Title: Dynamics and Relativity

Q: Inertial reference frames
A: \emph{Inertial frames} are frames of references in which equivalently
\begin{enumerate}
\item The frames themselves are not accelerating
\item Newton's laws hold
\item Acceleration of a particle is zero when the force impressed is zero
\end{enumerate}
Newton's first law states that inertial reference frames exist.

Q: Galilean transformations
A: These change in reference frames will generate a new inertial frame given a starting inertial frame.
\begin{itemize}
  \item Translations
    \[
      \mathbf{r}' = \mathbf{r} - \mathbf{r}_0
    \]
  \item Rotation (and reflection):
    \[
      \mathbf{r}' = R\mathbf{r}
    \]
    with $R\in O(3)$,
\item Boost
\[
  \mathbf{r}' = \mathbf{r} - \mathbf{v}t.
\]
\end{itemize}
All these transformations together generate the \emph{Galilean group} that describes the symmetry of Newtonian equations of motion.
The trivial Galilean transformation is $\vec x' =\lambda \vec x$, i.e. a change in units. Another sometimes included in the group is $t'=t+t_0$.

Q: The (Galilean) principle of relativity
A: The laws of physics are the same in all inertial frames.

Q: Newton's Second Law
A: \begin{law}
  The \emph{equation of motion} for a particle subject to a force $\mathbf{F}$ is
  \[
    \frac{\d \mathbf{p}}{\d t} = \mathbf{F},
  \]
  where $\mathbf{p} = m\mathbf{v} = m\dot{\mathbf{r}}$ is the (linear) momentum of the particle. We say $m$ is the (inertial) mass of the particle, which is a measure of its reluctance to accelerate.
\end{law}
Usually, $m$ is constant. Then
\[
  \mathbf{F} = m\mathbf{a} = m\ddot{\mathbf{r}}.
\]
Usually, $\mathbf{F}$ is specified as a function of $\mathbf{r}, \dot{\mathbf{r}}$ and $t$. Then we have a second-order ordinary differential equation for $\mathbf{r}$.
To determine the solution, we need to specify the initial values of $\mathbf{r}$ and $\dot{\mathbf{r}}$, ie. the initial position and velocity. The trajectory of the particle is then uniquely determined for all future (and past) times.

Q: Potential energy
A: Given a force field $F = F(x)$, we define the \emph{potential energy} to be a function $V(x)$ such that
  \[
    F = -\frac{\d V}{\d x}.
  \]
  or
  \[
    V = -\int F \;\d x.
  \]
  $V$ is defined up to a constant term. We usually pick a constant of integration such that the potential drops to $0$ at infinity.

Q: Conservation of Energy
A: For a conservative force, we have
\[
  m\ddot{x} = -\frac{\d V}{\d x}.
\]
  Then the total energy
  \[
    E = \frac{1}{2} m\dot{x}^2 + V(x)
  \]
  is conserved, ie. $\dot{E} = 0$.
\begin{proof}
  \begin{align*}
    \frac{\d E}{\d t} &= m\dot{x}\ddot{x} + \frac{\d V}{\d x}\dot{x}\\
    &= \dot{x}\left(m\ddot{x} + \frac{\d V}{\d x}\right)\\
    &= 0
  \end{align*}
\end{proof}
$\mathbf{F}$ is conservative iff
  \begin{align*}
    E &= T + V\\
    &= \frac{1}{2}m|\mathbf{v}|^2 + V(\mathbf{r})
  \end{align*}
  is conserved. For any particle moving under this force, the work done is equal to the change in potential energy, and is independent of the path taken between the end points. In particular, if we travel around a closed loop, no work is done.
\begin{proof}
  \begin{align*}
    \frac{\d E}{\d t} &= \frac{\d}{\d t}\left(\frac{1}{2}m\dot{\mathbf{r}}\cdot \dot{\mathbf{r}} + V\right)\\
    &= m\ddot{\mathbf{r}}\cdot \dot{\mathbf{r}} + \frac{\partial V}{\partial x_i}\frac{\d x_i}{\d t}\\
    &= (m\ddot{\mathbf{r}} + \nabla V)\cdot \dot{\mathbf{r}}\\
    &= 0
  \end{align*}
  So the energy is conserved. In this case, the work done is
  \[
    W = \int_C \mathbf{F}\cdot \d \mathbf{r} = -\int_C (\nabla V)\cdot \d \mathbf{r} = V(\mathbf{r}_1) - V(\mathbf{r}_2).
  \]
Conversely
\begin{align*}
W&=\int_C\vec F \cdot \d\vec x \\
&=\int_{t_1}^{t_2}\vec F \cdot \frac{\d\vec x}{\d t} \d t  \\
&=m\int_{t_1}^{t_2} \ddot{\vec{x}} \cdot \dot{\vec{x}} \ \d t \\
&=\frac{1}{2}m\int_{t_1}^{t_2} \frac{\d}{\d t} \left(\dot{\vec{x}} \cdot \dot{\vec{x}}\right) \d t \\
&=T(t_2)-T(t_1) \\
&=V(\vec x(t_2))-V(\vec x(t_1)) \\
\implies F&=-\nabla V
\end{align*}
\end{proof}

Q: Solving for the path of a particle through a field
A: \textbf{Analytical method} For a general potential energy $V(x)$, conservation of energy allows us to solve the problem formally:
\[
  E = \frac{1}{2}m\dot{x}^2 + V(x)
\]
Since $E$ is a constant, from this equation, we have
\begin{align*}
  \frac{\d x}{\d t} &= \pm \sqrt{\frac{2}{m}(E - V(x))}\\
  t - t_0 &= \pm \int \frac{\d x}{\sqrt{\frac{2}{m}(E - V(x))}}.
\end{align*}
To find $x(t)$, we need to do the integral and then solve for $x$.
\textbf{Graphical method} Draw the graph of $V(x)$ and imagine the particle falling.

Q: Equilibrium points in a potential
A: \begin{defi}[Equilibrium point]
  A particle is in \emph{equilibrium} if it has no tendency to move away. It will stay there for all time. Since $m\ddot{x} = -V'(x)$, the equilibrium points are the stationary points of the potential energy, ie.
  \[
    V'(x_0) = 0.
  \]
\end{defi}
Consider motion near an equilibrium point. We assume that the motion is small and we can approximate $V$ by a second-order Taylor expansion. Then we can write $V$ as
\[
  V(x) \approx V(x_0) + \frac{1}{2}V''(x_0)(x - x_0)^2.
\]
Then differentiating gives the equation of motion
\[
  m\ddot{x} = -V''(x_0)(x - x_0).\\
\]
If $V''(x_0) > 0$, then this is of the form of the harmonic oscillator. $V$ has a local minimum at $x_0$, and we say the equilibrium point is \emph{stable}. The particle oscillates with angular frequency
\[
  \omega = \sqrt{\frac{V''(x_0)}{m}}.
\]
If $V''(x_0) < 0$, then $V$ has a local maximum at $x_0$. In this case, the equilibrium point is unstable, and the solution to the equation is
\[
  x - x_0 \approx Ae^{\gamma t} + Be^{-\gamma t}
\]
for
\[
  \gamma = \sqrt{\frac{-V''(x_0)}{m}}.
\]
For almost all initial conditions, $A \not= 0$ and the particle will diverge from the equilibrium point, leading to a breakdown of the approximation.
If $V''(x_0) = 0$, then further work is required to determine the outcome.

Q: Simple pendulum
A: Suppose that the pendulum makes an angle $\theta$ with the vertical. Then the energy is
  \[
    E = T + V = \frac{1}{2}m\ell^2 \dot{\theta}^2 - mg\ell \cos\theta.
  \]
  Therefore $V\propto -\cos\theta$. We have a stable equilibrium at $\theta = 0 $, and unstable equilibrium at $\theta = \pi$.
 If $E > mg\ell$, then $\dot{\theta}$ never vanishes and the pendulum makes full circles.
  If $0 < E < mg\ell$, then $\dot{\theta}$ vanishes at $\theta = \pm \theta_0$ for some $0 < \theta_0 < \pi$ ie. $E = -mg\ell \cos\theta_0$. The pendulum oscillates back and forth. It takes a quarter of a period to reach from $\theta = 0$ to $\theta = \theta_0$. Using the general solution for a particle in a potential, oscillation period $P$ is given by
  \[
    \frac{P}{4} = \int_0^{\theta_0} = \frac{\d \theta}{\sqrt{\frac{2E}{m\ell^2} + \frac{2g}{\ell }\cos\theta}}.
  \]
  Since we know that $E = -mg\ell \cos \theta_0$, we know that
  \[
    \frac{P}{4} = \sqrt{\frac{\ell}{g}}\int_0^{\theta_0} \frac{\d \theta}{\sqrt{2\cos\theta - 2\cos\theta_0}}.
  \]
  The integral is difficult to evaluate in general, but for small $\theta_0$, we can use $\cos\theta \approx 1 - \frac{1}{2}\theta^2$. So
  \[
    P \approx 4\sqrt{\frac{\ell}{g}}\int_0^{\theta_0}\frac{\d \theta}{\sqrt{\theta_0^2 - \theta^2}} = 2\pi\sqrt{\frac{\ell}{g}}
  \]
  and is independent of the amplitude $\theta_0$. This is of course the result for the harmonic oscillator.

Q: Buckingham's Pi theorem
A: Suppose we believe that a physical quantity $Y$ depends on $3$ other physical quantities $X_1, X_2, X_3$, ie. $Y = Y(X_1, X_2, X_3)$. Let their dimensions be as follows:
\begin{itemize}
  \item $[Y] = L^aM^bT^c$
  \item $[X_i] = L^{a_i}M^{b_i}T^{c_i}$
\end{itemize}
Suppose further that we know that the relationship is a power law, ie.
\[
  Y = CX_1^{p_1}X_2^{p_2}X_3^{p_3},
\]
where $C$ is a dimensionless constant (ie. a pure number). Since the dimensions must work out, we know that
\begin{align*}
  a &= p_1a_1 + p_2a_2 + p_3a_3\\
  b &= p_1b_1 + p_2b_2 + p_3b_3\\
  c &= p_1c_1 + p_2c_2 + p_3c_3
\end{align*}
for which there is a unique solution provided that the dimensions of $X_1, X_2$ and $X_3$ are independent. So just by using dimensional analysis, we can figure out the relation between the quantities up to a constant.
In general, suppose we have many terms, and the dimensions of $X_i$ are not independent. We order the quantities so that the independent terms $[X_1], [X_2], [X_3]$ are at the front. For each of the remaining variables, form a dimensionless quantity $\lambda_i = X_iX_1^{q_1}X_2^{q_2}X_3^{q_3}$. Then the relationship must be of the form
\[
  Y = f(\lambda_4, \lambda_5, \cdots) X_1^{p_1}X_2^{p_2}X_3^{p_3}.
\]
where $f$ is an arbitrary function of the dimensionless variables.

