# 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

.

.

# 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

.

.

# 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

.

.

# 1990s, 12

— Me@2022-01-14 12:09:34 PM

.

.

# 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

.

.

# 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

.

.

# 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

.

.

# 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

.

.

# Entropy at the Beginning of Time, 1.1

Logical arrow of time, 10.2.1

.

Two distinguishable macrostates can both evolve into one indistinguishable macrostate.

— Me@2013-08-11 11:08 AM

.

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

.

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

.

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

.

a macrostate = a set of macroscopically-indistinguishable microstates

— Me@2022-01-09 07:43 AM

.

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

.

.

# 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.

.

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

.

.

# To realize is to realize, 1.5

.

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

.

Why self-fulfilling?

4. you have another half or not

.

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

.

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

.

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

— James Horner, Barry Mann, Cynthia Weil

.

.

# 1990s, 11

— Me@2022-01-06 12:14:03 PM

.

.

# 2.10 A spacetime orbifold in two dimensions, 6

A First Course in String Theory

.

(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 $\displaystyle{e^\lambda \equiv \sqrt{\frac{1+\beta}{1-\beta}}}$.

.

\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

.

.

# 時光起源

The Origin of Time

.

— Me@2022-01-04 12:23:25 PM

.

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.

.

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

.

.

— 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

.

.

2022.01.03 Monday ACHK

# 分支圖

.

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

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

— Me@2022-01-02 12:20:15 PM

.

.