac{\mu k}{J^{2}}}{\sqrt{\left(\frac{2 \mu E}{J^{2}}+\frac{\mu^{2} k^{2}}{J^{4}}\right)}}=\theta^{\prime}-\cos ^{-1} \frac{\frac{u J^{2}}{\mu k}-1}{\sqrt{\left(\frac{2 E J^{2}}{\mu k^{2}}+1\right)}} \notag\\
\frac{\frac{u J^{2}}{\mu k}-1}{\sqrt{\left(\frac{2 E J^{2}}{\mu k^{2}}+1\right)}}=\cos \left(\theta-\theta^{\prime}\right)\notag\\
\frac{u J^{2}}{\mu k}-1=\sqrt{\left(\frac{2 E J^{2}}{\mu k^{2}}+1\right)} \cos \left(\theta-\theta^{\prime}\right) \notag \\
u=\frac{\mu k}{J^{2}}\left[1+\sqrt{\left(\frac{2 E J^{2}}{\mu k^{2}}+1\right)} \cos \left(\theta-\theta^{\prime}\right)\right]
\end{align}
Substituting
\begin{align}
c &=\frac{\mu k}{J^{2}} \\
\varepsilon &=\sqrt{\left(1+\frac{2 E J^{2}}{\mu k^{2}}\right)}
\end{align}
Equation (2.34) gives\\
\begin{align}
	u &=c\left[1+\varepsilon \cos \left(\theta-\theta^{\prime}\right)\right] \notag \\
	\frac{1}{r} &=c\left[1+\varepsilon \cos \left(\theta-\theta^{\prime}\right)\right]
\end{align}
which is the equation of the conic with $\varepsilon$ as eccentricity and one focus as the origin. Thus the equation of the path of the two-body problem of reduced mass $\mu$ is always a conic section, which is the generalisation of Kepler's first law.\\
The nature of the conic dependes on the value of eccentricity given by eqn. (2.36).\\
 If $\in>1$, i.e., if $\sqrt{\left(\frac{2 E J^{2}}{\mu k^{2}}+1\right)}>1$ or $E>0$, the conic is hyperbola.\\
 If $\in=1$, i.e., if $\sqrt{\left(\frac{2 E J^{2}}{\mu k^{2}}+1\right)}=1$ or $E=0$, the conic is a parabola.\\
  If $\in<1$, i.e., if $\sqrt{\left(\frac{2 E^{2}}{\mu k^{2}}+1\right)}<1$ or $E<0$, the conic is an ellipse.\\
   If $\in=0$, i.e., if $\sqrt{\left(\frac{2 E^{2}}{\mu k^{2}}+1\right)}=0$ or $E=-\frac{\mu k^{2}}{2 J^{2}}$, the conic is a circle.\\
    In the case of elliptical orbits, when $\theta-\theta^{\prime}=0, r=r_{1}=$ perihelion,\\
    \begin{figure}[H]
    	\centering
    	\includegraphics[height=4cm,width=7cm]{diagram-20220217(3)-20220217170838-crop}
    	\caption{}
    	\label{}
    \end{figure}
    Then from equation (2.37) we have\\
    $$r_{1}=\frac{1}{c(1+\varepsilon)}$$
    When $\theta-\theta^{\prime}=\pi, r=r_{2}=$ aphelion, then eqn. (2.37) gives
    $$
    r_{2}=\frac{1}{c(1-\varepsilon)}
    $$
    The semi-major axis, which is one-half the sum of perihelion $r_{1}$ and aphelion $r_{2}$ is given by
    $$a=\frac{r_{1}+r_{2}}{2}=\frac{1}{2}\left[\frac{1}{c(1+\varepsilon)}+\frac{1}{c(1-\varepsilon)}\right]$$
    Substituting values of $c$ and $\varepsilon$ , we get
    \begin{align}
    &a=\frac{1}{\frac{\mu k}{J^{2}}\left\{1-\left(1+\frac{2 E J^{2}}{\mu k^{2}}\right)\right\}}=-\frac{k}{2 E} \notag \\
    &E=-\frac{k}{2 a} .
    \end{align}
  This shows that in the case of elliptical orbits the total energy depends solely on the major axis.\\
  \textbf{Deduction of II law}\\
  $$J=\mu r^{2} \dot{\theta}=\text { constant }$$
  This implies
  $$
  \frac{d A}{d t}=\frac{1}{2} r^{2} \dot{\theta}=\text { constant }
  $$
  which represents the areal velocity, i.e., the area swept out by the radius vector per unit time is constant. This means that the areas swept out by the radius vector in equal times are equal which is Kepler's II law.\\
  \textbf{Deduction of kepler's III Law}\\
If $T$ is the periodic time of describing the complete orbit, the area of the orbit is given by\\
\begin{align}
A &=\int_{0}^{T} \frac{d A}{d t} d t=\int_{0}^{T} \frac{1}{2} r^{2} \ddot{\theta} d t \notag \\
&=\int_{0}^{T} \frac{J}{2 \mu} d t \quad\left(\text { since } J=\mu r^{2} \dot{\theta}\right) \notag \\
&=\frac{J T}{2 \mu}
\end{align}
But area of the ellipse
\begin{equation}
\mathrm{A}=\pi a b
\end{equation}
where $a$ and $b$ are the semi-major and semi-minor axes of the ellipse respectively.
$$
\begin{array}{ll}
\text { Also } & b=a \sqrt{\left(1-\varepsilon^{2}\right)}=a \sqrt{\left(1-1-\frac{2 E J^{2}}{\mu k^{2}}\right)}=a \sqrt{\left(-\frac{2 E J^{2}}{\mu k^{2}}\right)} \\
\text { But } & E=-\frac{k}{2 a}
\end{array}
$$
Therefore
\begin{equation}
b=a \sqrt{\left(\frac{k J^{2}}{a \mu k^{2}}\right)}=a^{1 / 2} \sqrt{\left(\frac{J^{2}}{\mu k}\right)}
\end{equation}
Substituting value of $b$ in equation (2.40) we get
\begin{equation}
A=\pi a^{3 / 2} \sqrt{\left(\frac{J^{2}}{\mu k}\right)}
\end{equation}
Comparing equation (2.39 and 2.42)\\
\begin{align}
	&\frac{J T}{2 \mu}=\pi a^{3 / 2} \sqrt{\left(\frac{J^{2}}{\mu k}\right)} \notag \\
	&\frac{J^{2} T^{2}}{4 \mu^{2}}=\pi^{2} a^{3} \frac{J^{2}}{\mu k} \notag \\
	&T^{2}=4 \pi^{2} a^{3} \frac{\mu}{k} \notag \\
	&T^{2} \propto a^{3}
\end{align}
ie the square of the period of revolution of the planet around the sun is proportional to the cube of the semimajor axis, which is kepler's III law.
\section{Two body collisions}
When discussing conservation of momentum, we considered examples in which two
objects collide and stick together, and either there are no external forces acting in some
direction (or the collision was nearly instantaneous) so the component of the momentum
of the system along that direction is constant. We shall now study collisions between
objects in more detail. In particular we shall consider cases in which the objects do not
stick together. The momentum along a certain direction may still be constant but the
mechanical energy of the system may change. We will begin our analysis by considering
two-particle collision. We introduce the concept of the relative velocity between two
particles and show that it is independent of the choice of reference frame. We then show
that the change in kinetic energy only depends on the change of the square of the relative
velocity and therefore is also independent of the choice of reference frame. We will then
study one- and two-dimensional collisions with zero change in potential energy. In
particular we will characterize the types of collisions by the change in kinetic energy and analyze the possible outcomes of the collisions.
\subsection{Laboratary frame of reference}
 Let $\overrightarrow{\mathbf{R}}$ be the vector from the origin of frame $S$ to the origin of reference frame $S^{\prime}$. Denote the position vector of the $j^{\text {th }}$ particle with respect to the origin of reference frame $S$ by $\overrightarrow{\mathbf{r}}_{j}$ and similarly, denote the position vector of the $j^{\text {th }}$ particle with respect to the origin of reference frame $S^{\prime}$ by $\overrightarrow{\mathbf{r}}_{j}^{\prime}$ .\\
 \begin{figure}[H]
 	\centering
 	\includegraphics[height=3cm,width=5cm]{diagram-20220218-crop}
 	\caption{}
 	\label{}
 \end{figure}
The position vectors are related by
$$
\overrightarrow{\mathbf{r}}_{j}=\overrightarrow{\mathbf{r}}_{j}^{\prime}+\overrightarrow{\mathbf{R}}
$$
The relative velocity (call this the boost velocity) between the two reference frames is given by
$$
\overrightarrow{\mathbf{V}}=\frac{d \overrightarrow{\mathbf{R}}}{d t}
$$
Assume the boost velocity between the two reference frames is constant. Then, the relative acceleration between the two reference frames is zero,
$$
\overrightarrow{\mathbf{A}}=\frac{d \overrightarrow{\mathbf{V}}}{d t}=\overrightarrow{\mathbf{0}}
$$
When the equation is satisfied, the reference frames $S$ and $S^{\prime}$ are called relatively inertial reference frames.\\
Suppose the $j^{\text {th }}$ particle in Figure is moving; then observers in different reference frames will measure different velocities. Denote the velocity of $j^{\text {th }}$ particle in frame $S$ by $\overrightarrow{\mathbf{v}}_{j}=d \overrightarrow{\mathbf{r}}_{j} / d t$, and the velocity of the same particle in frame $S^{\prime}$ by $\overrightarrow{\mathbf{v}}_{j}^{\prime}=d \overrightarrow{\mathbf{r}}_{j}^{\prime} / d t$. Taking derivative, the velocities of the particles in two different reference frames are related according to
$$
\overrightarrow{\mathbf{v}}_{j}=\overrightarrow{\mathbf{v}}_{j}^{\prime}+\overrightarrow{\mathbf{V}}
$$
\subsection{ Center-of-mass Reference Frame}
Let $\overrightarrow{\mathbf{r}}_{c m}$ be the vector from the origin of frame $S$ to the center-of-mass of the system of particles, a point that we will choose as the origin of reference frame $S_{c m}$, called the center-of-mass reference frame. Denote the position vector of the $j^{\text {th }}$ particle with respect to origin of reference frame $S$ by $\overrightarrow{\mathbf{r}}_{j}$ and similarly, denote the position vector of the $j^{\text {th }}$ particle with respect to origin of reference frame $S_{c m}$ by $\overrightarrow{\mathbf{r}}_{j}^{\prime}$ \\
\begin{figure}[H]
	\centering
	\includegraphics[height=3cm,width=5cm]{diagram-20220218(1)-crop}
	\caption{}
	\label{}
\end{figure}
The position vector of the $j^{\text {th }}$ particle in the center-of-mass frame is then given by
$$
\overrightarrow{\mathbf{r}}_{j}^{\prime}=\overrightarrow{\mathbf{r}}_{j}-\overrightarrow{\mathbf{r}}_{c m} .
$$
The velocity of the $j^{\text {th }}$ particle in the center-of-mass reference frame is then given by
$$
\overrightarrow{\mathbf{v}}_{j}^{\prime}=\overrightarrow{\mathbf{v}}_{j}-\overrightarrow{\mathbf{v}}_{c m}
$$
There are many collision problems in which the center-of-mass reference frame is the most convenient reference frame to analyze the collision.\\
Consider a system consisting of two particles, which we shall refer to as particle 1 and particle $2 .$ We can determine the velocities of particles 1 and 2 in the center-of-mass,
as\\\\
$$\overrightarrow{\mathbf{v}}_{1}^{\prime}=\overrightarrow{\mathbf{v}}_{1}-\overrightarrow{\mathbf{v}}_{c m}=\overrightarrow{\mathbf{v}}_{1}-\frac{m_{1} \overrightarrow{\mathbf{v}}_{1}+m_{2} \overrightarrow{\mathbf{v}}_{2}}{m_{1}+m_{2}}=\frac{m_{2}}{m_{1}+m_{2}}\left(\overrightarrow{\mathbf{v}}_{1},-\overrightarrow{\mathbf{v}}_{2}\right)=\frac{\mu}{m_{1}} \overrightarrow{\mathbf{v}}_{1,2}$$
where $\overrightarrow{\mathbf{v}}_{12}=\overrightarrow{\mathbf{v}}_{1}-\overrightarrow{\mathbf{v}}_{2}$ is the relative velocity of particle 1 with respect to particle 2 . A similar result holds for particle 2 :\\\\
$$\overrightarrow{\mathbf{v}}_{2}^{\prime}=\overrightarrow{\mathbf{v}}_{2}-\overrightarrow{\mathbf{v}}_{c m}=\overrightarrow{\mathbf{v}}_{2}-\frac{m_{1} \overrightarrow{\mathbf{v}}_{1}+m_{2} \overrightarrow{\mathbf{v}}_{2}}{m_{1}+m_{2}}=-\frac{m_{1}}{m_{1}+m_{2}}\left(\overrightarrow{\mathbf{v}}_{1}-\overrightarrow{\mathbf{v}}_{2}\right)=-\frac{\mu}{m_{2}} \overrightarrow{\mathbf{v}}_{1,2}$$
The momentum of the system the center-of-mass reference frame is zero as we expect,
$$
m_{1} \overrightarrow{\mathbf{v}}_{1}^{\prime}+m_{2} \overrightarrow{\mathbf{v}}_{2}^{\prime}=\mu \overrightarrow{\mathbf{v}}_{12}-\mu \overrightarrow{\mathbf{v}}_{12}=\overrightarrow{\mathbf{0}}
$$
\subsection{ Characterizing Collisions}
In a collision, the ratio of the magnitudes of the initial and final relative velocities is called the coefficient of restitution and denoted by the symbol $e$,
$$
e=\frac{v_{B}}{v_{A}}
$$
If the magnitude of the relative velocity does not change during a collision, $e=1$, then the change in kinetic energy is zero. Collisions in which there is no change in kinetic energy are called elastic collisions,
$$\Delta K=0, \text{elastic collision }$$
If the magnitude of the final relative velocity is less than the magnitude of the initial relative velocity, $e<1$, then the change in kinetic energy is negative. Collisions in which the kinetic energy decreases are called inelastic collisions,
$$\Delta K<0, \text{inelastic collision }$$
If the two objects stick together after the collision, then the relative final velocity is zero, $e=0 .$ Such collisions are called totally inelastic. The change in kinetic energy can be written as\\
$$\Delta K=-\frac{1}{2} \mu v_{A}^{2}=-\frac{1}{2} \frac{m_{1} m_{2}}{m_{1}+m_{2}} v_{A}^{2}, \text { totally inelastic collision } .$$
If the magnitude of the final relative velocity is greater than the magnitude of the initial relative velocity, $e>1$, then the change in kinetic energy is positive. Collisions in which the kinetic energy increases are called superelastic collisions,
$$\Delta K>0, \textbf{superelastic collision}$$
\subsection{ Two-dimensional Elastic Collision in Laboratory Reference Frame}
Consider the elastic collision between two particles in which we neglect any external forces on the system consisting of the two particles. Particle 1 of mass $m_{1}$ is initially moving with velocity $\overrightarrow{\mathbf{v}}_{1, i}$ and collides elastically with a particle 2 of mass $m_{2}$ that is initially at rest. We shall refer to the reference frame in which one particle is at rest, 'the target', as the laboratory reference frame. After the collision particle 1 moves with velocity $\overrightarrow{\mathbf{v}}_{1, f}$ and particle 2 moves with velocity $\overrightarrow{\mathbf{v}}_{2, f}$, (Figure). The angles $\theta_{1, f}$ and $\theta_{2, f}$ that the particles make with the positive forward direction of particle 1 are called the laboratory scattering angles.\\
\begin{figure}[H]
	\centering
	\includegraphics[height=5cm,width=8cm]{class20}
	\caption{}
	\label{}
\end{figure}
Generally the initial velocity $\overrightarrow{\mathbf{v}}_{1, i}$ of particle 1 is known and we would like to determine the final velocities $\overrightarrow{\mathbf{v}}_{1, f}$ and $\overrightarrow{\mathbf{v}}_{2, f}$, which requires finding the magnitudes and directions of each of these vectors, $v_{1, f}, v_{2, f}, \theta_{1, f}$, and $\theta_{2, f} .$ These quantities are related by the two equations describing the constancy of momentum, and the one equation describing constancy of the kinetic energy. Therefore there is one degree of freedom that we must specify in order to determine the outcome of the collision. In what follows we shall express our results for $v_{1, f}, v_{2, f}$, and $\theta_{2, f}$ in terms of $v_{1, i}$ and $\theta_{1, f}$.\\
The components of the total momentum $\overrightarrow{\mathbf{p}}_{i}^{\mathrm{sys}}=m_{1} \overrightarrow{\mathbf{v}}_{1, i}+m_{2} \overrightarrow{\mathbf{v}}_{2, i}$ in the initial state are given by
$$
\begin{aligned}
p_{x, i}^{\mathrm{sys}} &=m_{1} v_{1, i} \\
p_{y, i}^{\mathrm{sys}} &=0 .
\end{aligned}
$$
The components of the momentum $\overrightarrow{\mathbf{p}}_{f}^{\mathrm{sys}}=m_{1} \overrightarrow{\mathbf{v}}_{1, f}+m_{2} \overrightarrow{\mathbf{v}}_{2, f}$ in the final state are given by
$$
\begin{aligned}
&p_{x, f}^{\mathrm{sys}}=m_{1} v_{1, f} \cos \theta_{1, f}+m_{2} v_{2, f} \cos \theta_{2, f} \\
&p_{y, f}^{\mathrm{sys}}=m_{1} v_{1, f} \sin \theta_{1, f}-m_{2} v_{2, f} \sin \theta_{2, f} .
\end{aligned}
$$
There are no any external forces acting on the system, so each component of the total momentum remains constant during the collision,
$$
\begin{aligned}
&p_{x, i}^{\text {sys }}=p_{x, f}^{\text {sys }} \\
&p_{y, i}^{\text {sys }}=p_{y, f}^{\text {sy }}
\end{aligned}
$$
substituting the values\\
$$\begin{gathered}
m_{1} v_{1, i}=m_{1} v_{1, f} \cos \theta_{1, f}+m_{2} v_{2, f} \cos \theta_{2, f} \\
0=m_{1} v_{1, f} \sin \theta_{1, f}-m_{2} v_{2, f} \sin \theta_{2, f}
\end{gathered}$$
rewriting the expressions we will get \\
\begin{align}
m_{2} v_{2, f} \cos \theta_{2, f}=m_{1}\left(v_{1, i}-v_{1, f} \cos \theta_{1, f}\right)\\
m_{2} v_{2, f} \sin \theta_{2, f}=m_{1} v_{1, f} \sin \theta_{1, f}
\end{align}
Squaring and adding and using the identity $\text { the identity } \cos ^{2} \theta+\sin ^{2} \theta=1 \text { yielding }$
$$v_{2, f}^{2}=\frac{m_{1}^{2}}{m_{2}^{2}}\left(v_{1, i}^{2}-2 v_{1, i} v_{1, f} \cos \theta_{1, f}+v_{1, f}^{2}\right)$$
The collision is elastic and therefore the system kinetic energy of is constant
$$
K_{i}^{\text {sys }}=K_{f}^{\text {sys }}
$$
$$
\frac{1}{2} m_{1} v_{1, i}^{2}=\frac{1}{2} m_{1} v_{1, f}^{2}+\frac{1}{2} m_{2} v_{2, f}^{2}
$$
substituting the value of $v_{2, f}^{2}$ in this equation\\
$$\frac{1}{2} m_{1} v_{1, i}^{2}=\frac{1}{2} m_{1} v_{1, f}^{2}+\frac{1}{2} \frac{m_{1}^{2}}{m_{2}}\left(v_{1, i}^{2}-2 v_{1, i} v_{1, f} \cos \theta_{1, f}+v_{1, f}^{2}\right)$$
$$0=\left(1+\frac{m_{1}}{m_{2}}\right) v_{1, f}^{2}-\frac{m_{1}}{m_{2}} 2 v_{1, i} v_{1, f} \cos \theta_{1, f}-\left(1-\frac{m_{1}}{m_{2}}\right) v_{1, i}^{2}$$
Let $\alpha=m_{1} / m_{2}$ then Equation can be written as
$$
0=(1+\alpha) v_{1, f}^{2}-2 \alpha v_{1, i} v_{1, f} \cos \theta_{1, f}-(1-\alpha) v_{1, i}^{2}
$$
The solution to this quadratic equation is given by
$$
v_{1, f}=\frac{\alpha v_{1, i} \cos \theta_{1, f} \pm\left(\alpha^{2} v_{1, i}^{2} \cos ^{2} \theta_{1, f}+(1-\alpha) v_{1, i}^{2}\right)^{1 / 2}}{(1+\alpha)}
$$
Divide equation (2.44 and 2.25) yields\\
$$\begin{gathered}
\frac{v_{2, f} \sin \theta_{2, f}}{v_{2, f} \cos \theta_{2, f}}=\frac{v_{1, f} \sin \theta_{1, f}}{v_{1, i}-v_{1, f} \cos \theta_{1, f}} \\
\tan \theta_{2, f}=\frac{v_{1, f} \sin \theta_{1, f}}{v_{1, i}-v_{1, f} \cos \theta_{1, f}} .
\end{gathered}$$
The relationship between the scattering angles is independent of the masses of the colliding particles. Thus the scattering angle for particle 2 is
$$\theta_{2, f}=\tan ^{-1}\left(\frac{v_{1, f} \sin \theta_{1, f}}{v_{1, i}-v_{1, f} \cos \theta_{1, f}}\right)$$
From (2.45) final velocity of the particle 1\\
$$v_{2, f}=\frac{v_{1, f} \sin \theta_{1, f}}{\alpha \sin \theta_{2, f}}$$
\begin{exercise}
	Object 1 with mass $m_{1}$ is initially moving with a speed $v_{1, i}=3.0 \mathrm{~m} \cdot \mathrm{s}^{-1}$ and collides elastically with object 2 that has the same mass, $m_{2}=m_{1}$, and is initially at rest. After the collision, object 1 moves with an unknown speed $v_{1, f}$ at an angle $\theta_{1, f}$ with respect to its initial direction of motion and object 2 moves with an unknown speed $v_{2, f}$, at an unknown angle $\theta_{2, f}$ (as shown in the Figure $15.10$ ). Find the final speeds of each of the objects and the angle $\theta_{2, f}$.\\
	\begin{figure}[H]
		\centering
		\includegraphics[height=5cm,width=8cm]{diagram-20220218(2)}
	\end{figure}
\end{exercise}
\begin{answer}
 Because the masses are equal, $\alpha=1$. We are given that $v_{1, i}=3.0 \mathrm{~m} \cdot \mathrm{s}^{-1}$ and $\theta_{1, f}=30^{\circ}$. Hence  $0=(1+\alpha) v_{1, f}^{2}-2 \alpha v_{1, i} v_{1, f} \cos \theta_{1, f}-(1-\alpha) v_{1, i}^{2}$ reduces  to
$$ v_{1, f}=v_{1, i} \cos \theta_{1, f}=\left(3.0 \mathrm{~m} \cdot \mathrm{s}^{-1}\right) \cos 30^{\circ}=2.6 \mathrm{~m} \cdot \mathrm{s}^{-1} $$
substitute this value in
$$\tan \theta_{2, f}=\frac{v_{1, f} \sin \theta_{1, f}}{v_{1, i}-v_{1, f} \cos \theta_{1, f}}$$
$$\begin{aligned}
	\theta_{2, f} &=\tan ^{-1}\left(\frac{v_{1, f} \sin \theta_{1, f}}{v_{1, i}-v_{1, f} \cos \theta_{1, f}}\right) \\
	\theta_{2, f} &=\tan ^{-1}\left(\frac{\left(2.6 \mathrm{~m} \cdot \mathrm{s}^{-1}\right) \sin \left(30^{\circ}\right)}{3.0 \mathrm{~m} \cdot \mathrm{s}^{-1}-\left(2.6 \mathrm{~m} \cdot \mathrm{s}^{-1}\right) \cos \left(30^{\circ}\right)}\right) \\
	&=60^{\circ} .
\end{aligned}$$
The above results for $v_{1, f}$ and $\theta_{2, f}$ may be substituted into either of the expressions in $m_{2} v_{2, f} \cos \theta_{2, f}=m_{1}\left(v_{1, i}-v_{1, f} \cos \theta_{1, f}\right)$  to find $v_{2, f}=1.5 \mathrm{~m} \cdot \mathrm{s}^{-1}$.
\end{answer}
\subsection{ Two-Dimensional Collision in Center-of-Mass Reference Frame}
Consider a collision between particle 1 of mass $m_{1}$ and velocity $\overrightarrow{\mathbf{v}}_{1, i}$ and particle 2 of mass $m_{2}$ at rest in the laboratory frame. Particle 1 is scattered elastically through a scattering angle $\Theta$ in the center-of-mass frame. The center-of-mass velocity is given by\\
$$\overrightarrow{\mathbf{v}}_{c m}=\frac{m_{1} \overrightarrow{\mathbf{v}}_{1, i}}{m_{1}+m_{2}}$$
In the center-of-mass frame, the momentum of the system of two particles is zero
$$
\overrightarrow{\mathbf{0}}=m_{1} \overrightarrow{\mathbf{v}}_{1, i}^{\prime}+m_{2} \overrightarrow{\mathbf{v}}_{2, i}^{\prime}=m_{1} \overrightarrow{\mathbf{v}}_{1, f}^{\prime}+m_{2} \overrightarrow{\mathbf{v}}_{2, f}^{\prime}
$$
Therefore
$$
\begin{aligned}
&\overrightarrow{\mathbf{v}}_{1, i}^{\prime}=-\frac{m_{2}}{m_{1}} \overrightarrow{\mathbf{v}}_{2, i}^{\prime} \\
&\overrightarrow{\mathbf{v}}_{1, f}^{\prime}=-\frac{m_{2}}{m_{1}} \overrightarrow{\mathbf{v}}_{2, f}^{\prime}
\end{aligned}
$$
The energy condition in the center-of-mass frame is
$$
\frac{1}{2} m_{1} v_{1, i}^{\prime 2}+\frac{1}{2} m_{2} v_{2, i}^{\prime 2}=\frac{1}{2} m_{1} v_{1, f}^{\prime 2}+\frac{1}{2} m_{2} v_{2, f}^{\prime 2} .
$$
substitute the value of velocities in this equation yields\\
$$v_{1, i}^{\prime}=v_{1, f}^{\prime}$$
(we are only considering magnitudes). Therefore
$$
v_{2, i}^{\prime}=v_{2, f}^{\prime} .
$$
Because the magnitude of the velocity of a particle in the center-of-mass reference frame is proportional to the relative velocity of the two particles, imply that the magnitude of the relative velocity also does not change\\
$$\left|\overrightarrow{\mathbf{V}}_{1,2, i}^{\prime}\right|=\left|\overrightarrow{\mathbf{V}}_{1,2, f}^{\prime}\right|$$
 Recall that the relative velocity is independent of the reference frame\\
 $$\overrightarrow{\mathbf{V}}_{1, i}-\overrightarrow{\mathbf{V}}_{2, i}=\overrightarrow{\mathbf{V}}_{1, i}^{\prime}-\overrightarrow{\mathbf{V}}_{2, i}^{\prime}$$
 In the laboratory reference frame $\overrightarrow{\mathbf{v}}_{2, i}=\overrightarrow{\mathbf{0}}$, hence the initial relative velocity is $\overrightarrow{\mathbf{v}}_{1,2, i}^{\prime}=\overrightarrow{\mathbf{v}}_{1,2, i}=\overrightarrow{\mathbf{v}}_{1, i}$, and the velocities in the center-of-mass frame of the particles are then
 $$\begin{gathered}
 \overrightarrow{\mathbf{v}}_{1, i}^{\prime}=\frac{\mu}{m_{1}} \overrightarrow{\mathbf{v}}_{1, i} \\
 \overrightarrow{\mathbf{V}}_{2, i}^{\prime}=-\frac{\mu}{m_{2}} \overrightarrow{\mathbf{v}}_{1, i}
 \end{gathered}$$
 Therefore the magnitudes of the final velocities in the center-of-mass frame are
 $$
 \begin{aligned}
 &v_{1, f}^{\prime}=v_{1, i}^{\prime}=\frac{\mu}{m_{1}} v_{1,2, i}^{\prime}=\frac{\mu}{m_{1}} v_{1,2, i}=\frac{\mu}{m_{1}} v_{1, i} . \\
 &v_{2, f}^{\prime}=v_{2, i}^{\prime}=\frac{\mu}{m_{2}} v_{1,2, i}^{\prime}=\frac{\mu}{m_{2}} v_{1,2, i}=\frac{\mu}{m_{2}} v_{1, i} .
 \end{aligned}
 $$
 \subsection{Relation between scattering angles in laboratory and centre of mass frame for particle undergoing elastic collision}
 Consider a particle of mass $m_1$ moving with velocity $\vec{u_1}$ in the laboratory frame and let it collide with particles of mass $m_2$ at rest ,the collision being perfectly elastic .After collision the incident particle moves with a velocity $\vec{v_1}$ making scattering angle $\theta_{1}$ with the initial direction and the target particle of mass $m_2$ moves with a velocity $\vec{v_2}$ making recoil angle $\theta_{2}$ with the initial direction of motion of $m_1$.The initial path of $m_1$ is along the X- axis and the plane containing $u_1$ and $v_1$ is the X-Y plane as ahown in figure.\\
 \begin{figure}[H]
 	\centering
 	\includegraphics[height=4cm,width=9cm]{collision}
 	\caption{}
 	\label{}
 \end{figure}
 Let $\vec{v_1^{\prime}}$ and $\vec{v_2^{\prime}}$ be the final velocities of the particle $m_1$ and $m_2$ after collision in the center of mass frame making an angle $\theta$ with the X-axis as shown in the figure.
 \begin{figure}[H]
 	\centering
 	\includegraphics[height=4cm,width=9cm]{diagram-20220221(14)-crop}
 	\caption{}
 	\label{}
 \end{figure}
 Then\\
 $$\vec{v_1^{\prime}}=\vec{v_1}-\vec{V}_{cm} \hspace{1cm} \vec{v_2^{\prime}}=\vec{v_2}-\vec{V}_{cm}$$
As $$\vec{u_2}=0 ,\quad \vec{V}_{cm}=\frac{m_1u_1}{m_1+m_2}$$
ie $\vec{V}_{cm}$ and $u_1$ have the same direction along X-axis.Therefore $\vec{V}_{cm}$ has no component along the Y-axis .The y component of the velocity of the particle of mass $m_1$ is the same in both frames.
\begin{equation}
v_1\sin\theta_1=v_1^{\prime}\sin\theta
\end{equation}
As the center of mass has a velocity $\vec{V}_{cm}$ along X-axis with respect to laboratory frame.
\begin{equation}
v_1\cos \theta_{1}=v_1^{\prime}\cos\theta +V_{cm}
\end{equation}
 Divide (1.46) by (1.47),we have \\
 $$\tan\theta_{1}=\frac{v_1^{\prime}\sin\theta}{v_1^{\prime}\cos\theta +V_{cm}}=\frac{\sin\theta}{\cos \theta +\frac{V_{cm}}{v_1^{\prime}}}$$
 But $$V_{\prime}=\frac{m_1}{m_1+m_2}u_1$$
 and $$v_1^{\prime}=\frac{m_2}{m_1+m_2}u_1$$
 Dividing we get $$\frac{V_{cm}}{v_1^{\prime}}=\frac{m_1}{m_2}$$
 $$\tan\theta_1=\frac{\sin\theta}{\cos\theta+\frac{m_1}{m_2}}$$
 Special cases (i)\textbf{when $m_1<<<<m_2$}\\
 in this case $m_1/m_2$ can be neglected and we have
 $$\tan\theta_{1}=\frac{\sin \theta}{\cos\theta}=\tan\theta$$
 Thus if the incident particle is very light as compared to the target particle ,the angle of scattering for the incident particle in the laboratory and center of mass system are very nearly equal.\\
 Case (ii)\textbf{when $m_1=m_2$}\\
 In this case $m_1/m_2$=1\\
 Hence $$\tan\theta_{1}=\frac{\sin\theta}{1+\cos \theta}=\tan(\theta/2)$$
 $$\theta_{1}=\theta/2$$
 Thus if the incident and target particle are of equal masses ,the angle of scattering in the laboratory system is the half the angle of scattering in the CM system.