Q: Central force
A: \begin{defi}[Central force]
  A \emph{central force} is a force with a potential $V(r)$ that depends only on the distance from the origin, $r = |\mathbf{r}|$.
\end{defi}
 Then
  \[
    \mathbf{F} = -\nabla V(r) = -\frac{\d V}{\d r} \hat{\mathbf{r}},
  \]
  where $\hat{\mathbf{r}} = \mathbf{r}/r$.
\begin{proof}
Considering the $i$th component
  \begin{align*}
    (\nabla V)_i &= \frac{\partial V}{\partial x_i}\\
& = \frac{\d V}{\d r} \frac{\partial r}{\partial x_i}\\
\end{align*}
\vspace{-1cm}
where  \[
    r^2 = x_1^2 + x_2^2 + x_3^2.
  \]
So
\begin{align*}
2r\frac{\partial r}{\partial x_i} &= 2x_i \\
\frac{\partial r}{\partial x_i} &= \frac{x_i}{r} = (\hat{\mathbf{r}})_i.
\end{align*}
Therefore  \[
 (\nabla V)_i= \frac{\d V}{\d r}(\hat{\mathbf{r}})_i. \]
\end{proof}

Q: Angular momentum and torque
A: The \emph{angular momentum} of a particle is
  \[
    \mathbf{L} = \mathbf{r}\times \mathbf{p} = m\mathbf{r}\times \dot{\mathbf{r}}.
  \]
Then
  \begin{align*}
    \frac{\d \mathbf{L}}{\d t} &= m\dot{\mathbf{r}} \times \dot{\mathbf{r}} + m\mathbf{r}\times \ddot{\mathbf{r}} \\
&= \mathbf{r}\times \mathbf{F} \\
&=\mathbf{G}
  \end{align*}
where $\vec G$ is the torque.
For a particle acted upon by a central force (taking the origin as the centre), $\mathbf{F}$ is parallel to $\mathbf{r}$ so the torque vanishes, i.e. the angular momentum is conserved.

Q: Gravitational potential and field
A: The \emph{gravitational potential} is the gravitational potential energy per unit mass. It is
  \[
    \Phi_g(r) = -\frac{GM}{r}.
  \]
So $V = m\Phi_g$.
  The \emph{gravitational field} is the force per unit mass,
  \[
    \mathbf{g} = -\nabla \Phi_g = -\frac{GM}{r^2}\hat{\mathbf{r}}.
  \]
So $\vec{F}=m\vec{g}$.

Q: Equation of motion in polar coordinates
A: We define \[
  \hat{\mathbf{r}} = \begin{pmatrix}\cos \theta\\ \sin \theta\end{pmatrix}, \quad \hat{\boldsymbol\theta} = \begin{pmatrix}-\sin \theta\\\cos\theta \end{pmatrix}.
\]
  \begin{align*}
    \frac{\d \hat{\mathbf{r}}}{\d \theta} &= \begin{pmatrix} -\sin \theta\\ \cos \theta\end{pmatrix} = \hat{\boldsymbol\theta}\\
    \frac{\d \hat{\boldsymbol\theta}}{\d \theta} &= \begin{pmatrix}-\cos \theta\\ -\sin \theta\end{pmatrix} = -\hat{\mathbf{r}}.
  \end{align*}
By the chain rule, we have
\[
  \frac{\d \hat{\mathbf{r}}}{\d t} = \dot{\theta}\hat{\boldsymbol\theta},\quad \frac{\d \hat{\boldsymbol\theta}}{\d t} = -\dot{\theta} \hat{\mathbf{r}}.
\]
We can now express the position, velocity and acceleration in this new polar basis. The position is given by
\[
  \mathbf{r} = r\hat{\mathbf{r}}.
\]
Taking the derivative gives the velocity as
\[
  \dot{\mathbf{r}} = \dot{r} \hat{\mathbf{r}} + r\dot{\theta} \hat{\boldsymbol\theta}.
\]
The acceleration is then
\begin{align*}
  \ddot{\mathbf{r}} &= \ddot{r} \hat{\mathbf{r}} + \dot{r}\dot{\theta}\hat{\boldsymbol\theta} + \dot{r}\dot{\theta}\hat{\boldsymbol\theta} + r\ddot{\theta}\hat{\boldsymbol\theta} - r\dot{\theta}^2\hat{\mathbf{r}}\\
  &= (\ddot{r} - r\dot{\theta}^2)\hat{\mathbf{r}} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\boldsymbol\theta}.
\end{align*}
So the equation of motion is
\begin{align*}
  \vec{F}/m= (\ddot{r} - r\dot{\theta}^2)\hat{\mathbf{r}} + (r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\boldsymbol\theta}.
\end{align*}

Q: Motion in a central force field
A: Newton's 2nd law in polar coordinates gives
\[
  m(\ddot{r} - r\dot{\theta}^2)\hat{\mathbf{r}} + m(r\ddot{\theta} + 2\dot{r}\dot{\theta})\hat{\boldsymbol\theta} = -\frac{\d V}{\d r}\hat{\mathbf{r}}.
\]
The $\theta$ component can be written as
\[
  \frac{1}{r}\frac{\d }{\d t}(mr^2 \dot{\theta}) = 0.
\]
But we know
\begin{align*}
  \mathbf{L} &= m\mathbf{r}\times \dot{\mathbf{r}}\\
  &= mr\hat{\mathbf{r}}\times (\dot{r}\hat{\mathbf{r}} + r\dot{\theta}\hat{\boldsymbol\theta})\\
  &= mr^2 \dot{\theta}\; \hat{\mathbf{r}}\times \hat{\boldsymbol\theta}\\
  &= mr^2 \dot{\theta}\; \hat{\mathbf{z}}
\end{align*}
so this says the angular momentum is conserved. Now eliminate $\dot{\theta}$ using $r^2\dot{\theta} = h$ to obtain
\[
  m\ddot{r} = -\frac{\d V}{\d r} + \frac{mh^2}{r^3} = -\frac{\d V_{\text{eff}}}{\d r},
\]
where
\[
  V_{\text{eff}}(r) = V(r) + \frac{mh^2}{2r^2}.
\]
We have now reduced the problem to 1D motion in an (effective) potential.
The total energy of the particle is
\begin{align*}
  E &= \frac{1}{2}m|\dot{\mathbf{r}}|^2 + V(r)\\
  &= \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) + V(r)\\
  &= \frac{1}{2}m\dot{r}^2 + \frac{mh^2}{2r^2} + V(r)\\
  &= \frac{1}{2}m\dot{r}^2 + V_{\text{eff}}(r).
\end{align*}

Q: Apsides
A: The points of minimum and maximum $r$ in an orbit are called the \emph{periapsis} and \emph{apoapsis}. They are collectively known as the \emph{apsides}.
        For an orbit around the Sun, the periapsis and apoapsis are known as the \emph{perihelion} and \emph{aphelion}.
      In particular, the perihelion and aphelion of the Earth are known as the \emph{perigee} and \emph{apogee}.

Q: Binet's equation
A: We introduce the new variable
\begin{notation}
  \[
    u = \frac{1}{r}.
  \]
\end{notation}
Then
\[
  \dot{r} = \frac{\d r}{\d \theta}\dot{\theta} = \frac{\d r}{\d \theta}\frac{h}{r^2} = -h\frac{\d u}{\d \theta},
\]
and
\[
  \ddot{r} = \frac{\d }{\d t}\left(-h \frac{\d u}{\d \theta}\right) = -h\frac{\d ^2 u}{\d \theta^2}\dot{\theta} = -h\frac{\d ^2u}{\d \theta^2}\frac{h}{r^2} = -h^2u^2\frac{\d^2 u}{\d \theta^2}.
\]
This doesn't look very linear with $u^2$, but it will help linearizing the equation when we put in other factors.
The radial equation of motion
\[
  m\ddot{r} - \frac{mh^2}{r^3} = F(r)
\]
then becomes
  \[
    -mh^2u^2\left(\frac{\d^2u}{\d \theta^2}+ u\right) = F\left(\frac{1}{u}\right).
  \]
This still looks rather complicated, but observe that for an inverse square force, $F(1/u)$ is proportional to $u^2$, and then the equation is linear!
In general, given an arbitrary $F(r)$, we aim to solve this second order ODE for $u(\theta)$. If needed, we can then work out the time-dependence via
\[
  \dot{\theta} = hu^2.
\]

