# A big scratch on the door

One of Intel’s top executives once told me that he finally understood how things had changed in microprocessor business when someone asked him if he would accept delivery on an expensive new automobile if it was in perfect condition except for a big scratch on the door. After all, the car would deliver the same fuel economy, same performance, same cargo capacity, same expected service lifetime. But no, he paid for a perfect product, and that’s what he expected.

— IEEE Computer

— 2002.07.08

.

.

2022.10.31 Monday ACHK

# CENG3430


75% -> vhdl
10% -> ps
15% -> transmission

~~~

EPROM: E = 0
G = 0
P = 1

~~~

Maths mid-term:
11/10/2011



.

.

# Euler problem 6.1

— by this meme’s creator

.

(defun range (max &key (min 0) (step 1))
(loop :for n :from min :below max :by step
collect n))

(defun sum-1-n (n)
(/ (* n (+ n 1)) 2))

(defun sum (lst)
(reduce #'+ lst))

(defun square (x)
(* x x))

(- (square (sum-1-n 100))
(sum (mapcar #'square (range 101 :min 1))))

; 25164150

.

— Me@2022-10-29 06:39:07 AM

.

.

# Ex 1.1: Motion on a Sphere, 2

Functional Differential Geometry

.

The metric for a unit sphere, expressed in colatitude $\displaystyle{\theta}$ and longitude $\displaystyle{\phi}$, is

$\displaystyle{g(u,v) = d\theta(u) d \theta(v) + (\sin \theta)^2 d \phi (u) d \phi (v)}$

Compute the Lagrange equations for motion of a free particle on the sphere …

~~~

(define ((L2 mass metric) place velocity)
(* 1/2
mass
((metric velocity velocity) place)))

(define ((Lc mass metric coordsys) state)
(let ((x (coordinates state))
(v (velocities state))
(e (coordinate-system->vector-basis
coordsys)))
((L2 mass metric)
((point coordsys) x) (* e v))))

(define the-metric
(literal-metric 'g R2-rect))

(define L
(Lc 'm the-metric R2-rect))

(L (up 't (up 'x 'y) (up 'vx 'vy)))

(show-expression
(L (up 't (up 'x 'y) (up 'v_x 'v_y))))


.

$\displaystyle{g(u,v) = d\theta(u) d \theta(v) + (\sin \theta)^2 d \phi (u) d \phi (v)}$

(show-expression
(L (up 't
(up 'theta 'phi)


.

When $\displaystyle{R = 1}$ and $\displaystyle{ [g] = \begin{bmatrix} 1 & 0 \\ 0 & (\sin \theta)^2 \end{bmatrix}}$,

$\displaystyle{ \frac{1}{2} m (\sin \theta)^2 \dot \phi^2 + 0 + \frac{1}{2} m \dot \theta^2}$

.

.

— Me@2022-10-27 10:30:50 AM

.

.

Posted in FDG

# Coherent light source

superkuh 17 hours ago [–]

Coherence isn’t what you think it is. It is not an “alignment in phase” of the sinusoid, like all the lay diagrams show. It isn’t even being the same frequency. In the early days of quantum physics the light sources were mercury arc lamps (muliple freqs) that achieved coherence by shining through tiny pinholes.

https://web.archive.org/web/20220820182938/http://amasci.com/miscon/coherenc.html

Coherence is being a point source. Stars, except for our sun since it is too close, are coherent light sources.

— Hacker News

.

.

2022.10.27 Thursday ACHK

# 1920s

Dear men,

If she won’t give you her 20s, don’t give her your 30s and 40s.

— Lovers’ Guide

— guideforlovers

.

Don’t give a woman your prime if she won’t give you hers.

A woman who spends her twenties sleeping around until her biological clock starts ticking isn’t worth a penny.

— ManlyConfidence

.

.

2022.10.26 Wednesday ACHK

# Presentation 基本原理 1.2.2.5

.

— Me@2010.09.05

.

— Me@2010.09.05

— Me@2022-10-25 02:58:36 PM

.

.

# The Sixth Sense, 2.2

Euler problem 5.2 | Folding an infinite list, 2

.

.

f = foldr1 lcm [1..20]

.

.

Most problems on Project Euler can be solved in three ways:

• with brute-force

• with an algorithm that solves a more general problem

• with a smart solution that requires pencil and paper at most

If you’re interested in a nice solution rather than fixing your code, try concentrating on the last approach …

— edited Oct 8, 2016 at 8:57

— huwr

— answered Dec 27, 2011 at 14:33

— Philip

— Stack Overflow

.

.

# 3.3 Electromagnetism in three dimensions

A First Course in String Theory

.

(a) Find the reduced Maxwell equations in three dimensions by starting with Maxwell’s equations and the force law in four dimensions, using the ansatz (3.11), and assuming that no field can depend on the $z$ direction.

