講道理 

情緒病 2 | 不思考之道 6

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年紀漸長,我越來越覺得「(太)喜歡講道理」,或者「(太)喜歡聽道理」,是一種情緒病。

— Me@2012.02.14

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2018.11.01 Thursday (c) All rights reserved by ACHK

Posted in OCD

Ken Chan 時光機 1.3

唔識就飛 8.4.2

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另一個技巧是,考試時「唔識就飛」:想不通的話,就立刻先做下一題,或者同一題的下一部分。

假設我現時記憶正確,Ken Chan 是第一個教我這個技巧的老師;我日校的純數學科老師,則是第二個(亦是最後一個)。

這個技巧,對我來說,實在太重要,因為,在兩年多後的純數學科高考中,如果我不是確切執行這個政策的話,我有極大機會會失手,繼而在當年,升不到大學。

那個故事,現不再詳述,因為已於另一篇文章「唔識就飛 8.4」中講過。請參考該文章。

— Me@2018-10-31 09:39:05 AM

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2018.10.31 Wednesday (c) All rights reserved by ACHK

如有 HKDSE physics 香港中學文憑試之物理問題,歡迎以右邊之電郵地址,聯絡本人。

funcall

In Common Lisp, apply can take any number of arguments, and the function given first will be applied to the list made by consing the rest of the arguments onto the list given last. So the expression

(apply #’+ 1 ’(2))

is equivalent to the preceding four. If it is inconvenient to give the arguments as
a list, we can use funcall, which differs from apply only in this respect. This expression

(funcall #’+ 1 2)

has the same effect as those above.

— p.13

— On Lisp

— Paul Graham

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Exercise 7.1

Define funcall.

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(defmacro our-funcall (f &rest p)
  `(apply ,f (list ,@p)))

— Me@2018-10-30 03:24:05 PM

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2018.10.30 Tuesday (c) All rights reserved by ACHK

Problem 14.5b2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(b) Consider the left R’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

We have 32 zero modes \displaystyle{\lambda_0^A} and 16 linear combinations behave as creation operators.

As usual half of the ground states have \displaystyle{(-1)^{F_L} = +1} and the other half have \displaystyle{(-1)^{F_L} = -1}.

Let \displaystyle{|R_\alpha \rangle_L} denote ground states with \displaystyle{(-1)^{F_L} = +1}.

