Length Contraction and Time Dilation

d_2018_09_26__20_58_04_PM_

Length Contraction and Time Dilation | Special Relativity Ch. 5

minutephysics

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I recommend this video.

Without it, I would have never realized that besides length contraction and time dilation, there are also distance dilation and time contraction.

— Me@2018-09-26 10:12:19 PM

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2018.09.26 Wednesday (c) All rights reserved by ACHK

Problem 14.5a3

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

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Paraphrasing the description of heterotic (closed) string theory:

  • right-moving part \sim an open superstring theory
    • NS sector: \alpha_{-r}^I, b_{-r}^I,~~~I = 2,3,...,9
    • R sector: \alpha_{-n}^I, d_{-n}^I,~~~I=2,3,...,9
    • “The standard GSO projection down to NS+ and R- applies.”

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  • left-moving part \sim a peculiar bosonic openstring theory
    • I = 2, 3, ..., 23: There are totally 24 transverse coordinates
      • 8 bosonic coordinates X^I with oscillators \bar \alpha_{-n}^I
      • 16 peculiar bosonic coordinates

        • can be replaced by 32 two-dimensional left-moving fermion fields, \lambda^A
        • \lambda^A (anti-commuting) fermion fields \to has NS' and R' sectors
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    • NS': oscillators \bar \alpha_{-n}^I, \lambda_{-r}^A act on the vacuum |NS' \rangle

      given (-1)^{F_L} |NS' \rangle_L = + |NS' \rangle_L \to defines the left NS'+ sector

      \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 \mathbb{Z} + \frac{1}{2}} r \lambda_{-r}^A \lambda_r^A}

    • R': oscillators \bar \alpha_{-n}^I, \lambda_{-n}^A act on R' ground states

      \displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \left( \bar \alpha_{-n}^I \bar \alpha_n^I + n \lambda_{-n}^A \lambda_n^A \right)}

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— Me@2018-09-20 09:51:17 PM

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2018.09.21 Friday (c) All rights reserved by ACHK

神的旨意 2.4

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。

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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。

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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」?

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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。

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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 。

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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

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2018.09.02 Sunday (c) All rights reserved by ACHK

Illusions destroyed, 2

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…

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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.

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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.

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— Why is nearing completion so demotivating?

— Hacker News

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2018.09.01 Saturday ACHK

Problem 14.5a2

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

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

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\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}

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— This answer is my guess. —

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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}

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\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]}

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Equation (13.116):

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

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\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}}

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\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}}

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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. —

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— Me@2018-09-01 06:05:29 AM

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2018.09.01 Saturday (c) All rights reserved by ACHK