Ex 1.32 Path functions and state functions, 2.1

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. …

~~~

1. Equation

$\displaystyle{\begin{aligned} \bar f &= f \circ \Gamma \\ \end{aligned}}$

is the definition of $\bar f$ . And equation

$\displaystyle{\begin{aligned} f &= \bar \Gamma (\bar f) \\ \end{aligned}}$

is the definition of $\bar \Gamma$ . So, in theory,

$\displaystyle{\begin{aligned} f &= \bar \Gamma (\bar f) \\ \end{aligned}}$

is just true by definition.

In practice, their initial inputs are both the same local tuple

$\displaystyle{\begin{aligned} &(t, q(t), v(t), \cdots, q^{(n)}(t)) \\ \end{aligned}}$ ,

which has only a finite number of components. The path $q$ in the process is generated by that initial input:

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

So the function $\Gamma$ would get you back the exact local tuple from the osculating path:

$\displaystyle{(t,q,v,\cdots, q^{(n)}) = \Gamma[\mathcal{O} (t,q,v, \cdots, q^{(n)}](t)}$

In other words, since for the functions $\displaystyle{f}$ and $\displaystyle{\bar \Gamma (\bar f)}$ , neither input is the path itself, the identity is exact.

Conceptually,

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

(define ((f-bar q) t)
  (f ((Gamma q) t))
 
;; (((f-bar q) 't)
;; == f((Gamma q) 't)
;; == f('t, (v 't), (a 't),...)

— Me@2024-10-16 10:34:35 AM

Physics Town

continues

Ex 1.32 Path functions and state functions, 2.1

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