# Accelerating Frames

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

## Linearly Accelerating Frame

#### Inertial Observer

The external observer will see the ball move in a straight line. $$L = T - U = \frac{1}{2}m \left(\dot{x}^2 + \dot{y}^2 \right)$$ which when processed will lead to: $$\ddot{y} = 0$$

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: $a_s$: $$\mathbf{r} = \mathbf{r}' + \frac{1}{2}\mathbf{a}_s t^2$$

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

In the inertial frame: $\mathbf{F} = m \mathbf{a}$, or, considering the two frames: $$\mathbf{F} = m \mathbf{a} = m \left(\mathbf{a}' + \mathbf{a}_s \right)$$ or, considering a new $\mathbf{F}'$: $$\mathbf{F}' = m \mathbf{a}' = \mathbf{F} - m \mathbf{a}_s$$

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

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: $$\mathbf{F} = 0$$ but $$\mathbf{F}' \neq 0 = 0 - m \mathbf{a}_s = m \mathbf{a}'$$ thus: $$\mathbf{a}' = -\mathbf{a}_s$$

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

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

### The pendulum in an accelerating frame

Find the Lagrangian for the non-inertial observer:

$$L = T' - U' = \frac{1}{2}m R^2 \dot{\theta}'^2 - m g_\textrm{eff} R \left(1 - \cos \theta' \right)$$

## Rotating Frames

Assume vector $\mathbf{A}$ is at rest in the rotating frame. As the frame rotates, the vector $\mathbf{A}$ will also rotate through an angle $d \phi$. The change in $\mathbf{A}$ can then be expressed by the cross-product: $$d\mathbf{A} = d \boldsymbol{\phi} \times \mathbf{A}$$

($d \boldsymbol{\phi}$ points out of the page)

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

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

Helpful to consider this an operator: $$\left. \frac{d}{dt} \right|_\textrm{in} = \left. \frac{d }{dt} \right|_\textrm{rot} + \boldsymbol{\omega} \times$$ 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 $\mathbf{r}$): $$\mathbf{v}_\textrm{in} = \mathbf{v}_\textrm{rot} + \boldsymbol{\omega} \times \mathbf{r}$$

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

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

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: $$L = \frac{1}{2}m \mathbf{v}_\textrm{rot}^2 + m \mathbf{v}_\textrm{rot} \cdot \left(\boldsymbol \omega \times \mathbf{r} \right) + \frac{1}{2}m \omega^2 r^2 - \frac{1}{2}m \left( \boldsymbol \omega \cdot \mathbf{r} \right)^2 - U(\mathbf{r})$$ This is clearly not just $\frac{1}{2}m \mathbf{v}_\textrm{rot}^2$

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

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

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

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

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

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

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

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

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

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

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

Finaaaaly, $$\mathbf{F}_\textrm{rot} = \mathbf{F}_\textrm{in} -m \boldsymbol{\omega} \times \left( \boldsymbol{\omega } \times \mathbf{r} \right)_\textrm{rot} - 2 m \left( \boldsymbol{\omega} \times \mathbf{v}_\textrm{rot} \right)_\textrm{rot} - m \left( \dot{\boldsymbol{\omega}} \times \mathbf{r} \right)_\textrm{rot}$$ 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?, $$\mathbf{F}_\textrm{rot} = \mathbf{F}_\textrm{in}\; \underbrace{- m \boldsymbol{\omega} \times \left( \boldsymbol{\omega } \times \mathbf{r} \right)_\textrm{rot} }_\textrm{centrifugal}\; \underbrace{- 2 m \left( \boldsymbol{\omega} \times \mathbf{v}_\textrm{rot} \right)_\textrm{rot}}_\textrm{Coriolis} \; \underbrace{- m \left( \dot{\boldsymbol{\omega}} \times \mathbf{r} \right)_\textrm{rot}}_\textrm{Euler}$$

## Pseudo-Forces

### Centrifugal

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

### Coriolis

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

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

### Euler

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

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

## Earth as rotating reference frame

Earth takes 24 hours to rotate once.

