The square root of the probability, 3.2.1

Eigenstates 3.3.2.1

.

According to Schrödinger, the Copenhagen interpretation implies that the cat remains both alive and dead until the state has been observed. Schrödinger did not wish to promote the idea of dead-and-live cats as a serious possibility; on the contrary, he intended the example to illustrate the absurdity of the existing view of quantum mechanics.

— Wikipedia on Schrödinger’s cat

.

Quantum mechanics seems to be unacceptable not because it is strange. Any new science must be strange because it must contain something never known and never expected before.

Quantum mechanics seems to be unacceptable not because it is strange, but because common quantum mechanics education, especially popular science, is so misleading that it makes quantum mechanics look bad; people falsely believe that quantum mechanics violates some of Aristotle’s 3 laws of logic:

1. Law of identity

2. Law of non-contradiction

3. Law of excluded middle

These 3 laws basically mean that

For any proposition $A$, either $A$ is true or $\text{NOT}~A$ is true, but not both.

.

Actually, quantum mechanics does NOT allow any violation of these logical laws. A quantum superposition state is NOT any “overlapping” of multiple physical states. A quantum superposition state is ONE single physical state.

Contrary to popular belief, Schrödinger created the thought experiment to illustrate that a quantum superposition state should NOT be regarded as any “overlapping” of multiple physical states.

To explain that, we should use the most basic quantum experiment, the double-slit experiment, instead of the cat experiment, because:

1.

Although we can regard the cat itself as a system of fundamental particles, we should not do so in this case. Instead, we should just regard the cat itself as one classical object (system).

If we regard the cat itself as a system of fundamental particles, the superposition quantum state will need to include also the classically-makes-NO-sense particle configurations as component eigenstates. In other words, besides the cat-alive state and cat-dead state, the superposition quantum state will need to also include, for example:

1.1   the state of “the cat transforms into a dog”;

1.2   the state of “the cat disappears”;

1.3   the state of “half of the cat becomes a computer and half becomes a harddisk”;

1.4   etc.

.

The major fault of the many-worlds interpretation of quantum mechanics is that it includes only the eigenstates (worlds) that make common sense.

2.1

2.2

— Me@2022-01-30 04:21:17 PM

.

.

2022.01.30 Sunday (c) All rights reserved by ACHK

億萬富翁

— Me@2011.10.11

.

— Me@2022.01.29

.

.

2022.01.29 Saturday (c) All rights reserved by ACHK

無拍之拖, 1.2

.

Pursuit of money, fame, and status generally does not produce a bumper crop of friends and friendly relatives in old age. Indeed, those goals are selfish and selfishness is not a way to make friends. Furthermore, money, fame, and status tend to attract false friends which may get in the way of your acquiring [of] real ones.

— John T. Reed

.

.

.

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

.

1. 相識於微時

1.1

.

.

1.2

.

「女朋友」為什麼叫「女朋友」，而不是「女敵人」或「女陌生」？

— Me@2022-01-28 09:29:54 AM

.

.

2022.01.28 Friday (c) All rights reserved by ACHK

1993

《紐約嫻情》是香港女歌手陳慧嫻，於 1993 年拍攝的音樂特輯；監製金廣誠，編導盧冰心；香港時間 1993 年 10 月 10 日（星期日）晚上 8 時至 9 時 05 分，於香港無綫電視翡翠台首播。

《千千闋歌》
《傻女》
《花店》
《Joe le Taxi》
《飄雪》
《碎花》
《人生何處不相逢》
《Jealousy》

《紅茶館》
《幾時再見?!》

— 文字在創用 CC 姓名標示-相同方式分享 3.0 協議之條款下提供，附加條款亦可能應用。

— 維基百科

.

When this music special premiered in 1993, my TV was showing it in the background while I was having my dinner. I was not watching it. So I did not know the content. But still, I remembered the atmosphere broadcast by it.

In an afternoon in 1996, it was broadcast again. I took a VCR cassette at once to record it. However, I was already a few minutes late, missing the first part of the show.

I watched this music special after every last exam.

In 200x, I bought a TV card just to transfer the music special from the VCR to my computer. However, I have already lost both the VCR cassette and the computer file.

— Me@2022-01-26 09:30:53 PM

.

