The Sixth Sense, 3

「機遇再生論」有兩大（潛）假設：

1. 宇宙，有無限長的未來。

（這對應於撲克比喻中，「可以洗牌無限次」的假設。）

2. 宇宙中的粒子數目有限；而它們的組合及排列數目，都是有限的。

（這對應於撲克比喻中，「只有 52 隻牌」和「只有有限個排列」( $52! \approx 8.07 \times 10^{67}$ )的假設。）

「機遇再生論」的第二個假設，同第一個假設一樣，都是疑點重重。

第一，宇宙的粒子總數，並不是常數。

「狹義相對論」加「量子力學」，等於「量子場論」。如果「量子場論」是正確的，真空中不斷有粒子生滅。

第二，即使假設，宇宙的粒子總數不變，隨著宇宙的膨脹，粒子可能狀態的數目，不斷變大。

第三，即使假設，字宙的體積固定，粒子數目有限，而又毋須考慮「量子力學」；粒子可能狀態數目，都可能不是有限的。

例如，即使只有一粒粒子，在一個邊長為一米的正立方體盒子之內，而宇宙只有那個盒子，沒有其他空間；

即使只考慮該粒子的位置，仍然有無限個可能態，因為，它可能在距離牆邊 0.1 米處、0.11 米處、0.111 米處，等等。

（問：空間未必可以，無限分割。假設空間可以無限分割，會導致「芝諾悖論」（Zeno’s paradoxes）。）

無錯。如果空間有最小的單位，不可無限無割，粒子在有限大空間中，可能位置的數目，則是有限。

第四，即使假設，字宙的體積固定，粒子數目有限，粒子可能狀態數目，都不是有限的。

宇宙最根本的物理定律，必須跟隨量子力學架構，經典物理定律只是，有時適用的近似。

（這裡，「經典」的意思，並不是「歷史悠久」，而是「非量子」。「經典物理」即是「不是建基於量子力學架構的，物理定律」，例如牛頓力學。）

如果你沒有忽略考慮，粒子的量子疊加態的話，你會發現，例如，即使只有一粒粒子，在一個邊長為一米的正立方體盒子之內，而宇宙只有那個盒子，沒有其他空間；

即使只考慮該粒子的位置，該粒子（宇宙）很可能地，有無限個態。

由於「機遇再生論」的兩大假設，都是「有待論證」，看來，想要靠「機遇再生論」來重生的話，有點難度。

究竟，有沒有其他方法，可以保存自己，擇日歸來呢？

— Me@2015.04.08

— Me@2017-12-09

— Me@2018-04-28

Importance, 3

Posted on April 26, 2018 by rpflee

People never understand what you do for them …

until you stop doing it!

