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\subsubsection*{Scaling (Bridgeman's Theorem)}

Dimensionful quantities can only appear in
equation as powers, for example $L^\alpha$ for
some $\alpha$ - can never have more complicated
functions. For example
\[ e^x = 1 + x + \frac{x^2}{2} + \cdots \]
$x$ must be dimensionless. \myskip
Suppose we want to compute a quantity $Y$ with
dimensions
\[ [Y] = M^\alpha L^\beta T^\gamma .\]
Let's suppose we wish to express $Y$ in terms of
other quantities $X_i$, $i = 1, \dots, n$. Let's
pick for example $X_1, X_2, X_3$. We assume that
they are \emph{dimensionally independent} (i.e.
we can build dimensions of length, mass and time
from them). Then, we can write
\[ Y = c X_1^{a_1} X_2^{a_2} X_3^{a_3} \]
where $c$ is a dimensionless constant. In general
it can be a dimensionless combination of $X_i$. In
the case where there are no dimensionless
combinations, $c$ is just a number.

\begin{example*}[Pendulum]
    We want to know period $T$. Obviously, $[T] =
    T$. It can depend upon:
    \begin{itemize}
        \item mass $m$, $[m] = M$
        \item length $l$, $[l] = L$
        \item gravity $g$, $[g] = LT^{-2}$
        \item starting angle $\theta_0$,
            $[\theta_0] = 0$, since $\theta_0
            \equiv \theta_0 + 2\pi \implies
            [\theta_0] = [2\pi] = 0$.
    \end{itemize}
    The only dimensionless quantity is $\theta_0$,
    \[ \implies T = c(\theta_0) g^{a_1} m^{a_2}
    l^{a_3} \]
    Take dimensions:
    \begin{align*}
        T
        &= [g^{a_1}][m^{a_2}][l^{a_3}] \\
        &= L^{a_1 + a_3} T^{-2a_1} M^{a_2}
    \end{align*}
    \[ \implies a_1 = -\half, \qquad a_2 = 0,
    \qquad a_3 = +\half .\]
    \[ \implies T = c(\theta_0) \sqrt{\frac{l}{g}} \]
    (often we write $T \sim \sqrt{\frac{l}{g}}$).
    So if we take two pendulums of different
    lengths $l_1$ and $l_2$, released at angle
    $\theta_0$, then
    \[ \frac{T_1}{T_2} = \sqrt{\frac{l_1}{l_2}} .\]
\end{example*}

\newpage
\section{Central Forces}

Study the three-dimensional motion of a particle
obeying
\[ m \ddot{\bf{x}} = -\nabla V(r) ; \qquad r :=
|\bf{x}| .\]
This could describe:
\begin{itemize}
    \item the position of a particle $\bf{x}$ in a
        fixed potential
    \item the separation $\bf{x}$ of two
        interacting particles where
        \[ m = \frac{m_1m_2}{m_1 + m_2} .\]
\end{itemize}

\begin{note*}
    When one particle is very heavy, this second
    case reduces to the first.
\end{note*}

\noindent
The angular momentum is conserved
\[ \bf{L} = m \bf{x} \times \dot{\bf{x}} \implies
\dfrac{\bf{L}}{t} = m \bf{x} \times \ddot{\bf{x}}
= -\bf{x} \times \nabla V ,\]
which is 0 since $\bf{x} \parallel \nabla V$, so
$\bf{L}$ is a fixed vector. But, by construction,
$\bf{L} \cdot \bf{x} = 0$. This is the equation
for a plane so all motion takes place in a
two-dimensional plane perpendicular to $\bf{L}$
(note $\bf{L} \cdot \bf{x} = 0$ too).
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/9e8725ba86a811ec.png}
\end{center}

\subsection{Polar Coordinates in the Plane}

A particle sits at coordinates
\[ x = r\cos\theta \qquad y = r\sin\theta \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/686b698686a911ec.png}
\end{center}
and we define unit vectors
\[ \hat{\bf{r}} =
\begin{pmatrix}
    \cos\theta \\
    \sin\theta
\end{pmatrix}
\qquad \hat{\bf{\theta}} =
\begin{pmatrix}
    -\sin\theta \\
    \cos\theta
\end{pmatrix}
\]
These form an orthonormal basis at each point, but
they depend on $\theta$.
\[ \dfrac{\hat{\bf{r}}}{\theta} =
\begin{pmatrix}
    -\sin\theta \\
    \cos\theta
\end{pmatrix}
= \hat{\bf{\theta}} \qquad
\dfrac{\hat{\bf{\theta}}}{\theta} =
\begin{pmatrix}
    -\cos\theta \\
    -\sin\theta
\end{pmatrix}
= -\hat{\bf{r}} .\]
The position of a particle is $\bf{x} = r
\hat{\bf{r}}$, so the velocity is
\[ \dot{\bf{x}} = \dot{r} \hat{\bf{r}} + r
\dfrac{\hat{\bf{r}}}{\theta} \dot{\theta} =
\dot{r} \hat{\bf{r}} + r \dot{\theta}
\hat{\bf{\theta}} \tag{4.1} \]
$\dot{\theta}$ is \emph{angular velocity}. The
acceleration is
\[ \ddot{\bf{x}} = \ddot{r} \hat{\bf{r}} + \dot{r}
\dfrac{\hat{\bf{r}}}{\theta} \dot{\theta} +
\dot{r} \dot{\theta} \hat{\bf{\theta}} + r
\ddot{\theta} \hat{\bf{\theta}} + r \dot{\theta}
\dfrac{\hat{\bf{\theta}}}{\theta} \dot{\theta} =
(\ddot{r} - r \dot{\theta}^2) \hat{\bf{r}} + (r
\ddot{\theta} + 2 \dot{r} \dot{\theta})
\hat{\bf{\theta}} \tag{4.2}\]

\begin{example*}[Circular Motion]
    A particle moving with constant angular speed
    in a circle has $\dot{r} = 0$, $\dot{\theta} =
    \omega$
    \[ (4.1) \implies \dot{\bf{x}} = r\omega
    \hat{\bf{\theta}} \]
    and so the speed is $|\dot{\bf{x}}| =
    r\omega$. The acceleration is
    \[ (4.2) \implies \ddot{\bf{x}} = -r \omega^2
    \hat{\bf{r}} \implies a = |\ddot{\bf{x}}| =
    r\omega^2 .\]
    If we want a particle to move like this,
    Newton's second law states that we
    must supply a \emph{centripetal force} toward
    to the origin.
\end{example*}