ParEdit

Posted on November 29, 2022 by rpflee

(autoload 'enable-paredit-mode
  "paredit" "Turn on pseudo-structural editing." t)

(add-hook 'emacs-lisp-mode-hook
      #'enable-paredit-mode)

(add-hook 'eval-expression-minibuffer-setup-hook   
      #'enable-paredit-mode)

(add-hook 'ielm-mode-hook
      #'enable-paredit-mode)

(add-hook 'lisp-mode-hook
      #'enable-paredit-mode)

(add-hook 'lisp-interaction-mode-hook
      #'enable-paredit-mode)

(add-hook 'scheme-mode-hook
      #'enable-paredit-mode)


(add-hook 'slime-repl-mode-hook
      (lambda () (paredit-mode +1)))

(defun override-slime-repl-bindings-with-paredit ()
   (define-key slime-repl-mode-map
     (read-kbd-macro paredit-backward-delete-key) nil))

(add-hook 'slime-repl-mode-hook
      'override-slime-repl-bindings-with-paredit)

— Me@2022-11-29 10:03:49 PM

Ex 1.2-1 Stationary States

Posted on November 28, 2022 by rpflee

Quantum Methods with Mathematica

Assume a wavefunction of the form psi[x, t] == f[t] psi[x] and perform a separation of variables on the wave equation.

Show that f[t] = E^(-I w t) where h w is the separation constant. Try the built-in function DSolve.

