## Position, velocity and acceleration

Often times it is desirable to discuss the path, $\mathbf{P}$ of some object as it moves through space. Such a trajectory has an existence independent of the coordinates selected to represent them.

Definitions of geometric velocity and acceleration

Formally we can define a trajectory to be a piecewise smooth embedding of a line into the four dimensional space-time, such that the tangent vector is always time-like. This means that there are only discretely many sharp turns, and that the curve is always moving forward in time in all frames.

Let us define a parameter $\tau$ such that $N\left(\frac{d\mathbf{P}}{d\tau}\right)=1$ everywhere, except for sharp turns along $\mathbf{P}$, where we require $\Delta\tau=0$. We can now define a geometric velocity as $\mathbf{u}=\frac{d\mathbf{P}}{d\tau}$.

A geometric acceleration can now be defined as $\mathbf{\mathbf{a}}=\frac{d\mathbf{u}}{d\tau}$. It should be noted that since the geometric velocity has constant magnitude, the geometric acceleration must be orthogonal to the velocity.

Frame Coordinates

We rarely work with raw geometric objects. Often we need to express things in terms of $\vec{v}$ and $\vec{a}$ relative to some fixed inertial frame of reference. First we observe:

$\frac{dt}{d\tau}=\left(1-\frac{v^2}{c^2}\right)^{-\frac{1}{2}}$

$\frac{d^2t}{d\tau^2}=\left(\frac{dt}{d\tau}\right)^4\frac{\vec{v}\cdot\vec{a}}{c^2}$

This now gives us:

$\mathbf{u}=\frac{dt}{d\tau}\left(1+\frac{\vec{v}}{c}j\right)$

$\mathbf{a}\equiv\frac{d\mathbf{u}}{d\tau}=\frac{d^2t}{d\tau^2}\left(1+\frac{\vec{v}}{c}j\right)+\frac{dt}{d\tau}\left(\frac{\vec{a}}{c}j\right)=\left(\frac{dt}{d\tau}\right)^3\frac{\vec{v}\cdot\vec{a}}{c^2}\mathbf{u}+\frac{dt}{d\tau}\left(\frac{\vec{a}}{c}j\right)$

Going the other direction is more involved. We can find $\frac{dt}{d\tau}$ by examining the scalar component of $\mathbf{u}$. Dividing the axial vector by this scalar now gives $\frac{\vec{v}}{c}$.

The scalar component of $\mathbf{a}$ gives us $\left(\frac{dt}{d\tau}\right)^4\frac{\vec{v}\cdot\vec{a}}{c^2}$. Multiplying this by $\frac{\vec{v}}{c}$ gives us a relativistic kinematic term that must be subtracted to account for velocity. The remaining axial vector is now  $\frac{dt}{d\tau}\left(\frac{\vec{a}}{c}j\right)$

Transformations between coordinates

Given $x'=q(x+V)\overline{q}$, it is often useful to be able to translate from the representation of a geometric object in $x$ to a representation in $x'$.

Since $q$ and $V$ are constant, the transformations for $\mathbf{u}$ and  $\mathbf{a}$ are somewhat trivial.

$\mathbf{u}'=q\mathbf{u}\overline{q}$

$\mathbf{a}'=q\mathbf{a}\overline{q}$

The problem of going from a representation using $\vec{v}$ and $\vec{a}$ to a similar representation in another reference frame then involves three steps. First we convert these to geometric form. Next we convert the geometric form to the new frame. Finally we decompose the new geometric object into the desired representation.

Translations and rotations preserve $\frac{dt}{d\tau}$, so only boosts present significant difficulties.

Let us consider a point along path $\mathbf{P}$ with velocity $\vec{v}$ and acceleration $\vec{a}$ in frame $\mathbf{F}$. What is the value of $\vec{v}'$ and $\vec{a}'$ in frame $\mathbf{F}'$, with the same origin as $\mathbf{F}$, but moving with velocity $\vec{V}$?

Step 1: Find $\mathbf{u}$ and $\mathbf{a}$

Let $\gamma_p\equiv\frac{dt}{d\tau}=\left(1-\frac{\vec{v}\cdot\vec{v}}{c^2}\right)^\frac{-1}{2}$.

$\mathbf{u}=\gamma_p\left(1+\frac{\vec{v}}{c}j\right)$

$\mathbf{a}=\gamma_p^3\frac{\vec{v}\cdot\vec{a}}{c^2}\mathbf{u}+\gamma_p\frac{\vec{a}}{c}j$

Step 2: Convert to the new frame.

Let $\tanh\varphi=\frac{|\vec{V}|}{c}$

$q=\overline{q}=\cosh\frac{\varphi}{2}-\frac{\vec{V}}{c}\sinh\frac{\varphi}{2}j$