Q: Shapes of orbits under attractive, central inverse square law forces
A: For $k = GM$ or $\displaystyle k = -\frac{Qq}{4\pi\varepsilon_0 m}$ we have
\[
  V(r) = \frac{mk}{r},\quad F(r) = -\frac{mk}{r^2}.
\]
Binet's equation gives
\begin{align*}
  \frac{\d^2 u}{\d \theta^2} + u &= \frac{k}{h^2} \\
u &= \frac{k}{h^2} + A\cos(\theta - \theta_0).
\end{align*}
If $A = 0$, then $u$ is constant, and the orbit is circular. Otherwise, $u$ reaches a maximum at $\theta = \theta_0$. This is the periapsis. Choose coordinates such that the periapsis is at $\theta = 0$. Then
  \[
    r = \frac{\ell}{1 + e\cos \theta},\tag{$*$}
  \]
  with $\ell = h^2/k$ and $e = Ah^2/k$. This is a conic, with a focus (the sun) at the origin. In Cartesians
\[
  (1 - e^2)x^2 + 2e\ell x + y^2 = \ell^2.\tag{$\dagger$}.
\]
\begin{itemize}
  \item Ellipse: ($0\leq e < 1$). $r$ is bounded by
    \[
      \frac{\ell}{1 + e} \leq r \leq \frac{\ell}{1 - e}.
    \]
    ($\dagger$) gives an ellipse centered on $(-ea, 0)$,
    \[
      \frac{(x + ea)^2}{a^2} + \frac{y^2}{b^2} = 1,
    \]
    where $\displaystyle a = \frac{\ell}{1 - e^2}$ and $\displaystyle b = \frac{\ell}{\sqrt{1 - e^2}}\leq a$.
    $a$ and $b$ are the semi-major and semi-minor axis. $\ell$ is the \emph{semi-latus rectum}. If $e = 0$, then $a = b = \ell$ and the ellipse is a circle.
  \item Hyperbola: ($e > 1$). For $e > 1$, $r\to \infty$ as $\theta \to \pm \alpha$, where $\alpha = \cos^{-1}(1/e)\in (\pi/2, \pi)$. Then ($\dagger$) is a hyperbola, center $(ea, 0)$,
    \[
      \frac{(x - ea)^2}{a^2} - \frac{y^2}{b^2} = 1,
    \]
    with $\displaystyle a = \frac{\ell}{e^2 - 1}$, $\displaystyle b = \frac{\ell}{\sqrt{e^2 - 1}}$.
    This corresponds to an unbound orbit that is deflected (scattered) by an attractive force. $b$ is the semi-minor axis and \emph{impact parameter}. The asymptote is $y = \frac{b}{a}(x - ea)$, or
    \begin{align*}
      x \sqrt{e^2 - 1} - y &= eb \\
(x, y) \cdot \left(\frac{\sqrt{e^2 - 1}}{e}, -\frac{1}{e}\right) &= b
    \end{align*}
 which is of the form $\mathbf{r} \cdot \mathbf{n} = b$.
  \item Parabola: ($e = 1$). Then ($*$) becomes
    \[
      r = \frac{\ell}{1 + \cos \theta}.
    \]
    We see that $r\to \infty$ as $\theta \to \pm \pi$. ($\dagger$) becomes $y^2 = \ell(\ell - 2x)$. The trajectory is similar to that of a hyperbola.
\end{itemize}

Q: Energy of orbits under inverse square law forces
A: \begin{align*}
  E &= \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) - \frac{mk}{r}\\
  &= \frac{1}{2}mh^2\left(\left(\frac{\d u}{\d \theta}\right)^2 + u^2\right) - mku
\end{align*}
Substitute $\displaystyle u = \frac{1}{\ell}(1 + e\cos \theta)$ and $\displaystyle \ell = \frac{h^2}{k}$, and it simplifies to
\[
  E = \frac{mk}{2\ell}(e^2 - 1),
\]
which is independent of $\theta$, as it must be.
Orbits are bounded for $e < 1\iff E<0$. Unbounded orbits have $e > 1\iff E>0$. A parabolic orbit has $e = 1$, $E = 0$, and is ``marginally bound''.
Note that the condition $E > 0$ is equivalent to $|\dot{\mathbf{r}}| > \sqrt{\frac{2GM}{r}} = v_{\mathrm{esc}}$, which means you have enough kinetic energy to escape orbit.

Q: Kepler's laws of planetary motion
A: \begin{law}[Kepler's first law]
  The orbit of each planet is an ellipse with the Sun at one focus.
\end{law}
\begin{proof}
Planets have low energy so form low eccentricity conics.
\end{proof}
\begin{law}[Kepler's second law]
  The line between the planet and the sun sweeps out equal areas in equal times.
\end{law}
\begin{proof}
The area swept out by moving $\d \theta$ is $\d A = \frac{1}{2}r^2\;\d \theta$ (area of sector of circle). So
\[
  \frac{\d A}{\d t} = \frac{1}{2}r^2\dot{\theta} = \frac{h}{2} = \text{const}.
\]
by the conservation of angular momentum for central forces.
\end{proof}
\begin{law}[Kepler's third law]
  The square of the orbital period is proportional to the cube of the semi-major axis, or
  \[
    P^2 \propto a^3.
  \]
\end{law}
\begin{proof}
The total area of the ellipse is $A = \pi ab = \frac{h}{2}P$ (by K2). But $b^2 = a^2( 1 - e^2)$ and $h^2 = k\ell = ka(1 - e^2)$. So
\[
  P^2 = \frac{(2\pi)^2a^4(1 - e^2)}{ka(1 - e^2)} = \frac{(2\pi)^2 a^3}{k}.
\]
\end{proof}

Q: Orbit under a repulsive, inverse square law, central force
A: Under a repulsive force, the potential and force are given by
\[
  V(r) = +\frac{mk}{r}, \quad F(r) = +\frac{mk}{r^2}.
\]
The solution is now
\[
  u = -\frac{k}{h^2} + A\cos (\theta - \theta_0).
\]
wlog, we can take $A \geq 0, \theta_0 = 0$. Then
\[
  r = \frac{\ell}{e\cos \theta - 1}
\]
with
\[
  \ell = \frac{h^2}{k}, \quad e = \frac{Ah^2}{k}.
\]
We know that $r$ and $\ell$ are positive. So we must have $e \geq 1$. Then $r\to \infty$ as $\theta \to \pm \alpha$, where $\alpha = \cos^{-1}(1/e)$. This is the orbit is a hyperbola
\[
  \frac{(x - ea)^2}{a^2} - \frac{y^2}{b^2} = 1,
\]
with $a = \frac{\ell}{e^2 - 1}$ and $b = \frac{\ell}{\sqrt{e^2 - 1}}$. The particle is deflected by $O$. We can characterize the path of the particle by the impact parameter $b$ and the incident speed $v$ s.t. $h = bv$.
Define the scattering angle $\beta = \pi - 2\alpha$. Then
\begin{align*}
  \frac{1}{e} &= \cos \alpha = \cos \left(\frac{\pi}{2} - \frac{\beta}{2}\right) = \sin\left(\frac{\beta}{2}\right) \\
b &= \frac{\ell}{\sqrt{e^2 - 1}} = \frac{(bv)^2}{k}\tan \frac{\beta}{2} \\
\beta &= 2\tan^{-1}\left(\frac{k}{bv^2}\right)
\end{align*}
We see that if we have a small impact parameter, ie. $b \ll k/v^2$, then we can have a scattering angle approaching $\pi$, as seen in Rutherford's gold foil experiment.

Q: Equation of motion for rotating frames
A: Let $\{\mathbf{e}_i\}$ and $\{\mathbf{e}_i'\}$ be basis vectors of $S$ and $S'$ respectively, where $S'$ has relative angular velocity $\omega \hat{\mathbf{z}}$.
As the basis vectors are at rest in $\mathbf{S}'$
\[
  \left(\frac{\d \mathbf{e}_i'}{\d t}\right)_S = \boldsymbol\omega\times \mathbf{e}_i'.
\]
Then using the product rule
\begin{align*}
\mathbf{r}(t)&= \sum r_i'(t) \mathbf{e}_i' \\
  \implies\left(\frac{\d \mathbf{r}}{\d t}\right)_S &= \sum \frac{\d r_i'}{\d t}\mathbf{e}_i' + \sum r_i'\boldsymbol\omega \times \mathbf{e}_i' \\
&= \left(\frac{\d \mathbf{r}}{\d t}\right)_{S'} + \boldsymbol\omega \times \mathbf{r}.
\end{align*}
We apply this formula a second time, and allow $\boldsymbol\omega$ to depend on time. Then we have
\begin{align*}
  \left(\frac{\d ^2\mathbf{r}}{\d t^2}\right)_S &= \left(\left(\frac{\d }{\d t}\right)_{S'} + \boldsymbol\omega\times \right)\left(\left(\frac{\d \mathbf{r}}{\d t}\right)_{S'} + \boldsymbol\omega \times \mathbf{r}\right).\\
  &= \left(\frac{\d^2 \mathbf{r}}{\d t^2}\right)_{S'} + 2\boldsymbol \omega \times \left(\frac{\d \mathbf{r}}{\d t}\right)_{S'} + \dot{\boldsymbol\omega} \times \mathbf{r} + \boldsymbol\omega\times(\boldsymbol\omega\times \mathbf{r})
\end{align*}
Since $S$ is inertial, Newton's Second Law gives
  \[
    m\left(\frac{\d^2 \mathbf{r}}{\d t^2}\right)_{S'} = \mathbf{F} - 2m\boldsymbol\omega \times \left(\frac{\d \mathbf{r}}{\d t}\right)_{S'} - m\dot{\boldsymbol\omega}\times \mathbf{r} - m\boldsymbol\omega \times (\boldsymbol\omega \times \mathbf{r}).
  \]

