3.1 Lorentz covariance for motion in electromagnetic fields, 1

A First Course in String Theory

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The Lorentz force equation (3.5) can be written relativistically as

\displaystyle{\frac{d p_\mu}{ds} = \frac{q}{c} F_{\mu \nu} \frac{d x^\nu}{ds}},

where \displaystyle{p_{\mu}} is the four-momentum.

(a) Check explicitly that this equation reproduces (3.5) when \displaystyle{\mu} is a spatial index.

(b) What does (1) gives when \displaystyle{\mu = 0}? Does it make sense?

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Eq. (3.5):

\displaystyle{\frac{d \vec p}{dt} = q \left( \vec E + \frac{\vec v}{c} \times \vec B \right)}

Eq. (2.20):

\displaystyle{ds \equiv \sqrt{ds^2}}    if    \displaystyle{ds^2 > 0}

Eq. (2.21):

\displaystyle{-ds^2 = \eta_{\mu \nu} dx^\mu dx^\nu}

The spacetime interval \displaystyle{ds^2} is Lorentz invariant. If \displaystyle{ds^2 > 0}, we have Eq. (2.27) and (2.28):

\displaystyle{\begin{aligned}     ds &= c dt_p \\     ds &= c dt \sqrt{1 - \beta^2}     \end{aligned}}

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\displaystyle{\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)

\displaystyle{    \begin{aligned}     \frac{d p_1}{dt}     &= \frac{q}{c} F_{1 \nu} \frac{d x^\nu}{dt} \\    &= \frac{q}{c} \left( F_{1 0} \frac{d x^0}{dt} + F_{1 1} \frac{d x^1}{dt} + F_{1 2} \frac{d x^2}{dt} + F_{1 3} \frac{d x^3}{dt} \right) \\    \frac{d p_x}{dt} &= \frac{q}{c} \left( E_x c \frac{d t}{dt} + (0) \frac{d x^1}{dt} + B_z \frac{d x^2}{dt} - B_y \frac{d x^3}{dt} \right) \\    &= q \left( E_x + \frac{1}{c} \left( \vec v \times \vec B \right)_x \right) \\    \end{aligned}}

\displaystyle{    \begin{aligned}     \frac{d p_0}{dt}     &= \frac{q}{c} \left( F_{0 0} \frac{d x^0}{dt} + F_{0 1} \frac{d x^1}{dt} + F_{0 2} \frac{d x^2}{dt} + F_{0 3} \frac{d x^3}{dt} \right) \\    &= \frac{q}{c} \left( 0 \frac{d x^0}{dt} - E_x \frac{d x^1}{dt} - E_y \frac{d x^2}{dt} - E_z \frac{d x^3}{dt} \right) \\    \frac{d }{dt} \left( \frac{-E}{c} \right)&= - \frac{q}{c} \left( \vec v \cdot \vec   E \right) \\    \frac{d E}{dt}  &= \vec v \cdot \vec F_E \\      \end{aligned}}

— Me@2022-08-04 04:17:59 PM

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2022.08.06 Saturday (c) All rights reserved by ACHK

Bravado

Anyone who expects to create, be it as scientist or artist, scholar or writer, needs self-confidence, even bravado. How else can one dare to imagine understanding what no one else has understood, discovering what no one else has discovered? Where does this confidence come from? Fortunately, every young person is blessed with some of it. It is part of human character. What of the girl or boy who reads about Newton and Maxwell and Bohr and Einstein and says, “I want to build on what they have built; I want to add to the sum of knowledge about the most basic laws of nature”?

– Geons, Black Holes & Quantum Foam, p. 84

– 2000
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2022.08.03 Wednesday (c) All rights reserved by ACHK

CSCI3420


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砌摩打

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Design:
1: make car moving
2: make car stop itself
3: follow the track
4: change direction
5: tuning

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CSCI3420 Project Phase 2

Deadline: 23:59

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2022.08.03 Wednesday (c) All rights reserved by ACHK

排列組合 1.1

nCr, 0

這段改編自 2010 年 7 月 27 日的對話。

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\displaystyle{n!}

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1.1  意思:

\displaystyle{n} 個人 \displaystyle{n} 個座位的話,有多少種坐法?

