The problem of induction 3.1.2

Posted on December 18, 2018 by rpflee

Square of opposition

“everything has a pattern”?

“everything follows some pattern” –> no paradox

“everything follows no pattern” –> paradox

— Me@2012.11.05

My above statements are meaningless, because they lack a precise meaning of the word “pattern”. In other words, whether each statement is correct or not, depends on the meaning of “pattern”.

In common usage, “pattern” has two possible meanings:

1. “X has a pattern” can mean that “X has repeated data“.

Since the data set X has repeated data, we can simplify X’s description.

For example, there is a die. You throw it a thousand times. The result is always 2. Then you do not have to record a thousand 2’s. Instead, you can just record “the result is always 2”.

2. “X has a pattern” can mean that “X’s are totally random, in the sense that individual result cannot be precisely predicted“.

Since the data set X is totally random, we can simplify the description using probabilistic terms.

For example, there is a die. You throw it a thousand times. The die lands on any of the 6 faces 1/6 of the times. Then you do not have to record those thousand results. Instead, you can just record “the result is random” or “the die is fair”.

— Me@2018-12-18 12:34:58 PM

Best 1.2

Posted on December 17, 2018 by rpflee

The point is not to eliminate a risk, which is always impossible, but to minimize it, which is always possible.

— Me@2018-12-10 02:41:14 PM

迷宮直升機 5.3

Posted on December 17, 2018 by rpflee

Ken Chan 時光機 1.4.3

…

其實，那種情形，對現為中四的你而言，並不只是一個「假設」，因為，你現正修讀， G.Maths（核心數學）和 A.Maths（附加數學）。

「附加數」比「基礎數」而言，艱深非常，大部分人也覺得，十分辛苦。但是，亦正正是因為「附加數艱深非常」，你才會覺得「核心數容易萬分」。

那又為什麼，會有這個現象呢？

（問：程度高了？）

無錯。水平高了。

試想想，你參加了一個走迷宮比賽—與另外幾位參賽者，鬥快逃出同一個迷宮。

在那個情況下，「勤力」一點，跑快一些，是沒有什麼大作用的；因為，你只要在其中一個分岔路口，做錯決定，你就已經可以，前功盡廢。

走迷宮的致勝之道—在現埸極速逃出的方法是，在事前用大量時間準備，一架直升機，在比賽期間，把你釣出迷宮。

即使沒有那麼多資源，至起碼，在比賽開始前，你就要準備好，該迷宮的平面圖。

（問：無論是直升機，或者是鳥瞰圖，好像都是犯規？）

那要視乎，具體是哪一個遊戲。

試想想，中學生自修大學程度的課程，有沒有犯規？

（問：似乎沒有。但是，那又好像十分辛苦。）

一般而言，世間少有不勞而獲的東西。鉅大的好處，很多時也需要付出，鉅大的代價。

還有，如果策略妥當，閱讀大學程度書籍，所需的額外時間，未必如想像中的那麼多。

比喻說，假設你修讀 A.Maths（附加數學）的唯一目的是，提升 G.Maths（核心數學）。那樣，你的「附加數」成績，並不需要保證，名列前茅。

反而，即使在最壞情況，你的「附加數」成績不合格，只要你曾經用心研習過，「附加數」也會令你覺得，「核心數」容易了很多。

同理，如果你閱讀大學物理書籍的主要目的是，提升中學物理的話，你並不需要完成，整本「大學物理」的課文和習題。

反而，最重要的是，你嘗試過「大學物理」中，部分題目的難度；那就足以令你感到，中學的版本，顯淺非常。

— Me@2018-12-17 07:05:34 PM

The Android NDK Tools

Posted on December 16, 2018 by rpflee

EQL5-Android | Common Lisp for Android App Development 2018

After installing the Android SDK Tools by installing Android Studio, originally, you would be able to install the Android NDK through Android Studio’s SDK Manager.

d_2018_12_16__17_27_49_PM_

However, since EQL5-Android requires an old version of the NDK (version 10e), you have to download the NDK from the Android’s official NDK webpage.

d_2018_12_16__16_22_31_PM_

In Ubuntu, move the file android-ndk-r10e-linux-x86_64.zip to the same location as the Android SDK and then unzip it.

