Mathematics, 2

Posted on September 3, 2018 by rpflee

Approximately speaking, mathematics is whatever language with the maximum precision.

— Me@2012-04-16

Self de-centralization, 3

Posted on September 2, 2018 by rpflee

Life is a process of self de-centralization. If you do not follow that path. You feel unhappy.

— Me@2011.08.19

神的旨意 2.4

Posted on September 2, 2018 by rpflee

…

魔：為什麼「全能」者不可「全惡」？

甲：你如果是「全能」，就毋須問我這個問題。

如果你是「全惡」，你的構成部分，就不能相處。而「你」，作為一個整體，並不會存在；必須散落成一大堆，獨立的部分而存在。而各個分部，各自內裡必有善部，才可能凝聚，不再分裂。

魔：即使我不是「全惡」；即使我有所謂「善部」，你難保我「惡部」的法力，大於「善部」？

甲：

第一，即使假設是那樣，那也沒有大意義，因為，你總不能完全忽略，你善部的旨意。

如果你的惡部，完全不理善部地作惡，那就即是，你的善部名存實亡。沒有善部，「你」必會分裂。

第二，除了你存在以外，我也存在；其他生命體也存在。

善的會合作，那是定義。善的會合作，去抵擋你的惡。

雖然，惡部有「破壞容易過建設」的優勢，但是，善部也可以應用「破壞容易過建設」，去破壞惡部的破壞計劃。

第三，惡人自有惡人磨：

相似的人，因為各種原因，傾向身處相近的地方，簡稱「物以類聚」。

壞人的身邊，通常是其他壞人。壞人最怕的，往往是其他壞人。

最終對付你的人，是你自己。最終對付你惡部的人，是你自己的惡部。

— Me@2018-09-02 03:05:45 PM

Illusions destroyed, 2

Posted on September 1, 2018 by rpflee

Ask HN: Why is nearing completion so demotivating?

534 points by danschumann 3 months ago

So I’ve been working on animation software for over two years. Part of me is very excited for launch so I can have money again (I’ve been freelancing a minimum amount these last two years, and went car-less, moved, cut lifestyle into a third). I should be wholeheartedly excited, but I’m feeling tired and generally sluggish regarding the project. I still make consistent progress, but it takes a lot of will power.

Part of me thinks it might be an aversion to sales. Part of me thinks this could have been built up so much in my head that anything short of overnight millions would be a disappointment (though I would be happy with 1500 bucks a month), part of me thinks I might be scared of success (or scared of surpassing my parents) (media attention), part of me fears the attacks that might come with success (having something to lose), part of it is the un-fun-ness of mature projects where the focus is on polish and bugs rather than broad new features, and part of me is scared of commitment: if I succeed I have to stick with this (freedom value), part of me wonders what will happen when more people become involved, if I will be able to maintain my creative direction, since I’m scratching my own itch. Part of me wonders if diet and exercise isn’t a factor.

A combination, likely…

mikekchar 3 months ago [-]

When your project is finished, the dream is dead and the reality is born. The death of a dream is like the death of a friend. It’s probably been with you for a long time — longer even than the length of the project. A dream is the manifestation of what’s possible. When it is over, the possible diminishes very quickly and you are left with what actually is. Will people respond well to your project — in the dream stage it is possible; everything is possible. In the reality stage, it will only be what it is.

So while it’s common to think of a release as a birth of something new, realise that you also have a significant loss. You will mourn that loss. Give yourself some emotional space to deal with the mourning.

riantogo 3 months ago [-]

This is exactly it. Tens of my personal projects have died in this stage. It was always much easier to move on to the next dream. There is always the next big problem that could use a solution. Why not build when it is what we do best? Rinse, repeat.

I took a break from side projects for several years but recently got back to it and couple weeks back finished building. It is the same story all over again. Same feeling. I’m dreading what comes next.

— Why is nearing completion so demotivating?

— Hacker News

2018.09.01 Saturday ACHK

Problem 14.5a2

Posted on September 1, 2018 by rpflee

Counting states in heterotic SO(32) string theory | A First Course in String Theory

(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

$\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

— This answer is my guess. —

Equation at Problem 14.5:

$\displaystyle{\alpha' M_L^2}$

$\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{8} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$
$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right]}$

Equation (13.116):

$\displaystyle{\sum_{k \in \mathbf{Z}^+_{\text{odd}}} k = \frac{1}{12}}$

$\displaystyle{\begin{aligned} &\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A \\ &= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right] \\ &= - \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ &= - \frac{1}{2} \sum_{r = 1, 3, ...} r + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ &= - \frac{1}{24} + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \end{aligned}}$