Equate h w to the Energy by evaluating the [expected] value of hamiltonian[V] in the state psi[x, t].

~~~

Remove["Global`*"]


hbar := \[HBar]

H[V_] @ psi_  := -hbar^2/(2m) D[psi,{x,2}] + V psi



psi[x_,t_] := f[t] psi[x]

I hbar D [psi[x,t],t] == H[V] @ psi[x, t]

I hbar D [psi[x,t],t] / psi[x,t] == H[V] @ psi[x,t] / psi[x,t]

$\displaystyle{i \hbar \psi (x) f'(t)=V f(t) \psi (x)-\frac{\hbar ^2 f(t) \psi ''(x)}{2 m}}$

$\displaystyle{\frac{i \hbar f'(t)}{f(t)}=\frac{V f(t) \psi (x)-\frac{\hbar ^2 f(t) \psi ''(x)}{2 m}}{f(t) \psi (x)}}$

E1 := I hbar D [psi[x,t],t] / psi[x,t] == H[V] @ psi[x,t] / psi[x,t]

Simplify[E1]

$\displaystyle{\frac{1}{2} \hbar \left(\frac{\hbar \psi ''(x)}{m \psi (x)}+\frac{2 i f'(t)}{f(t)}\right)=V}$

E2 := - 1/2 hbar hbar (D[D[psi[x],x],x]/(m psi[x])) == hbar omega

DSolve[E2, psi[x], x]


E3 := 1/2 hbar 2 i D[f[t],t] / f[t] == hbar omega

DSolve[E3, f[t], t]

$\displaystyle{\left\{\left\{\psi (x)\to c_1 \cos \left(\frac{\sqrt{2} \sqrt{m} \sqrt{\omega } x}{\sqrt{\hbar }}\right)+c_2 \sin \left(\frac{\sqrt{2} \sqrt{m} \sqrt{\omega } x}{\sqrt{\hbar }}\right)\right\}\right\}}$

$\displaystyle{\left\{\left\{f(t)\to c_1 e^{\frac{\omega t}{i}}\right\}\right\}}$


k

psi[x_] := c E^(I k x)

psi[x]

f[t_] := E^(-I omega t)

f[t]

psi[x_,t_] := f[t] psi[x]

psi[x,t]

$\displaystyle{ \left\{k,c e^{i k x},e^{-i \omega t},c e^{i k x-i \omega t}\right\} }$

E4 := Conjugate[psi[x,t]] H[0] @ psi[x,t]

E4

E5 := Simplify[E4]

E5

k := Sqrt[2 m omega / hbar]

Refine[E5, {Element[{c, omega, m, t, hbar, k, x}, Reals]}]

$\displaystyle{ \frac{c k^2 \hbar ^2 c^* \exp \left(-i \left(-(\omega t-k x)^*-k x+\omega t\right)\right)}{2 m} }$

$\displaystyle{ = c^2 \omega \hbar }$

E6 := Conjugate[psi[x,t]] psi[x,t]

Simplify[E6]

$\displaystyle{ c c^* \exp \left(-i \left(-(\omega t-k x)^*-k x+\omega t\right)\right) }$

$\displaystyle{ = c^2 }$

$\displaystyle{\begin{aligned} \langle E \rangle &= \frac{\int_{-\infty}^{\infty} \psi^* H_{V=0} \psi dx}{\int_{-\infty}^{\infty} \psi^* \psi dx} \\ \\ &= \frac{c^2 \omega \hbar \int_{-\infty}^{\infty} dx}{c^2 \int_{-\infty}^{\infty} dx} \\ \\ &= \omega \hbar \\ \end{aligned}}$

— Me@2022-11-26 07:17:29 PM

Why does the universe exist? 7.2

Posted on November 28, 2022 by rpflee

“There is nothing in that region of space”

and

“there is nothing outside the universe”

have different meanings.

there is nothing except quantum fluctuations in that region of space

= the best detector detects nothing but quantum fluctuations

there is nothing outside the universe

= whatever detected, label the whole collection as “the universe”

“There is nothing outside the universe” does not (!!!) mean that “we go outside the universe to keep searching, but find nothing”.

— Me@2012-10-15 08:33:01 AM

— Me@2022-11-27 09:09:53 PM

Going through hell

Posted on November 27, 2022 by rpflee

If you’re going through hell, keep going.

— Winston Churchill

2022.11.27 Sunday ACHK

「超級 nCr」== 分三組

Posted on November 26, 2022 by rpflee

排列組合 1.7 | nCr, 0.7

Euler problem 8.3

Posted on November 24, 2022 by rpflee

Directory[]

mString := Import["n.txt"]

nString := StringDelete[mString, "\n" | "\r"]

nList := Map[FromDigits, Characters[nString]]

take13[lst_] := Times @@ Take[lst,13]

Fmax13n[lst_, n_] := If[Length[lst] < 13,
                        n,
                        With[{t13 = take13[lst]},
                            If[n > t13,
                                Fmax13n[Rest[lst], n],
                                Fmax13n[Rest[lst], t13]]]]

Fmax13n[nList, 0]

Wmax13n[lst_, n_] := Which[
                        Length[lst] < 13, n,
