No reading, 2

Posted on July 12, 2022 by rpflee

Teaching is useless, 4

[Projections start to stare at Ariadne as they walk by]

Ariadne: Why are they all looking at me?

Cobb: Cause my subconscious feels that someone else is creating this world. The more you change things, the quicker the projections start to converge on you.

Ariadne: Converge?

Cobb: They sense the foreign nature of the dreamer. They attack, like white blood cells fighting an infection.

Ariadne: What, they’re gonna attack us?

Cobb: No, no….Just you.

— Inception

Most people hate being taught because they don’t want their personal world to be interfered with.

— Me@2016-05-29 06:29:40 AM

星期日休息

Posted on July 12, 2022 by rpflee

在生活事務上，有些夫妻男方做主導，即「夫唱婦隨」；有些則女方做主導，即「婦唱夫隨」。

一方主導的關係，不健康，因為不平等。平等的交往，雙方才可坦誠相對。

可改為「雙方不主導」和「雙方主導」。

有客觀標準的事項，按該道理來行事，雙方不主導。例如，男方懂電腦，女方不太懂。那樣，購買電腦時，由男方選擇。

而只是風格喜好的情節，則雙方主導。逢星期一、三、五，由男方話事；逢二、四、六，由女方。

— Me@2022-07-12 12:33:24 PM

1.7 Evolution of Dynamical State, 2.2

Posted on July 9, 2022 by rpflee

Structure and Interpretation of Classical Mechanics

$\displaystyle{ \begin{aligned} \partial_1 L \circ \Gamma[q] &= D ( \partial_2 L \circ \Gamma[q]) \\ \\ &= \partial_0 ( \partial_2 L \circ \Gamma[q]) Dt + \partial_1 ( \partial_2 L \circ \Gamma[q]) Dq + \partial_2 ( \partial_2 L \circ \Gamma[q]) Dv \\ \\ &= \partial_0 \partial_2 L \circ \Gamma[q] + ( \partial_1 \partial_2 L \circ \Gamma[q]) Dq + (\partial_2 \partial_2 L \circ \Gamma[q]) D^2 q \\ \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} (\partial_2 \partial_2 L \circ \Gamma[q]) D^2 q &= \partial_1 L \circ \Gamma[q] - \partial_0 \partial_2 L \circ \Gamma[q] - (\partial_1 \partial_2 L \circ \Gamma[q]) Dq \\ \\ D^2 q &= \left[ \partial_2 \partial_2 L \circ \Gamma[q] \right]^{-1} \left\{ \partial_1 L \circ \Gamma[q] - \partial_0 \partial_2 L \circ \Gamma[q] - (\partial_1 \partial_2 L \circ \Gamma[q]) Dq \right\} \\ \\ \end{aligned}}$