 \newpage
 \begin{abox}
 	Practice set 1
 	\end{abox}
 \begin{enumerate}
 	\item The acceleration due to gravity $(g)$ on the surface of Earth is approximately $2.6$ times that on the surface of Mars. Given that the radius of Mars is about one half the radius of Earth, the ratio of the escape velocity on Earth to that on Mars is approximately
 	{\exyear{NET JUNE 2011}}
 \begin{tasks}(2)
 	\task[\textbf{A.}] $1.1$
 	\task[\textbf{B.}]$1.3$
 	\task[\textbf{C.}]$2.3$
 	\task[\textbf{D.}]$5.2$
 \end{tasks}
	\item Two particles of identical mass move in circular orbits under a central potential $V(r)=\frac{1}{2} k r^{2}$. Let $l_{1}$ and $l_{2}$ be the angular momenta and $r_{1}, r_{2}$ be the radii of the orbits respectively. If $\frac{l_{1}}{l_{2}}=2$, the value of $\frac{r_{1}}{r_{2}}$ is:
	{\exyear{NET DEC 2011}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\sqrt{2}$
	\task[\textbf{B.}]$1 / \sqrt{2}$
	\task[\textbf{C.}] 2
	\task[\textbf{D.}] $1 / 2$
\end{tasks}

