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\subsubsection*{Inertial versus Gravitational Mass}

Mass has appeared in a couple of formulae:
\[ \bf{F} = m_I \ddot{\bf{x}} \]
\[ \bf{F} = -\frac{GM_G m_G}{r^2} \hat{\bf{r}} \]
here $m_I$ denotes \emph{inertial mass} and $m_G$
denotes \emph{gravitational mass}. \myskip
Conceptually, these are very different quantities.
Experimentally, we know that $m_I = m_G$ (to 1
part in $10^{13}$) - only explained by Einstein's
general theory of relativity.

\subsection{Electromagnetism}

The force experienced by a particle with electric
charge $q$ is
\[ \bf{F} = q (\bf{E}(\bf{x}) + \dot{\bf{x}}
\times \bf{B}(\bf{x})) .\]
($E$ is \emph{electric field} and $B$ is
\emph{magnetic field}.) \emph{This is the Lorentz
Force Law}. \myskip
The electron has a charge $-1.6 \times 10^{-19}
\mathsf{C}$ ($\mathsf{C} = \text{Coulomb}$).
$\bf{E}$ and $\bf{B}$ are functions of space and
(in principle) time. We'll only consider time
independent fields, where we can write $\bf{E}$ as
\[ \bf{E} = -\nabla \phi \]
for some \emph{electric potential} $\phi \implies$
the electric force is conservative.

\begin{claim*}
    For $t$-independent $\bf{E}(\bf{x})$ and
    $\bf{B}(\bf{x})$ the energy
    \[ E = \half m \dot{\bf{x}} \cdot \dot{\bf{x}}
    + q\phi(\bf{x}) \]
    is conserved.
\end{claim*}

\begin{proof}
    \begin{align*}
        \dfrac{E}{t}
        &= m \dot{\bf{x}} \cdot \ddot{\bf{x}} + q
        \nabla \phi \cdot \dot{\bf{x}} \\
        &= \dot{\bf{x}} (m \ddot{\bf{x}} + q\nabla
        \phi) \\
        &= q \dot{\bf{x}} \cdot (\dot{\bf{x}}
        \times \bf{B}(\bf{x})) \\
        &= 0
        && \text{since perpendicular}
    \end{align*}
\end{proof}

\subsubsection*{Point Charges}

A fixed particle of charge $Q$ produces an
electric field
\[ \bf{E} = -\nabla \left( \frac{Q}{4\pi\eps_0 r}
\right) = \frac{Q}{4\pi\eps_0}
\frac{\hat{\bf{r}}}{r^2} \]
$\eps_0$ is the \emph{permitivity of free space},
and is equal to approximately $8.85 \times
10^{-12} \mathsf{m^{-3}kg^{-1}s^2C^2}$. The
resulting force on a particle of charge $q$,
$\bf{F} = q \bf{E}$, is the \emph{Coulomb force}.
\[ \bf{F} = q \bf{E} = \frac{qQ}{4\pi\eps_0}
\frac{\hat{\bf{r}}}{r^2} \]
(Note that the signs of $q$ and $Q$ tell you
whether the force is attractive or repulsive.)

\subsubsection*{Motion in Constant $\bf{B}$}

Set $\bf{E} = \bf{0}$ and consider
$\bf{B}(0,0,B)$, $\bf{F} = q \dot{\bf{x}} \times
\bf{B}$. In components,
\begin{align*}
    m \ddot{x}
    &= q B \dot{y} \tag{1} \\
    m \ddot{y}
    &= -q B \dot{x} \tag{2} \\
    m \ddot{z}
    &= 0 \tag{3}
\end{align*}
so constant velocity in $z$ direction. \\
Trick: let $\xi = x + iy$, then
\[ (1) + i(2) \implies m \ddot{\xi} = -iq
B\dot{\xi} \implies \xi = \alpha e^{i\omega t} +
\beta \]
with $\omega :+ \frac{qB}{m}$, the \emph{cyclotron
frequency}. For initial conditions, pick at $t =
0$, $\bf{x} = 0$, $\dot{\bf{x}} = (0, -v, 0)$,
then at $t = 0$, $\xi = 0$ and $\dot{\xi} = -iv$.
\[ \implies \xi = \frac{v}{w} (e^{i\omega t} - 1)
.\]
\[ \implies x = \frac{v}{w} (\cos \omega t - 1) ,\]
\[ y = -\frac{v}{w} \sin \omega t \]
\[ \implies \left( x + \frac{v}{w} \right)^2 + y^2
= \left( \frac{v}{w} \right)^2 .\]
The particle moves around the circle with period
$T = \frac{2\pi}{\omega} = \frac{2\pi m}{qB}$.

\begin{note*}
    $T$ is independent of $v$. But the faster the
    particle, the bigger the circle.
\end{note*}

\subsubsection*{A Comment on Solving Using Vectors}

Sometimes better off keeping everything in vector
form.
\[ m \ddot{\bf{x}} = q \dot{\bf{x}} \times \bf{B} \]
\[ \implies m \ddot{\bf{x}} \cdot \bf{B} =
q(\dot{\bf{x}} \times \bf{B}) \cdot \bf{B} = 0 .\]
i.e. constant velocity in the $\bf{B}$ direction.
This is only one equation. We get two others by
taking the cross product of the top equation with
$\bf{B}$. In this case, it's best to integrate
first
\[ m \dot{\bf{x}} = q \bf{x} \times \bf{B} + \bf{C} \]
now take $\times \bf{B}$ of this:
\[ m \dot{\bf{x}} \times \bf{B} = q(\bf{x} \times
\bf{B}) \times \bf{B} + \bf{C} \times \bf{B} \]
and plug into the original equation.