Problem 2.3b2

Prove that a metric tensor is symmetric.

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Assume \displaystyle{\eta_{\alpha\beta} \neq \eta_{\beta\alpha}}. Because it’s irrelevant what letter we use for our indices,

\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}.

Then

\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2} (\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}}

So only the symmetric part of \displaystyle{\eta_{\alpha\beta}} would survive the sum. As such we may as well take \displaystyle{\eta_{\alpha\beta}} to be symmetric in its definition.

— edited Jun 15 ’15 at 22:48

— rob

— answered Jun 15 ’15 at 17:52

— FenderLesPaul

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— Why is the metric tensor symmetric?

— Physics StackExchange

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1.

\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}

means that

\displaystyle{\sum_{\alpha, \beta} \eta_{\alpha\beta}dx^{\alpha}dx^{\beta}=\sum_{\alpha, \beta}\eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}

So in

\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}},

we cannot cancel out \displaystyle{dx^{\alpha}dx^{\beta}} on both sides. In other words, we do NOT assume that \displaystyle{\eta_{\alpha\beta} = \eta_{\beta\alpha}} in the first place.

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2.

\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2} (\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}}

means that

\displaystyle{\sum_{\alpha, \beta}\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}\sum_{\alpha, \beta}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2} \sum_{\alpha, \beta}(\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}}

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3. “… only the symmetric part of \displaystyle{\eta_{\alpha\beta}} would survive the sum” means that only the sum \displaystyle{\left(\eta_{\alpha\beta} + \eta_{\beta\alpha}\right)} is physically meaningful.

— Me@2020-08-14 03:34:05 PM

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2020.08.14 Friday (c) All rights reserved by ACHK

Spacetime rate

Every motion in space is also a motion in time.

The speed of light is the upper limit of the spacetime rate.

— Me@2012-04-28 12:42:08 PM

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An object cannot change this position without changing its time coordinate.

In short, there is no instantaneous motion.

— Me@2020-08-12 05:26:20 PM

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2020.08.12 Wednesday (c) All rights reserved by ACHK

機遇創生論 1.6.3

一萬個小時 2 | 十年 3

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任何一門通用知識,起點都是「學海無涯,唯勤是岸」,終點都是「學海無涯,回頭是岸」。

「通用知識」的意思是,每人日常也需要知道的東西,例如,健康、財政、人際、時間管理等。

As we get older, generic reading becomes less and less useful. We then gain new knowledge mostly by personal life experience and directed reading.

— paraphrasing John T. Reed

那是指一般「通用知識」。

directed reading ~ research

至於「專業知識」,即是「技術細節」,結構和「通用知識」雖有相似之處,但不大一樣。

任何一門專業知識,由初學至成熟,則要大概十年的刻意用功。

這裡「成熟」的意思是「可用」;而「可用」的意思是「可以用來維生」。亦即是話,任何一門專業知識,由初學至可以用來作,長久穩定的賺錢工具,需要大概一萬個小時的有意識磨煉。

這裡的「十年」或「一萬個小時」,並不是精細的估計,而是數量級估計。換句話說,可以是七年、十二年,但絕不會是一年。

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這就是「專業知識」和「通用知識」結構相似的地方——都是由最初的「學海無涯,唯勤是岸」,發展到「學海無涯,回頭是岸」。

而結構不相似的地方就是,對於「專業知識」而言,由「唯勤是岸」到「回頭是岸」的過程,需要長很多的時間(十年一萬小時)。二來,對於「專業知識」而言,「回頭是岸」(純熟到可以用來維生)只是里程碑而已,而不是終點。

— Me@2020-08-11 03:09:13 PM

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2020.08.11 Tuesday (c) All rights reserved by ACHK

Ex 1.8.2.4 Implementation of $\delta$

Structure and Interpretation of Classical Mechanics

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Verify the product rule of variation (Equation 1.23) using the
scmutils software library:

\displaystyle{\delta_\eta \left(f [q] g [q] \right)  = \left( \delta_\eta f[q] \right) g[q]  + f[q] \delta_\eta g[q]}

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(define (((delta eta) f) q)
  (define (g epsilon)
    (f (+ q (* epsilon eta))))
  ((D g) 0))

(define q (literal-function 'q (-> Real (UP Real))))

(define eta (literal-function 'eta (-> Real (UP Real))))

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(define (f q)
   (compose (literal-function 'f
              (-> (UP Real (UP* Real) (UP* Real)) Real))
            (Gamma q)))