.

2022.01.26 Wednesday ACHK

2.10 Extra dimension and statistical mechanics

A First Course in String Theory

.

Write a double sum that represents the statistical mechanics partition function $\displaystyle{Z(a, R)}$ for the quantum mechanical system considered in Section 2.10. Note that $\displaystyle{Z(a, R)}$ factors as $\displaystyle{Z(a, R) = Z(a) \tilde{Z}(R)}$.

~~~

Eq. (2.118):

\displaystyle{ \begin{aligned} - \frac{\hbar^2}{2m} \frac{1}{\psi(x)} \frac{d^2 \psi(x)}{dx^2} - \frac{\hbar^2}{2m} \frac{1}{\phi(x)} \frac{d^2 \phi(x)}{dx^2} &= E \\ \end{aligned}}

\displaystyle{ \begin{aligned} \psi_{k, l} (x,y) &= \psi_k (x) \phi_l (y) \\ \end{aligned}}

Eq. (2.119):

\displaystyle{ \begin{aligned} \psi_k (x) &= c_k \sin \left( \frac{k \pi x}{a} \right) \\ \end{aligned}}

\displaystyle{ \begin{aligned} \frac{d^2 \psi_k (x)}{dx^2} &= - \left( \frac{k \pi}{a} \right)^2 \psi_k (x) \\ \end{aligned}}

\displaystyle{ \begin{aligned} - \frac{\hbar^2}{2m} \frac{1}{\psi(x)} \frac{d^2 \psi(x)}{dx^2} - \frac{\hbar^2}{2m} \frac{1}{\phi(x)} \frac{d^2 \phi(x)}{dx^2} &= E \\ \frac{\hbar^2}{2m} \left( \frac{k \pi}{a} \right)^2 + \frac{\hbar^2}{2m} \left( \frac{l}{R} \right)^2 &= E \\ \end{aligned}}

\displaystyle{ \begin{aligned} E &= \frac{\hbar^2}{2m} \left[ \left( \frac{k \pi}{a} \right)^2 + \left( \frac{l}{R} \right)^2 \right] \\ \end{aligned}}

.

[guess]

Index $\displaystyle{k = 1, 2, \dotsb}$ but not negative integers because, for example, $k = 1$ and $k=-1$ give physically identical states

\displaystyle{ \begin{aligned} \psi_{-1} (x) &= c_{-1} \sin \left( \frac{- \pi x}{a} \right) \\ \end{aligned}} and \displaystyle{ \begin{aligned} \psi_{1} (x) &= c_{1} \sin \left( \frac{\pi x}{a} \right) \\ \end{aligned}}.

The two wave functions give the same probability density distribution, if $c_{-1} = c_{1}$.

However, that is not the case for

\displaystyle{ \begin{aligned} \phi_l(y) &= a_l \sin \left(\frac{ly}{R}\right) + b_l \cos \left(\frac{ly}{R} \right) \\ \end{aligned}}.

So $l$ should have also negative integers as possible values: $l = \dotsb, -2, -1, 0, 1, 2, \dotsb$.

[guess]

.

Eq. (2.120):

\displaystyle{ \begin{aligned} \phi_l(y) &= a_l \sin \left(\frac{ly}{R}\right) + b_l \cos \left(\frac{ly}{R} \right) \\ \end{aligned}}

Eq. (2.121):

\displaystyle{ \begin{aligned} \phi_l(y) &= \phi_l(y+2\pi R) \\ \end{aligned}}

.

\displaystyle{ \begin{aligned} Z &= \sum_{k} \sum_{l} e^{- \beta E_{k,l}} \\ &= \sum_{k} \sum_{l} \exp\left\{- \beta \left( \frac{\hbar^2}{2m} \right) \left[ \left(\frac{k \pi}{a} \right)^2 + \left(\frac{l}{R}\right)^2 \right]\right\} \\ &= Z(a) \tilde{Z}(R) \\ \end{aligned}}

\displaystyle{ \begin{aligned} Z(a) &= \sum_{k=1}^\infty \exp\left[- \beta \left( \frac{\hbar^2}{2m} \right) \left(\frac{k \pi}{a} \right)^2 \right] \\ \tilde Z (R) &= \sum_{l=-\infty}^\infty \exp\left[- \beta \left( \frac{\hbar^2}{2m} \right) \left(\frac{l}{R}\right)^2 \right] \\ &= \sum_{l=-\infty}^{-1} \left( \dotsb \right) + \sum_{l=0} \left( \dotsb \right) + \sum_{l=1}^\infty \left( \dotsb \right) \\ &= 1 + 2 Z(R \pi) \\ \end{aligned}}

— Me@2022-01-19 08:45:05 PM

.

.

2022.01.26 Wednesday (c) All rights reserved by ACHK

Inertial mass, 2

.

.

mass ~ (a kind of) potential energy

— Me@2022-01-25 12:25 AM

.

$\displaystyle{a = \frac{F_{\text{net}}}{m}}$

The bigger the value of the inertia mass, the more difficult it is to accelerate that object.

.

$\displaystyle{m = m_0 + m_\text{extra}}$

$\displaystyle{m_\text{extra} = \frac{E}{c^2}}$

The higher the energy content of an object, the more difficult it is to accelerate it.

— Me@2022-01-09 12:34 PM

.

.

2022.01.25 Tuesday (c) All rights reserved by ACHK

A Whole New World

.

Do not ask your children to strive for extraordinary lives. Such striving may seem admirable, but it is the way of foolishness. Help them instead to find the wonder and the marvel of an ordinary life. Show them the joy of tasting tomatoes, apples and pears. Show them how to cry when pets and people die. Show them the infinite pleasure in the touch of a hand. And make the ordinary come alive for them. The extraordinary will take care of itself.

— William Martin

.

I can show you the world
Shining, shimmering, splendid