Q: Centrifugal force
A: \begin{align*}
  -m\boldsymbol\omega \times (\boldsymbol\omega \times \mathbf{r}) &= -m\big((\boldsymbol\omega \cdot \mathbf{r})\boldsymbol\omega - (\boldsymbol\omega \cdot \boldsymbol\omega)\mathbf{r}\big) \\
&=m\omega^2((\hat{\boldsymbol\omega} \cdot \hat{\boldsymbol\omega}) \mathbf{r} - (\hat{\boldsymbol\omega} \cdot \mathbf{r})\hat{\boldsymbol\omega}) \\
&= m\omega^2 \mathbf{r}_{\bot}
\end{align*}
where $\mathbf{r}_{\bot} = \mathbf{r} - (\mathbf{r}\cdot \hat{z})\hat{z}$ is the projection of the position in the $x-y$ plane. So the centrifugal force is directed away from the axis of rotation.
Now using the Leibniz property
\begin{align*}
  \mathbf{r}_\bot \cdot \mathbf{r}_\bot &= \mathbf{r}\cdot \mathbf{r} - (\mathbf{r} \cdot \hat{\boldsymbol\omega})^2\\
  \implies \nabla(|\mathbf{r}_\bot|^2) &= 2\mathbf{r} - 2(\mathbf{r}\cdot \hat{\boldsymbol\omega})\hat{\boldsymbol\omega} = 2\mathbf{\boldsymbol\omega}_\bot.
\end{align*}
So
\[
  -m\boldsymbol\omega\times(\boldsymbol\omega\times \mathbf{r}) = -\nabla\left(-\frac{1}{2}m\omega^2|\mathbf{r}_\bot|^2\right) = -\nabla\left(-\frac{1}{2}m|\boldsymbol\omega\times \mathbf{r}|^2\right).
\]
Thus the centrifugal force is a conservative (fictitious) force.
This adds to the \emph{effective gravity}
\[
  \mathbf{g}_{\text{eff}} = \mathbf{g} + \omega^2 \mathbf{r}_\bot.
\]
This gravity will not be vertically downwards. Consider a point $P$ at latitude $\lambda$ on the surface of a spherical planet of radius $R$. Take axes $\hat{\mathbf{x}}$ East (into the page), $\hat{\mathbf{y}}$ (locally) North and $\hat{\mathbf{z}}$ up.
At $P$, we have
\begin{align*}
  \mathbf{r} &= R\hat{\mathbf{z}}\\
  \mathbf{g} &= -g\hat{\mathbf{z}}\\
  \boldsymbol\omega &= \omega(\cos\lambda \hat{\mathbf{y}} + \sin \lambda \hat{\mathbf{z}})
\end{align*}
So
\begin{align*}
  \mathbf{g}_{\mathrm{eff}} &= \mathbf{g} + \omega^2 \mathbf{r}_\bot \\
  &= -g\hat{\mathbf{z}} + \omega^2 R\cos\lambda(\cos\lambda\hat{\mathbf{z}} - \sin\lambda\hat{\mathbf{y}})\\
  &= -\omega^2 R\cos\lambda\sin\lambda\hat{\mathbf{y}} - (g - \omega^2 R\cos^2\lambda)\hat{\mathbf{z}}.
\end{align*}
So the angle $\alpha$ between $\mathbf{g}$ and $\mathbf{g}_{\text{eff}}$ is given by
\[
  \tan \alpha = \frac{\omega^2 R\cos\lambda\sin\lambda}{g - \omega^2R\cos^2\lambda}.
\]
This is 0 at the equator and the poles, and greatest when you are halfway between.

Q: The Coriolis force
A: Writing $\mathbf{v} = \left(\frac{\d \mathbf{r}}{\d t}\right)_{S'}$, we can write the force as
\[
  -2m\boldsymbol\omega\times \mathbf{v}.
\]
Since this force is always perpendicular to the velocity, it does no work.
Consider motion parallel to the Earth's surface. We only care about the effect of the Coriolis force on the horizontal trajectory, and ignore the vertical component that is tiny compared to gravity.
So we only take the vertical component of $\boldsymbol\omega$, $\omega\sin\lambda\hat{\mathbf{z}}$. The horizontal velocity $\mathbf{v} = v_x \hat{\mathbf{x}} + v_y \hat{\mathbf{y}}$ generates a horizontal Coriolis force:
\[
  -2m\omega\sin\lambda\hat{\mathbf{z}}\times \mathbf{v} = 2m\omega\sin\lambda(v_y \hat{\mathbf{x}} - v_x \hat{\mathbf{y}}).
\]
In the Northern hemisphere $(0 < \lambda < \pi/2)$, this causes a deflection towards the right. In the Southern Hemisphere, the deflection is to the left. The effect vanishes at the equator.

Q: Equation of motion of the centre of mass
A: Newton's Second Law for particle $i$ is
\[
  m_i \ddot{\mathbf{r}}_i = \dot{\mathbf{p}}_i = \mathbf{F}_i,
\]
where $\mathbf{F}_i$ is the total force acting on particle $i$. We can write $\mathbf{F}_i$ as
\[
  \mathbf{F}_i = \mathbf{F}_i^{\text{ext}} + \sum_{j = 1}^N \mathbf{F}_{ij},
\]
where $\mathbf{F}_{ij}$ is the force on particle $i$ by particle $j$, and $\mathbf{F}_i^{\text{ext}}$ is the external force on $i$, which comes from particles outside the system.
We define the \emph{centre of mass}
  \[
    \mathbf{R} = \frac{1}{M}\sum_{i = 1}^N m_i\mathbf{r}_i.
  \]
Then the \emph{total linear momentum} is
  \[
    \mathbf{P} = \sum_{i = 1}^N \mathbf{p}_i = \sum_{i = 1}^N m_i \dot{\mathbf{r}}_i = M\dot{\mathbf{R}}.
  \]
  Differentiating gives
 \begin{align*}
    M\ddot{\mathbf{R}} &= \dot{\mathbf{P}}\\
    &= \sum_{i = 1}^N \dot{\mathbf{p}}_i\\
    &= \sum_{i = 1}^N \mathbf{F}_i^{\text{ext}} + \sum_{i = 1}^N\sum_{j = 1}^N \mathbf{F}_{ij}\\
    &= \mathbf{F} + \frac{1}{2}\sum_i \sum_j\underbrace{(\mathbf{F}_{ij} + \mathbf{F}_{ji})}_{=0\textrm{ by N3}}\\
    &= \mathbf{F}.
  \end{align*}
In particular, $\vec{F}=0\implies \dot{\vec{P}}=0$, i.e, momentum is conserved.

Q: Conservation of angular momentum for a system of particles
A: The \emph{total angular momentum} of the system about the origin is
  \[
    \mathbf{L} = \sum_i \mathbf{r}_i \times \mathbf{p}_i.
  \]
Differentiating gives
\begin{align*}
  \dot{\mathbf{L}} &= \sum_i \mathbf{r}_i \times \dot{\mathbf{p}}_i + \dot{\mathbf{r}}_i \times \mathbf{p}_i\\
  &= \sum_i \mathbf{r}_i \times \left(\mathbf{F}_i^{\text{ext}} + \sum_j F_{ij}\right) + m(\dot{\mathbf{r}}_i\times \dot{\mathbf{r}}_i)\\
  &= \sum_i \mathbf{r}_i\times \mathbf{F}_i^{\text{ext}} + \sum_i \sum_j \mathbf{r}_i \times \mathbf{F}_{ij}\\
  &= \sum_i \mathbf{G}_i^{\text{ext}} + \frac{1}{2}\sum_i\sum_j (\mathbf{r}_i \times \mathbf{F}_{ij} + \mathbf{r}_j\times \underbrace{\mathbf{F}_{ji}}_{-\vec{F}_{ij}})\\
  &= \mathbf{G} + \frac{1}{2}\sum_i\sum_j\underbrace{(\mathbf{r}_i - \mathbf{r}_j)\times \mathbf{F}_{ij}}_{||\textrm{ by strong N3}}\\
  &= \mathbf{G},
\end{align*}
where
  \[
    \mathbf{G} = \sum_i \mathbf{r}_i \times \mathbf{F}_i^{\text{ext}}
  \]
is the total external torque.
So the total angular momentum is conserved if $\mathbf{G} = \mathbf{0}$.

Q: Motion relative to the center of mass
A: Write $\mathbf{r}_i = \mathbf{R} + \mathbf{r}_i^c$, where $\mathbf{r}_i^c$ is the position of particle $i$ relative to the center of mass.
We first obtain two useful equalities:
\[
  \sum_i m_i\mathbf{r}_i^c = \sum m_i \mathbf{r}_i - \sum m_i \mathbf{R} = M\mathbf{R} - M\mathbf{R} = \mathbf{0}.
\]
Differentiating gives
\[
  \sum_i m_i \dot{\mathbf{r}}_i^c = \mathbf{0}.
\]
Using these equalities, we can express the angular momentum and kinetic energy in terms of $\mathbf{R}$ and $\mathbf{r}_i^c$ only:
\begin{align*}
  \mathbf{L} &= \sum_i m_i(\mathbf{R} + \mathbf{r}_i^c) \times (\dot{\mathbf{R}} + \dot{\mathbf{r}}_i^c)\\
  &= \sum_i m_i \mathbf{R}\times \dot{\mathbf{R}} + \mathbf{R}\times \sum_i m_i \dot{\mathbf{r}}_i^c + \sum_i m_i \mathbf{r}_i^c\times \dot{\mathbf{R}} + \sum_i m_i \mathbf{r}_i^c \times \dot{\mathbf{r}}_i^c\\
  &= M\mathbf{R}\times \dot{\mathbf{R}} + \sum_i m_i \mathbf{r}_i^c \times \dot{\mathbf{r}}_i^c\displaybreak[0]\\
  T&= \frac{1}{2}\sum_i m_i|\dot{\mathbf{r}}_i|^2\\
  &= \frac{1}{2}\sum_i m_I (\dot{\mathbf{R}} + \dot{\mathbf{r}_i}^c)\cdot (\dot{\mathbf{R}} + \dot{\mathbf{r}}_i^c)\\
  &= \frac{1}{2}\sum_i m_i \dot{\mathbf{R}}\cdot \dot{\mathbf{R}} + \dot{\mathbf{R}} \cdot \sum_i m_i \dot{\mathbf{r}}_i^c + \frac{1}{2}\sum_i m_i \dot{\mathbf{r}}_i^c\cdot \dot{\mathbf{r}}_i^c\cdot \dot{\mathbf{r}}_i^c\\
  &=\frac{1}{2}M|\dot{\mathbf{R}}|^2 + \frac{1}{2}\sum_i m_i|\dot{\mathbf{r}}_i^c|^2.
\end{align*}
If the forces are conservative in the sense that
\[
  \mathbf{F}_i^{\text{ext}} = -\nabla_i V_i(\mathbf{r}_i),
\]
and
\[
  \mathbf{F}_{ij} = -\nabla_i V_{ij} (\mathbf{r}_i - \mathbf{r}_j),
\]
where $\nabla_i$ is the gradient with respect to $\mathbf{r}_i$, then energy is conserved in the form
\[
  E = T + \sum_i V_i(\mathbf{r}_i) + \frac{1}{2}\sum_i\sum_j V_{ij}(\mathbf{r}_i - \mathbf{r}_j).
\]

