Accelerating Frames

What do we do if we happen to be in a non-inertial frame?

Linearly Accelerating Frame

Inertial Observer

Let $\mathbf{r}$ be the position in the inertial (external) frame, and $\mathbf{r}'$ in the ship (non-inertial) frame. These can be related through the acceleration of the ship w.r.t the external frame: $a_s$: \begin{equation} \mathbf{r} = \mathbf{r}' + \frac{1}{2}\mathbf{a}_s t^2 \end{equation}

Velocities are likewise found to be (through differentiation w.r.t $t$): \begin{equation} \mathbf{v} = \mathbf{v}' + \mathbf{a}_s t \end{equation} and acceleration: \begin{equation} \mathbf{a} = \mathbf{a}' + \mathbf{a}_s \end{equation}

So, for observers inside the non-inertial frame: \begin{equation} \mathbf{F}' = \mathbf{F} + \mathbf{F}_\textrm{pseudo} \end{equation}

This observed force, $\mathbf{F}'$ is the sum of any real forces and the Pseudo-Forces created by the accelerating reference frame.

Example: effective gravity for the accelerating ship: \begin{equation} \mathbf{F} = 0 \end{equation} but \begin{equation} \mathbf{F}' \neq 0 = 0 - m \mathbf{a}_s = m \mathbf{a}' \end{equation} thus: \begin{equation} \mathbf{a}' = -\mathbf{a}_s \end{equation}

In this simple case, we can call our pseudo-force, the effective gravity: \begin{equation} \mathbf{g}_\textrm{eff} = - \mathbf{a}_s \end{equation} Everything can be done that same as before if we include this in our set-up.

What about the Lagrangian approach? \begin{equation} L = T' - U' \end{equation} where $T'$ is the kinetic energy in the non-inertial frame, and $U'$ would include and pseudo-potentials: \begin{equation} L = \frac{1}{2}m \left( \dot{x}'^2 + \dot{y}'^2 \right) - m g_\textrm{eff} y' \end{equation}

Rotating Frames

Now, what if $\mathbf{A}$ is also changing in the rotating frame? \begin{equation} d \mathbf{A}_\textrm{in} = d \mathbf{A}_\textrm{rot} + d \boldsymbol{\phi} \times \mathbf{A} \end{equation}

Next, consider the time derivatives: \begin{equation} \left. \frac{d \mathbf{A}}{dt} \right|_\textrm{in} = \left. \frac{d \mathbf{A}}{dt} \right|_\textrm{rot} + \boldsymbol{\omega} \times \mathbf{A} \end{equation} where $\omega = \frac{d\phi}{dt}$

Helpful to consider this an operator: \begin{equation} \left. \frac{d}{dt} \right|_\textrm{in} = \left. \frac{d }{dt} \right|_\textrm{rot} + \boldsymbol{\omega} \times \end{equation} This can operate on any vector and will transform from one coordinate system to another: inertial -> rotating.

Now we express the velocity in the inertial frame in terms of the velocity in the rotating frame (let our position be $\mathbf{r}$): \begin{equation} \mathbf{v}_\textrm{in} = \mathbf{v}_\textrm{rot} + \boldsymbol{\omega} \times \mathbf{r} \end{equation}

Using this, we can construct the Lagrangian: \begin{equation} L = \frac{1}{2}m \mathbf{v}^2 - U(\mathbf{r}) \end{equation}

Now for some vector math: $$ \begin{aligned} \frac{1}{2}m \left(\mathbf{v}_\textrm{rot} + \boldsymbol{\omega} \times \mathbf{r} \right)^2 = & \frac{1}{2}m \mathbf{v}_\textrm{rot}^2 + m \mathbf{v}_\textrm{rot} \cdot \left( \boldsymbol \omega \times \mathbf{r} \right) + \frac{1}{2}m \left( \boldsymbol \omega \times \mathbf{r}\right)^2 \\ = & \frac{1}{2}m \mathbf{v}_\textrm{rot}^2 + m \mathbf{v}_\textrm{rot} \cdot \left(\boldsymbol \omega \times \mathbf{r} \right) + \frac{1}{2}m \omega^2 r^2 - \frac{1}{2}m \left( \boldsymbol \omega \cdot \mathbf{r} \right)^2 \end{aligned} $$

Thus, in the rotating frame, were we to construct a Lagrangian based on our measurements, we would obtain: \begin{equation} L = \frac{1}{2}m \mathbf{v}_\textrm{rot}^2 + m \mathbf{v}_\textrm{rot} \cdot \left(\boldsymbol \omega \times \mathbf{r} \right) + \frac{1}{2}m \omega^2 r^2 - \frac{1}{2}m \left( \boldsymbol \omega \cdot \mathbf{r} \right)^2 - U(\mathbf{r}) \end{equation} This is clearly not just $\frac{1}{2}m \mathbf{v}_\textrm{rot}^2$

Find the equations of motion using the Lagrangian: \begin{equation} \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{r}_\textrm{rot}^i} \right) = \frac{\partial L}{\partial r_\textrm{rot}^i} \end{equation}

