# Ex 1.18 Bead on a triaxial surface, 2

Structure and Interpretation of Classical Mechanics

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A bead of mass m moves without friction on a triaxial ellipsoidal surface. In rectangular coordinates the surface satisfies

$\displaystyle{\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1}$

for some constants a, b, and c. Identify suitable generalized coordinates, formulate a Lagrangian, and find Lagrange’s equations.

~~~

[guess]

The generalized coordinates:

\displaystyle{\begin{aligned} x &= a \sin(\theta )\cos(\varphi),\\ y &= b \sin(\theta )\sin(\varphi),\\ z &= c \cos(\theta), \end{aligned}}

where

${\displaystyle 0\leq \theta \leq \pi ,\qquad 0\leq \varphi <2\pi .}$

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{\displaystyle {\begin{aligned} \dot x &= a \dot \theta \cos(\theta) \cos(\varphi) - a \dot \varphi\sin(\theta ) \sin(\varphi) \\ \dot y &= b \dot \theta \cos(\theta) \sin(\varphi) + b \dot \varphi \sin(\theta) \cos(\varphi) \\ \dot z &= - c \dot \theta \sin(\theta) \end{aligned}}}


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

;

(define ((e->r a b c) local)
(let ((q (coordinate local)))
(let ((theta (ref q 0))
(phi (ref q 1)))
(let ((x (* a (sin theta) (cos phi)))
(y (* b (sin theta) (sin phi)))
(z (* c (cos theta))))
(up x y z)))))

;

(define ((L-rect m) local)
(let ((q (coordinate local))
(v (velocity local)))
(* 1/2 m (square v))))

(define (L-e m a b c)
(compose (L-rect m) (F->C (e->r a b c))))

;

(show-expression
((L-e 'm 'a 'b 'c)
(->local 't
(up 'theta 'phi)

;

(show-expression
(((Lagrange-equations
(L-e 'm 'a 'b 'c))
(up (literal-function 'theta) (literal-function 'phi)))
't))



[guess]

— Me@2021-03-01 06:22:50 PM

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# Trajectory

It is not possible to derive Schrödinger’s equation from “anything we know”.

— R. P. Feynman

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The most confusing part in the Quantum Mechanics is [the concept of] Trajectory.

There exist[s] no fixed path for a particle to go from Point A to Point B. This is clearly visible from [the] Interference Experiment.

So, the approach here is to work with deductive reasoning. We eliminate the possible region/paths which [are] impossible to be followed.

To do this we assume that Energy Conservation Relation is valid for Quantum Mechanics too. So, those regions where particle[s] [violate] this law automatically [get] eliminated.

Then, we guess [the] State Function[s] for certain conditions i.e. how it should be in certain cases, then build an energy conservation equation with that. We will shortly demonstrate how Schrodinger itself reached the conclusion.

— Why can’t the Schrödinger equation be derived?

— Abhas Kumar Sinha

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2021.02.28 Sunday ACHK

# Deus ex machina 4

This [Hegel’s philosophy] illustrates an important truth, namely, that the worse your logic, the more interesting the consequences to which it gives rise.

― A History of Western Philosophy

― Bertrand Russell

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A good way to have an intellectual stimulation is to let go of a part of logic and see how the story evolves.

For example, when you let go of the time logic, you will get a time travel story.

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let go of a part of logic

~ perturb the reality

~ be thought provocative

~ follow a new set of rules/physical laws

~ 迷魂

— Me@2016-01-29 8:47 AM

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We need to distinguish between real philosophical insights and bewitchments.

— Me@2016-02-02 04:29:34 PM

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… because both can give you the same feeling of thought-provocative-ness.

— Me@2021-02-27 08:55:27 PM

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# 如此仙子 1.3

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— Me@2010.01.28

— Me@2021-02-27 08:40:29 AM

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# Spinister

~ 1988 – 1990

The Transformers of this size were sold 37 HKD each at the Watsons near my home. That was much more expensive than an ordinary size one, which costed only around 25 HKD.

One time, mom brought me to ToyRUs. I found that the Transformers there were much cheaper. This one costed only about 29 HKD.

I did not mind buying a Decepticon.

A much more beautiful one was released last year.

— Me@2021-02-25 12:38:38 PM

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# Problem 2.6c

A First Course in String Theory

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2.5 Constructing $\displaystyle{T^2/\mathbb{Z}_3}$ orbifold

(c) Determine the three fixed points of the $\displaystyle{\mathbb{Z}_3}$ action on the torus. Show that the orbifold $\displaystyle{T^2/\mathbb{Z}_3}$ is topologically a two-dimensional sphere, naturally presented as a triangular pillowcase with seamed edges and corners at the fixed points.

~~~

[guess]

To find the fixed points, we consider the cases when

$\displaystyle{z + m + n e^{i \pi/3} = e^{2 \pi i/3} z}$,

where $\displaystyle{m,n \in \mathbb{Z}}$.

