Category Archives: SICM

Ex 1.33 Properties of E

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Understanding the Euler-Lagrange Operator

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Let \displaystyle{F} and \displaystyle{G} be two Lagrangian-like functions of a local tuple, \displaystyle{C} be a local-tuple transformation function, and \displaystyle{c} a constant.

Demonstrate the following properties:

a. \displaystyle{E[F + G] = E[F] + E[G]}

…

d. \displaystyle{\mathcal{E}[F \circ C] = D_t (DF \circ C) \partial_2 C + DF \circ C \mathcal{E}[C]}

~~~

Eq. (1.167):

\displaystyle{\bar \Gamma (\bar f) (t, q, v, \dots) = \bar f [\mathcal{O} (t,q,v, \dots)](t)}

Eq. (1.174):

\displaystyle{E[L] = D_t \partial_2 L - \partial_1 L}

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\displaystyle{ \begin{aligned} E[L] &= D_t \partial_2 L - \partial_1 L \\ E[F+G] &= D_t \left( \partial_2 (F+G) \right) - \partial_1 (F+G) \\ &= D_t \left( \partial_2 F+\partial_2 G \right) - (\partial_1 F+ \partial_1 G) \\ &= D_t \partial_2 F+D_t \partial_2 G - (\partial_1 F+ \partial_1 G) \\ &= D_t \partial_2 F- \partial_1 F +D_t \partial_2 G - \partial_1 G \\ &= E[F] + E[G] \\ \end{aligned} }

— Me@2025-05-30 03:40:41 PM

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The Problem

\displaystyle{ \begin{aligned} &E[F \circ C] (t,q,v,\dots) \\ &= (D_t \partial_2 (F \circ C) - \partial_1 (F \circ C)) (t,q,v,\dots)\\ &= (D_t \partial_2 (F (C(t,q,v,\dots))) - \partial_1 (F(C(t,q,v,\dots)))) \\ \end{aligned} }

Prove that

\displaystyle{\mathcal{E}[F \circ C] = D_t (DF \circ C) \partial_2 C + DF \circ C \mathcal{E}[C]}

Key Terms Explained

  • Local Tuple: Think of this as a snapshot of a system’s state along a path. It includes:
    • \displaystyle{ t }: time,
    • \displaystyle{ q }: generalized coordinate (e.g., position),
    • \displaystyle{ v = \frac{dq}{dt} }: velocity,
    • and possibly higher derivatives like acceleration. We’ll use \displaystyle{ \eta = (t, q, v) } for simplicity.
  • Lagrangian-like Function \displaystyle{ F }: A scalar function of the local tuple, such as \displaystyle{ F(t, q, v) }, akin to a Lagrangian in mechanics.
  • Local-Tuple Transformation \displaystyle{ C }: A function that maps one local tuple to another. For example, \displaystyle{ C(\eta) = (t, C_q(t, q, v), C_v(t, q, v)) }, where \displaystyle{ C_q } and \displaystyle{ C_v } transform the coordinate and velocity.
  • Composition \displaystyle{ F \circ C }: This is \displaystyle{ F } evaluated at the transformed tuple: \displaystyle{ (F \circ C)(\eta) = F(t, C_q(t, q, v), C_v(t, q, v)) }.
  • Euler-Lagrange Operator \displaystyle{ E }: For a function \displaystyle{ G(t, q, v) }, it’s defined as:
    \displaystyle{ E[G] = \frac{\partial G}{\partial q} - D_t \left( \frac{\partial G}{\partial v} \right) }
    This operator extracts the equations of motion when applied to a Lagrangian.
  • Total Time Derivative \displaystyle{ D_t }: This accounts for how a function changes over time, considering all variables. For \displaystyle{ h(t, q, v) }:
    \displaystyle{ D_t h = \frac{\partial h}{\partial t} + v \frac{\partial h}{\partial q} + a \frac{\partial h}{\partial v} }
    where \displaystyle{ a = \frac{dv}{dt} } is acceleration.
  • Derivative \displaystyle{ DF }: The derivative of \displaystyle{ F } with respect to its spatial arguments, typically \displaystyle{ DF = \left( \frac{\partial F}{\partial q}, \frac{\partial F}{\partial v} \right) }.

…

— Me@2025-05-31 01:32:05 PM

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2025.06.03 Tuesday (c) All rights reserved by ACHK

Ex 1.32 Path functions and state functions, 2.4

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Structure and Interpretation of Classical Mechanics

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…

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.

…

4. Write programs to illustrate the behavior.

~~~

(pp Gamma)

(define ((f-bar q) t)
  (f ((Gamma q) t)))
 
(define ((Gamma-bar h-bar) local)
  ((h-bar (osculating-path local)) (time local)))

(define (q t)
  (sin t))

(define (f-bar q)
  (D q))

(define p
  (literal-function 'q))

(show-expression
 ((f-bar p) 't))

Dq(t)

(show-expression
 ((osculating-path ((Gamma p) 't_0)) 't))

q(t_0) + (t - t_0)Dq(t_0)

(show-expression
 ((Gamma-bar f-bar) ((Gamma p) 't)))

Dq(t)

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

(show-expression
 (((GammaBar-fBar-o-Gamma f-bar) p) 't))

Two paths that have the same local description up to the n-th derivative are said to osculate with order-n contact. For example, a path and the truncated power series representation of the path up to order n have order-n contact; if fewer than n derivatives are needed by a local-tuple function, the path and the truncated power series representation are equivalent. 