How many ground states \displaystyle{|R_\alpha \rangle_L} are there?

~~~

— This answer is my guess. —

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\ \end{aligned}}

p.315 “Being zero modes, these creation operators do not contribute to the mass-squared of the states. Postulating a unique vacuum \displaystyle{|0 \rangle}, the creation operators allow us to construct \displaystyle{16 = 2^4} degenerate Ramond ground states.”

Following the same logic:

Postulating a unique vacuum \displaystyle{|0 \rangle}, the creation operators allow us to construct \displaystyle{2^{16}} degenerate Ramond ground states.

Therefore, there are \displaystyle{2^{15}} ground states \displaystyle{|R_\alpha \rangle_L}.

— This answer is my guess. —

— Me@2018-10-29 03:11:07 PM

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2018.10.29 Monday (c) All rights reserved by ACHK

Visualizing higher dimensions

The trick of visualizing higher dimension is: not to visualize it.

— Wikipedia

— Me@2011.08.19

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Besides trying to visualize, there are other methods to understand higher dimensions.

— Me@2018-10-28 04:28:01 PM

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What is the meaning of visualization?

— Me@2018-09-02 4:35 pm

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feel ~ receive all the data at once

(This definition is not totally correct, but is useful in the meantime.)

visual ~ feel at once through eyes

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you can visualize a 3D object ~ you can see all of a 3D object at once

you cannot visualize a 4D object ~ you cannot see all of a 4D object at once

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Actually, you can only visualize a 2D object, such as a square.

You cannot visualize a 3D object, such as a cube.

That’s why the screen of any computer monitor is 2 dimensional, not 3.

— Me@2018-10-28 04:32:41 PM

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2018.10.28 Sunday (c) All rights reserved by ACHK

A pretty girl, 2

Women are directly fitted for acting as the nurses and teachers of our early childhood by the fact that they are themselves childish, frivolous and short-sighted; in a word, they are big children all their life long–a kind of intermediate stage between the child and the full-grown man, who is man in the strict sense of the word. See how a girl will fondle a child for days together, dance with it and sing to it; and then think what a man, with the best will in the world, could do if he were put in her place.

— Of Women

— Arthur Schopenhauer

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When the elderly Schopenhauer sat for a sculpture portrait by the Prussian sculptor Elisabet Ney in 1859, he was much impressed by the young woman’s wit and independence, as well as by her skill as a visual artist. After his time with Ney, he told Richard Wagner’s friend Malwida von Meysenbug, “I have not yet spoken my last word about women. I believe that if a woman succeeds in withdrawing from the mass, or rather raising herself above the mass, she grows ceaselessly and more than a man.

— Wikipedia on Arthur Schopenhauer

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Anybody can look at a pretty girl and see a pretty girl. An artist can look at a pretty girl and see the old woman she will become. A better artist can look at an old woman and see the pretty girl that she used to be. But a great artist — a master — and that is what Auguste Rodin was — can look at an old woman, portray her exactly as she is… and force the viewer to see the pretty girl she used to be…. and more than that, he can make anyone with the sensitivity of an armadillo, or even you, see that this lovely young girl is still alive, not old and ugly at all, but simply prisoned inside her ruined body. He can make you feel the quiet, endless tragedy that there was never a girl born who ever grew older than eighteen in her heart…. no matter what the merciless hours have done to her. Look at her, Ben. Growing old doesn’t matter to you and me; we were never meant to be admired — but it does to them. Look at her! (UC)

— Robert A. Heinlein

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2018.10.27 Saturday ACHK

凌晨舊戲 2

這段改編自 2010 年 4 月 18 日的對話。

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(問:你閱讀過很多有關「瀕死經驗」的文章?)

可以這樣說。

如果你閱讀那些文章的話,要小心一點,因為那類文章良莠不齊——當中有些文章發人深省,有些則謊話連篇

(問:那你怎樣分辨,「瀕死經驗」的文章之中,哪些是真,哪些為假?)

看看文中所說的,合不合理。

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例如,你怎樣知道,我說的話,是真還是假?

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合情合理,自圓其說的,就有機會真;無情無理,自相矛盾的,則必定為假。

——

如果,合情合理之餘,我感覺到它說的道理,遠超過我當時境界的話,那道理的可信程度,就再高一點。

(問:那亦可能只是你的誤會。按常理,程度較低的人不會知道,所謂「程度高的人或說話」的真假。

例如,如果你沒有學過物理,當一個物理學家,介紹核電廠如何運作時,其實,你並不會百分百肯定,他講的東西,是真還是假;除非,他的言論,明顯自相矛盾;那樣,你就可以立刻肯定,那是假的。)

所以,我只提及「可信度」,而沒有一口斷定「真假」。

其實,有更日常生活的例子:

我因為不是醫生,醫學知識遠低於醫生,有病時才會考慮去醫生。而正正是因為,我的醫學知識遠低於醫生,我並沒有能力,可以肯定一個醫生的建議是正確的。但是,我總不有能因為那樣,而任何情況下,都不去看醫生。

判斷一位醫生可不可信,唯有靠(他人或自己)的實證。

合理的做法是,如果一個醫生的建議,通常令你的病情舒緩的話,他的說話,就為之「可信」。相反,如果一個醫生的治療,只是「間中」令你病情舒緩,而「間中」到一個程度,令你覺得那跟隨機沒有分別的話,他的說話,就為之「不可信」。