where $\displaystyle{\left[ \partial_2 \partial_2 L \circ \Gamma \right]}$ is a structure that can be represented by a symmetric square matrix, so we can compute its inverse.

~~~

[guess]

The Lagrange equation:

$\displaystyle{ \begin{aligned} D \left( \frac{\partial}{\partial \dot q_1} L \circ \Gamma[\begin{bmatrix} q_1 \\ q_2 \end{bmatrix}] \right) - \left(\frac{\partial}{\partial q_1} L \circ \Gamma[\begin{bmatrix} q_1 \\ q_2 \end{bmatrix}]]\right) &= 0 \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} D \left( \frac{\partial}{\partial \dot q_2} L \circ \Gamma[\begin{bmatrix} q_1 \\ q_2 \end{bmatrix}]] \right) - \left(\frac{\partial}{\partial q_2} L \circ \Gamma[\begin{bmatrix} q_1 \\ q_2 \end{bmatrix}]]\right) &= 0 \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial}{\partial \dot q_1} L (t, q_1(t), \dot q_1(t), q_2(t), \dot q_2(t)) \right) - \frac{\partial}{\partial q_1} L (t, q_1(t), \dot q_1(t), q_2(t), \dot q_2(t)) &= 0 \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial}{\partial \dot q_2} L (t, q_1(t), \dot q_1(t), q_2(t), \dot q_2(t)) \right) - \frac{\partial}{\partial q_2} L (t, q_1(t), \dot q_1(t), q_2(t), \dot q_2(t)) &= 0 \end{aligned}}$

$\displaystyle{ \begin{aligned} D \left( \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \end{bmatrix} L \circ \Gamma[q_1, q_2] \right) - \left( \begin{bmatrix} \frac{\partial}{\partial q_1} \\ \frac{\partial}{\partial q_2} \end{bmatrix} L \circ \Gamma[q_1, q_2]\right) &= 0 \\ D \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) - \left( \vec \partial_1 L \circ \Gamma[q_1, q_2]\right) &= 0 \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \vec \partial_1 L \circ \Gamma[q] &= D ( \vec \partial_2 L \circ \Gamma[q]) \\ \\ \begin{bmatrix} \frac{\partial}{\partial q_1} \\ \frac{\partial}{\partial q_2} \end{bmatrix} L \circ \Gamma[q_1, q_2] &= D \left( \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \end{bmatrix} L \circ \Gamma[q_1, q_2] \right) \\ \\ \frac{\partial}{\partial q_1} L \circ \Gamma[q_1, q_2] &= \partial_0 \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) Dt \\ &+ \partial_{q_1} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) D q_1 + \partial_{v_1} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) D v_1 \\ &+ \partial_{q_2} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) D q_2 + \partial_{v_2} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) D v_2 \\ \frac{\partial}{\partial q_2} L \circ \Gamma[q_1, q_2] &= ... \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{\partial}{\partial q_1} L \circ \Gamma[q_1, q_2] &= \partial_0 \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) Dt \\ &+ \frac{\partial}{\partial q_1} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) D q_1 + \frac{\partial}{\partial q_2} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) D q_2 \\ &+ \frac{\partial}{\partial \dot q_1} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) D^2 q_1 + \frac{\partial}{\partial \dot q_2} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) D^2 q_2 \\ \frac{\partial}{\partial q_2} L \circ \Gamma[q_1, q_2] &= \partial_0 \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[q_1, q_2] \right) Dt \\ &+ \frac{\partial}{\partial q_1} \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[q_1, q_2] \right) D q_1 + \frac{\partial}{\partial q_2} \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[q_1, q_2] \right) D q_2 \\ &+ \frac{\partial}{\partial \dot q_1} \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[q_1, q_2] \right) D^2 q_1 + \frac{\partial}{\partial \dot q_2} \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[q_1, q_2] \right) D^2 q_2 \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \vec \partial_1 L \circ \Gamma[q_1, q_2] &= \partial_0 \left(\vec \partial_2 L \circ \Gamma[q_1, q_2] \right) Dt \\ &+ \frac{\partial}{\partial q_1} \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D q_1 + \frac{\partial}{\partial q_2} \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D q_2 \\ &+ \frac{\partial}{\partial \dot q_1} \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D^2 q_1 + \frac{\partial}{\partial \dot q_2} \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D^2 q_2 \\ \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \begin{bmatrix} \frac{\partial}{\partial q_1} \\ \frac{\partial}{\partial q_2} \end{bmatrix} L \circ \Gamma[q_1, q_2] &= \partial_0 \left( \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \end{bmatrix} L \circ \Gamma[q_1, q_2] \right) Dt \\ &+ \frac{\partial}{\partial q_1} \left( \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \end{bmatrix} L \circ \Gamma[q_1, q_2] \right) D q_1 + \frac{\partial}{\partial q_2} \left( \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \end{bmatrix} L \circ \Gamma[q_1, q_2] \right) D q_2 \\ &+ \frac{\partial}{\partial \dot q_1} \left( \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \end{bmatrix} L \circ \Gamma[q_1, q_2] \right) D^2 q_1 + \frac{\partial}{\partial \dot q_2} \left( \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \end{bmatrix} L \circ \Gamma[q_1, q_2] \right) D^2 q_2 \\ \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{\partial}{\partial \dot q_1} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[\vec q] \right) D^2 q_1 + \frac{\partial}{\partial \dot q_2} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[\vec q] \right) D^2 q_2 &= \frac{\partial}{\partial q_1} L \circ \Gamma[\vec q] - \partial_0 \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[\vec q] \right) Dt \\ &- \frac{\partial}{\partial q_1} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[\vec q] \right) D q_1 - \frac{\partial}{\partial q_2} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[\vec q] \right) D q_2 \\ \\ \frac{\partial}{\partial \dot q_1} \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[\vec q] \right) D^2 q_1 + \frac{\partial}{\partial \dot q_2} \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[\vec q] \right) D^2 q_2 &= \frac{\partial}{\partial q_2} L \circ \Gamma[\vec q] - \partial_0 \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[\vec q] \right) Dt \\ &- \frac{\partial}{\partial q_1} \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[\vec q] \right) D q_1 - \frac{\partial}{\partial q_2} \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[\vec q] \right) D q_2 \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} &\begin{bmatrix} \frac{\partial}{\partial \dot q_1} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[\vec q] \right) & \frac{\partial}{\partial \dot q_2} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[\vec q] \right) \\ \frac{\partial}{\partial \dot q_1} \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[\vec q] \right) & \frac{\partial}{\partial \dot q_2} \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[\vec q] \right) \end{bmatrix} \begin{bmatrix} D^2 q_1 \\ D^2 q_2 \end{bmatrix} \\ &= \begin{bmatrix} \frac{\partial}{\partial q_1} L \circ \Gamma[\vec q] - \partial_0 \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[\vec q] \right) Dt - \frac{\partial}{\partial q_1} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[\vec q] \right) D q_1 - \frac{\partial}{\partial q_2} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[\vec q] \right) D q_2 \\ \\ \frac{\partial}{\partial q_2} L \circ \Gamma[\vec q] - \partial_0 \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[\vec q] \right) Dt - \frac{\partial}{\partial q_1} \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[\vec q] \right) D q_1 - \frac{\partial}{\partial q_2} \left(\frac{\partial}{\partial \dot q_2} L \circ \Gamma[\vec q] \right) D q_2 \\ \end{bmatrix} \end{aligned}}$

$\displaystyle{ \begin{aligned} &\left(\begin{bmatrix} \frac{\partial}{\partial \dot q_1} \left(\frac{\partial}{\partial \dot q_1} \right) & \frac{\partial}{\partial \dot q_2} \left(\frac{\partial}{\partial \dot q_1} \right) \\ \frac{\partial}{\partial \dot q_1} \left(\frac{\partial}{\partial \dot q_2} \right) & \frac{\partial}{\partial \dot q_2} \left(\frac{\partial}{\partial \dot q_2} \right) \end{bmatrix} L \circ \Gamma[\vec q] \right)\begin{bmatrix} D^2 q_1 \\ D^2 q_2 \end{bmatrix} \\ &= \begin{bmatrix} \frac{\partial}{\partial q_1} \\ \frac{\partial}{\partial q_2} \end{bmatrix} L \circ \Gamma[\vec q] - \left( \partial_0 \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \end{bmatrix} L \circ \Gamma[\vec q] \right) Dt - \left( \begin{bmatrix} \frac{\partial}{\partial q_1} \left(\frac{\partial}{\partial \dot q_1} \right) & \frac{\partial}{\partial q_2} \left(\frac{\partial}{\partial \dot q_1} \right) \\ \\ \frac{\partial}{\partial q_1} \left(\frac{\partial}{\partial \dot q_2} \right) & \frac{\partial}{\partial q_2} \left(\frac{\partial}{\partial \dot q_2} \right) \\ \end{bmatrix} L \circ \Gamma[\vec q] \right) \begin{bmatrix} D q_1 \\ D q_2 \end{bmatrix} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} &\left( \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \\ \end{bmatrix} \begin{bmatrix} \frac{\partial}{\partial \dot q_1} & \frac{\partial}{\partial \dot q_2} \\ \end{bmatrix} L \circ \Gamma[\vec q] \right)\begin{bmatrix} D^2 q_1 \\ D^2 q_2 \end{bmatrix} \\ &= \begin{bmatrix} \frac{\partial}{\partial q_1} \\ \frac{\partial}{\partial q_2} \end{bmatrix} L \circ \Gamma[\vec q] - \left( \partial_0 \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \end{bmatrix} L \circ \Gamma[\vec q] \right) Dt - \left( \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \end{bmatrix} \begin{bmatrix} \frac{\partial}{\partial q_1} & \frac{\partial}{\partial q_2} \end{bmatrix} L \circ \Gamma[\vec q] \right) \begin{bmatrix} D q_1 \\ D q_2 \end{bmatrix} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} &\left( \vec \partial_2 \vec \partial_2^T L \circ \Gamma[\vec q] \right) D^2 \vec q \\ &= \vec \partial_1 L \circ \Gamma[\vec q] - \left( \partial_0 \vec \partial_2 L \circ \Gamma[\vec q] \right) - \left( \vec \partial_2 \vec \partial_1^T L \circ \Gamma[\vec q] \right) D \vec q \\ \end{aligned}}$

[guess]

— Me@2022-07-07 05:20:38 PM

1.7 Evolution of Dynamical State, 2.1

Posted on July 9, 2022 by rpflee

Structure and Interpretation of Classical Mechanics

where $\displaystyle{\left[ \partial_2 \partial_2 L \circ \Gamma \right]}$ is a structure that can be represented by a symmetric square matrix, so we can compute its inverse.