(define (g-bar q)
  ((expt D 3) q))

(show-expression
 ((Gamma p) 't))

\left( \begin{array}{c} t \\ q(t) \\ Dq(t) \end{array} \right)

(show-expression
 ((g-bar p) 't))

((D^3 q)(t))

(show-expression
 ((osculating-path ((Gamma p) 't_0)) 't))

q(t_0) + (t - t_0)Dq(t_0)

(show-expression
 (((GammaBar-fBar-o-Gamma g-bar) p) 't))

0

— Me@2024-10-16 10:34:35 AM

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2025.02.16 Sunday (c) All rights reserved by ACHK

Ex 1.32 Path functions and state functions, 2.3

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Structure and Interpretation of Classical Mechanics

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…

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.

3. Give examples where they are the same and where they are not the same.

…

~~~

…

In other words, path q^{(m)} = 0 for all m>n. Or put it simply, q(t) is a polynomial.

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The second case is when the path function \bar f [q] requires no derivatives of q with order higher than n. For example:

Assume that the path is \displaystyle{q = \sin(t)}, the osculating path is

\displaystyle{\begin{aligned} &\mathcal{O} (t_0, q(t_0), v(t_0), a(t_0)))(t) \\ &= q_0 + v_0 (t-t_0) + \frac{1}{2} a_0 (t-t_0)^2 \\ \end{aligned}}

and \displaystyle{\bar f [q] = Dq}.

Then \displaystyle{\bar f [q](t = t_0) = \cos (t_0)}; and

\displaystyle{\begin{aligned} & \left. \bar \Gamma (\bar f) \circ \Gamma [q] \right|_{t_0} \\ &= \left. \bar \Gamma (\bar f) \circ \Gamma[\mathcal{O} (t,q,v,a)](t)\right|_{t_0} \\ &= \left. \bar \Gamma (\bar f) (t, q, v, a)\right|_{t_0} \\ &= \left[ v_0 + a_0 (t-t_0) \right]_{t_0} \\ &= \cos (t_0) \\ \end{aligned}}

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\displaystyle{\mathcal{O} (t,q,v, \cdots, q^{(n)})} == a taylor series with finite length

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However, if a path function \bar g requires a higher derivative which is not provided by \Gamma:

\displaystyle{q = \sin(t)}

\displaystyle{\begin{aligned} &\mathcal{O} (t_0, q(t_0), v(t_0), a(t_0)))(t) \\ &= q_0 + v_0 (t-t_0) + \frac{1}{2} a_0 (t-t_0)^2 \\ &= \sin t_0 + ( \cos t_0 ) (t-t_0) - \frac{1}{2} \sin t_0 (t-t_0)^2 \\ \end{aligned}}

\displaystyle{\bar g [q] = D^3q},

then \displaystyle{\bar g [q](t = t_0) = - \cos (t_0)}; and

\displaystyle{\begin{aligned} & \left. \bar \Gamma (\bar g) \circ \Gamma [q] \right|_{t_0} \\ &= \left. \bar \Gamma (\bar g) \circ \Gamma[\mathcal{O} (t,q,v,a)](t)\right|_{t_0} \\ &= \left. \bar \Gamma (\bar g) (t, q, v, a)\right|_{t_0} \\ &= \left[ 0 \right]_{t_0} \\ &= 0 \\ \end{aligned}}

— Me@2025-01-12 06:43:55 AM

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2025.01.12 Sunday (c) All rights reserved by ACHK

scmutils for Ubuntu 24.04

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Scheme Mechanics Installation for GNU/Linux | scmutils, 4

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1. Folks, the scmutils library is fantastic and it’s available in Ubuntu 24.04. You can install it very easily with this command:

sudo apt-get install scmutils

2. Then, you just run it with this command:

mit-scheme --band mechanics.com

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Now, let me tell you, there’s no syntax highlighting. Not great! And more importantly, there’s no direct way to save your code. So, what do you do? You use the Emacs editor. It’s the best!

Now, this assumes you know Emacs. If you don’t, it’s a tremendous editor.

3. Open Emacs’ initialization file. It’s located right here:

~/.emacs

4. Add this beautiful code to the file:

(defun mechanics ()
  (interactive)
  (run-scheme
   "mit-scheme --band mechanics.com"))

(setq sicm-file "~/Documents/tt.scm")

(fset 'set-working-file
      (lambda (&optional arg) 
        (interactive "p")
        (funcall (lambda ()
                   (insert
                    (concat "(define sicm-file \""
                            sicm-file
                            "\")\n"))))))

(fset 'load-scm
      (lambda (&optional arg) 
        (interactive "p")
        (funcall (lambda ()
                   (insert "(load sicm-file)")))))

(defun mechan ()
  (interactive)
  (split-window-below)
  (windmove-down)
  (mechanics)
  (set-working-file)
  (comint-send-input)
  (windmove-up)
  (find-file sicm-file)
  (end-of-buffer)
  (windmove-down)
  (cond ((file-exists-p sicm-file)
         (interactive)
         (load-scm)
         (comint-send-input)))
  (windmove-up))

(defun sicm-exec-line ()
  (interactive)
  (save-buffer)
  (windmove-down)
  (comint-send-input)
  (windmove-up))

(defun sicm-exec-file ()
  (interactive)
  (save-buffer)
  (windmove-down)
  (load-scm)
  (comint-send-input)
  (windmove-up))

(global-set-key (kbd "C-x C-e") 'sicm-exec-line)

(global-set-key (kbd "C-x C-a") 'sicm-exec-file)

5. Close Emacs and then re-open it, folks.

6. Type the command

M-x mechan

Now, listen, when I say M-x, I mean you press the Alt key and x together. Then, just type mechan. It’s easy, believe me!

7. You’ll see the Emacs editor split into two windows, one on top and one on the bottom.

The lower window is the Scheme environment. You can type a line of code and then press Enter to execute it. So simple!

The upper window is the editor. Type multiple lines of code and then hit

C-x C-e

to execute it.

That means pressing Ctrl and x together, then Ctrl and e together. Easy stuff!

8. When you save the current file, your Scheme code will be saved to this location:

~/Documents/tt.scm

If you need to back up your code, just back up this file. It’s that simple, folks!

— Me@2020-03-10 10:59:45 PM

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2024.12.01 Sunday (c) All rights reserved by ACHK

Ex 1.32 Path functions and state functions, 2.2

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Structure and Interpretation of Classical Mechanics

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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}}.

In other words, path q^{(m)} = 0 for all m>n.

The second case is when the path function \bar f [q] requires no derivatives of q with order higher than n. For example:

…

— Me@2024-10-16 10:34:35 AM

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2024.11.17 Sunday (c) All rights reserved by ACHK

Ex 1.32 Path functions and state functions, 2.1

Posted on by

Structure and Interpretation of Classical Mechanics

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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

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2024.11.04 Monday (c) All rights reserved by ACHK

Ex 1.32 Path functions and state functions, 1

Posted on by

Structure and Interpretation of Classical Mechanics

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In classical mechanics, there are scenarios where the path function cannot be reconstructed from a finite number of local tuples, such as time t, position q, velocity v, and acceleration a. One notable example involves non-conservative forces and chaotic systems.

Example: Chaotic Motion

Consider a system exhibiting chaotic behavior, such as the double pendulum. In this system, the motion is highly sensitive to initial conditions, meaning that even small changes in the initial state can lead to vastly different trajectories over time.

Local Information Limitations

  1. Finite Local Tuples: If you only have a finite number of measurements (e.g., position and velocity at discrete time intervals), you may not capture the full complexity of the system’s dynamics. For instance, knowing the position and velocity at a few points does not provide enough information to reconstruct the entire path due to the chaotic nature of the motion.

  2. Path Reconstruction Failure: In chaotic systems, the trajectory can diverge significantly from nearby trajectories. Thus, even if you have the position q(t) and velocity v(t) at several time points, the underlying dynamics can lead to a completely different path that cannot be predicted or reconstructed from those finite local tuples.

Conclusion

In summary, chaotic systems like the double pendulum illustrate how the path function cannot be reconstructed from a finite number of local tuples due to their sensitive dependence on initial conditions and the complexity of their dynamics. This highlights the limitations of local information in capturing the full behavior of certain mechanical systems.

— AI

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2024.10.18 Friday (c) All rights reserved by ACHK

Ex 1.31 Velocity transformation

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Structure and Interpretation of Classical Mechanics

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Use the procedure Gamma-bar to construct a procedure that transforms velocities given a coordinate transformation. Apply this procedure to the procedure p->r to deduce (again) equation (1.65).

