Momentum NOT as Spatial Energy, 1.2

能量空間版 1.2

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那樣,能量平方 E^2 就去了左邊,成了數式的主角。

\displaystyle{ \begin{aligned} \frac{E^2}{c^2} &= \mathbf p \cdot \mathbf p + m^2 c^2 \\ \end{aligned}}

意思就可以講成:「能量(平方)E^2 有兩個構成部:動態(平方)部分 \mathbf p \cdot \mathbf p 和靜態(平方)部分 m^2 c^2。」

在這個上文下理中,(相對論版)三維動量,其實就是能量中的動態部分。運動,必在空間中。所以動態部分,就是「空間部分」。

而質量,則是能量中的靜態部分。所以,質量又稱「靜止能」。靜止,代表「不在空間中運動」;可以視為「只在時間方向移動」。所以靜態部分,就是「時間部分」。

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把能量平方 \displaystyle{\frac{E^2}{c^2}} 的兩部分 \displaystyle{\mathbf{p} \cdot \mathbf{p}}\displaystyle{m^2 c^2},分別命名為「動態部分」和「靜態部分」,並無不妥。但是,再把它們標籤為「空間部分」和「時間部分」,則不恰當。

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同理,你可以把能量一次方 E 的兩部分,分別命名為「動態部分」\displaystyle{E_k} 和「靜態部分」\displaystyle{E_r}

\displaystyle{\begin{aligned}  E_k &= {\sqrt {p^{2}c^{2}+m^{2}c^{4}}}-mc^{2} \\  E_r &= m c^2 \\ \\  E &= E_{k} + E_{r} \\  \end{aligned}}

但是,你不應該把它們,標籤為「空間部分」和「時間部分」。原因是,能量 E 本身,就已經是時空動量 p 中的「時間部分」。

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之前講過,能量 E 是時空動量 p 中,時間方向的分量。

\displaystyle { p = \left(p^{0},p^{1},p^{2},p^{3}\right) = \left({E \over c},p_{x},p_{y},p_{z}\right) = \left({E \over c},\bf p \right) }

所以,你可以把能量 E,稱為「時間動量」。

但是,如果你再把能量 E,自己內部的兩部分 E_kE_r,標籤為「空間能量」和「時間能量」的話,它們就成了「空間時間動量」和「時間時間動量」。

這兩名字既奇怪,亦失準。

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這兩標籤不正確的原因是,它們不似「能量 E 是『動量在時間方向的分量』(簡稱『時間動量』)」般,是客觀術詞數學事實,因為,它真的是四次元時空向量,四格中的最左一格:

\displaystyle { p = \left({E \over c},p_{x},p_{y},p_{z}\right)}

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能量 E 自己內部的兩部分 E_kE_r,分別標籤為「動態部分」和「靜態部分」,亦是有根有據物理事實,因為,在物件靜止不動 (v = 0) 時,動態部分 E_k 真的等於零。它就是所謂的「動能」。

換句話說,動能 E_k,只會在物體有運動 (v \ne 0) 時,才會出現。

而「靜態能量」E_r,就是當物件,即使靜止不動 (v = 0) 時,身上仍然持有的能量。它就是所謂的「質量」。

\displaystyle{\begin{aligned}  E &= E_{k} + E_{r} \\ \\  E_r &= m c^2 \\  E_k &= {\sqrt {p^{2}c^{2}+m^{2}c^{4}}}-mc^{2} \\ \\    E_k  &= {\frac {p^{2}}{2m}}-{\frac {p^{4}}{8m^{3}c^{2}}} + \dotsb \\  &= {\frac {1}{2}}mv^{2}+{\frac {3}{8}}m{\frac {v^{4}}{c^{2}}} + \dotsb \\  \end{aligned}}

但是,如果再把「動態能量E_k 和「靜態能量E_r,標籤為「空間能量」和「時間能量」的話,那就沒有根據不是事實;極其量,只是主觀喜好文學修辭

首先,能量 E 的已知身份是「時間動量」。它本身就是,時間的特性。與它直接有關的,是時間。

而能量 E 與空間的關係,是間接的。

能量 E 與空間有關的原因是,它與時間有關,而時間又和空間有關。

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相對論的重點是,時空是會互動的,各自也不能獨立運行。時間和空間,必定會影響對方。而那些數學細節,就是「時空畢氏定理」。