                        t13 = take13[lst];
                            n > t13, Wmax13n[Rest[lst], n],
                            True, Wmax13n[Rest[lst], t13]]

Wmax13n[nList, 0]

Fmax13n[nList, 0] - Wmax13n[nList, 0]

— colorized by palette fm

— Me@2022-11-24 05:51:56 PM

Problem 14.5d1.2.2

Posted on November 23, 2022 by rpflee

A First Course in String Theory

The generating function is an infinite product:

$\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}$

$\displaystyle{\begin{aligned} &f_{L, NS+}(x) \\ &= a_{NS+} (r) x^r \\ &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ \end{aligned}}$

To evaluate the infinite product, you can use Mathematica (or its official free version Wolfram Engine) with the following commands:

TeXForm[
    HoldForm[
        (1/x)*Product[
                 (1+x^(r-1/2))^32/(1-x^r)^8,
                 {r, 1, Infinity}]]]

f[x_] := (1/x)*Product[
                 (1+x^(r-1/2))^32/(1-x^r)^8,
                 {r, 1, Infinity}]

Print[f[x]]

TeXForm[f[x]]

TeXForm[Series[f[x], {x,0,3}]]




                     1        32
    QPochhammer[-(-------), x]
                  Sqrt[x]
------------------------------------
        1    32                    8
(1 + -------)   x QPochhammer[x, x]
     Sqrt[x]

$\displaystyle{ \begin{aligned} &f_{L, NS+}(x) \\ \end{aligned}}$

$\displaystyle{ \approx \frac{1}{x}+\frac{32}{\sqrt{x}}+504+5248 \, \sqrt{x}+40996 \, x+258624 \, x^{\frac{3}{2}}+1384320 \, x^{2}+6512384 \, x^{\frac{5}{2}} + ...}$

— Me@2022-11-23 04:40:28 PM

Why does the universe exist? 7.1

Posted on November 22, 2022 by rpflee

Why is there something instead of nothing?

Why is there the universe?

The existence of the universe is not a property of the universe itself.

Instead, it is a property of the system that the universe is in.

However, there is no bigger system that contains the universe, because by the definition of the word “universe”, the universe contains everything.

So the question “why is there the universe” should be transformed to “why is there something instead of nothing?”

For example, the watch exists because the watchmaker has made it.

Similarly, if someone has created the universe, we can say that that someone is the cause of the existence of the universe.

However, there is no “someone” outside the universe, because of the definition of the word “universe”.

The universe has no outside.

— Me@2012-10-15 08:33:01 AM

— Me@2022-11-21 09:41:23 PM

Computing Note, 2

Posted on November 21, 2022 by rpflee

30112002

1. A secret is not a secret if it is known by more than one person.

2022.11.21 Monday ACHK

還是覺得你最差

Posted on November 21, 2022 by rpflee

這段改編自 2010 年 4 月 24 日的對話。

每人都有對自己的最先責任，和最終責任。其他人對你，則沒有必然責任。你對其他人，亦沒有。

那是一個好重要的常識。那是 Stephen Covey 在他那本《First Things First》中教的。

記住，一個人（甲）並不能夠改變，另一個人（乙）；除非，乙自己想改變。

當乙自己想改變時，不知道如何執行，你可以教他方法；或者執行時過份緩慢，你可以教他加速。

但是，如果乙本身沒有意願，去改變自己的話，那就一切已失去，不可以再追。

老師的責任，是把學生的學習動機，乘大一千倍。

只要有學習動機，無論是多麼小，乘大一千倍後，都可以威力驚人。

但是，如果學習動機是零的話，零乘一千還是零。

— Me@2010.01.29

— Me@2022-11-20 07:46:26 PM

Importance, 2.2

Posted on November 19, 2022 by rpflee

Euler problem 8.2

import Data.Char

max13 lst = max13n lst 0
  where
    max13n lst n | (length lst) < 13 = n
                 | n > take13        = max13n (tail lst) n
                 | otherwise         = max13n (tail lst) take13
      where
        take13 = product (take 13 lst)

str <- readFile "n.txt"

max13 (map (fromIntegral . digitToInt) . concat . lines $ str)

— Me@2022-11-19 12:04:41 PM

Ex 2.0

Posted on November 18, 2022 by rpflee

Functional Differential Geometry