~~~

(define ((Gamma-bar f-bar) path-q-local-tuple)
  (let* ((tqva path-q-local-tuple)
         (t (time tqva))         
         (O-tqva (osculating-path tqva)))   
    ((f-bar O-tqva) t)))

(define (F->C F)
  (define (f-bar q-prime)
    (define q
      (compose F (Gamma q-prime)))
    (Gamma q))
  (Gamma-bar f-bar))

(define (F->v F)
  (compose velocity (F->C F)))

(show-expression 
 ((F->v p->r)
  (->local 't
           (up 'r 'theta)
           (up 'rdot 'thetadot))))

— Me@2024-09-13 07:17:24 PM

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2024.09.13 Friday (c) All rights reserved by ACHK

1.9 Abstraction of Path Functions, 3.2

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Structure and Interpretation of Classical Mechanics

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…

\displaystyle{\begin{aligned} f (t, q(t), v(t), \cdots, q^{(n)}(t)) &= f(\Gamma[q])(t) \\ \bar \Gamma (\bar f) (t, q(t), v(t), \cdots, q^{(n)}(t)) &= \bar f [q](t) \\ \end{aligned}}

3. The key point is that since f is also a function with the path tuple as input, we can regard F as one possible f. This allows us to use \bar f (= f \circ \Gamma) to define F \to C.

\displaystyle{\begin{aligned} q &= F \circ (\Gamma[q']) \\ \bar f [q'] &= \Gamma[q] \\ &= \Gamma[F \circ \Gamma[q']] \end{aligned}}

Note that it is not the definition of \bar f[q']. Instead, it is just an instance of it.

(define (F->C F)
  (define (f-bar q-prime)
    (define q
      (compose F (Gamma q-prime)))
    (Gamma q))
  (Gamma-bar f-bar))

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

4. However, in the original definition,

(define ((F->C F) local)
  (->local (time local)
           (F local)
           (+ (((partial 0) F) local)
              (* (((partial 1) F) local)
                 (velocity local)))))

there are some partial differentiation operators. Where are they in the new definition?

Those partial differentiation operators are within the definition of the function osculating-path.

