Quantum Gravity

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We sometimes say that string theory is the only consistent theory of quantum gravity. It’s the only game in town. This is an observation mostly based on various types of circumstantial evidence. Whenever you try something that deviates from string/M-theory, you run into inconsistencies. Sometimes you don’t run into inconsistencies but something else happens. Many good ideas that were thought to be “competitors” to string theory were shown to be just aspects of some (usually special) solutions to string theory (noncommutative geometry, CFT, matrix models, and even the Hořava-Lifshitz class of theories have been found to be parts of string theory), and so on. And decades of attempts to find a truly inequivalent competing theory have utterly failed. That’s not a complete proof of their absence, either, but it is evidence that shouldn’t be completely ignored.

But that doesn’t mean that the statement that every consistent theory of quantum gravity has to be nothing else than another approach to string/M-theory is just an expression of vague feelings, a guesswork, or a partial wishful thinking. We don’t have the “most complete proof” of this assertion yet – this fact may be partly blamed on the absence of the completely universal, most rigorous definition of both “quantum gravity” and “string theory”. But there exist partial proofs and this paper is an example.

— Every theory of quantum gravity is a part of string theory: a partial proof

— A successful test in \(AdS_3\)

— Lubos Motl

2015.05.06 Wednesday ACHK

玄悟慧能 1.4

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這段改編自 2010 年 4 月 18 日的對話。

剛才我講「人生注定」問題時，猜想答案是「兩者之間」，但卻仍未知道，是「兩者之間」的哪一處。

或者，現實是以某種特別形式，存在於「完全自由」和「完全注定」之間 ——　特別到根本不是「之間」。其中一個可能是，「自由」和「注定」是同一樣「東西」的兩個方面。

而另一個可能是，它們是相容重疊的。比喻說，如果我宣稱，有一塊有限大的田地，是沒有邊緣的，你會覺得沒有可能。「有限大」的土地，又怎可能「沒有邊緣」呢？

原來是有可能的。地球表面，視為「一塊土地」的話，就是正正「有限而無邊」的，因為它是一個球體的表面。

我懷疑，「自由」和「注定」，雖然表面上，像「有限」和「無邊」般，貌似有衝突，但是在它們的內心深處，其實互相兼容。

至於怎樣兼容法，我暫時不知道。

— Me@2015.05.03

2015.05.05 Tuesday (c) All rights reserved by ACHK

Quick Calculation 14.2

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A First Course in String Theory
 
 

