Quick Calculation 14.7

Posted on June 29, 2015 by rpflee

A First Course in String Theory

Count the number of graviton, Kalb-Ramond, and dilation states in ten dimensions. Add these numbers up and confirm you get 64.

~~~

Equation (13.69):

$\sum_{I,J} \hat S_{IJ} {a_1^I}^\dagger {\bar a}_1^{J \dagger} | p^+, \vec p_T \rangle$

p.292 “… the states (13.69) represent one-particle graviton states … ”

Equation (13.64):

In the closed string state space, the general state of fixed momentum at the massless level is

$\sum_{I,J} R_{IJ} {a_1^I}^\dagger {\bar a}_1^{J \dagger} | p^+, \vec p_T \rangle$

Equation (13.68):

$R_{IJ} = \hat S_{IJ} + A_{IJ} + S' \delta_{IJ}$

$S' = \frac{S}{D-2}$

Equation (10.108):

For the symmetric traceless part, $\hat S_{IJ}$ (with the size $(D-2) \times (D-2)$ ), the number of independent components is

$n(D) = \frac{1}{2} (D - 2) (D - 1) - 1$

The $-1$ at the end reflects the fact that if the trace is zero, the last component of the trace is not independent anymore.

In ten dimensions ( $D = 10$ ),

$n(D) = \frac{1}{2} (8) (9) -1 = 35$

p.292 For $A_{IJ}$ (a skew-symmetric matrix), the number of independent components is

$n(D) = \frac{1}{2} (D - 3) (D - 2) = 28$

p.293 “The oscillator part of (13.71) has no free indices ( $I$ is summed over), so it represents one state. It corresponds to [an] one-particle state of a massless scalar field. This field is called the dilation.”

— Me@2015-06-29

2015.06.29 Monday (c) All rights reserved by ACHK

Predestination

Posted on June 25, 2015 by rpflee

Predestination is a 2014 Australian science fiction film, written and directed by Michael and Peter Spierig.

— Wikipedia on Predestination (film)

2015.06.25 Thursday ACHK

避免犯錯 2

Posted on June 23, 2015 by rpflee

懂得從自己錯誤中學習的，叫做聰明；

懂得從別人錯誤中學習的，叫做智慧。

—

2015.06.23 Tuesday ACHK

自由決定 2

Posted on June 23, 2015 by rpflee

這段改編自 2010 年 4 月 18 日的對話。

甚至，再有一個可能是，它們極端地兼容 — 其中一方直情建基於，另外一方。

前人提出過，如果沒有「注定」（物理等自然定律），人或其他生命體，就根本不可能有「自由意志」。

例如，你想拿起一隻茶杯。因為你有自由意志，所以可以由腦部下指令，訊號由神經線傳達到手部，拿起茶杯。手部正正是因為是「注定」的，即是受制於自然定律，才保證必會執行，腦部的指令。

試想想，如果手部未必根據自然定律來行事，它就不一定會執行，你心中的目標。那樣，你（腦部）反而就沒有自由意志，因為手部的動作根本是隨機的，不一定會把你（例如「拿起茶杯」）的意志，化成現實。

如果所有東西也是注定的，你就沒有自由。如果所有東西也是隨機（不注定）的，你也沒有自由。換句話說，「自由」是既建基於「注定」，亦建基於「不注定」，缺一不可。

所以，「自由」和「注定」的關係是，「自由」建基於「注定」。

它們的關係，有點像「著佐權」（copyleft）和「著作權」（copyright）。

字面上，兩者相反 — 前者是「反版權」，後者是「版權」。實際上，「著佐權」不是「反版權」。「著佐權」建基於「著作權」。正如我在幾個星期之前所講：

「

Copyleft 的確切執行，建基於完善的 copyright 制度。

當一個地區的著作權（copyright）制度，尚未成熟時，著佐權（copyleft）不能有效發展。

」

— Me@2015-06-21

Euler problem 27

Posted on June 20, 2015 by rpflee

Haskell

------------------------------


primes = 2 : filter (null . tail . primeFactors) [3,5..]
primeFactors n = factor n primes
                 where
                   factor n (p:ps)
                       | p*p > n = [n]
                       | n `mod` p == 0 = p : factor (n `div` p) (p:ps)
                       | otherwise = factor n ps
list_max (x:xs) = lmf x xs
    where 
      lmf x []                    = x
      lmf x xs | (x >= (head xs)) = lmf x (tail xs)
               | otherwise        = lmf (head xs) (tail xs)
elemInd y [] = -1
elemInd y (x:xs) = ei 0 y (x:xs)
                    where 
                      ei n y (x:xs) 
                          | (x == y)  = n
                          | (null xs) = -1
                          | otherwise = ei (n+1) y xs
                                    
b_list = takeWhile (<= 1000) primes
isPrime x | x <= 1    = False
          | otherwise = null $ tail (primeFactors x)
cPrimeLen [a, b] = [m, a, b]
    where m = length $ takeWhile (== True) $ map (isPrime . (\n -> n^2 + a*n + b)) [0..]
p27_list = [cPrimeLen [a, b] | a <- [-n..n], b <- b_list]    where n = 1000
p27_n = map head p27_list
p27 = p27_list !! (elemInd (list_max p27_n) p27_n)
-- 71, a == -61, b == 971

------------------------------

— Me@2015-06-19 10:01:22 PM

Quick Calculation 14.4

Posted on June 13, 2015 by rpflee

A First Course in String Theory

How many states are there at $N^\perp = \frac{3}{2}$ ?