(define ((Gamma-bar f-bar) path-q-local-tuple)
  (let* ((tqva path-q-local-tuple)
         (t (time tqva))         
         (O-tqva (osculating-path tqva)))   
    ((f-bar O-tqva) t)))

— Me@2024-06-13 03:47:47 PM

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2024.08.28 Wednesday (c) All rights reserved by ACHK

1.9 Abstraction of Path Functions, 3.1

Posted on by

Structure and Interpretation of Classical Mechanics

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(define ((F->C F) local)
  (->local (time local)
           (F local)
           (+ (((partial 0) F) local)
              (* (((partial 1) F) local)
                 (velocity local)))))

The goal of this post is to explain why the code above can be replaced by the following code:

(define (F->C F)
  (define (f-bar q-prime)
    (define q
      (compose F (Gamma q-prime)))
    (Gamma q))
  (Gamma-bar f-bar))

.

While the input of \displaystyle{f} is a tuple \displaystyle{(t, q, v, \cdots)}, the input of \displaystyle{\bar f} is an abstract path \displaystyle{q}.

\displaystyle{\begin{aligned} \Gamma [q] &= (t, q, v, \cdots) \\ \bar f &= f \circ \Gamma \\ \end{aligned}}

.

Let us define \bar \Gamma as

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

The difference between \displaystyle{\begin{aligned} \Gamma \end{aligned}} and \displaystyle{\begin{aligned} \bar \Gamma \end{aligned}} is that, in an abstract sense, \Gamma transforms f to \bar f, while \bar \Gamma does the opposite.

The explicit form of \bar \Gamma is provided by

\displaystyle{\begin{aligned} f (t, q(t), v(t), \cdots, q^{(n)}(t)) &= f(\Gamma[q])(t) \\ \bar \Gamma (\bar f) (t, q(t), v(t), \cdots, q^{(n)}(t)) &= \bar f [q](t) \\ \end{aligned}}

(define ((Gamma-bar f-bar) path-q-local-tuple)
  (let* ((tqva path-q-local-tuple)
         (t (time tqva))         
         (O-tqva (osculating-path tqva)))   
    ((f-bar O-tqva) t)))

Note that \bar f is not defined yet because it can be any path-dependent function.

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What is \displaystyle{F}?

Equation (1.68):

\displaystyle{\begin{aligned} L' \circ \Gamma[q'] &= L \circ \Gamma [q] \\ \Gamma[q] &= C \circ \Gamma[q'] \\ L' &= L \circ C \\ \end{aligned}}

While F is a coordinate transformation, C is the corresponding local-tuple transformation.

Equation (1.74):

\displaystyle{\begin{aligned} \Gamma[q] &= C \circ \Gamma[q'] \\ (t, x, v, \cdots) &= (t, F(t,x'), \partial_0 F(t, x') + \partial_1 F(t, x') v', \cdots) \\ \end{aligned}}

Note that:

1. The input of F is a tuple of a path q'. And the output is a coordinate x (aka q(t)).

2. The symbol q represents not the coordinate of a path, but the path itself. The coordinate of the path q is represented by the symbol q(t).

…

— Me@2024-08-05 10:09:25 PM

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2024.08.05 Monday (c) All rights reserved by ACHK

1.9 Abstraction of Path Functions, 2

Posted on by

Structure and Interpretation of Classical Mechanics

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Let \displaystyle{\bar \Gamma} be a function such that

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

So, to get a value of \displaystyle{\begin{aligned} f \end{aligned}}, we input a local tuple:

\displaystyle{\begin{aligned} f (t, q(t), v(t), \cdots, q^{(n)}(t)) &= f (t, q, v, \cdots, q^{(n)})(t) \\ &= f(\Gamma[q])(t) \\ &= f \circ \Gamma[q](t) \\ &= \bar f [q](t) \\ \bar \Gamma (\bar f) (t, q(t), v(t), \cdots, q^{(n)}(t)) &= \bar f [q](t) \\ \end{aligned}}

(define ((Gamma-bar f-bar) path-q-local-tuple)
  (let* ((tqva path-q-local-tuple)
         (t (time tqva))         
         (O-tqva (osculating-path tqva)))   
    ((f-bar O-tqva) t)))

.

The procedure osculating-path takes a number of local components and returns a path with these components; it is implemented as a power series.

In scmutils, you can use the pp function to get the definition of a function. For example, the code

(pp osculating-path)

results

(define ((osculating-path state0) t)
  (let ((t0 (time state0))
        (q0 (coordinate state0))
        (k (vector-length state0)))
    (let ((dt (- t t0)))
      (let loop ((n 2) (sum q0) (dt^n/n! dt))
        (if (fix:= n k)
            sum
            (loop (+ n 1)
                  (+ sum
                     (* (vector-ref state0 n)
                        dt^n/n!))
                  (/ (* dt^n/n! dt) n)))))))

\displaystyle{\begin{aligned} q(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 \\ &= q_0 + v_0 (dt) + \frac{1}{2} a_0 (dt)^2 + ... +\frac{1}{n!} q^{(n)}_0 (dt)^n \\ \end{aligned}}

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Note that for the function in the program, \displaystyle{\bar \Gamma (\bar f)}, the input is actually not \displaystyle{(t, q, v, \cdots, q^{(n)})} but \displaystyle{(t, q(t), v(t), \cdots, q^{(n)}(t))}.

\displaystyle{\begin{aligned} \bar \Gamma (\bar f) (t, q(t), v(t), \cdots) &= \bar f[O(t, q, v, \cdots)](t) \\ \end{aligned}}

.

