Quick Calculation 3.8

A First Course in String Theory

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Show that this condition fixes uniquely \displaystyle{\alpha = \gamma = 1/2}, and \displaystyle{\beta = - 3/2}, thus reproducing the result in (3.90).

~~~

Eq. (3.93):

\displaystyle{l_P = (G)^\alpha (c)^\beta (\hbar)^\gamma}

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\displaystyle{l_P = \left( \frac{l_p^3}{m_p t_P^2} \right)^\alpha \left( \frac{l_P}{t_P} \right)^\beta \left( \frac{m_P l_P^2}{t_P} \right)^\gamma}

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\displaystyle{\begin{aligned}   3 \alpha + \beta + 2\gamma &= 1 \\   -\alpha + \gamma &= 0 \\   - 2 \alpha - \beta - \gamma &= 0 \\   \end{aligned}}

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var('a b c')

solve([3*a+b+2*c==1, -a+c==0, -2*a-b-c==0], a, b, c)

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\displaystyle{\begin{aligned}   \alpha &= \frac{1}{2} \\ \\  \beta &= \frac{-3}{2} \\ \\  \gamma &= \frac{1}{2} \\ \\  \end{aligned}}

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Eq. (3.90):

\displaystyle{l_P = \sqrt{\frac{G \hbar}{c^3}}}

— Me@2022-06-23 10:46:22 AM

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2022.06.23 Thursday (c) All rights reserved by ACHK

Schrodinger cat’s misunderstanding

Schrodinger’s cat, 3.2

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In 1935, Erwin Schrödinger devised a well-known thought experiment, now known as Schrödinger’s cat, which highlighted this dissonance between quantum mechanics and classical physics.

The main point of the Schrödinger’s cat thought experiment is NOT to prove that there should also be superposition for macroscopic objects. Instead, the main point of the thought experiment is exactly the opposite—to prove that regarding a superposition state as a physical state leads to logical contradiction.

— Me@2022-06-15 07:19:36 PM

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2022.06.22 Wednesday (c) All rights reserved by ACHK

數學教育 7.5.3

A Fraction of Algebra, 2.3

這段改編自 2010 年 4 月 24 日的對話。

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大多數情況反而是,遇到不佳的教師,因為大部分教師,也不懂教學。那就導致,在某一年的數學知識,出現斷層。數學作為理科之心,累積性最高。不幸斷了一層的話,就會同時斷了,之後的每一層。

隨著越來越陌生,那些對數學先天的純真喜愛,也變得不再復見。與數學,再見也不會是朋友。

我比較幸運,在中學時代,遇到一些合理的數學教師。即使在中六七的預科中,一科「應用數學」的教師,未如理想;另一科「純數學」的教師,卻驚為天人。

而我的物理科,則沒有那麼幸運。

留意,我這裡問的是,

你的日校物理教師,有沒有教,實質內容呢?

而不是

你的日校物理教師,教得好不好?

因為「好不好」有時,只是主觀的感受,未必能化成,實質的知識和成績。

我當年就是就正正,犯下這個錯誤。

我中三和中四時覺得,日校物理教師十分好,因為他無論是講物理故事,或物理概念,都十分精采;精采到一個地步是,他啓發了我,對物理的興趣。

但是,當時不知何故,我大部分 MC(多項選擇題)都不懂做。

教師好,而我成績差,很明顯是我的問題。

那是一個無知年輕人的想法。

事隔多年,我終於知道,那不是我的問題。原來我的日校物理教師,極少跟我們研究題目。每類題目應該如何理解,如何運算,要運用什麼概念或技巧等,都是必須教的東西,因為,極少人可以,無師自通。

根本九成九的要點,在題目的解答過程之中。而我的日校教師,偏偏沒有教。很大可能是,其實連他自己也不太懂。

— Me@2022-06-21 11:51:20 AM

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2022.06.21 Tuesday (c) All rights reserved by ACHK

C-h ?


C-h w command-name

C-h k key-sequence

C-h f function-name

— Me@2022-06-18 12:52:10 PM

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2022.06.19 Sunday (c) All rights reserved by ACHK

Ex 1.27 Identifying total time derivatives

Structure and Interpretation of Classical Mechanics

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From equation (1.112), we see that \displaystyle{G} must be linear in the generalized velocities

\displaystyle{    G(t, q, v) = G_0(t, q, v) + G_1(t, q, v) v    }

where neither \displaystyle{G_1} nor \displaystyle{G_0} depend on the generalized velocities: \displaystyle{\partial_2 G_1 = \partial_2 G_0 = 0}.

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So if \displaystyle{G} is the total time derivative of \displaystyle{F} then

\displaystyle{    \partial_0 G_1 = \partial_1 G_0    }

For each of the following functions, either show that it is not a total time derivative or produce a function from which it can be derived.

