## 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.

## Lorentz transformations in Space Time Algebra (STA)

[If mathjax doesn’t display properly for you, click here for a PDF of this post]

## Motivation.

One of the remarkable features of geometric algebra are the complex exponential sandwiches that can be used to encode rotations in any dimension, or rotation like operations like Lorentz transformations in Minkowski spaces. In this post, we show some examples that unpack the geometric algebra expressions for Lorentz transformations operations of this sort. In particular, we will look at the exponential sandwich operations for spatial rotations and Lorentz boosts in the Dirac algebra, known as Space Time Algebra (STA) in geometric algebra circles, and demonstrate that these sandwiches do have the desired effects.

## Theorem 1.1: Lorentz transformation.

The transformation
\label{eqn:lorentzTransform:580}
x \rightarrow e^{B} x e^{-B} = x’,

where $$B = a \wedge b$$, is an STA 2-blade for any two linearly independent four-vectors $$a, b$$, is a norm preserving, that is
\label{eqn:lorentzTransform:600}
x^2 = {x’}^2.

### Start proof:

The proof is disturbingly trivial in this geometric algebra form
\label{eqn:lorentzTransform:40}
\begin{aligned}
{x’}^2
&=
e^{B} x e^{-B} e^{B} x e^{-B} \\
&=
e^{B} x x e^{-B} \\
&=
x^2 e^{B} e^{-B} \\
&=
x^2.
\end{aligned}

### End proof.

In particular, observe that we did not need to construct the usual infinitesimal representations of rotation and boost transformation matrices or tensors in order to demonstrate that we have spacetime invariance for the transformations. The rough idea of such a transformation is that the exponential commutes with components of the four-vector that lie off the spacetime plane specified by the bivector $$B$$, and anticommutes with components of the four-vector that lie in the plane. The end result is that the sandwich operation simplifies to
\label{eqn:lorentzTransform:60}
x’ = x_\parallel e^{-B} + x_\perp,

where $$x = x_\perp + x_\parallel$$ and $$x_\perp \cdot B = 0$$, and $$x_\parallel \wedge B = 0$$. In particular, using $$x = x B B^{-1} = \lr{ x \cdot B + x \wedge B } B^{-1}$$, we find that
\label{eqn:lorentzTransform:80}
\begin{aligned}
x_\parallel &= \lr{ x \cdot B } B^{-1} \\
x_\perp &= \lr{ x \wedge B } B^{-1}.
\end{aligned}

When $$B$$ is a spacetime plane $$B = b \wedge \gamma_0$$, then this exponential has a hyperbolic nature, and we end up with a Lorentz boost. When $$B$$ is a spatial bivector, we end up with a single complex exponential, encoding our plane old 3D rotation. More general $$B$$’s that encode composite boosts and rotations are also possible, but $$B$$ must be invertible (it should have no lightlike factors.) The rough geometry of these projections is illustrated in fig 1, where the spacetime plane is represented by $$B$$.

fig 1. Projection and rejection geometry.

What is not so obvious is how to pick $$B$$’s that correspond to specific rotation axes or boost directions. Let’s consider each of those cases in turn.

## Theorem 1.2: Boost.

The boost along a direction vector $$\vcap$$ and rapidity $$\alpha$$ is given by
\label{eqn:lorentzTransform:620}
x’ = e^{-\vcap \alpha/2} x e^{\vcap \alpha/2},

where $$\vcap = \gamma_{k0} \cos\theta^k$$ is an STA bivector representing a spatial direction with direction cosines $$\cos\theta^k$$.

### Start proof:

We want to demonstrate that this is equivalent to the usual boost formulation. We can start with decomposition of the four-vector $$x$$ into components that lie in and off of the spacetime plane $$\vcap$$.
\label{eqn:lorentzTransform:100}
\begin{aligned}
x
&= \lr{ x^0 + \Bx } \gamma_0 \\
&= \lr{ x^0 + \Bx \vcap^2 } \gamma_0 \\
&= \lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap + \lr{ \Bx \wedge \vcap} \vcap } \gamma_0,
\end{aligned}

where $$\Bx = x \wedge \gamma_0$$. The first two components lie in the boost plane, whereas the last is the spatial component of the vector that lies perpendicular to the boost plane. Observe that $$\vcap$$ anticommutes with the dot product term and commutes with he wedge product term, so we have
\label{eqn:lorentzTransform:120}
\begin{aligned}
x’
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap } \vcap } \gamma_0
e^{\vcap \alpha/2 }
e^{\vcap \alpha/2 }
+
\lr{ \Bx \wedge \vcap } \vcap \gamma_0
e^{-\vcap \alpha/2 }
e^{\vcap \alpha/2 } \\
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap } \vcap } \gamma_0
e^{\vcap \alpha }
+
\lr{ \Bx \wedge \vcap } \vcap \gamma_0.
\end{aligned}