~~~

This answer is my guess.

There are $(D-2)$ transverse (i.e. non-lightcone) coordinates. When $D = 10$ , the number of transverse coordinates is 8. So there are 8 states with $N^\perp = \frac{1}{2}$ .

For $N^\perp = \frac{3}{2}$ , there are 3 indices $I$ , $J$ , and $K$ . So there are totally 24 states.

— Me@2015.05.26 03:56 PM

The above answer is incorrect.

Equation (14.38) $\left(\alpha'M^2 = 1,~~N^\perp = \frac{3}{2}\right)$ :

$\{ \alpha_{-1}^I b_{-1/2}^J, b_{-3/2}^I, b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times 8$ states for

$\alpha_{-1}^I b_{-1/2}^J |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8$ states for

$b_{-3/2}^I |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $\frac{8 \times 7 \times 6}{3!}$ states for

$b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K |NS \rangle \otimes |p^+, \vec p_T \rangle$ .

So the total number of states is 64 + 8 + 56 = 128.

You can check this answer against Equation (14.67):

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

— Me@2015-08-21 08:15:50 AM

2015.06.13 Saturday (c) All rights reserved by ACHK

EPR paradox, 5.2

Posted on June 13, 2015 by rpflee

With a spacelike separation, it makes no absolute sense to say which of Alice and Bob causes the wave function to collapse.

— Me@2012.04.10

天空堤壩

Posted on June 12, 2015 by rpflee

Talent at its best and character at its worst

Lord Acton said we should judge talent at its best and character at its worst. For example, if you write one great book and ten bad ones, you still count as a great writer — or at least, a better writer than someone who wrote eleven that were merely good. Whereas if you’re a quiet, law-abiding citizen most of the time but occasionally cut someone up and bury them in your backyard, you’re a bad guy.

— Paul Graham

才能方面，如果一個人的最高點是可以接受的話，你就可以錄用他，即使他有其他任何才能缺點；

品德方面，如果一個人的最低點是不可接受的話，你就千萬不要錄用他，即使他有其他任何品德優點。

— Me@2010.03.06

才能方面，一個太陽可以照亮整個天空；

品德方面，一條裂縫可以摧毀整個堤壩。

— Me@2010.03.06

Speaking 6.3

Posted on June 10, 2015 by rpflee

這段改編自 2010 年 8 月 11 日的對話。

亦即是話，你在準備時，想像到「火、水、人」之中，任何兩者之間，有什麼關係，就即管（只管）收集起來。然後，把那堆關係排列妥當，形成一個上文下理。

例如，你可以這樣開始：

「

火，代表熱情。

水，代表冷漠。

人，就代表人生。

人生，就正正處於冷漠和熱情之間。

」

然後，你就加以舉例。只要你舉例舉得動聽，令人覺得你是「有腦人士」就可以。

「

其實，即使你內心十分熱情，如果你外表太有熱情，反而會阻礙工作進展。例如，即使你十分重視考試成績，你在考試作答期間，也應保持冷靜。太過激動，反而會傷害成績。

所以有些情況下，你需要『內熱外冷』。

但是，亦有些情況下，你應該相反，要做到『內冷外熱』。例如，雖然你平時內心十分平靜，但是如果遇到一些情境，你發現如果不發脾氣，就不能立刻制止，某些人做壞事的話，你就應該發脾氣，即使你不喜歡，因為那是劇情的需要，兩害取其輕。

」

最後，講出結論：

「

所以，在人生遇到的眾多處境之中，有時應該『外熱內熱』；有時應該『外冷內熱』；有時應該『外熱內冷』；有時應該『外冷內冷』。

至於什麼時候需要怎麼樣，不能一概而論，要隨機應變。

你不能期望，可以在年輕時，就準確地掌握到，因為，那需要人生的磨練。

」

你發覺，我並沒有講什麼實質的東西。

（HYC：但那假扮到十分有文有路。）

那只是一些簡單的技巧。只要看多幾題例子，你都可以做到。

— Me@2015.06.10

Euler problem 26

Posted on June 5, 2015 by rpflee

Haskell

——————————


rems n d xs 
    | (r == 0)    = xs ++ [0]
    | (elem r xs) = xs ++ [r] ++ [-1]
    | otherwise   = (rems (10*r) d (xs ++ [r]))
    where r = n `rem` d
remainders n d = rems n d []
recur_cyc (x:xs) 
    | (elem 0 ys)          = 0
    | (not (elem (-1) ys)) = -2        
    | (not (elem x xs))    = recur_cyc xs                          
    | otherwise            = (length ys) - 2
    where ys = (x:xs)
elemInd y [] = -1
elemInd y (x:xs) = ei 0 y (x:xs)
                    where 
                      ei n y (x:xs) 
                          | (x == y)  = n
                          | (null xs) = -1
                          | otherwise = ei (n+1) y xs                                     
list_max [x] = x
list_max (x:xs) = max x $ list_max xs
recur_list = map (\x -> (recur_cyc (remainders 1 x))) [1..1000] 
p26 = 1 + (elemInd (list_max recur_list) recur_list)