~~~


(define R2 (make-manifold R^n 2))

(define U (patch 'origin R2))


(define R2-rect (coordinate-system 'rectangular U))

(define R2-polar (coordinate-system 'polar/cylindrical U))


(define R2-rect-chi (chart R2-rect))

(define R2-rect-chi-inverse (point R2-rect))

(define R2-polar-chi (chart R2-polar))

(define R2-polar-chi-inverse (point R2-polar))


(show-expression
 ((compose R2-polar-chi R2-rect-chi-inverse)
  (up 'x_0 'y_0)))

(show-expression
 ((D (compose R2-rect-chi R2-polar-chi-inverse))
  (up 'r_0 'theta_0)))

(define R2->R (-> (UP Real Real) Real))

(define f
  (compose (literal-function 'f-rect R2->R) R2-rect-chi))



(define R2-rect-point (R2-rect-chi-inverse (up 'x_0 'y_0)))

(define corresponding-polar-point
  (R2-polar-chi-inverse
   (up (sqrt (+ (square 'x_0) (square 'y_0)))
       (atan 'y_0 'x_0))))


(f R2-rect-point)

(f corresponding-polar-point)

(show-expression
 (f R2-rect-point))

(show-expression
 (f corresponding-polar-point))

— Me@2022-11-18 11:22:36 AM

Undo

Posted on November 16, 2022 by rpflee

time ~ change

time travel ~ undo the change

— Me@2016-07-20 06:16:34 AM

Default as Power

Posted on November 15, 2022 by rpflee

權力來源 3

political power

~ market power

~ the power due to being the default

~ network effect

— Me@2022-11-09 04:02:00 PM

nCr == 分兩組

Posted on November 14, 2022 by rpflee

排列組合 1.6 | nCr, 0.6

Pier, 1.2

Posted on November 12, 2022 by rpflee

Euler problem 8.1

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

(defun file-get-contents (filename)
  (with-open-file (stream filename)
    (let ((contents (make-string
                     (file-length stream))))
      (read-sequence contents stream)
      contents)))

(defun file-get-lines (filename)
  (with-open-file (stream filename)
    (loop :for line = (read-line stream nil)
          :while line
          :collect line)))

(file-get-lines #P"n.txt")

(defun string-to-list (the-string)
  (loop :for char :across the-string
        :collect char))

(defun char-to-integer-list (char-list)
  (mapcar #'digit-char-p char-list))

(let ((the-digits (char-to-integer-list
           (string-to-list
            (remove #\newline
                    (file-get-contents #P"n.txt"))))))

  (loop :for i :from 0 :to (- (length the-digits) 13)
        :maximize (apply #'*
                    (subseq
                      the-digits i (+ i 13)))))