—

2018.04.26 Thursday ACHK

Problem 14.4a4

Posted on April 25, 2018 by rpflee

Closed string degeneracies | A First Course in String Theory

(a) State the values of $\alpha' M^2$ and give the degeneracies for the first five mass levels of the closed bosonic string theory.

~~~

…

$\|$	$~\text{Number of states}$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} \| p^+, \vec p_T \rangle~\|$	$~\left[ \frac{(D-2)(D-1)}{2} \right]^2$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} \| p^+, \vec p_T \rangle~\|$	$~\left[ \frac{(D-2)(D-1)}{2} \right](D-2)$
${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} \| p^+, \vec p_T \rangle~\|$	$~(D-2)\left[ \frac{(D-2)(D-1)}{2} \right]$
${a_2^I}^\dagger \bar a_2^{K\dagger} \| p^+, \vec p_T \rangle~\|$	$~(D-2)^2$

Can we create a formula for the number of states?

$\left[ \frac{(D-2)(D-1)}{2} \right]^2 + \left[ \frac{(D-2)(D-1)}{2} \right](D-2) + (D-2)\left[ \frac{(D-2)(D-1)}{2} \right] + (D-2)^2$
$= ...$
$= (D-2)^2\left\{ \frac{1}{2} \frac{(D-1)^2}{2} + D \right\}$
$= 104976$
$= 324^2$
$= \left[ \frac{(D-2)(D-1)}{2} + (D-2) \right]^2$

The result is the same as the square of the coefficients of $x$ in Equation (14.63) on page 318.

$\frac{1}{2} \alpha' M^2~\|$	$N~\|$	$~\bar N~$	$\|~\text{Number of states}$
$-2~\|$	$0~\|$	$~0~$	$\|~1$
$0~\|$	$1~\|$	$~1~$	$\|~(D-2)^2$
$2~\|$	$2~\|$	$~2~$	$\|~(D-2)^2\left\{ \frac{1}{2} \frac{(D-1)^2}{2} + D \right\}$
$4~\|$	$3~\|$	$~3~$	$\|~3200^2$
$8~\|$	$4~\|$	$~4~$	$\|~25650^2$

— Me@2018-04-25 05:13:04 PM

Quantum decoherence 8

Posted on April 24, 2018 by rpflee

12. On the other hand, consistent histories are just a particular convenient framework to formulate physical questions in a certain way; the only completely invariant consequence of this formalism is the Copenhagen school’s postulate that physics can only calculate the probabilities, they follow the laws of quantum mechanics, and when decoherence is taken into account, to find both the quantum/classical boundary as well as the embedding of the classical limit within the full quantum theory, some questions about quantum systems follow the laws of classical probability theory (and may be legitimately asked) while others don’t (and can’t be asked)[.]

— Decoherence is a settled subject

— Lubos Motl

2018.04.24 Tuesday ACHK

Intuition

Posted on April 24, 2018 by rpflee

Unsourced variant:

The intellect has little to do on the road to discovery. There comes a leap in consciousness, call it intuition or what you will, and the solution comes to you and you do not know how or why. All great discoveries are made in this way.

The earliest published version of this variant appears to be The Human Side of Scientists by Ralph Edward Oesper (1975), p. 58, but no source is provided, and the similarity to the “Life Magazine” quote above suggests it’s likely a misquote.

In response to statement “You once told me that progress is made only by intuition, and not by the accumulation of knowledge.”

It’s not as simple as that. Knowledge is necessary, too. An intuitive child couldn’t accomplish anything without some knowledge.

There will come a point in everyone’s life, however, where only intuition can make the leap ahead, without ever knowing precisely how. One can never know why, but one must accept intuition as a fact.

— Albert Einstein

2018.04.24 Tuesday ACHK

Inception 16.4.2

Posted on April 24, 2018 by rpflee

潛行凶間 16.4.2

這段改編自 2010 年 8 月 13 日的對話。

…

例如，你在夢裡，正在期待著夢中故事的結局時，就不小心地醒了。你試過沒有？

（CPK：試過臨知道答案前，就醒了。）

你留意，《潛行凶間》就刻意拍到那樣—在觀眾就要知道結果時，就停了。換句話說，導演 Christopher Nolan 刻意把這部以夢為主題的電影，拍到有如夢境一樣。

那是他的習慣。例如，他的另一部電影—《死亡魔法》—以魔術為主題。導演就把它拍到彷彿魔術一般，由始至終，一路戲弄著觀眾。

所以，千萬不要連續看幾部，Christopher Nolan 的電影。看了一部已後，最好先隔一段時間，待心情平伏以後，才看下一部。

至於《潛行凶間》結尾，主角是否還在夢境之中？

是或否，兩個解釋，二選其一，都可以講得通。

你記不記得，戲中有一句，大概有以下的意思：「在夢中，通常都不會知道，自己正在發夢。只會在蘇醒時，才發現剛才，自己在發夢；才驚覺剛才，劇情不合理。」

你在看《潛行凶間》時，都有同樣的感覺。在戲院看時，覺得一切都合理，又有詳細解釋。但是，看完後，回家時，才驚覺劇情，有些地方不妥，講來講去講不通。其中一個例子是，陀螺會不會停止，並不能用來推斷，主角是否仍在夢中。

如果陀螺永不停止，那必定是夢中。但是，如果陀螺倒了下來，也不一定是現實，因為，即使在夢中，造夢者都可以令陀螺停止。

— Me@2018-04-24 11:36:31 AM

Some pundits have argued that the top was not in fact Cobb’s totem, rendering the discussion irrelevant. They say that the top was Mal’s totem; Cobb’s was his wedding ring, as he can be seen wearing it whenever he is in a dream and without it whenever he isn’t. As he hands his passport to the immigration officer, his hand is shown with no ring; thus he was conclusively in reality when seeing his children. Furthermore, the children were portrayed by different actors, indicating they had aged.

— Wikipedia on Inception

Wedding

Posted on April 23, 2018 by rpflee

d_2018_04_23__15_07_04_PM_

We’re starting our lives together.

Let’s spend all our money in one day.

—

2018.04.23 Monday ACHK

Problem 14.4a3

Posted on April 23, 2018 by rpflee

Closed string degeneracies | A First Course in String Theory

(a) State the values of $\alpha' M^2$ and give the degeneracies for the first five mass levels of the closed bosonic string theory.

