# Build cross-compiled ECL for Android, 3

EQL5-Android | Common Lisp for Android App Development 2018

.

After successfully running the command ./1-make-ecl-host.sh, when I tried to run the command ./2-make-ecl-android.sh, I got the following errors:

— Me@2018-12-29 11:21:46 PM

.

.

# Problem 14.5c8

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) Calculate the total number of states in heterotic string theory (bosons plus fermions) at $\displaystyle{\alpha' M^2 =4}$.

~~~

.

— This answer is my guess. —

.

When $\displaystyle{\alpha' M^2 =4}$,

\displaystyle{\begin{aligned} \alpha' M_L^2 &= 1 \\ \alpha' M_R^2 &= 1 \end{aligned}}

~~~

The left NS’+ sector:

\displaystyle{\begin{aligned} \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 \\ \end{aligned}}

.

The right-moving NS+ states:

\displaystyle{\begin{aligned} \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, \\ \end{aligned}}

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

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

~~~

spacetime bosons:

$\displaystyle{NS'+ \otimes NS+}$

\displaystyle{\begin{aligned} \left( \{ \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 \right) \otimes \left( \{ \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 \right) \end{aligned}}

$\displaystyle{R'+ \otimes NS+}$

\displaystyle{\begin{aligned} \left( |R_\alpha \rangle_L \right) \otimes \left( \{ \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 \right) \end{aligned}}

.

spacetime fermions:

$\displaystyle{NS'+ \otimes R-}$

\displaystyle{\begin{aligned} \left( \{ \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 \right) \otimes \left( \alpha_{-1}^{I'} |R_{a} \rangle,~d_{-1}^{I'} |R_{\bar a} \rangle \right) \\ \end{aligned}}

$\displaystyle{R'+ \otimes R-}$

\displaystyle{\begin{aligned} \left( |R_\alpha \rangle_L \right) \otimes \left( \alpha_{-1}^{I} |R_{a} \rangle,~d_{-1}^{I} |R_{\bar a} \rangle \right) \\ \end{aligned}}

.

— This answer is my guess. —

— Me@2018-12-28 11:12:59 PM

.

.

# The problem of induction 3.3

“Everything has no patterns” (or “there are no laws”) creates a paradox.

.

If “there are 100% no first order laws”, then it is itself a second order law (the law of no first-order laws), allowing you to use probability theory.

In this sense, probability theory is a second order law: the law of “there are 100% no first order laws”.

In this sense, probability theory is not for a single event, but statistical, for a meta-event: a collection of events.

Using meta-event patterns to predict the next single event, that is induction.

.

Induction is a kind of risk minimization.

— Me@2012-11-05 12:23:24 PM

.

.

# understand \neq agree

$\displaystyle{\text{understand} \neq \text{agree}}$

.

Seek first to understand, then to be understood

— The Habit 5

— The 7 Habits of Highly Effective People

— Stephen Covey

.

To understand does not mean to agree.

— Stephen Covey

.

.

2018.12.28 Friday by ACHK

# Ken Chan 時光機 2.1

— Me@2018-12-27 03:50:13 PM

.

.

# Build cross-compiled ECL for Android, 2

EQL5-Android | Common Lisp for Android App Development 2018

.

The step 2 is to “build cross-compiled ECL for Android”. It should have been straightforward to follow the instructions in the README page of the EQL5-Android project.

However, when trying to run the command ./1-make-ecl-host.sh, I got the following error messages:

The error was caused by the fact that my Ubuntu 18.04’s 64-bit gcc toolchain could not compile any source code to create 32-bit executables.

The solution is to run the following command to install the 32-bit gcc toolchain first:

sudo apt-get install g++-multilib libc6-dev-i386

— Me@2018-12-25 09:51:02 PM

.

.

# Problem 14.5c7

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

.

c) … Write out the massless states of the theory (bosons and fermions) and describe the fields associated with the bosons.

~~~

.

— This answer is my guess. —

.

spacetime bosons:

$NS'+ \otimes NS+$

\displaystyle{\begin{aligned} \left( \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} |NS' \rangle_L \right) \otimes \left( b_{-1/2}^J~|NS \rangle \otimes |p^+, \vec p_T \rangle \right) \end{aligned}}

\displaystyle{\begin{aligned} = \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} b_{-1/2}^J |NS' \rangle_L \otimes \left( |NS \rangle \otimes |p^+, \vec p_T \rangle \right) \end{aligned}}

.

What is the nature of each of the indices $I, J, A, B$?

The vector index $J$ runs over eight values.

— c.f. p.323 A First Course in String Theory (Second Edition)

$\displaystyle{I = 2,3,...,9}$

$\displaystyle{A, B = 1, 2, ..., 32}$

— c.f. the blog post Problem 14.5a3

— Me@2018-12-24 10:04:52 PM

.

