radial representation

Radial vector representation, momentum, and angular momentum.

September 8, 2023 math and physics play , , , , , ,

[Click here for a PDF version of this post], and here for a video version of this post.



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

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},
\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
&= \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’}.
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
&= 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 }.
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
\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,
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 \( L^2 \le 0 \).



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


&= \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’ }.


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
&= \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}}.


[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

September 7, 2023 math and physics play , , , , ,


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.

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