~~~

…

But for non-massless states, this probably is not true anymore:

$\|$	$~\text{Number of states}$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} \| p^+, \vec p_T \rangle~\|$	$~\left[ \frac{(D-2)(D-1)}{2} \right]^2$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} \| p^+, \vec p_T \rangle~\|$	$~\left[ \frac{(D-2)(D-1)}{2} \right](D-2)$
${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} \| p^+, \vec p_T \rangle~\|$	$~(D-2)\left[ \frac{(D-2)(D-1)}{2} \right]$
${a_2^I}^\dagger \bar a_2^{K\dagger} \| p^+, \vec p_T \rangle~\|$	$(D-2)^2$

So the total number of states for $\frac{1}{2} \alpha' M^2 = 2$ ( $N = \bar N = 2$ ) is

$\left[ \frac{(D-2)(D-1)}{2} \right]^2 + \left[ \frac{(D-2)(D-1)}{2} \right](D-2)$
$+ (D-2)\left[ \frac{(D-2)(D-1)}{2} \right] + (D-2)^2$

Should $D$ be 10 or 26?

p.324 “Out of 26 left-moving bosonic coordinates of the bosonic factor only ten of them are matched by the right-moving bosonic coordinates of the superstring factor.”

$D$ should be 26 for bosonic strings. So the total number of states is

$\frac{1}{2} \alpha' M^2~\|$	$~\text{Number of states}$
$2~\|$	$~104976$

.
What does the difference of this part and Section 14.6 come from?

This part is for bosonic closed string, while Section 14.6 is for bosonic open string. There is no $\bar N$ to consider in Section 14.6.

p.290 “A basis vector $| \lambda, \bar \lambda \rangle$ belongs to the state space if and only if it satisfies the level-matching constraint”

$N^\perp = \bar N^\perp$ .

Can we create a formula for the number of states?

…

— Me@2018-04-23 11:31:16 AM

Is logic empirical?

Posted on April 22, 2018 by rpflee

Each of the logics is analytic.

“Which logic is the best for describing the world” is synthetic.

— Me@2012-04-14 11:32:36 AM

Almost nothing new

Posted on April 22, 2018 by rpflee

The most valuable insights are both general and surprising. F = ma for example. But general and surprising is a hard combination to achieve. That territory tends to be picked clean, precisely because those insights are so valuable.

…

Because these start out so general, you only need a small delta of novelty to produce a useful insight.

A small delta of novelty is all you’ll be able to get most of the time. Which means if you take this route your ideas will seem a lot like ones that already exist. Sometimes you’ll find you’ve merely rediscovered an idea that did already exist. But don’t be discouraged. Remember the huge multiplier that kicks in when you do manage to think of something even a little new.

…

It’s not true that there’s nothing new under the sun. There are some domains where there’s almost nothing new. But there’s a big difference between nothing and almost nothing, when it’s multiplied by the area under the sun.