而「時空畢氏定理」的時空動量版本,就是

\displaystyle{ \begin{aligned}  p \cdot p &= -(p^0)^2 + (p^1)^2 + (p^2)^2 + (p^3)^2 \\  &= -(\frac{E}{c})^2 + (p_x)^2 + (p_y)^2 + (p_z)^2 \\ \\    p \cdot p &= -m^{2}c^{2} \\ \\    -m^{2}c^{2}  &= -(\frac{E}{c})^2 + (p_x)^2 + (p_y)^2 + (p_z)^2 \\    -m^{2}c^{2}  &= -(\frac{E}{c})^2 + |\mathbf{p}|^2 \\    \end{aligned} }

時空動量 p 的長度平方 p \cdot p,對於任何粒子而言,在任何情況下,都等於 \left(- m^{2}c^{2}\right)。由於這個硬性規定,能量(時間動量)E 和三次元動量(空間動量)\displaystyle { \left(p_{x},p_{y},p_{z}\right)} 並不能獨立變動。任何一方的數值改變,必定牽動另一方。

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\displaystyle{\begin{aligned}  E &= E_{k} + E_{r} \\  \end{aligned}}

無論靜止或運動,能量數值 \displaystyle{ E = \sqrt{|\mathbf{p}|^2 c^2 + m^2 c^4} } 都不會零;意思是,無論靜止或運動,物體必在時間中「移動」。(三次元動量 \mathbf{p},則在物體沒有運動,即靜止時,立刻為零。)

記住,無論是總能量 E,或其動靜分部 (E_k, E_r) 部的任何一個,都是時間的「產物」;沒有所謂的「純空間」分部。所以,只把靜能(質量)E_{r} 標籤為「時間能量」,卻把動能 E_k 叫作「空間能量」,暗示其為「純空間能量」,並不恰當。

換句話說,把身份已知為「時間動量」的能量 E,再分拆為「時間能量」和「空間能量」,並不合理:

\displaystyle{\begin{aligned}  E &= E_{k} + E_{r} \\  \end{aligned}}

能量 E = 空間能量 + 時間能量

正確的分類和命名是:

\displaystyle { p = \left(p^{0},p^{1},p^{2},p^{3}\right) = \left({E \over c},p_{x},p_{y},p_{z}\right) = \left({E \over c},\bf p \right) }

時空動量 =(時動量, x 空動量,y 空動量,z 空動量)
時空動量 =(能量,動量)

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\displaystyle{ \begin{aligned} \frac{E^2}{c^2} &= \mathbf p \cdot \mathbf p + m^2 c^2 \\ \end{aligned}}

能量平方 = 動量平方 + 質量平方
能量平方 = 動態部分 + 靜態部分

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\displaystyle{\begin{aligned}  E_k &= {\frac {p^{2}}{2m}}-{\frac {p^{4}}{8m^{3}c^{2}}} + \dotsb \\ &= {\frac {1}{2}}mv^{2}+{\frac {3}{8}}m{\frac {v^{4}}{c^{2}}} + \dotsb \\ \\    E_r &= m c^2 \\  E_k &= {\sqrt {p^{2}c^{2}+m^{2}c^{4}}}-mc^{2} \\ \\    E &= E_{k} + E_{r} \\  \end{aligned}}

能量 = 運動能量 + 靜止能量
能量 = 動能 + 質量

— Me@2022-01-20 11:33:48 AM

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2022.01.21 Friday (c) All rights reserved by ACHK

Momentum as Spatial Energy

能量空間版

這段改編自 2021 年 12 月 5 日的對話。

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In special relativity, four-momentum is the generalization of the classical three-dimensional momentum to four-dimensional spacetime. Momentum is a vector in three dimensions; similarly four-momentum is a four-vector in spacetime. The contravariant four-momentum of a particle with relativistic energy E and three-momentum p = (p_x, p_y, p_z) = \gamma m \bf v, where \bf v is the particle’s three-velocity and \gamma the Lorentz factor, is

\displaystyle{p=\left(p^{0},p^{1},p^{2},p^{3}\right)=\left({E \over c},p_{x},p_{y},p_{z}\right).}

The quantity m \bf v of above is ordinary non-relativistic momentum of the particle and m its rest mass. The four-momentum is useful in relativistic calculations because it is a Lorentz covariant vector. This means that it is easy to keep track of how it transforms under Lorentz transformations.