~~~

[guess]

a. \displaystyle{G(t, x, v_x) = m v_x}

\displaystyle{    \begin{aligned}     G_0 &= 0 \\    G_1 &= m \\ \\    \partial_0 G_1 &= 0 \\     \partial_1 G_0 &= 0 \\ \\     \end{aligned}     }

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\displaystyle{    \begin{aligned}     \partial_0 F &= 0 \\    F &= k_0(x, v_x) \\ \\    \partial_1 F &= m \\     F &= m x + k_1(t, v_x) \\     \end{aligned}     }

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Let

\displaystyle{    \begin{aligned}     k_0(x, v_x) &= mx \\     k_1(t, v_x) &= 0 \\     \end{aligned}     }

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Then

\displaystyle{    \begin{aligned}     F &= mx \\     \end{aligned}     }

[guess]

— Me@2022-06-17 05:10:38 PM

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2022.06.17 Friday (c) All rights reserved by ACHK

Schrodinger’s cat, 3.1

It is natural to ask why ordinary everyday objects and events do not seem to display quantum mechanical features such as superposition. Indeed, this is sometimes regarded as “mysterious”, for instance by Richard Feynman. In 1935, Erwin Schrödinger devised a well-known thought experiment, now known as Schrödinger’s cat, which highlighted this dissonance between quantum mechanics and classical physics.

The modern view is that this mystery is explained by quantum decoherence. A macroscopic system (such as a cat) may evolve over time into a superposition of classically distinct quantum states (such as “alive” and “dead”). However, the state of the cat is entangled with the state of its environment (for instance, the molecules in the atmosphere surrounding it). If one averages over the quantum states of the environment—a physically reasonable procedure unless the quantum state of all the particles making up the environment can be controlled or measured precisely—the resulting mixed quantum state for the cat is very close to a classical probabilistic state where the cat has some definite probability to be dead or alive, just as a classical observer would expect in this situation.

Quantum superposition is exhibited in fact in many directly observable phenomena, such as interference peaks from an electron wave in a double-slit experiment. Superposition persists at all scales, provided that coherence is shielded from disruption by intermittent external factors. The Heisenberg uncertainty principle states that for any given instant of time, the position and velocity of an electron or other subatomic particle cannot both be exactly determined. A state where one of them has a definite value corresponds to a superposition of many states for the other.

— Wikipedia on Quantum superposition

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It is natural to ask why ordinary everyday objects and events do not seem to display quantum mechanical features such as superposition. Indeed, this is sometimes regarded as “mysterious”, for instance by Richard Feynman.

Superposition is not “mysterious”. It is “mysterious” only if you regard “a superposition state” as a physical state.

Only observable states are physical states. Any observable, microscopic or macroscopic, is NOT a superposition.

A superposition is NOT observable, even in principle; because the component states of a superposition are physically-indistinguishable mathematical states, aka macroscopically-indistinguishable microscopic states.

(Those component states, aka eigenstates, are observable and distinguishable once the corresponding measuring device is allowed.)

They are indistinguishable because the distinction is not defined in terms of the difference between different potential experimental or observational results.

Actually, the distinction is not even definable, because the corresponding measuring device is not allowed in the experimental design yet.

— Me@2022-06-15 11:51:22 AM

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2022.06.16 Thursday (c) All rights reserved by ACHK

超整理法

日本著名的經濟學者〈野口悠紀雄〉歸納他個人在整理資料、安排時間的經驗,提出各項實際做法,先後出版《「超級」整理法》、《「超級」整理法續集.時間篇》,連獲好評。

從小老師、父母都教導我們,「整理,一定要先分類」。但如果整理資料的目的,是為了日後需要時快速找到,那麼,對個人而言,「分類」是不可能,而且徒勞無功的。

分門別類時,最常碰到「牆頭草」問題。例如,在設定「土地」類、「稅」類等檔案後,如果出現一份關於「土地課稅」的資料,到底該分到哪一邊?有不少人將無法歸類的資料,分到「其他」檔案,不消數週,「其他」檔的資料很可能多到無法收拾。