Noting that $$\vcap^2 = 1$$, we may expand the exponential in hyperbolic functions, and find that the boosted portion of the vector expands as
\label{eqn:lorentzTransform:260}
\begin{aligned}
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0 e^{\vcap \alpha}
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0 \lr{ \cosh\alpha + \vcap \sinh \alpha} \\
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \lr{ \cosh\alpha – \vcap \sinh \alpha} \gamma_0 \\
&=
\lr{ x^0 \cosh\alpha – \lr{ \Bx \cdot \vcap} \sinh \alpha} \gamma_0
+
\lr{ -x^0 \sinh \alpha + \lr{ \Bx \cdot \vcap} \cosh \alpha } \vcap \gamma_0.
\end{aligned}

We are left with
\label{eqn:lorentzTransform:320}
\begin{aligned}
x’
&=
\lr{ x^0 \cosh\alpha – \lr{ \Bx \cdot \vcap} \sinh \alpha} \gamma_0
+
\lr{ \lr{ \Bx \cdot \vcap} \cosh \alpha -x^0 \sinh \alpha } \vcap \gamma_0
+
\lr{ \Bx \wedge \vcap} \vcap \gamma_0 \\
&=
\begin{bmatrix}
\gamma_0 & \vcap \gamma_0
\end{bmatrix}
\begin{bmatrix}
\cosh\alpha & – \sinh\alpha \\
-\sinh\alpha & \cosh\alpha
\end{bmatrix}
\begin{bmatrix}
x^0 \\
\Bx \cdot \vcap
\end{bmatrix}
+
\lr{ \Bx \wedge \vcap} \vcap \gamma_0,
\end{aligned}

which has the desired Lorentz boost structure. Of course, this is usually seen with $$\vcap = \gamma_{10}$$ so that the components in the coordinate column vector are $$(ct, x)$$.

## Theorem 1.3: Spatial rotation.

Given two linearly independent spatial bivectors $$\Ba = a^k \gamma_{k0}, \Bb = b^k \gamma_{k0}$$, a rotation of $$\theta$$ radians in the plane of $$\Ba, \Bb$$ from $$\Ba$$ towards $$\Bb$$, is given by
\label{eqn:lorentzTransform:640}
x’ = e^{-i\theta} x e^{i\theta},

where $$i = (\Ba \wedge \Bb)/\Abs{\Ba \wedge \Bb}$$, is a unit (spatial) bivector.

### Start proof:

Without loss of generality, we may pick $$i = \acap \bcap$$, where $$\acap^2 = \bcap^2 = 1$$, and $$\acap \cdot \bcap = 0$$. With such an orthonormal basis for the plane, we can decompose our four vector into portions that lie in and off the plane
\label{eqn:lorentzTransform:400}
\begin{aligned}
x
&= \lr{ x^0 + \Bx } \gamma_0 \\
&= \lr{ x^0 + \Bx i i^{-1} } \gamma_0 \\
&= \lr{ x^0 + \lr{ \Bx \cdot i } i^{-1} + \lr{ \Bx \wedge i } i^{-1} } \gamma_0.
\end{aligned}

The projective term lies in the plane of rotation, whereas the timelike and spatial rejection term are perpendicular. That is
\label{eqn:lorentzTransform:420}
\begin{aligned}
x_\parallel &= \lr{ \Bx \cdot i } i^{-1} \gamma_0 \\
x_\perp &= \lr{ x^0 + \lr{ \Bx \wedge i } i^{-1} } \gamma_0,
\end{aligned}

where $$x_\parallel \wedge i = 0$$, and $$x_\perp \cdot i = 0$$. The plane pseudoscalar $$i$$ anticommutes with $$x_\parallel$$, and commutes with $$x_\perp$$, so
\label{eqn:lorentzTransform:440}
\begin{aligned}
x’
&= e^{-i\theta/2} \lr{ x_\parallel + x_\perp } e^{i\theta/2} \\
&= x_\parallel e^{i\theta} + x_\perp.
\end{aligned}