Q: Two body problem
A: The center of mass is at
\[
  \mathbf{R} = \frac{1}{M} (m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2),
\]
where $M = m_1 + m_2$.
Define the separation vector (or relative position vector)
\[
  \mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2.
\]
Then we write everything in terms of $\mathbf{R}$ and $\mathbf{r}$:
\[
  \mathbf{r}_1 = \mathbf{R} + \frac{m_2}{M}\mathbf{r},\quad \mathbf{r}_2 = \mathbf{R} - \frac{m_1}{M}\mathbf{r}.
\]
Since the external force $\mathbf{F} = \mathbf{0}$, we have $\ddot{\mathbf{R}} = \mathbf{0}$, ie. the center of mass moves uniformly.
Meanwhile,
\begin{align*}
  \ddot{\mathbf{r}} &= \ddot{\mathbf{r}}_1 - \ddot{\mathbf{r}}_2\\
  &= \frac{1}{m_1} \mathbf{F}_{12} - \frac{1}{m_2}\mathbf{F}_{21}\\
  &= \left(\frac{1}{m_1} + \frac{1}{m_2}\right) \mathbf{F}_{12}
\end{align*}
We can write this as
\[
  \mu \ddot{\mathbf{r}} = \mathbf{F}_{12}(\mathbf{r}),
\]
where
\[
  \mu = \frac{m_1m_2}{m_1 + m_2}
\]
is the \emph{reduced mass}.
It can be shown that
\begin{align*}
  \mathbf{L} &= M\mathbf{R} \times \dot{\mathbf{R}} + \mu \mathbf{r}\times \dot{\mathbf{r}}\\
  T &= \frac{1}{2} M|\dot{\mathbf{R}}|^2 + \frac{1}{2}\mu |\dot{\mathbf{r}}|^2 \\
E&=T+V_{12}(\vec{r})
\end{align*}
by expressing $\mathbf{r}_1$ and $\mathbf{r}_2$ in terms of $\mathbf{r}$ and $\mathbf{R}$.

Q: Variable mass problem
A: Apply Newton's second law in the form \[
  \mathbf{F}=\frac{\d \mathbf{p}}{\d t}.
\]

Q: Rocket Equation
A: Fuel is ejected from a rocket exhaust at velocity $-u$ relative to the rocket. Therefore the change in momentum of the system is
\begin{align*}
p(t+\delta t)-p(t)&= m(t + \delta t)v(t + \delta t) + [m(t) - m(t + \delta t)][v(t) - u(t) + O(\delta t)] - m(t)v(t)\\
   \delta p&= (m + \dot{m}\delta t + O(\delta t^2))(v + \dot{v} \delta t + O(\delta t^2)) - \dot{m}\delta t(v - u) + O(\delta t^2) - mv\\
  &= (\dot{m}v + m\dot{v} - \dot{m}v + \dot{m}u)\delta t + O(\delta t^2)\\
  &= (m\dot{v} + \dot{m}u)\delta t + O(\delta t^2).
\end{align*}
Newton's second law gives
\[
  \lim_{\delta t\to 0} \frac{\delta p}{\delta t} = F
\]
where $F$ is the external force on the rocket. So we obtain
  \[
   F= m\frac{\d v}{\d t} + u\frac{\d m}{\d t}.
  \]

Q: Rigid body
A: A \emph{rigid body} is an extended object, consisting of $N$ particles that are constrained such that the distance between any pair of particles, $|\mathbf{r}_i - \mathbf{r}_j|$, is fixed.

Q: Moment of inertia
A: The \emph{moment of inertia} of a rigid body about the rotation axis $\hat{\boldsymbol\omega}$ is
  \[
    I  = \sum_{i = 1}^N m_i |\hat{\boldsymbol\omega}\times \mathbf{r}_i|^2.
  \]

Q: Rotational Kinetic Energy and Angular Momentum for a purely rotating system of particles
A: Consider a system in which all $N$ particles rotate about the same axis with the same angular velocity:
\[
  \dot{\mathbf{r}}_i = \boldsymbol\omega\times \mathbf{r}_i.
\]
This ensures that
\[
  \frac{\d }{\d t}|\mathbf{r}_i - \mathbf{r}_j|^2 = 2(\dot{\mathbf{r}}_i - \dot{\mathbf{r}}_j)\cdot (\mathbf{r}_i - \mathbf{r}_j) = 2\big(\boldsymbol\omega\times (\mathbf{r}_i - \mathbf{r}_j)\big) \cdot (\mathbf{r}_i - \mathbf{r}_j) = 0,
\]
as required for a rigid body.
Then the rotational kinetic energy is
\begin{align*}
  T &= \frac{1}{2}\sum_{i = 1}^N m_i|\dot{\mathbf{r}_i}|^2 \\
&= \frac{1}{2}\sum_{i = 1}^N m_i|\boldsymbol\omega\times \mathbf{r}_i|^2 \\
&= \frac{1}{2}\sum_{i = 1}^N m_i|\hat{\boldsymbol\omega}\times \mathbf{r}_i|^2\boldsymbol\omega^2 \\
&= \frac{1}{2}I\omega^2.
\end{align*}
  The \emph{angular momentum} is
  \begin{align*}
    \mathbf{L} &= \sum_i m_i \mathbf{r}_i \times \dot{\mathbf{r}}_i \\
& = \sum_i m_i \mathbf{r}_i \times (\boldsymbol\omega \times \mathbf{r}_i).
  \end{align*}
Thus
\begin{align*}
  \mathbf{L} \cdot \hat{\boldsymbol\omega} &= \omega \sum_i m_i \hat{\boldsymbol\omega}\cdot (\mathbf{r}_i \times (\hat{\boldsymbol\omega} \times \mathbf{r}_i))\\
  &= \omega \sum_i m(\hat{\boldsymbol\omega}\times \mathbf{r}_i)\cdot (\hat{\boldsymbol\omega} \times \mathbf{r}_i)\\
  &= I\omega.
\end{align*}
Using vector identities, we have
\[
  \mathbf{L} = \sum_i m_i\big((\mathbf{r}_i\cdot \mathbf{r}_i)\boldsymbol \omega - (\mathbf{r}_i \cdot \boldsymbol\omega)\mathbf{r}_i\big)= \mathsf{I}\boldsymbol \omega
\]
where $\mathsf{I}$ is the \emph{inertia tensor}. This is represented by a symmetric matrix with components
\[
  \mathsf{I}_{jk} = \sum_i m_i(|\mathbf{r}_i|^2 \delta_{jk} - (\mathbf{r}_i)_j(\mathbf{r}_i)_k).
\]
If the body rotates about a \emph{principal axis}, ie. one of the three orthogonal eigenvectors of $\mathsf{I}$, then $\mathbf{L}$ will be parallel to $\boldsymbol\omega$. Usually, the principal axes lie on the axes of rotational symmetry of the body.

Q: Perpendicular axis theorem
A: For a two-dimensional object (a lamina), and three perpendicular axes $x, y, z$ through the same spot, with $z$ normal to the plane,
  \[
    I_z = I_x + I_y,
  \]
  where $I_z$ is the moment of inertia about the $z$ axis.
\begin{proof}
  Let $\rho$ be the mass per unit volume. Then
  \begin{align*}
    I_x &= \int \rho y^2 \;\d A\\
    I_y &= \int \rho x^2 \;\d A\\
    I_z &= \int \rho (x^2 + y^2)\;\d A = I_x + I_y.
  \end{align*}
\end{proof}

Q: Parallel axis theorem
A: If a rigid body of mass $M$ has moment of inertia $I^C$ about an axis passing through the center of mass, then its moment of inertia about a parallel axis a distance $d$ away is
  \[
    I = I^C + Md^2.
  \]
\begin{proof}
Choose such that the center of mass is at the origin and the two rotation axes are $x = y =0$ and $x = d, y = 0$. Then
  \[
    I^C = \int \rho (x^2 + y^2) \;\d V,
  \]
  and
  \[
    \int \rho \mathbf{r}\;\d V = \mathbf{0}.
  \]
  So
  \begin{align*}
    I &= \int \rho((x - d)^2 + y^2)\;\d V\\
    &=\int \rho (x^2 + y^2)\;\d V - 2d\int \rho x\;\d V + \int d^2 \rho\;\d V\\
    &= I^c + 0 + Md^2\\
    &= I^c + Md^2.
  \end{align*}
\end{proof}

Q: Motion of a rigid body relative to a point inside it
A: Let $\mathbf{r}_i = \mathbf{R} + \mathbf{r}_i^c$. Then
\[
  \dot{\mathbf{r}}_i = \dot{\mathbf{R}} + \dot{\mathbf{r}}_i^c.
\]
If the body rotates with angular velocity $\boldsymbol\omega$ about the center of mass, then
\[
  \dot{\mathbf{r}}_i^c = \boldsymbol\omega \times \mathbf{r}_i^c.
\]
Combining these we have
\[
  \dot{\mathbf{r}_i} = \dot{\mathbf{R}} + \boldsymbol \omega \times \mathbf{r}_i^c = \dot{\mathbf{r}} + \boldsymbol\omega\times (\mathbf{r}_i - \mathbf{R}).
\]
We know the total KE of the system is
\begin{align*}
  T &= \frac{1}{2}M|\dot{\mathbf{R}}|^2 + \frac{1}{2}\sum_i m_i |\dot{\mathbf{r}}_i^c|^2\\
  &= \underbrace{\frac{1}{2}M|\dot{\mathbf{R}}|^2}_{\text{translational KE}} + \underbrace{\frac{1}{2}I^c\omega^2}_{\text{rotational KE}}.
\end{align*}
Thus for any point $Q$, with position vector $\mathbf{Q}(t)$ that moves with the rigid body
\[
  \dot{\mathbf{Q}} = \dot{\mathbf{R}} + \boldsymbol\omega\times (\mathbf{Q} - \mathbf{R}).
\]
Then we can write
\begin{align*}
  \dot{\mathbf{r}}_i &= \dot{\mathbf{R}} + \boldsymbol\omega\times (\mathbf{r}_i - \mathbf{R})\\
  &= \dot{\mathbf{Q}} - \boldsymbol\omega\times (\mathbf{Q} - \mathbf{R}) + \boldsymbol\omega \times (\mathbf{r}_i - \mathbf{R})\\
  &= \dot{\mathbf{Q}} + \boldsymbol\omega\times (\mathbf{r}_i - \mathbf{Q}).
\end{align*}
So as before
\begin{align*}
  T &= \frac{1}{2}M|\dot{\mathbf{Q}}|^2 + \frac{1}{2}\sum_i m_i |\dot{\mathbf{r}}_i^Q|^2\\
  &= \underbrace{\frac{1}{2}M|\dot{\mathbf{Q}}|^2}_{\text{translational KE}} + \underbrace{\frac{1}{2}I^Q\omega^2}_{\text{rotational KE}}.
\end{align*}
Therefore the motion can be considered as a translation of $Q$, together with rotation about $Q$ with angular velocity $\boldsymbol\omega$.