	\item A planet of mass $m$ moves in the inverse square central force field of the Sun of mass $M$. If the semi-major and semi-minor axes of the orbit are $a$ and $b$, respectively, the total energy of the planet is:
	{\exyear{NET DEC 2011}}
\begin{tasks}(2)
	\task[\textbf{A.}] $-\frac{G M m}{a+b}$
	\task[\textbf{B.}]$-G M m\left(\frac{1}{a}+\frac{1}{b}\right)$
	\task[\textbf{C.}]$-\frac{G M m}{a}\left(\frac{1}{b}-\frac{1}{a}\right)$
	\task[\textbf{D.}]$-G M m\left(\frac{a-b}{(a+b)^{2}}\right)$
\end{tasks}
	\item A planet of mass $m$ moves in the gravitational field of the Sun (mass $M$ ). If the semimajor and semi-minor axes of the orbit are $a$ and $b$ respectively, the angular momentum of the planet is
	{\exyear{NET DEC 2012}}
\begin{tasks}(2)
	\task[\textbf{A.}]$\sqrt{2 G M m^{2}(a+b)}$
	\task[\textbf{B.}]$\sqrt{2 G M m^{2}(a-b)}$
	\task[\textbf{C.}]$\sqrt{\frac{2 G M m^{2} a b}{a-b}}$
	\task[\textbf{D.}]$\sqrt{\frac{2 G M m^{2} a b}{a+b}}$
\end{tasks}
	\item A planet of mass $m$ and an angular momentum $L$ moves in a circular orbit in a potential, $V(r)=-k / r$, where $k$ is a constant. If it is slightly perturbed radially, the angular frequency of radial oscillations is
	{\exyear{NET JUNE 2013}}
\begin{tasks}(2)
	\task[\textbf{A.}] $m k^{2} / \sqrt{2} L^{3}$
	\task[\textbf{B.}]$m k^{2} / L^{3}$
	\task[\textbf{C.}]$\sqrt{2} m k^{2} / L^{3}$
	\task[\textbf{D.}]$\sqrt{3} m k^{2} / L^{3}$
\end{tasks}
	\item The radius of Earth is approximately $6400 \mathrm{~km}$. The height $h$ at which the acceleration due to Earth's gravity differs from $g$ at the Earth's surface by approximately $1 \%$ is
	{\exyear{NET DEC 2014}}
\begin{tasks}(2)
	\task[\textbf{A.}] $64 \mathrm{~km}$
	\task[\textbf{B.}] $48 \mathrm{~km}$
	\task[\textbf{C.}]$32 \mathrm{~km}$
	\task[\textbf{D.}]$16 \mathrm{~km}$
\end{tasks}
	\item The probe Mangalyaan was sent recently to explore the planet Mars. The inter-planetary part of the trajectory is approximately a half-ellipse with the Earth (at the time of launch),Sun and Mars (at the time the probe reaches the destination) forming the major axis. Assuming that the orbits of Earth and Mars are approximately circular with radii $R_{E}$ and $R_{M}$, respectively, the velocity (with respect to the Sun) of the probe during its voyage when it is at a distance
	\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=5cm]{diagram-20210926(22)-crop(1)}
	\end{figure}
	$r\left(R_{E}<<r<<R_{M}\right) \text { from the Sun, neglecting the effect of Earth and Mars, is }$
	{\exyear{NET DEC 2014}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\sqrt{2 G M \frac{\left(R_{E}+R_{M}\right)}{r\left(R_{E}+R_{M}-r\right)}}$
	\task[\textbf{B.}]$\sqrt{2 G M \frac{\left(R_{E}+R_{M}-r\right)}{r\left(R_{E}+R_{M}\right)}}$
	\task[\textbf{C.}]$\sqrt{2 G M \frac{R_{E}}{r R_{M}}}$
	\task[\textbf{D.}]$\sqrt{\frac{2 G M}{r}}$
\end{tasks}
	\item After a perfectly elastic collision of two identical balls, one of which was initially at rest, the velocities of both the balls are non zero. The angle $\theta$ between the final, velocities (in the lab frame) is
	{\exyear{NET DEC 2016}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\theta=\frac{\pi}{2}$
	\task[\textbf{B.}]$\theta=\pi$
	\task[\textbf{C.}]$0<\theta \leq \frac{\pi}{2}$
	\task[\textbf{D.}] $\frac{\pi}{2}<\theta \leq \pi$
\end{tasks}
	\item Consider circular orbits in a central force potential $V(r)=-\frac{k}{r^{n}}$, where $k>0$ and $0<n<2$. If the time period of a circular orbit of radius $R$ is $T_{1}$ and that of radius $2 R$ is $T_{2}$, then $\frac{T_{2}}{T_{1}}$
	{\exyear{NET DEC 2016}}
\begin{tasks}(2)
	\task[\textbf{A.}] $2^{\frac{n}{2}}$
	\task[\textbf{B.}]$2^{\frac{2}{3} n}$
	\task[\textbf{C.}]$2^{\frac{n}{2}+1}$
	\task[\textbf{D.}]$2^{n}$
\end{tasks}
	\item A ball weighing $100 \mathrm{gm}$, released from a height of $5 \mathrm{~m}$, bounces perfectly elastically off a plate. The collision time between the ball and the plate is $0.5 \mathrm{~s}$. The average force on the plate is approximately
	{\exyear{NET JUNE 2017}}
\begin{tasks}(2)
	\task[\textbf{A.}] $3 N$
	\task[\textbf{B.}]$2 N$
	\task[\textbf{C.}]$5 N$
	\task[\textbf{D.}]$4 N$
\end{tasks}