(define (F->C F)
  (define (f-bar q-prime)
    (define q
      (compose F (Gamma q-prime)))
    (Gamma q))
  (Gamma-bar f-bar))

(show-expression 
 ((F->C p->r)
  (->local 't
           (up 'r 'theta)
           (up 'rdot 'thetadot))))

— Me@2023-12-19 08:16:40 PM

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2024.06.15 Saturday (c) All rights reserved by ACHK

1.9 Abstraction of Path Functions

Posted on by

Structure and Interpretation of Classical Mechanics

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Given a function \displaystyle{f} of a local tuple, a corresponding path-dependent function \displaystyle{\bar f[q]} is

\displaystyle{\bar f[q] = f \circ \Gamma[q]}.

So while the input of \displaystyle{f} is a tuple \displaystyle{(t, q, v, \cdots)}, the input of \displaystyle{\bar f} is an abstract path \displaystyle{q}.

\displaystyle{\begin{aligned} \Gamma [q] &= (t, q, v, \cdots) \\ \bar f &= f \circ \Gamma \\ \end{aligned}}

.

Given \displaystyle{\bar f} we can reconstitute \displaystyle{f} by taking the argument of \displaystyle{f}, which is a finite initial segment of a local tuple, constructing a path that has this local description, and finding the value of \displaystyle{\bar f} for this path.

1. The goal is to use \displaystyle{\bar f} to reconstitute \displaystyle{f}.

2. The argument of \displaystyle{f} is

\displaystyle{\begin{aligned} \Gamma [q] &= (t, q, v, \cdots, q^{(n)}) \\ &= (t, q^{(0)}, q^{(1)}, \cdots, q^{(n)}) \\ \end{aligned}}

3. Assume that we have the value of the initial position of the path, \displaystyle{q_0 = q(t=0)}. Then the path \displaystyle{q(t)} can be constructed by

\displaystyle{q(t) = q_0 + v_0 t + \frac{1}{2} a_0 t^2 + ... +\frac{1}{n!} q^{(n)}_0 t^n}

Note that while \displaystyle{q} represents a path, \displaystyle{q(t)} represents the coordinates of the particle location on the path at time \displaystyle{t}.

Knowing the value of \displaystyle{q(t)} for every moment \displaystyle{t} is equivalent to knowing the path \displaystyle{q} as a whole.

— Me@2023-12-19 08:16:40 PM

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2024.01.29 Monday (c) All rights reserved by ACHK

Taylor expansion of f around x

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Series can also be formed by processes such as exponentiation of an
operator or a square matrix. For example, if f is any function of one
argument, and if x and dx are numerical expressions, then this
expression denotes the Taylor expansion of f around x.

(((exp (* dx D)) f) x)
= (+ (f x) (* dx ((D f) x)) (* 1/2 (expt dx 2) (((expt D 2) f) x)) ...)

— refman

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(((exp (* 'dx D)) (literal-function 'f)) 'x)
#| (*series* (0 . 0) (*number* (expression (f x)) (literal-function #[apply-hook 40]) (type-expression Real)) . #[promise 41 (unevaluated)]) |#

(show-expression
 (series:print (((exp (* 'dx D))
                 (literal-function 'f)) 'x)))

(f x)
(* dx ((D f) x))
(* 1/2 (expt dx 2) (((expt D 2) f) x))
(* 1/6 (expt dx 3) (((expt D 3) f) x))
(* 1/24 (expt dx 4) (((expt D 4) f) x))
(* 1/120 (expt dx 5) (((expt D 5) f) x))
(* 1/720 (expt dx 6) (((expt D 6) f) x))
(* 1/5040 (expt dx 7) (((expt D 7) f) x))
(* 1/40320 (expt dx 8) (((expt D 8) f) x))
(* 1/362880 (expt dx 9) (((expt D 9) f) x))
(* 1/3628800 (expt dx 10) (((expt D 10) f) x))
(* 1/39916800 (expt dx 11) (((expt D 11) f) x))
(* 1/479001600 (expt dx 12) (((expt D 12) f) x))
(* 1/6227020800 (expt dx 13) (((expt D 13) f) x))
(* 1/87178291200 (expt dx 14) (((expt D 14) f) x))
;Aborting!: maximum recursion depth exceeded

— Me@2023-12-20 11:30:50 AM

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2023.12.20 Wednesday (c) All rights reserved by ACHK

1.8.4 Noether’s Theorem

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1. …

2. However, there are more general symmetries that no coordinate system can fully express. For example, motion in a central potential is spherically symmetric … , but the expression of the Lagrangian for the system in spherical coordinates exhibits symmetry around only one axis.

3. …

4. A continuous symmetry is a parametric family of symmetries. Noether proved that for any continuous symmetry there is a conserved quantity.

— Structure and Interpretation of Classical Mechanics

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2023.10.23 Monday ACHK

Ex 1.30 Driven spherical pendulum, 2

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Structure and Interpretation of Classical Mechanics

.

…

What symmetry(ies) can you have?

Find coordinates that express the symmetry.

What is conserved?