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當然,「瀕死經驗」並沒有所謂「實證」。你總不能叫我,刻意製造「瀕死經驗」,冒生命危險,去檢驗其真偽。

但是,留意,即使是剛才「判別醫生說話真偽」的例子,所講的「實證」,其實也只是,間接的實證。如果你要直接的實證,你就真的需要,先入醫學院,受訓成為醫生,才接近百分百地肯定,其他醫生的建議,是真還是假。

但是,那成本太高。在一般情況下,常人也不應那麼誇張。利用一個醫生的治療準繩度,來釐定其可信度,才是合理可行的方法。

——

同理,如果某人(甲)宣稱,經歷過「瀕死經驗」,而你又想「判別其真假」的話,你可以用以下的幾個方法。

(留意,這裡的「判別其真假」只是「評價其可信度」的簡寫。剛才你也提到,要百分百地證明,某些言論是真的,是沒有可能的。)

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第一,甲所講的瀕死經驗,和其他曾經瀕死人士的講法,有何異同。

第二,會不會甲其實根本未曾有過瀕死經驗,只是道聽途說,人云亦云而已。

這一點非常難考證。以下的則比較容易觀察。

第三,瀕死經驗中,他所領悟到的道理,合情合理之餘,我感覺到它說的道理,遠超過一般道理的境界的話,那道理的可信程度,就再高一點。

繼而,甲版本的瀕死經驗,可信度亦會高一些;因為,如果甲所講述的「境界極高道理」,其實是他自己的創作的話,一般而言,他並沒有動機,去把功勞賦予一個,虛構的瀕死經驗。

(問:但是,道理又怎樣為之「境界極高」呢?)

這個問題,跟前一個問題性質一樣,所以,答案跟前一個答案相若,都是用「好樹結好果」這個大原則。

— Me@2018-10-26 08:48:21 PM

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2018.10.26 Friday (c) All rights reserved by ACHK

A Road to Common Lisp

tumba 57 days ago [-]

My advice is this: as you learn Common Lisp and look for libraries, try to suppress the voice in the back of your head that says “This project was last updated six years ago? That’s probably abandoned and broken.” The stability of Common Lisp means that sometimes libraries can just be done, not abandoned, so don’t dismiss them out of hand.

I have found this to be true in my own experience. The perception of stagnation is, however, a common initial objection to folks working in CL for the first time.

mike_ivanov 57 days ago [-]

My personal problem with CL libraries is not that, but rather the lack of documentation. More often than not, there is no documentation at all, not even a readme file.. It feels like some library authors simply don’t care. I’d say this attitude has a negative impact on how people perceive viability of those libraries — and by extension, of the language.

armitron 56 days ago [-]

A lot of libraries that don’t have separate documentation in the form of HTML/PDF/README.. are actually well-documented at source level in the form of docstrings.
Since Common Lisp is an interactive programming language and is meant to be used interactively (think Smalltalk, not Python) it is common practice to (interactively) load a library in a Common Lisp image and explore it (interactively). One can see what symbols are exported from packages, what these symbols are used for and (interactively) retrieve their documentation. All of this takes place within the editing environment (ideally Emacs) in a rapid feedback loop (again think Smalltalk, not Python) that feels seamless and tremendously empowering.

stevelosh 56 days ago [-]

I agree with this — it’s a real problem. My only solution has been to try to be the change I want to see in the world and document all of my own libraries before I officially release them.

— A Road to Common Lisp

— Hacker News

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2018.10.24 Wednesday ACHK

Problem 14.5b

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(b) Consider the left R’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

— This answer is my guess. —

The naive mass formula in the left R’ sector:

\displaystyle{ \begin{aligned}  \alpha' M_L^2 &= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\  &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\  \end{aligned}}

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\displaystyle{ \begin{aligned}  \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A  &= \frac{1}{2} \sum_{n = -1, -2, ...} n \lambda_{-n}^A \lambda_n^A + \frac{1}{2} \sum_{n = 1, 2, ...} n \lambda_{-n}^A \lambda_n^A  \\  \end{aligned}}

p.316 Equation (14.51):

\displaystyle{ \begin{aligned}  \frac{1}{2} \sum_{n = -1. -2, ...} n \lambda_{-n}^A \lambda_n^A &= \frac{1}{2} \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{1}{24} (D - 2) \\  \end{aligned}}

\displaystyle{ \begin{aligned}  \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A  &= \left[ \frac{1}{2} \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{1}{24} \left[(D - 2)\right]_A \right] + \frac{1}{2} \sum_{n = 1, 2, ...} n \lambda_{-n}^A \lambda_n^A \\  &= \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{32}{24} \\  &= \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{4}{3} \\ \\  \end{aligned}}

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\displaystyle{ \begin{aligned}  \alpha' M_L^2  &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\  &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \left[ \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{4}{3} \right] \\  &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\  \end{aligned}}

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\displaystyle{ \begin{aligned}  \alpha' M_L^2  &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\  \end{aligned}}

— This answer is my guess. —

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2018.10.24 Wednesday (c) All rights reserved by ACHK

Relational quantum mechanics

EPR paradox, 10