	\item Which of the following figures best describes the trajectory of a particle moving in a repulsive central potential $V(r)=\frac{a}{r}(a>0$ is a constant)?
	{\exyear{NET JUNE 2018}}
\begin{tasks}(2)
	\task[\textbf{A.}]\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=5cm]{diagram-20210926(47)-crop}
	\end{figure}
	\task[\textbf{B.}]\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=5cm]{diagram-20210926(48)-crop}
	\end{figure}
	\task[\textbf{C.}]\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=5cm]{diagram-20210926(49)-crop}
	\end{figure}
	\task[\textbf{D.}]\begin{figure}[H]
		\centering
		\includegraphics[height=3cm,width=5cm]{diagram-20210926(50)-crop}
	\end{figure}
\end{tasks}
	\item A particle of mass $m$ moves in a central potential $V(r)=-\frac{k}{r}$ in an elliptic orbit $r(\theta)=\frac{a\left(1-e^{2}\right)}{1+e \cos \theta}$, where $0 \leq \theta<2 \pi$ and $a$ and $e$ denote the semi-major axis and eccentricity, respectively. If its total energy is $E=-\frac{k}{2 a}$, the maximum kinetic energy is
	{\exyear{NET JUNE 2018}}
\begin{tasks}(2)
	\task[\textbf{A.}] $E\left(1-e^{2}\right)$
	\task[\textbf{B.}]$E \frac{(e+1)}{(e-1)}$
	\task[\textbf{C.}]$E /\left(1-e^{2}\right)$
	\task[\textbf{D.}]$E \frac{(e-1)}{(e+1)}$
\end{tasks}
	\item In the attractive Kepler problem described by the central potential $V(r)=\frac{-k}{r}($ where $k$ is a positive constant), a particle of mass $m$ with a non-zero angular momentum can never reach the centre due to the centrifugal barrier. If we modify the potential to
	$$
	V(r)=-\frac{k}{r}-\frac{\beta}{r^{3}}
	$$
	one finds that there is a critical value of the angular momentum $\ell_{c}$ below which there is no centrifugal barrier. This value of $\ell_{c}$ is
	{\exyear{NET DEC 2018}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\left[12 \mathrm{~km}^{2} \beta\right]^{1 / 2}$
	\task[\textbf{B.}]$\left[12 \mathrm{~km}^{2} \beta\right]^{-1 / 2}$
	\task[\textbf{C.}]$\left[12 \mathrm{~km}^{2} \beta\right]^{1 / 4}$
	\task[\textbf{D.}]$\left[12 \mathrm{~km}^{2} \beta\right]^{-1 / 4}$
\end{tasks}
 \end{enumerate}
\colorlet{ocre1}{ocre!70!}
\colorlet{ocrel}{ocre!30!}
\setlength\arrayrulewidth{1pt}
\begin{table}[H]
	\centering
	\arrayrulecolor{ocre}