— General and Surprising

— Paul Graham

2018.04.22 Sunday ACHK

機遇再生論 1.9.2

Posted on April 22, 2018 by rpflee

…

但是，未來時間是否無限長？

或者說，宇宙的壽命，是否無限呢？

可以參考的數據有：

宇宙現在的年齡，大概是只有十三億年（ $13.799 \times 10^9$ ）。

（問：那和宇宙壽命有無限，沒有直接關係。）

無錯。但那可以凸顯 $10^{10^{50}}$ 是多麼的大。

$10^{10^{50}}$ 大概是，宇宙現時年齡的 $10^{10^{50} - 10}$ 倍。

另外，即使假設了宇宙本身是，永在不滅的，你仍然可追問，物質粒子的壽命，又是否無限呢？

暫時，物理學家仍不知道，質子的壽命是否有限。

他們根據一些理論運算和實驗結果，估計質子壽命，大概有 $10^{29}$ 至 $10^{36}$ 年。但那仍然小於 $10^{10^{50}}$ 很多很多。

$10^{36} \ll 10^{10^{50}}$

（問：「宇宙」這個詞語的定義是「一切」。我們現時以為的「宇宙」，未必是真正的「宇宙」，因為，我們已知的「一切」，並非必定是，真正的「一切」。真正的「宇宙」，真正的「一切」，應連未知的部分，也包括在內。

所以，可能，真正宇宙的年齡，遠大於十三億年；可能，「 $10^{10^{50}}$ 年」對於真正宇宙來說，仍然是微不足道。）

無錯。未知永比已知多。而正正是這個理由，你既不可以假設，宇宙保證永在，亦不可以假設，宇宙必定有盡。

所以，「機遇再生論」的兩大假設的第一個——宇宙永在，並非必為正確（，除非你還有，額外的理據）。

「機遇再生論」有兩大（潛）假設：

1. 宇宙，有無限長的未來。

（這對應於撲克比喻中，「可以洗牌無限次」的假設。）

2. 宇宙中的粒子數目有限；而它們的組合及排列數目，都是有限的。

（這對應於撲克比喻中，「只有 52 隻牌」和「只有有限個排列」( $52! \approx 8.07 \times 10^{67}$ )的假設。）

「機遇再生論」的第二個假設，同第一個假設一樣，都是疑點重重。

…

— Me@2018-04-22 02:48:21 PM

Inspiring words

Posted on April 20, 2018 by rpflee

If you need inspiring words [for you to do something],

don’t do it.

— Elon Musk

2018.04.20 Friday ACHK

Quick Calculation 14.6b

Posted on April 20, 2018 by rpflee

A First Course in String Theory

Construct explicitly all the states with $\alpha' M^2=2$ and count them, verifying that there are indeed a total of 3200 states. You may find the counting formula in Problem 12.11 useful.

~~~

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

p.318 “For open bosonic strings $\alpha' M^2 = N^\perp - 1$ , …”

When $\alpha' M^2 = 2, N^\perp = 3$ , the cases are:

1. three $a_1^\dagger$ ‘s:

$N(3,24) = 2600$

2. one $a_2^\dagger$ and one $a_1^\dagger$ :

$24 \times 24 = 576$

3. one $a_3^\dagger$ :

Total number of possible states for $N^\perp = 3$ is 3200.

— Me@2018-04-20 02:45:35 PM

History 2.2

Posted on April 19, 2018 by rpflee

time ~ change

fact ~ no change

— Me@2017-12-31 11:43 AM

Time dissipates to shining ether the solid angularity of facts.

— Ralph Waldo Emerson

Life as a recursion, 5

Posted on April 18, 2018 by rpflee

recursion ~ act as if

self fulfilling ~ act as if

— Me@2011

Inception 16.4

Posted on April 18, 2018 by rpflee

潛行凶間 16.4

這段改編自 2010 年 8 月 13 日的對話。

（CPK：在自己的夢裡面，是不是真的可以，想怎樣就怎樣？）

有時可以。但只是有時。

剛才講過，

「

《潛行凶間》中的意念，你可以假想，有七成是真的。

」

亦即是話，有些部分不是真的。例如，現實中（暫時）並沒有那「夢境同步」機器。

又例如，現實中，你並不可以完全操控著，你的夢境。你可以控制到一點點，但不會完全控制到。

例如，你在夢裡，正在期待著夢中故事的結局時，就不小心地醒了。你試過沒有？

（CPK：試過臨知道答案前，就醒了。）

那是他的習慣。例如，他的另一部電影—《死亡魔法》—以魔術為主題。導演就把它拍到彷彿魔術一般，由始至終，一路戲弄著觀眾。

— Me@2018-04-18 02:48:05 PM

Sharp

Posted on April 17, 2018 by rpflee

d_2018_04_17__17_25_06_PM_

It’s easy to look sharp when you haven’t done any work.

—

2018.04.17 Tuesday ACHK

Problem 14.4a2

Posted on April 17, 2018 by rpflee

Closed string degeneracies | A First Course in String Theory

(a) State the values of $\alpha' M^2$ and give the degeneracies for the first five mass levels of the closed bosonic string theory.