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\displaystyle{ p\cdot p=\eta _{\mu \nu }p^{\mu }p^{\nu }=p_{\nu }p^{\nu }=-{E^{2} \over c^{2}}+|\mathbf {p} |^{2}=-m^{2}c^{2}}

— Wikipedia on Four-momentum

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之前我講,動能(KE)是時間方向的動量(momentum)。

那是錯的。

正確的講法是,動量(momentum)是總能量(total energy )的空間方向分量。(這裡,總能量(total energy )的數值,已包含了動能(KE)部分。)

動量(momentum)不是真身,總能量(total energy )才是。

那都是錯的,因為,這裡有歧義。

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在這裡,「動量」(momentum)這個詞語,有三個可能的意思:

1. Newtonian momentum 牛頓動量

\displaystyle{m \bf v}

2. Three-momentum 相對論三維動量

\displaystyle{\mathbf{p} = \gamma m \bf v}, \displaystyle{\gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}

3. Four-momentum 時空動量(相對論四維動量)

\displaystyle {  p   = \left(p^{0},p^{1},p^{2},p^{3}\right)  = \left({E \over c},p_{x},p_{y},p_{z}\right)  = \left({E \over c},\bf p \right)  }

如果「動量」是指,時空動量 p 的話,總能量 E 的確是,時間方向的分量。(而時空動量 p 的空間分量,則是相對論三維動量 \bf p。)

留意,這裡的 E 是總能量,而不是動能 KE。

在這個背景下,「動量」和「能量」被統一成「時空動量」。換句話說,動量 \bf p 和能量 E 其實是,時空動量 p 中的,兩個部分。動量 \bf p 就是,時空動量 p 中是「空間動量」;而能量 E,則是「時間動量」。

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但是,如果現在考慮的「動量」,不是指時空動量 p 本身,而是它的數值 \displaystyle{\| p \|}(或稱「向量長度」)的話,則會得到這些公式:

\displaystyle{ \begin{aligned}   \| p \| &= \sqrt{p \cdot p} \\ \\  p \cdot p &= -(p^0)^2 + (p^1)^2 + (p^2)^2 + (p^3)^2 \\   \end{aligned}}

第零分量 p^0 的平方 (p^0)^2 的平方,之前要加負號的原因是,它並非純空間上的畢氏定理,而是時空版本的畢氏定理。而在這「時空畢氏定理」中,凡是屬於時間方向的,必須加負號。

那樣,時空動量數值平方 \displaystyle{ p \cdot p },就會等於:

\displaystyle{   \begin{aligned}   p \cdot p   &= -(p^0)^2 + (p^1)^2 + (p^2)^2 + (p^3)^2 \\  &= -(\frac{E}{c})^2 + (p_x)^2 + (p_y)^2 + (p_z)^2 \\  \end{aligned}  }

在這裡,如果沿用剛才,「能量 E 是,時空動量 \displaystyle{   \begin{aligned}   p &= \left({E \over c},p_{x},p_{y},p_{z}\right) \\   \end{aligned}  } 中,在時間方向的分量」講話風格的話,我們可以把

\displaystyle{   \begin{aligned}   \| p \|  &= \sqrt{p \cdot p}   &= \sqrt{- \left(\frac{E}{c}\right)^2 + (p_x)^2 + (p_y)^2 + (p_z)^2} \\  \end{aligned}  }

說成:「能量 E 的數值,構成了時空動量 p 的數值 \| p \| 的一部分。」換句話說,時空動量長度 \| p \|,由能量 E 和三維動量 \left( p_x, p_y, p_z \right) 兩部分組成。

這個講法雖然可以,但有少許弱點。

平時同「構成」或「組成」這字眼時,通常是指貢獻。但是,在 \| p \| 的公式中,能量平方 E^2 之前的,是負號;即是負累,不是貢獻。所以,很多時,物理學家會將該公式,調成

\displaystyle{   \begin{aligned}   \left(\frac{E}{c}\right)^2 &= (p_x)^2 + (p_y)^2 + (p_z)^2 -  p \cdot p \\   \left(\frac{E}{c}\right)^2 &= |\mathbf {p} |^{2} -  p \cdot p \\   \end{aligned}  }

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留意一個有趣的事實:

\displaystyle{   \begin{aligned}   p \cdot p &= -m^{2}c^{2} \\   \end{aligned}  }

時空動量 p 的長度平方 \| p \|^2,即是 p \cdot p,在任何情況下,對於任何粒子而言,都等於 \left(- m^{2}c^{2}\right)。(平方竟然為負數的原因是,這裡的「平方」只是比喻,並非平時數學中的二次方。)

那樣,能量平方 E^2 就去了左邊,成了數式的主角。

\displaystyle{ \begin{aligned} \frac{E^2}{c^2} &= \mathbf p \cdot \mathbf p + m^2 c^2 \\ \end{aligned}}

