*****************************
Non-inertial reference frames
*****************************

.. math::
   \def\dd#1{\,\text{d}}
   \def\ddt#1{\frac{\text{d} #1}{\text{d}t}}
   \def\ddf#1#2{\frac{\text{d} #1}{\text{d} #2}}
   \def\dddf#1#2#3{\frac{\text{d}^{#3} #1}{\text{d} #2^{#3}}}
   \def\cp{\times}

Rotating Frames
===============

Consider a frame :math:`S_1` which is rotating with angular velocity, :math:`\omega`, about its :math:`z`-axis with respect to an inertial frame, :math:`S_0`, such that the axes coincide when :math:`t=0`.
As such there will be an angle between the sympathetic axes of :math:`\theta = \omega t` at any given time.

Now consider a vector :math:`\vec{A}_1` in :math:`S_1` which becomes 

.. math:: A_0 = {\mathsf{R}}(\omega t) \vec{A}_1

i.e. :math:`\vec{A}` as seen in :math:`S_0` rotates about the :math:`z` axis, corresponding to the rotation of the :math:`S_1` axes.
The explicit form of :math:`{\mathsf{R}}(\theta)` is then

.. math::

   {\mathsf{R}} = 
   \begin{bmatrix}
     \cos(\omega t) & - \sin(\omega t) & 0 \\
   \sin(\omega t)   & \cos(\omega t)   & 0 \\
   0              & 0              & 1
   \end{bmatrix}

In general :math:`A_1` may vary with :math:`t`, i.e. :math:`A_1` is not fixed in :math:`S_1`.
Its rate of change in :math:`S_1` is not simply given by :math:`A_1`.

This seems confusing at first but due to the fact that both the vector and the coordinate axes in :math:`S_1` are changing with time.
For example, a fixed vector in :math:`S_1` has :math:`\dot{A}_1=0` but :math:`\dot{A}_0 \neq 0`.

The rate of change of the axes introduces an additional term to velocities in :math:`S_0`; letting

.. math::

   \begin{aligned}
   \dd{\vec{A}_0} &= \mathop{{\left[ \df{{\mathsf{R}}}(\omega t) \right] \vec{A}_1}}\limits_{\text{From axis rotation.}} + {\mathsf{R}}(\omega t) \dd{\vec{A}_1} \\
   &= \vec{\omega} \cp \vec{A}_0 \dd{t} + {\mathsf{R}}(\omega t) \dd{A_1} \\
   &= \vec{\omega} \cp \left[ {\mathsf{R}}(\omega t) \vec{A}_1 \right] \dd{t} + {\mathsf{R}}(\omega t) \dd{A_1}
   \end{aligned}

Since

.. math::

   \left| \dd{{\mathsf{R}}} \vec{A}_1 \right| = \omega \dd{t} \left| \vec{A}_0(\omega t) \right|
                                              = \left| \omega \cp \vec{A}_0 \right| \dd{t}

Where the second part follows from the change in :math:`\vec{A}_0` being perpendicular to :math:`\vec{A}_0` for the infinitessimal interval :math:`\dd{t}`.

The quantity :math:`\vec{\omega} \cp \vec{A}_0` points in the direction of :math:`\dd{R} \vec{A}_1`, so given that the rotation is about the same axis as the finite rotation :math:`{\mathsf{R}}(\omega t)`,

.. math::

   \omega \times \qty( {\mathsf{R}} \vec{A}_1) = {\mathsf{R}} \qty[ \vec{\omega}
   \cp \vec{A}_1]

 This implies that the order of rotations is irrelevant, so

.. math::

   \begin{aligned}
     \dd{{\mathsf{R}}} \vec{A}_1 &= {\mathsf{R}}(\omega t) \qty[ \vec{\omega} \cp \vec{A}_1] \dd{t} \nonumber \\
   \implies \dd{\vec{A}_0} &= {\mathsf{R}}(\omega t) \qty[ \dd{\vec{A}_1} + \vec{\omega} \cp \vec{A}_1] \dd{t} \nonumber \\
   \label{eq:2}
   \dot{\vec{A}}_0 &= {\mathsf{R}}( \omega t ) \qty[ \dot{\vec{A}}_1 + \vec{\omega} \cp \vec{A}_1 ]\end{aligned}

The Coriolis and Centrifugal Forces
===================================

Let a particle have positions :math:`\vec{x}_0` and :math:`\vec{x}_1` in
the frames :math:`S_0` and :math:`S_1` respectively, with :math:`S_1`
rotating relative to :math:`S_0` according to the transformation
:math:`{\mathsf{R}}(\omega t)`, so

.. math::

   \begin{aligned}
     \vec{x}_0 &= {\mathsf{R}}(\omega t) \vec{x}_1 \\
   &= {\mathsf{R}}(\omega t) \qty[ \dot{\vec{x}}_1 + \vec{\omega} \cp \vec{x}_1 ] &\omit\hfill \text{ (by eq. \eqref{eq:2})} \\
   \implies \vec{v}_0 &= {\mathsf{R}}(\omega t) \qty[ \vec{v}_1 + \vec{\omega} \cp \vec{x}_1 ]\end{aligned}

 Letting :math:`\vec{A}_0 = \vec{v}_0`, and
:math:`\vec{A}_1 = \vec{v}_1 + \vec{\omega} \cp \vec{x}_1`, then