— colorized by palette fm

— Me@2022-11-12 04:50:19 PM

3.3 Electromagnetism in three dimensions, 2

Posted on November 10, 2022 by rpflee

A First Course in String Theory

(b) Repeat the analysis of three-dimensional electromagnetism starting with the Lorentz covariant formulation. Take $A^\mu = (\Phi, A^1, A^2)$ , examine $F_{\mu \nu}$ , the Maxwell equations (3.34), and the relativistic form of the force law derived in Problem 3.1.

~~~

$A^\mu = (\Phi, A^1, A^2)$

Eq. (3.20):

$F_{\mu \nu} = \begin{bmatrix} 0 & -E_x & -E_y & -E_z=0 \\ E_x & 0 & B_z & -B_y =0\\ E_y & -B_z & 0 & B_x = 0\\ E_z=0 & B_y=0 & -B_x=0 & 0\\ \end{bmatrix}$

Eq. (3.33):

$F^{\mu \nu} = \begin{bmatrix} 0 & E_x & E_y & E_z=0 \\ -E_x & 0 & B_z & -B_y =0\\ -E_y & -B_z & 0 & B_x = 0\\ -E_z=0 & B_y=0 & -B_x=0 & 0\\ \end{bmatrix}$

Eq. (3.34):

$\begin{aligned} \frac{\partial F^{\mu \nu}}{\partial x^\nu} &= \frac{1}{c} j^\mu \\ \end{aligned}$

$\begin{aligned} \frac{\partial F^{0 0}}{\partial x^0} + \frac{\partial F^{0 1}}{\partial x^1} + \frac{\partial F^{0 2}}{\partial x^2} + \frac{\partial F^{0 3}}{\partial x^3} &= \frac{1}{c} j^0 \\ \frac{\partial F^{1 0}}{\partial x^0} + \frac{\partial F^{1 1}}{\partial x^1} + \frac{\partial F^{1 2}}{\partial x^2} + \frac{\partial F^{1 3}}{\partial x^3} &= \frac{1}{c} j^1 \\ \frac{\partial F^{2 0}}{\partial x^0} + \frac{\partial F^{2 1}}{\partial x^1} + \frac{\partial F^{2 2}}{\partial x^2} + \frac{\partial F^{2 3}}{\partial x^3} &= \frac{1}{c} j^2 \\ \frac{\partial F^{3 0}}{\partial x^0} + \frac{\partial F^{3 1}}{\partial x^1} + \frac{\partial F^{3 2}}{\partial x^2} + \frac{\partial F^{3 3}}{\partial x^3} &= \frac{1}{c} j^3 \\ \end{aligned}$

$\begin{aligned} \frac{\partial F^{0 0}}{\partial x^0} + \frac{\partial F^{0 1}}{\partial x^1} + \frac{\partial F^{0 2}}{\partial x^2} &= \frac{1}{c} j^0 \\ \frac{\partial F^{1 0}}{\partial x^0} + \frac{\partial F^{1 1}}{\partial x^1} + \frac{\partial F^{1 2}}{\partial x^2} &= \frac{1}{c} j^1 \\ \frac{\partial F^{2 0}}{\partial x^0} + \frac{\partial F^{2 1}}{\partial x^1} + \frac{\partial F^{2 2}}{\partial x^2} &= \frac{1}{c} j^2 \\ \end{aligned}$

$\begin{aligned} \frac{\partial 0}{\partial x^0} + \frac{\partial E_x}{\partial x^1} + \frac{\partial E_y}{\partial x^2} &= \frac{1}{c} j^0 \\ \frac{\partial (-E_x)}{\partial x^0} + \frac{\partial 0}{\partial x^1} + \frac{\partial B_z}{\partial x^2} &= \frac{1}{c} j^1 \\ \frac{\partial (-E_y)}{\partial x^0} + \frac{\partial (-B_z)}{\partial x^1} + \frac{\partial 0}{\partial x^2} &= \frac{1}{c} j^2 \\ \end{aligned}$

$\begin{aligned} \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} &= \rho \\ - \frac{1}{c} \frac{\partial E_x}{\partial t} + \frac{\partial B_z}{\partial y} &= \frac{1}{c} j_x \\ - \frac{1}{c} \frac{\partial E_y}{\partial t} - \frac{\partial B_z}{\partial x} &= \frac{1}{c} j_y \\ \end{aligned}$

$\begin{aligned} \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} &= \rho \\ \frac{\partial B_z}{\partial y} &= \frac{1}{c} j_x + \frac{1}{c} \frac{\partial E_x}{\partial t} \\ - \frac{\partial B_z}{\partial x} &= \frac{1}{c} j_y + \frac{1}{c} \frac{\partial E_y}{\partial t} \\ \end{aligned}$

P. (3.1):

$\displaystyle{ \begin{aligned} \frac{d p_\mu}{ds} &= \frac{q}{c} F_{\mu \nu} \frac{d x^\nu}{ds} \\ \frac{d p_\mu}{ds} \left( \frac{ds}{dt} \right) &= \frac{q}{c} F_{\mu \nu} \frac{d x^\nu}{ds} \left( \frac{ds}{dt} \right) \\ \frac{d p_\mu}{dt} &= \frac{q}{c} F_{\mu \nu} \frac{d x^\nu}{dt} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d p_0}{dt} &= \frac{q}{c} F_{0 0} \frac{d x^0}{dt} + \frac{q}{c} F_{0 1} \frac{d x^1}{dt} + \frac{q}{c} F_{0 2} \frac{d x^2}{dt} + \frac{q}{c} F_{0 3} \frac{d x^3}{dt} \\ \frac{d p_1}{dt} &= \frac{q}{c} F_{1 0} \frac{d x^0}{dt} + \frac{q}{c} F_{1 1} \frac{d x^1}{dt} + \frac{q}{c} F_{1 2} \frac{d x^2}{dt} + \frac{q}{c} F_{1 3} \frac{d x^3}{dt} \\ \frac{d p_2}{dt} &= \frac{q}{c} F_{2 0} \frac{d