意思就可以講成:「能量(平方)E^2 由兩個構成部:動態(平方)部分 \mathbf p \cdot \mathbf p 和靜態(平方)部分 m^2 c^2。」

在這個上文下理中,(相對論版)三維動量,其實就是能量中的動態部分。運動,必在空間中。所以動態部分,就是「空間部分」。

而質量,則是能量中的靜態部分。所以,質量又稱「靜止能」。靜止,代表「不在空間中運動」;可以視為「只在時間方向移動」。所以靜態部分,就是「時間部分」。

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概念上,精簡而言,能量是動量的一部分——能量 E 是時間方向的動量(momentum)。精準來說,能量 E 是時空動量 \displaystyle { p = \left({E \over c},p_{x},p_{y},p_{z}\right)} 中,時間方向的分量。

數值上,精簡而言,動量是能量的一部分——動量(momentum)是總能量(total energy )的空間部分。精準來說,能量 E 是的數值中,包括了(相對論版)三次元動量 \displaystyle{\mathbf{p}} 的貢獻。

\displaystyle{ \begin{aligned} \frac{E^2}{c^2} &= \mathbf p \cdot \mathbf p + m^2 c^2 \\ \\    \frac{E}{c} &= \sqrt{|\mathbf p |^2 + m^2 c^2} \\ \end{aligned}}

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留意,暫時從來沒有提及過,動能 KE。剛才只是說,能量平方 E^2 可以分成兩部分——動態部分(空間部分)和靜態部分(時間部分)。

\displaystyle{ \begin{aligned} \frac{E^2}{c^2} &= \mathbf p \cdot \mathbf p + m^2 c^2 \\ \end{aligned}}

但是,暫時從來沒有提及過,要將能量 E 本身,分成兩部分。

\displaystyle{ \begin{aligned} \frac{E}{c} &= \dotsb + \dotsb \\ \end{aligned}}

動能 KE 是我們在,企圖將能量一次方 E,分成動靜兩部分時,出現的副產物。

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剛才提過,因為在能量平方的公式 \displaystyle{ \left( \frac{E^2}{c^2} = \mathbf p \cdot \mathbf p + m^2 c^2 \right) } 中,質量平方 \displaystyle{ \left( m^2 c^2 \right)}代表靜態部分,所以,質量又名「靜止能」。

而質量可稱為「靜止能」的另一個理據是,如果物體靜止,它的三次元動量 | \bf p |,就會等於零。那樣,

\displaystyle{ \begin{aligned} \frac{E^2}{c^2}     &= \mathbf p \cdot \mathbf p + m^2 c^2 \\     &= 0 + m^2 c^2 \\     \end{aligned}}

亦即是話,

\displaystyle{ \begin{aligned}     \frac{E^2}{c^2} &= m^2 c^2 \\ \\     E &= m c^2 \\     \end{aligned}}

所以,「質量 m c^2 是『靜止能』」的意思是,物體靜止時,身上僅有的能量數值,就是質量 m c^2

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然後,透過這個「靜能」,就可以定義到,所謂的「動能 KE」。

清晰起見,「質量 mc^2」命名為「靜能 E_r」:

\displaystyle{ \begin{aligned} E_r &= m c^2 \\ \end{aligned}}

總能量 E

\displaystyle{ \begin{aligned} \frac{E}{c} &= \sqrt{|\mathbf p |^2 + m^2 c^2} \\ \end{aligned}}

簡化起見,暫只考慮直線(即是一次元空間)上的運動。那樣,三次元動量 \bf p,就會降級成一次元動量 p

\displaystyle{ \begin{aligned}   \frac{E}{c} &= \sqrt{p^2 + m^2 c^2} \\ \\  E &= c \sqrt{p^2 + m^2 c^2} \\   \end{aligned}}

然後,我們把總能量 E,減去靜能 E_r 部分;把餘量命名為 E_k

\displaystyle{     \begin{aligned}     E_k &= E - E_r \\     &= \sqrt{p^2 c^2 + m^2 c^4} - m c^2 \\     \end{aligned}}

物理量 E_k 的真正全名是「總能量 E 減去靜能 E_r 部分後的餘額」。由於這名字實在太長,我們簡稱它為「動能」;意思是,原本靜止的物體,因為運動而多了出來的能量。

原本靜止的物件,如果要令它,以速度 v 運動,你就需要給予它,動能 E_k 數值的能量。

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留意,這個所謂的「動量 E_k」,是由人工合成,再由後天標籤而成的。

\displaystyle{ \begin{aligned} E_k &= \sqrt{p^2 c^2 + m^2 c^4} - m c^2 \\ \end{aligned}}

