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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.

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.