However
\label{eqn:lorentzTransform:460}
\begin{aligned}
\lr{ \Bx \cdot i } i^{-1}
&=
\lr{ \Bx \cdot \lr{ \acap \wedge \bcap } } \bcap \acap \\
&=
\lr{\Bx \cdot \acap} \bcap \bcap \acap
-\lr{\Bx \cdot \bcap} \acap \bcap \acap \\
&=
\lr{\Bx \cdot \acap} \acap
+\lr{\Bx \cdot \bcap} \bcap,
\end{aligned}

so
\label{eqn:lorentzTransform:480}
\begin{aligned}
x_\parallel e^{i\theta}
&=
\lr{
\lr{\Bx \cdot \acap} \acap
+
\lr{\Bx \cdot \bcap} \bcap
}
\gamma_0
\lr{
\cos\theta + \acap \bcap \sin\theta
} \\
&=
\acap \lr{
\lr{\Bx \cdot \acap} \cos\theta

\lr{\Bx \cdot \bcap} \sin\theta
}
\gamma_0
+
\bcap \lr{
\lr{\Bx \cdot \acap} \sin\theta
+
\lr{\Bx \cdot \bcap} \cos\theta
}
\gamma_0,
\end{aligned}

so
\label{eqn:lorentzTransform:500}
x’
=
\begin{bmatrix}
\acap & \bcap
\end{bmatrix}
\begin{bmatrix}
\cos\theta & – \sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\Bx \cdot \acap \\
\Bx \cdot \bcap \\
\end{bmatrix}
\gamma_0
+
\lr{ x \wedge i} i^{-1} \gamma_0.

Observe that this rejection term can be explicitly expanded to
\label{eqn:lorentzTransform:520}
\lr{ \Bx \wedge i} i^{-1} \gamma_0 =
x –
\lr{ \Bx \cdot \acap } \acap \gamma_0

\lr{ \Bx \cdot \acap } \acap \gamma_0.

This is the timelike component of the vector, plus the spatial component that is normal to the plane. This exponential sandwich transformation rotates only the portion of the vector that lies in the plane, and leaves the rest (timelike and normal) untouched.

## Problem: Verify components relative to boost direction.

In the proof of thm. 1.2, the vector $$x$$ was expanded in terms of the spacetime split. An alternate approach, is to expand as
\label{eqn:lorentzTransform:340}
\begin{aligned}
x
&= x \vcap^2 \\
&= \lr{ x \cdot \vcap + x \wedge \vcap } \vcap \\
&= \lr{ x \cdot \vcap } \vcap + \lr{ x \wedge \vcap } \vcap.
\end{aligned}

Show that
\label{eqn:lorentzTransform:360}
\lr{ x \cdot \vcap } \vcap
=
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0,

and
\label{eqn:lorentzTransform:380}
\lr{ x \wedge \vcap } \vcap
=
\lr{ \Bx \wedge \vcap} \vcap \gamma_0.

Let $$x = x^\mu \gamma_\mu$$, so that
\label{eqn:lorentzTransform:160}
\begin{aligned}
x \cdot \vcap
&=
\gpgradeone{ x^\mu \gamma_\mu \cos\theta^b \gamma_{b 0} } \\
&=
x^\mu \cos\theta^b \gpgradeone{ \gamma_\mu \gamma_{b 0} }
.
\end{aligned}

The $$\mu = 0$$ component of this grade selection is
\label{eqn:lorentzTransform:180}
=
-\gamma_b,

and for $$\mu = a \ne 0$$, we have
\label{eqn:lorentzTransform:200}
=
-\delta_{a b} \gamma_0,

so we have
\label{eqn:lorentzTransform:220}
\begin{aligned}
x \cdot \vcap
&=
x^0 \cos\theta^b (-\gamma_b)
+
x^a \cos\theta^b (-\delta_{ab} \gamma_0 ) \\
&=
-x^0 \vcap \gamma_0

x^b \cos\theta^b \gamma_0 \\
&=
– \lr{ x^0 \vcap + \Bx \cdot \vcap } \gamma_0,
\end{aligned}

where $$\Bx = x \wedge \gamma_0$$ is the spatial portion of the four vector $$x$$ relative to the stationary observer frame. Since $$\vcap$$ anticommutes with $$\gamma_0$$, the component of $$x$$ in the spacetime plane $$\vcap$$ is
\label{eqn:lorentzTransform:240}
\lr{ x \cdot \vcap } \vcap =
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0,

as expected.

For the rejection term, we have
\label{eqn:lorentzTransform:280}
x \wedge \vcap
=
x^\mu \cos\theta^s \gpgradethree{ \gamma_\mu \gamma_{s 0} }.