~~~

[guess]

Eq. (1.110):

$\displaystyle{ \left(D \Gamma[q] \right)(t) = \left( 1, Dq(t), D^2 q(t), ... \right) = \begin{bmatrix} 1 \\ Dq(t) \\ D^2 q(t) \\ ... \\ \end{bmatrix} \\ }$

$\displaystyle{ \Gamma[q](t) = \left( t, q(t), D q(t), D^2 q(t), ... \right) = \begin{bmatrix} t \\ q(t) \\ D q(t) \\ D^2 q(t) \\ ... \\ \end{bmatrix} \\ }$

$\displaystyle{ \Gamma[q] = \begin{bmatrix} I \\ q \\ D q \\ D^2 q \\ ... \\ \end{bmatrix} \\ }$ ,

where $I(t) = t$ .

$\displaystyle{ \Gamma[q_1, q_2](t) = \left( t, \begin{bmatrix} q_1(t) \\ q_2(t) \end{bmatrix}, D \begin{bmatrix} q_1(t) \\ q_2(t) \end{bmatrix}, D^2 \begin{bmatrix} q_1(t) \\ q_2(t) \end{bmatrix}, ... \right) = \begin{bmatrix} t \\ \begin{bmatrix} q_1(t) \\ q_2(t) \end{bmatrix} \\ D \begin{bmatrix} q_1(t) \\ q_2(t) \end{bmatrix} \\ D^2 \begin{bmatrix} q_1(t) \\ q_2(t) \end{bmatrix} \\ ... \\ \end{bmatrix} \\ }$

The Lagrange equation:

$\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial}{\partial \dot q} L (t, q(t), \dot q(t)) \right) - \frac{\partial}{\partial q} L (t, q(t), \dot q(t)) &= 0 \end{aligned}}$

$\displaystyle{ \begin{aligned} D \left( \frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) - \left(\frac{\partial}{\partial q_1} L \circ \Gamma[q_1, q_2]\right) &= 0 \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} D \left( \frac{\partial}{\partial \dot q_2} L \circ \Gamma[q_1, q_2] \right) - \left(\frac{\partial}{\partial q_2} L \circ \Gamma[q_1, q_2]\right) &= 0 \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \vec \partial_1 L \circ \Gamma[q] &= D ( \vec \partial_2 L \circ \Gamma[q]) \\ \\ \begin{bmatrix} \frac{\partial}{\partial q_1} \\ \frac{\partial}{\partial q_2} \end{bmatrix} L \circ \Gamma[q_1, q_2] &= D \left( \begin{bmatrix} \frac{\partial}{\partial \dot q_1} \\ \frac{\partial}{\partial \dot q_2} \end{bmatrix} L \circ \Gamma[q_1, q_2] \right) \\ \\ \frac{\partial}{\partial q_1} L \circ \Gamma[q_1, q_2] &= D \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) \\ \frac{\partial}{\partial q_2} L \circ \Gamma[q_1, q_2] &= D \left( \frac{\partial}{\partial \dot q_2} L \circ \Gamma[q_1, q_2] \right) \\ \\ \frac{\partial}{\partial q_1} L \circ \Gamma[q_1, q_2] &= \partial_0 \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) Dt \\ &+ \partial_{q_1} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) D q_1 + \partial_{v_1} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) D v_1 \\ &+ \partial_{q_2} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) D q_2 + \partial_{v_2} \left(\frac{\partial}{\partial \dot q_1} L \circ \Gamma[q_1, q_2] \right) D v_2 \\ \frac{\partial}{\partial q_2} L \circ \Gamma[q_1, q_2] &= ... \\ \end{aligned}}$

This part is wrong.