— 「超級」整理.超高效率

— 天下雜誌173期

— 孫曉萍

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在野口教授的另一本書《超上班法》中,他提到,應該用年份月份來分類。

— Me@2022-06-14 03:54:57 PM

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2022.06.14 Tuesday ACHK

烹飪事實

這段改編自 2021 年 12 月 14 日的對話。

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費曼與他的太太(琳)有一個協定,盡力不向對方說謊。

琳不幸患上當時的絕症(肺結核)。家人叫費曼暫時不要,將消息告訴琳。家人在場時,費曼對琳謊稱「沒有事」。

但是,家人離開後,費曼即向琳講出實情。琳的第一反應是傷感,但不是因為知道,自己身患不治之症;亦不是因為知道,費曼欺騙了她;而是因為,她覺得費曼剛才被迫講大話,十分委屈。

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即使是「善意謊言」,也不應使用,除非當刻實在迫不得已:真話會令人嚴重受傷,甚至死亡;然後,稍後如果事主的情況安全許可,你就在最短時間內,還原真相給他。

「善意謊言」的最大問題是,「善意」不代表「善行」,更不代表「善果」。到病人最終知道真相時,只加倍傷心。

謊言如借錢,最終必須連本帶利地歸還。謊言不如借錢,因為金錢欠債,可以透過還錢來化解;但是,誠信有損,沒有任何方法修補。

他日你公佈好消息時,病人也不會相信,因為,她仍會覺得,你只是安慰她,不是講真相。

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往地獄之路,往往由善意建築而成的;往天堂之路,則是由善行。

好樹結好果。帶來「善果」的,才是「善行」。

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愛情上,那些原為了避免,引起誤會的所謂「善意謊言」,往往會觸發,更大的誤會。

典型例子有,你去見一位異性老朋友,與她聚舊。雖然只是見朋友,但為免令女朋友誤會,你不告訴她。

到你女朋友發現時,她會質疑,你為何不知會她。明明是光明正大的活動,卻因為所謂的「善意謊言」,而變得不明不白。

然後,你女朋友再會猜想,你還有什麼其他事項隱瞞呢?

你一個所謂「善意謊言」,引致無窮盡的誤會。

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正確的相處技巧是,毋須技巧;如實相告便是。

「我明日下午見,一位異性老朋友。你一同去嗎?」

你女朋友反而會放心,因為,她知道你的政策是「有嗰句,講嗰句」。

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你可以包裝事實,令人聽得舒適一點。包裝了的事實,恰尚好處的話,仍然可以是真相。但你不可以歪曲事實,形成假相。

正如,你可以烹飪食物,令其成熟;亦可以為其加調味料。但是,你不可以下毒。

— Me@2022-06-12 01:08:49 PM

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2022.06.13 Monday (c) All rights reserved by ACHK

Quick Calculation 3.7

A First Course in String Theory

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The force \displaystyle{\vec F} on a test charge \displaystyle{q} in an electric field \displaystyle{\vec E} is \displaystyle{\vec F = q \vec E}. What are the units of charge in various dimensions?

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Eq. (3.74):

\displaystyle{E(r) = \frac{\Gamma\left( \frac{d}{2} \right)}{2 \pi^{\frac{d}{2}}} \frac{Q}{r^{d-1}}}

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\displaystyle{     \begin{aligned}         \vec F &= q \vec E \\     &=  \frac{\Gamma\left( \frac{d}{2} \right)}{2 \pi^{\frac{d}{2}}} \frac{qQ}{r^{d-1}} \\ \\ \\      [\vec F] &=  \left[ \frac{\Gamma\left( \frac{d}{2} \right)}{2 \pi^{\frac{d}{2}}} \frac{qQ}{r^{d-1}} \right] \\     &=  \left[ \frac{1}{1} \frac{q^2}{r^{d-1}} \right] \\     &=  \left[ \frac{q^2}{r^{d-1}} \right] \\     \end{aligned}}

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\displaystyle{     \begin{aligned}             [r^{d-1} \vec F] &= \left[ q^2 \right] \\ \\        \left[ q \right] &= [\sqrt{r^{d-1} \vec F}] \\ \\    &= \sqrt{m^{d-1} N} \\ \\    \end{aligned}}