$\displaystyle{(e^{2 \pi i/3} - 1) z = m + n e^{i \pi/3}}$

$\displaystyle{z = \frac{m + n e^{i \pi/3}}{e^{2 \pi i/3} - 1}}$

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When $\displaystyle{m, n = 0}$,

$\displaystyle{z = 0}$

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When $\displaystyle{m = 0; n = 1}$,

$\displaystyle{z = \frac{e^{i \pi/3}}{e^{2 \pi i/3} - 1} = \frac{-i}{\sqrt{3}}}$

When $\displaystyle{m = 1; n = 0}$,

$\displaystyle{z = \frac{1}{e^{2 \pi i/3} - 1} = \frac{1}{\sqrt{3}} e^{- 5 i \pi/6}}$

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When $\displaystyle{m = -2; n = 1}$,

$\displaystyle{z = \frac{-2 + 1 e^{i \pi/3}}{e^{2 \pi i/3} - 1} = 1}$

When $\displaystyle{m = -1; n = -1}$,

$\displaystyle{z = \frac{-1 - 1 e^{i \pi/3}}{e^{2 \pi i/3} - 1} = e^{i \pi/3}}$

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In the fundamental domain, the 3 fixed points are:

$\displaystyle{z = 0}$

when $\displaystyle{z = R(z)}$;

$\displaystyle{z = 1}$

when $\displaystyle{T_2 \circ T_1^{-1} \circ T_1^{-1} (z) = R(z)}$;

$\displaystyle{z = e^{i \pi/3}}$

when $\displaystyle{T_2^{-1} \circ T_1^{-1} (z) = R(z)}$.

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Duplicate the fundamental triangle to create a fundamental parallelogram.

If we label the some edges as $B$ instead of $A$, the fundamental parallelogram will have a sphere topology $\displaystyle{ ABB^{-1}A^{-1} }$.

However, it is not exactly the same as a sphere topology, because a sphere topology would not have the $A=B$ identification.

[guess]

— Me@2021-02-23 03:44:57 PM

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# Particle indistinguishability is the major source of quantum effects, 1.2

However, this definition of “every trajectory is well-defined” has a problem.

If the trajectory concept cannot predict correct experiment results, “the trajectory concept is broken” is only one of the possible causes.

In other words, how can you know the non-classical results (aka quantum effects) are not due to other factors?

— Me@2021-02-15 05:03:20 PM

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This question is exactly what the Bell tests designed for.

No. It is not correct. A Bell test can check whether the trajectory concept is well-defined, but not whether “the trajectory concept is broken” is the major source of quantum randomness.

However, it is the undefinable trajectory concept that makes the superposition, which is a unique and major feature of quantum mechanics.

— Me@2021-02-07 06:03:53 PM

— Me@2021-02-15 10:24:17 PM

— Me@2021-02-21 05:14:55 PM

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To date, all Bell tests have found that the hypothesis of local hidden variables is inconsistent with the way that physical systems behave.

— Wikipedia on Bell test

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Source of quantumness

~ the indistinguishability of cases

~ the individual trajectory of individual particles cannot be well-defined

~ the indistinguishability of particles

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~ “individual” particle has no individuality

~ “individual” particle has no individual identity

— Me@2021-02-06 4:03 PM

— Me@2021-02-15 9:14 PM

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# Logan

Logan is a 2017 American superhero film starring Hugh Jackman as the titular character.

— Wikipedia on Logan (film)

— Me@2017-06-12 09:24:16 PM

— Me@2021-02-21 12:43:45 PM

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The movies X-Men (2000) and X2 (2003) are prerequisites.

— Me@2021-02-21 12:43:45 PM

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2021.02.21 Sunday ACHK

# 大種子論, 2.4

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心靈作業系統

多重自我
虛擬機器

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「心靈作業系統」和「多重自我」的關係是：一個人腦對應於一部電腦，而「多重自我」中的每個「人格」，就是一個「心靈作業系統」，對應於該電腦的其中一個作業系統。

— Me@2021-02-19 07:25:39 PM

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# 1994, 2

— Me@2021-02-18 06:24:37 PM

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# Ex 1.18 Bead on a triaxial surface

Structure and Interpretation of Classical Mechanics

.

A bead of mass m moves without friction on a triaxial ellipsoidal surface. In rectangular coordinates the surface satisfies

