Ask users

gallerdude 3 months ago

There’s a good moral here. Everytime they had a question, they asked their users. Users don’t lie.

atYevP 3 months ago

Yev from Backblaze here -> That’s right! For better or worse they usually tell it like they see it, and that helps us inform decisions!

— How Backblaze Got Started (2017)

— Hacker News

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2018.05.14 Monday ACHK

Problem 14.4b1.3

Closed string degeneracies | A First Course in String Theory

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(b) State the values of \alpha' M^2 and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

— This answer is my guess. —

Since for NS, the first 5 levels’ degeneracies are 8, 36, 128, 402, 1152, the degeneracies of (NS, NS) are 8^2, 36^2, 128^2, 402^2, 1152^2.

This is incorrect, for there are no (NS, NS) states. Instead, you should consider (NS+, NS+).

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Since for NS+, the first 5 levels’ degeneracies are 8, 128, 1152, 7680, 42112, the degeneracies of (NS+, NS+) are 8^2, 128^2, 1152^2, 7680^2, 42112^2.

p.317 Consider the relationship of the degeneracy of R+ and that of R-:

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How about the first 5 levels of R+?

The degeneracies are the same as those of R-.

p.317 Equation (14.54) “The appearance of an equal number of bosonic and fermionic states at every mass level is a signal of supersymmetry. This is, however, supersymmetry on the world-sheet.”

Equation (14.71):

f_{R-}(x) = 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ...

p.321 “Indeed, the integer mass-squared levels in the NS generating function (14.67) have degeneracies that match those of (14.71) for the R- sector.”

— This answer is my guess. —

— Me@2018-05-14 02:51:55 PM

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2018.05.14 Monday (c) All rights reserved by ACHK

Van der Waals equation 1.1

Why do we add, and not subtract, the correction term for pressure in [Van der Waals] equation?

Since the pressure of real gases is lesser than the pressure exerted by (imaginary) ideal gases, shouldn’t we subtract some correction term to account for the decrease in pressure?

I mean, that’s what we have done for the volume correction: Subtracted a correction term from the volume of the container V since the total volume available for movement is reduced.

asked Sep 30 ’16 at 15:20
Ram Bharadwaj

— Physics Stackexchange

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Ideal gas law:

P_{\text{ideal}} V_{\text{ideal}} = nRT

However, since in a real gas, there are attractions between molecules, so the measured value of pressure P is smaller than that in an ideal gas:

P_{\text{measured}} = P_{\text{real}}

P_{\text{measured}} < P_{\text{ideal gas}}

Also, since the gas molecules themselves occupy some space, the measured value of the volume V is bigger that the real gas really has:

V_{\text{measured}} > V_{\text{real}}

P_{\text{ideal}} V_{\text{ideal}} = nRT

If we substitute P_{\text{measured}} onto the LHS, since P_{\text{measured}} < P_{\text{ideal}}, the LHS will be smaller than the RHS:

P_{\text{measured}} V_{\text{ideal}} < nRT

So in order to maintain the equality, a correction term to the pressure must be added:

\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) V_{\text{ideal}} = nRT

P_{\text{ideal}} V_{\text{ideal}} = nRT

If we substitute V_{\text{measured}} onto the LHS, since that volume is bigger that actual volume available for the gas molecules to move, the LHS will be bigger than the RHS:

P_{\text{ideal}} V_{\text{measured}} > nRT

So in order to maintain the equality, a correction term to the pressure must be subtracted:

P_{\text{ideal}} \left(V_\text{measured}-nb\right) = nRT

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In other words,

V_{\text{measured}} > V_{\text{real}}

V_{\text{ideal}} = V_{\text{real}}

V_{\text{measured}} > V_{\text{ideal}}

— Me@2018-05-13 03:37:18 PM

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Why? I still do not understand.

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How come

P_{\text{measured}} = P_{\text{real}}

but

V_{\text{measured}} \ne V_{\text{real}}?

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How come

V_{\text{real}} = V_{\text{ideal}}

but

P_{\text{real}} \ne P_{\text{ideal}}?

— Me@2018-05-13 03:22:54 PM

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The above is wrong.

The “real volume” V_{\text{real}} has 2 possible different meanings.

One is “the volume occupied by a real gas”. In other words, it is the volume of the gas container.