The $$\mu = 0$$ term clearly contributes nothing, leaving us with:
\label{eqn:lorentzTransform:300}
\begin{aligned}
\lr{ x \wedge \vcap } \vcap
&=
\lr{ x \wedge \vcap } \cdot \vcap \\
&=
x^r \cos\theta^s \cos\theta^t \lr{ \lr{ \gamma_r \wedge \gamma_{s}} \gamma_0 } \cdot \lr{ \gamma_{t0} } \\
&=
\lr{ \gamma_r \wedge \gamma_{s} } \gamma_0 \gamma_{t0}
} \\
&=
-x^r \cos\theta^s \cos\theta^t \lr{ \gamma_r \wedge \gamma_{s}} \cdot \gamma_t \\
&=
-x^r \cos\theta^s \cos\theta^t \lr{ -\gamma_r \delta_{st} + \gamma_s \delta_{rt} } \\
&=
x^r \cos\theta^t \cos\theta^t \gamma_r

x^t \cos\theta^s \cos\theta^t \gamma_s \\
&=
\Bx \gamma_0
– (\Bx \cdot \vcap) \vcap \gamma_0 \\
&=
\lr{ \Bx \wedge \vcap} \vcap \gamma_0,
\end{aligned}

as expected. Is there a clever way to demonstrate this without resorting to coordinates?

## Problem: Rotation transformation components.

Given a unit spatial bivector $$i = \acap \bcap$$, where $$\acap \cdot \bcap = 0$$ and $$i^2 = -1$$, show that
\label{eqn:lorentzTransform:540}
\lr{ x \cdot i } i^{-1}
=
\lr{ \Bx \cdot i } i^{-1} \gamma_0
=
\lr{\Bx \cdot \acap } \acap \gamma_0
+
\lr{\Bx \cdot \bcap } \bcap \gamma_0,

and
\label{eqn:lorentzTransform:560}
\lr{ x \wedge i } i^{-1}
=
\lr{ \Bx \wedge i } i^{-1} \gamma_0
=
x –
\lr{\Bx \cdot \acap } \acap \gamma_0

\lr{\Bx \cdot \bcap } \bcap \gamma_0.

Also show that $$i$$ anticommutes with $$\lr{ x \cdot i } i^{-1}$$ and commutes with $$\lr{ x \wedge i } i^{-1}$$.

This problem is left for the reader, as I don’t feel like typing out my solution.

The first part of this problem can be done in the tedious coordinate approach used above, but hopefully there is a better way.

For the last (commutation) part of the problem, here is a hint. Let $$x \wedge i = n i$$, where $$n \cdot i = 0$$. The result then follows easily.

## Geometric algebra notes collection split into two volumes

I’ve now split my (way too big) Exploring physics with Geometric Algebra into two volumes:

Each of these is now a much more manageable size, which should facilitate removing the redundancies in these notes, and making them more properly book like.

Also note I’ve also previously moved “Exploring Geometric Algebra” content related to:

• Lagrangian’s
• Hamiltonian’s
• Noether’s theorem

into my classical mechanics collectionÂ (449 pages).

## Motivation

In [2] the Schwartz inequality

\label{eqn:qmSchwartz:20}
\boxed{
\braket{a}{a}
\braket{b}{b}
\ge \Abs{\braket{a}{b}}^2,
}

is used in the derivation of the uncertainty relation. The proof of the Schwartz inequality uses a sneaky substitution that doesn’t seem obvious, and is even less obvious since there is a typo in the value to be substituted. Let’s understand where that sneakiness is coming from.

## Without being sneaky

My ancient first year linear algebra text [1] contains a non-sneaky proof, but it only works for real vector spaces. Recast in bra-ket notation, this method examines the bounds of the norms of sums and differences of unit states (i.e. $$\braket{a}{a} = \braket{b}{b} = 1$$.)

\label{eqn:qmSchwartz:40}
\braket{a – b}{a – b}
= \braket{a}{a} + \braket{b}{b} – \braket{a}{b} – \braket{b}{a}
= 2 – 2 \textrm{Re} \braket{a}{b}
\ge 0,

so
\label{eqn:qmSchwartz:60}
1 \ge \textrm{Re} \braket{a}{b}.

Similarily

\label{eqn:qmSchwartz:80}
\braket{a + b}{a + b}
= \braket{a}{a} + \braket{b}{b} + \braket{a}{b} + \braket{b}{a}
= 2 + 2 \textrm{Re} \braket{a}{b}
\ge 0,

so
\label{eqn:qmSchwartz:100}
\textrm{Re} \braket{a}{b} \ge -1.

This means that for normalized state vectors

\label{eqn:qmSchwartz:120}
-1 \le \textrm{Re} \braket{a}{b} \le 1,

or
\label{eqn:qmSchwartz:140}
\Abs{\textrm{Re} \braket{a}{b}} \le 1.

