# Consistent histories, 7

In quantum mechanics, the consistent histories (also referred to as decoherent histories) approach is intended to give a modern interpretation of quantum mechanics, …

— Wikipedia on Consistent histories

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It needs to be decoherent in order to be consistent.

— Me@2017-08-08 01:25:54 PM

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decoherent ~ no quantum superposition

consistent ~ classical logic can apply

— Me@2020-05-30 03:52:15 PM

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# 反貼士搵笨大行動 1.4

「無知」即是「缺乏足夠資料」；

「無知」不是「愚蠢」。

— Me@2020-04-25 08:08:39 AM

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# Problem 2.2c

A First Course in String Theory

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2.2 Lorentz transformations for light-cone coordinates.

Consider coordinates $\displaystyle{x^\mu = ( x^0, x^1, x^2, x^3 )}$ and the associated light-cone coordinates $\displaystyle{x^\mu = ( x^+, x^-, x^2, x^3 )}$. Write the following Lorentz transformations in terms of the light-cone coordinates.

(c) A boost with velocity parameter $\displaystyle{\beta}$ in the $\displaystyle{x^3}$ direction. \displaystyle{ \begin{aligned} \begin{bmatrix} c t' \\ z' \\ x' \\ y' \end{bmatrix} &= \begin{bmatrix} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} c\,t \\ z \\ x \\ y \end{bmatrix} \\ \end{aligned} }

~~~ \displaystyle{ \begin{aligned} \begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} &= \begin{bmatrix} \gamma & 0 & 0 & -\beta \gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\beta \gamma & 0 & 0 & \gamma \\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix} \\ \end{aligned} } \displaystyle{ \begin{aligned} \begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & \sqrt{2} & 0 \\ 0 & 0 & 0 & \sqrt{2} \\ \end{bmatrix} \begin{bmatrix} (x^+)' \\ (x^-)' \\ y' \\ z' \end{bmatrix} &= \begin{bmatrix} \gamma & 0 & 0 & -\beta \gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\beta \gamma & 0 & 0 & \gamma \\ \end{bmatrix} \begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & \sqrt{2} & 0 \\ 0 & 0 & 0 & \sqrt{2} \\ \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \\ y \\ z \end{bmatrix} \\ \end{aligned} } \displaystyle{ \begin{aligned} \begin{bmatrix} (x^+)' \\ (x^-)' \\ y' \\ z' \end{bmatrix} &= \frac{1}{2} \begin{bmatrix} \gamma + 1 & \gamma - 1 & 0 & -\sqrt{2}\,\beta\,\gamma \\ \gamma - 1 & \gamma + 1 & 0 & -\sqrt{2}\,\beta\,\gamma \\ 0 & 0 & 2 & 0 \\ - \sqrt{2}\,\beta\,\gamma & - \sqrt{2}\,\beta\,\gamma & 0 & 2 \gamma \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \\ y \\ z \end{bmatrix} \\ \end{aligned} }

— Me@2020-04-19 11:52:09 PM

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# Omnipotence 4.2

When responding to the question “can X create a stone that it cannot lift”, another flawed argument is

X can create the stone that it cannot lift but it chooses not to create it. So there is no stone it cannot lift yet. So X has not failed the omnipotence test.

This argument is wrong.

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When we ask “can X choose to create a stone that it cannot lift”, we are discussing whether X has an ability. When we discuss ability, it is always about a potential, a possibility.

Y is able to do action B

always means that

“Y does B” is possible,

which is equivalent to

“Y does B” is not contradictory to any logical laws nor physical laws.

“Whether Y has already done B or will do B” is not the point.

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If we allow such “Y can do B but it chooses not to” argument, then anyone is omnipotent. For example,

Can you fly?

I can fly but I choose not to. So even though you have never seen me flying and will never see me flying, it is not because I cannot fly; it is just because I choose not to.

Can you choose to fly?

I can choose to fly but I choose not to choose to fly.

This type of arguments make the word “can“ meaningless.

— Me@2020-03-30 06:52:58 AM

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# Superpower

Impossible self, 2 | 電腦輪迴觀 4.2

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ability 能力

~ you can access your own time to do things

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power 權力

~ weighted ability

~ amplified ability

~ super ability

~ 超過一人之力

~ you can access also our people’s time to do things

— Me@2020-04-17 09:00:44 PM

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Freedom is being able to make decisions that affect mainly you; power is being able to make decisions that affect others more than you. If we confuse power with freedom, we will fail to uphold real freedom.

— Freedom or Power?

— Bradley M. Kuhn and Richard M. Stallman

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So the term “superpower” is actually redundant, because “power” already means “super-ability”, the ability to access more than one person’s time.

— Me@2020-04-18 07:41:36 PM

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In daily life, “power” is often used as a synonym of “ability”. In this sense, “superpower” is the ability to access more than one person’s time.

In this context, with our people’s agreement, other people are your superpower. Also, you are our people’s superpower.

— Me@2020-04-18 03:58:32 PM

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# 太極滅世戰

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「間書原理」的意思，其實是「陽之極為陰；陰之極為陽」。但那不易理解，所以，我在十多年前，舉了「間書」的例子：

— Me@2003-2004

— 改編自李天命先生

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（問：「不理成績」而又要「盡情發揮」？自相矛盾也？）

. Wikipedia
public domain image

— Me@2020-04-13 06:58:18 PM

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# CSS, 3


blockquote {
font-family: Helvetica;
font-style: normal;
color: #4f7499;
background: #EAEFF3;
border-left: solid 2px #9ab3cb;
margin-left: 20px;
}

#infinite-handle {
display: none;
}

.infinite-scroll #nav-below {
display: block;
}

.infinite-scroll #content {
margin-bottom: 0;
}

.wp-caption .wp-caption-text:before {
display:none;
}

.wp-caption .wp-caption-text {
text-align:center;
}



— Me@2020-03-14 04:30:08 PM

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2020.03.14 Saturday ACHK

# Ex 1.8.2.1 Implementation of $\delta$

Structure and Interpretation of Classical Mechanics

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b. Use your delta procedure to verify the properties of $\displaystyle{\delta}$ listed in exercise 1.7 for simple functions such as implemented by the procedure f:

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


This implements an n-degree-of-freedom path-dependent function that depends on the local tuple of the path at each moment. You can define a literal two-dimensional path by

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


You should compute both sides of the equalities and subtract the results. The answer should be zero.