For the states in the form

$\displaystyle{ \bar \alpha_{-1}^I b_{-1/2}^J |NS' \rangle_L \otimes \left( |NS \rangle \otimes |p^+, \vec p_T \rangle \right)}$,

they

carry two independent vector indices $I$, $J$ that run over eight values. There are therefore 64 bosonic states. Just like the massless states in bosonic closed string theory[,] they carry two vector indices. We therefore get a graviton, a Kalb-Ramond field, and a dilation:

(NS+, NS+) massless fields: $g_{\mu \nu}, B_{\mu \nu}, \phi$.

— p.323 A First Course in String Theory (Second Edition)

.

Then how about the states in the form

\displaystyle{\begin{aligned} \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B b_{-1/2}^J |NS' \rangle_L \otimes \left( |NS \rangle \otimes |p^+, \vec p_T \rangle \right) \end{aligned}}?

What kinds of fields do they represent?

— Me@2018-12-24 10:42:03 PM

.

— This answer is my guess. —

— Me@2018-12-23 11:16:56 PM

.

.

# Afshar experiment, 2

Double slit experiment, 8.2

.

In the double slit experiment, the screen is used to detect interference pattern itself, causing the photon wavefunctions to “collapse”.

In the Afshar experiment, there is no classically definite position for a photon when the photon passes “through” the vertically wire slits. So there is no interference patterns “formed”, unless you put some kind of screen afterwards. [Me@2015-07-21 10:59 PM: i.e. making the observation, c.f. delayed choice experiment]

— Me@2012-04-09 12:19:52 AM

.

Being massless, they cannot be localized without being destroyed…

— Photon dynamics in the double-slit experiment

— Wikipedia on Photon

.

.

# Universe City

.

A university is a city.

— Me@2011.08.23

.

University ~ Universe City

— Me@2011.09.02

.

.

# 宇宙大戰 1.1

PhD, 2.3 | 故事連線 1.1.5 | 碩士 3.3

.

（問：你好似講到，人類那麼危險？）

.

.

.

（問：人類真的那麼危險嗎？）

.

.

（問：我也遇過類似的情境。

.

.

（問：那樣，如果要「複雜地說」呢？）

.

— Me@2018-12-20 11:06:49 PM

.

.

# Build cross-compiled ECL for Android

EQL5-Android | Common Lisp for Android App Development 2018

.

The step 2 is to “build cross-compiled ECL for Android”. It should have been straightforward to follow the instructions in the README page of the EQL5-Android project.

However, when trying to run the command ./1-make-ecl-host.sh, I got the following error messages:

A more serious problem was that I could not even locate the error log file config.log.

.

Another abnormality was that while the process claimed that it was

Creating directory 'build',

the directory build was actually not created at all.

So I manually created that directory before running the command ./1-make-ecl-host.sh again. Only then, I could find the error log file config.log in the build directory, after the failure of the compilation.

— Me@2018-12-19 10:50:31 PM

.

.

# Problem 14.5c6

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?

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

~~~

.

— This answer is my guess. —

.

The left NS’+ sector:

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

The left R’+ sector has no massless states.

The right-moving NS+ states:

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

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

\displaystyle{\begin{aligned} \alpha'M^2=0,~~~&N^\perp = 0:~~~~&|R_a \rangle \\ \end{aligned}}

~~~

Since R’+ has no massless states:

spacetime bosons:

$NS'+ \otimes NS+$

\displaystyle{\begin{aligned} \left( \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} |NS' \rangle_L \right) \otimes \left( b_{-1/2}^I~|NS \rangle \otimes |p^+, \vec p_T \rangle \right) \end{aligned}}

\displaystyle{\begin{aligned} = \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} b_{-1/2}^I |NS' \rangle_L \otimes \left( |NS \rangle \otimes |p^+, \vec p_T \rangle \right) \end{aligned}}

spacetime fermions:

$NS'+ \otimes R-$

\displaystyle{\begin{aligned} \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} |NS' \rangle_L \otimes |R_a \rangle \\ \end{aligned}}

.

— This answer is my guess. —

— Me@2018-12-18 07:46:15 PM

.

.

# The problem of induction 3.1.2

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

.

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

.

.

Posted in OCD

# 迷宮直升機 5.3

Ken Chan 時光機 1.4.3

.

「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」。

.

（問：程度高了？）

.

.

（問：無論是直升機，或者是鳥瞰圖，好像都是犯規？）

.

.

（問：似乎沒有。但是，那又好像十分辛苦。）

.

— Me@2018-12-17 07:05:34 PM

.

.

# The Android NDK Tools

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.

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.

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

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

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

.

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

.

.

（問：即是表裡不一？）

.

.

（問：那應該怎麼辦？

.

.

（問：你好似講到，人類那麼危險？）

.

.

— Me@2018-12-13 10:33:47 PM

.

.