運算操作而言,動量 E_k 就是那麼的論盡;並沒有精簡的版本,除非取其近似值。

理論架構來說,動量 E_k 不是主角,只為配角;重要性並不如:

1. Four-momentum 時空動量

\displaystyle{p=\left(p^{0},p^{1},p^{2},p^{3}\right)=\left({E \over c},p_{x},p_{y},p_{z}\right)}

2. Total energy 總能量

\displaystyle{E = p^0 c}

3. Three-momentum (3-space momentum) 相對論三次元空間動量

\displaystyle{\mathbf{p} = \left(p^{1},p^{2},p^{3}\right) = \left( p_x, p_y, p_z \right)}

\displaystyle{\mathbf{p} = \gamma m \mathbf{v} = \frac{m \mathbf{v}}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}

這三位主角,各自會守恆不變:

1. 總能量 \displaystyle{E} 會因為,時間的均勻性而守恆。「時間均勻性」的意思是,物理定律不會隨時間改變。

2. 空間動量 \displaystyle{\mathbf{p}} 會因為,空間的均勻性而守恆。「空間均勻性」的意思是,宇宙間任何兩個不同地方,物理定律必為相同。

3. 結果,因為時間分量 \displaystyle{\frac{E}{c}} 和空間分量 \displaystyle{\mathbf{p}} 都守恆,時空動量 \displaystyle{p=\left({E \over c},\mathbf{p}\right)} 作為整體,亦會守恆。

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但是,動能 E_k,則從來沒有,對應的守恆定律,所以用處較少。動能 E_k 作為「未必守恆量」,只能作大配角。

\displaystyle{\begin{aligned}     E_{\text{k}}   &= {\sqrt {p^{2}c^{2}+m^{2}c^{4}}}-mc^{2} \\  &= {\frac {p^{2}}{2m}}-{\frac {p^{4}}{8m^{3}c^{2}}} + \dotsb \\ \\  E_{\text{k}} &\approx {\frac {p^{2}}{2m}}-{\frac {p^{4}}{8m^{3}c^{2}}}    \end{aligned}}

— Me@2022-01-16 05:39:11 PM

— Me@2022-01-17 01:17:51 PM

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I would flip it and say momentum [is] energy in the space-like direction. When at rest (four velocity u = (c, {\bf 0})), you are moving entirely in the time direction. Four momentum is:

\displaystyle{p_{\mu} = mu_{\mu} = (mc^2, {\bf 0})}

For a observer boosted to velocity \displaystyle{ \bf v }:

\displaystyle{p_{\mu} = mu_{\mu} =\gamma(mc^2, m{\bf v}c)=(mc^2+T, {\bf p}c)}

[Y]our rest mass (energy) now appears as momentum (and the timelike term has kinetic energy added to it).

That last equality may make it clear that when in motion (c=1):

\displaystyle{m \rightarrow m+T}

\displaystyle{{\bf 0}\rightarrow {\bf 0} + {\bf p}}

which could be interpreted as “Kinetic energy is momentum in the time-like direction”.

— answered Apr 25 ’18 at 1:13

— JEB

— Physics Stack Exchange

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2022.01.17 Monday (c) All rights reserved by ACHK

無拍之拖, 2

婚姻不幸福,缺乏的不是愛,而是友誼。

— 尼采

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It is not a lack of love, but a lack of friendship that makes unhappy marriages.

— Friedrich Nietzsche

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2022.01.16 Sunday ACHK

原來是你, 2.2

相聚零刻 2.4 | 尋覓 2.2.3.6.4

這段改編自 2010 年 10 月 14 日的對話。

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你所中意的人,可能一直在你身邊,只不過是,雙方暫未察覺,或者時機尚未成熟。所以,找到另一半的,其中一個可能劇情是:「竟然是你!」

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心動為戀

戀愛的起點是,竟然是你

心定為愛

戀愛的終點是,依然是你

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你所追求的東西,可能一直在你身邊,只不過是,雙方未有發現;到發現的時候,就可能已經,花了幾年的時間。

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能力範圍內,應盡量避免浪費那幾年,因為,人的青春,尤其是女仔,就只有那幾年。

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避免浪費的主要方法有:

1. 相識於微時 

2. 不要神化異性,要平民化

3. 不要幻化愛情,要現實化

— Me@2022-01-15 11:51:28 AM

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2022.01.16 Sunday (c) All rights reserved by ACHK