Another is “the volume available for a real gas’ molecules to move”.

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To avoid confusion, we should define

V_{\text{real}} \equiv V_{\text{measured}}

P_{\text{real}} \equiv P_{\text{measured}}

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Or even better, avoid the terms P_{\text{real}} and V_{\text{real}} altogether. Instead, just consider the relationship between (P_{\text{ideal}}, P_{\text{measured}}) and that between (V_{\text{ideal}}, V_{\text{measured}}).

Whether X_{\text{measured}} is bigger or smaller than X_{\text{ideal}} ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

— Me@2018-05-13 04:15:34 PM

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2018.05.13 Sunday (c) All rights reserved by ACHK

時空兌換率

這段改編自 2015 年的對話。

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我的相對論教授說,所謂

E = m c^2

在某些意思之下,沒有那麼特別,因為,你可以把它看成,貨幣的兌換。

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E = c^2 m

能量 =(光速二次方)\times 質量

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1 \text{USD} \approx 8 \times 1 \text{HKD}

1 美元 \approx 8 \times 1 港元

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公式中的 c^2(光速平方),角色其實正正就是,能量 E 和質量 m 之間的「貨幣兌換率」。

(而光速 c,則是時間和空間的兌換率。)

— Me@2018-05-11 09:10:00 PM

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2018.05.11 Friday (c) All rights reserved by ACHK

Sleep apnea

erentz 3 months ago

Slightly tangential to the study here but I’ve been going down the sleep apnea rabbit hole in recent months and I strongly encourage folks to investigate their sleep. A lot of people have sleep apnea and don’t realize it. You don’t have to snore to have it. A surprising statistic I found was 20-30% of people with ADHD have sleep apnea. A lot of people may be treating symptoms of sleep apnea like ADHD and high blood pressure with medications while ignoring the root cause. My experience with this has been that doctors are surprisingly ignorant. They’ll happy prescribe you medications for anxiety, ADHD, blood pressure for years, and never think to ask you about your sleep. Do some of your own investigations or ask about it if you have any suspicions.

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copperx 3 months ago

I’m going to add myself as a data point. After suffering years of feeling sleepy regardless of how much I slept and frequently unmotivated and withdrawn, I was diagnosed with severe sleep apnea. A few years ago I was given Dexedrine, Ritalin, Adderall, and I even tried self-medicating with modafinil to ameliorate the symptoms of what doctors thought was ADHD with varying levels of success; but the drug-free treatment of sleep apnea with a BiPAP machine got rid of all of these issues; in addition, I feel about ten years younger.

— Sleep and Mortality: A Population-Based 22-Year Follow-Up Study (2007)

— Hacker News

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2018.05.09 Wednesday ACHK

Problem 14.4b1.2

Closed string degeneracies | A First Course in String Theory

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(b) State the values of \alpha' M^2 and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

NS+ equations of (14.38):

\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,

For N^\perp = \frac{5}{2}, the number of states is

8^2 + \left[ \frac{(8)(7)}{2!} + 8 \right] (8) + 8^2
+ 8 + \frac{(8)(7)(6)}{3!} + 8 + (8) \left[ \frac{(8)(7)}{2!} \right] + \frac{(8)(7)(6)(5)(4)}{5!}
= 1152

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Since \alpha' M^2 = N^\perp - \frac{1}{2}, when N^\perp = \frac{5}{2}, \alpha' M^2 = 2.

Equation (14.67):

f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

Equation (14.66):

f_{NS} (x) = \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8

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p.321

If we take f_{NS} (x) in (14.66) and change the sign inside each factor in the numerator

Equation (14.72):

\frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8

the _only_ effect is changing the sign of each term in the generating function whose states arise with an odd number of fermions.

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\prod_{n=1}^\infty \left( \frac{1}{1-x^n} \right)^8 is the boson contribution.

\prod_{n=1}^\infty \left( 1+x^{n-\frac{1}{2}} \right)^8 is the fermion contribution.

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Turning Equation (14.67) into

f_{NS?} (x)
= - \frac{1}{\sqrt{x}} + 8 - 36 \sqrt{x} + 128 x - 402 x \sqrt{x} + 1152 x^2 - ...

is equivalent to turning all \sqrt{x} into - \sqrt{x}:

f_{NS?} (x)
= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8

Me@2015.08.29 12:49 PM: Somehow, \sqrt{x} represents “contribution from fermions”.

