﻿ Further Central Force Details & Derivations | Primer on Geometric Algebra | David Hestenes

## Further Central Force Details & Derivations

Let's summarize and fill in some derivations from this section. We start with
the equation of motion (Newton's 2nd Law for an attractive inverse square force):

$\Large{m{\bf{\dot v}}\,\,\, = \,\,\, - \,\,\frac{k}{{{r^2}}}\,\,{\bf{\hat r}}}$

We can use this to model the usual cases of a planet attracted by gravity to a massive sun, a satellite bound to Earth's gravity, or perhaps a small iron puck attracted to a strong permanent magnet fastened to an air hockey table. These models assume the center of force (our origin in the equation) is fixed (the one body problem), although the two body problem can be transformed into this equation using center of mass considerations.

Next, we define
the angular momentum of the particle and show it is a constant of motion:

$\Large{\begin{array}{l}{\bf{L}}\,\,\, \equiv \,\,\,m{\bf{r}} \wedge {\bf{v}}\,\,\, = \,\,\,m{\bf{r}} \wedge \left( {\dot r{\bf{\hat r}} + r{\bf{\dot {\hat r}}}} \right)\,\,\, = \,\,\,m{r^2}\,{\bf{\hat r\dot {\hat r}}}\\\,~~~~~~~~~~\left( {{\rm{because}}\,\,\,{{{\bf{\hat r}}}^2} = \,\,1\,\,\,\, \Rightarrow \,\,\,\,{\bf{\hat r}} \cdot {\bf{\dot {\hat r}}} = \,\,0} \right)\\\\{\bf{\dot L}}\,\,\, = \,\,\,\frac{d}{{dt}}\left( {m{\bf{r}} \wedge {\bf{v}}} \right)\,\,\, = \,\,\,m{\bf{v}} \wedge {\bf{v}}\,\, + \,\,m{\bf{r}} \wedge {\bf{\dot v}}\\\,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \,\,\,0\,\, + \,\,m{\bf{r}} \wedge \,\,\left( { - \,\,\frac{k}{{m{r^2}}}\,\,{\bf{\hat r}}} \right)\,\,\, = \,\,\,0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\bf{L}}\,\,\, = \,\,\,{\rm{constant}}\end{array}}$

Now that we see L is one constant of motion, we can find another using the equation of motion and the expression above for L:

$\Large{\frac{d}{{dt}}\left( {{\bf{Lv}} - k{\bf{\hat r}}} \right)\,\,\, = \,\,\,{\bf{L\dot v}} - k{\bf{\dot {\hat r}}}\,\,\, = \,\,\,{\bf{L}}\left( { - \,\,\frac{k}{{m{r^2}}}\,\,{\bf{\hat r}}} \right)\,\, - \,\,k\left( {\frac{{{\bf{\hat rL}}}}{{m{r^2}}}} \right)\,\,\, = \,\,\,0}$

because the bivector L anticommutes with any vector in its plane. We'll use that constant to
define the eccentricity and see that it is another constant of motion. It helps to have centuries of mathematical efforts by others to build upon, so that now we can use GA and hindsight to find elegant expressions and an efficient line of thought to produce results. In addition to artistic enjoyment, clarity and elegance often lead to new insights.

$\Large{{\bf{Lv}} - k{\bf{\hat r}}\,\,\, = \,\,\,{\rm{constant}}\,\,\, \equiv \,\,\,k{\boldsymbol{\varepsilon}} }$

We have two constants of motion. You probably know the
total energy is another, although it is not independent of these two. We can derive the standard expression by dotting the equation of motion with velocity v:

$\Large{\begin{array}{l}0\,\,\, = \,\,\,{\bf{v}} \cdot \left( {m{\bf{\dot v}}\,\, + \,\,\frac{k}{{{r^2}}}\,\,{\bf{\hat r}}} \right)\,\,\, = \,\,\,\frac{d}{{dt}}\left( {\frac{1}{2}m{\bf{vv}}} \right)\,\, + \,\,\frac{k}{{{r^3}}}\frac{d}{{dt}}\left( {\frac{1}{2}{\bf{rr}}} \right)\,\,\, = \,\,\,\frac{d}{{dt}}\left( {\frac{1}{2}m{v^2}} \right)\,\, + \,\,\frac{k}{{{r^3}}}\frac{d}{{dt}}\left( {\frac{1}{2}{r^2}} \right)\\\\~~~~ = \,\,\,\frac{d}{{dt}}\left( {\frac{1}{2}m{v^2}} \right)\,\, + \,\,\frac{k}{{{r^2}}}\frac{{dr}}{{dt}}\,\,\, = \,\,\,\frac{d}{{dt}}\left( {\frac{1}{2}m{v^2}\,\, - \,\,\frac{k}{r}} \right)\end{array}}$

Therefore, the total energy (
kinetic plus potential) is conserved:

$\Large{E\,\,\, \equiv \,\,\,\frac{1}{2}m{v^2}\,\, - \,\,\frac{k}{r}\,\,\, = \,\,\,{\rm{constant}}\,\,\, = \,\,\,K\,\, + \,\,V(r)}$

It isn't an independent constant, however, since we can write
energy in terms of eccentricity and angular momentum:

$\Large{\begin{array}{l}{k^2}{\varepsilon ^2}\,\,\, = \,\,\,\left( {{\bf{Lv}} - k{\bf{\hat r}}} \right)\left( {{\bf{Lv}} - k{\bf{\hat r}}} \right)\,\,\, = \,\,\,{k^2}\,\, - \,\,{{\bf{L}}^2}{v^2}\,\, - \,\,k\left( {{\bf{\hat rLv}}\,\, + \,\,{\bf{Lv\hat r}}} \right)\\\\ ~~~~~~~~~ = \,\,\,{k^2}\,\, + \,\,{L^2}{v^2}\,\, + \,\,k\left( {{\bf{L\hat rv}}\,\, - \,\,{\bf{Lv\hat r}}} \right)\,\,\, = \,\,\,{k^2}\,\, + \,\,{L^2}{v^2}\,\, + \,\,\frac{{2k{\bf{L}}}}{{mr}}\left( {m{\bf{r}} \wedge {\bf{v}}} \right)\\\\ ~~~~~~~~~ = \,\,\,{k^2}\,\, + \,\,{L^2}{v^2}\,\, - \,\,\frac{{2k{L^2}}}{{mr}}\,\,\, = \,\,\,{k^2}\,\, + \,\,\frac{{2{L^2}}}{m}\left( {\frac{1}{2}m{v^2}\,\, - \,\,\frac{k}{r}} \right)\,\,\, = \,\,\,{k^2}\,\, + \,\,\frac{{2{L^2}}}{m}\,E\\\\{\rm{Therefore}}:\,\,\,\,\,\,\,\,\,\,E\,\,\, = \,\,\,\frac{{m{k^2}}}{{2{L^2}}}\,\left( {{\varepsilon ^2}\, - \,\,1} \right)\end{array}}$

Now we'll find the
orbit, the shape of the space curve followed by the particle. That is, we'll get an expression for r(θ), where θ will be the angle between our current position vector and the constant eccentricity (a convenient reference direction). That's pretty useful, because then we know the shape of the orbit and can find the velocity corresponding to any position, and (see NFCM) can even derive Kepler's laws of planetary motion. But, it isn't a trajectory, which would be an expression for r(t). For that, we would need to find the angle as a function of time. As shown in NFCM (section 4-4), Kepler found an expression for time as a function of angle, but it's a transcendental equation, and hence not reversible by algebraic means. To get the orbit, dot the eccentricity definition with the direction of the particle:

$\Large{\begin{array}{l}k{\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}\,\,\, = \,\,\,\left( {{\bf{Lv}} - k{\bf{\hat r}}} \right) \cdot {\bf{\hat r}}\,\,\, = \,\,\,{\left\langle {{\bf{Lv\hat r}}} \right\rangle _0}\, - \,\,k\,\,\, = \,\,\,{\bf{Lv}} \wedge {\bf{\hat r}}\,\, - \,\,k\,\,\, = \,\,\,\frac{{ - \,{{\bf{L}}^2}}}{{mr}}\,\, - \,\,k\\\\\\~~~~~~~r\,\,\, = \,\,\,\frac{{{L^2}}}{{mk}}\left( {\frac{1}{{1\, + \,{\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}}}} \right)\,\,\, = \,\,\, \pm \,\,\frac{{{L^2}}}{{m\left| k \right|}}\left( {\frac{1}{{1\, + \,{\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}}}} \right)\,\,\, \equiv \,\,\, \pm \,\,\frac{\ell }{{1\, + \,{\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}}}\end{array}}$

Remember
k would be negative for a repulsive force. The numerator in the final expression, the semi-latus rectum, is always positive since by convention it is defined to be the distance from the center of force to the particle when the particle direction is perpendicular to the eccentricity vector. For an attractive force, we can write the standard polar formula for conic sections:

$\Large{r(\theta )\,\,\, = \,\,\,\frac{\ell }{{1\, + \,\varepsilon \cos (\theta )}}}$

Since we know the angle just grows as the orbit is traversed and
L tells us which way the orbit is being traced, we can get an expression for the vector position as a function of angle:

$\Large{{\bf{r}}\left( {\theta = \hat {\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}} \right)\,\,\, = \,\,\,\frac{{{L^2}/\,\,\left( {mk} \right)}}{{1 + \varepsilon \cos (\theta )}}\hat {\boldsymbol{\varepsilon}} {e^{{\bf{\hat L}}\theta }}}$

The
hodograph formula also follows from the eccentricity definition:

$\Large{\begin{array}{l}{\bf{Lv}}\,\,\, = \,\,\,k{\bf{\hat r}}\,\, + \,\,k{\boldsymbol{\varepsilon}} \\\\{\bf{v}}\,\,\, = \,\,\,k{{\bf{L}}^{ - 1}}\left( {{\bf{\hat r}}\,\, + \,\,{\boldsymbol{\varepsilon}} } \right)\,\,\, = \,\,\,\frac{k}{{{L^2}}}\,\,\left( {{\bf{\hat r}}\,\, + \,\,{\boldsymbol{\varepsilon}} } \right)\,\,{\bf{L}}\,\,\, \equiv \,\,\,k{{\bf{L}}^{ - 1}}{\bf{\hat r}}\,\, + \,\,{\bf{u}}\\\\{\left( {{\bf{v}}\, - \,{\bf{u}}} \right)^2}\,\,\, = \,\,\,{k^2}{{\bf{L}}^{ - 1}}{\bf{\hat r}}{{\bf{L}}^{ - 1}}{\bf{\hat r}}\,\,\, = \,\,\,\frac{{{k^2}}}{{{L^2}}}\end{array}}$

This gives us velocity as a function of direction and shows that all possible velocities can be visualized using a circle of radius |
k|/L.

(Click here for an example of elliptical motion and an interactive orbit/hodograph diagram.)