— Me@2018-12-16 09:25:10 PM

Problem 14.5c5

Posted on December 15, 2018 by rpflee

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

c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

~~~

We have the following well-known bosonic string mass formulae:

open string:

$\displaystyle{\alpha' M^2 = N - a}$

closed string:

$\displaystyle{\frac{1}{4} \alpha' M^2 = N_L - a}$
$\displaystyle{\frac{1}{4} \alpha' M^2 = N_R - a}$

p.55

— Solutions to K. Becker, M. Becker, J. Schwarz String Theory And M-theory

— Mikhail Goykhman

How come there is an extra $\displaystyle{\frac{1}{4}}$ at the beginning of the closed string formula?

p.322

$\displaystyle{\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2}$

$\displaystyle{\begin{aligned} \alpha' M_L^2 &= \alpha' M_R^2 \\ \frac{1}{2} \alpha' M^2 &= \alpha' M_L^2 + \alpha' M_R^2 \\ \alpha' M^2 &= 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2 \\ \end{aligned}}$

— Me@2018-12-15 08:59:18 PM

Wavefunction of a single photon

Posted on December 14, 2018 by rpflee

Photon dynamics in the double-slit experiment, 3

What equation describes the wavefunction of a single photon?

The Schrödinger equation describes the quantum mechanics of a single massive non-relativistic particle. The Dirac equation governs a single massive relativistic spin-½ particle. The photon is a massless, relativistic spin-1 particle.

What is the equivalent equation giving the quantum mechanics of a single photon?

— edited Jun 3 ’13 at 19:42

— Ben Crowell

— asked Nov 9 ’10 at 20:38

— nibot

There is no quantum mechanics of a photon, only a quantum field theory of electromagnetic radiation. The reason is that photons are never non-relativistic and they can be freely emitted and absorbed, hence no photon number conservation.

— answered Nov 10 ’10 at 20:00

— Igor Ivanov

You can also say that the wavefunction of a photon is defined as long as the photon is not emitted or absorbed. The wavefunction of a single photons is used in single-photon interferometry, for example. In a sense, it is not much different from the electron, where the wave-function start to be problematic when electrons start to be created or annihilated…

– Frédéric Grosshans Nov 17 ’10 at 10:19

— Physics StackExchange

2018.12.14 Friday ACHK

The fallacy of average

Posted on December 14, 2018 by rpflee

Those in the print media who dismiss the writing online because of its low average quality are missing an important point: no one reads the average blog. In the old world of channels, it meant something to talk about average quality, because that’s what you were getting whether you liked it or not. But now you can read any writer you want. So the average quality of writing online isn’t what the print media are competing against. They’re competing against the best writing online. And, like Microsoft, they’re losing.

— What Business Can Learn from Open Source

— Paul Graham

— Me@2011.08.23

2018.12.14 Friday ACHK

PhD, 2.2

Posted on December 14, 2018 by rpflee

故事連線 1.1.4 | 碩士 3.2

這段改編自 2010 年 4 月 18 日的對話。

…

有很多人的「個人形象」和「公眾形象」，都是差天共地的。

（問：即是表裡不一？）

有時，甚至是「表裡相反」。

（問：那應該怎麼辦？

除了在本科生年代，修讀心目中候選導師的課以外，還可以有什麼「測試」？）

可以試試找他，做你的本科畢業論文的導師。但是，這個風險仍然太大。我不太建議。

風險較小的方法有，詢問一下，他現時的研究生。他們最知道，該教授的真正面目，究竟是真材實學，還是欺世盜名。

還有，即使他真材實學，也不代表他肯花時間，用心教導研究生。他會不會那樣做，只有他以前或現時的研究生，才會知道。

（問：你好似講到，人類那麼危險？）

因為事實上，人類的確是，那麼危險。

剛才所講，有關選擇碩士或博士論文導師，所需的技巧，背後的精神，其實是通用的—同時適用於你將來選擇公司、上司、生意合作伙伴、配偶，等等。

選擇錯誤，同樣是有改變一生的後果。

— Me@2018-12-13 10:33:47 PM

The Android SDK Tools

Posted on December 12, 2018 by rpflee

EQL5-Android | Common Lisp for Android App Development 2018

After installing Qt, check whether its version is 5.9 or later.

d_2018_12_12__14_28_48_PM_

If so, install the Android SDK Tools by installing Android Studio.

If you use Ubuntu 18.04 or later, you can use the snap command to install Android Studio.

d_2018_12_12__22_11_20_PM_

Besides installing Android Studio, it also automatically updates Android Studio regularly.

— Me@2018-12-12 02:21:45 PM

Slope parameter

Posted on December 11, 2018 by rpflee

Problem 14.5c4 | Counting states in heterotic SO(32) string theory

c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

~~~

What is the meaning of $\displaystyle{\alpha'}$ ?