-- 983

——————————

— Me@2015-06-05 03:25:44 PM

Quick Calculation 14.6

Posted on June 4, 2015 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.

~~~

$\alpha' M^2 = 2~~~\to~~~N^\perp = 3$

The first possibility group includes the cases of one $a_3^\dagger$ . Since there are 24 species, this group has 24 possibilities.

The second possibility group includes the cases of one $a_1^\dagger$ with one $a_2^\dagger$ . (Their order is not important because interchanging the order would create the same state anyway.) Since there are 24 species for each of $a_1^\dagger$ and $a_2^\dagger$ , this group has $24 \times 24$ possibilities.

The third possibility group includes the cases of three $a_1^\dagger$ ‘s. We need to further divide this group into 3 sub-groups.

For the first sub-group, three $a_1^\dagger$ ‘s are all different:

$a_{1A}^\dagger a_{1B}^\dagger a_{1C}^\dagger$

Since there are 24 species for each $a_1^\dagger$ , there should be $24 \times 23 \times 22$ possibilities.

However, since their order is not important, there are repetitions among those possibilities. To eliminate repetitions, we divide the number with $3!$ . So, this subgroup has

$\frac{24 \times 23 \times 22}{6}$

possibilities.

For the second sub-group, two of the three $a_1^\dagger$ ‘s are identical:

$a_{1A}^\dagger a_{1B}^\dagger a_{1B}^\dagger$

~~There should be $24 \times 24$ possibilities.~~Me@2018-04-20 12:18:38 PM

There should be $24 \times 23$ possibilities.

For the third sub-group, all three are identical. In other words, there are 24 possibilities.

— Me@2015-06-04 10:19:30 PM

Truth values in Quantum Mechanics

Posted on June 3, 2015 by rpflee

It’s pretty much the same mistake that all the “realists” are doing all the time. They are imagining that they can assign truth values to all questions that could be in principle answered by experiments. (In some way, they give answers “No” or “zero” to all the other questions that were actually not addressed by the measurement.) But quantum mechanics prohibits that. If one assigns classical truth values (or real values) to some operators, one can no longer assign truth values (or real values) to “complementary” (not mutually commuting) operators and questions they represent. Instead of the correct statement that “the value of $N_a$ isn’t determined if one measures $L$ instead”, they say that it is zero which is just wrong.

— No particle upon a quantum field means no information

— Lubos Motl

2015.06.03 Wednesday ACHK

注定外傳 1.1

Posted on June 3, 2015 by rpflee

問過去的事情是不是注定，其實沒有意思，因為過去的事情已為固定，否則就不會稱為「過去」。

問未來的事情是不是注定，亦沒有意思，因為未來的事情尚未固定，否則就不會稱為「未來」。

換句話說，如果你在事後問，一件事是否注定，答案必為「是」。如果你在事前問，一件事是否注定，答案必為「不是」，因為它尚未發生，還有其他的可能性。

（問：那如果有某一件將來的事，改變到的機會率極小，近乎是零，該件事算不算是注定的呢？

例如，「明天太陽都會，由東方升起」是否注定呢？）

雖為未來之事，但發生的機會率極高，所以合理的答案，應為「那是注定的」。你問得很好。我原本的講法，的確有漏洞。

「過去」和「將來」的分別，並不是

「

~~前者『完全注定』，後者『完全不注定』。~~

」

「過去」和「將來」的真正的分別，應該是

「

前者『完全注定』，後者『不完全注定』。

」

「不完全注定」有兩個意思。

其一，將來的眾多事情中，有些事情是注定的，有些事情不是。

其二，將來的眾多事情中，有些事情本身是，部分注定和部分不注定的。亦即是話，有些個別事情的發生原因，既有注定的元素，亦有不注定的元素。

而過去的事情，則是完全固定的。

（問：但是，有沒有一個可能是，其實，將來所有事情的所有方面，都是注定的。所謂的「將來」，其實和「過去」一樣，都完全是固定的。

還有，你的講法的另一個不當之處是，當一個人問，某一件事是不是注定的，即使問的時間，是在該事件發生之後，他問題的意思，當然是「那件事，在事前是否注定的？」

否則，問者就是一個傻子。試想想，又怎會有人問，「那件事，在事後是否注定」呢？）

— Me@2015.05.26

Physics Town

continues

Monthly Archives: June 2015