$\displaystyle{ \begin{aligned} \alpha' M_L^2 &= \frac{-7}{48} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ \end{aligned}}$

If we define $N^\perp$ in the way similar to equation (14.37), we have

$\displaystyle{ \begin{aligned} \alpha' M_L^2 &= \frac{-7}{48} + N^\perp \\ \end{aligned}}$

— This answer is my guess. —

— Me@2018-09-01 06:05:29 AM

The square root of the probability

Posted on August 19, 2018 by rpflee

Probability amplitude in Layman’s Terms

What I understood is that probability amplitude is the square root of the probability … but the square root of the probability does not mean anything in the physical sense.

Can any please explain the physical significance of the probability amplitude in quantum mechanics?

edited Mar 1 at 16:31
nbro

asked Mar 21 ’13 at 15:36
Deepu

Part of you problem is

“Probability amplitude is the square root of the probability […]”

The amplitude is a complex number whose amplitude is the probability. That is $\psi^* \psi = P$ where the asterisk superscript means the complex conjugate. ${}^{[1]}$ It may seem a little pedantic to make this distinction because so far the “complex phase” of the amplitudes has no effect on the observables at all: we could always rotate any given amplitude onto the positive real line and then “the square root” would be fine.

But we can’t guarantee to be able to rotate more than one amplitude that way at the same time.

More over, there are two ways to combine amplitudes to find probabilities for observation of combined events.

When the final states are distinguishable you add probabilities:

$P_{dis} = P_1 + P_2 = \psi_1^* \psi_1 + \psi_2^* \psi_2$

When the final state are indistinguishable, ${}^{[2]}$ you add amplitudes:

$\Psi_{1,2} = \psi_1 + \psi_2$

and

$P_{ind} = \Psi_{1,2}^*\Psi_{1,2} = \psi_1^*\psi_1 + \psi_1^*\psi_2 + \psi_2^*\psi_1 + \psi_2^* \psi_2$

The terms that mix the amplitudes labeled 1 and 2 are the “interference terms”. The interference terms are why we can’t ignore the complex nature of the amplitudes and they cause many kinds of quantum weirdness.

${}^1$ Here I’m using a notation reminiscent of a Schrödinger-like formulation, but that interpretation is not required. Just accept $\psi$ as a complex number representing the amplitude for some observation.

${}^2$ This is not precise, the states need to be “coherent”, but you don’t want to hear about that today.

edited Mar 21 ’13 at 17:04
answered Mar 21 ’13 at 16:58

dmckee

— Physics Stack Exchange

Number, Time, Money, 3

Posted on August 15, 2018 by rpflee

time ~ change

time interval

~ amount of change

~ quota of change

— Me@2017-09-19 02:48:22 PM

— Me@2018-04-22 06:00:05 PM

神的旨意 2.3

Posted on August 13, 2018 by rpflee

神：我是神，而你只是人。神的心思，凡人不一定能夠理解。所以，作為凡人的你，應該心存謙卑，完全信任你的神，奉行神的旨意。

甲：不合理呀！

如果要殺害無辜，才可上天堂，而不殺害無辜，會被罰落地獄的話，那代表著，你天堂的居民，都是壞人；而你地獄的居民，反而是善人。

那樣，你的所謂「天堂」，根本是我的地獄。而你雖然自封為「神」，卻根本是我的「魔」。

魔：我是魔又如何？我的法力遠高於你。我可以令你求生不得，求死不能。

甲：會有真正善良的神，去保護我。

魔：難保我的法力高過善神。「魔」就不可以是全能的嗎？

甲：如果是全能，你怎會有動機，去虐待凡人呢？既是全能者，又怎會那麼無聊，損人不利己？

另外，如果你法力高強，乃至「全能」，你就不可能是「全惡」的。你必定有善部和惡部。你的善部，就是善神。

魔：為什麼「全能」者不可「全惡」？

…

— Me@2018-08-13 11:54:47 AM

The Jacobian of the inverse of a transformation

Posted on August 9, 2018 by rpflee

The Jacobian of the inverse of a transformation is the inverse of the Jacobian of that transformation

In this post, we would like to illustrate the meaning of

the Jacobian of the inverse of a transformation = the inverse of the Jacobian of that transformation

by proving a special case.

Consider a transformation $\mathscr{T}: \bar{x}^i=\bar{x}^i (x^1,x^2)$ , which is an one-to-one mapping from unbarred $x^i$ ‘s to barred $\bar{x}^i$ coordinates, where $i=1, 2$ .