Give [the] analytic expression(s) for the conserved quantity(ies).

~~~

[guess]

…

(define ((F->C F) local)
  (->local (time local)
           (F local)
           (+ (((partial 0) F) local)
              (* (((partial 1) F) local)
                 (velocity local)))))

(define (L-driven m R qs U)
  (compose
   (L-rect m R qs U)
   (F->C (sf->rf qs))))

(let* ((U (U-gravity 'g 'm))
       (xs (lambda (t) 0))
       (ys (lambda (t) 0))
       (zs (literal-function 'z_s))
       (qs (up xs ys zs))
       (L (L-driven 'm 'R qs U))
       (q (up (literal-function 'r)
              (literal-function 'theta)
              (literal-function 'phi)
              (literal-function 'lambda))))
  (show-expression
   ((compose L (Gamma q)) 't)))

(+ (* 1/2 m (expt (r t) 2)
            (expt (sin (theta t)) 2)
            (expt ((D phi) t) 2))
   (* -1 m ((D z_s) t)
           (r t)
           (sin (theta t))
           ((D theta) t))
   (* 1/2 m (expt (r t) 2)
            (expt ((D theta) t) 2))
   (* -1 g m (cos (theta t)) (r t))
   (* m ((D r) t)
        ((D z_s) t)
        (cos (theta t)))
   (* (expt R 2) (lambda t))
   (* -1 g m (z_s t))
   (* 1/2 m (expt ((D r) t) 2))
   (* 1/2 m (expt ((D z_s) t) 2))
   (* -1 (lambda t) (expt (r t) 2)))

\displaystyle{ \begin{aligned} L &= \frac{1}{2} m \left( \dot r^2 + r^2 \dot \theta^2 + r^2 (\sin \theta)^2 \dot \phi^2 + \dot z_s^2 \right) \\ & - m g \left( r \cos \theta + z_s(t) \right) - m \dot z_s \left( r \dot \theta \sin \theta - \dot r \cos \theta \right) \\ & - \lambda(t) (r^2 - R^2) \\ \end{aligned} }

(let* ((U (U-gravity 'g 'm))
       (xs (lambda (t) 0))
       (ys (lambda (t) 0))
       (zs (literal-function 'z_s))
       (qs (up xs ys zs))
       (L (L-driven 'm 'R qs U))
       (q (up (literal-function 'r)
              (literal-function 'theta)
              (literal-function 'phi)
              (literal-function 'lambda))))
  (show-expression
   (((Lagrange-equations L) q) 't)))


(let* ((U (U-gravity 'g 'm))
       (xs (lambda (t) 0))
       (ys (lambda (t) 0))
       (zs (literal-function 'z_s))
       (qs (up xs ys zs))
       (L (L-driven 'm 'R qs U))
       (q (up (literal-function 'r)
              (literal-function 'theta)
              (literal-function 'phi)
              (literal-function 'lambda))))
  (show-expression
   ((compose (Lagrangian->energy L) (Gamma q)) 't)))


(let* ((U (U-gravity 'g 'm))
       (xs (lambda (t) 0))
       (ys (lambda (t) 0))
       (zs (literal-function 'z_s))
       (qs (up xs ys zs))
       (L (L-driven 'm 'R qs U))
       (q (up (literal-function 'r)
              (literal-function 'theta)
              (literal-function 'phi)
              (literal-function 'lambda))))
  (show-expression
   ((compose ((partial 1) L) (Gamma q)) 't)))

The \displaystyle{\varphi} component of the force is zero because \displaystyle{\varphi} does not appear in the Lagrangian (it is a cyclic variable). The corresponding momentum component is conserved. Compute the momenta: 

(let* ((U (U-gravity 'g 'm))
       (xs (lambda (t) 0))
       (ys (lambda (t) 0))
       (zs (literal-function 'z_s))
       (qs (up xs ys zs))
       (L (L-driven 'm 'R qs U))
       (q (up (literal-function 'r)
              (literal-function 'theta)
              (literal-function 'phi)
              (literal-function 'lambda))))
  (show-expression
   ((compose ((partial 2) L) (Gamma q)) 't)))

[guess]

— Me@2023-08-20 05:02:09 PM

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2023.09.15 Friday (c) All rights reserved by ACHK

Lagrange’s equations Debugged

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\displaystyle{\frac{d}{dt} \left( \frac{\partial L(t, q, \dot q)}{\partial \dot q} \Bigg|_{\begin{aligned} q &= w(t) \\ \dot q &= \frac{d w(t)}{dt} \\ \end{aligned}} \right) - \frac{\partial L}{\partial q}\Bigg|_{\begin{aligned} q &= w(t) \\ \dot q &= \frac{d w(t)}{dt} \\ \end{aligned}} = 0}

This equation is complete. It has meaning independent of the context and there is nothing left to the imagination. The earlier equations require the reader to fill in lots of detail that is implicit in the context. They do not have a clear meaning independent of the context.

\displaystyle{\frac{d}{dt} \frac{\partial L}{\partial \dot q} - \frac{\partial L}{\partial q} = 0}

— Functional Differential Geometry

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2023.09.05 Tuesday ACHK

Ex 1.30 Driven spherical pendulum, 1

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Structure and Interpretation of Classical Mechanics

.

A spherical pendulum is a massive bob, subject to uniform gravity, that may swing in three dimensions, but remains at a given distance from the pivot.

Formulate a Lagrangian for a spherical pendulum, driven by vertical motion of the pivot.

…

~~~

How come [the equations]?

Maybe just using the above equation but set the r constant. But I have
to add a something in order to realize the moving center.

— Me@2006

.