(define (g q)
   (compose (literal-function 'g
              (-> (UP Real (UP* Real) (UP* Real)) Real))
            (Gamma q)))

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(define (f_times_g q) (* (f q) (g q)))

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(define LHS ((((delta eta) f_times_g) q) 't))

(define RHS (+ (* ((((delta eta) f) q) 't) ((g q) 't))
               (* ((f q) 't) ((((delta eta) g) q) 't))))

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(print-expression LHS)

(show-expression LHS)

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\displaystyle{g D \eta \partial_2 f + g \eta \partial_1 f + f D \eta \partial_2 g + f \eta \partial_1 g}

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(print-expression RHS)

(show-expression RHS)

(- LHS RHS)

— Me@2020-08-06 07:23:27 PM

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2020.08.06 Thursday (c) All rights reserved by ACHK

The square root of the probability, 2

Mixed states, 4.2

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Superposition in quantum mechanics is a complex number superposition.

— Me@2017-08-02 02:56:23 PM

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Superposition in quantum mechanics is not a superposition of probabilities.

Instead, it is a superposition of probability amplitudes, which have complex number values.

Probability amplitude, in a sense, is the square root of probability.

— Me@2020-08-04 03:38:43 PM

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2020.08.04 Tuesday (c) All rights reserved by ACHK

大腦物理性損毀不是比喻, 2

權力來源 2.1

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大多數人相信沒用,苟且偷生,什麼都不做,少數人去做,就真的沒用,而且還要被大多數人背叛出賣。大多數人相信有用,去做一點點,就真的有用。所以有用沒用不是獨立於人的行為的客觀事實,而純粹是人選擇的結果,這是典型的博弈問題。您國命運就是您國人選擇的結果,也就是所謂報應,不能怪任何別人[。]

— 李穎

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2020.08.04 Tuesday ACHK

太極滅世戰 2.1

似水流年 2.2

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還記得童年時不太開心,好像沒有太多事情,令我快樂。換句話說,即是過得沉悶。一方面喜歡讀書,另一方面卻緊張成績,終日惶恐。

近乎唯一娛樂就是看電視。那時,我還沒有電腦,更加沒有互聯網。那時,沒有什麼金錢買玩具。就算有新玩具,我有時會想:「怎麼辦呢? 很快又會厭倦。」

中二開始,發覺對數學幾有興趣。

中途又有些所謂的嗜好,例如集郵,其實沒有意義。相對沒有那麼無聊的,有電腦遊戲。我的摯愛有 Simcity 2000 和 Chrono Trigger 等。但那也只是暫時的快樂,仍未能為我生活,帶來意義;遊戲完結時,感到額外的空虛。

中四上物理時,不知何故,精神上,有一點清涼暢快的感覺。那就代表著,疑似喜歡物理。但是,不知何故,幾乎凡是需要物理思考的題目,我都不懂得做。

一九九六年,逛書店時,看到有一本書叫《時間簡史》。我同學說:「這書(內容)就是你最喜愛的那些東西。」那時,我尚未有觸動心靈的感覺,所以不以為意,沒有購買。

那年暑假,看了香港太空館天象廳的《輪椅中的宇宙》,十分震撼。我大概記得,那是我自己一個去看,沒有朋友一起。

dsc025651

從紀錄片中,我認識了霍金,知道了《時間簡史》的來源。不久之後,我買了《時間簡史》,企圖一口氣讀完它。過程中,我有惶恐不安的感覺。原因不是書中內容,而是我沒有閱讀的習慣。換句話說,《時間簡史》令我開始了,培養閱讀的習慣。

《輪椅中的宇宙》和《時間簡史》令我發現,我的人生目標是物理。

但是,我自中三開始,那些物理題目,真是不太懂做呀。空有目標,沒有執行之道。

其實,我那時沒有也辦法,因為,我日校物理教師,講物理講得十分有趣,而我又有留心聽講。如果在這個情況下,我仍然不懂的話,那大概是我的智力問題。

幸好,後來發現,那不是事實。至起碼,中五會考物理,並不需要高智力,也可以奪得 A 級成績。十多年後,在自己有足夠的讀書和教學經驗後,我發現,我當年不懂做題目,主因是日校物理教師,只談理論,不會教題目細節。甚至,有些課題,例如電學,連他自己也根本不太認識。

一九九六年的那個暑假,我開始參加 Ken Chan 的物理補習班,開始一步一步的,解決了「空有目標,沒有執行之道」這問題。

有了「做物理學家」這長遠目標後,我感到人生不再一樣,雖然,我讀書的日常生活,仍然十分混亂,

— Me@2020-07-27 07:33:45 AM

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2020.08.03 Monday (c) All rights reserved by ACHK

Problem 2.3b1

A First Course in String Theory

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2.3 Lorentz transformations, derivatives, and quantum operators.

