## 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.

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

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

## Lemma 1.1: Radial length derivative.

The derivative of a spherical length $$r$$ can be expressed as
\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
2 r \frac{dr}{dt} = 2 \Bx \cdot \frac{d\Bx}{dt},

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

### End proof.

Application of the chain rule to \ref{eqn:radialderivatives:20} is straightforward
\Bx’ = r’ \mathbf{\hat{r}} + r \mathbf{\hat{r}}’,

but we don’t know the form for $$\mathbf{\hat{r}}’$$. We could proceed with a niave expansion of
\frac{d}{dt} \lr{ \frac{\Bx}{r} },

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

We used our lemma in the last step above, and after distribution, find
\Bx’ = r’ \mathbf{\hat{r}} + \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \wedge \Bx’ }.

Comparing to \ref{eqn:radialderivatives:100}, we see that
r \mathbf{\hat{r}}’ = \mathbf{\hat{r}} \lr{ \mathbf{\hat{r}} \wedge \Bx’ }.

We see that the radial unit vector derivative is proportional to the rejection of $$\mathbf{\hat{r}}$$ from $$\Bx’$$
\mathbf{\hat{r}}’ = \inv{r} \mathrm{Rej}_{\mathbf{\hat{r}}}(\Bx’) = \inv{r^3} \Bx \lr{ \Bx \wedge \Bx’ }.

The vector $$\mathbf{\hat{r}}’$$ is perpendicular to $$\mathbf{\hat{r}}$$ for any parameterization of it’s orientation, or in symbols
\mathbf{\hat{r}} \cdot \mathbf{\hat{r}}’ = 0.

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{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}

Observe that $$p_r = m r’$$ is the radial component of the momentum. It is natural to introduce a bivector valued angular momentum operator
L = \Br \wedge \Bp,

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

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{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}

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

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

## Problem:

\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}

## Problem:

Show that \ref{eqn:radialderivatives:200} can be expressed as a triple vector cross product
\mathbf{\hat{r}}’ = \inv{r^3} \lr{ \Bx \cross \Bx’ } \cross \Bx,

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

# References

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

## New video: Velocity and angular momentum with geometric algebra

In this video, we compute velocity in a radial representation $$\mathbf{x} = r \mathbf{\hat{r}}$$.

We use a scalar radial coordinate $$r$$, and leave all the angular dependence implicitly encoded in a radial unit vector $$\mathbf{\hat{r}}$$.

We find the geometric algebra structure of the $$\mathbf{\hat{r}}’$$ in two different ways, to find

$$\mathbf{\hat{r}}’ = \frac{\mathbf{\hat{r}}}{r} \left( \mathbf{\hat{r}} \wedge \mathbf{\hat{x}}’ \right),$$

then derive the conventional triple vector cross product equivalent for reference:

$$\mathbf{\hat{r}}’ = \left( \mathbf{\hat{r}} \times \mathbf{\hat{x}}’ \right) \times \frac{\mathbf{\hat{r}}}{r}.$$

We then compute kinetic energy in this representation, and show how a bivector-valued angular momentum $$L = \mathbf{x} \wedge \mathbf{p}$$, falls naturally from that computation, where we have

$$\frac{m}{2} \mathbf{v}^2 = \frac{1}{2 m} {(m r’)}^2 – \frac{1}{2 m r^2 } L^2.$$

Prerequisites: calculus (derivatives and chain rule), and geometric algebra basics (vector multiplication, commutation relationships for vectors and bivectors in a plane, wedge and cross product equivalencies, …)

Errata: at around 4:12 I used $$\mathbf{r}$$ instead of $$\mathbf{x}$$, then kept doing so every time after that when the value for $$L$$ was stated.

As well as being posted to Google’s censorship-tube, this video can also be found on odysee.