Here, $i$ implies $x$, $y$, and $z$ in the rotating frame.

\begin{equation} L = \frac{1}{2}m \mathbf{v}_\textrm{rot}^2 + m \mathbf{v}_\textrm{rot} \cdot \left(\boldsymbol \omega \times \mathbf{r} \right) + \frac{1}{2}m \omega^2 r^2 - \frac{1}{2}m \left( \boldsymbol \omega \cdot \mathbf{r} \right)^2 - U(\mathbf{r}) \end{equation}

So, taking the time derivative of $\frac{\partial L}{\partial \dot{r}_\textrm{rot}^i}$: \begin{equation} \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{r}_\textrm{rot}^i} \right) = m \ddot{r}_\textrm{rot}^i + m \left(\boldsymbol{\omega} \times \mathbf{v}_\textrm{rot} \right)^i + m \left( \frac{d \boldsymbol{\omega}}{dt } \times \mathbf{r} \right)^i \end{equation}

\begin{equation} L = \frac{1}{2}m \mathbf{v}_\textrm{rot}^2 + m \mathbf{v}_\textrm{rot} \cdot \left(\boldsymbol \omega \times \mathbf{r} \right) + \frac{1}{2}m \omega^2 r^2 - \frac{1}{2}m \left( \boldsymbol \omega \cdot \mathbf{r} \right)^2 - U(\mathbf{r}) \end{equation}

For the R.H.S. ($\frac{\partial L}{\partial r_\textrm{rot}^i}$): first rewrite one term as: \begin{equation} m \mathbf{v}_\textrm{rot} \cdot \left( \boldsymbol{\omega} \times \mathbf{r} \right) = m \mathbf{r} \cdot \left(\mathbf{v}_\textrm{rot} \times \boldsymbol{\omega} \right) \end{equation}

Now the Lagrangian is: \begin{equation} L = \frac{1}{2}m \mathbf{v}_\textrm{rot}^2 +m \mathbf{r} \cdot \left(\mathbf{v}_\textrm{rot} \times \boldsymbol{\omega} \right) + \frac{1}{2}m \omega^2 r^2 - \frac{1}{2}m \left( \boldsymbol \omega \cdot \mathbf{r} \right)^2 - U(\mathbf{r}) \end{equation}

\begin{equation} L = \frac{1}{2}m \mathbf{v}_\textrm{rot}^2 +m \mathbf{r} \cdot \left(\mathbf{v}_\textrm{rot} \times \boldsymbol{\omega} \right) + \frac{1}{2}m \omega^2 r^2 - \frac{1}{2}m \left( \boldsymbol \omega \cdot \mathbf{r} \right)^2 - U(\mathbf{r}) \end{equation}

Leads to: \begin{equation} \frac{\partial L }{\partial r_\textrm{rot}^i} = m \left( \mathbf{v}_\textrm{rot} \times \boldsymbol{\omega} \right)_\textrm{rot}^i + m \omega^2 r_\textrm{rot}^i - m \left( \boldsymbol{\omega} \cdot \mathbf{r} \right)\omega^i - \frac{\partial U(\mathbf{r})}{\partial r_\textrm{rot}^i} \end{equation}

Thus our equation of motion in the rotating system is: \begin{equation} m \mathbf{a} = m \omega^2 \mathbf{r}_\textrm{rot} - m \left( \boldsymbol{\omega} \cdot \mathbf{r} \right) \boldsymbol{\omega}_\textrm{rot} - 2 m \left( \boldsymbol{\omega} \times \mathbf{v}_\textrm{rot} \right)_\textrm{rot} - m \left( \frac{d\boldsymbol{\omega}}{dt} \times \mathbf{r} \right)_\textrm{rot} - \nabla_\textrm{rot} U(\mathbf{r}) \end{equation}

Combine the first two terms: \begin{equation} m \omega^2 \mathbf{r}_\textrm{rot} - m \left( \boldsymbol{\omega} \cdot \mathbf{r} \right) \boldsymbol{\omega}_\textrm{rot} = -m \boldsymbol{\omega} \times \left( \boldsymbol{\omega } \times \mathbf{r} \right)_\textrm{rot} \end{equation} (using a vector identity.)