\sum_{s_1, s_2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle

Prove that “only the part of the wavefunction \psi that is antisymmetric under the simultaneous exchange p_1 \leftrightarrow p_2 and s_1 \leftrightarrow s_2 contributes…”

~~~

We divide the proof into two parts. In the first part, we prove that being antisymmetric under p_1 \leftrightarrow p_2 is a must. Then, in the second part, we use the same method to prove that being antisymmetric under s_1 \leftrightarrow s_2 is also a must.

First, we separate \psi into two parts:

\psi (p_1, s_1; p_2, s_2) = \psi_{sp} + \psi_{ap},

where \psi_{sp} is symmetric under p_1 \leftrightarrow p_2 and \psi_{ap} is anti-symmetric.

Wavefunctions \psi_{sp} and \psi_{ap} can be defined as

\psi_{sp} = \frac{1}{2} \left[ \psi (p_1, s_1; p_2, s_2) + \psi (p_2, s_1; p_1, s_2) \right]
\psi_{ap} = \frac{1}{2} \left[ \psi (p_1, s_1; p_2, s_2) - \psi (p_2, s_1; p_1, s_2) \right]

Now, we consider the contribution of the symmetric part, \psi_{sp} (p_1, s_1; p_2, s_2).

\int d \vec p \psi_{sp} (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle
= \frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle
+ \frac{1}{2} \int d \vec p \psi (p_2, s_1; p_1, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle

By interchanging the dummy variables p_1 and p_2 in the second term, we have

...
= \frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle
+ \frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_2, s_1}^\dagger f_{p_1, s_2}^\dagger | \Omega \rangle
=\frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) \left[ f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger + f_{p_2, s_1}^\dagger f_{p_1, s_2}^\dagger \right] | \Omega \rangle
= 0

So we have proved that the part symmetric under p_1 \leftrightarrow p_2 does not contribute.

Using the same method, we can also prove that the part symmetric under s_1 \leftrightarrow s_2 also does not contribute.

— Me@2015-05-02 01:10:27 PM
 
 
 
2015.05.02 Saturday (c) All rights reserved by ACHK

Speaking 6

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這段改編自 2010 年 8 月 11 日的對話。

.

至於口試，很大程度上，是要觀察那個考生，有沒有頭腦。其實，即使不是考試，而是日常生活，你只要花大概五分鐘，跟一個人對話，就可以大概感受到，他的智力或學養，暫時達到哪個水平。

換句話說，只要你智力不太低，你的口試表現，自然不會太差，所以不用太擔心。

當然，你可能會說，高考中，「個人短講」的題目，有時十分天馬行空，不知如何是好。

但是，正正是因為這個問題，我才會事先強調，在考試時，考官的主要目標，並不是要透過你的說話，去學到些什麼；而是要透過你的說話，去評價你的智力學養，從而給予分數。所以，面對空泛題目時，不單不要太過害怕，而且更反而要，深明其好處。

如果你現在有空泛題目的例子，你可以叫我示範一下，如何透過它們，去展示自己的腦袋實力。

— Me@2015.05.02

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2015.05.02 Saturday (c) All rights reserved by ACHK

Ideal clock 1.3

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An ideal clock is a clock (i.e., recurrent process) that makes the most other recurrent processes periodic.

— Wikipedia on Clock

The ideal clock itself is a meta-clock — a property of a set of clocks.

— Me@2015-04-16 9:35 AM

If you compare the accuracies of two clocks (A and B) by comparing each of them with a third clock (C), then I can ask, “How can you know that C is more accurate than both A and B?”

By defining the ideal clock as a meta-clock, we avoid infinite regress.

— Me@2015-04-27 11:36:59 AM

2015.04.29 Wednesday (c) All rights reserved by ACHK

Ideal clock 1.2

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A clock is a recurrent process and a counter.

A good clock is one which, when used to measure other recurrent processes, finds many of them to be periodic.

An ideal clock is a clock (i.e., recurrent process) that makes the most other recurrent processes periodic.

— Wikipedia on Clock

The ideal clock itself is a meta-clock — a property of a set of clocks.

— Me@2015-04-16 9:35 AM

2015.04.27 Monday (c) All rights reserved by ACHK

The greatest gift

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.

The greatest gift is a portion of thyself.

— Ralph Waldo Emerson

.

The greatest gift you can give someone is your time because when you are giving someone your time, you are giving them a portion of your life that you will never get back.