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Relational quantum mechanics (RQM) is an interpretation of quantum mechanics which treats the state of a quantum system as being observer-dependent, that is, the state is the relation between the observer and the system. This interpretation was first delineated by Carlo Rovelli in a 1994 preprint, and has since been expanded upon by a number of theorists. It is inspired by the key idea behind special relativity, that the details of an observation depend on the reference frame of the observer, and uses some ideas from Wheeler on quantum information.

,,,

Relational solution

In RQM, an interaction between a system and an observer is necessary for the system to have clearly defined properties relative to that observer. Since the two measurement events take place at spacelike separation, they do not lie in the intersection of Alice’s and Bob’s light cones. Indeed, there is no observer who can instantaneously measure both electrons’ spin.

The key to the RQM analysis is to remember that the results obtained on each “wing” of the experiment only become determinate for a given observer once that observer has interacted with the other observer involved. As far as Alice is concerned, the specific results obtained on Bob’s wing of the experiment are indeterminate for her, although she will know that Bob has a definite result. In order to find out what result Bob has, she has to interact with him at some time {\displaystyle t_{3}} in their future light cones, through ordinary classical information channels.

The question then becomes one of whether the expected correlations in results will appear: will the two particles behave in accordance with the laws of quantum mechanics? Let us denote by {\displaystyle M_{A}(\alpha )} the idea that the observer {\displaystyle A} (Alice) measures the state of the system {\displaystyle \alpha} (Alice’s particle).

So, at time {\displaystyle t_{2}}, Alice knows the value of {\displaystyle M_{A}(\alpha )}: the spin of her particle, relative to herself. But, since the particles are in a singlet state, she knows that

{\displaystyle M_{A}(\alpha )+M_{A}(\beta )=0,}

and so if she measures her particle’s spin to be {\displaystyle \sigma }, she can predict that Bob’s particle ( {\displaystyle \beta } ) will have spin {\displaystyle -\sigma }. All this follows from standard quantum mechanics, and there is no “spooky action at a distance” yet. From the “coherence-operator” discussed above, Alice also knows that if at {\displaystyle t_{3}} she measures Bob’s particle and then measures Bob (that is asks him what result he got) — or vice versa — the results will be consistent:

{\displaystyle M_{A}(B)=M_{A}(\beta )}

Finally, if a third observer (Charles, say) comes along and measures Alice, Bob, and their respective particles, he will find that everyone still agrees, because his own “coherence-operator” demands that

{\displaystyle M_{C}(A)=M_{C}(\alpha )} and {\displaystyle M_{C}(B)=M_{C}(\beta )}

while knowledge that the particles were in a singlet state tells him that

{\displaystyle M_{C}(\alpha )+M_{C}(\beta )=0.}

Thus the relational interpretation, by shedding the notion of an “absolute state” of the system, allows for an analysis of the EPR paradox which neither violates traditional locality constraints, nor implies superluminal information transfer, since we can assume that all observers are moving at comfortable sub-light velocities. And, most importantly, the results of every observer are in full accordance with those expected by conventional quantum mechanics.

— Wikipedia on Relational quantum mechanics

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2018.10.22 Monday ACHK

Ken Chan 時光機 1.2

多項選擇題 7.2 | Multiple Choices 7.2

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重點是,無論「勤力」還是「懶惰」,也只是手段,不是目的本身。

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二來,即使有時,「勤力」是正確的;那又應該如何「勤力」呢?

我的老師中,大概只有兩位,曾經講過一點讀書方法。其中一位,就正正是 Ken Chan。(另一位則是,我日校的「純數學」的老師(程兆海)。)

我記憶所及,Ken Chan 教過的溫習技巧,又不是真的很多,因為他並沒有,(例如)花一個課堂的時間去講。但是,就憑他有講的一點點,就已經令我,受用無窮。

Ken Chan 所提及,其中一個技巧是,在物理科,如果新學一個課題,就應該先做,大量該個課題的 MC(多項選擇題)。那樣,你就可以極速釐清,該個課題中的新概念。

註:必須為考試局所出的,以往公開試題目,而不是坊間出版社的練習。

— Me@2018-10-20 11:46:04 AM

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2018.10.20 Saturday (c) All rights reserved by ACHK

如果需要聯絡本人,可用右邊的電郵地址。

SLIME

SLIME, the Superior Lisp Interaction Mode for Emacs, is an Emacs mode for developing Common Lisp applications. SLIME originates in an Emacs mode called SLIM written by Eric Marsden. It is developed as an open-source public domain software project by Luke Gorrie and Helmut Eller. Over 100 Lisp developers have contributed code to SLIME since the project was started in 2003. SLIME uses a backend called Swank that is loaded into Common Lisp.

— Wikipedia on SLIME

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C-x o

Window-Move to other

C-x C-e

Evaluate last expression

C-c C-r

Evaluate region

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2018.10.19 Friday (c) ACHK

Problem 14.5a4

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

The GSO projection here keeps the states with \displaystyle{(-1)^{F_L} = + 1}; this defines the left NS’+ sector.

Write explicitly and count the states we keep for the three lowest mass levels, indicating the corresponding values of \displaystyle{\alpha' M_L^2}. [This is a long list.]