$\displaystyle{\begin{aligned} \vec \partial_1 L \circ \Gamma[q_1, q_2] &= \partial_0 \left(\vec \partial_2 L \circ \Gamma[q_1, q_2] \right) Dt \\ &+ \frac{\partial}{\partial q_1} \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D q_1 + \frac{\partial}{\partial q_2} \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D q_2 \\ &+ \frac{\partial}{\partial \dot q_1} \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D^2 q_1 + \frac{\partial}{\partial \dot q_2} \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D^2 q_2 \\ \\ &= \partial_0 \left(\vec \partial_2 L \circ \Gamma[q_1, q_2] \right) Dt \\ &+ \begin{bmatrix} \frac{\partial}{\partial q_1} & \frac{\partial}{\partial q_2} \\ \end{bmatrix} \begin{bmatrix} \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D q_1 \\ \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D q_2 \\ \end{bmatrix} + \begin{bmatrix} \frac{\partial}{\partial \dot q_1} & \frac{\partial}{\partial \dot q_2} \\ \end{bmatrix} \begin{bmatrix} \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D^2 q_1 \\ \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D^2 q_2 \\ \end{bmatrix} \\ \\ \end{aligned}}$

$\displaystyle{\begin{aligned} &= \partial_0 \left(\vec \partial_2 L \circ \Gamma[q_1, q_2] \right) Dt \\ &+ \begin{bmatrix} \frac{\partial}{\partial q_1} & \frac{\partial}{\partial q_2} \\ \end{bmatrix} \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D \begin{bmatrix} q_1 \\ q_2 \\ \end{bmatrix} + \begin{bmatrix} \frac{\partial}{\partial \dot q_1} & \frac{\partial}{\partial \dot q_2} \\ \end{bmatrix} \left( \vec \partial_2 L \circ \Gamma[q_1, q_2] \right) D^2 \begin{bmatrix} q_1 \\ q_2 \\ \end{bmatrix} \\ \\ \vec \partial_1 L \circ \Gamma[\vec q] &= \partial_0 \left(\vec \partial_2 L \circ \Gamma[q_1, q_2] \right) + \vec \partial_1^T \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) D \vec q + \vec \partial_2^T \left( \vec \partial_2 L \circ \Gamma[\vec q] \right) D^2 \vec q \\ \end{aligned}}$

This can be proven wrong by just checking the dimensions of the matrix products.

The source of the errors is the second step. It puts the column matrix $\displaystyle{\vec \partial_2 = \begin{bmatrix} \frac{\partial}{\partial \dot q_1} & \frac{\partial}{\partial \dot q_2} \end{bmatrix}^T}$ into each equation. Instead, each row of $\displaystyle{\vec \partial_2}$ should match only one of the two equations.

[guess]

— Me@2022-07-07 05:20:38 PM

Schrodinger’s cat, 3.5

Posted on July 7, 2022 by rpflee

This description is wrong.

Quantum superposition is exhibited in fact in many directly observable phenomena, such as interference peaks from an electron wave in a double-slit experiment.

— Wikipedia on Quantum superposition

Whatever you observe, it is not a superposition.

If no left-right detector is allowed, the moving-left/right variable is an unobservable. What you observe, instead, is the dot on the final screen.

In other words, the observable is the final position of a particle when it reaches the final screen. And the final screen itself acts as the detector for that observable.

Superposition is unobservable due to the lack of definition (of the distinction between different states) of the corresponding physical variable, due to the fact that no corresponding detector, such as the left-right detector, is allowed in the experimental design.

physical phenomena

~ observable events

Superposition

~ logically unobservable, since not yet defined

~ not yet defined, since logically undefinable

~ logically undefinable, since no corresponding detector is allowed

A superposition state is not an observable state. In other words, it is not a physical state. Then what is the point of considering it?

Although a superposition state is not a physical state, it is a mathematical state that can be used to calculate the probabilities of different possible physical-states/observable-events.

— Me@2022-07-06 06:00:55 PM

ELEG 1110

Posted on July 6, 2022 by rpflee

ELEG 1110 – Basic Circuit Theory – Assignment 1

Due Date: 27/9/2011 (Tue)

Please submit your assignments in the collection box before the due time.

Please do not send soft copy to Prof. Choy or tutors.

The collection box is located on 3rd floor of Ho Sin Hang building, near room 321.