How come $\displaystyle{\alpha' = \frac{1}{2}}$ ?

— Me@2018-12-07 10:43:10 PM

If we consider a rigidly rotating open string, $\alpha'$ is the proportionality constant that relates the angular momentum $J$ of the string, measured in units of $\hbar$ , to the square of its energy $E$ . More explicitly,

$\displaystyle{ \frac{J}{\hbar} = \alpha' E^2 }$

— p.68

$\displaystyle{ J = \frac{1}{ 2 \pi T_0 c} E^2 }$

As anticipated, the angular momentum is proportional to the square of the energy of the string. Comparing with equation (8.69) we deduce that

$\displaystyle{ \begin{aligned} \alpha' &= \frac{1}{2 \pi T_0 \hbar c} \\ T_0 &= \frac{1}{2 \pi \alpha' \hbar c} \end{aligned}}$

These equations relate the slope parameter $\alpha'$ to the string tension $T_0$ .

— p.69

— A First Course in String Theory

The problem of induction 3.2

Posted on December 10, 2018 by rpflee

The meaning of induction is that

we regard, for example, that

“AAAAA –> the sixth is also A”

is more likely than

“AA –> the second is also A”

We use induction to find “patterns”. However, the induced results might not be true. Then, why do we use induction at all?

There is everything to win but nothing to lose.

— Hans Reichenbach

If the universe has some patterns, we can use induction to find those patterns.

But if the universe has no patterns at all, then we cannot use any methods, induction or else, to find any patterns.

However, to find patterns, besides induction, what are the other methods?

What is meaning of “pattern-finding methods other than induction”?

— Me@2012.11.05

— Me@2018.12.10

Miserable in comfort

Posted on December 9, 2018 by rpflee

Money, if it does not bring you happiness, will at least help you be miserable in comfort.

– Helen Gurley Brown

2018.12.09 Sunday ACHK

迷宮直昇機 5.2

Posted on December 9, 2018 by rpflee

Ken Chan 時光機 1.4.2

…

但是，他的學校不正常——在中四和中五的校內考試測驗中，不斷地考核高考課程。所以，他在中四時代開始，已經要鑽研，高考程度和大學程度的物理。

結果，到真正會考時，由於「只會」考核中五程度的東西，他會突然覺得十分容易。

其實，那種快感不難體會。

假設，你現在是中學四年級。

你想像，如果突然之間，那份中四的數學卷，換成中一程度的數學卷。你會有什麼感受？

你不單會覺得如釋重負，而且會有信心，有機會取滿分；即使，那份中一測驗卷的課題，在中一以後，從來未刻意溫習過。

（問：我又不敢說，我一定會滿分。）

我沒有說「一定」。我只是說「覺得有機會」。

試想想，現為中四的你，面對一份中四的數學卷，你夠不夠膽說「有機會取滿分」？

（問：不太敢，因為，那幾乎沒有可能。但是，我明白你的意思。現在要我做回中一的測驗卷，又真的會覺得，一定會高分，可能會滿分。）

其實，那種情形，對現為中四的你而言，並不只是一個「假設」，因為，你現正修讀， G.Maths（核心數學）和 A.Maths（附加數學）。

「附加數」比「基礎數」而言，艱深非常，大部分人也覺得十分辛苦。但是，亦正正是因為「附加數艱深非常」，你才會覺得「核心數容易萬分」。

那又為什麼，會有這個現象呢？

— Me@2018-12-09 09:03:02 PM

EQL5-Android

Posted on December 8, 2018 by rpflee

Common Lisp for Android App Development 2018

The first step to set up EQL5-Android is to install Qt.

d_2018_12_08__21_55_22_PM_

In Ubuntu, if you do not need the most updated Qt, you can just install the Qt-Creator using apt-get.