~~~

\displaystyle{\begin{aligned} \nabla \cdot \mathbf {E} &= \rho \\ \nabla \cdot \mathbf {B} &= 0 \\ \nabla \times \mathbf {E} &= - \frac{1}{c} {\frac {\partial \mathbf {B} }{\partial t}} \\ \nabla \times \mathbf {B} &= \frac{1}{c} \mathbf {J} + \frac{1}{c} {\frac {\partial \mathbf {E} }{\partial t}} \\ \end{aligned}}

.

Eq. (3.11):

\displaystyle{\begin{aligned} E_z &= 0 \\ B_x &= 0 \\ B_y &= 0 \\ \end{aligned}}

.

\displaystyle{\begin{aligned} \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} &= \rho \\ \frac{\partial B_z}{\partial z} &= 0 \\ \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y} &= - \frac{1}{c} {\frac {\partial B_z }{\partial t}} \\ \nabla \times \mathbf {B} &= \frac{1}{c} \mathbf {J} + \frac{1}{c} {\frac {\partial \mathbf {E} }{\partial t}} \\ \end{aligned}}

.

\displaystyle{\begin{aligned} \frac{\partial B_z }{\partial y} - \frac{\partial B_y }{\partial z} &= \frac{1}{c} j_x + \frac{1}{c} {\frac {\partial E_x }{\partial t}} \\ \frac{\partial B_x }{\partial z} - \frac{\partial B_z }{\partial x} &= \frac{1}{c} j_y + \frac{1}{c} {\frac {\partial E_y }{\partial t}} \\ \frac{\partial B_y }{\partial x} - \frac{\partial B_x }{\partial y} &= \frac{1}{c} j_z + \frac{1}{c} {\frac {\partial E_z }{\partial t}} \\ \end{aligned}}

.

\displaystyle{\begin{aligned} \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}}

.

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

.

\displaystyle{\begin{aligned} \frac{d p_x}{dt} &= q \left( E_x + \frac{1}{c} (v_y B_z - v_z B_y) \right) \\ \frac{d p_y}{dt} &= q \left( E_y - \frac{1}{c} (v_x B_z - v_z B_x) \right) \\ \frac{d p_z}{dt} &= q \left( E_z + \frac{1}{c} (v_x B_y - v_y B_x) \right) \\ \end{aligned}}

.

\displaystyle{\begin{aligned} \frac{d p_x}{dt} &= q \left( E_x + \frac{1}{c} v_y B_z \right) \\ \frac{d p_y}{dt} &= q \left( E_y - \frac{1}{c} v_x B_z \right) \\ \frac{d p_z}{dt} &= 0 \\ \end{aligned}}

.

— Me@2022-10-22 04:17:10 PM

.

.

# Swampland

In physics, the term swampland refers to effective low-energy physical theories which are not compatible with string theory, in contrast to the so-called “string theory landscape” of compatible theories. In other words, the swampland is the set of consistent-looking theories with no consistent ultraviolet completion in string theory.