Q: Equations of motion in the centre of mass frame
A: We know that in any inertial frame
\begin{align*}
  \dot{\mathbf{P}} &= \mathbf{F}\quad \text{(total external force)}\\
  \dot{\mathbf{L}} &= \mathbf{G}\quad \text{(total external torque)}.
\end{align*}
If
\[
  m_i \ddot{\mathbf{r}}_i = \mathbf{F}_i,
\]
then
\[
  m_i \ddot{\mathbf{r}}_i^c = \mathbf{F}_i - m_i \ddot{\mathbf{R}}.
\]
So there is a fictitious force $-m_i \ddot{\mathbf{R}}$ in the center-of-mass frame. But the total torque of the fictitious forces about the center of mass is
\[
  \sum_i \mathbf{r}_i^c \times \left(-m_i \ddot{\mathbf{R}}\right) = -\left(\sum m_i \mathbf{r}_i^c\right) \times \ddot{\mathbf{R}} = 0\times \dot{\mathbf{R}} = 0.
\]
So in the centre of mass frame, the centre of mass is stationary and $ \dot{\mathbf{L}} = \mathbf{G}$.

Q: Motion of a system of particles in a uniform gravitational field
A: In a uniform gravitational field $\mathbf{g}$, the total gravitational force and torque are the same as those that would act on a single particle of mass $M$ located at the center of mass (which is also the \emph{center of gravity}):
\[
  \mathbf{F} = \sum_i \mathbf{F}_i^{\mathrm{ext}} = \sum_i m_i \mathbf{g} = M\mathbf{g},
\]
and
\[
  \mathbf{G} = \sum_i \mathbf{G}_i^{\mathrm{ext}} = \sum_i \mathbf{r}_i \times (m_i \mathbf{g}) = \sum m_i \mathbf{r}_i \times \mathbf{g} = M\mathbf{R}\times \mathbf{g}.
\]
In particular, the gravitational torque about the center of mass vanishes: $\mathbf{G}^c = \mathbf{0}$.
Similarly, the gravitational potential in a uniform $\mathbf{g}$ is
\[
  \varphi_g = -\mathbf{r}\cdot \mathbf{g}
\]
(since $\mathbf{g} = -\nabla \phi_g$).
So
\begin{align*}
  V^{\mathrm{ext}} &= \sum_i V_i^{\mathrm{ext}}\\
  &= \sum_i m_i (- \mathbf{r}_i \cdot \mathbf{g})\\
  &= M (-\mathbf{R}\cdot \mathbf{g}).
\end{align*}

Q: Sliding vs. rolling
A: Consider a cylinder or sphere of radius $a$.
In general, the motion consists of a translation of the center of mass (with velocity $v$) plus a rotation about the center of mass (with angular velocity $\omega$).
The horizontal (slip) velocity at the point of contact is
\[
  v_{\text{slip}} = v - a\omega \quad v\neq 0, \omega\neq 0.
\]
For a pure sliding motion $v - a\omega \not = 0$: the point of contact moves relative to the surface and kinetic friction may occur.
For a pure rolling motion $v - a\omega = 0$: the point of contact is stationary. This is the no-slip condition. Energy is conserved as the frictional force creates no movement so does no work.
The rolling body can alternatively be considered to be rotating instantaneously about the point of contact (with angular velocity $\omega$) and not translating.

Q: The Lorentz transformation in the standard configuration (in one spatial dimension)
A: As the frames are inertial, straight lines in $(x, t)$ must map into straight lines in $(x', t')$. Therefore the relationship must be linear.
Now $x = vt$ must map into $x'= 0$ so
\[
  x' = \gamma(x - vt),\tag{1}
\]
for some factor $\gamma(|v|)$ by symmetry.
Also $x'=-vt$ must map to $x=0$ so
\[
  x = \gamma(x' + vt'),\tag{2}.
\]
We require $ x = ct$ to map to $x' = ct'$ to agree on $c$. Substitute into (1) and (2)
\begin{align*}
  ct' &= \gamma(c - v)t\\
  ct &= \gamma(c + v)t'.
\end{align*}
Multiply the two equations together and cancel $tt'$ to obtain
\[
  c^2 = \gamma^2(c^2 - v^2).
\]
So
\[
  \gamma = \sqrt{\frac{c^2}{c^2 - v^2}} = \frac{1}{\sqrt{1 - (v/c)^2}}.
\]
is the Lorentz factor. Eliminate $x'$ between (1) and (2) to obtain
\[
  x = \gamma(\gamma(x - vt) + vt').
\]
Rearranging for $t'$
\[
  t' = \gamma t - (1 - \gamma^{-2})\frac{\gamma x}{v} = \gamma\left(t - \frac{v}{c^2}x\right).
\]
Inverting these linear maps ($v\mapsto -v$)
\begin{align*}
  x &= \gamma(x' + vt')\\
  t &= \gamma\left(t' + \frac{v}{c^2}x'\right).
\end{align*}
We claim $y' = y, z' = z$ works.
For a light ray travelling in the $x$ direction in $S$:
\begin{align*}
  x &= ct,\quad y = 0,\quad z = 0 \\
\frac{x'}{t'} &= \frac{\gamma(x - vt)}{\gamma(t - vx/c^2)} = \frac{(c - v)t}{(1 - v/c)t} = c.
\end{align*}
For a light ray travelling in the $y$ direction in $S$:
\begin{align*}
  x &= 0,\quad y = ct,\quad z = 0 \\
 \frac{x'}{t'} &= \frac{\gamma(x - vt)}{\gamma(t - vx/c^2)} = -v \\
\frac{y'}{t'} &= \frac{y}{\gamma(t - vx/c^2)} = \frac{c}{\gamma}\\
  z' &= 0 \\
\frac{\sqrt{x'^2 + y'^2}}{t'} &= \sqrt{v^2 + \gamma^{-2}c^2} = c.
\end{align*}

Q: Space-time diagram for standard configuration frames
A: The $ct'$ axis corresponds to $x' = 0$, ie. $x = vt$ or $ct=cx/v$.   [/latex]
[latex]
The $x'$ axis corresponds to $t' = 0$, ie. $t = vx/c^2$ or $ct=vx/c$.

Q: Minkowski spacetime
A: Each point $P$ represents an \emph{event}, labelled by coordinates $(ct, x)$.
A particle traces out a world line in this spacetime.

Q: Lines of simultaneity
A: Lines of simultaneity in $S$ are horizontal dashed lines.
However, events that are simultaneous in $S'$ have equal values of $t'$, and so lie on lines
\[
  ct - \frac{v}{c}x = \text{constant}.
\]
[/latex]
[latex]
$S$ thinks $P_1$ and $P_2$ happened at the same time, while $S'$ thinks $P_2$ happens first.

Q: Causality
A: Although different people may disagree on the temporal order of events, the consistent ordering of cause and effect can be ensured.
[/latex]
[latex]
We can draw a \emph{light cone} that denotes the regions in which things can be influenced by $P$. These are the regions of space-time light (or any other particle) can possibly travel to. $P$ can only influence events within its \emph{future light cone}, and \emph{be influenced} by events within its \emph{past light cone}.
All observers agree that $Q$ occurs after $P$. Different observers may disagree on the temporal ordering of $P$ and $R$. However, since nothing can travel faster than light, $P$ and $R$ cannot influence each other. Since everyone agrees on how fast light travels, they also agree on the light cones, and hence causality.

Q: Time dilation
A: Suppose we have a clock that is stationary in $S'$ (which travels at constant velocity $v$ with respect to inertial frame $S$) ticks at constant intervals $\Delta t'$.
Lorentz transformation gives
\[
  t = \gamma\left(t' + \frac{v}{c^2}x'\right).
\]
Since $x' =$ constant for the clock, we have
\[
  \Delta t = \gamma \Delta t' > \Delta t'.
\]
By the principle of relativity, all clocks must measure the same time dilation, or else we can compare the two clocks and know if we are ``moving''.

Q: Length contraction
A: A rod of proper length $L'$ has rest frame $S'$.
The lines $x' = 0$ and $x' = L'$ map into $x = vt$ and $x = vt + L'/\gamma$. So the length in $S$ is $L = L'/\gamma < L'$. Therefore moving objects are contracted in the direction of motion.
Now consider a train of length $2L$, a platform of length $L$, relative speed $v$, $\gamma = 2$.
For the system of observers on the platform $S$, the train contracts to a length $2L/\gamma = L$. So it fits.
But for the system of observers on the train $S'$, the platform contracts to length $L/\gamma = L/2$, which is much too short! [/latex]
[latex]
This can be explained by the difference of lines of simultaneity, since length is the distance between front and back \emph{at the same time}.