	\begin{tabular}{|p{1.5cm}|p{1.5cm}||p{1.5cm}|p{1.5cm}|}
		\hline
		\multicolumn{4}{|c|}{\textbf{Answer key}}\\\hline\hline
		\rowcolor{ocrel}Q.No.&Answer&Q.No.&Answer\\\hline
		1&\textbf{c}&2&\textbf{a}\\\hline
		3&\textbf{a}&4&\textbf{d}\\\hline
		5&\textbf{b}&6&\textbf{c}\\\hline
		7&\textbf{b}&8&\textbf{a}\\\hline
		9&\textbf{c}&10&\textbf{d}\\\hline
		11&\textbf{c}&12&\textbf{b}\\\hline
		13&\textbf{c}&&\\\hline
	\end{tabular}
\end{table}

\newpage
\begin{abox}
	Practice set 2
	\end{abox}
\begin{enumerate}
	\item In a central force field, the trajectory of a particle of mass $m$ and angular momentum $L$ in plane polar coordinates is given by,\\
	$$\frac{1}{r}=\frac{m}{l^{2}}(1+\varepsilon \cos \theta)$$
	where, $\varepsilon$ is the eccentricity of the particle's motion. Which one of the following choice for $\varepsilon$ gives rise to a parabolic trajectory?
	{\exyear{GATE 2012}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\varepsilon=0$
	\task[\textbf{B.}]$\varepsilon=1$
	\task[\textbf{C.}] $0<\varepsilon<1$
	\task[\textbf{D.}] $\varepsilon>1$
\end{tasks}
	\item A particle of unit mass moves along the $x$-axis under the influence of a potential, $V(x)=x(x-2)^{2}$. The particle is found to be in stable equilibrium at the point $x=2$. The time period of oscillation of the particle is
	{\exyear{GATE 2012}}
\begin{tasks}(2)
	\task[\textbf{A.}] $\frac{\pi}{2}$
	\task[\textbf{B.}]$\pi$
	\task[\textbf{C.}]$\frac{3 \pi}{2}$
	\task[\textbf{D.}] $2 \pi$
\end{tasks}
	\item $\text { A particle of mass } m \text { is in a potential given by }$
	$$V(r)=-\frac{a}{r}+\frac{a r_{0}^{2}}{3 r^{3}}$$
	where $a$ and $r_{0}$ are positive constants. When disturbed slightly from its stable equilibrium position it undergoes a simple harmonic oscillation. The time period of oscillation is
	{\exyear{GATE 2014}}
\begin{tasks}(2)
	\task[\textbf{A.}] $2 \pi \sqrt{\frac{m r_{0}^{3}}{2 a}}$
	\task[\textbf{B.}]$2 \pi \sqrt{\frac{m r_{0}{ }^{3}}{a}}$
	\task[\textbf{C.}]$2 \pi \sqrt{\frac{2 m r_{0}^{3}}{a}}$
	\task[\textbf{D.}]$4 \pi \sqrt{\frac{m r_{0}^{3}}{a}}$
\end{tasks}

	\item A planet of mass $m$ moves in a circular orbit of radius $r_{0}$ in the gravitational potential $V(r)=-\frac{k}{r}$, where $k$ is a positive constant. The orbit angular momentum of the planet is
	{\exyear{GATE 2014}}
\begin{tasks}(2)
	\task[\textbf{A.}] $2 r_{0} \mathrm{~km}$
	\task[\textbf{B.}]$\sqrt{2 r_{0} \mathrm{~km}}$
	\task[\textbf{C.}]$r_{0} k m$
	\task[\textbf{D.}]$\sqrt{r_{0} k m}$
\end{tasks}
	\item An interstellar object has speed $v$ at the point of its shortest distance $R$ from a star of much larger mass $M$. Given $v^{2}=2 G M / R$, the trajectory of the object is
	{\exyear{GATE 2018}}
\begin{tasks}(2)
	\task[\textbf{A.}] circle
	\task[\textbf{B.}]ellipse
	\task[\textbf{C.}]parabola
	\task[\textbf{D.}]hyperbola
\end{tasks}
\end{enumerate}
\colorlet{ocre1}{ocre!70!}
\colorlet{ocrel}{ocre!30!}
\setlength\arrayrulewidth{1pt}
\begin{table}[H]
	\centering
	\arrayrulecolor{ocre}