By definition, the Jacobian matrix J of $\mathscr{T}$ is

$J= \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{pmatrix}$

Now we consider the the inverse of the transformation $\mathscr{T}$ :

$\mathscr{T}^{-1}: x^i=x^i(\bar{x}^1,\bar{x}^2)$

By definition, the Jacobian matrix $\bar{J}$ of this inverse transformation, $\mathscr{T}^{-1}$ , is

$\bar{J}= \begin{pmatrix} \displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^1}{\partial \bar{x}^2}} \\ \displaystyle{\frac{\partial x^2}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^2}{\partial \bar{x}^2}} \end{pmatrix}$

On the other hand, the inverse of Jacobian $J$ of the original transformation $\mathscr{T}$ is

$J^{-1}=\displaystyle{\frac{1}{ \begin{vmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{vmatrix} }} \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} & \displaystyle{-\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{-\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} \end{pmatrix}$

If $\bar{J} = J^{-1}$ , their (1, 1)-elementd should be equation:

$\displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}}\stackrel{?}{=}\displaystyle{\frac{1}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}-\displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} }} \bigg( \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \bigg)$

Let’s try to prove that.

Consider equations

$\bar{x}^1 = \bar{x}^1(x^1,x^2)$

$\bar{x}^2 = \bar{x}^2(x^1,x^2)$

Differentiate both sides of each equation with respect to $\bar{x}^1$ , we have:

$A := 1=\displaystyle{\frac{\partial \bar{x}^1}{\partial \bar{x}^1}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}$

$B := 0 = \displaystyle{\frac{\partial \bar{x}^2}{\partial \bar{x}^1}=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}$

$A \times \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}:~~~~~C := \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}}$

$B \times \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}:~~~~~D := \displaystyle{0=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}}$

$D-C:$

$\displaystyle{ \frac{\partial \bar{x}^2}{\partial x^2}= \bigg( \frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial \bar{x}^1}{\partial x^2}\bigg) \frac{\partial x^1}{\partial \bar{x}^1}}$ ,

results

$\displaystyle{ \frac{\partial x^1}{\partial \bar{x}^1}}=\frac{\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial \bar{x}^2}{\partial x^1}}}$

…

— Me@2018-08-09 09:49:51 PM

Problem 14.5a1

Posted on August 7, 2018 by rpflee

Counting states in heterotic $SO(32)$ string theory | A First Course in String Theory

(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

$\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

What is normal-ordering?

Put all the creation operators on the left.

What for?

p.251 “It is useful to work with normal-ordered operators since they act in a simple manner on the vacuum state. We cannot use operators that do not have a well defined action on the vacuum state.”

“The vacuum expectation value of a normal ordered product of creation and annihilation operators is zero. This is because, denoting the vacuum state by $|0\rangle$ , the creation and annihilation operators satisfy”

$\displaystyle{\langle 0 | \hat{a}^\dagger = 0 \qquad \textrm{and} \qquad \hat{a} |0\rangle = 0}$

— Wikipedia on Normal order

— This answer is my guess. —

$\displaystyle{\sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^-} \bar \alpha_{-n}^I \bar \alpha_n^I + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^+} \bar \alpha_{n}^I \bar \alpha_{-n}^I + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^+} \left[ \bar \alpha_{n}^I \bar \alpha_{-n}^I - \bar \alpha_{-n}^I \bar \alpha_{n}^I + \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^+} \left[ \bar \alpha_{n}^I, \bar \alpha_{-n}^I \right] + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$= \displaystyle{\sum_{n \in \mathbf{Z}^+} n \eta^{II} + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I}$

c.f. p.251:

$\displaystyle{\sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^+} n \eta^{II} + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I}$

$\displaystyle{= \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I}$

Equation at Problem 14.5:

$\displaystyle{\alpha' M_L^2}$

$\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{1}{2} \left[ \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{24} (D - 2) + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{8} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$D = 10$

$\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \sum_{r = - \frac{1}{2}, - \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} (-r) \lambda_{r}^A \lambda_{-r}^A + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \lambda_{r}^A \lambda_{-r}^A + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ \lambda_{-r}^A, \lambda_r^A \right]}$

Equation (14.29):

$\displaystyle{\left\{ b_r^I, b_s^J \right\} = \delta_{r+s, 0} \delta^{IJ}}$

$\displaystyle{b_r^I b_s^J = - b_s^I b_r^J + \delta_{r+s, 0} \delta^{IJ}}$

$\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \lambda_{r}^A \lambda_{-r}^A + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \left( - \lambda_{-r}^A \lambda_r^A + \delta_{r-r, 0} \delta^{AA} \right) + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ \lambda_{-r}^A \lambda_r^A + \lambda_{-r}^A \lambda_r^A - 1 \right]}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right]}$