(b) Show that the objects \displaystyle{\frac{\partial}{\partial x^\mu}} transform under Lorentz transformations in the same way as the \displaystyle{a_\mu} considered in (a) do. Thus, partial derivatives with respect to conventional upper-index coordinates \displaystyle{x^\mu} behave as a four-vector with lower indices – as reflected by writing it as \displaystyle{\partial_\mu}.

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\displaystyle{ \begin{aligned}  (x')^\mu &= L^\mu_{~\nu} x^\nu \\ \frac{\partial}{\partial (x')^\mu}  &= \frac{\partial x^\nu}{\partial (x')^\mu} \frac{\partial}{\partial x^\nu} \\  &= \frac{\partial x^0}{\partial (x')^\mu} \frac{\partial}{\partial x^0}  + \frac{\partial x^1}{\partial (x')^\mu} \frac{\partial}{\partial x^1}  + \frac{\partial x^2}{\partial (x')^\mu} \frac{\partial}{\partial x^2}  + \frac{\partial x^3}{\partial (x')^\mu} \frac{\partial}{\partial x^3}  \\  \end{aligned}}

The Lorentz transformation:

\displaystyle{ \begin{aligned}  (x')^\mu &= L^\mu_{~\nu} x^\nu \\  \end{aligned}}

Lowering the indices to create covariant vectors:

\displaystyle{ \begin{aligned}  x_\mu &= \eta_{\mu \nu} x^\nu \\  \end{aligned}}

In matrix form, covariant vectors are represented by row vectors:

\displaystyle{ \begin{aligned}  \left[ x_\mu \right] &= \left( [\eta_{\mu \nu}] [x^\nu] \right)^T \\  \end{aligned}}

Change the subject:

\displaystyle{ \begin{aligned}  \left[ x_\mu \right]^T &= [\eta_{\mu \nu}] [x^\nu] \\  [\eta_{\mu \nu}] [x^\nu] &= \left[ x_\mu \right]^T  \\  [x^\nu] &= [\eta_{\mu \nu}]^{-1} \left[ x_\mu \right]^T  \\  \end{aligned}}

With \displaystyle{ \begin{aligned}  \eta^{\mu \nu} &\stackrel{\text{\tiny def}}{=} \left[ \eta_{\mu \nu} \right]^{-1}   \\  \end{aligned}}, we have:

\displaystyle{ \begin{aligned}  \left[ x^\nu \right] &= \left[ \eta^{\mu \nu} \right] \left[ x_\mu \right]^T  \\  \end{aligned}}

\displaystyle{ \begin{aligned}  x^\nu &= x_\mu \eta^{\mu \nu}  \\  \end{aligned}}

Now we lower the indices in order to find the Lorentz transformation for the covariant components:

\displaystyle{ \begin{aligned}  (x')^\mu &= L^\mu_{~\nu} x^\nu \\  \eta^{\rho \mu} x_\rho &= L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\  x_\rho &= \eta_{\rho \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\  \end{aligned}}

— Me@2020-07-21 10:46:32 AM

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2020.07.22 Wednesday (c) All rights reserved by ACHK

Quantum entanglement, 4

What’s sneaky about quantum mechanics is that the whole system can be in a pure state which when restricted to each subsystem gives a mixed state, and that these mixed states are then correlated (necessarily, as it turns out). That’s what “entanglement” is all about.

The first way things get trickier in quantum mechanics is that something we are used to in classical mechanics fails. In classical mechanics, pure states are always dispersion-free — that is, for every observable, the probability measure assigned by the state to that observable is a Dirac delta measure, that is, the observable has a 100% chance of being some specific value and a 0% chance of having any other value. (Consider the example of the dice, with the observable being the number of dots on the face pointing up.) In quantum mechanics, pure states need NOT be dispersion-free. In fact, they usually aren’t.