— Me@2018-12-08 09:18:19 PM

Problem 14.5c3

Posted on December 7, 2018 by rpflee

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

At any mass level $\displaystyle{\alpha' M^2 = 4k}$ of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = k}$ with the right-moving NS+ states with $\displaystyle{\alpha' M_R^2 = k}$ .

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = k}$ with the right-moving R- states with $\displaystyle{\alpha' M_R^2 = k}$ .

c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

~~~

Open String:

$\displaystyle{ \begin{aligned} N^\perp &= \sum_{p=1}^{\infty} \alpha_{-p}^I \alpha_p^I \\ L_n^{\perp} &= \frac{1}{2} \sum_{-\infty}^{\infty} \alpha_{n-p}^I \alpha_{p}^I~~~(n \ne 0) \\ L_0^{\perp} &= \alpha' p^I p^I + N^\perp \end{aligned} }$

Closed String:

$\displaystyle{ \begin{aligned} N^\perp &= \sum_{p=1}^{\infty} \alpha_{-p}^I \alpha_p^I \\ \bar N^\perp &= \sum_{p=1}^{\infty} \bar \alpha_{-p}^I \bar \alpha_p^I \\ L_0^\perp &= \frac{\alpha'}{4} p^I p^I + N^\perp \\ \bar L_0^\perp &= \frac{\alpha'}{4} p^I p^I + \bar N^\perp \end{aligned} }$

— A First Course in String Theory

We have the following well-known bosonic string mass formulae $\displaystyle{\alpha' = \frac{1}{2}}$ :

open string:

$\displaystyle{\frac{1}{2} M^2 = N - a}$

closed string:

$\displaystyle{\frac{1}{8} M^2 = N_L - a}$
$\displaystyle{\frac{1}{8} M^2 = N_R - a}$

p.55

— Solutions to K. Becker, M. Becker, J. Schwarz String Theory And M-theory

— Mikhail Goykhman

What is the meaning of $\displaystyle{\alpha'}$ ?

How come $\displaystyle{\alpha' = \frac{1}{2}}$ ?

— Me@2018-12-07 10:43:10 PM

Double slit experiment, 8

Posted on November 27, 2018 by rpflee

Although the screen itself is a photon position detector, it gets no which-way information. So it can get an interference pattern.

— Me@2012-04-09 7:24:23 PM

Infinite copying, 6

Posted on November 26, 2018 by rpflee

如何拯救眾生 6 | 無限複製 6 | 無友不如己者 6 |

How to save the world, 6

Information survives by people copying it.

— John Baez

— Me@2011.07.02

2018.11.26 Monday ACHK

PhD, 2.1

Posted on November 25, 2018 by rpflee

故事連線 1.1.3 | 碩士 3.1

這段改編自 2010 年 4 月 18 日的對話。

（安：你的意思時，不鼓勵年青人，攻讀研究院？）

不是。

研究工作，有其獨特的至尊好處。我反而覺得，每個人都應花兩年時間，經歷一次。換句話說，你可以考慮，先讀一個研究式的碩士，然後才盤算，自己適不適合再攻讀博士。

不過，你要留意，無論是攻讀碩士還是博士，你在之前選擇指導教授時，都要格外小心。你要保證，你的指導教授，有品德和有才能，幫你把與論文課題沒有直接關係的工作，全部推開。當然，那樣可靠的教授，萬中無一。

— Me@2012.08.20

在從來未與一位教授共事過的情況下，就選他為自己的碩士或博士論文導師，是十分危險的事。

那足以令你，自此不能再於學術界發展，抱憾終生。

（問：在那位教授成為你的論文導師前，又怎會共事過呢？）

可以在之前，有過其他形式的工作關係，直接或間接地，知道他的才德如何。

（問：你意思時，在本科時，就選修他的課？）

可以那樣說。聽他的講課和做他所予之功課，就可以知道他，工作態度認不認真、思考清不清晰和顧不顧及他人感受。但是，那也只是第一重的「測試」，先決而未充分；因為，那只是他的「公眾形象」而已。當他作為你的論文導師，你作為他的研究生時，你再也不是，和他的「公眾形象」相處了。

有很多人的「個人形象」和「公眾形象」，都是差天共地的。

（問：即是表裡不一？）

有時，甚至是「表裡相反」。

（問：那應該怎麼辦？）

…

— Me@2018-11-24 08:17:17 PM

Common Lisp for Android App Development 2018

Posted on November 23, 2018 by rpflee

An REPL called “CL REPL” is available in the Google Play Store. But itself is not for developing standalone Android apps, unless those apps are Common Lisp source code files only.

However, “CL REPL” itself is an open source GUI app using Common Lisp and Qt. So by learning and using its source, in principle, we can create other Android apps using Common Lisp with Qt.

The library that “CL REPL” uses is EQL5-Android.

— Me@2018-11-23 04:07:54 PM

Problem 14.5c2

Posted on November 22, 2018 by rpflee

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

At any mass level $\displaystyle{\alpha' M^2 = 4k}$ of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = k}$ with the right-moving NS+ states with $\displaystyle{\alpha' M_R^2 = k}$ .

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = 4k}$ with the right-moving R- states with $\displaystyle{\alpha' M_R^2 = k}$ .

c) Are there tachyonic states in heterotic string theory?