—

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2015.04.26 Sunday ACHK

玄悟慧能（審美篇）

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這段改編自 2010 年 4 月 18 日的對話。

.

其實，「審美」並不是，完全沒有客觀標準的。

其中一個客觀標準，就是「黃金分割」（黃金比例）。

This file is made available under the Creative Commons CC0 1.0 Universal Public Domain Dedication.

詳情請參閱 BBC 的紀錄片《The Human Face》。

但是，你又可以追問，為什麼人腦會傾向感覺到，「黃金分割」就是美？

— Me@2015.04.23

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2015.04.26 Sunday (c) All rights reserved by ACHK

重複 2

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這段改編自 2010 年 8 月 11 日的對話。

.

如果

1. 你真的有神人指導嶄新的讀書方法，或者高人提供重大的學術資料；

2. 而你原本的成績只是差了一點點，只要下年每科升一個等級，就可以加入到你心儀的大學中，至愛的學系；

3. 再加上，你覺得即使在短短數十年的人生中，花足足一整年的額外時間成本，都算値得的話，

你就可以考慮重讀。

.

但是，如果

1. 你已經用了老師、同學、朋友等，傳授給你的所有讀書心得；

2. 而你又想像不到，你在重讀年和第一年的學術生活，或者精神世界，風格會有什麼大分別的話，

你就千萬不要重讀。

.

記住，同一組輸入於同一個系統中，就會有同一組對應的輸出。亦即是話，如果重讀年的讀書風格，和第一年的差不多，你重讀年的高考成績，都會和第一年的不相上下。

千萬不要以為，在沒有高人（或者天書）的指點下，同一堆東西，只要讀多一次，功力就會倍增。正如，如果你唱歌難聽，並不會因為你唱多幾次，就唱得悅耳；如果你跑步太慢，並不會因為你跑多一公里，就突然變快。

要有大進步，一定要有專人訓練，而那個「專人」，最理想是教練，亦可以是自己，但是一定要是嶄新的自己。

總而言之，沒有新的輸入，就不會有新的輸出。

— Me@2015.04.17

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2015.04.19 Sunday (c) All rights reserved by ACHK

Godel 19

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… But it so happens that you just used second-order logic, because you talked about groups or collections of entities, whereas first-order logic only talks about individual entities.

— Standard and Nonstandard Numbers

— Eliezer Yudkowsky

— Less Wrong

2015.04.17 Friday ACHK

Replay

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stephengillie 64 days ago

In gaming, the concept is called a “replay”, where instead of recording the pixels on the screen in every frame, they instead record all inputs processed on every frame, and just replay them thru the same engine. The action is technically idempotent in the game world.

Where this breaks down is when features get updated between revisions. If your game patched the “jump” function to increase upward momentum from 1.1 m/s to 1.13 m/s, the Replay would be incorrect. You would be jumping onto platforms you couldn’t get up to before, moving faster, maybe even dodging enemy attacks that hit you when you played that match.

The human neuroprocessor is always changing and growing, always revising itself. Thus memories replay incorrectly. You apply old feelings to new mental patterns, and sometimes they lead to weird places. Or sometimes you mistake something easy for being difficult, because your memory data is out-of-date for your current processes. 

— Hacker News

2015.04.16 Thursday ACHK

機遇再生論 1.4

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「

『機遇再生論』的大概意思是，所有可能發生的事情，例如重生，只要等足夠長的時間，總會發生。

」

但是，即使避開了「無限」，用了「足夠長」，仍然會有其他問題。「足夠長」這個詞語雖然不算違法，但是十分空泛，空泛到近乎沒有意義。

試想想，怎樣才為之「足夠長」呢？

.

以前在本網誌中提及過，凡是科學句子，都一定要有「可否證性」。因為凡是科學句子，都對世界有所描述，所以必為「經驗句」，不是「重言句」。凡是「經驗句」，必定有機會錯。換而言之，無論正確的機會率有多高，都不會是百分百。

因此，要測試某一句說話，是不是「科學句子」，你可以檢查一下，它有沒有「可否證性」。「可否證性」的意思是，如果一句「科學句子」有意義，你就可以講得出，至少在原則上，它在什麼情況下，為之錯。

例如，

「

甲在過身之後，一千億年內會重生。

」

是句「科學句」（經驗句），因為你知道在什麼情境下，可以否證到它 —— 如果你在甲過身後，等了一千億年，甲還未重生的話，那句就為之錯。

但是，

「

甲在過身之後，只要等足夠長的時間，必會重生。

」

則沒有任何科學意義。

— Me@2015.04.08

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2015.04.15 Wednesday (c) All rights reserved by ACHK

重複

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這段改編自 2010 年 8 月 11 日的對話。

.

（CYW：如果這次高考，到了最後，都考不上大學，我應不應該重讀，下年再考多一次？）

沒有一定的答案，因為要考慮的因素實在有太多，一定要看上文下理，不宜妄下判斷。

大方向是，你重讀的那一年，讀書情境會不會有大改變？

.

你要問一問自己，你知不知道，第一次高考失敗的主要原因是什麼？在下學年的重讀中，你可不可以把該原因刪除？

如果你只有空泛的答案，你重讀也只會有空泛的成績。

.

如果你第一次高考不成功，是因為缺乏某些讀書方法，或者學術內容，而在下個學年開始前，剛好有高人開始指點一些心得，再加上你感受到那些心得，明顯是對症下藥的話，你重讀就有意思，因為成績有很大的機會，會大大提升。

相反，如果你的讀書系統沒有大分別，那樣，重讀以後的成績，都會差不多；即使進步，也只會是一點點。

至於那一點點足不足夠，則因人而異。

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重點是，千萬不要在不明所以的情況下，選擇重讀。相反，你應該詳細調查和考慮（包括重讀在內的）所有升學方案。權衡輕重後，你發覺重讀是「眾害取其輕」的話，才選擇重讀。

— Me@2015.04.10

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2015.04.13 Monday (c) All rights reserved by ACHK