~~~

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

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

(define eta (literal-function 'eta))

(define q (literal-function 'q))

(print-expression ((((delta eta) f) q) 't))


— Patrick Eli Catach

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(print-expression ((((delta eta) f) q) 't))

(+ (* ((D eta) t) (((partial 2) f) (up t (q t) ((D q) t))))
(* (eta t) (((partial 1) f) (up t (q t) ((D q) t)))))

(show-expression ((((delta eta) f) q) 't)) — Me@2020-04-11 12:01:04 PM

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# Omnipotence 4.1

For all, 3 | Omnipotence

For all, 3.2 | Omnipotence 2

You can find them by searching “omnipotence” using this blog’s search box.

— Me@2020-04-08 03:17:34 PM

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If X is omnipotent, X can create a stone that it cannot lift. Then X is not omnipotent, because there is a stone it cannot lift. So omnipotence is a self-contradictory concept.

What if we define omnipotence not as “being able to do anything” but as “being able to do anything except logical self-contradictory ones“?

In order words, omnipotence means that being able to do anything logically possible. Omnipotence does not mean that being able to do also logically impossible things.

This re-definition is not useful, because the original meaning of “being omnipotent” already is “being able to do anything except logical self-contradictory ones“.

There is no re-definition needed. You can only say that the re-definition clarifies the original meaning of “being omnipotent”. However, this clarification cannot eliminate the self-contradictory nature of the meaning of “omnipotence” itself. For example, the following argument is wrong.

If X is omnipotent, “X can create a stone that it cannot lift” is self-contradictory because it is contradictory to “X is omnipotent”.

Since “X can create a stone that it cannot lift” is logically impossible, it should not be a requirement of being omnipotent.

This argument is wrong because:

1. “X can create a stone that it cannot lift” is not SELF-contradictory.

2. “X can create a stone that it cannot lift” is not logically impossible, because, for example, even a human being can create an object that he cannot lift. For example, human beings can create a car that no single person can lift.