Developments in string theory suggest that the string theory landscape of false vacua is vast, so it is natural to ask if the landscape is as vast as allowed by consistent-looking effective field theories. Some authors, such as Cumrun Vafa, suggest that is not the case and that the swampland is in fact much larger than the string theory landscape.

— Wikipedia on Swampland (physics)

.

Recall that Vafa’s Swampland is a giant parameter space of effective field theories that cannot be realized within a consistent theory of quantum gravity i.e. within string/M-theory. Only a tiny island inside this Swampland, namely the stringy Landscape, is compatible with quantum gravity. String/M-theory makes lots of very strong predictions – namely that we don’t live in the Swampland. We have to live in the special hospitable Landscape.

— Vafa, quintessence vs Gross, Silverstein

— The Reference Frame

— Luboš Motl

.

.

2022.10.21 Friday ACHK

# 以原則為中心

.

When your MindOS is aligned with PRINCIPLES, MAGIC expands.

— Me@2016-12-08 06:17:32 AM

.

PRINCIPLES

~ natural laws

.

— Me@2022-10-15 11:47:03 PM

.

.

# 排列組合 1.4

nCr, 0.4

.

— Me@2022-10-19 12:26:30 PM

.

.

# Euler problem 5.1

1930s, 3

.

— meme creator

.

(defun range (max &key (min 0) (step 1))
(loop :for n :from min :below max :by step
collect n))

(defmacro lcm-lst (lst)
(apply #'lcm ,lst))

(lcm-lst (range 21 :min 2))

.

— palette fm

— Me@2022-10-17 05:21:23 PM

.

.

# Ex 1.28 Kinetic energy contains terms that are linear

Structure and Interpretation of Classical Mechanics

.

An analogous result holds when the $f_\alpha$‘s depend explicitly on time.

a. Show that in this case the kinetic energy contains terms that are linear in the generalized velocities.

~~~

\displaystyle{\begin{aligned} \mathbf{v_\alpha} &= \partial_0 f_\alpha (t,q) + \partial_1 f_\alpha (t,q) v \\ T(t,q,v) &= \frac{1}{2} \sum_\alpha m_\alpha v^2_\alpha \\ v_\alpha &= |\mathbf{v}_\alpha| \\ v &= \text{generalized velocity} \\ \mathbf{v}_\alpha &= \text{rectangular velocity} \\ \end{aligned}}

.

\displaystyle{\begin{aligned} &T(t,q,v) \\ &= \frac{1}{2} \sum_\alpha m_\alpha |\partial_0 f_\alpha (t,q) + \partial_1 f_\alpha (t,q) v|^2 \\ &= \frac{1}{2} \sum_\alpha m_\alpha \left \{ [\partial_0 f_\alpha (t,q)]^2 + 2 \partial_0 f_\alpha (t,q) \partial_1 f_\alpha (t,q) v + [\partial_1 f_\alpha (t,q) v]^2 \right \} \\ \end{aligned}}

— Me@2022-10-15 11:17:59 AM

.

.

# Computing Note

21122002 not mine but quotation

Yet, in principle, computers cannot be asked to discover proofs, except in very restricted areas of mathematics–such as elementary Euclidean geometry–where the set of theorems happens to be recursive, as was proved by Tarski.

… the prime number theorem was first suggested as the result of extensive hand calculations on the prime numbers up to 3,000,000 by the Swiss mathematician Leonhard Euler (1707-83), a process that would have been greatly facilitated by the availability of a modern computer.

Thus Godel was able to assert that the set of theorems of mathematics is recursively enumerable, and, more recently, the American linguist Noam Chomsky (b. 1928) could say that the set of grammatical sentences of a natural language, such as English, is recursively enumerable.

Godel’s incompleteness theorem is that the consistency of mathematics can be proved only in a language which is stronger than the language of mathematics itself. Yet, formalism is not dead- …

— Encyclopædia Britannica

.

.

2022.10.13 Thursday ACHK

# 2003 AI Quantum Mechanics

2004 AI MSc Physics/Teacher/Programmer

2005 Physics M.Phil.

2003to2008

~ Me@2002

.

This history has unfolded with 70% accuracy.

— Me@2022-10-12 05:45:41 PM

.

.

# 眾害取其輕 10.2

The least of all evils, 10.2

.

「眾害之最輕」有時可以零，有時不可以零。你只能客觀面對，不能主觀判斷。換句話說，「眾害之最輕」不一定「可以零」，亦不一定「不可以零」。

.

1. 理解後背誦

2. 有用的東西就背

3. 無用的東西就不背

.

.

— Me@2022-10-11 11:39:33 PM

.

.

# Euler problem 4.2

.

Find the largest palindrome made from the product of two 3-digit numbers.

g = [(y, z, y*z) | y<-[100..999], z<-[y..999], f==y*z]
where
f = maximum [x | y<-[100..999], z<-[y..999],
let x=y*z, let s=show x, s==reverse s]

.

— Me@2022-10-10 10:09:53 PM

.

.

# Ex 1.1: Motion on a Sphere

Functional Differential Geometry

.

The metric for a unit sphere, expressed in colatitude $\displaystyle{\theta}$ and longitude $\displaystyle{\phi}$, is

$\displaystyle{g(u,v) = d\theta(u) d \theta(v) + (\sin \theta)^2 d \phi (u) d \phi (v)}$

Compute the Lagrange equations for motion of a free particle on the sphere and convince yourself that they describe great circles. For example, consider the motion on the equator $\displaystyle{\theta = \pi/2}$ and motion on a line of longitude ($\displaystyle{\phi}$ is constant).