1.2.1  算式:

\displaystyle{n! = (n)(n-1)(n-2) \cdots (3)(2)(1)}

1.2.2  由來:

第一個位,有 \displaystyle{n} 個可能的人選;第一個位被坐後,第二個位只有 \displaystyle{(n-1)} 個,可能的人選;如此類推,直到最後一個位,被餘下的唯一個人,選了為此。

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1.3 

留意,\displaystyle{n} 是多少,就有多少項。

\displaystyle{n! = (n)(n-1)(n-2) \cdots (3)(2)(1)}

例如,\displaystyle{5!},就有 5 項相乘;

\displaystyle{\begin{aligned}     5! &= (5)(4)(3)(2)(1) \\ \\     \end{aligned}}

\displaystyle{3!},就有 3 項;等等。

\displaystyle{\begin{aligned}     3! &= (3)(2)(1) \\    \end{aligned}}

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1.4.0  零的階乘,\displaystyle{0!},還未有定義,因為,算式 \displaystyle{n! = (n)(n-1) \cdots (2)(1)} 中的 \displaystyle{n},只可以是正整體,不可以零。

階乘零,\displaystyle{0!},需要額外定義,因為會常用到。那樣,\displaystyle{0!} 應該定義為,什麼數值呢?

1.4.1  既然 \displaystyle{n!}意思是「\displaystyle{n} 個人 \displaystyle{n} 個座位,有多少種坐法」,那樣,你就可以視,\displaystyle{0!} 的意思是「\displaystyle{0} 個人 \displaystyle{0} 個座位,有多少種坐法」;那明顯是一,因為,那個情況之下,只有一個「坐法」,就是「沒有人又沒有位」這個唯一的可能性。

\displaystyle{0! = 1}

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1.4.2  另外一種設計定義的想法是,由 \displaystyle{n!}算式取靈感。

既然 \displaystyle{n} 是多少,\displaystyle{n!} 就有多少項相乘,那樣,零的階乘,\displaystyle{0!},理應只有零項相乘,\displaystyle{~a^0~}

\displaystyle{0! = a^0}

但是,「零項相乘」的數值,又應該是什麼呢?

「零項相乘」即是「乘了等如沒有乘」,所以是一,因為任何數乘了一,效果都等於沒有乘。

\displaystyle{a^0 = 1}

(「零項相乘」的正式學名是,「空積」或「零項積」。)

\displaystyle{0! = 1}

— Me@2022-08-02 02:41:43 PM

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2022.08.02 Tuesday (c) All rights reserved by ACHK

Mega Man Zero 3

Euler problem 1

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(proclaim '(optimize speed))

(reduce #'+ '(1 2 3 4))

; 10

(loop :for n :below 10 :collect n)

; (0 1 2 3 4 5 6 7 8 9)

(describe :below)

(defun range (max &key (min 0) (step 1))
   (loop :for n :from min :below max :by step
      collect n))
      
(- (+ (* 3 (reduce #'+ (range 334 :min 1 :step 1)))
      (* 5 (reduce #'+ (range 200 :min 1 :step 1))))
   (* 15 (reduce #'+ (range 67 :min 1 :step 1))))
   
; 233168

(defun sum-1-n (n)
  (/ (* n (+ n 1)) 2))
  
(- (+ (* 3 (sum-1-n 333))
      (* 5 (sum-1-n 199)))
   (* 15 (sum-1-n 66)))
   
; 233168

— Me@2022-08-01 03:29:01 PM

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2022.08.01 Monday (c) All rights reserved by ACHK

Jupyter Notebook

SICMUtils, 3

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The goal of this post to setup Jupyter Notebook.

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1. Read and follow the exact steps of my post titled “SICMUtils“.

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2. To install the Jupyter Notebook software in Ubuntu, use this command:

sudo apt-get install sagemath-jupyter 

3. Try to open the SageMath program.

4. It will open a Jupyter notebook page.

5. Click the “New” button at the top right corner and then select “SageMath“.

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6. Type

1 + 1

onto the input line.

7. Hit the keys shift-enter to get the output

2

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8. Input

Integrate(x^3, x)

9. Hit shift-enter:

NameError: name 'Integrate' is not defined

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10. Input

from sage.symbolic.integration.integral import *

indefinite_integral(x^3, x)

11. Hit shift-enter:

1/4*x^4

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12. Input

%display latex

13. Hit shift-enter.

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14. Input

indefinite_integral(x^3, x)

15. Hit shift-enter:

\displaystyle{\frac{1}{4} \, x^{4}}

— Me@2022-07-30 12:18:50 PM

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2022.08.01 Monday (c) All rights reserved by ACHK