Then someone might keep arguing that

But if X is omnipotent, “X can create a stone that it cannot lift” means that “X is omnipotent and X can create a stone it cannot lift”, which is logically impossible. So “X cannot create a stone that it cannot lift” does not make X non-omnipotent.

In other words, “whether X can create a stone that it cannot lift” should not be the requirement of the omnipotence test.

The argument is wrong, because what we are questioning is

Can someone X be omnipotent?

or

Is omnipotence logically possible?

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Remember:

“Being logically possible” means “not self-contradictory”.

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If “X is omnipotent” is true,

then “X can create a stone that it cannot lift” is true.

Then “there is a stone that X cannot lift” is true.

Then “X is not omnipotent” is true.

But “X is not omnipotent” is contradictory to the assumption “X is omnipotent“.

So “X is omnipotent” is self-contradictory.

So the question “whether an entity X can be omnipotent and create a stone that it cannot lift” is illegitimate because “an entity X is omnipotent” is logically impossible in the first place. It should not be placed within a question.

Note that our omnipotent test is

“whether an entity X can create a stone that it cannot lift”,

NOT “whether an entity X can be omnipotent and create a stone that it cannot lift”,

NOR “whether an omnipotent entity X can create a stone that it cannot lift”.

— Me@2020-03-30 06:52:58 AM

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# Time Travel OCD

Once you have DECIDED that you are not going to fear, you will not fear because you know that you are not going to fear.

— Me@2009.09.21

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# 反貼士搵笨大行動 1.3

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（問：那你即是贊成補習？）

1. 「無知」即是「缺乏足夠資料」；

2. 「無知」不是「愚蠢」。

— Me@2020-03-31 04:28:23 PM

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# Pier It was a picture I drawn in 1997. It was the summer holiday after my HKCEE public exam. It was a drawing trip with my school’s Art Club. The location was Sai Kung Pier.

As far as I remember, I haven’t drawn since then. Perhaps I do not like to draw alone. Hope that one day I could find my girlfriend finally so that I have someone to draw with.

— Me@2020-03-24 10:37:58 PM

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# Problem 2.2b

A First Course in String Theory

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2.2 Lorentz transformations for light-cone coordinates.

Consider coordinates $\displaystyle{x^\mu = ( x^0, x^1, x^2, x^3 )}$ and the associated light-cone coordinates $\displaystyle{x^\mu = ( x^+, x^-, x^2, x^3 )}$. Write the following Lorentz transformations in terms of the light-cone coordinates.