A second, subtler way things get trickier in quantum mechanics concerns systems made of parts, or subsystems. Every observable of a subsystem is automatically an observable for the whole system (but not all observables of the whole system are of that form; some involve, say, adding observables of two different subsystems). So every state of the whole system gives rise to, or as we say, “restricts to,” a state of each of its subsystems. In classical mechanics, pure states restrict to pure states. For example, if our system consisted of 2 dice, a pure state of the whole system would be something like “the first die is in state 2 and the second one is in state 5;” this restricts to a pure state for the first die (state 2) and a pure state for the second die (state 5). In quantum mechanics, it is not true that a pure state of a system must restrict to a pure state of each subsystem.

It is this latter fact that gave rise to a whole bunch of quantum puzzles such as the Einstein-Podolsky-Rosen puzzle and Bell’s inequality. And it is this last fact that makes things a bit tricky when one of the two subsystems happens to be you. It is possible, and indeed very common, for the following thing to happen when two subsystems interact as time passes. Say the whole system starts out in a pure state which restricts to a pure state of each subsystem. After a while, this need no longer be the case! Namely, if we solve Schroedinger’s equation to calculate the state of the system a while later, it will necessarily still be a pure state (pure states of the whole system evolve to pure states), but it need no longer restrict to pure states of the two subsystems. If this happens, we say that the two subsystems have become “entangled.”

— December 16, 1993

— This Week’s Finds in Mathematical Physics (Week 27)

— John Baez

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2020.07.19 Sunday ACHK

追求, 2

We talked half the night, and in the middle of talk became lovers.

— Bertrand Russell

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2020.07.18 Saturday ACHK

機遇創生論 1.6.2

這段改編自 2010 年 4 月 18 日的對話。

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那就有如在該門知識中,你已經找齊七粒龍珠,「神龍」出現了,你可以實現任何願望。

所以,其實不用太緊張。與其懼怕有所遺漏,倒不如活在當下,享受獲取知識的過程。根本,並沒有需要,去窮盡「所有」的書籍。正如,學習英文時,你有需要學懂,英文字典中的所有詞匯嗎?

The man who grasps principles can successfully select his own methods. The man who tries methods, ignoring principles, is sure to have trouble.

— Ralph Waldo Emerson

任何一門知識,起點都是「學海無涯,唯勤是岸」,終點都是「學海無涯,回頭是岸」。

「通用知識」的意思是,每人日常也需要知道的東西,例如,健康、財政、人際、時間管理等。

As we get older, generic reading becomes less and less useful. We then gain new knowledge mostly by personal life experience and directed reading.

— paraphrasing John T. Reed

那是指一般「通用知識」。

至於「專業知識」,即是「技術細節」,結構和「通用知識」雖有相似之處,但不大一樣。

— Me@2020-07-16 06:14:51 PM

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2020.07.17 Friday (c) All rights reserved by ACHK

反貼士搵笨大行動 1.5

中六時,我日校的同學中,有些在中五時和我一樣,都是補 Ken Chan 的物理班。升上中六後,他們大部分也補 MC Chan 的物理班。我在中六時則沒有補習。我在中六升中七的暑假,才去上 MC Chan 的物理班。

我的那個年代,有很多補習班也以「貼中考試題目」作為招徠。有一次上課時,MC Chan 說:

如果到咗呢個時候,你仍然相信,世間上有『貼士』嘅話,咁你都無資格讀大學啦。

(如果到了這個時候,你仍然相信,世間上有『貼士』的話,你大概沒有資格讀大學吧。)

他說,出題人員有著極高的月薪(當年七萬多港元)。他們根本沒有任何動機去貪污。

他又解釋,他當年所謂的「貼中」題目,是如何達成的。當年他在高考前,預測題目會有「光電效應」。結果,有一題長題目,真的是考「光電效應」。而方法則很簡單,就是每年也預測同一個課題。他每年也預測「光電效應」,總有一年會「貼中」。然後補習社就可以,大肆宣傳。

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其實,根本不用他提點,一般人只要冷靜一點,也可以想到,「貼士」近乎沒有可能。

第一,如果有一位補習導師的「貼題」命中率,真的十分高的話,他為何還未被廉政公署拘捕呢?

第二,或者有人會為他辯護:「可能他有超能力,能知過去未來。運用超能力,是完全合法的。」

如果是那樣的話,他在公開試前,開幾堂就可以;為何一科要上那麼多堂,收我那麼多學費?

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即使命中了題目,那代表你會懂答嗎?

即使命中了題目,那代表單憑該題的分數,你就會有上乘成績嗎?

— Me@2020-07-10 04:13:13 PM

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2020.07.10 Friday (c) All rights reserved by ACHK