	\begin{tabular}{|p{1.5cm}|p{1.5cm}||p{1.5cm}|p{1.5cm}|}
		\hline
		\multicolumn{4}{|c|}{\textbf{Answer key}}\\\hline\hline
		\rowcolor{ocrel}Q.No.&Answer&Q.No.&Answer\\\hline
		1&\textbf{b}&2&\textbf{b}\\\hline
		3&\textbf{a}&4&\textbf{d}\\\hline
		5&\textbf{c}&&\\\hline
	\end{tabular}
\end{table}
\newpage
\begin{abox}
	Practice set 3
	\end{abox}
\begin{enumerate}
		\item  A particle moves under a central potential $V(r)=\frac{-k}{r^{m}} .$ What should be value of $m$ for its orbit to be stable.
	\begin{answer}
		\begin{align*}
		\text{Corresponding is }f(r)&=-\frac{\partial V}{\partial r}=\frac{-m k}{r^{m+1}}=-m k r^{-(m+1)}\\
		\text{	We know that for }f&=-k r^{n}\text{ condition for stability is $n>-3$.}
		\intertext{Therefore for orbit to be stable under given potential we must have.}
		-(m+1)>-3&\text{ or }m+1<3 \quad \therefore m<2
		\end{align*}
	\end{answer}
	\item  Equation of the orbit of a particle moving under central force is $r \theta=\beta$, where $\beta$ is a constant. Find the force acting on the particle.
\begin{answer}
	\begin{align*}
	r \theta=\beta,\text{ therefore, }u&=\frac{1}{r}=\frac{\theta}{\beta} \quad \therefore \frac{\partial^{2} u}{\partial \theta^{2}}=0\\
	\text{	Differential equation of orbit, }\frac{\partial^{2} u}{\partial \theta^{2}}+u&=\frac{-m f(r)}{L^{2} u^{2}}\\
	\therefore 0+u=\frac{-m f(r)}{L^{2} u^{2}} \quad \therefore \quad f(r)&=\frac{-L^{2} u^{3}}{m}=\frac{-L^{2}}{m r^{3}}
	\end{align*}
\end{answer}
\item  Equation of orbit of a particle moving under central force is $r^{n}=a \cos n \theta .$ Find the force on the particle.
\begin{answer}
	\begin{align}
	\text{	Given, }r^{n}=a \cos n \theta,\text{ therefore, }u^{n}=\frac{1}{a \cos n \theta}&=\frac{1}{a} \sec n \theta \label{25}\\
	\text{ Taking $\ln$ both sides we get, }n \ln u&=\ln \left(\frac{1}{a}\right)+\ln \sec n \theta\notag\\
	\text{Differentiating w.r.t. $\theta$ we get, }\frac{n}{u} \frac{\partial u}{\partial \theta}&=n \tan n \theta \quad \therefore \frac{\partial u}{\partial \theta}=u \tan n \theta\notag\\
	\text{Differentiating again w.r.t. $\theta$ we get }\frac{\partial^{2} u}{\partial \theta^{2}}&=\frac{\partial u}{\partial \theta} \tan n \theta+u n \sec ^{2} n \theta\notag\\&=u \tan ^{2} n \theta+u n \sec ^{2} n \theta\notag\\
	\text{ Differential equation of orbit is }\frac{\partial^{2} u}{\partial \theta^{2}}+u&=\frac{-m f(r)}{L^{2} u^{2}}\notag\\
	u \tan ^{2} n \theta+u n \sec ^{2} n \theta+u&=\frac{-m f(r)}{L^{2} u^{2}}, u \sec ^{2} n \theta+u n \sec ^{2} n \theta=\frac{-m f(r)}{L^{2} u^{2}}\notag\\
	\therefore f(r)&=-\frac{L^{2} u^{3}(1+n) \sec ^{2} n \theta}{m}\notag\\
	\text{from (\ref{25}) }\sec ^{2} n \theta&=a^{2} u^{2 n}\notag\\
	\therefore f(r)&=\frac{-L^{2}(n+1) a^{2} u^{2 n+3}}{m}\notag\\
	\therefore \quad f(r)&=\frac{-L^{2}(n+1) a^{2}}{m} \cdot \frac{1}{r^{2 n+3}}\notag\\
	\text{Or }
	f(r) \propto \frac{1}{r^{2 n+3}}&
	\text{and the force is attractive in nature.}\notag
	\end{align}
\end{answer}
	\item  For a particle moving under gravitational force pericentre distance in parabolic orbit is $r_{p}$ while the radius of the circular orbit with same angular momentum is $r_{c}$. What is relation between $r_{p}$ and $r_c$ ?
\begin{answer}
	Pericentre distance is the minimum distance $(\cos \theta=\max =1)$ and for parabolic orbit $e=1 .$
	\begin{align*}
	\text{Therefore, }r_{\min }&=\frac{l}{1+e \cos \theta}=\frac{l}{2}=r_{p}\\
	\text{for circular orbit $e=0$, therefore }r_{c}&=\frac{l}{1+e \cos \theta}=l \quad \therefore r_{p}=\frac{r_{c}}{2}
	\end{align*}
\end{answer}
\item  Ratio of maximum to minimum speed of a planet revolving around the sun in an elliptical orbit is $2: 1$, What is eccentricity of the orbit?
\begin{answer}
	\begin{align*}
	\text{	Given }\frac{v_{\max }}{v_{\min }}=\frac{2}{1} \therefore \frac{\sqrt{\frac{G M}{a}\left(\frac{1+e}{1-e}\right)}}{\sqrt{\frac{G M}{a}\left(\frac{1-e}{1+e}\right)}}=\frac{2}{1} \quad\text{ or }\quad \frac{1+e}{1-e}=\frac{2}{1} \quad \therefore e=\frac{1}{3}
	\end{align*}
\end{answer}
\item  A planet is revolving around the sun in a circular orbit. Due to some reason the speed of the planet suddenly becomes double. What is new orbit of the planet.
\begin{answer}
	\begin{align*}
	\text{	Orbital speed of the planet is }&\sqrt{\frac{G M}{r}}\\
	\text{New speed of the planet }&=2 \sqrt{\frac{G M}{r}}\\
	\text{Therefore, new energy of the planet }&=\frac{1}{2} m v^{2}-\frac{G M m}{r}=\frac{1}{2} m \cdot \frac{4 G M}{r}-\frac{G M m}{r}=\frac{2 G M m}{r}>0
	\intertext{	Total energy of the planet becomes positive on doubling its speed therefore new orbit of the planet will be hyperbolic.}
	\end{align*}
\end{answer}
\item  Two masses constrained to move in a horizontal plane collide. Given initially $m_{1}=$ $85 \mathrm{gms}, m_{2}=200 \mathrm{gms} ; u_{1}=6.48 \mathrm{cms} / \mathrm{sec}$ and $u_{2}=-6.78 \mathrm{cms} / \mathrm{sec}$, find the velocity of centre of mass.
\begin{answer}
\begin{align*}
 \intertext{The velocity of centre of mass is given by}
\vec{V}_{\mathrm{cm}}&=\frac{m_{1} \overrightarrow{u_{1}}+m_{2} \overrightarrow{u_{2}}}{m_{1}+m_{2}}\\
\vec{V}_{\mathrm{cm}}&=\frac{85 \times 6.48+200 \times(-6.78)}{85+200}=2.82 \mathrm{~cm} / \mathrm{sec} \text{in the direction of motion of $m_{2}$}
\end{align*}
\end{answer}
\item Two particles each of mass 2kg are moving with velocities $2\vec{i}+4\vec{j}$m/s and $5\vec{i}+6\vec{j}$m/s respectively.Find the kinetic energy of the system relative to the center of mass.
\begin{answer}
	\begin{align*}
	\text { Given } m_{1}&=m_{2}=2 \mathrm{~kg} . ; u_{1}=3 \hat{i}+4 \hat{j} ; \quad u_{2}=5 \hat{i}+6 \hat{j}\\
	\text{velocity of centre of mass}\quad  \vec{V}_{cm}&=\frac{m_{1} \overrightarrow{u_{1}}+m_{2} \overrightarrow{u_{2}}}{m_{1}+m_{2}}=\frac{2(3 \hat{i}+4 \hat{j})+2(5 \hat{i}+6 \hat{j})}{2+2}=4 \hat{i}+5 \hat{j}\\
	\intertext{Velocity of $m_1$ in centre of mass frame}
	\vec{u}_1^{\prime}&=\vec{u}_1-\vec{V}_{cm}\\
	\vec{u}_1^{\prime}&=3\hat{i}+4\hat{j}-4\hat{i}-5\hat{j}=-\hat{i}-\hat{j}\\
	\intertext{Velocity of $m_2$ in centre of mass frame}
	\vec{u}_2^{\prime}&=\vec{u}_2-\vec{V}_{cm}\\
	\vec{u}_2^{\prime}&=5\hat{i}+6\hat{j}-4\hat{i}-5\hat{j}\\
	\intertext{Kinetic energy relative to centre of mass before collision}
	&=\frac{1}{2}m_1u_1^{\prime 2}+\frac{1}{2}m_2u_2^{\prime 2}\\
	&=\frac{1}{2}m_1|\sqrt{(-1)^2+(-1)^2}|^2+\frac{1}{2}m_2|\sqrt{(1)^2+(1)^2}|^2\\
	&=2+2=4 Joule
	\end{align*}
\end{answer}
\item A particle of mass $m_{1}$ moving with a velocity $\overrightarrow{u_{1}}$ is elastically scattered from another particle of mass $m_{2}$. After collision the two particles move in opposite directions with the same speed. Find the relation between the two masses.
\begin{answer}
	Ans. Let $\overrightarrow{v_{1}}$ and $\overrightarrow{v_{2}}$ be the velocities of the two particles after collision, then\\ $\overrightarrow{v_{1}}=-\overrightarrow{v_{2}}$ \\
	The particle of mass $m_{2}$ is at rest\\
	$\therefore$ Linear momentum before collision $=m_{1} \overrightarrow{u_{1}}$\\
+	Linear momentum after collision $=m_{1} \overrightarrow{v_{1}}+m_{2} \o
verrightarrow{v_{2}}=m_{1} \overrightarrow{v_{1}}-m_{2} \overrightarrow{v_{1}}=\left(m_{1}-m_{2}\right) \overrightarrow{v_{1}}$\\
	According to the law of conservation of linear momentum
	\begin{align*}
	m_{1} \overrightarrow{u_{1}} &=\left(m_{1}-m_{2}\right) \overrightarrow{v_{1}} \\
	\overrightarrow{v_{1}} &=\frac{m_{1}}{m_{1}-m_{2}} \overrightarrow{u_{1}}
	\end{align*}
	Considering magnitudes only $\left|\overrightarrow{v_{1}}\right|=\frac{m_{1}}{m_{1}-m_{2}}\left|\overrightarrow{u_{1}}\right|$\\
	 According to the law of conservation of energy
	\begin{align*}
		&\frac{1}{2} m_{1}\left|\overrightarrow{u_{1}}\right|^{2}=\frac{1}{2} m_{1}\left|\overrightarrow{v_{1}}\right|^{2}+\frac{1}{2} m_{2}\left|\overrightarrow{v_{1}}\right|^{2} \\
		&=\frac{1}{2}\left(m_{1}+m_{2}\right)\left|\overrightarrow{v_{1}}\right|^{2}=\frac{1}{2} \frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)^{2}} m_{1}^{2}\left|\overrightarrow{u_{1}}\right|^{2} \\
		&\left(m_{1}-m_{2}\right)^{2}=\left(m_{1}+m_{2}\right) m_{1} \\
		&m_{1}^{2}+m_{2}^{2}-2 m_{1} m_{2}=m_{1}^{2}+m_{1} m_{2} \\
		&m_{2}^{2}=3 m_{1} m_{2} \quad \text { or } \quad m_{2}=3 m_{1}
	\end{align*}
\end{answer}
\end{enumerate}

Classical Mechanics  -CSIR/chapter/central force motions.tex


\section{Preliminaries: decision diagrams and stabilizer states \label{sec:preliminaries}}
%\todo[inline]{mention terminology `low' and `high' for 0-edge and 1-edge}

%Here, we briefly introduce two methods to manipulate and succinctly represent quantum states: decision diagrams and stabilizer states.
%For an introduction to quantum computation, see appendix~\ref{sec:quantum-nutshell}.

The computational unit of quantum computers are quantum bits or qubits.
A single-qubit state is a complex vector $(\alpha_0, \alpha_1)^{T} \in \mathbb{C}^2$ with norm 1, usually written in Dirac notation as $\alpha_0 \ket{0} + \alpha_1 \ket{1}$.
Two quantum states which differ only by a complex multiple are considered equal.
The joint state $\ket{\phi}$ of $n$ quantum bits can be written as
\begin{equation}
    \label{eq:quantum-state-expansion}
    \sum_{x_1, x_2, \dots, x_n \in \set{0,1}} f(x_1, x_2, \dots, x_n) \ket{x_1}\otimes\ket{x_2}\otimes \dots \otimes\ket{x_n}\vspace{-1em}
\end{equation}
for a function $f: \{0, 1\}^n \rightarrow \mathbb{C}$, where $\otimes$ denotes the tensor product.
An example two-qubit state is $\left( \ket{0} \otimes \ket{0} + i \ket{1} \otimes \ket{1}\right) / \sqrt{2} = \frac{1}{\sqrt{2}} (1,0,0,i)^T$.
Alternatively to \autoref{eq:quantum-state-expansion}, we can recursively describe an $n$-qubit quantum state $\ket{\phi}$ for $n>1$ as
\begin{equation}
\ket{\phi} = \alpha_0 \ket{0} \otimes \ket{\phi_0} + \alpha_1 \ket{1} \otimes \ket{\phi_1}, \text{where $\ket{\phi_0}, \ket{\phi_1}$ are $n$-1-qubit states and $\alpha_0,\alpha_1  \in \mathbb{C}$.}
    \label{eq:quantum-state-recursive}
\end{equation}
An Algebraic Decision Diagram (\add) represents a function of the form $f\colon \{0, 1\}^n \rightarrow \mathbb{C}$, and thus also a quantum state via \autoref{eq:quantum-state-expansion}, see \autoref{fig:qmdd-isoqmdd-exposition} for an example.
%A succint representation of functions of the form $f\colon \{0, 1\}^n \rightarrow \mathbb{C}$, and thus of quantum states via \autoref{eq:quantum-state-expansion}, has been achieved by the use of Algebraic Decision Diagrams (\adds), see \autoref{fig:qmdd-isoqmdd-exposition} for an example.
An \add is a rooted directed acyclic graph (DAG), which has a leaf node for each unique value in the image of $f$, i.e., in $\{f(\vec{x}) \mid \vec{x} \in \{0, 1\}^n \}$.
Each path from the root to a leaf visits nodes representing the variables $x_1, x_2, x_3, x_4$; one variable at each level of the diagram.
The value $f(x_1, \dots, x_n)$ is found by traversing such a path, following the \concept{low edge} (dashed line)
 when $x_i=0$, and the  \concept{high edge} (solid line) when $x_i=1$;
 so, e.g., $f(1,1,1,0) = -i$ in \autoref{fig:qmdd-isoqmdd-exposition}.
Hence every node in an \add, not only the root node, can be said to represent a function.