~~~

(define ((Lfree mass) state)
(* 1/2 mass (square (velocity state))))

(define ((sphere->R3 R) state)
(let ((q (coordinate state)))
(let ((theta (ref q 0)) (phi (ref q 1)))
(up (* R (sin theta) (cos phi))
(* R (sin theta) (sin phi))
(* R (cos theta))))))

(define ((F->C F) local)
(up (time local)
(F local)
(+ (((partial 0) F) local)
(* (((partial 1) F) local)
(velocity local)))))

(define (Lsphere m R)
(compose (Lfree m) (F->C (sphere->R3 R))))

(show-expression
((Lsphere 'm 'R)
(up 't
(up 'theta 'phi)

(show-expression
(((Lagrange-equations
(Lsphere 'm 'R))
(up
(literal-function 'theta)
(literal-function 'phi)))
't))
`

\displaystyle{ \begin{aligned} - \sin \theta (D \phi)^2 \cos \theta + D^2 \theta &= 0 \\ 2 \sin \theta D \theta D \phi \cos \theta + D^2 \phi (\sin \theta)^2 &= 0 \\ \\ D^2 \theta &= (D \phi)^2 \cos \theta \sin \theta \\ D( D \phi (\sin \theta)^2) &= 0 \\ \\ \end{aligned}}

.

So

\displaystyle{ \begin{aligned} D \phi (\sin \theta)^2 &\equiv C \\ \\ \end{aligned}}

for some constant $C$.

.

Since \displaystyle{ \begin{aligned} D \phi (\sin \theta)^2 &= 0 \\ \\ \end{aligned}} for some $\theta$,

\displaystyle{ \begin{aligned} D \phi (\sin \theta)^2 &\equiv 0 \\ \\ \end{aligned}}

This is equivalent to setting up the coordinate system such that the initial value of $\theta$ equals zero.

.

Also, since \displaystyle{ \begin{aligned} (\sin \theta)^2 &\ne 0 \\ \\ \end{aligned}} for some $\theta$,

\displaystyle{ \begin{aligned} D \phi &\equiv 0 \\ \\ \end{aligned}}

.

— Me@2022-10-08 04:56:27 PM

.

.