Ex 1.22 Driven pendulum, 2.1

Structure and Interpretation of Classical Mechanics

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Show that the Lagrangian (1.89) …

~~~

[guess]

The Lagrangian (1.89):

Formally, we can reproduce Newton’s equations with the Lagrangian:

\displaystyle{ L(t;x, F; \dot x, \dot F)}

\displaystyle{= \sum_\alpha \frac{1}{2} m_\alpha \dot{\mathbf{x}_\alpha}^2  - V(t, x) - \sum_{\{ \alpha, \beta | \alpha < \beta, \alpha \leftrightarrow \beta \}} \frac{F_{\alpha \beta}}{2 l_{\alpha \beta}} \left[ (\mathbf{x}_\beta - \mathbf{x}_\alpha)^2 - l_{\alpha \beta}^2 \right] }

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(define (KE-particle m v)
  (* 1/2 m (square v)))

(define ((extract-particle pieces) local i)
  (let* ((indices (apply up (iota pieces (* i pieces))))
         (extract (lambda (tuple)
                    (vector-map (lambda (i)
                                  (ref tuple i))
                                indices))))
    (up (time local)
        (extract (coordinate local))
        (extract (velocity local)))))

(define (U-constraint q0 q1 F l)
  (* (/ F (* 2 l))
     (- (square (- q1 q0))
        (square l))))

(define ((U-gravity g m) q)
  (let* ((y (ref q 1)))
    (* m g y))) 

(define ((L-driven-free m l x_s y_s U) local)
  (let* ((extract (extract-particle 2))
	 
     (p (extract local 0))
     (q (coordinate p))
     (qdot (velocity p))
     
     (F (ref (coordinate local) 2)))
  
    (- (KE-particle m qdot)
       (U q)
       (U-constraint (up (x_s (time local)) (y_s (time local)))
		     q
		     F
		     l))))

(let* ((U (U-gravity 'g 'm))
       (x_s (literal-function 'x_s))
       (y_s (literal-function 'y_s))
       (L (L-driven-free 'm 'l x_s y_s U))
       (q-rect (up (literal-function 'x)
		           (literal-function 'y)
		           (literal-function 'F))))
  (show-expression
   ((compose L (Gamma q-rect)) 't)))

\displaystyle{     L     =     \frac{1}{2} m \left[(Dx)^2 + (Dy)^2 \right] - mgy     - \frac{F}{2l} \left[ (x-x_s)^2 + (y-y_s)^2 - l^2 \right] }

[guess]

— Me@2022-01-13 01:19:34 PM

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2022.01.14 Friday (c) All rights reserved by ACHK

Fall of Cybertron

Transformers: Fall of Cybertron is a third-person shooter video game based on the Transformers franchise, developed by High Moon Studios and published by Activision. It is the sequel to the 2010 video game Transformers: War for Cybertron, and directly follows the events of that game, as the Autobots struggle to defeat their Decepticon foes in a civil war for their home planet of Cybertron.

The game tells the story of the Transformers, fictional robotic life forms, and the final days of conflict on their home planet of Cybertron. An origins subplot for the Dinobots is also told, reimagined from the Transformers: Generation 1 continuity. Other subplots also tell an adapted story for several characters. Some of the voice cast from the 1984 series The Transformers return to reprise their roles, including Peter Cullen as Autobot leader Optimus Prime and Gregg Berger as Grimlock. Other actors return to reprise their roles from Transformers: War for Cybertron.

— Wikipedia on Transformers: Fall of Cybertron

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2022.01.12 Wednesday ACHK

Entropy at the Beginning of Time, 1.2

Logical arrow of time, 10.2.2

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If at the beginning, the universe had a high entropy, it was at a macrostate corresponding to many indistinguishable microstates.

That description is self-contradictory, because “two macroscopically-indistinguishable microstates” is meaningful only if they were once macroscopically distinguishable before.

That is not possible for the state(s) at the beginning of the universe, because at that moment, there was no “before”.

So it is meaningless to label the universe’s beginning macrostate as “a state corresponding to many indistinguishable microstates”.

Instead, we should label the universe’s beginning state as “a state corresponding to one single microstate”.

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For example, assume that the universe was at the macrostate \displaystyle{A} at the beginning; and the \displaystyle{A} is corresponding to two macroscopically-indistinguishable microstates \displaystyle{a_1} and \displaystyle{a_2}.

Although microstates \displaystyle{a_1} and \displaystyle{a_2} are macroscopically-indistinguishable, we can still label them as “two” microstates, because they have 2 different histories — history paths that are macroscopically distinguishable.

However, for the beginning of the universe, there was no history. So it is meaningless to label the state as “a macrostate with two (or more) possible microstates”.

So we should label that state not only as one single macrostate but also as one single microstate.

In other words, that state’s entropy value should be defined to be zero.

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If in some special situation, it is better to label the universe’s beginning state as “a state with non-zero entropy”, that state will still have the smallest possible entropy of the universe throughout history.

So it is not possible for the universe to have “a high entropy” at the beginning.

— Me@2022-01-08 02:38 PM

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2022.01.09 Sunday (c) All rights reserved by ACHK