$\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= - \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ b_{-r}^A b_r^A + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= - \frac{1}{2} \sum_{r = 1, 3, ...} r + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ b_{-r}^A b_r^A + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= \left[ - \frac{1}{24} + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left( b_{-r}^A b_r^A + \lambda_{-r}^A \lambda_r^A \right) \right]}$

— This answer is my guess. —

— Me@2018-08-06 10:23:48 PM

Universal wave function, 20

Posted on July 21, 2018 by rpflee

The physical (synthetic) universal wave function logically cannot be found by any local observers.

The definition of “universe” is “all the things”. So there is no outside.

A global observer has to be outside the universe.

However, a mathematical (analytic) universal function is possible.

It applies to theoretical/model universe, which can be used to develop interpretations of quantum mechanics and successively approximate the physical universe.

— Me@2012-04-16

The fault of optimism

Posted on July 18, 2018 by rpflee

The illusion of peace

The seemingly rational world is rendered by exceptional good parents.

The seemingly peaceful time is provided by exceptional expense of defense.

— Me@2011.08.19

神的旨意 2.2

Posted on July 18, 2018 by rpflee

這段改編自 2010 年 4 月 18 日的對話。

合情合理，自圓其說的，就有機會真；無情無理，自相矛盾的，則必定為假。

又例如，如果現在神明顯靈在你面前，你怎樣判斷，那真的是「神明」？

「祂」既可能其實是「邪靈」，亦可能只是你自己的幻覺而已。

假設，現在神明顯靈，向你（甲）頒下旨意：

神：你必須殺害無辜，將來死後，才可以上天堂；否則，我會罰你下地獄。

甲：我不想作惡。

神：那不是你想不想的問題。你必須奉行神的旨意。

甲：但是，為什麼要我殺害無辜呢？

神：我是神，而你只是人。神的心思，凡人不一定能夠理解。所以，作為凡人的你，應該心存謙卑，完全信任你的神，奉行神的旨意。

甲：不合理呀！

如果要殺害無辜，才可上天堂，而不殺害無辜，會被罰落地獄的話，那代表著，你天堂的居民，都是壞人；而你地獄的居民，反而是善人。

那樣，你的所謂「天堂」，根本是我的地獄。而你雖然自封為「神」，卻根本是我的「魔」。

— Me@2018-07-16 07:51:33 PM

Chain Rule of Differentiation

Posted on July 15, 2018 by rpflee

Consider the curve $y = f(x)$ .

$\displaystyle{\frac{d}{dx}}$ is an operator, meaning “the slope of the tangent of”. So the expression $\displaystyle{\frac{dy}{dx}}$ , meaning $\displaystyle{\frac{d}{dx} (y)}$ , is not a fraction.

In order words, it means the slope of the tangent of the curve $y = f(x)$ at a point, such as point $A$ in the graph.

d_2018_07_15__21_31_32_PM_

The symbol $dx$ has no relation with the symbol $\displaystyle{\frac{dy}{dx}}$ . It means $\Delta x$ as shown in the graph. In other words,

$dx = \Delta x$

The symbol $dy$ also has no relation with the symbol $\displaystyle{\frac{dy}{dx}}$ . It means the vertical distance between the current point $A(x_0, y_0)$ , where $y_0 = f(x_0)$ , and the point $C$ on the tangent line $y = mx + c$ , where $m$ is the slope of the tangent line. In other words,

$dy = m~dx$

$\displaystyle{dy = \left[ \left( \frac{d}{dx} \right) y \right] dx}$

The relationship of $\Delta y$ and $dy$ is that

$\displaystyle{\Delta y = \frac{dy}{dx} \Delta x + \text{higher order terms}}$

$\displaystyle{\Delta y = \frac{dy}{dx} dx + \text{higher order terms}}$

$\Delta y = dy + \text{higher order terms}$

Similarly, for functions of 2 variables:

$\displaystyle{\Delta f(x,y) = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + \text{higher order terms}}$

$\displaystyle{df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy}$

For functions of 3 variables:

$\displaystyle{df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz}$

$\displaystyle{\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} + \frac{\partial f}{\partial z}\frac{dz}{dt}}$

— Me@2018-07-15 09:30:29 PM

Problem 14.4b2

Posted on July 14, 2018 by rpflee

Closed string degeneracies | A First Course in String Theory

(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 IIB closed superstrings

Equation (14.85)

$(NS+, NS+), (NS+, R-), (R-, NS+), (R-, R-)$

— Me@2015.09.16 06:08 AM: Should be the same. But I am not sure whether I have missed something.