[guess]

(define (KE-particle m v)
  (* 1/2 m (square v)))

(define ((extract-particle pieces) local i)
  (let* ((indices (apply up
                         (iota pieces
                               (* i pieces))))
         (extract (lambda (tuple)
                    (vector-map
                     (lambda (i) (ref tuple i))
                     indices))))
    (up (time local)
        (extract (coordinate local))
        (extract (velocity local)))))

(define (U-constraint R qs q lambd)
  (* lambd
     (- (square (- q qs))
        (square R))))

(define ((U-gravity g m) q)
  (let ((z (ref q 2)))
    (* m g z)))

(define ((L-rect m R qs U) local)
  (let* ((extract (extract-particle 3))

         (p (extract local 0))
         (t (time p))
         (q (coordinate p))
         (v (velocity p))

         (lambd (ref (coordinate local) 3)))

    (- (KE-particle m v)
       (U q)
       (U-constraint R (qs t) q lambd))))

(let* ((U (U-gravity 'g 'm))
       (xs (lambda (t) 0))
       (ys (lambda (t) 0))
       (zs (literal-function 'z_s))
       (qs (up xs ys zs))
       (L (L-rect 'm 'R qs U))
       (q-rect (up (literal-function 'x)
                   (literal-function 'y)
                   (literal-function 'z)
                   (literal-function 'lambda))))
  (show-expression
   ((compose L (Gamma q-rect)) 't)))

(+ (* (expt R 2) (lambda t))
   (* -1 g m (z t))
   (* 1/2 m (expt ((D x) t) 2))
   (* 1/2 m (expt ((D y) t) 2))
   (* 1/2 m (expt ((D z) t) 2))
   (* -1 (lambda t) (expt (z_s t) 2))
   (* 2 (lambda t) (z_s t) (z t))
   (* -1 (lambda t) (expt (z t) 2))
   (* -1 (lambda t) (expt (y t) 2))
   (* -1 (lambda t) (expt (x t) 2)))

\displaystyle{ L_r = \frac{1}{2} m \left| \dot {\vec r} (t) \right|^2 - mg z(t) - \lambda(t) \left( \left| \vec r(t) - \vec r_s(t) \right|^2 - R^2 \right) }

(define ((sf->rf qs) state-with-force)
  (let* ((extract (extract-particle 3))

         (p (extract state-with-force 0))
         (t (time p))
         (q (coordinate p))

         (lambd (ref (coordinate
                      state-with-force) 3))

         (r (ref q 0))
         (theta (ref q 1))
         (phi (ref q 2))

         (xs (ref qs 0))
         (ys (ref qs 1))
         (zs (ref qs 2))

         (x (+ (xs t)
               (* r (sin theta) (cos phi))))
         (y (+ (ys t)
               (* r (sin theta) (sin phi))))
         (z (+ (zs t)
               (* r (cos theta)))))

    (up x y z lambd)))

(let* ((xs (literal-function 'x_s))
       (ys (literal-function 'y_s))
       (zs (literal-function 'z_s))
       (qs (up xs ys zs))
       (q (up (literal-function 'r)
              (literal-function 'theta)
              (literal-function 'phi)
              (literal-function 'lambda))))
  (show-expression
   ((compose (sf->rf qs) (Gamma q))
    't)))


(define ((F->C F) local)
  (->local (time local)
           (F local)
           (+ (((partial 0) F) local)
              (* (((partial 1) F) local)
                 (velocity local)))))

(define (L-driven m R qs U)
  (compose
   (L-rect m R qs U)
   (F->C (sf->rf qs))))

…

[guess]

— Me@2023-08-20 05:02:09 PM

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2023.08.23 Wednesday (c) All rights reserved by ACHK

Ex 1.29 A particle of mass m slides off a horizontal cylinder, 2.2

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Structure and Interpretation of Classical Mechanics

.