Tell me, princess, now when did
You last let your heart decide?

I can open your eyes
Take you wonder by wonder
Over, sideways and under
On a magic carpet ride

A whole new world
A new fantastic point of view
No one to tell us, “No”
Or where to go
Or say we’re only dreaming

— Aladdin’s A Whole New World

.

.

2022.01.24 Monday ACHK

超時空接觸 3

.

.

.

.

— Me@2022-01-23 11:34:20 PM

.

Your world is your world.

— Ludwig Wittgenstein

.

— Me@2009.09.16

.

.

2022.01.24 Monday (c) All rights reserved by ACHK

Momentum NOT as Spatial Energy, 1.2

.

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

.

.

\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}}

.

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

.

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

.

\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}}

.

\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} }

.

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

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

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

.

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

.

\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}}

— Me@2022-01-20 11:33:48 AM

.

.

2022.01.21 Friday (c) All rights reserved by ACHK

Momentum as Spatial Energy

.

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.

.

.

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

— Wikipedia on Four-momentum

.

.

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）

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

.

\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}}

\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} }

\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} }

\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} }

.

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

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

.

\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}}

.

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

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

.

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

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

.

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

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

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

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

.

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

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

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

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

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。

.

\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

.

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

.

.

2022.01.17 Monday (c) All rights reserved by ACHK

無拍之拖, 2

— 尼采

.

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

— Friedrich Nietzsche

.

.

2022.01.16 Sunday ACHK

原來是你, 2.2

.

.

.

.

.

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

.

.

2022.01.16 Sunday (c) All rights reserved by ACHK

Ex 1.22 Driven pendulum, 2.1

Structure and Interpretation of Classical Mechanics

.

Show that the Lagrangian (1.89) …

~~~

[guess]

The Lagrangian (1.89):

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

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

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

.

(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

.

.

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

.

.

2022.01.12 Wednesday ACHK

理智與感情, 2

.

.

~ 客氣

~ 保持距離

— Me@2016-09-03 02:58:13 PM

.

.

2022.01.12 Wednesday (c) All rights reserved by ACHK

機構邪惡

.

— 尼采

.

「機構」先天邪惡，其中一個核心原因是，在機構中，下決定的，和承受（該決定的）後果的，往往不是同一個人。

— Me@2022-01-11 05:43:42 PM

.

.

2022.01.11 Tuesday (c) All rights reserved by ACHK

Entropy at the Beginning of Time, 1.2

Logical arrow of time, 10.2.2

.

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”.

.

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.

.

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

.

.

2022.01.09 Sunday (c) All rights reserved by ACHK