# Eikonal Blog

## 2010.07.20

### Vector bases for 3D kinematics

Filed under: mathematics — Tags: — sandokan65 @ 16:02

## Relevant sets of basis vectors

There are four relevant sets of orthonormal basis vectors (=triads) here:

• ${\cal B}_1 :\equiv \{\underline{e}_x,\underline{e}_y,\underline{e}_x\}$ – orts corresponding to the Cartesian coordinates $\{x,y,z\}$,
• ${\cal B}_3 :\equiv \{\underline{n},\underline{p},\underline{q}\}$, defined by:
• $\underline{n} :\equiv \frac1{r}\underline{r} = (\cos(\theta)\cos(\varphi),\cos(\theta)\sin(\varphi),\sin(\theta)) ,$
• $\underline{p} :\equiv \frac{\partial}{\partial \theta} \underline{n} = (-\sin(\theta)\cos(\varphi),-\sin(\theta)\sin(\phi),\cos(\theta)),$
• $\underline{q} :\equiv \frac1{\cos(\theta)}\frac{\partial}{\partial \varphi} \underline{n} = (-\sin(\varphi),\cos(\phi),0),$

Note that here: $\underline{n}\times\underline{p} = - \underline{q}$, ie. that this basis is left-handed.

• ${\cal B}_3 :\equiv \{\underline{n},\underline{m},\underline{k}\}$ – radius-based basis, defined by:
• $\underline{n} :\equiv \frac1{r}\underline{r},$
• $\underline{m} :\equiv \hbox{Normalized}(\underline{n}')$,
• $\underline{k} :\equiv \underline{n}\times\underline{m}$.
• ${\cal B}_4 :\equiv \{\underline{T},\underline{O},\underline{B}\}$ – tangent-based basis (“frame”), defined by:
• $\underline{T} :\equiv \hbox{Normalized}(\underline{r}') = \frac1{v}\underline{v} = \frac{r'}{v} \underline{n} + \frac{r \Phi'}{v} \underline{m}$,
• $\underline{N} :\equiv \hbox{Normalized}(\underline{T}')$
• $\underline{B} :\equiv \underline{T}\times\underline{N}$.

where $\underline{v} :\equiv \underline{r}' = r' \underline{n} + r \Phi' \underline{m}$ and $v(t) = \sqrt{r'^2+r^2\Phi'^2}$.
Note that $\underline{N} \ne \underline{n}$.

## Triad kinematic formulas

${\cal B}_2$:

• $\underline{n}' = \theta' \underline{p} + \cos(\theta)\varphi' \underline{q}$,
• $\underline{p}' = -\theta' \underline{n} - \sin(\theta)\varphi' \underline{q}$,
• $\underline{q}' = -\cos(\theta) \theta' \underline{n} + \sin(\theta)\varphi' \underline{p}$.

${\cal B}_3$:

• $\underline{n}' = \Phi' \underline{m}$,
• $\underline{m}' = -\Phi' \underline{n} + \Theta' \underline{k}$,
• $\underline{k}' = -\Theta' \underline{m}$.

Note that $\Phi$ and $\Theta$ are not the same as $\varphi$ and $\theta$. They are defined by these equations.

A vector $\underline{A} = A_n\underline{n}+A_m\underline{m}+A_k\underline{k}$ evolves according to:
$\underline{A}' = [A'_n - \Phi' A_m]\underline{n} + [A'_m + \Phi' A_n - \Theta' A_k]\underline{m} + [A'_k + \Theta' A_m]\underline{k}$.

${\cal B}_4$ is guided by the Frenet–Serret formulas:

• $\frac{d}{ds}\underline{T} = \kappa \underline{N}$,
• $\frac{d}{ds}\underline{N} = -\kappa \underline{T} + \tau \underline{B}$,
• $\frac{d}{ds}\underline{B} = -\tau \underline{N}$.

where $s(t) :\equiv \int_0^t dt' v(t')$.

## Relations between triads

${\cal B}_3$ in terms of ${\cal B}_2$:

• $\underline{n} = \underline{n}$,
• $\underline{p} = \frac{\theta'}{\Phi'} \underline{m} + \cos(\theta)\frac{\varphi'}{\Phi'} \underline{k}$,
• $\underline{q} = \cos(\theta) \frac{\varphi'}{\Phi'} \underline{m} - \frac{\theta'}{\Phi'} \underline{k}$.

${\cal B}_2$ in terms of ${\cal B}_3$:

• $\underline{n} = \underline{n}$,
• $\underline{m} = \frac{\theta'}{\Phi'} \underline{p} + \cos(\theta)\frac{\varphi'}{\Phi'} \underline{q}$,
• $\underline{k} = \cos(\theta) \frac{\varphi'}{\Phi'} \underline{p} - \frac{\theta'}{\Phi'} \underline{q}$.