A particle of mass m slides off a horizontal cylinder of radius R in a uniform gravitational field with acceleration g. If the particle starts close to the top of the cylinder with zero initial speed, with what angular velocity does it leave the cylinder?

~~~

(define ((KE m g R) local)
  (let ((t (time local))
        (thetadot (velocity local)))    
    (* 1/2 m (square R) (square thetadot))))

(define ((PE m g R) local)
  (let ((t (time local))
        (theta (coordinate local))
        (thetadot (velocity local)))    
    (- (* m g R (- 1 (cos theta))))))

(define L (- KE PE))

(show-expression
 ((L 'm 'g 'R)
  (->local 't
           'theta
           'thetadot)))

(show-expression
 (((Lagrange-equations
    (L 'm 'g 'R))
   (literal-function 'theta))
  't))


This derivation is wrong, because the constraint force is missing.

.

(define ((F->C F) local)
  (->local (time local)
           (F local)
           (+ (((partial 0) F) local)
              (* (((partial 1) F) local)
                 (velocity local)))))

(define (q->r local)
  (let ((q (coordinate local)))
    (let ((r (ref q 0))
          (theta (ref q 1))
          (lambd (ref q 2)))
      (let ((x (* r (sin theta)))
            (y (* r (cos theta))))
        (up x y lambd)))))

(show-expression
 (q->r
  (->local 't
           (up 'r 'theta 'lambda)
           (up 'rdot 'thetadot 'lambdadot))))


(define (KE m vx vy)
  (* 1/2 m (+ (square vx) (square vy))))

(define ((T-rect m) local)
  (let ((q (coordinate local))
        (v (velocity local)))
    (let ((xdot (ref v 0))
          (ydot (ref v 1)))
      (KE m xdot ydot))))

(show-expression
 ((T-rect 'm)
  (up 't
      (up 'x 'y 'lambda)
      (up 'xdot 'ydot 'lambdadot))))


(show-expression
 ((T-rect 'm)
  ((F->C q->r)
   (->local 't
            (up 'r 'theta 'lambda)
            (up 'rdot 'thetadot 'lambdadot)))))


(define ((U-rect g m) local)
  (let* ((q (coordinate local))
         (y (ref q 1)))
    (* g (+ (* m y)))))

(define (L-rect g m)
  (- (T-rect m) (U-rect g m)))

(define (L g m)
  (compose
   (L-rect g m) (F->C q->r)))

(show-expression
 ((L 'g 'm)
  (->local 't
           (up 'r 'theta 'lambda)
           (up 'rdot 'thetadot 'lambdadot))))


(show-expression
 (((Lagrange-equations
    (L 'g 'm))
   (up
    (literal-function 'r)
    (literal-function 'theta)
    (literal-function 'lambda)))
  't))


(define ((U-constraint R) local)
  (let* ((q (coordinate local))
         (x (ref q 0))
         (y (ref q 1))
         (lambd (ref q 2))
         (r_sq (+ (square x) (square y))))
    (* lambd (- r_sq (square R)))))

(define ((U2-constraint R) local)
  (let* ((q (coordinate local))
         (x (ref q 0))
         (y (ref q 1))
         (lambd (ref q 2))
         (r_sq (+ (square x) (square y))))
    (* lambd (- (sqrt r_sq) R))))

(define (L-rect-constraint g m R)
  (- (T-rect m) (+ (U-rect g m) (U-constraint R))))

(define (L2-rect-constraint g m R)
  (- (T-rect m) (+ (U-rect g m) (U2-constraint R))))

(show-expression
 ((L-rect-constraint 'g 'm 'R)
  ((F->C q->r)
   (->local 't
            (up 'r 'theta 'lambda)
            (up 'rdot 'thetadot 'lambdadot)))))

(show-expression
 ((L2-rect-constraint 'g 'm 'R)
  ((F->C q->r)
   (->local 't
            (up 'r 'theta 'lambda)
            (up 'rdot 'thetadot 'lambdadot)))))


(show-expression
 (((Lagrange-equations
    (compose (L-rect-constraint 'g 'm 'R) (F->C q->r)))
   (up
    (literal-function 'r)
    (literal-function 'theta)
    (literal-function 'lambda)))
  't))

(show-expression
 (((Lagrange-equations
    (compose (L2-rect-constraint 'g 'm 'R) (F->C q->r)))
   (up
    (literal-function 'r)
    (literal-function 'theta)
    (literal-function 'lambda)))
  't))


(show-expression
 (((Lagrange-equations
    (compose (L-rect-constraint 'g 'm 'R) (F->C q->r)))
   (up
    (lambda (t) 'R)
    (literal-function 'theta)
    (literal-function 'lambda)))
  't))

— Me@2023-07-12 10:00:19 PM

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2023.07.13 Thursday (c) All rights reserved by ACHK

Ex 1.29 A particle of mass m slides off a horizontal cylinder, 2.1

Posted on by

Structure and Interpretation of Classical Mechanics

.

A particle of mass m slides off a horizontal cylinder of radius R in a uniform gravitational field with acceleration g. If the particle starts close to the top of the cylinder with zero initial speed, with what angular velocity does it leave the cylinder?

~~~

.

Kinetic energy:

\displaystyle{\begin{aligned} T(r(t), \theta(t)) &= \frac{1}{2} m \left( v_x^2 + v_y^2 \right) \\ &= \frac{1}{2} m \left[ \left( \frac{d}{dt} r \sin \theta \right)^2 + \left( \frac{d}{dt} r \cos \theta \right)^2 \right] \\ &= \frac{1}{2} m \left[ \left( \dot r \sin \theta + r \dot \theta \cos \theta \right)^2 + \left( \dot r \cos \theta - r \dot \theta \sin \theta \right)^2 \right] \\ &= \frac{1}{2} m \left[ \dot r^2 + r^2 \dot \theta^2 \right] \\ \end{aligned}}

Potential energy:

\displaystyle{\begin{aligned} V(r(t), \theta(t)) &= mgr \cos \theta \\ \end{aligned}}

Constraint:

\displaystyle{\begin{aligned} \phi(t) &= r(t) - R \\ &= 0 \\ \end{aligned}}

.

The Lagrangian:

\displaystyle{\begin{aligned} L(r(t), \theta(t)) &= T - V + \lambda \phi \\ &= \frac{1}{2} m ( \dot r^2 + r^2 \dot \theta^2 ) - mg r \cos \theta + \lambda (r - R) \\ \end{aligned}}

The Lagrange equations:

\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot r} \right) - \frac{\partial L}{\partial r} &= 0 \\ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot \theta} \right) - \frac{\partial L}{\partial \theta} &= 0 \\ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot \lambda} \right) - \frac{\partial L}{\partial \lambda} &= 0 \\ \end{aligned}}

.

\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot r} \right) - \frac{\partial L}{\partial r} &= 0 \\ \frac{d}{dt} m \dot r - \left( m r \dot \theta^2 - mg \cos \theta + \lambda \right) &= 0 \\ m \ddot r &= m r \dot \theta^2 - mg \cos \theta + \lambda \\ \end{aligned}}

.

\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot \theta} \right) - \frac{\partial L}{\partial \theta} &= 0 \\ \frac{d}{dt} \left( m r^2 \dot \theta \right) - mgr \sin \theta &= 0 \\ m \left( 2 r \dot r \dot \theta + r^2 \ddot \theta \right) - mgr \sin \theta &= 0 \\ \end{aligned}}

.

\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot \lambda} \right) - \frac{\partial L}{\partial \lambda} &= 0 \\ - ( r - R ) &= 0 \\ r(t) &= R \\ \end{aligned}}

.

By the constraint \displaystyle{r(t) = R},

\displaystyle{ \begin{aligned} \dot r(t) &= 0 \\ \ddot r(t) &= 0 \\ \end{aligned}}

— Me@2023-06-25 07:47:21 PM

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2023.06.28 Wednesday (c) All rights reserved by ACHK