x^0}{dt} + \frac{q}{c} F_{2 1} \frac{d x^1}{dt} + \frac{q}{c} F_{2 2} \frac{d x^2}{dt} + \frac{q}{c} F_{2 3} \frac{d x^3}{dt} \\ \frac{d p_3}{dt} &= \frac{q}{c} F_{3 0} \frac{d x^0}{dt} + \frac{q}{c} F_{3 1} \frac{d x^1}{dt} + \frac{q}{c} F_{3 2} \frac{d x^2}{dt} + \frac{q}{c} F_{3 3} \frac{d x^3}{dt} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d p_0}{dt} &= \frac{q}{c} F_{0 0} \frac{d x^0}{dt} + \frac{q}{c} F_{0 1} \frac{d x^1}{dt} + \frac{q}{c} F_{0 2} \frac{d x^2}{dt} \\ \frac{d p_1}{dt} &= \frac{q}{c} F_{1 0} \frac{d x^0}{dt} + \frac{q}{c} F_{1 1} \frac{d x^1}{dt} + \frac{q}{c} F_{1 2} \frac{d x^2}{dt} \\ \frac{d p_2}{dt} &= \frac{q}{c} F_{2 0} \frac{d x^0}{dt} + \frac{q}{c} F_{2 1} \frac{d x^1}{dt} + \frac{q}{c} F_{2 2} \frac{d x^2}{dt} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d p_0}{dt} &= q (0) + \frac{q}{c} \left( - E_x \frac{d x}{dt} - E_y \frac{d y}{dt} \right) \\ \frac{d p_1}{dt} &= q E_x + \frac{q}{c} \left( (0) \frac{d x}{dt} + B_z \frac{d y}{dt} \right) \\ \frac{d p_2}{dt} &= q E_y + \frac{q}{c} \left( - B_z \frac{d x}{dt} + (0) \frac{d y}{dt} \right) \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d E}{dt} &= \vec v \cdot \vec F_E \\ \frac{d p_x}{dt} &= q \left( E_x + \frac{v_y}{c} B_z \right) \\ \frac{d p_y}{dt} &= q \left( E_y - \frac{v_x}{c} B_z \right) \\ \end{aligned}}$

— Me@2022-11-08 03:46:01 PM

Black hole mass can’t be

Posted on November 8, 2022 by rpflee

A singularity doesn’t have mass. Mass is a property of an object that exists in time. A (spacelike, e.g. Schwarzschild) singularity is not an object that exists in time. A singularity is a moment in time when time ends along with mass. Furthermore, a black hole does not have a center. The geometry of the Schwarzschild spacetime inside the horizon is an infinitely long 3-cylinder with a quickly shrinking circumference. Also, no black hole solution is valid inside the horizon, because all solutions assume a static metric, but it is not static inside the horizon.

— safesphere

— May 20, 2019 at 10:38

And if you wanted to say that the whole mass $M$ is obtained from the singularity, you won’t be able to get a good calculation because the integral over the singularity would be singular. Moreover, the space and time are really interchanged inside the black hole (the signs of the components $grr$ and $gtt$ get inverted for $r < 2GM$ ) so the exercise is in no way equivalent to a simple 3D volume integral of $M \delta(x) \delta(y) \delta(z)$ . The Schwarzschild singularity, to pick the "simplest" black hole, is a moment in time, not a place in space. It is the final moment of life for the infalling observers. In a locally (conformally) Minkowski patch near the singularity with some causally Minkowskian coordinates $t,x,y,z$ and $r = |(x,y,z)|$ , the Schwarzschild singularity looks like a $t=t_f$ hypersurface, not as $r=0$ .

— Black hole mass can't be visualized as a property of the black hole interior

— Lubos Motl

2022.11.08 Tuesday ACHK

Know-how, 2

Posted on November 7, 2022 by rpflee

technical/skills

~ a collection of exceptions

~ something you cannot totally derive

— Me@2016-03-16 04:54:02 PM

天助自助者，自助人恆助之

Posted on November 6, 2022 by rpflee

這段改編自 2021 年 12 月 15 日的對話。

Don’t spend your time making other people happy. Other people being happy is their problem. It’s not your problem. If you are happy, it makes other people happy.

— @naval

你在人生中，有一大部分問題，都是來自於，你以為你有責任，去解決其他人的問題。

在那愚善的年代，我不明白這一句。直到近年，我才深切體會到，它的意思。另一個版本是：

每人都有對自己的最先責任，和最終責任。其他人對你，則沒有必然責任。你對其他人，亦沒有。

當你覺得你一定、必須、任何時候，也有責任要，幫助任何他人時，那就是你摧毁自己（和他人）的人生之始。合理的版本是，

你先獨善其身；然後，行有餘力的話，就在有些時候，幫助有些應該幫助的人，如果當事人同意的話。

記住，起點是，你有責任，去解決你自己的問題，但是，你並沒有責任，去解決其他人的問題。有時，間中幫助人，並不是因為責任，而是基於愛心或情義。而「有時候，在你沒有責任幫時，仍然選擇幫」，正正是人性光輝之處。

精確一點用字的話：必須做的工作，為之「責任」；可做可不做的，為之「愛心」或「情義」。

既然可幫可不幫，那樣，什麼時候應該幫呢？

— Me@2022-11-05 12:54:58 PM

Physics Town

continues

Monthly Archives: November 2022