Q: Composition of velocities
A: A particle moves with constant velocity $u'$ in frame $S'$, which moves with velocity $v$ relative to $S$.
In $S'$
\[
  x' = u't'.
\]
In $S$
\[
  u = \frac{x}{t} = \frac{\gamma(x' + vt')}{\gamma(t' + (v/c^2) x')} = \frac{u't' + vt'}{t' + (v/c^2)u't'} = \frac{u' + v}{1 + u'v/c^2}.
\]
This is the formula for the relativistic composition of of velocities.
The inverse transformation is found by swapping $u$ and $u'$, and swapping the sign of $v$, ie.
\[
  u' = \frac{u - v}{1 - uv/c^2}.
\]

Q: The invariant intervals
A: Consider events $P$ and $Q$ with coordinates $(ct_1, x_1)$ and $(ct_2, x_2)$ separated by $\Delta t = t_2 - t_1$ and $\Delta x = x_2 - x_1$.
  The \emph{invariant interval} or \emph{spacetime interval} between $P$ and $Q$ is defined as
  \[
    \Delta s^2 = c^2 \Delta t^2 - \Delta x^2.
  \]
  Note that this quantity $\Delta s^2$ can be both positive or negative --- so $\Delta s$ might be imaginary!
\begin{prop}
  All inertial observers agree on the value of $\Delta s^2$.
\end{prop}
\begin{proof}
  \begin{align*}
    c^2 \Delta t'^2 - \Delta x'^2 &= c^2 \gamma^2 \left(\Delta t - \frac{v}{c^2}\Delta x\right)^2 - \gamma^2 (\Delta x - v\Delta t)^2\\
    &= \gamma^2 \left(1 - \frac{v^2}{c^2}\right)(c^2 \Delta t - \Delta x^2)\\
    &= c^2\Delta t - \Delta x^2.
  \end{align*}
\end{proof}
In three spatial dimensions,
\[
  \Delta s^2 = c^2\Delta t^2 - \Delta x^2 - \Delta y^2 - \Delta z^2.
\]
We take this as the ``distance'' between the two points. For two infinitesimally separated events, we have
\begin{defi}[Line element]
  The \emph{line element} is
  \[
    \d s^2 = c^2 \d t^2 - \d x^2 - \d y^2 - \d z^2.
  \]
\end{defi}
  Events with $\Delta s^2 > 0$ are \emph{timelike separated}. It is possible to find inertial frames in which the two events occur in the same position, and are purely separated by time. Timelike-separated events lie within each other's light cones and can influence one another.
  Events with $\Delta s^2 < 0$ are \emph{spacelike separated}. It is possible to find inertial frame in which the two events occur in the same time, and are purely separated by space. Spacelike-separated events lie out of each other's light cones and cannot influence one another.
  Events with $\Delta s^2 = 0$ are \emph{lightlike} or \emph{null separated}. In all inertial frames, the events lie on the boundary of each other's light cones. eg. different points in the trajectory of a photon are lightlike separated, hence the name.

Q: The Lorentz group
A: The coordinates of an event $P$ in frame $S$ can be written as a \emph{4-vector} (ie. 4-component vector) $X$. We write
\[
  X =
  \begin{pmatrix}
    ct\\
    x\\
    y\\
    z
  \end{pmatrix}
\]
The invariant interval between the origin and $P$ can be written as an inner product
\[
  X\cdot X = X^T\eta X = c^2t^2 - x^2 - y^2 - z^2,
\]
where
\[
  \eta =
  \begin{pmatrix}
    1 & 0 & 0 & 0\\
    0 & -1 & 0 & 0\\
    0 & 0 & -1 & 0\\
    0 & 0 & 0 & -1
  \end{pmatrix}.
\]
4-vectors with $X\cdot X > 0$ are called timelike, and those $X \cdot X < 0$ are spacelike. If $X\cdot X = 0$, it is lightlike or null.
A Lorentz transformation is a linear transformation of the coordinates from one frame $S$ to another $S'$, represented by a $4\times 4$ tensor (``matrix''):
\[
  X' = \Lambda X
\]
Lorentz transformations can be defined as those that leave the inner product invariant:
\[
  (\forall X)(X'\cdot X' = X\cdot X),
\]
which implies the matrix equation
\[
  \Lambda^T\eta \Lambda = \eta.\tag{$*$}
\]
These also preserve $X\cdot Y$ if $X$ and $Y$ are both 4-vectors.
Two classes of solution to this equation are:
\[
  \Lambda =
  \begin{pmatrix}
    1 & 0 & 0 & 0\\
    0\\
    0 & & R\\
    0
  \end{pmatrix},
\]
where $R$ is a $3\times 3$ orthogonal matrix, which rotates (or reflects) space and leaves time intact; and
\[
  \Lambda =
  \begin{pmatrix}
    \gamma & -\gamma \beta & 0 & 0\\
    -\gamma\beta & \gamma & 0 & 0\\
    0 & 0 & 1 & 0\\
    0 & 0 & 0 & 1
  \end{pmatrix},
\]
where $\beta = \frac{v}{c}$, and $\gamma = 1/\sqrt{1 - \beta^2}$. Here we leave the $y$ and $z$ coordinates intact, and apply a Lorentz boost along the $x$ direction.
The set of all matrices satisfying equation $(*)$ form the \emph{Lorentz group} $O(1, 3)$. It is generated by rotations and boosts, as defined above, which includes the absurd spacial reflections and time reversal.
The subgroup with $\det \Lambda = +1$ is the \emph{proper Lorentz group} $SO(1, 3)$.
The subgroup that preserves spatial orientation and the direction of time is the \emph{restricted Lorentz group} $SO^+(1, 3)$. [Note that this is different from $SO(1, 3)$, since if you do \emph{both} spacial reflection and time reversal, the determinant of the matrix is still positive. We want to eliminate those as well!]

Q: Rapidity
A: Focus on the upper left $2\times 2$ matrix of Lorentz boosts in the $x$ direction. Write
\[
  \Lambda[\beta] =
  \begin{pmatrix}
    \gamma & -\gamma\beta\\
    -\gamma\beta & \gamma
  \end{pmatrix}
  ,\quad
  \gamma = \frac{1}{\sqrt{1 - \beta^2}}.
\]
Combining two boosts in the $x$ direction, we have
\[
  \Lambda[\beta_1]\Lambda[\beta_2] =
  \begin{pmatrix}
    \gamma_1 & -\gamma_1\beta_1\\
    -\gamma_1\beta_1 & \gamma_1
  \end{pmatrix}
  \begin{pmatrix}
    \gamma_2 & -\gamma_2\beta_2\\
    -\gamma_2\beta_2 & \gamma_2
  \end{pmatrix}
  = \Lambda\left[\frac{\beta_1 + \beta_2}{1 + \beta_1\beta_2}\right]
\]
after some messy algebra. This is just the velocity composition formula as before.
We define the \emph{rapidity} of a Lorentz boost, $\phi$, such that
  \[
    \beta = \tanh \phi,\quad \gamma = \cosh\phi,\quad \gamma\beta=\sinh \phi.
  \]
Then
\[
  \Lambda[\beta] =
  \begin{pmatrix}
    \cosh \phi & -\sinh \phi\\
    -\sinh \phi & \cosh \phi
  \end{pmatrix}
  = \Lambda(\phi).
\]
The rapidities add like rotation angles:
\[
  \Lambda(\phi_1)\Lambda(\phi_2) = \Lambda(\phi_1 + \phi_2).
\]
Lorentz boosts are simply \emph{hyperbolic} rotations in spacetime!

Q: Proper time
A: The \emph{proper time} $\tau$ is defined such that
  \[
    \Delta \tau = \frac{\Delta s}{c}
  \]
  $\tau$ is the time experienced by the particle, ie. the time in the particles rest frame.
Infinitesimal changes are related by
\[
  \d \tau = \frac{\d s}{c} = \frac{1}{c}\sqrt{c^2\;\d t^2 - |\d \mathbf{x}|^2} = \sqrt{1 - \frac{|\mathbf{u}|^2}{c^2}}\;\d t.
\]
Thus
\[
  \frac{\d t}{\d \tau} = \gamma_u
\]
with
\[
  \gamma_u = \frac{1}{\sqrt{1 - \frac{|\mathbf{u}|^2}{c^2}}}.
\]
The total time experienced by the particle along a segment of its world line is thus
\[
  T = \int \;\d \tau = \int\frac{1}{\gamma_u}\;\d t.
\]

Q: Position 4-vector and 4-velocity
A: The \emph{position 4-vector} is
  \[
    X(\tau) =
    \begin{pmatrix}
      ct(\tau)\\
      \mathbf{x}(\tau)
    \end{pmatrix}.
  \]
  Its \emph{4-velocity} is defined as
  \[
    U = \frac{\d X}{\d \tau} =
    \begin{pmatrix}
      c\frac{\d t}{\d \tau}\\
      \frac{\d \mathbf{x}}{\d \tau}
    \end{pmatrix}
    = \frac{\d t}{\d \tau}
    \begin{pmatrix}
      c\\
      \mathbf{u}
    \end{pmatrix} = \gamma_u
    \begin{pmatrix}
      c\\
      \mathbf{u}
    \end{pmatrix},
  \]
  where $\mathbf{u} = \frac{\d \mathbf{x}}{\d t}$.
Frames $S$ and $S'$ are related by $X' = \Lambda X$ and $\tau$ is Lorentz invariant so $U' = \Lambda U$. We know (and can easily check) that the inner product $U\cdot U = U' \cdot U'$ is Lorentz invariant. In the rest frame of the particle, $U = (c, 0)$. So $U\cdot U = c^2$.

Q: 4-vector
A: A \emph{4-vector} is a 4-component vectors that transforms appropriately under a Lorentz transformation, ie. $X' = \Lambda X$.
We write $X^\mu$ for $\mu = 0, 1, 2, 3$.

Q: Aberration
A: In frame $S$, consider a particle moving with speed $u$ at an angle $\theta$ to the $x$ axis in the $xy$ plane.
Its 4-velocity is
\[
  U =
  \begin{pmatrix}
    \gamma_u c\\
    \gamma_u u\cos \theta\\
    \gamma_u u\sin \theta\\
    0
  \end{pmatrix}, \quad \gamma_u = \frac{1}{\sqrt{1 - u^2/c^2}}.
\]
With frames $S$ and $S'$ in standard configuration
\[
  U' = \begin{pmatrix}
    \gamma_{u'} c\\
    \gamma_{u'} u'\cos \theta'\\
    \gamma_{u'} u'\sin \theta'\\
    0
  \end{pmatrix}
  =
  \begin{pmatrix}
    \gamma_v & -\gamma_v v/c & 0 & 0\\
    -\gamma_{v} v/c & \gamma_v & 0 & 0\\
    0 & 0 & 1 & 0\\
    0 & 0 & 0 & 1
  \end{pmatrix}
  \begin{pmatrix}
    \gamma_u c\\
    \gamma_u u\cos \theta\\
    \gamma_u u\sin \theta\\
    0
  \end{pmatrix}
\]
The ratio of the first to second line ($\times c$) gives
\[
  u'\cos \theta' = \frac{u\cos \theta - v}{1 - \frac{uv}{c^2}\cos \theta},
\]
just like the composition of parallel velocities.
The ratio of the third to second line gives
\[
  \tan \theta' = \frac{u\sin \theta}{\gamma_v(u\cos \theta - v)},
\]
which describes \emph{aberration}, a change in the direction of motion of a particle due to the motion of the observer.