(Actually, a little bit less: sidereal day is 23h56'04'') \begin{aligned} \Omega & = \frac{2\pi}{24 \; \textrm{h} \times 3600 \; \textrm{s/h}} \left(\frac{366.5}{365.5} \right) \\ & = 7.292 \times 10^{-5} \; \textrm{s}^{-1} \end{aligned}

The Centrifugal pseudo-force will be directed outward and proportional to the distance from the rotation axis $\rho$. $$\mathbf{F}_\textrm{cent} = - m \boldsymbol{\omega} \times \left( \boldsymbol{\omega } \times \mathbf{r} \right)_\textrm{rot}$$ can be simplified to: $$F = m \Omega^2 \rho$$ Taking the radius at the equator to be 6378 km: $$6.374 \times 10^{6} \; \textrm{m} \times \left( 7.292 \times 10^{-5} \; \textrm{s}^{-1}\right)^2 = 0.033914 \; \textrm{m}/\textrm{s}^2$$ Which is about 0.3% of the normal $g$.

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

Since these are all vectors, we can see the effects on a plumb-bob hanging in the north hermisphere: $$\mathbf{F}_\textrm{G-eff} = \mathbf{F}_G - \mathbf{F}_\textrm{cf}$$

## Coriolis

Express $\Omega$ in terms of cartiesian vectors: $$\boldsymbol{\Omega} = \Omega \left(\hat{\bf{y}} \cos \lambda + \hat{\bf{z}} \sin \lambda \right)$$

A particle will have velocity components: $$\mathbf{v} = \dot{x} \hat{\bf{x}} + \dot{y} \hat{\bf{y}} + \dot{z} \hat{\bf{z}}$$

Thus we can compute the Coriolis pseudo-force vector: \begin{align} \mathbf{F}_\textrm{Cor} & = -2 m \boldsymbol{\Omega} \times \mathbf{v} \\ & = 2 m \Omega \left[ \left( \dot{y} \sin \lambda - \dot{z} \cos \lambda \right) \hat{\bf{x}} - \dot{x} \sin \lambda \; \hat{\bf{y}} + \dot{x} \cos \lambda \; \hat{\bf{z}}\right] \end{align}

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

or, in terms of the vector: $\hat{\boldsymbol{\theta}}$ $$\mathbf{F}_\textrm{Cor, horiz} = -2 m \Omega \sin \lambda v \hat{\boldsymbol{\theta}}$$

1. The Coriolis force therefore pushes moving objects the right of their velocity vector if $\sin \lambda \gt 0$ (i.e. northern hemisphere)
2. The Coriolis force therefore pushes moving objects the left of their velocity vector if $\sin \lambda \lt 0$ (i.e. southern hemisphere)

### Foucault Pendulum

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

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: $$m \ddot{\mathbf{r}} = \mathbf{T} + m \mathbf{g} + 2 m \dot{\mathbf{r}} \times \boldsymbol{\Omega}$$

As usual, let's restrict ourselves to small oscillations

This let's us say that $$T \approx mg$$ since $T_z = T \cos \beta \approx T$

Next, we need to examine the x and y components

Looking at the figure: $$\frac{T_x}{T} = -\frac{x}{L} \;\; \textrm{and} \;\; \frac{T_y}{T} = -\frac{y}{L}$$ therefore: $$T_x = \frac{-mgx}{L} \;\; \textrm{and} \;\; T_y = \frac{-mgy}{L}$$

$$m \ddot{\mathbf{r}} = \mathbf{T} + m \mathbf{g} + 2 m \dot{\mathbf{r}} \times \boldsymbol{\Omega}$$

Then we have two equations of motion, after some algebra \begin{align} \ddot{x} & = \frac{-gx}{L} + 2 \dot{y} \Omega \cos \theta \\ \ddot{y} & = \frac{-gy}{L} - 2 \dot{x} \Omega \cos \theta \end{align}

$\theta$ is the colatitude of the experiment.

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

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

This implies two indepedent solutions

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

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

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

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

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