Non-inertial reference frames

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

Rotating Frames

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

Now consider a vector \(\vec{A}_1\) in \(S_1\) which becomes

\[A_0 = {\mathsf{R}}(\omega t) \vec{A}_1\]

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

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

In general \(A_1\) may vary with \(t\), i.e. \(A_1\) is not fixed in \(S_1\). Its rate of change in \(S_1\) is not simply given by \(A_1\).

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

The rate of change of the axes introduces an additional term to velocities in \(S_0\); letting

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

Since

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

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

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

\[ \begin{align}\begin{aligned} \omega \times \qty( {\mathsf{R}} \vec{A}_1) = {\mathsf{R}} \qty[ \vec{\omega} \cp \vec{A}_1]\\This implies that the order of rotations is irrelevant, so\end{aligned}\end{align} \]
\[\begin{split}\begin{aligned} \dd{{\mathsf{R}}} \vec{A}_1 &= {\mathsf{R}}(\omega t) \qty[ \vec{\omega} \cp \vec{A}_1] \dd{t} \nonumber \\ \implies \dd{\vec{A}_0} &= {\mathsf{R}}(\omega t) \qty[ \dd{\vec{A}_1} + \vec{\omega} \cp \vec{A}_1] \dd{t} \nonumber \\ \label{eq:2} \dot{\vec{A}}_0 &= {\mathsf{R}}( \omega t ) \qty[ \dot{\vec{A}}_1 + \vec{\omega} \cp \vec{A}_1 ]\end{aligned}\end{split}\]

The Coriolis and Centrifugal Forces

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

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

\(\vec{A}_1 = \vec{v}_1 + \vec{\omega} \cp \vec{x}_1\), then

\[ \begin{align}\begin{aligned}\begin{split} \begin{aligned} \dot{\vec{v}}_0 &= {\mathsf{R}}(\omega t) \qty[ \dot{\vec{v}}_1 + \vec{\omega} \cp \dot{\vec{x}} + \vec{\omega} \cp \qty( \vec{v}_1 + \vec{\omega} \cp \vec{x}_1 ) ] \\ &= {\mathsf{R}}(\omega t) \qty[ \dot{\vec{v}}_1 + 2 \vec{\omega} \cp \vec{v}_1 + \vec{\omega} \cp ( \vec{\omega} \cp \vec{x}_1 )]\end{aligned}\end{split}\\Since :math:`\vec{a}_i = \dot{\vec{v}}_i` is the acceleration observed\end{aligned}\end{align} \]

in the frame \(S_i\), then

\[ \begin{align}\begin{aligned} \vec{a}_1 = {\mathsf{R}}(-\omega t) \vec{a}_0 - 2 \vec{\omega} \cp \vec{v}_1 - \vec{\omega} \cp ( \vec{\omega} \cp \vec{x}_1 )\\So the acceleration is not the same in each frame, and since in the\end{aligned}\end{align} \]

individual frames Newton’s laws are valid, we see two additional forces,

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

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, \(\lambda\), and longitude, \(\beta\). We also have a rotation vector, \(\vec{\Omega}\). A point close to the surface is specified by three coordinates: altitude, latitude, and longitude, and its velocity has the vector

\[ \begin{align}\begin{aligned}\vec{v} = \qty( v~{long}, v~{lat}, v~{alt})\\an object near the Earth’s surface will experience a Coriolis force,\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\vec{f}~{cor} = - 2m \vec{\Omega} \cp \vec{v}\\and we have\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\vec{\Omega} = \qty( 0, \Omega \cos(\lambda), \Omega \sin(\lambda) )\\Thus\end{aligned}\end{align} \]
\[\begin{split}\begin{aligned} \vec{f}~{cor} &= -2m \qty( 0, \Omega \cos(\lambda), \Omega \sin(\lambda) ) \cp \qty( v~{long}, v~{lat}, v~{alt}) \nonumber\\ &= - 2 m \Omega \begin{pmatrix} v~{alt} \cos(\lambda) - v~{lat} \sin(\lambda)\\ v~{long} \sin(\lambda) \\ - v~{long} \cos(\lambda) \end{pmatrix} \nonumber \\ & \text{ if } v~{alt} = 0 \nonumber \\ &= -2m \Omega \begin{pmatrix} - v~{lat} \sin(\lambda) \\ v~{long} \sin(\lambda) \\ 0 \end{pmatrix} \nonumber \\ &= - 2m \qty(0 , 0 , \Omega \sin(\lambda) ) \cp \qty( v~{long}, v~{lat}, 0) \nonumber \\ \therefore \vec{f}~{cor} &= - 2 m \vec{\Omega}~{alt} \cp \vec{v}\end{aligned}\end{split}\]

The Foucalt Pendulum

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

\[ \begin{align}\begin{aligned} \label{eq:1} \vec{a}_1 = {\mathsf{R}} \qty[ \vec{a}_2 + 2 \vec{\omega}_2 \cp \vec{v}_2 + \vec{\omega}_2 \cp ( \vec{\omega}_2 \cp \vec{x}_2 )]\\where :math:`\vec{x}_i`, :math:`\vec{v}_i`, and :math:`\vec{a}_i` are\end{aligned}\end{align} \]

respectively the position, velocity, and accleration in the \(i\)th frame. Assuming both \(\vec{\Omega}\) and \(\vec{\omega}_2\) are small, then the centrifugal term can be neglected, and

\[ \begin{align}\begin{aligned}\vec{a}_1 = {\mathsf{R}} \qty[ \vec{a}_2 + 2 \vec{\omega}_2 \cp \vec{v}_2 ]\\we know\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned}\vec{a}_1 = \vec{g} - 2 \vec{\Omega}~{alt} \cp \vec{v}_2\\For :math:`\vec{g}` the acceleration due to gravity, so, given that the\end{aligned}\end{align} \]

pendulum undergoes horizontal motion, and we can neglect its vertical motion, ignoring quadratic terms,

\[ \begin{align}\begin{aligned}\vec{v}_1 \approx {\mathsf{R}} \vec{v}_2\\and so\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \vec{g} - 2 {\mathsf{R}} \qty[ \vec{\Omega}~{alt} \cp \vec{v}_2 ] = {\mathsf{R}} \qty[ \vec{a}_2 + 2 \vec{\omega}_2 \cp \vec{v}_2 ]\\thus, multiplying both sides by :math:`{\mathsf{R}}^{-1}`,\end{aligned}\end{align} \]
\[ \begin{align}\begin{aligned} \label{eq:4} \vec{a}_2 = {\mathsf{R}}^{-1} \vec{g}\\and so the acceleration of the bob is simply the acceleration due to\end{aligned}\end{align} \]

gravity transformed into a different frame, and in \(S_2\) 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.