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12.7 The Second Differential

The goal of this section is to see why the following theorem is true.
Theorem.
The second ruler differential is identically zero
\[ \begin{align} \bar{\diff} \bar{\diff} z &= 0 \end{align} \]
where \( z \depends (x, y) \) is any function.
Before we can explain this theorem, we'll need a couple formulas for differentials. The first formula should be familiar from our work with metrics.
Formula.
The differential of a function \( z \depends (x, y) \) is the ruler
\[ \begin{align} \bar{\diff} z &: \Ruler_{(x, y)} \\ \bar{\diff} z &= a \mult \bar{\standard}_{x} + b \mult \bar{\standard}_{y} \end{align} \]
whose parts
\[ \begin{align} a &= \overset{x}{\diff} z & &\andSpaced & b &= \overset{y}{\diff} z \end{align} \]
are the partials of \( z \).
Partials are the parts of the differential, whether that's the metric differential or the ruler differential. We'll also need a formula for the differential of a ruler.
Formula.
The differential of a ruler \( r : \Ruler_{(x, y)} \) is the polyruler
\[ \begin{align} \bar{\diff} r &: \PolyRuler_{(x, y)}^{2} \\ \bar{\diff} r &= \Bigl(\overset{x}{\diff} b - \overset{y}{\diff} a\Bigr) \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} \end{align} \]
when we've written \( r = a \mult \bar{\standard}_{x} + b \mult \bar{\standard}_{y} \) in standard form.
Why?
We compute.
\[ \begin{align} \bar{\diff} r &= \bar{\diff} \bigl(a \mult \bar{\standard}_{x} + b \mult \bar{\standard}_{y}\bigr) \\ &= \bar{\diff} a \mult \bar{\standard}_{x} + \bar{\diff} b \mult \bar{\standard}_{y} \\ &= \Bigl( \overset{x}{\diff} a \mult \bar{\standard}_{x} + \overset{y}{\diff} a \mult \bar{\standard}_{y} \Bigr) \mult \bar{\standard}_{x} + \Bigl( \overset{x}{\diff} b \mult \bar{\standard}_{x} + \overset{y}{\diff} b \mult \bar{\standard}_{y} \Bigr) \mult \bar{\standard}_{y} \\ &= \overset{x}{\diff} a \mult \bar{\standard}_{x} \mult \bar{\standard}_{x} + \overset{y}{\diff} a \mult \bar{\standard}_{y} \mult \bar{\standard}_{x} + \overset{x}{\diff} b \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} + \overset{y}{\diff} b \mult \bar{\standard}_{y} \mult \bar{\standard}_{y} \\ &= \overset{y}{\diff} a \mult \bar{\standard}_{y} \mult \bar{\standard}_{x} + \overset{x}{\diff} b \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} \\ &= - \overset{y}{\diff} a \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} + \overset{x}{\diff} b \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} \\ &= \Bigl(\overset{x}{\diff} b - \overset{y}{\diff} a\Bigr) \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} \end{align} \]
We're now ready to explain our theorem!
Why?
The first ruler differential
\[ \begin{align} \bar{\diff} z &= a \mult \bar{\standard}_{x} + b \mult \bar{\standard}_{y} \end{align} \]
has partials for parts
\[ \begin{align} a &= \overset{x}{\diff} z & b &= \overset{y}{\diff} z. \end{align} \]
Taking the second differential, we find
\[ \begin{align} &\bar{\diff} \bar{\diff} z = \Bigl(\overset{x}{\diff} b - \overset{y}{\diff} a\Bigr) \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} = \Bigl(\overset{x}{\diff} \overset{y}{\diff} z - \overset{y}{\diff} \overset{x}{\diff} z\Bigr) \mult \bar{\standard}_{x} \mult \bar{\standard}_{y}. \end{align} \]
But this is zero because we can reorder the partials
\[ \begin{align} \overset{y}{\diff} \overset{x}{\diff} z &= \overset{x}{\diff} \overset{y}{\diff} z. \end{align} \]
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