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## Motivation.

In my last couple GA YouTube videos, circular and spherical coordinates were examined.

This post is a text representation of a new video that follows up on those two videos.

We found the form of the unit vector derivatives in both cases.

\begin{equation}\label{eqn:radialderivatives:20}

\Bx = r \mathbf{\hat{r}},

\end{equation}

leaving the angular dependence of \( \mathbf{\hat{r}} \) unspecified. We want to find both \( \Bv = \Bx’ \) and \( \mathbf{\hat{r}}’\).

## Derivatives.

## Lemma 1.1: Radial length derivative.

\begin{equation*}

\frac{dr}{dt} = \mathbf{\hat{r}} \cdot \frac{d\Bx}{dt}.

\end{equation*}

### Start proof:

We write \( r^2 = \Bx \cdot \Bx \), and take derivatives of both sides, to find

\begin{equation}\label{eqn:radialderivatives:60}

2 r \frac{dr}{dt} = 2 \Bx \cdot \frac{d\Bx}{dt},

\end{equation}

or

\begin{equation}\label{eqn:radialderivatives:80}

\frac{dr}{dt} = \frac{\Bx}{r} \cdot \frac{d\Bx}{dt} = \mathbf{\hat{r}} \cdot \frac{d\Bx}{dt}.

\end{equation}

### End proof.

Application of the chain rule to \ref{eqn:radialderivatives:20} is straightforward

\begin{equation}\label{eqn:radialderivatives:100}

\Bx’ = r’ \mathbf{\hat{r}} + r \mathbf{\hat{r}}’,

\end{equation}

but we don’t know the form for \( \mathbf{\hat{r}}’ \). We could proceed with a niave expansion of

\begin{equation}\label{eqn:radialderivatives:120}

\frac{d}{dt} \lr{ \frac{\Bx}{r} },

\end{equation}

but we can be sneaky, and perform a projective and rejective split of \( \Bx’ \) with respect to \( \mathbf{\hat{r}} \). That is

\begin{equation}\label{eqn:radialderivatives:140}

\begin{aligned}

\Bx’

&= \mathbf{\hat{r}} \mathbf{\hat{r}} \Bx’ \\

&= \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \Bx’ } \\

&= \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \cdot \Bx’ + \mathbf{\hat{r}} \wedge \Bx’} \\

&= \mathbf{\hat{r}} \lr{ r’ + \mathbf{\hat{r}} \wedge \Bx’}.

\end{aligned}

\end{equation}

We used our lemma in the last step above, and after distribution, find

\begin{equation}\label{eqn:radialderivatives:160}

\Bx’ = r’ \mathbf{\hat{r}} + \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \wedge \Bx’ }.

\end{equation}

Comparing to \ref{eqn:radialderivatives:100}, we see that

\begin{equation}\label{eqn:radialderivatives:180}

r \mathbf{\hat{r}}’ = \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \wedge \Bx’ }.

\end{equation}

We see that the radial unit vector derivative is proportional to the rejection of \( \mathbf{\hat{r}} \) from \( \Bx’ \)

\begin{equation}\label{eqn:radialderivatives:200}

\mathbf{\hat{r}}’ = \inv{r} \mathrm{Rej}_{\mathbf{\hat{r}}}(\Bx’) = \inv{r^3} \Bx \lr{ \Bx \wedge \Bx’ }.

\end{equation}

The vector \( \mathbf{\hat{r}}’ \) is perpendicular to \( \mathbf{\hat{r}} \) for any parameterization of it’s orientation, or in symbols

\begin{equation}\label{eqn:radialderivatives:220}

\mathbf{\hat{r}} \cdot \mathbf{\hat{r}}’ = 0.

\end{equation}

We saw this for the circular and spherical parameterizations, and see now that this also holds more generally.

## Angular momentum.

Let’s now write out the momentum \( \Bp = m \Bv \) for a point particle with mass \( m \), and determine the kinetic energy \( m \Bv^2/2 = \Bp^2/2m \) for that particle.

The momentum is

\begin{equation}\label{eqn:radialderivatives:320}

\begin{aligned}

\Bp

&= m r’ \mathbf{\hat{r}} + m \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \wedge \Bv } \\

&= m r’ \mathbf{\hat{r}} + \inv{r} \mathbf{\hat{r}} \lr{ \Br \wedge \Bp }.