Please pay attention to submit the assignment to the collection box located at the second row, not the distribution box in the first row.

2022.07.06 Wednesday ACHK

愛情畢業證書（兩張）

Posted on July 5, 2022 by rpflee

相聚零刻 1.2.3

如果愛情是一個朴實的話，那你就一定要，在成熟的時候，［才］去採摘。如果在此之前採摘，那這個果子採下來之後呢，它就不會再成長了；直接爛掉。

— 為什麽遇到喜歡的女生，千萬別去表白！

— 楚兒戀愛說

實情並非：

你原本沒有中學的學問。在頒發中學畢業證書的那一刻，突然集齊中學的知識。

而是相反：

當你的學識達到，準大學的程度時，中學才頒畢業證書。

同樣原理，實情並非：

你倆原本不是情侶。透過表白，你倆有機會，𣊬間變成情侶。

而是相反：

你倆的感情累積已達到，情侶的階段時，表白可有可無。

表白的目的，不是開始建立愛情，而是確認已知情侶。

表白的作用是，愛情里程碑。

正如，旅行時拍照，只是為了作歷史紀錄，而不是「拍照促成旅行」；即使沒有拍照，旅行仍在。

又正如，行李箱用來放行李；即使沒有行李箱，行李仍在，只是不便攜帶。

已為情侶之人，確認為情侶的目的是，穩定感情，從而再發展，至最終成夫妻。

沒有打算成夫妻，表白又有何用？

愛情的畢業證書，有兩張。第二張是結婚證書。

— Me@2022-07-05 12:18:26 PM

One decision

Posted on July 4, 2022 by rpflee

You’re always one decision away from a totally different life.

2022.07.04 Monday ACHK

Quick Calculation 3.10

Posted on July 3, 2022 by rpflee

A First Course in String Theory

Equation (3.96) has content: if you move a particle along a closed loop in a static gravitational field, the net work that you do against the gravitational field is zero.