~~~

p.314 “Let us declare that number to be minus one, thus making the ground states fermionic:”

Equation (14.39):

\displaystyle{(-1)^F |NS \rangle \otimes |p^+, \overrightarrow{p}_T \rangle = - |NS \rangle \otimes |p^+, \overrightarrow{p}_T \rangle}

Equation (14.40):

\displaystyle{(-1)^F |\lambda \rangle = -(-1)^{\sum_{r,J} \rho_{r,J}} |\lambda \rangle}

p.315 “So all the states with integer \displaystyle{N^{\perp}} have \displaystyle{(-1)^F = -1}; they are fermionic states.”

However, in this problem:

“The left NS’ sector is built with oscillators \displaystyle{\bar \alpha_{-n}^I} and \displaystyle{\lambda_{-r}^A} acting on the vacuum \displaystyle{|NS' \rangle_L}, declared to have \displaystyle{(-1)^{F_L} = + 1}:”

\displaystyle{(-1)^{F_L} |NS' \rangle_L = + |NS' \rangle_L}

So all the states with integer \displaystyle{N^{\perp}} have \displaystyle{(-1)^F = +1}.

— This answer is my guess. —

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ \end{aligned}}

If we define N^\perp in the way similar to equation (14.37), we have

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + N^\perp \\ \end{aligned}}

\displaystyle{\begin{aligned}  \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&|NS' \rangle_L, \\  \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&\{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \}|NS' \rangle_L, \\  \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, ... \} \\ & & \{ ..., \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L \\  \end{aligned}}

Let \displaystyle{N(n, k) = {n + k - 1 \choose k - 1}}, the number of ways to put n indistinguishable balls into k boxes.

\displaystyle{\begin{aligned}  \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&1 \\  \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&8 + \frac{32 \times 31}{2} = 504 \\  \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\left( \frac{8 \times 8}{2} + \frac{8}{2} \right) = 36, 8 \times \left( \frac{32 \times 31}{2} \right) = 3968, 32 \times 32 = 1024, {32 \choose 4} = 35960 \\  \end{aligned}}

\displaystyle{\begin{aligned}  \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&N(2,8) = 36, 8 \times {32 \choose 2} = 3968, 32 \times 32 = 1024, {32 \choose 4} = 35960 \\  \alpha'M^2=1,~~~&N^\perp = 2:~~~~~& 36 + 3968 + 1024 + 35960 = 40988 \\  \end{aligned}}

— This answer is my guess. —

— Me@2018-10-14 03:25:08 PM

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2018.10.16 Tuesday (c) All rights reserved by ACHK

How far away is tomorrow?

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The cumulative part of spacetime is time.

It is the cumulative nature of time [for an macroscopic scale] that makes the time a minus in the spacetime interval formula?

\displaystyle{\Delta s^{2} = - (c \Delta t)^{2} + (\Delta x)^{2} + (\Delta y)^{2} + (\Delta z)^{2}}

— Me@2011.09.21

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Space cannot be cumulative, for two things at two different places at the same time cannot be labelled as “the same thing”.

— Me@2013-06-12 11:41 am

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There is probably no directly relationship between the minus sign and the cumulative nature of time.

Instead, the minus sign is related to fact that the larger the time distance between two events, the causally-closer they are.

— Me@2018-10-13 12:46 am

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Recommended reading:

d_2018_10_13__20_54_50_PM_

— Distance and Special Relativity: How far away is tomorrow?

— minutephysics

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2018.10.13 Saturday (c) All rights reserved by ACHK

凌晨舊戲

這段改編自 2010 年 4 月 18 日的對話。

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(問:你閱讀過很多有關「瀕死經驗」的文章?)

可以這樣說。

如果你閱讀那些文章的話,要小心一點,因為那類文章良莠不齊——當中有些文章發人深省,有些則謊話連篇

(問:那你怎樣分辨,「瀕死經驗」的文章之中,哪些是真,哪些為假?)