— Me@2022-06-08 11:09:27 AM

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Lorentz–Heaviside units (or Heaviside–Lorentz units) constitute a system of units (particularly electromagnetic units) within CGS, named for Hendrik Antoon Lorentz and Oliver Heaviside. They share with CGS-Gaussian units the property that the electric constant \displaystyle{\epsilon_0} and magnetic constant \displaystyle{\mu_0} do not appear, having been incorporated implicitly into the electromagnetic quantities by the way they are defined. Heaviside-Lorentz units may be regarded as normalizing \displaystyle{\epsilon_0 = 1} and \displaystyle{\mu_0 = 1}, while at the same time revising Maxwell’s equations to use the speed of light \displaystyle{c} instead.

Heaviside–Lorentz units, like SI units but unlike Gaussian units, are rationalized, meaning that there are no factors of \displaystyle{4 \pi} appearing explicitly in Maxwell’s equations. That these units are rationalized partly explains their appeal in quantum field theory: the Lagrangian underlying the theory does not have any factors of \displaystyle{4 \pi} in these units. Consequently, Heaviside-Lorentz units differ by factors of \displaystyle{\sqrt{4\pi}} in the definitions of the electric and magnetic fields and of electric charge. They are often used in relativistic calculations, and are used in particle physics. They are particularly convenient when performing calculations in spatial dimensions greater than three such as in string theory.

— Wikipedia on Lorentz–Heaviside units

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2022.06.08 Wednesday (c) All rights reserved by ACHK

Emacs, 3

jarcane on Oct 10, 2014 | next [–]

“They used a manual someone had written which showed how to extend Emacs, but didn’t say it was a programming. So the secretaries, who believed they couldn’t do programming, weren’t scared off.”

easytiger on Oct 10, 2014 | parent | next [–]

I was reading a book on a long train journey from Paris to Nice by train that I read usually once a year. My girlfriend couldn’t understand why I was reading a kids book as it had an elephant on the front. I told her it wasn’t, but was in fact one of the most profound books about teaching you to think in a new way.

So I let her read it and she got about half way through and she totally got it and loved it. No harder than doing a crossword or a sudoku for the first time.

The book is: little-schemer

— My Lisp Experiences and the Development of GNU Emacs (2002)

— Hacker News

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2022.06.07 Tuesday ACHK

飯餸湯

這段改編自 2021 年 12 月 13 日的對話。

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最近發現,原來有些女仔,喜歡煮飯煮餸或煲湯,給心上人。換句話說,如果她親自弄東西給你吃,而你又是靚仔的話,她有八成機會喜歡你。

甚至,她是原本不懂烹飪,而特意去學習,目的只是為你製作美食。那樣,她就有九成機會,已愛上了你。

(但是,如果你不是俊男,那就只是兩場誤會,通常。)

換句話說,如果你對她,沒有愛情上的意思,就千萬不要接受,她的飯餸湯示愛。如果你已經有,女朋友或妻子,就要更加小心。

— Me@2022-06-06 05:12:48 PM

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2022.06.06 Monday (c) All rights reserved by ACHK

3Blue1Brown

(defun 3b1b ()

    (interactive)

    (setq is-python-mode (string= major-mode "python-mode"))

    (if (not is-python-mode)
        (print "This is not a python file.")


        (print buffer-file-name)

        (setq the-command (format "%s %s %s" 
                                "manim -p"
                                buffer-file-name  
                                "JustAShape")) 

        (print the-command)                        

        (shell-command the-command) 
    )  
)

(global-set-key (kbd "C-p") '3b1b)

(global-set-key (kbd "C-/") 'comment-region)

(global-set-key (kbd "C-.") 'uncomment-region)

— Me@2022-06-05 04:00:37 PM

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2022.06.05 Sunday (c) All rights reserved by ACHK

Ex 1.26 Lagrange equations for total time derivatives

Structure and Interpretation of Classical Mechanics

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Let \displaystyle{F(t, q)} be a function of \displaystyle{t} and \displaystyle{q} only, with total time derivative