\end{aligned}

\end{equation}

Observe that \( p_r = m r’ \) is the radial component of the momentum. It is natural to introduce a bivector valued angular momentum operator

\begin{equation}\label{eqn:radialderivatives:340}

L = \Br \wedge \Bp,

\end{equation}

splitting the momentum into a component that is strictly radial and a component that lies purely on the surface of a spherical surface in momentum space. That is

\begin{equation}\label{eqn:radialderivatives:360}

\Bp = p_r \mathbf{\hat{r}} + \inv{r} \mathbf{\hat{r}} L.

\end{equation}

Making use of the fact that \( \mathbf{\hat{r}} \) and \( \mathrm{Rej}_{\mathbf{\hat{r}}}(\Bx’) \) are perpendicular (so there are no cross terms when we square the momentum), the

kinetic energy is

\begin{equation}\label{eqn:radialderivatives:380}

\begin{aligned}

\inv{2m} \Bp^2

&= \inv{2m} \lr{ p_r \mathbf{\hat{r}} + \inv{r} \mathbf{\hat{r}} L }^2 \\

&= \inv{2m} p_r^2 + \inv{2 m r^2 } \mathbf{\hat{r}} L \mathbf{\hat{r}} L \\

&= \inv{2m} p_r^2 – \inv{2 m r^2 } \mathbf{\hat{r}} L^2 \mathbf{\hat{r}} \\

&= \inv{2m} p_r^2 – \inv{2 m r^2 } L^2 \mathbf{\hat{r}}^2,

\end{aligned}

\end{equation}

where we’ve used the anticommutative nature of \( \mathbf{\hat{r}} \) and \( L \) (i.e.: a sign swap is needed to swap them), and used the fact that \( L^2 \) is a scalar, allowing us to commute \( \mathbf{\hat{r}} \) with \( L^2 \). This leaves us with

\begin{equation}\label{eqn:radialderivatives:400}

E = \inv{2m} \Bp^2 = \inv{2m} p_r^2 – \inv{2 m r^2 } L^2.

\end{equation}

Observe that both the radial momentum term and the angular momentum term are both strictly postive, since \( L \) is a bivector and \( L^2 \le 0 \).

## Problems.

## Problem:

Find \ref{eqn:radialderivatives:200} without being sneaky.

## Answer

\begin{equation}\label{eqn:radialderivatives:280}

\begin{aligned}

\mathbf{\hat{r}}’

&= \frac{d}{dt} \lr{ \frac{\Bx}{r} } \\

&= \inv{r} \Bx’ – \inv{r^2} \Bx r’ \\

&= \inv{r} \Bx’ – \inv{r} \mathbf{\hat{r}} r’ \\

&= \inv{r} \lr{ \Bx’ – \mathbf{\hat{r}} r’ } \\

&= \inv{r} \lr{ \mathbf{\hat{r}} \mathbf{\hat{r}} \Bx’ – \mathbf{\hat{r}} r’ } \\

&= \inv{r} \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \Bx’ – r’ } \\

&= \inv{r} \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \Bx’ – \mathbf{\hat{r}} \cdot \Bx’ } \\

&= \inv{r} \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \wedge \Bx’ }.

\end{aligned}

\end{equation}

## Problem:

Show that \ref{eqn:radialderivatives:200} can be expressed as a triple vector cross product

\begin{equation}\label{eqn:radialderivatives:230}

\mathbf{\hat{r}}’ = \inv{r^3} \lr{ \Bx \cross \Bx’ } \cross \Bx,

\end{equation}

## Answer

While this may be familiar from elementary calculus, such as in [1], we can show follows easily from our GA result

\begin{equation}\label{eqn:radialderivatives:300}

\begin{aligned}

\mathbf{\hat{r}}’

&= \inv{r} \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \wedge \Bx’ } \\

&= \inv{r} \gpgradeone{ \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \wedge \Bx’ } } \\

&= \inv{r} \gpgradeone{ \mathbf{\hat{r}} I \lr{ \mathbf{\hat{r}} \cross \Bx’ } } \\

&= \inv{r} \gpgradeone{ I \lr{ \mathbf{\hat{r}} \cdot \lr{ \mathbf{\hat{r}} \cross \Bx’ } + \mathbf{\hat{r}} \wedge \lr{ \mathbf{\hat{r}} \cross \Bx’ } } } \\

&= \inv{r} \gpgradeone{ I^2 \mathbf{\hat{r}} \cross \lr{ \mathbf{\hat{r}} \cross \Bx’ } } \\

&= \inv{r} \lr{ \mathbf{\hat{r}} \cross \Bx’ } \cross \mathbf{\hat{r}}.

\end{aligned}

\end{equation}

# References

[1] S.L. Salas and E. Hille. *Calculus: one and several variables*. Wiley New York, 1990.