Prove the above statement.

~~~

Eq. (3.96):

$\displaystyle{\vec g = - \nabla V_g}$

By the gradient theorem (aka the fundamental theorem of calculus for line integrals):

$\displaystyle{\int _{\gamma }\nabla \varphi (\mathbf {r} )\cdot \mathrm {d} \mathbf {r} =\varphi \left(\mathbf {q} \right)-\varphi \left(\mathbf {p} \right)}$

…

— Me@2022-07-03 04:27:19 PM

Schrodinger’s cat, 3.4

Posted on July 3, 2022 by rpflee

A macroscopic system (such as a cat) may evolve over time into a superposition of classically distinct quantum states (such as “alive” and “dead”).

— Wikipedia on Quantum superposition

The components of a superposition must be indistinguishable states.

A superposition is neither an AND state nor an OR state.

AND or OR are only possible for more than one state.

AND or OR are only possible for at least 2 (distinguishable) states.

The cat is not in a superposition state of “alive” and “dead”.

Instead, it is in a mixed state of “alive” and “dead”.

A mixed state is an OR state (of at least 2 distinguishable states).

— Me@2022-07-03 11:02:24 AM

超整理法, 2

Posted on July 2, 2022 by rpflee

兩項重點必須遵守：一是取出用過的檔案不要放回原位，而放進最左邊，如此日復一日，常用的資料集中在左邊，被「推擠」到最右邊的，多半是不常用的資料。二是將同學錄、電話簿、保證書、說明書等要永久或長期保存的資料，放在書架最右邊，標上特別記號。

「野口式推擠建檔」的優點包括，輕鬆隨手地就可以把桌上的資料整理好，不像以前要剪貼、分類，弄得人仰馬翻；最近常用的檔案都集中在左邊，立刻可以找到所需資料，而靠近右邊，比較不常用的資料，可以定期查看，丟掉不用的資料。

— 「超級」整理．超高效率

— 天下雜誌 173 期

— 孫曉萍

2022.07.02 Saturday ACHK

物理避數學；數學避思考

Posted on July 2, 2022 by rpflee

數學教育 8.1

這段改編自 2010 年 4 月 24 日的對話。

科技是一套，避開科學的系統。

科技的存在，是為了讓不懂科學的人，也能享受，科學的成果。

例如，即使你不懂寫程式，也可以使用電腦。

科學的起點是物理。物理是一套，避開數學的系統。

物理的存在，是為了讓人在，毋須全懂數學的情況下，也能享受，數學的成果。

例如，在物理科中「簡諧運動」，原本需要懂數學中的「三角學」和「微分方程」，才可以處理得到。但是，在中學的物理教科書中，會教你把「簡諧運動」，看成某個「圓周運動」的投影；那樣，你可以在不太懂，「三角學」和「微分方程」的情況下，獲得那些運算成果。

所以，如果一題物理題目，你竟然需要，大量的數學運算才能完成，有很大機會是，你根本不懂，該題的物理原理，導致沒有工具，去簡化（甚至避開）運算。

數學的存在，是為了讓人在，盡量避開思考的情況下，也能享受，思考的成果。

例如，你在學過乘法的定義和原理後，就會背誦乘數表，從而毋須再刻意思考，也可以利用到，乘法的成果。

那樣，既然數學是一個「盡量避開思考」的系統，為什麼還要學呢？

完全不學數學，就可以完全避開思考。那不是更好嗎？

學過數學思考，才會有能力判斷，哪時需要思考，哪時不需要。

而有時「不需要思考」，正正是因為在那些情況，你知道可以用，哪些內在或外在工具，去避開思考之餘，而獲取成果。

如果你從來未學過乘法，即使給你計數機，你也不會用它來做乘法，因為，你根本不知道，有「乘法」這個數學概念工具。

— Me@2022-07-02 12:02:59 PM

Car

Posted on July 1, 2022 by rpflee

— Me@2022-07-01 04:29:08 PM

1.7 Evolution of Dynamical State

Posted on June 30, 2022 by rpflee

Lagrange’s equations are ordinary differential equations that the path must
satisfy. They can be used to test if a proposed path is a realizable path of the
system. However, we can also use them to develop a path, starting with initial
conditions.

Assume that the state of a system is given by the tuple $\displaystyle{(t, q, v)}$ . If we are
given a prescription for computing the acceleration $\displaystyle{a = A(t, q, v)}$ , then

$\displaystyle{D^2 q = A \circ \Gamma[q]}$

and we have as a consequence

$\displaystyle{D^3 q = D( A \circ \Gamma[q]) = D_t A \circ \Gamma[q]}$

and so on.

So the higher-derivative components of the local tuple are given by functions $\displaystyle{D_t A, D_t^2 A, \dots}$ . Each of these functions depends on lower-derivative components of the local tuple. All we need to deduce the path from the state is a function that gives the next-higher derivative component of the local description from the state. We use the Lagrange equations to find this function.

— Structure and Interpretation of Classical Mechanics

Eq. (1.113):

$\displaystyle{ D_t F \circ \Gamma[q] = D(F \circ \Gamma[q]) }$

Eq. (1.114):

$\displaystyle{ \begin{aligned} D_t F (t, q, v, a, ...) &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ... \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} D_t F \circ \Gamma[q] (t) &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v(t) + \partial_2 F(t, q, v, a, ...) a(t) + ... \\ \end{aligned} }$

The Lagrange equation:

$\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial}{\partial \dot q} L (t, q(t), \dot q(t)) \right) - \frac{\partial}{\partial q} L (t, q(t), \dot q(t)) &= 0 \end{aligned}}$

$\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ \end{aligned}}$

where $\displaystyle{\left[ \partial_2 \partial_2 L \circ \Gamma \right]}$ is a structure that can be represented by a symmetric square matrix, so we can compute its inverse.

— Me@2022-06-30 11:33:27 AM

Schrodinger’s cat, 3.3

Posted on June 29, 2022 by rpflee

The modern view is that this mystery is explained by quantum decoherence.

Quantum decoherence is useful, but NOT necessary.

It is useful for the self-consistency checking of quantum mechanics.

Some microscopic states are expressed as (mathematical) superpositions of macroscopic-indistinguishable-if-no-measuring-device-is-allowed states.

Two states are called “macroscopic-distinguishable” only if they result in two different physical phenomena. In other words, the distinction must be observable.

an eigenstate

~ an observable (at least in principle) state

~ a physical state

.

a superposition state

~ an unobservable (even in principle) state

~ a mathematical (but not physical) state

We define microscopic states and events in terms of macroscopic states and events. A consistent theory must be able to deduce (explain or predict) macroscopic states and events from those microscopic states and events.

— Me@2022-06-15 07:40:37 PM

Eigenstates 3

Posted on June 28, 2022 by rpflee

an eigenstate

~ an state identical to the overall average

In analogy, in the equation

$\displaystyle{\frac{12+13+14}{3}=13}$ ,

the number 13 appears both on the left (as one of the component numbers) and on the right (as the overall average).

In this sense, the number 13 is an “eigenstate”.

— Me@2016-08-25 01:36 AM

— Me@2022-06-28 08:19 PM

An eigenstate has a macroscopic equivalence.

— Me@2016-08-29 06:10:21 PM

An eigenstate is a microstate that has a corresponding macrostate.

An eigenstate is a mathematical state which is also a physical state.

An eigenstate is an observable state.

— Me@2022-06-28 07:36:40 PM

好心分手

Posted on June 28, 2022 by rpflee

Importance, 4.1

這段改編自 2021 年 12 月 14 日的對話。

no good deed goes unpunished

Beneficial actions often go unappreciated or are met with outright hostility.

If they are appreciated, they often lead to additional requests.

— Wiktionary

有時，災民會埋怨救援的人。

救援者往往是，災難之後，災民第一批見到的人，所以，有些災民會認定，那些災難是救援者造成的。

例如，在電影《美國隊長3：英雄內戰》中，美國政府要追究「復仇者聯盟」；認定「紐約之戰」中的二千多人傷亡，是「復仇者」們造成的。

實情是，沒有他們的抵抗，外星人所引起的傷亡，會以百萬計。

— Me@2022-06-27 04:10:30 PM

Why not both?

Posted on June 24, 2022 by rpflee

Singles 3

2022.06.24 Friday ACHK

Quick Calculation 3.8

Posted on June 23, 2022 by rpflee

A First Course in String Theory

Show that this condition fixes uniquely $\displaystyle{\alpha = \gamma = 1/2}$ , and $\displaystyle{\beta = - 3/2}$ , thus reproducing the result in (3.90).