%\adds for functions $f$ on length-$n$ bitstrings can be defined recursively on $n$: in case $n=1$, then the \add has a single node with two outgoing edge, labelled $0$ and $1$ and pointing to the leaf node holding the value $f(0)$ and $f(1)$ respectively.
%If $f(0) = f(1)$, then the \add only has a single leaf node and both edges point to that node.
%If $n>1$, then and \add for $f$ is constructed by first constructing \adds for the two functions $f_0: (x_2, x_3, \dots, x_n) \mapsto f(0, x_2, x_3, \dots, x_n)$ and $f_1: (x_2, x_3, \dots, x_n) \mapsto f(1, x_2, x_3, \dots, x_n)$.
%An \add for $f_0$ is then made by adding a fresh node and a $0$-edge ($1$-edge) pointing to the root node of the \add of $f_0$ (the \add of $f_1$).
%If the \adds of $f_0$ and $f_1$ are identical, then remove one of them and make both edges point to the remaining one.
%Thus, in an \add, the value $f(x_1, \dots, x_n)$ is found by traversing the path $x_1 \rightarrow x_2 \rightarrow \dots \rightarrow x_n$, e.g. $f(1,1,0) = -1/\sqrt{2}$ in \autoref{fig:add-qmdd-example}.
%Although generally, functions $f: \{0, 1\}^n \rightarrow \mathbb{C}$ have exponentially many partial assignments, many functions encountered in practice have only a polynomial number of unique subfunctions, and therefore can be represented by polynomial-size \add.

A \emph{partial assignment} $(x_1=a_1,\ldots, x_k=a_k)$ to the variables induces a \emph{subfunction} $f_a$, defined as $f_a(x_{k+1},\ldots, x_n)\defn f(a_1,\ldots, a_k,x_{k+1},\ldots, x_n)$.
Two \add nodes representing the same subfunction can be \emph{merged}, i.e., one node is deleted and its incident edges are rerouted to the other.
When all eligible nodes have been merged, an \add is \emph{reduced}.
The nodes of a reduced \add are in one-to-one correspondence with the unique \emph{subfunctions} of $f$.
An \add is a \emph{canonical} representation: a given function has exactly one reduced \add.
\adds can represent both states and matrices, and there are algorithms which multiply a matrix with a vector in \add form.
An \add can represent any quantum state, using exponentially many nodes in the worst case.
A given quantum gate can be compiled into an \add, thus allowing one to simulate any quantum circuit by repeatedly multiplying a gate's matrix with a state. %\todo[inline]{@Lieuwe: David asks for ref here}
%For example, the functions $(x_2, x_2, \dots, x_n) \mapsto f(0, x_2, x_3, \dots, x_n)$ and $(x_4, x_5, \dots, x_n) \mapsto f(0, 1, 1, x_4, x_5, \dots, x_n)$ are subfunctions of $f$.
%At a node in an \add corresponding to the $k$-th variable, the two outgoing arrows correspond to the partial assignments $x_k=0$ and $x_k=1$.

The Quantum Multi-valued Decision Diagram (\qmdd) \cite{miller2006qmdd} improves on the \add representation by also merging two nodes when they represent functions $f,g$ that are related by $f=\lambda\cdot g$ for some $\lambda\in\mathbb C^{\ast}$.
\autoref{fig:qmdd-isoqmdd-exposition} gives an example.
Each edge is labelled with a weight. % (the normalization constant $\lambda$).%$^{\ref{fn:norm}}$
To read a value of $f$, traverse the \qmdd from the root to the leaf just as in an \add, and multiply the weights of all edges on that path.
Like \adds, \qmdds are a canonical representation and can be used to simulate any quantum circuit.

The single-qubit Pauli operators are
$2\times 2$ unitary matrices:
\begin{equation*}
\id[2] \defn \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix},
X \defn \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix},
Y \defn \begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix},
Z \defn \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}
\label{eq:pauli-matrices}
\end{equation*}
where $i$ is the complex unit.
An $n$-qubit operator of the form $P_n\otimes\cdots\otimes P_1$ is called a \emph{Pauli string} if $P_j$ are single-qubit Pauli operators.
The $n$-qubit Pauli strings generate a nonabelian group $\Pauli_n$ (under matrix multiplication), consisting of all operators of the form $\lambda P_n\otimes\cdots\otimes P_1$ with $\lambda \in \{\pm 1, \pm i\}$.
Note that the high indices are on the left, in keeping with the custom that the least significant (qu)bit is the first qubit, and the leftmost operator $P_n$ acts on the most significant qubit.
Pauli operators $A, B$ either commute ($A\cdot B = B\cdot A$) or anticommute ($A \cdot B = - B \cdot A$).

In contrast to decision diagrams, the stabilizer formalism forms a subset of quantum computation that is efficiently simulatable.
A stabilizer state on $n$ qubits can be prepared from the state $\ket{0}^{\otimes n}$ by repeatedly applying any of the following gates (generators of the Clifford set):
\begin{equation}
    H \defn
        \frac{1}{\sqrt{2}}
        \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}
            ,
        S
        \defn
        \begin{pmatrix} 1 & 0\\ 0 & i \end{pmatrix},
            \textnormal{CNOT} \defn
        \begin{pmatrix}
            1 & 0 & 0 & 0\\
            0 & 1 & 0 & 0\\
            0 & 0 & 0 & 1\\
            0 & 0 & 1 & 0
        \end{pmatrix}.
        \label{eq:clifford-generators}
\end{equation}
There exist $2^{\Theta(n^2)}$ stabilizer states on $n$ qubits \cite{aaronson2008improved}; examples are $\ket{00}$ and $(\ket{00} + \ket{11}) / \sqrt{2}$.
A strict subset of stabilizer states is the set of graph states \cite{hein2006entanglement}.
The relationship between graph states and stabilizer states has been extensively investigated in, e.g., \cite{nest2004graphical,nest2005local}
%\todo{Move to S4.1?}
For an undirected graph $G=(V,E)$, the graph state $\ket{G}$ is the following state on $n$ qubits, where $Z_{jk}$ denotes the controlled $Z$-gate between qubits $j$ and $k$.
\begin{equation}
    \frac{1}{\sqrt{2}^n} \prod_{(j, k) \in E} Z_{jk} \left(\ket{0} + \ket{1}\right)^{\otimes n}, \text{ where e.g. }
        Z_{1,2} =
        \begin{pmatrix}
            1 & 0 & 0 & 0\\
            0 & 1 & 0 & 0\\
            0 & 0 & 1 & 0\\
            0 & 0 & 0 & -1
        \end{pmatrix} \text{ for } n=2.
	\label{eq:graph-state-definition}
\end{equation}
%Although not every stabilizer state is a graph state, any stabilizer state can be converted into one by local Cliffords (LC)~\cite{nest2004graphical}.
%Moreover, since LC-equivalence between two graphs is efficient~\cite{nest2005local}, so is LC-equivalence between stabilizer states.\todo{Tim: Vedran mentioned this and I added it here, but I am not so sure this is the right place to mention this... All Paulis are CLiffords so checking local-Pauli equivalence is easy for stabilizer states, but we consider arbitrary states...}
An $n$-qubit stabilizer state $\ket{\phi}$ is uniquely specified by the set $S$ of Pauli operators $A \in \Pauli_n$ for which $A\ket{\phi} = \ket{\phi}$.
This set $S$ is an abelian group of $2^n$ elements, succinctly represented by $n$ independent generators.
Since each Pauli generator takes $\mathcal{O}(n)$ space to represent, an $n$-qubit stabilizer is represented by $\mathcal O(n^2)$ bits.
Updating the stabilizer generating set after application of one of the gates from eq.~\eqref{eq:clifford-generators} or a single-qubit computational-basis measurement can be done in polynomial time \cite{gottesman1998heisenberg}.
Also, we note that multiplying two $n$-qubit Pauli strings can be done in $O(n)$ time by using the property of the tensor product $\otimes$ that $(a\otimes b) \cdot (c\otimes d) = (a\cdot c) \otimes (b\cdot d)$.

Stabilizer-rank based methods~\cite{bravyi2016trading,bravyi2017improved,bravyi2019simulation, huang2019approximate,kocia2018stationary,kocia2020improved} extend this approach to families of Clifford circuits with arbitrary input states $\ket{\phi_n}$, enabling the simulation of universal quantum computation in general~\cite{bravyi2005universal}.
By decomposing $\ket{\phi_n}$ as linear combination of $\chi$ stabilizer states, the measurement outcome probabilities can be computed in time $\oh(\chi \cdot \textnormal{poly}(n))$, where the least $\chi$ is referred to as the \concept{stabilizer rank}.
Therefore, stabilizer-rank based methods are efficient for a family of input states $\ket{\phi_n}$ with a stabilizer rank polynomially growing~in~$n$.

In this work we will also consider stabilizer groups of states which are not stabilizer states.
In general, we will refer to an abelian subgroup of $\Pauli_n$, not containing $-\id[2]^{\otimes n}$, as an $n$-qubit \emph{stabilizer subgroup}, which generally has $\leq n$ generators.
Such objects are also studied in the context of simulating mixed states \cite{audenaert2005entanglement} and quantum error correction~\cite{gottesman1997stabilizer}.
Examples of stabilizer subgroups are $\{\id[2]\}$ for $\ket{0} + e^{i\pi/4}\ket{1}$, $\langle -Z\rangle$ for $\ket{1}$ and $\langle X \otimes X\rangle$ for $(\ket{00} + \ket{11}) + 2(\ket{01} + \ket{10})$.