$\displaystyle{\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1}$

for some constants a, b, and c. Identify suitable generalized coordinates, formulate a Lagrangian, and find Lagrange’s equations.

~~~

[guess]

The generalized coordinates:

{\displaystyle {\begin{aligned}x&=a\sin(\theta )\cos(\varphi),\\y&=b\sin(\theta )\sin(\varphi),\\z&=c\cos(\theta),\end{aligned}}\,\!}

where

${\displaystyle 0\leq \theta \leq \pi ,\qquad 0\leq \varphi <2\pi .}$

.

{\displaystyle {\begin{aligned} \dot x &= a \dot \theta \cos(\theta) \cos(\varphi) - a \dot \varphi\sin(\theta ) \sin(\varphi) \\ \dot y &= b \dot \theta \cos(\theta) \sin(\varphi) + b \dot \varphi \sin(\theta) \cos(\varphi) \\ \dot z &= - c \dot \theta \sin(\theta) \end{aligned}}}


(define theta (literal-function 'theta))

(define phi (literal-function 'phi))

;

(define (x t) (* 'a (sin (theta t)) (cos (phi t))))

((D x) 't)

(show-expression ((D x) 't))

;

(define (y t) (* 'b (sin (theta t)) (sin (phi t))))

(show-expression (y 't))

((D y) 't)

(show-expression ((D y) 't))



[guess]

— Me@2021-02-16 07:20:25 AM

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# Particle indistinguishability is the major source of quantum effects, 1.1

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If particles are distinguishable, there is no quantum-ness.

Why?

— Me@2021-02-06 4:00 PM

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If there is no particle indistinguishability, all trajectories are distinguishable, then there is no case indistinguishability.

In other words, if every trajectory is well-defined, there is no indistinguishability of cases, even when no detector is installed.

Why?

— Me@2021-02-06 4:01 PM

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In other words, how to define “every trajectory is well-defined” when no detector is installed?

— Me@2021-02-15 5:00 PM

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Thought experiment:

In the double-slit experiment, turn on the detector. Then observe the pattern on the final screen.

Next, tune down the detector’s accuracy/resolution a little bit. Repeat the experiment. Observing the pattern again.

Keep repeating the experiment with a little bit lower detector accuracy/resolution at each iteration.

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If a non-classical pattern never appears on the final screen, we can say that each trajectory is well-defined.

In other words, if the trajectory concept can predict correct experiment results, we say that the trajectory concept is well-defined.

— Me@2021-02-06 4:02 PM

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It is because the final screen itself is a kind of detector, although not a position detector.

— Me@2021-02-06 05:07:21 PM

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So it is a kind of Bell-type experiment.

— Me@2021-02-07 06:03:53 PM

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However, this definition of “every trajectory is well-defined” has a problem.

If the trajectory concept cannot predict correct experiment results, “the trajectory concept is broken” is only one of the possible causes.

In other words, how can you know the non-classical results (aka quantum effects) are not due to other factors?

— Me@2021-02-15 05:03:20 PM

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# Publish! 11

Why do we like to share insights?

So that those insights can have a future.

— Me@2011.08.08

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So that those ideas can continue to exist.

So that those ideas can be upgradable.

— Me@2021-02-15 09:00:18 AM

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# 如此仙子 1.2

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If you begin by sacrificing yourself to those you love, you will end by hating those to whom you have sacrificed yourself.

— George Bernard Shaw

— Me@2021-01-31 05:29:02 PM

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# 2005, 2

— Pinta (software)

— Me@2021-02-13 12:08:52 PM

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# Problem 2.6

A First Course in String Theory

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2.5 Constructing $\displaystyle{T^2/\mathbb{Z}_3}$ orbifold

(a) A fundamental domain, with its boundary, is the parallelogram with corners at $\displaystyle{z = 0, 1}$ and $\displaystyle{e^{i \pi/3}}$. Where is the fourth corner? Make a sketch and indicate the identifications on the boundary. The resulting space is an oblique torus.

(b) Consider now an additional $\displaystyle{\mathbb{Z}_3}$ identification

$\displaystyle{z \sim R(z) = e^{2 \pi i/3} z}$

To understand how this identification acts on the oblique torus, draw the short diagonal that divides the torus into two equilateral triangles. Describe carefully the $\displaystyle{{Z}_3}$ action on each of the two triangles (recall that the action of $\displaystyle{R}$ can be followed by arbitrary action with $\displaystyle{T_1}$, $\displaystyle{T_2}$, and their inverses).

[guess]

(a)

$\displaystyle{z = 1 + e^{\frac{i \pi}{3}}}$

(b)

[guess]

— Me@2021-02-11 06:03:36 PM

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# Quantum information makes classical information consistent, 1.2

Consistent histories, 10.2 | Cosmic computer, 2.2

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Quantum mechanics is a theory of classical information.

— Me@2021-02-03 07:48:01 AM

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Quantum mechanics is a theory of measurement results.

Quantum mechanics explains why measurement results are always consistent with each other.

— Me@2021-02-11 11:10:17 AM

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# 搜神記

— 林夕

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2021.02.10 Wednesday ACHK

# 大種子論

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心靈作業系統

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（安：我們原本的「心靈作業系統」，似乎不是講這些東西。

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（安：你那個「多重自我」理論，和「心靈作業系統」理論，又可不可以合到體呢？）

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（安：但那不能同時運行兩個系統。）

— Me@2021-02-08 12:34:31 AM

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— Me@2011.02.24

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… 後來改稱「解離性身分疾患」（dissociative identity disorder，DID） …

— 維基百科

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