\displaystyle{D_t F = \partial_0 F + \partial_1 F \dot Q}

Show explicitly that the Lagrange equations for \displaystyle{D_t F} are identically zero, …

~~~

[guess]

The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ \end{aligned}}

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

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Eq. (1.114):

\displaystyle{  \begin{aligned}  D_t F (t, q, v, a, ...)   &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ...   \\   \end{aligned}  }

\displaystyle{  \begin{aligned}  D_t F \circ \Gamma[q] (t)  &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v(t) + \partial_2 F(t, q, v, a, ...) a(t) + ...   \\   \end{aligned}  }

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Put \displaystyle{ D_t F (t, q, Dq, ...) } into the Lagrange equation:

\displaystyle{     \begin{aligned}     &\frac{d}{dt} \left( \frac{\partial}{\partial \dot q}     D_t F (t, q, Dq, ...)  \right) - \frac{\partial}{\partial q}     D_t F (t, q, Dq, ...) \\ \\    &= \frac{d}{dt} \left[ \frac{\partial}{\partial \dot q}     \left( \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ... \right) \right] \\    &- \frac{\partial}{\partial q}     \left( \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ... \right)   \\     \\    &=\frac{d}{dt} \left[        \partial_2 \partial_0 F  + \partial_2 (v \partial_1 F)   + \partial_2 (a \partial_2 F) + ...  \right] \\    &- \left[    \partial_1 \partial_0 F  + \partial_1 (v \partial_1 F)   + \partial_1 (a \partial_2 F) + ... \right]   \\    \\    &= \partial_0 \left[          \partial_1 \partial_0 F  + \partial_1 (v \partial_1 F)   + \partial_1 (a \partial_2 F) + ...  \right] \\     &+ \partial_1 \left[          \partial_1 \partial_0 F  + \partial_1 (v \partial_1 F)   + \partial_1 (a \partial_2 F) + ...   \right] v \\    &+ \partial_2 \left[          \partial_1 \partial_0 F  + \partial_1 (v \partial_1 F)   + \partial_1 (a \partial_2 F) + ...   \right] a + ... \\    &- \left[      \partial_1 \partial_0 F  + \partial_1 (v \partial_1 F)   + \partial_1 (a \partial_2 F) + ...  \right]   \\        \\    &= ... \\     \end{aligned}    }

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Assume that \displaystyle{\frac{\partial \dot F}{\partial \dot q} = \frac{\partial F}{\partial q}}:

\displaystyle{     \begin{aligned}     &\frac{d}{dt} \left( \frac{\partial}{\partial \dot q}     D_t F (t, q, Dq, ...)  \right) - \frac{\partial}{\partial q}     D_t F (t, q, Dq, ...) \\ \\    &= \frac{d}{dt} \left( \frac{\partial  \dot F }{\partial \dot q}       \right) - \frac{\partial}{\partial q} \frac{d F}{dt}  \\ \\    &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \frac{\partial}{\partial q} \frac{dF}{dt} \\ \\    &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \frac{\partial}{\partial q} \left( \frac{\partial F}{\partial t}     + \frac{\partial F}{\partial q} \dot q     + \frac{\partial F}{\partial \dot q} \ddot q + ...  \right) \\ \\ \end{aligned}}

\displaystyle{\begin{aligned}  &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \left( \frac{\partial \frac{\partial F}{\partial q}}{\partial t}    + \frac{\partial \frac{\partial F}{\partial q}}{\partial q} \dot q     + \frac{\partial F}{\partial q} \frac{\partial}{\partial q} \dot q     + \frac{\partial \frac{\partial F}{\partial q}}{\partial \dot q} \ddot q     + \frac{\partial F}{\partial \dot q} \frac{\partial}{\partial q}    \ddot q     + ... \right) \\ \\     &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) - \left( \frac{\partial \frac{\partial F}{\partial q}}{\partial t}    + \frac{\partial \frac{\partial F}{\partial q}}{\partial q} \dot q     + \frac{\partial F}{\partial q} (0)    + \frac{\partial \frac{\partial F}{\partial q}}{\partial \dot q} \ddot q     + \frac{\partial F}{\partial \dot q} (0)    + ... \right) \\ \\     &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) -      \left( \frac{\partial \frac{\partial F}{\partial q}}{\partial t}     + \frac{\partial \frac{\partial F}{\partial q}}{\partial q} \dot q     + \frac{\partial \frac{\partial F}{\partial q}}{\partial \dot q} \ddot q + ...  \right) \\ \\    &= \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right) -      \frac{d}{dt} \left( \frac{\partial F}{\partial q} \right)  \\ \\    &= 0 \\ \\     \end{aligned}    }

