Appendix G. Rulers in Space

Let's sketch how our theory of rulers extends to three-dimensional space. In \( (x, y, z) \)-coordinate space, we have three standard rulers.

\[ \begin{align} \bar{\standard}_{x} &: \Ruler_{(x, y, z)} & \bar{\standard}_{y} &: \Ruler_{(x, y, z)} & \bar{\standard}_{z} &: \Ruler_{(x, y, z)} \end{align} \]

We have an anti-commutative multiplication

\[ \begin{align} s \mult r &= - r \mult s \end{align} \]

where \( r : \Ruler_{(x, y, z)} \) and \( s : \Ruler_{(x, y, z)} \) are any two rulers. We can write polyrulers in standard forms.

degree	polyruler standard form
0	\( a \)
1	\( a \mult \bar{\standard}_{x} + b \mult \bar{\standard}_{y} + c \mult \bar{\standard}_{z} \)
2	\( a \mult \bar{\standard}_{y} \mult \bar{\standard}_{z} + b \mult \bar{\standard}_{z} \mult \bar{\standard}_{x} + c \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} \)
3	\( a \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} \mult \bar{\standard}_{z} \)

We produce polyrulers with degree three as the product of the rulers. The polyruler

\[ \begin{align} &\bar{\standard}_{x} \mult \bar{\standard}_{y} \mult \bar{\standard}_{z} \end{align} \]

is the volume polyruler for \( (x, y, z) \)-coordinate space.

Formulas.

The following formulas describe how to measure a polyvector with a polyruler.

\[ \begin{align} &\langle{a_r}\mathbin{|}{a_v}\rangle \\ &\qquad = a_r a_v \\ &\bigl\langle{ a_r \mult \bar{\standard}_{x} + b_r \mult \bar{\standard}_{y} + c_r \mult \bar{\standard}_{z} }\mathbin{\big|}{ a_v \mult \vec{\standard}_{x} + b_v \mult \vec{\standard}_{y} + c_v \mult \vec{\standard}_{z} }\bigr\rangle \\ &\qquad = a_r a_v + b_r b_v + c_r c_v \\ &\bigl\langle{ a_r \mult \bar{\standard}_{y} \mult \bar{\standard}_{z} + b_r \mult \bar{\standard}_{z} \mult \bar{\standard}_{x} + c_r \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} }\mathbin{\big|}{ a_v \mult \vec{\standard}_{y} \mult \vec{\standard}_{z} + b_v \mult \vec{\standard}_{z} \mult \vec{\standard}_{x} + c_v \mult \vec{\standard}_{x} \mult \vec{\standard}_{y} }\bigr\rangle \\ &\qquad = a_r a_v + b_r b_v + c_r c_v \\ &\bigl\langle{ a_r \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} \mult \bar{\standard}_{z} }\mathbin{\big|}{ a_v \mult \vec{\standard}_{x} \mult \vec{\standard}_{y} \mult \vec{\standard}_{z} }\bigr\rangle \\ &\qquad = a_r a_v \end{align} \]

Grad, Curl, Div, and All That

The ruler differential \( \bar{\diff} \) satisfies all abstract differential laws, and we interpret:

\[ \begin{align} \bar{\diff} x &= \bar{\standard}_{x}, & \bar{\diff} y &= \bar{\standard}_{y}, & \bar{\diff} z &= \bar{\standard}_{z}. \end{align} \]

We may calculate the differential of polyrulers using the familiar laws

\[ \begin{align} \bar{\diff} \bigl(p \mult \bar{\standard}_{x}\bigr) &= \bar{\diff} p \mult \bar{\standard}_{x} \\ \bar{\diff} \bigl(p \mult \bar{\standard}_{y}\bigr) &= \bar{\diff} p \mult \bar{\standard}_{y} \\ \bar{\diff} \bigl(p \mult \bar{\standard}_{z}\bigr) &= \bar{\diff} p \mult \bar{\standard}_{z} \end{align} \]

where \( p : \PolyRuler_{(x, y, z)}^{} \) is any polyruler. Or, if we prefer partials, we can find the differential using the following formulas.

Formulas.

A 0-polyruler \( a \) has differential

\[ \begin{align} &\overset{x}{\diff} a \mult \bar{\standard}_{x} + \overset{y}{\diff} a \mult \bar{\standard}_{y} + \overset{z}{\diff} a \mult \bar{\standard}_{z}. \end{align} \]

A 1-polyruler \( a \mult \bar{\standard}_{x} + b \mult \bar{\standard}_{y} + c \mult \bar{\standard}_{z} \) has differential

\[ \begin{align} &\Bigl(\overset{y}{\diff} c - \overset{z}{\diff} b\Bigr) \mult \bar{\standard}_{y} \mult \bar{\standard}_{z} + \Bigl(\overset{z}{\diff} a - \overset{x}{\diff} c\Bigr) \mult \bar{\standard}_{z} \mult \bar{\standard}_{x} + \Bigl(\overset{x}{\diff} b - \overset{y}{\diff} a\Bigr) \mult \bar{\standard}_{x} \mult \bar{\standard}_{y}. \end{align} \]

A 2-polyruler \( a \mult \bar{\standard}_{y} \mult \bar{\standard}_{z} + b \mult \bar{\standard}_{z} \mult \bar{\standard}_{x} + c \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} \) has differential

\[ \begin{align} &\Bigl(\overset{x}{\diff} a + \overset{y}{\diff} b + \overset{z}{\diff} c\Bigr) \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} \mult \bar{\standard}_{z}. \end{align} \]

These formulas let us demonstrate an important fact: the second differential is identically zero

\[ \begin{align} \bar{\diff} \bar{\diff} p &= 0 \end{align} \]

for a polyruler \( p : \PolyRuler_{(x, y, z)}^{} \) of any degree!