(b) A rotation with angle $\displaystyle{\theta}$ in the $\displaystyle{x^1, x^2}$ plane. \displaystyle{ \begin{aligned} \begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} &= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix} \\ \end{aligned} }

~~~ \displaystyle{ \begin{aligned} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 & 0 \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} (x^+)' \\ (x^-)' \\ y' \\ z' \end{bmatrix} &= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 & 0 \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \\ y \\ z \end{bmatrix} \\ \begin{bmatrix} (x^+)' \\ (x^-)' \\ y' \\ z' \end{bmatrix} &= \frac{1}{2} \begin{bmatrix} \cos\theta + 1 & 1 - \cos\theta & -\sqrt{2} \sin\theta & 0 \\ 1 - \cos\theta & \cos\theta + 1 & \sqrt{2} \sin\theta & 0 \\ \sqrt{2} \sin{\theta} & -\sqrt{2} \sin{\theta} & 2 \cos{\theta} & 0 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \\ y \\ z \end{bmatrix} \end{aligned} }

— Me@2020-03-22 10:16:09 PM

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# Logical arrow of time, 7.2

Microscopically, there is no time arrow.

— Me@2011.06.23

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No. There is weak force.

— Me@2011.07.22

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Myth: The arrow of time is a consequence of CP-symmetry violation.

The weak nuclear interactions violate the CP symmetry which is equivalent to saying that they violate the T symmetry. Is it the reason why eggs don’t unbreak? Of course not. There are two basic ways to see why. First, the weak interactions much like all other interactions preserve the CPT symmetry – there is extensive theoretical as well as experimental evidence supporting this assertion. And the CPT symmetry would be enough to show that eggs break as often as unbreak. More precisely, eggs break as often as mirror anti-eggs unbreak. ;-)

— Myths about the arrow of time

— Lubos Motl

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Weak force’s T-symmetry-violation has nothing to do with the time arrow.

In other words, microscopic time arrow has nothing to do with the macroscopic time arrow.

— Me@2020-03-21 07:56:01 PM

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About T-violation and the arrow of time: the simple answer is that the weak interactions are perfectly unitary, even if they are not T-invariant. They don’t affect the entropy in any way, so they don’t help with the arrow of time.

A bit more carefully: if you did want to explain the arrow of time using microscopic dynamics, you would have to argue that there exist more solutions to the equations of motion in which entropy grows than solutions in which entropy decreases. But CPT invariance is enough to guarantee that that’s not true. For any trajectory (or ensemble of trajectories, or evolution of a distribution function) in which the entropy changes in one way, there is another trajectory (or set…) in which the entropy changes in precisely the opposite way: the CPT conjugate. Such laws of physics do not in and of themselves pick out what we think of as the arrow of time.

People talk about the “arrow of time of the weak interactions,” but ask yourself: in which direction does it point? There just isn’t any direct relationship to entropy.

— Sean Carroll

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~ 愚蠢化

— Me@2011.06.23

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# 機遇創生論 1.4

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1. 不懂道理

2. 懂道理但缺乏足夠情報

「反白論」的意思是，現時的地球人間，極多的情況下，名義與實情都是相反的。

As a rule of thumb, the more qualifiers there are before the name of a country, the more corrupt the rulers. A country called The Socialist People’s Democratic Republic of X is probably the last place in the world you’d want to live.

–- Paul Graham

「反白論」的應用在於，掌握以後，大大減少了，你被世人欺騙得到的比例。

— Me@2020-03-19 09:34:39 PM

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# scmutils, 3

Structure and Interpretation of Classical Mechanics

Scheme Mechanics Installation for GNU/Linux

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This post assumes that you have already installed the scmutils library and been able to open it using the standard editor Emacs.

If not, go to the bottom of this post to click the category scmutils, so that you can see all the posts in this scmutils series. Then go to the post “scmutils, 2.3.2” to follow the installation and setup instructions.

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After installing and setting up the scmutils library, you can start to use it. However, what if you want to close the Emacs editor? How to save your scheme program before closing Emacs?

By default, you cannot. So I have written a small program to help. Here is the installation instruction:

1. Go to the end of the .emacs file. Add the following code, if it does not already exist:


(defun mechanics()
(interactive)
(run-scheme
"/usr/local/scmutils/mit-scheme/bin/scheme --library
/usr/local/scmutils/mit-scheme/lib"
))




(fset 'set-working-dir
(lambda (&optional arg) "Keyboard macro."
(interactive "p")
(kmacro-exec-ring-item
(quote ("(set-working-directory-pathname!
\"~/Documents/\")\n" 0 "%d")) arg)))

(lambda (&optional arg) "Keyboard macro."
(interactive "p")
(kmacro-exec-ring-item
(quote ("(load \"tt.scm\")" 0 "%d")) arg)))

(defun mechan ()
(interactive)
(split-window-below)
(windmove-down)
(mechanics)
(set-working-dir)
(comint-send-input)
(windmove-up)
(find-file "~/Documents/tt.scm")
(end-of-buffer)
(windmove-down)
(cond ((file-exists-p "~/Documents/tt.scm")
(interactive)
(comint-send-input)))
(windmove-up)
)

(defun cxce ()
(interactive)
(save-buffer)
(windmove-down)
(comint-send-input)
(windmove-up)
)

(global-set-key (kbd "C-x C-e") 'cxce)



3. Close Emacs. Re-open it.

4. Type the command

M-x mechan


The command M-x means pressing the Alt key and x together. Then type the word mechan.

5. You will see the Emacs editor is split into two windows, one up and one down.

The lower window is the scheme environment. You can type a line of code and the press Enter to execute it.

The upper window is the editor. You can type multiple lines of code and the type

C-x C-e


to execute it. The command C-x C-e means pressing Ctrl and x together and then Ctrl and e. 6. Your scheme code is saved to the following file

"~/Documents/tt.scm"


In case you need to backup your code, backup this file.

— Me@2020-03-10 10:59:45 PM

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