~~~

p.288 Equation (13.48):

$M^2 = \frac{2}{\alpha'} \left( N^\perp + \bar N^\perp - 2 \right)$

p.290 “A basis vector $| \lambda, \bar \lambda \rangle$ belongs to the state space _if and only if_ it satisfies the level-matching constraint”

$N^\perp = \bar N^\perp$

$\frac{1}{2} \alpha' M^2~\|$	$~\text{Number of states}$
$-2~\|$	$~1$
$0~\|$	$~(D-2)^2$

$\frac{1}{2} \alpha' M^2~\|$	$~\text{Number of states}$
$2~\|$	$...$

${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} | p^+, \vec p_T \rangle$
${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} | p^+, \vec p_T \rangle$

Assuming the signs of the wave functions do not matter:

$\|$	$\|~\text{Number of states}$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger}~\|$	$\left[ \frac{(D-2)(D-1)}{2} \right]^2$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} \| p^+, \vec p_T \rangle~\|$
${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} \| p^+, \vec p_T \rangle~\|$

p.291 How come the number of the components of the matrix represents the number of states?

p.292 For massless states, we have only one ${a_1^{I}}^\dagger$ and $\bar a_1^{J\dagger}$ , where $a$ and $\bar a$ cannot interchange. So $I$ and $J$ also cannot interchange. In other words, in Equation (13.69), there is no double count. All the states are independent.

But for non-massless states, this probably is not true anymore:

$\|$	$~\text{Number of states}$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} \| p^+, \vec p_T \rangle~\|$	$~\left[ \frac{(D-2)(D-1)}{2} \right]^2$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} \| p^+, \vec p_T \rangle~\|$	$~\left[ \frac{(D-2)(D-1)}{2} \right](D-2)$
${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} \| p^+, \vec p_T \rangle~\|$	$~(D-2)\left[ \frac{(D-2)(D-1)}{2} \right]$

— Me@2018-04-17 04:47:31 PM

$\frac{1}{2} \alpha' M^2~\|$	$N~\|$	$~\bar N~$	$\|~\text{Number of states}$
$-2~\|$	$0~\|$	$~0~$	$\|~1$
$0~\|$	$1~\|$	$~1~$	$\|~(D-2)^2$
$2~\|$	$2~\|$	$~2~$	$\|~(D-2)^2\left\{ \frac{1}{2} \frac{(D-1)^2}{2} + D \right\}$
$4~\|$	$3~\|$	$~3~$	$\|~3200^2$
$8~\|$	$4~\|$	$~4~$	$\|~25650^2$

$\|$	$\|~\text{Number of states}$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger}~\|$	$\left[ \frac{(D-2)(D-1)}{2} \right]^2$
${a_1^I}^\dagger {a_1^J}^\dagger \bar a_2^{K\dagger} \| p^+, \vec p_T \rangle~\|$
${a_2^I}^\dagger \bar a_1^{K\dagger} \bar a_1^{L\dagger} \| p^+, \vec p_T \rangle~\|$

Physics Town

continues

Monthly Archives: April 2018

The Sixth Sense, 3

Story, 7

機遇再生論 1.10

Importance, 3

Problem 14.4a4

Quantum decoherence 8

Intuition

Inception 16.4.2

Wedding

Problem 14.4a3

Is logic empirical?

Almost nothing new

機遇再生論 1.9.2

Inspiring words

Quick Calculation 14.6b

History 2.2

Life as a recursion, 5

Inception 16.4

Sharp

Problem 14.4a2