.. math::

   \begin{aligned}
     \dot{\vec{v}}_0 &= {\mathsf{R}}(\omega t) \qty[ \dot{\vec{v}}_1 + \vec{\omega} \cp \dot{\vec{x}} + \vec{\omega} \cp \qty( \vec{v}_1 + \vec{\omega} \cp \vec{x}_1 ) ] \\ 
   &= {\mathsf{R}}(\omega t) \qty[ \dot{\vec{v}}_1 + 2 \vec{\omega} \cp \vec{v}_1 + \vec{\omega} \cp ( \vec{\omega} \cp \vec{x}_1 )]\end{aligned}

 Since :math:`\vec{a}_i = \dot{\vec{v}}_i` is the acceleration observed
in the frame :math:`S_i`, then

.. math::

   \vec{a}_1 = {\mathsf{R}}(-\omega t) \vec{a}_0 - 2 \vec{\omega} \cp
   \vec{v}_1 - \vec{\omega} \cp ( \vec{\omega} \cp \vec{x}_1 )

 So the acceleration is not the same in each frame, and since in the
individual frames Newton’s laws are valid, we see two additional forces,

.. math::

   \begin{aligned}
     \label{eq:3}
     \vec{f}~{cor} &= - 2 m \vec{\omega} \cp \vec{v}_1 \\
   \vec{f}~{cen} &= - m \vec{\omega} \cp \qty( \vec{\omega} \cp \vec{x}_1 )\end{aligned}

Respectively the Coriolis and Centrifugal forces.

Motion at the Earth’s Surface
=============================

The coordinate system on the Earth’s surface is defined using two
coordinates, latitude, :math:`\lambda`, and longitude, :math:`\beta`. We
also have a rotation vector, :math:`\vec{\Omega}`. A point close to the
surface is specified by three coordinates: altitude, latitude, and
longitude, and its velocity has the vector

.. math:: \vec{v} = \qty( v~{long}, v~{lat}, v~{alt})

 an object near the Earth’s surface will experience a Coriolis force,

.. math:: \vec{f}~{cor} = - 2m \vec{\Omega} \cp \vec{v}

 and we have

.. math:: \vec{\Omega} = \qty( 0, \Omega \cos(\lambda), \Omega \sin(\lambda) )

 Thus

.. math::

   \begin{aligned}
   \vec{f}~{cor} &= -2m  \qty( 0, \Omega \cos(\lambda), \Omega \sin(\lambda) ) \cp \qty( v~{long}, v~{lat}, v~{alt}) \nonumber\\
   &= - 2 m \Omega \begin{pmatrix} v~{alt} \cos(\lambda) - v~{lat} \sin(\lambda)\\ v~{long} \sin(\lambda) \\ - v~{long} \cos(\lambda)
   \end{pmatrix} \nonumber \\
   & \text{ if } v~{alt} = 0 \nonumber \\
   &= -2m \Omega
   \begin{pmatrix}
     - v~{lat} \sin(\lambda) \\ v~{long} \sin(\lambda) \\ 0 
   \end{pmatrix} \nonumber \\
   &= - 2m \qty(0 , 0 , \Omega \sin(\lambda) ) \cp \qty( v~{long}, v~{lat}, 0) \nonumber \\
   \therefore \vec{f}~{cor} &= - 2 m \vec{\Omega}~{alt} \cp \vec{v}\end{aligned}

The Foucalt Pendulum
====================

The plane of a pendulum’s motion will rotate over time in a rotating
reference frame. Let :math:`S_1` be the rotating reference frame of the
Earth, and :math:`S_2` be a frame rotating with angular velocity
:math:`\omega_2
\propto \Omega~{alt}`. The accelerations in the two frames satisfy

.. math::

   \label{eq:1}
     \vec{a}_1 = {\mathsf{R}} \qty[ \vec{a}_2 + 2 \vec{\omega}_2 \cp \vec{v}_2 + \vec{\omega}_2 \cp ( \vec{\omega}_2 \cp \vec{x}_2 )]

 where :math:`\vec{x}_i`, :math:`\vec{v}_i`, and :math:`\vec{a}_i` are
respectively the position, velocity, and accleration in the
:math:`i`\ th frame. Assuming both :math:`\vec{\Omega}` and
:math:`\vec{\omega}_2` are small, then the centrifugal term can be
neglected, and

.. math:: \vec{a}_1 = {\mathsf{R}} \qty[ \vec{a}_2 + 2 \vec{\omega}_2 \cp \vec{v}_2 ]

 we know

.. math:: \vec{a}_1 = \vec{g} - 2 \vec{\Omega}~{alt} \cp \vec{v}_2

 For :math:`\vec{g}` the acceleration due to gravity, so, given that the
pendulum undergoes horizontal motion, and we can neglect its vertical
motion, ignoring quadratic terms,

.. math:: \vec{v}_1 \approx {\mathsf{R}} \vec{v}_2

 and so

.. math::

   \vec{g} - 2 {\mathsf{R}} \qty[ \vec{\Omega}~{alt} \cp \vec{v}_2 ] =
   {\mathsf{R}} \qty[ \vec{a}_2 + 2 \vec{\omega}_2 \cp \vec{v}_2 ]

 thus, multiplying both sides by :math:`{\mathsf{R}}^{-1}`,

.. math::

   \label{eq:4}
     \vec{a}_2 = {\mathsf{R}}^{-1} \vec{g}

 and so the acceleration of the bob is simply the acceleration due to
gravity transformed into a different frame, and in :math:`S_2` it
appears as if only gravity acts, so this represents a frame which is
rotating to counteract the rotation of the pendulum’s plane, thus
demonstrating that the plane is rotating.