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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# 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

— 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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# 時空兌換率

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$E = m c^2$

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$E = c^2 m$

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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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（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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# 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:

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

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

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

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

.

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

— Me@2018-05-08 08:50:32 PM

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# 技術細節 3

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— 尼采

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1. 應用

2. 傳播

3. 研究

— 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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# 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):

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

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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.

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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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# 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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# Story, 7

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story ~ a storage device for pieces of experience

— Me@2017-09-25 11:37:10 PM

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# 機遇再生論 1.10

「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。

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「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。

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（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。）

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— Me@2015.04.08

— Me@2017-12-09

— Me@2018-04-28

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# Importance, 3

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People never understand what you do for them …

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until you stop doing it!

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2018.04.26 Thursday 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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# 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

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（CPK：試過臨知道答案前，就醒了。）

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— 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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