Ex 1.32 Path functions and state functions, 2.2

Structure and Interpretation of Classical Mechanics

1. The local-tuple function $f$ is the same as the local-tuple function $\bar \Gamma (\bar f)$ where $\bar f[q] = f \circ \Gamma [q]$ .

2. On the other hand, the path function $\bar f[q]$ and the path function $\bar \Gamma (\bar f) \circ \Gamma [q]$ are not necessarily the same. Explain.

…

~~~

1. …

To summarize:

In theory where a series can have infinite terms, functions $f$ and $\bar \Gamma (\bar f)$ are identical because it is just the definition of $\bar \Gamma$ .

In practice where a series can have only finite terms, they are still identical because their inputs are the same exact local tuples.

2. However, for $\bar f [q]$ and $\bar \Gamma (\bar f) \circ \Gamma [q]$ , although they are both path functions with the same path $q$ as input, while the function $\bar f$ processes the path $q$ directly, the function $\bar \Gamma (\bar f) \circ \Gamma$ would first turn the path $q$ into a local tuple $\Gamma[q]$ , which in practice would have only a finite number of components.

(define ((GammaBar-fBar-o-Gamma f-bar) q)
  (compose (Gamma-bar f-bar) (Gamma q)))

Then it would use the function osculating-path to generate the path.

(define ((Gamma-bar f-bar) local)
  ((f-bar (osculating-path local)) (time local)))

In other words, the path being used is the osculating path

$\displaystyle{\mathcal{O} (t_0, q(t_0), v(t_0), \cdots, q^{(n)}(t_0)))}$ ,

instead of the original path $q$ itself. Therefore, functions $\bar f [q]$ and $\bar \Gamma (\bar f) \circ \Gamma [q]$ do not have to be equal. They are identical in the following two cases:

The first case is when

$\displaystyle{\mathcal{O} (t_0, q(t_0), v(t_0), \cdots, q^{(n)}(t_0))) \equiv q}$ ,

where

$\displaystyle{\begin{aligned} &\mathcal{O} (t_0, q(t_0), v(t_0), \cdots, q^{(n)}(t_0)))(t) \\ &= q_0 + v_0 (t-t_0) + \frac{1}{2} a_0 (t-t_0)^2 + ... +\frac{1}{n!} q^{(n)}_0 (t-t_0)^n \\ \end{aligned}}$ .