Entropy at the Beginning of Time, 1.1

Logical arrow of time, 10.2.1

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Two distinguishable macrostates can both evolve into one indistinguishable macrostate.

— Me@2013-08-11 11:08 AM

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Note that, tautologically, any system can be at only one single macrostate at any particular time.

So the statement actually means that it is possible for two identical systems at different macrostates evolve into the same later macrostate.

— Me@2022-01-08 03:12 PM

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But the opposite is not possible. Two indistinguishable macrostates is actually, by definition, one macrostate. It cannot evolve into two distinguishable macrostates.

One single macrostate is logically impossible to be corresponding to two different possible later macrostates.

— Me@2022-01-08 01:29 PM

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If the beginning universe state had a high entropy, by definition, it was at a macroscopic state with many possible macroscopically-indistinguishable microstates.

However, if it is really the state of the universe at the beginning, it is, by definition, a single microstate, because “different microstates” is meaningful only if they were once distinguishable.

— Me@2013-08-11 01:42 PM

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a macrostate = a set of macroscopically-indistinguishable microstates

— Me@2022-01-09 07:43 AM

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The meaning of “entropy increases” is that state \displaystyle{S_1} and state \displaystyle{S_2} both evolve into state \displaystyle{S_3}.

But for the beginning of the universe, there were no multiple possible macrostates that the beginning state could be evolved from.

— Me@2013-08-11 01:44 PM

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2022.01.09 Sunday (c) All rights reserved by ACHK

魚目混珠 1.2

Pure evil does no harm, because if someone is purely evil, everyone will know that and avoid him.

It is the evilness of a good man that creates big harm.

The evilness of great man creates the biggest harm.

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An organization cannot be purely evil.

Anything purely evil cannot be big, because being big requires consistency, which requires good.

— Me@2011.10.11

— Me@2022-01-08

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2022.01.08 Saturday (c) All rights reserved by ACHK

To realize is to realize, 1.5

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The ultimate self-fulfilling prophecies:

1. free will or not

2. god or no god

3. afterlife or not

4. future spouse exists or not

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Why self-fulfilling?

4. you have another half or not

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If you act as if your future wife exists, you have a much higher chance of finding her, because you have actually been providing the environment for her to exist.

If you act as if your future wife exists, compared with assuming not having one, you live completely differently. You will make yourself financially stable; you will earn and save money for your future wife and children. You will make yourself healthy so that you can live long enough to finally find her; and then stay with her for decades. You will make and meet genuine friends, for you know that one of them is either your future wife or the one who introduces her to you.

If you know that your future wife exists, you will not have any romantic relationship with anyone that is certainly not your future wife, no matter how beautiful she is. Actually, you will not start any romantic relationship with anyone at all, unless she is potentially your future wife.

You will keep writing letters, sending them to the future; to your future wife. Once your future wife realizes that those letters may be actually for her, she will find you to confirm.

— Me@2021-02-19 06:32:05 AM

— Me@2021-03-02 04:39:33 PM

— Me@2022-01-07 10:00:29 AM

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If you act as if your future wife exists, you know that the goal is not to find the “best” lady, but to find your future wife.

Your wife is the key to the lock of your life. Your goal is not to find the “best” key in the world, but to find the key of your life.

It is not to find the “best” key, but to find your key.

— Me@2021-12-03 12:03 PM

— Me@2021-12-05 03:22 AM

— Me@2022-01-07 12:30 PM

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Somewhere out there
beneath the pale moonlight
Someone’s thinking of me
and loving me tonight

— James Horner, Barry Mann, Cynthia Weil

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2022.01.07 Friday (c) All rights reserved by ACHK

2.10 A spacetime orbifold in two dimensions, 6

A First Course in String Theory

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(d) Consider the two curves \displaystyle{x^+ x^- = a^2} for some fixed \displaystyle{a}. The identification (2) makes each of these curves into a circle. Find the invariant circumference of this circle by integrating the appropriate root of \displaystyle{ds^2} between two neighboring identified points. Give your answers in terms of \displaystyle{a} and \displaystyle{\lambda}. Answer: \displaystyle{\sqrt{2} a \lambda}.