~~~

— This answer is my guess. —

The left NS’+ sector:

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

The left R’+ sector:

$\displaystyle{\begin{aligned} (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\ (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_\alpha \rangle_L \\ \alpha'M^2=2,~~~&N^\perp = 1:~~~~~&\alpha_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\ \end{aligned}}$

The right-moving NS+ states:

NS+ equations of (14.38):

$\displaystyle{\begin{aligned} \alpha'M^2=0, ~~~&N^\perp = \frac{1}{2}: &b_{-1/2}^I~&|NS \rangle \otimes |p^+, \vec p_T \rangle, \\ \alpha'M^2=1, ~~~&N^\perp = \frac{3}{2}: &\{ \alpha_{-1}^I b_{\frac{-1}{2}}^J, b_{\frac{-3}{2}}^I, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K \}~&|NS \rangle \otimes |p^+, \vec p_T \rangle, \\\ \alpha'M^2=2, ~~~&N^\perp = \frac{5}{2}: &\{\alpha_{-2}^I b_{\frac{-1}{2}}^J, \alpha_{-1}^I \alpha_{-1}^J b_{\frac{-1}{2}}^K, \alpha_{-1}^I b_{\frac{-3}{2}}^J, \alpha_{-1}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M, ...\}~& \\ &&\{ ..., b_{\frac{-5}{2}}^I, b_{\frac{-3}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M b_{\frac{-1}{2}}^N \}~&|NS \rangle \otimes |p^+, \vec p_T \rangle \end{aligned}}$

The R- states (that used as right-moving states):

Mass levels of R- and R+ (Equations 14.54):

$\displaystyle{\begin{aligned} \alpha'M^2=0,~~~&N^\perp = 0:~~~~&|R_a \rangle~~&||~~|R_{\bar a} \rangle \\ \alpha'M^2=1,~~~&N^\perp = 1:~~~~&\alpha_{-1}^I |R_{a} \rangle,~d_{-1}^I |R_{\bar a} \rangle ~~&||~~ ... \\ \alpha'M^2=2,~~~&N^\perp = 2:~~~~&\{ \alpha_{-2}^I,~\alpha_{-1}^I \alpha_{-1}^J,~d^I_{-1} d^J_{-1} \} |R_{a} \rangle,~&|| \\ &&\{\alpha_{-1}^I d_{-1}^J,~d_{-2}^I \} |R_{\bar a} \rangle~~&||~~ ... \\ \end{aligned}}$

There are no tachyonic states in heterotic string theory, since neither of the right-moving parts (NS+ and R-) has states with $\displaystyle{\begin{aligned} \alpha' M^2 < 0\end{aligned}}$ .

— This answer is my guess. —

— Me@2018-11-22 12:00:30 PM

Physics Town

continues

The problem of induction 3.1.2

Best 1.2

迷宮直升機 5.3

The Android NDK Tools

Problem 14.5c5

Wavefunction of a single photon

The fallacy of average

PhD, 2.2

The Android SDK Tools

Slope parameter

The problem of induction 3.2

Miserable in comfort

迷宮直昇機 5.2

EQL5-Android

Problem 14.5c3

Double slit experiment, 8

Infinite copying, 6

PhD, 2.1

Common Lisp for Android App Development 2018

Problem 14.5c2