Note: Here

• $\Phi'^2 :\equiv \theta'^2 + \cos(\theta)^2 \varphi'^2$,
• $\Theta' :\equiv \sin(\theta) \varphi' \left(1+\frac{\theta'^2}{\Phi'^2}\right) + \cos(\theta) \frac{\theta''\varphi'-\theta'\varphi''}{\Phi'^2}$.

Note: if we replace $t$ with $\Phi$, then

• $\left(\frac{d\theta}{d\Phi}\right)^2 + \cos(\theta)^2 \left(\frac{d\varphi}{d\Phi}\right)^2=1$,
• $\frac{d\Theta}{d\Phi} = \sin(\theta) \left(\frac{d\varphi}{d\Phi}\right) \left[1+\left(\frac{d\theta}{d\Phi}\right)^2\right] + \cos(\theta) \left[\frac{d^2\theta}{d\Phi^2}\frac{d\varphi}{d\Phi}-\frac{d\theta}{d\Phi}\frac{d^2\varphi}{d\Phi'^2}\right]$.

Related:

### Conservative motion of a 3D particle in potential-like force fields

Filed under: Uncategorized — Tags: , — sandokan65 @ 09:20

Let’s consider the right-hand oriented orthonormal triad ${\cal B} :\equiv \{\underline{n},\underline{m},\underline{k}\}$ ($\underline{n}\cdot\underline{m}= \underline{n}\cdot\underline{k} = \underline{m}\cdot\underline{k}= 0$ and $\underline{n}^2 = \underline{m}^2 = \underline{k}^2 = 1$) assigned to a curve $c$ in 3-dimensional Euclidean space ${\Bbb R}^3$. Let this curve be parametrized by the “time” parameter $t$ associated to the “current” position $\underline{r}(t)$ of a particle moving along that curve. Instead of defining the triad ${\cal T}$ as a tangential vectors (i.e. the tangential, normal and bi-normal vectors) to the $c$ at $\underline{r}$, define it based on the radius vector $\underline{r}$ by $\underline{r} \equiv: \underline{n} r$.

The triad vectors satisfy following evolution equations:

• $\underline{n}' \equiv: \varphi' \underline{m}$,
• $\underline{m}' \equiv: - \varphi' \underline{n} + \theta' \underline{k}$,
• $\underline{k}' = -\theta' \underline{m}$,

where $\phi(t)$ and $\theta(t)$ are two angles defining the position of the particle.

In following we will look at the several choices for the force field $\underline{F}$ that guides the motion of the particle via Newton’s equation $\underline{r}'' = \underline{F}(\underline{r}, \underline{r}')$.

## The simplest case

For $\underline{r}'' = f(r) \underline{r}$ one finds:

• $r'' = f(r) r + \frac{l^2}{r^3}$ (radial equation of motion),
• $r^2\phi' = l (= const|_{t})$ (preservation of angular momentum),
• $\theta' \equiv 0$ (i.e. the trajectory is planar).

## A simple generalization

In force field $\underline{F} = f(r) \underline{r} + g(r) \underline{r}'$ one finds:

• $r'' = f(r) r + g(r) r' + r \varphi'^2$ (radial equation of motion),
• $l(t) :\equiv r^2\phi'$ (definition of non-constant angular momentum),
• $l' = g(r) l$ (equation guiding the evolution of the angular momentum),
• $\theta' \equiv 0$, i.e. the trajectory is (still) planar.

This system can be somehow simplified by use of $\varphi$ instead of $t$, and by introduction of the Binet variable $u(\varphi) :\equiv \frac1{r(t)}$:

• $- \frac{d^2 u}{d\varphi^2} = u + \frac{f[u]}{l^2 u^3}$
• $\frac{d l}{d\varphi} = \frac{g[u]}{u^2}$.

with $f[u] :\equiv f(r)$, $g[u] :\equiv g(r)$, $\frac{d}{dt} = l u^2 \frac{d}{d\varphi}$, $r'(t) = - l \frac{d u}{d\varphi}$ and $r''(t) = - l u^2 \frac{d}{d\varphi}\left(l \frac{d u}{d\varphi}\right)$.

## Further generalization

For $\underline{r}'' = f(r) \underline{r} + g(r) \underline{r}' + h(r) \underline{r}\times\underline{r}'$ we get:

• $r'' = f r + g r' + r \varphi'^2$,
• $l :\equiv r^2 \varphi'$,
• $l' = g(r) l$,
• $\theta' = h(r) r$ – the trajectory is not planar any more.

which, in Binet coordinates has following form

• $- \frac{d^2 u}{d\varphi^2} = u + \frac{f[u]}{l^2 u^3}$
• $\frac{d l}{d\varphi} = \frac{g[u]}{u^2}$.
• $\frac{d \theta}{d\varphi} = \frac{h[u]}{l u^3}$.