看看文中所說的,合不合理。

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例如,你怎樣知道,我說的話,是真還是假?

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合情合理,自圓其說的,就有機會真;無情無理,自相矛盾的,則必定為假。

我當年閱讀,有關「瀕死經驗」的文章,是因為有極大的興趣。

而我當年閱讀,很多有關「瀕死經驗」的文章,則是因為那時是,我的研究生時代;在工作上,有著極大的不幸。閱讀「瀕死經驗」的文章,可以暫時抽離當時的生活,以作減壓。

那段時期,我大部晚上,也不太願睡覺;往往拖到凌晨三點鐘才睡。我十分不想,明日的來臨。我不願面對,明日的事務。

當然,那十分不健康,不宜鼓勵。

— Me@2018-10-12 05:48:23 AM

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2018.10.12 Friday (c) All rights reserved by ACHK

Insomnia

如夢初醒 4.2 | 讀書與睡眠 4.2

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DrPhish 5 months ago [-]

This is exactly the Zen moment that fixed my lifelong insomnia: lying in bed relaxed with my eyes closed, even if I’m not asleep, leaves me feeling worlds better in the morning than fretting about not being able to sleep. It’s something that happened when I forgot about it completely. Once I realized I could literally just lie in bed for 8 hours and be OK, I was able to let go.

It’s like a trap that closes tighter about you the more you struggle, and releases if you allow yourself to relax[.]

Once I really got that, sleep became regular and uninterrupted[.]

laumars 5 months ago [-]

I discovered this recently too. Even in the instances when I still don’t get to sleep (eg my 1 year old coming into bed and wriggling in her sleep waking me) I’ve still felt more refreshed when I’ve relaxed rather than laid in bed wound up. I’m not into new age / whatever stuff but I like to think of it as night time meditation as while it’s still not as good as sleep it’s still a great deal better than full blown insomnia! And best of all, sometimes it clears your mind enough to actually sleep.

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— Thinking of yourself as an insomniac may be a part of the problem

— Hacker News

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2018.10.11 Thursday ACHK

Problem 14.5a4

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

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\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

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— This answer is my guess. —

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\displaystyle{\alpha' M_L^2}

\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \frac{1}{2} \left[ \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \frac{-1}{24} \left[ (D - 2) \right]_{I= 2, 3, ..., 9} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

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\displaystyle{\alpha' M_L^2 = \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}

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Equation (14.34):

\displaystyle{\frac{1}{2} \sum_{r=- \frac{1}{2}, -\frac{3}{2}} r b_{-r}^I b_{r}^I = \frac{1}{2} \sum_{r=\frac{1}{2}, \frac{3}{2}} r b_{-r}^I b_{r}^I - \frac{1}{48} (D - 2)}

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\displaystyle{\alpha' M_L^2}

\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \frac{1}{2} \sum_{r = - \frac{1}{2}, - \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}

\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \left[ \frac{1}{2} \sum_{r=\frac{1}{2}, \frac{3}{2}} r \lambda_{-r}^A \lambda_{r}^A - \frac{1}{48} \left[D - 2\right]_A \right]}

\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A - \frac{32}{48}}

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\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ \end{aligned}}

— This answer is my guess. —

— Me@2018-10-10 05:38:08 PM

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2018.10.10 Wednesday (c) All rights reserved by ACHK

Existence and Description

Bertrand Russell, “Existence and Description”

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§1 General Propositions and Existence

“Now when you come to ask what really is asserted in a general proposition, such as ‘All Greeks are men’ for instance, you find that what is asserted is the truth of all values of what I call a propositional function. A propositional function is simply any expression containing an undetermined constituent, or several undetermined constituents, and becoming a proposition as soon as the undetermined constituents are determined.” (24a)

“Much false philosophy has arisen out of confusing propositional functions and propositions.” (24b)

A propositional function can be necessary (when it is always true), possible (when it is sometimes true), and impossible (when it is never true).

“Propositions can only be true or false, but propositional functions have these three possibilities.” (24b)

“When you take any propositional function and assert of it that it is possible, that it is sometimes true, that gives you the fundamental meaning ‘existence’…. Existence is essentially a property of a propositional function. It means that the propositional function is true in at least one instance.” (25a)

— Brandon C. Look

— University Research Professor and Chair

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2018.10.07 Sunday ACHK