Writing out the unit vectors explicitly, that last inequality is

\label{eqn:qmSchwartz:180}
\Abs{ \textrm{Re} \braket{ \frac{a}{\sqrt{\braket{a}{a}}} }{ \frac{b}{\sqrt{\braket{b}{b}}} } } \le 1,

squaring and rearranging gives

\label{eqn:qmSchwartz:200}
\Abs{\textrm{Re} \braket{a}{b}}^2 \le
\braket{a}{a}
\braket{b}{b}.

This is similar to, but not identical to the Schwartz inequality. Since $$\Abs{\textrm{Re} \braket{a}{b}} \le \Abs{\braket{a}{b}}$$ the Schwartz inequality cannot be demonstrated with this argument. This first year algebra method works nicely for demonstrating the inequality for real vector spaces, so a different argument is required for a complex vector space (i.e. quantum mechanics state space.)

## Arguing with projected and rejected components

Notice that the equality condition in the inequality holds when the vectors are colinear, and the largest inequality holds when the vectors are normal to each other. Given those geometrical observations, it seems reasonable to examine the norms of projected or rejected components of a vector. To do so in bra-ket notation, the correct form of a projection operation must be determined. Care is required to get the ordering of the bra-kets right when expressing such a projection.

Suppose we wish to calculation the rejection of $$\ket{a}$$ from $$\ket{b}$$, that is $$\ket{b – \alpha a}$$, such that

\label{eqn:qmSchwartz:220}
0
= \braket{a}{b – \alpha a}
= \braket{a}{b} – \alpha \braket{a}{a},

or
\label{eqn:qmSchwartz:240}
\alpha =
\frac{\braket{a}{b} }{ \braket{a}{a} }.

Therefore, the projection of $$\ket{b}$$ on $$\ket{a}$$ is

\label{eqn:qmSchwartz:260}
\textrm{Proj}_{\ket{a}} \ket{b}
= \frac{\braket{a}{b} }{ \braket{a}{a} } \ket{a}
= \frac{\braket{b}{a}^\conj }{ \braket{a}{a} } \ket{a}.

The conventional way to write this in QM is in the operator form

\label{eqn:qmSchwartz:300}
\textrm{Proj}_{\ket{a}} \ket{b}
= \frac{\ket{a}\bra{a}}{\braket{a}{a}} \ket{b}.

In this form the rejection of $$\ket{a}$$ from $$\ket{b}$$ can be expressed as

\label{eqn:qmSchwartz:280}
\textrm{Rej}_{\ket{a}} \ket{b} = \ket{b} – \frac{\ket{a}\bra{a}}{\braket{a}{a}} \ket{b}.

This state vector is normal to $$\ket{a}$$ as desired

\label{eqn:qmSchwartz:320}
\braket{a}{b – \frac{\braket{a}{b} }{ \braket{a}{a} } a }
=
\braket{a}{ b} – \frac{ \braket{a}{b} }{ \braket{a}{a} } \braket{a}{a}
=
\braket{a}{ b} – \braket{a}{b}
= 0.

How about it’s length? That is

\label{eqn:qmSchwartz:340}
\begin{aligned}
\braket{b – \frac{\braket{a}{b} }{ \braket{a}{a} } a}{b – \frac{\braket{a}{b} }{ \braket{a}{a} } a }
&=
\braket{b}{b} – 2 \frac{\Abs{\braket{a}{b}}^2}{\braket{a}{a}} +\frac{\Abs{\braket{a}{b}}^2 }{ \braket{a}{a}^2 } \braket{a}{a} \\
&=
\braket{b}{b} – \frac{\Abs{\braket{a}{b}}^2}{\braket{a}{a}}.
\end{aligned}

Observe that this must be greater to or equal to zero, so

\label{eqn:qmSchwartz:360}
\braket{b}{b} \ge \frac{ \Abs{ \braket{a}{b} }^2 }{ \braket{a}{a} }.

Rearranging this gives \ref{eqn:qmSchwartz:20} as desired. The Schwartz proof in [2] obscures the geometry involved and starts with

\label{eqn:qmSchwartz:380}
\braket{b + \lambda a}{b + \lambda a} \ge 0,

where the “proof” is nothing more than a statement that one can “pick” $$\lambda = -\braket{b}{a}/\braket{a}{a}$$. The Pythagorean context of the Schwartz inequality is not mentioned, and without thinking about it, one is left wondering what sort of magic hat that $$\lambda$$ selection came from.

# References

[1] W Keith Nicholson. Elementary linear algebra, with applications. PWS-Kent Publishing Company, 1990.

[2] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.