14.3 Path Integrals
Let's see how we take an integral along a path.
Definition.
Suppose we have a ruler, a path, and a time interval
\[
\begin{align}
\bar{\diff} z &: \Ruler_{(x, y)}
&
p &: \Path_{(x, y)}
&
i &: \Interval_{t}
\end{align}
\]
where \( z \depends (x, y) \) is a function. We define the path integral
\[
\begin{align}
\Bigl\langle{\bar{\diff} z}\mathrel{\overset{p}{\textstyle{\int}}}{i}\Bigr\rangle &: \Number
\\
\Bigl\langle{\bar{\diff} z}\mathrel{\overset{p}{\textstyle{\int}}}{i}\Bigr\rangle &= \Bigl\langle{z}\mathrel{\overset{p}{\textstyle{\int}}}{\boundary i}\Bigr\rangle
\end{align}
\]
by trading the differential for a boundary.
A path integral has many ingredients. An example should help us see how they all work together.
Example.
Consider the path integral
\[
\begin{align}
&\biggl\langle{\bar{\diff} z}\mathrel{\overset{p}{\displaystyle{\int}}}{i}\biggr\rangle
\end{align}
\]
where
\[
\begin{align}
z &\depends (x, y)
&
p &: \Path_{(x, y)}
&
i &: \Interval_{t}
\\
z &= x y
&
&\left\{\begin{aligned} x_p &= 3 t \\ y_p &= t \end{aligned}\right.
&
i &= [1, 4]_{t}.
\end{align}
\]
We calculate.
\[
\begin{align}
\biggl\langle{\bar{\diff} z}\mathrel{\overset{p}{\displaystyle{\int}}}{i}\biggr\rangle &= \biggl\langle{z}\mathrel{\overset{p}{\displaystyle{\int}}}{\boundary [1, 4]_{t}}\biggr\rangle
\\
&= \biggl\langle{z}\mathrel{\overset{p}{\displaystyle{\int}}}{[4]_{t}}\biggr\rangle - \biggl\langle{z}\mathrel{\overset{p}{\displaystyle{\int}}}{[1]_{t}}\biggr\rangle
\\
&= \biggl\langle{x y}\mathrel{\overset{p}{\displaystyle{\int}}}{[4]_{t}}\biggr\rangle - \biggl\langle{x y}\mathrel{\overset{p}{\displaystyle{\int}}}{[1]_{t}}\biggr\rangle
\\
&= \biggl\langle{3 t^2}\mathrel{\displaystyle{\int}}{[4]_{t}}\biggr\rangle - \biggl\langle{3 t^2}\mathrel{\displaystyle{\int}}{[1]_{t}}\biggr\rangle
\\
&= 48 - 3
\\
&= 45
\end{align}
\]
The path integral of a differential \( \bar{\diff} z \) computes a change in height! Let's visualize a path integral using a contour map.
Example.
Consider the path \( p \) drawn on the contour map for a function \( z \depends (x, y) \).
Our path begins at height \( z = 2 \) and ends at height \( z = 4 \). The path integral with ruler \( \bar{\diff} z \) computes the change in height.
\[
\begin{align}
\biggl\langle{\bar{\diff} z}\mathrel{\overset{p}{\displaystyle{\int}}}{i}\biggr\rangle &= \biggl\langle{z}\mathrel{\overset{p}{\displaystyle{\int}}}{\boundary i}\biggr\rangle = 4 - 2 = 2
\end{align}
\]