~~~

Eq. (3.93):

$\displaystyle{l_P = (G)^\alpha (c)^\beta (\hbar)^\gamma}$

$\displaystyle{l_P = \left( \frac{l_p^3}{m_p t_P^2} \right)^\alpha \left( \frac{l_P}{t_P} \right)^\beta \left( \frac{m_P l_P^2}{t_P} \right)^\gamma}$

$\displaystyle{\begin{aligned} 3 \alpha + \beta + 2\gamma &= 1 \\ -\alpha + \gamma &= 0 \\ - 2 \alpha - \beta - \gamma &= 0 \\ \end{aligned}}$

var('a b c')

solve([3*a+b+2*c==1, -a+c==0, -2*a-b-c==0], a, b, c)

$\displaystyle{\begin{aligned} \alpha &= \frac{1}{2} \\ \\ \beta &= \frac{-3}{2} \\ \\ \gamma &= \frac{1}{2} \\ \\ \end{aligned}}$

Eq. (3.90):

$\displaystyle{l_P = \sqrt{\frac{G \hbar}{c^3}}}$

— Me@2022-06-23 10:46:22 AM

Schrodinger cat’s misunderstanding

Posted on June 22, 2022 by rpflee

Schrodinger’s cat, 3.2

In 1935, Erwin Schrödinger devised a well-known thought experiment, now known as Schrödinger’s cat, which highlighted this dissonance between quantum mechanics and classical physics.

The main point of the Schrödinger’s cat thought experiment is NOT to prove that there should also be superposition for macroscopic objects. Instead, the main point of the thought experiment is exactly the opposite—to prove that regarding a superposition state as a physical state leads to logical contradiction.

— Me@2022-06-15 07:19:36 PM

Physics Town

continues

No reading, 2

星期日休息

1.7 Evolution of Dynamical State, 2.2

1.7 Evolution of Dynamical State, 2.1

Schrodinger’s cat, 3.5

ELEG 1110

愛情畢業證書（兩張）

One decision

Quick Calculation 3.10

Schrodinger’s cat, 3.4

超整理法, 2

物理避數學；數學避思考

Car

1.7 Evolution of Dynamical State

Schrodinger’s cat, 3.3

Eigenstates 3

好心分手

Why not both?

Quick Calculation 3.8

Schrodinger cat’s misunderstanding