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Prove that \displaystyle{\frac{\partial \dot F}{\partial \dot q} = \frac{\partial F}{\partial q}}.

\displaystyle{\begin{aligned}            &\frac{\partial \dot F}{\partial \dot q} \\ \\      &=     \frac{\partial }{\partial \dot q} \left( \frac{\partial F}{\partial t}     + \frac{\partial F}{\partial q} \dot q + \frac{\partial F}{\partial \dot q} \ddot q + ... \right)   \\ \\    &= \frac{\partial }{\partial \dot q} \frac{\partial F}{\partial t}     + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial q} \right) \dot q   + \frac{\partial F}{\partial q} \frac{\partial }{\partial \dot q}\dot q      + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial \dot q} \right) \ddot q    + \frac{\partial F}{\partial \dot q} \frac{\partial }{\partial \dot q}\ddot q     + ... \\ \\        &= \frac{\partial }{\partial \dot q} \frac{\partial F}{\partial t}     + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial q} \right) \dot q   + \frac{\partial F}{\partial q} (1)      + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial \dot q} \right) \ddot q    + \frac{\partial F}{\partial \dot q} (0)     + ... \\ \\      &= \frac{\partial }{\partial \dot q} \frac{\partial F}{\partial t}     + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial q} \right)\dot q   + \frac{\partial F}{\partial q}       + \left( \frac{\partial }{\partial \dot q}\frac{\partial F}{\partial \dot q}\right) \ddot q         + ... \\ \\            &=       \left[ \frac{\partial \frac{\partial }{\partial \dot q} F}{\partial t}     + \left( \frac{\partial \frac{\partial }{\partial \dot q} F}{\partial q} \right)\dot q         + \left( \frac{\partial \frac{\partial }{\partial \dot q} F}{\partial \dot q}\right) \ddot q      + ... \right]      + \frac{\partial F}{\partial q}     \\ \\          &= \frac{d}{dt} \frac{\partial F }{\partial \dot q}     + \frac{\partial F}{\partial q}      \\ \\    \end{aligned}}

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If \displaystyle{ L' = L + D_t F } depends on \displaystyle{ \{t, q, Dq\} } only, then \displaystyle{ F } will depend on \displaystyle{ \{t, q\} } only.

\displaystyle{\begin{aligned}     \frac{\partial \dot F}{\partial \dot q}     &= \frac{d}{dt} \frac{\partial F }{\partial \dot q} + \frac{\partial F}{\partial q} \\ \\     &= \frac{d}{dt} (0) + \frac{\partial F}{\partial q} \\ \\     &= \frac{\partial F}{\partial q} \\ \\     \end{aligned}}

[guess]

— Me@2022-06-03 09:04:37 PM

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2022.06.05 Sunday (c) All rights reserved by ACHK

PS2

CPU: Core 2 Duo 3.2 GHz / Core-I series
GPU: GeForce 9600 GT
2GB RAM

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2022.06.03 Friday ACHK

電腦病毒

這段改編自 2010 年 10 月 14 日的對話。

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很多時,「解決」問題最好的方法是,令到自己毋須,解決那個問題。

例如,你現在的問題是,本年是你的高考年。每天回家後,理應操練以往試題。但實情是,每天回家後,也會忍不住,打開電腦一會兒。而那「一會兒」卻往往會,橫跨數小時。

解決的方法是,訓練到自己每天回家後,忍得住不打開電腦。(我自己就在預科那兩年間,從來沒有開電腦,幾乎。)

更好的方法是,令到自己根本,毋須這個訓練。那就是每天下課後,留在學校完成了,自己給自己的試題操練目標,才回家。(這裡假設了,操練題目數小時後,回家仍然不會太晚。)

既然「每天回家後,也會忍不住,打開電腦一會兒」,倒不如索性,在完成功課前,暫時刪除了「每天回家後」這情節,那就完全避開了,「也會忍不住,打開電腦一會兒」這劇情。

跟壞人相處的最好方法是,不要跟他相處。不跟壞人相處的話,就毋須研究「如何跟壞人相處」。

完成操練試題前,暫時不讓「每天回家後的自己」出現,你就毋須研究,如何駕馭對付他。

— Me@2022-06-02 11:52:12 AM

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2022.06.02 Thursday (c) All rights reserved by ACHK