$f_{NS+}(x) = 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ...$

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

$f_{NS-}(x) = \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x^{\frac{3}{2}} + 3064 x^{\frac{5}{2}} + ...$

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

— Me@2018-07-14 09:41:10 PM

Pointer state

Posted on July 3, 2018 by rpflee

Eigenstates 3

In quantum Darwinism and similar theories, pointer states are quantum states that are less perturbed by decoherence than other states, and are the quantum equivalents of the classical states of the system after decoherence has occurred through interaction with the environment.

— Wikipedia on Pointer state

In calculation, if a quantum state is in a superposition, that superposition is a superposition of eigenstates.

However, real superposition does not just includes states that make macroscopic senses.

That is the major mistake of the many-worlds interpretation of quantum mechanics.

— Me@2017-12-30 10:24 AM

— Me@2018-07-03 07:24 PM

Mirror selves, 5.2

Posted on July 2, 2018 by rpflee

Anatta 3.3 | 無我 3.3

You fight for existence, for being alive.

However, your existence is not “yours”.

The existence of you, is not your property.

The existence of you, is a property of the group you are in.

The existence of you, is a property of other people.

To meaningfully say the statement “I exist”, you have to specify you exist with respect to whom.

To exist, you have to specify to exist in which people’s world.

— Me@2018-05-22 7:43 AM

神的旨意 2.1

Posted on June 30, 2018 by rpflee

這段改編自 2010 年 4 月 18 日的對話。

（問：你閱讀過很多有關「瀕死經驗」的文章？）

可以這樣說。

如果你閱讀那些文章的話，要小心一點，因為那類文章良莠不齊——當中有些文章發人深省，有些則謊話連篇。

（問：那你怎樣分辨，「瀕死經驗」的文章之中，哪些是真，哪些為假？）

看看文中所說的，合不合理。

例如，你怎樣知道，我說的話，是真還是假？

合情合理，自圓其說的，就有機會真；無情無理，自相矛盾的，則必定為假。

又例如，如果現在神明顯靈在你面前，你怎樣判斷，那真的是「神明」？

「祂」既可能其實是「邪靈」，亦可能只是你自己的幻覺而已。

— Me@2018-06-28 10:23:28 PM

Block spacetime, 9

Posted on June 26, 2018 by rpflee

motohagiography 42 days ago [-]

I once saw a fridge magnet that said “time is natures way of making sure everything doesn’t happen all at once,” and it’s stuck with me.

The concept of time not being “real,” can be useful as an exercise for modelling problems where to fully explore the problem space, you need to decouple your solutions from needing them to occur in an order or sequence.

From an engineering perspective, “removing” time means you can model problems abstractly by stepping back from a problem and asking, what are all possible states of the mechanism, then which ones are we implementing, and finally, in what order. This is different from the relatively stochastic approach most people take of “given X, what is the necessary next step to get to desired endstate.”

More simply, as a tool, time helps us apprehend the states of a system by reducing the scope of our perception of them to sets of serial, ordered phenomena.

Whether it is “real,” or an artifact of our perception is sort of immaterial when you can choose to reason about things with it, or without it. A friend once joked that math is what you get when you remove time from physics.

I look forward to the author’s new book.

— Gödel and the unreality of time

— Hacker News

2018.06.26 Tuesday ACHK

Quick Calculation 14.8.2

Posted on June 26, 2018 by rpflee

A First Course in String Theory

What sector(s) can be combined with a left-moving NS- to form a consistent closed string sector?

~~~

There are no mass levels in NS+, R+, or R- that can match those in NS-. So NS- can be paired only with NS-:

$(NS-, NS-)$

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

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

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

$g (\sqrt{x}) + g (-\sqrt{x})$
$= 2(1 + 36 x + 402 x^{2} + 3064 x^{3} + ...)$

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

— Me@2018-06-26 07:36:41 PM

Physics Town

continues

Author Archives: rpflee

Mathematics, 2

Self de-centralization, 3

神的旨意 2.4

Illusions destroyed, 2

Problem 14.5a2

The square root of the probability

Number, Time, Money, 3

神的旨意 2.3

The Jacobian of the inverse of a transformation

Problem 14.5a1

Universal wave function, 20

The fault of optimism

神的旨意 2.2

Chain Rule of Differentiation

Problem 14.4b2

Pointer state

Mirror selves, 5.2

神的旨意 2.1

Block spacetime, 9

Quick Calculation 14.8.2

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