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Me@2015.08.29 12:50 PM: If you still cannot understand, try replace all \sqrt{x} with y.

f_{NS+} (x) = \frac{1}{2} \left( f_{NS} - f_{NS?} \right)

f_{NS} (x)
= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

f_{NS?} (x)
= - \frac{1}{\sqrt{x}} + 8 - 36 \sqrt{x} + 128 x - 402 x \sqrt{x} + 1152 x^2 - ...

f_{NS+} (x)
= \frac{1}{2} \left( f_{NS} - f_{NS?} \right)
= \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x \sqrt{x} + ...

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It is _not_ correct. Just consider it as \left(\sqrt{x} \to -\sqrt{x} \right) is not correct, since the beginning factor \frac{1}{\sqrt{x}} is not considered yet.

Instead, we should present in the following way:

f_{NS} (x)
= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= \frac{1}{\sqrt{x}} g_{NS}(x)
= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...

g (\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 1 + 8 \, \sqrt{x} + 36 \, x + 128 \, x^{\frac{3}{2}} + 402 \, x^{2} + 1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...

g (-\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 1 -8 \, \sqrt{x} + 36 \, x -128 \, x^{\frac{3}{2}} + 402 \, x^{2} -1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...

.

g (\sqrt{x}) - g (-\sqrt{x})
= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 - \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8
= 16 \, \sqrt{x} + 256 \, x^{\frac{3}{2}} + 2304 \, x^{\frac{5}{2}} + 15360 \, x^{\frac{7}{2}} + ...

.

f_{NS+}(x) = \frac{1}{2 \sqrt{x}} \left[ g (\sqrt{x}) - g (-\sqrt{x}) \right]
= \frac{1}{2 \sqrt{x}} \left[ \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 - \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8 \right]
= \frac{1}{2 \sqrt{x}} \left[ 16 \, \sqrt{x} + 256 \, x^{\frac{3}{2}} + 2304 \, x^{\frac{5}{2}} + 15360 \, x^{\frac{7}{2}} + ... \right]
= 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ...

— Me@2018-05-08 08:50:32 PM

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2018.05.08 Tuesday (c) All rights reserved by ACHK

技術細節 3

親歷其境 3

這段改編自 2010 年 4 月 18 日的對話。

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一切文學,余愛以血書者。

— 尼采

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剛才講到,我想去升學那間大學的物理系,其獨特優秀之處。

世界上,只有極少數大學,提供超弦理論的課程。而且,那個碩士課程,不是純修課式。它為期兩年:第一年是上課,第二年是研究。

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對於我現在這個年紀,單純知識的吸收(簡稱「學習」),沒有大意義。對我有大意義的是,研究和製造。

或者說,即使一個人對學術有極大興趣,十分喜歡學習,都不應做「永恆學生」。「永恆學生」在這裡,作負面意思用。「永恆學生」的特徵是,知識多而能力差,簡稱「高分低能」。

留意,所謂「知識多而能力差」,其實就即是,知識流於表面,未詳細到可以應用。例如,即使你對中國象棋中的每一隻,行走的方法,非常熟悉,那並不代表,你「懂」怎樣下棋。

要真正下到棋,你要靠不斷嘗試,長期的實戰。那樣,你才可以獲得下一個層次的知識;令自己對於象棋,知得足夠詳細,詳細到足以有效實行。

有時,知易行難,是因為知得不夠深入;

有時,行而無效,是因為行得不夠徹底。

永久單純地,吸收新知識,而沒有下一個層次的發展,是沒有意義的。或者說,知識上,只有輸入而沒有輸出,是病態。

所謂「下一個層次的發展」,可以是:

1. 應用

隨著越來越多次的應用,你於該個領域的知識,再不是依書直說,而是真才實學。

2. 傳播

如果你要傳播知識,必須靠說話或文章。

無論用兩者的哪一個,你必須透過自己的「語言」。換句話說,你必須為舊知識,創造一個新的包裝;即使,有時,你是不自覺地那樣做,未必意識到自己正在從是創作。

還有,有時,你亦會無意中,有知識上的新發現—在為舊知識創造新包裝時,竟然引發了新知識。

3. 研究

透過舊知識,發掘新知識。

— Me@2018-05-05 10:06:15 AM

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Know the subject—At the Dale Carnegie public speaking class, which I highly recommend—they say anyone can make a good speech if he or she has _earned the right_ to speak on the subject in question. How do you earn the right? By living through the subject or by doing extensive research on it—which is arguably another form of living through it. Same principle applies to how-to writing. You cannot do high school student research. That is, find 21 facts and write an essay that consists solely of those 21 facts. Rather you need the proverbial iceberg of unused facts under the “tip” that is your book. The good news is virtually everyone has lived through something that fits that criterion. And, with a year or so, we can all research something that interest us to the point where we can write about it. The key to bad writing is assigned topics—the standard of high school and college teachers.

— John T. Reed

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2018.05.06 Sunday (c) All rights reserved by ACHK

The best software

somethingsimple 3 months ago

> Software is the currency that we pay to solve problems, which is our actual goal. We should endeavor to build as little software as possible to solve our problems.

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ScottBurson 3 months ago

My point today is that, if we wish to count lines of code, we should not regard them as “lines produced” but as “lines spent”: the current conventional wisdom is so foolish as to book that count on the wrong side of the ledger. — E. W. Dijkstra

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That’s why the best choice of software is often no software …

— Coding Horror

— by Jeff Atwood

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— Write dumb code

— Hacker News

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2018.05.01 Tuesday ACHK

Problem 14.4b1.1

Closed string degeneracies | A First Course in String Theory

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(b) State the values of \alpha' M^2 and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

Type IIA closed superstrings

p.322

In closed superstring theories spacetime bosons arise from the (NS, NS) sector and also from the (R, R) sector, since this sector is “doubly” fermionic. The spacetime fermions arise from the (NS, R) and (R, NS) sectors.

p.322 \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

Equation (14.77):

\text{type IIA}:~~~(NS+, NS+), ~(NS+, R+),~ (R-, NS+), ~ (R-, R+)

.

What is the difference of the meanings of R+ and R-?

R+ states are world-sheet bosonic states.

p.316

It thus follows that all eight | R_a \rangle states are fermionic and all | R_{\bar a} \rangle are bosonic.

Be careful:

Here, “fermionic”/”bosonic” refers to the world-sheet fermions/bosons, not the spacetime ones.

p.320

Identifying | R_a \rangle as spacetime fermions and | R_{\bar a} \rangle as spacetime bosons is not an alternative either, since spacetime bosons cannot carry a spinor index.

.

How come the R+ cannot be the left-moving part?

p.320

A strategy then emerges. Since all states in the R sector have a spinor index, we will only attempt to get spacetime fermions from this sector. We also recognize that all fermions must arise from states with the same value of (-1)^F.

Me@2015.09.11 10:36 AM: In other words, it is a convention:

Following Gliozzi, Scherk, and Olive (GSO) we proceed to truncate the Ramond sector down to the set of states with (-1)^F = -1.

— Me@2018-05-01 05:59:53 PM

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2018.05.01 Tuesday (c) All rights reserved by ACHK

The Sixth Sense, 3

Mirror selves, 2 | Anatta 3.2 | 無我 3.2

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You cannot feel your own existence or non-existence. You can feel the existence or non-existence of (such as) your hair, your hands, etc.

But you cannot feel the existence or non-existence of _you_.

— Me@2018-03-17 5:12 PM

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Only OTHER people or beings can feel your existence or non-existence.

— Me@2018-04-30 11:29:08 AM

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2018.04.30 Monday (c) All rights reserved by ACHK

機遇再生論 1.10

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。

.

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。

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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。

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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

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2018.04.28 Saturday (c) All rights reserved by ACHK

Problem 14.4a4

Closed string degeneracies | A First Course in String Theory

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(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

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Can we create a formula for the number of states?

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\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

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2018.04.25 Wednesday (c) All rights reserved by ACHK

Quantum decoherence 8

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

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2018.04.24 Tuesday ACHK

Intuition

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

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2018.04.24 Tuesday ACHK

Inception 16.4.2

潛行凶間 16.4.2

這段改編自 2010 年 8 月 13 日的對話。

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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。

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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。

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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。

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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。

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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。

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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

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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

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2018.04.24 Tuesday (c) All rights reserved by ACHK