Q: 4-momentum and relativistic energy
A: The \emph{4-momentum} of a particle of mass $m$ is
  \[
    P = mU = m\gamma_u
    \begin{pmatrix}
      c\\
      \mathbf{u}
    \end{pmatrix}
  \]
  The 4-momentum of a system of particles is the sum of the 4-momentum of the particles, and is conserved in the absence of external forces.
  The spatial components of $P$ are the \emph{relativistic 3-momentum},
  \[
    \mathbf{p} = m\gamma_u \mathbf{u}.,
  \]
Expand the first component for $|\mathbf{u}| < c$:
\[
  P^0 = m\gamma c = \frac{mc}{\sqrt{1 + |\mathbf{u}|^2/c^2}} = \frac{1}{c}\left(mc^2 + \frac{1}{2}m|\mathbf{u}|^2 + \cdots\right).
\]
So we interpret $P$ as
\[
  P =
  \begin{pmatrix}
    E/c\\
    \mathbf{p}
  \end{pmatrix}
\]
where $E$ is the \emph{relativistic energy} which is the sum of the rest energy $mc^2$ and kinetic energy $m(\gamma_u-1)c^2$.
Since $P\cdot P = \frac{E^2}{c^2} - |\mathbf{p}|^2$ is a Lorentz invariant (lengths of 4-vectors are always Lorentz invariant) and equals $m^2 c^2$ in the particle's rest frame, we have the general relation between energy and momentum
\[
  E^2 = |\mathbf{p}|^2 c^2 + m^2c^4.
\]

Q: Massless particles
A: For massless particles, energy and momentum are related by
\[
  E^2 = |\mathbf{p}|^2 c^2.
\]
So
\[
  E = |\mathbf{p}|c.
\]
Thus
\[
  P = \frac{E}{c}
  \begin{pmatrix}
    1\\
    \mathbf{n}
  \end{pmatrix},
\]
where $\mathbf{n}$ is a unit (3-)vector in the direction of propagation.
This shows the equivalence of the \emph{de Broglie relation} and \emph{Planck's relation}
\[
  |\mathbf{p}| = \frac{h}{\lambda} \iff  E = \frac{hc}{\lambda} = h\nu.
\]
Note $P\cdot P = 0$. So massless particles have light-like (or null) trajectories, and no proper time can be defined for such particles.

Q: Newton's second law in special relativity
A: The \emph{4-force} is
  \[
    F = \frac{\d P}{\d \tau}.
  \]
It is related to the 3-force $\mathbf{F}$ by
\[
  F = \gamma_u
  \begin{pmatrix}
    \mathbf{F}\cdot \mathbf{u}/c\\
    \mathbf{F}
  \end{pmatrix}.
\]
Considering first components gives
\[
  \frac{\d E}{\d \tau} = \gamma_u \mathbf{F}\cdot \mathbf{u} \Rightarrow \frac{\d E}{\d t} = \mathbf{F}\cdot \mathbf{u}
\]
and the last three give
\[
  \frac{\d \mathbf{p}}{\d \tau} = \gamma_u \mathbf{F} \Rightarrow \frac{\d \mathbf{p}}{\d t} = \mathbf{F}.
\]
Alternatively, for a particle of mass $m$,
\[
  F =mA
\]
where
\begin{align*}
  A &= \frac{\d U}{\d \tau} \\
&=\gamma_u \frac{\d U}{\d t} \\
&=\gamma_u \begin{pmatrix} \dot{\gamma_u}c \\ \gamma_u\vec{a}+\dot{\gamma_u}\vec{u}\end{pmatrix}
\end{align*}
is the 4-acceleration and obviously $\mathbf{\mathbf{a}} = \frac{\d \mathbf{u}}{\d t}$ and $\dot{\gamma}_u = \gamma_u^3 \frac{\mathbf{a}\cdot \mathbf{u}}{c^2}$.
In the instantaneous rest frame of a particle, $\mathbf{u} = \mathbf{0}$ and $\gamma_u = 1$. So
\[
  U =
  \begin{pmatrix}
    c\\
    \mathbf{0}
  \end{pmatrix}, \quad
  A =
  \begin{pmatrix}
    0\\
    \mathbf{a}
  \end{pmatrix}
\]
Thus $\mathbf{U}\cdot \mathbf{A} = 0$ is Lorentz invariant.

Q: Center of momentum frame
A: The \emph{center of momentum (CM) frame}, or \emph{zero momentum frame}, is an inertial frame in which the total 3-momentum is $\sum \mathbf{p} = 0$.
This exists for particles with mass.
Using conservation of $4-$momentum in the CM frame is a good way to solve problems.

Q: Particle decay
A: A particle of mass $m_1$ decays into two particles of masses $m_2$ and $m_2$.
We have
\[
  P_1 = P_2 + P_3.
\]
ie.
\begin{align*}
  E_1 &= E_2 + E_3\\
  \mathbf{p}_1 &= \mathbf{p}_2 + \mathbf{p}_3.
\end{align*}
In the CM frame (ie. the rest frame of the original particle),
\begin{align*}
  E_1 = m_1 c^2 &= \sqrt{|\mathbf{p}_2|^2 c^2 + m_2^2c^4} + \sqrt{|\mathbf{p}_3|^2 c^2 + m_2^2 c^4}\\
  &\geq m_2c^2 + m_3 c^2.
\end{align*}
So decay is possible only if
\[
  m_1 \geq m_2 + m_3.
\]

Q: Particle scattering
A: In the laboratory frame $S$, suppose that particle 1 travels with speed $u$ and collides with particle 2 (at rest). Assume they have equal rest mass. Then $P_1+P_2=P_3+P_4$, where
$P_1 =(    m\gamma_u c, m\gamma_u u,0,0),$
$P_2 =(mc,0,0,0),$
$P_3=(m\gamma_w c, m\gamma_w w\cos\theta, m\gamma_w w\sin\theta,0),$
$P_4=(m\gamma_w c,m\gamma_w w\cos\phi,-m\gamma_w w\sin\phi)$.
In the CM frame $S'$,
\[
  \mathbf{p}'_1 + \mathbf{p}'_2 = 0 = \mathbf{p}_3' + \mathbf{p}'_4.
\]
Then we can choose axes such that
\[
  P_1' =
  \begin{pmatrix}
    m\gamma_v c\\
    m\gamma_v v\\
    0\\
    0
  \end{pmatrix},\quad
  P_2' =
  \begin{pmatrix}
    m\gamma_v c\\
    -m\gamma_v v\\
    0\\
    0
  \end{pmatrix}
\]
and after the collision we must have
\[
  P_3' =
  \begin{pmatrix}
    m\gamma_v c\\
    m\gamma_v v\cos \theta'\\
    m\gamma_v v\sin \theta'\\
    0
  \end{pmatrix},\quad
  P_4' =
  \begin{pmatrix}
    m\gamma_v c\\
    -m\gamma_v v\cos \theta'\\
    -m \gamma_v v\sin \theta'\\
    0
  \end{pmatrix}.
\]
We then use the Lorentz transformation to return to the laboratory frame $S$. The relative velocity of the frames is $v$. So the Lorentz transform is
\[
  \Lambda =
  \begin{pmatrix}
    \gamma_v & \gamma_v v/c & 0 & 0\\
    \gamma_v v/c & \gamma_v & 0 & 0\\
    0 & 0 & 1 & 0\\
    0 & 0 & 0 & 1
  \end{pmatrix}
\]
but we know
so equating expressions for $P_1^0$, $P_1^1$ and dividing gives
\[
  u = \frac{2v}{1 + v^2/c^2},
\]
(cf. velocity composition formula).
Similarly, the ratios $P_3^3/P_3^2$ and $P_4^3/P_4^2$ give
\[
  \tan \theta = \frac{\sin \theta'}{\gamma_v (1 + \cos \theta')} = \frac{1}{\gamma_v}\tan \frac{\theta'}{2},
\]
and
\[
  \tan \phi = \frac{\sin \theta'}{\gamma_v(1 - \cos \theta')} = \frac{1}{\gamma_v}\cot \frac{\theta'}{2}.
\]
Multiplying these expressions together, we obtain
\[
  \tan \theta\tan \phi = \frac{1}{\gamma_v^2}.
\]
In the Newtonian limit, $\tan \theta\tan \phi = 1$, ie. the outgoing trajectories are perpendicular in $S$.

Q: Particle Creation
A: Collide two particles of mass $m$ fast enough, and you create an extra particle of mass $M$.
\[
  P_1 + P_2 = P_3 + P_4 + P_5.
\]
In the CM frame
\[
  P_1 + P_2 =
  \begin{pmatrix}
    2m\gamma_v c\\
    \mathbf{0}
  \end{pmatrix}.
\]
After the creation
\[
  P_3 + P_4 + P_5 =
  \begin{pmatrix}
    (E_3 + E_4 + E_5)/c\\
    \mathbf{0}
  \end{pmatrix}
\]
So
\[
  2m\gamma_v c^2 = E_3 + E_4 + E_5 \geq 2mc^2 + Mc^2.
\]
So in order to create this new particle, we must have
\[
  \gamma_v \geq 1 + \frac{M}{2m}.
\]
Alternatively, it occurs only if the initial kinetic energy in the CM frame satisfies
\[
  2(\gamma_v - 1)mc^2 \geq Mc^2.
\]
If we transform to a frame in which the initial speeds are $u$ and 0 (ie. stationary target), then
\[
  u = \frac{2v}{1 + v^2/c^2}
\]
and so
\[
  \gamma_u = 2\gamma_v^2 - 1.
\]
So we require
\[
  \gamma_u \geq 2\left(1 + \frac{M}{2m}\right)^2 - 1 = 1 + \frac{2M}{m} + \frac{M^2}{2m}.
\]
This means that the initial kinetic energy in this frame must be
\[
  m(\gamma_u - 1)c^2 = \left(2 + \frac{M}{2}\right)Mc^2,
\]
which could be much larger than $Mc^2$ so colliding head on is usually better.