Any $n$-qubit Pauli string can (modulo factor $\in \{\pm 1, \pm i\}$) be written as $(X^{x_n} Z^{z_n}) \otimes \dots \otimes (X^{x_1} Z^{z_1})$ for bits $x_j, z_j, 1 \leq j \leq n$.
We can therefore write an $n$-qubit Pauli string $P$ as length-$2n$ binary vector
\[
    (\underbrace{x_n, x_{n-1}, \dots x_1}_{\textnormal{X block}} | \underbrace{z_n, z_{n-1}, \dots, z_1}_{\textnormal{Z block}})
    ,
\]
where we added the horizontal bar ($|$) only to guide the eye.
We will refer to such vectors as \emph{check vectors}.
For example $X \sim (1, 0)$ and $Z \otimes Y \sim (0, 1 | 1, 1)$ \cite{aaronson2008improved}.
A set of $k$ Pauli strings thus can be written as $2n\times k$ binary matrix, often called \emph{check matrix}, e.g.
\[
    \begin{pmatrix}
        X &\otimes& X &\otimes& X\\
        \id[2] &\otimes& Z &\otimes& Y
    \end{pmatrix}
    \sim
    \begin{pmatrix}
        1& 1& 1& |& 0& 0& 0\\
        0& 0& 1& |& 0& 1& 1
    \end{pmatrix}
    .
\]
This equivalence induces an ordering on Pauli strings following the lexicographic ordering on bit strings. %, defined as $\vec{y} \leq \vec{z}$ if either $\vec{y} = \vec{z}$ or else if $\vec{y}$ $\vec{z}$ agree on the first $k$ elements but $\vec{y}_{k+1} < \vec{z}_{k+1}$, where $0<1$.
For example, $X<Y$ because $(1|0) < (1|1)$ and $Z\otimes \id[2] < Z \otimes X$ because $(0 0 | 1 0) < (0 1 | 1 0)$.
Furthermore, if $P, Q$ are Pauli strings corresponding to binary vectors $\vec{x}^P, \vec{z}^P$ and $\vec{x}^Q, \vec{z}^Q$, then
\[
P \cdot Q \propto
\bigotimes_{j=1}^n
\left(X^{x^P_j} Z^{z^P_j}\right) \left(X^{x^Q_j} Z^{z^Q_j}\right)
=
\bigotimes_{j=1}^n
\left(X^{x^P_j \oplus x^Q_j } Z^{z^P_j \oplus z^Q_j} \right)
\]
and therefore the group of $n$-qubit Pauli strings with multiplication (disregarding factors) is group isomorphic to the vector space $\{0, 1\}^{2n}$ with bitwise addition (i.e., exclusive or; `xor').
Consequently, many efficient algorithms for linear-algebra problems carry over to sets of Pauli strings.
In particular, if $G = \{g_1, \dots, g_k\}$ are length$-2n$ binary vectors (/ $n$-qubit Pauli strings) with $k\leq n$, then we can efficiently perform the following operations.
\begin{description}
%    \item[\emph{Orthogonalization:}] convert $G$ to a (potentially smaller) independent set, using the Gram-Schmidt procedure, in $O(n^3)$ time.
    \item[\emph{RREF:}] bring $G$ into a reduced-row echelon form (RREF) using Gauss-Jordan elimination (both standard linear algebra notions) where each row (in check matrix form) has strictly more leading zeroes than the row above.
        The RREF is achievable by $O(k^2)$ row additions (/~multiplications modulo factor) and thus $O(k^2 \cdot n)$ time (see \cite{berg2020circuit} for a similar algorithm).
        In the RREF, the first $1$ after the leading zeroes in a row is called a `pivot'.
    \item[\emph{Independent Set}] convert $G$ to a (potentially smaller) independent set by performing the RREF procedure and discarding resulting all-zero rows.
    \item[\emph{Membership:}] determining whether a given a vector (/~Pauli string) $h$ has a decomposition in elements of $G$.
        This task can be reduced to independence by first getting $G^{\textnormal{RREF}}$ by applying RREF, followed by adding $h$ to $G$ and performing the Independent-Set procedure. The result has $|G^{\textnormal{RREF}}|$ rows if $h\in \langle G\rangle$, and $|G^{\textnormal{RREF}}| + 1$ rows otherwise.
    \item[\emph{Intersection:}] determine all Pauli strings which, modulo a factor, are contained in both $G_A$ and $G_B$, where $G_A, G_B$ are generator sets for $n$-qubit stabilizer subgroups.
        This can be achieved using the Zassenhaus algorithm \cite{LUKS1997335} in time $O(n^3)$.
    \item[\emph{Division remainder:}] given a vector $h$  (/~Pauli string $h$), determine \mbox{$ h^{\textnormal{rem}} := \min_{g\in \langle G\rangle} \{ g  h\}$} (minimum in the lexicographic ordering) where $\oplus$ denotes bitwise XOR (/~factor-discarding multiplication).
        We do so in the check matrix picture by bringing $G$ into RREF, and then making the check vector of $h$ contain as many zeroes as possible by adding rows from $G$:
	\begin{algorithmic}[1]
        \For{column index $j=1$ to $2n$}
        \If{$h_j = 1$ and $G$ has a row $g_i$ with its pivot at position $j$}
		$h := h \oplus g_i$
        \EndIf
        \EndFor
	\end{algorithmic}
        The resulting  $h$ is $h^{\textnormal{rem}}$.
This algorithm's runtime is dominated by the RREF step; $O(n^3)$.
        %\todo{It is hard to follow. An example might help. Perhaps indeed explain where you need it?}
%        \todo[inline]{Tim: this algorithm is obvious but I have not been able to find a reference.... Need to move to elsewhere?}
% AL: I like this overview. If it is trivial let's leave it here. People might still refer to it, and we avoid claiming trivial stuff.
\end{description}

In this work, we will consider the group of $n$-qubit Pauli operators $\lambda P_n \otimes \dots P_n$ for arbitrary $\lambda \in \mathbb{C}-\{0\}$, denoted as $\paulilim_n$.
Since each stabilizer $\lambda P \in \paulilim_n$ has factor $\lambda =\pm 1$ (follows from
$ (\lambda P)\ket{\phi} = (\lambda P)^2 \ket{\phi} =  \lambda^2 \id \ket{\phi}  =\ket{\phi}$, hence $\lambda^2 = 1$), the stabilizer subgroups in $\paulilim_n$ are the same as in $\pauli_n$.
As extension of the check matrix form to $A \in \paulilim_n$, we write $A = r \cdot e^{i\theta} \cdot P_n \otimes ...\otimes P_1$, for $r \in \mathbb{R}_{> 0}$ and $\theta\in [0, 2\pi)$ and represent $A$ by a length-$(2n+2)$ vector where the last entries store $r$ and $\theta$, e.g.:
\[
    \begin{pmatrix}
        3X &\otimes& X &\otimes& X\\
        -\frac{1}{2} i\id[2] &\otimes& Z &\otimes& Y
    \end{pmatrix}
    \sim
    \begin{pmatrix}
        1& 1& 1& |& 0& 0& 0& |&3 & 0\\
        0& 0& 1& |& 0& 1& 1& |&\frac{1}{2} & \frac{3\pi}{2}
    \end{pmatrix}
\]
where we used $3 = 3\cdot e^{i\cdot 0}$ and $-\frac{1}{2}i = \frac{1}{2} \cdot e^{3\pi i/2}$.
The ordering on real numbers induces a lexicographic ordering (from left to right) on such extended check vectors, for example $(1, 1, | 0, 0 | 3, \frac{1}{2} ) < (1, 1 | 1, 0 | 2, 0)$.
Let us stress that the factor encoding $(r, \theta)$ is less significant than the Pauli string encoding $(x_n, \dots, x_1 | z_n, \dots, z_1)$.
As a consequence, we can greedily determine the minimum of two Pauli operators.
%We state this remark separately because we will explicitly use it for proving correctness of our algorithms in~\autoref{sec:choose-canonical-isomorphism-pauli}--\ref{sec:pauli-isomorphism-detection}.

%\def\rp{r}
%\def\thetap{\theta}
%\def\rq{r'}
%\def\thetaq{\theta'}
%\begin{remark}
%    \label{remark:ordering}
%    Checking which of two Pauli operators $\rp e^{i\thetap} P, \rq e^{i\thetaq} P' \in \paulilim_n$ is smaller can be performed in two greedy steps: first, declare $\rp e^{i\thetap} P <  \rq e^{i\thetaq} P'$ if $P < P'$.
%    Otherwise (i.e., if $P = P'$), then proceed to comparing the factors and declare $\rp e^{i\thetap} P <  \rq e^{i\thetaq} P'$ if and only if $(\rp, \thetap) < (\rq, \thetaq)$.
%\end{remark}

Finally, we emphasize that the algorithms above rely on row addition, which is a commutative operation.
%are correct because Pauli operators with factor-ignoring multiplication are group isomorphic to binary vectors with xor.
Since conventional (i.e., factor-respecting) multiplication of Pauli operators is not commutative, the algorithms above are not straightforwardly applicable to (nonabelian subgroups of) $\paulilim_n$.
(For abelian subgroups of $\paulilim_n$, such as stabilizer subgroups \cite{aaronson2008improved}, the algorithms still do work.)
Fortunately, since Pauli strings either commute or anti-commute, row addition may only yield an factors up to the $\pm$ sign, not the resulting Pauli strings.
This feature, combined with the stipulated order assigning least significance to the factor,
enables us to invoke the algorithms above as subroutine, with postpocessing to obtain the correct factor.
We will do so in~\autoref{sec:choose-canonical-isomorphism-pauli}--\ref{sec:pauli-isomorphism-detection}.
%
%
%, which may yield incorrect output (for example, the intersection of $\langle Z \rangle$ and $\langle -Z \rangle$ is $\{\id[2]\}$, while if we discard scalars, the Intersection algorithm above would also return $Z$ as element in the intersection).
%However, because $\paulilim_n$ with factor-respecting multiplication is a non-abelian group while row addition is, there is no commutative row addition operation that preserves factors in the Pauli picture.