~~~

Identification (2):

\displaystyle{(x^+, x^-) \sim \left( e^{-\lambda} x^+, e^{\lambda} x^- \right)}, where \displaystyle{e^\lambda \equiv \sqrt{\frac{1+\beta}{1-\beta}}}.

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\displaystyle{  \begin{aligned}  ds &= \sqrt{- ds^2} \\  \int_{(x^+, x^-)}^{(e^{-\lambda} x^+, e^{\lambda} x^-)} ds     &= \int_{x^+}^{e^{-\lambda} x^+} \sqrt{2 \frac{a^2}{(x^+)^2} (dx^+)^2}  \\     &= \sqrt{2} a \int_{x^+}^{e^{-\lambda} x^+} \sqrt{\frac{1}{(x^+)^2}} dx^+  \\   \end{aligned}}

Choose a segment on which \displaystyle{x^+ > 0} .

\displaystyle{  \begin{aligned}  \int_{(x^+, x^-)}^{(e^{-\lambda} x^+, e^{\lambda} x^-)} ds     &= \sqrt{2} a \int_{x^+}^{e^{-\lambda} x^+} \frac{1}{x^+} dx^+  \\     &= \sqrt{2} a \ln \left|\frac{e^{-\lambda} x^+}{x^+} \right| \\     &= - \sqrt{2} a \lambda \\   \end{aligned}}

The lower limit should be smaller than the upper limit.

\displaystyle{  \begin{aligned}  \left| \int_{(x^+, x^-)}^{(e^{-\lambda} x^+, e^{\lambda} x^-)} ds \right|    &= \sqrt{2} a \int_{e^{-\lambda} x^+}^{x^+} \frac{1}{x^+} dx^+  \\     &= \sqrt{2} a \lambda \\   \end{aligned}}

— Me@2022-01-05 02:40:45 PM

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2022.01.05 Wednesday (c) All rights reserved by ACHK

時光起源

The Origin of Time

這段改編自 2021 年 12 月 5 日的對話。

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昨日的那段影片說,個別的粒子,其實都是質量零;以光速行走,感受不到時間。

但是,把一堆粒子圈成一件東西的話,那件東西整體而言,就會有質量;就會低於光速行走;就會感受到時間。

變相來說,時間的來源就是,你將一堆粒子組成一個物件、一個身份、一個自我時,那個自我就會,感受到時間。

— Me@2022-01-04 12:23:25 PM

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To form an object (an observer), the component particles cannot all always move at light speed in the same direction, for that would prevent the object as a whole from feeling time.

Anything moving at the speed of light cannot feel the passage of time. If a set of particles all moves at light speed in the same direction all the time, they cannot feel time either as individuals or as a whole; so they cannot form an “object”.

To form an object (an observer), the component particles need to interact. So some component particles need to move in other directions sometimes.

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An object requires an internal structure to exist and evolve. The component particles need to interact in order to evolve as a single identity. So different particles need to move in different directions sometimes. As a result, the component particles as a whole, aka “the object”, will move slower than light.

— Me@2021-12-08 08:09:17 AM

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2022.01.05 Wednesday (c) All rights reserved by ACHK

Already here 3

他肯定可以看到,未來演算可知的結局。

但是,未來是碎片化的。

不真正到達未來那一刻,誰也不知道會發生什麼。

— 無縫連接《駭客任務:復活》 37.43

— 大聰看電影

— Me@2021-12-26 11:20:40 PM

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2022.01.03 Monday ACHK

分支圖

這段改編自 2021 年 12 月 9 日的對話。

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條件概率(conditional probability)的公式中,會除以「已知」。

\displaystyle{P(A|B)={{P(B|A)*P(A)} \over {P(B)}}}

而「除以『已知』」的目的就是,改變 tree diagram 的起點。這一句是,我學了機會率之後的十幾年間,原本都不知道的東西。在 2010 年,第一次教時,我才領悟得到。

This file is made available under the Creative Commons CC0 1.0 Universal Public Domain Dedication.

樹形圖的起點,機率必為一,因為「樹形圖起點」的定義,就正正是「已知」。

樹形圖的重要性在於,多複雜的題目,只要你用樹形圖來思考,通常都可以明白到。

甚至,如果你問最根本的問題「乘法從何而來」,其實,都需要用樹形圖來解釋。例如,\displaystyle{3 \times 2} 是什麼意思呢?

三大分支的每支上,再有 2 小分支的話,就總共有 6 個終點。

— Me@2022-01-02 12:20:15 PM

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2022.01.03 Monday (c) All rights reserved by ACHK