Finaaaaly, \begin{equation} \mathbf{F}_\textrm{rot} = \mathbf{F}_\textrm{in} -m \boldsymbol{\omega} \times \left( \boldsymbol{\omega } \times \mathbf{r} \right)_\textrm{rot} - 2 m \left( \boldsymbol{\omega} \times \mathbf{v}_\textrm{rot} \right)_\textrm{rot} - m \left( \dot{\boldsymbol{\omega}} \times \mathbf{r} \right)_\textrm{rot} \end{equation} where $\mathbf{F}_\textrm{in} = - \nabla_\textrm{rot} U(\mathbf{r})$ is the sum of real forces acting in the inertial frame

What are these pseudo-forces?, \begin{equation} \mathbf{F}_\textrm{rot} = \mathbf{F}_\textrm{in}\; \underbrace{- m \boldsymbol{\omega} \times \left( \boldsymbol{\omega } \times \mathbf{r} \right)_\textrm{rot} }_\textrm{centrifugal}\; \underbrace{- 2 m \left( \boldsymbol{\omega} \times \mathbf{v}_\textrm{rot} \right)_\textrm{rot}}_\textrm{Coriolis} \; \underbrace{- m \left( \dot{\boldsymbol{\omega}} \times \mathbf{r} \right)_\textrm{rot}}_\textrm{Euler} \end{equation}

Pseudo-Forces

Centrifugal

\begin{equation} - m \boldsymbol{\omega} \times \left( \boldsymbol{\omega } \times \mathbf{r} \right)_\textrm{rot} \end{equation}

Coriolis

\begin{equation} - 2 m \left( \boldsymbol{\omega} \times \mathbf{v}_\textrm{rot} \right) \end{equation}

The Coriolis force acts when the object is moving in a direction that is not parallel to $\boldsymbol{\omega}$

Euler

\begin{equation} - m \left( \boldsymbol{\dot{\omega}} \times \mathbf{r} \right)_\textrm{rot} \end{equation}

This pseudo-force only arises when $\omega$ is changing, i.e. the speed of rotation is changing.

[Sim from https://sciencesims.com/sims/rotating-ref-frame/]

Earth as rotating reference frame

Thus, the effective gravity would be: \begin{equation} g_\textrm{eff} = g - \Omega^2 \rho \end{equation} The value of $\rho$ depends on latitude, and you can easily see how at the poles, this contribution should be zero.

Coriolis

If there is no vertical motion, i.e. $\dot{z} = 0$: \begin{equation} \mathbf{F}_\textrm{Cor, horiz} = 2 m \Omega \sin \lambda \left( \dot{y} \hat{\bf{x}} - \dot{x} \hat{\bf{y}} \right) \end{equation} but $\mathbf{v} = (\dot{x},\dot{y}) = \left( v \cos \theta, v \sin \theta \right)$ so: \begin{equation} \mathbf{F}_\textrm{Cor, horiz} = 2 m \Omega \sin \lambda v \left( \sin \theta \hat{\bf{x}} - \cos \theta \hat{\bf{y}} \right) \end{equation}

Foucault Pendulum

The forces in the rotating frame of the Earth: \begin{equation} m \ddot{\mathbf{r}} = \mathbf{T} + m \mathbf{g}_0 + m \left(\boldsymbol{\Omega} \times \mathbf{r} \right) \times \boldsymbol{\Omega} + 2 m \dot{\mathbf{r}} \times \boldsymbol{\Omega} \end{equation}

Combine $\mathbf{g}_0$ and $m \left(\boldsymbol{\Omega} \times \mathbf{r} \right) \times \boldsymbol{\Omega}$ to express in terms of the observed gravitational force: \begin{equation} m \ddot{\mathbf{r}} = \mathbf{T} + m \mathbf{g} + 2 m \dot{\mathbf{r}} \times \boldsymbol{\Omega} \end{equation}

A little more cleaning up and we have: $$ \begin{align} \ddot{x} - 2 \dot{y} \Omega_z + \omega_0^2 x = 0 \\ \ddot{y} + 2 \dot{x} \Omega_z + \omega_0^2 y = 0 \\ \end{align} $$

Let's solve these. $$ \begin{align} \ddot{x} - 2 \dot{y} \Omega_z + \omega_0^2 x = 0 \\ \ddot{y} + 2 \dot{x} \Omega_z + \omega_0^2 y = 0 \\ \end{align} $$

Start by defining a complex number: \begin{equation} \eta = x + iy \end{equation} And multiply the $\ddot{y}$ equation by $i$, then add it to the $\ddot{x}$ equation: \begin{equation} \ddot{\eta} + 2 i \Omega_z \dot{\eta} + \omega_0^2 \eta = 0 \end{equation}

Now we have a second-order, linear, homogeneous differential equation.

This implies two indepedent solutions

Try: $\eta(t) = e^{-i\alpha t}$

Leads to: \begin{equation} \alpha^2 - 2 \Omega_z \alpha - \omega_0^2 = 0 \end{equation} or \begin{equation} \alpha = \Omega_z \pm \sqrt{\Omega_z^2+ \omega_0^2} \end{equation}

Thus: \begin{equation} \eta = e^{-i\Omega_z t}\left(C_1 e^{i \omega_0 t} + C_2 e^{-i \omega_0 t} \right) \end{equation}

With $x = A$ and $y = 0$ at $t = 0$, the pendulum is released from rest ($v_{x0} = v_{y0} = 0$) and we can obtain: \begin{equation} C_1 = C_2 = \frac{A}{2} \end{equation}

\begin{eqnarray} \eta(t) & = & x(t) + iy(t) \\ & = & A e^{-i \Omega_z t} \cos \omega_0 t \end{eqnarray}