Physical laws are low-energy approximations to reality, 1.5

… difficult, as you have to heat up [the system] …

… messenger …

… collider particle …

… cool it down to discover new physics …

— Me@2019-06-27 11:23:18 PM

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Ex 1.7 Properties of $\delta$

Let $\displaystyle{F}$ be a path-independent function and $\displaystyle{g}$ be a path-dependent function; then

$\displaystyle{\delta_\eta h[q] = \left( DF \circ g[q] \right) \delta_\eta g[q]~~~~~\text{with}~~~~~h[q] = F \circ g[q].~~~~~(1.26)}$

— 1.5.1 Varying a path

— Structure and Interpretation of Classical Mechanics

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Prove that

$\displaystyle{\delta_\eta F \circ g[q] = \left( DF \circ g[q] \right) \delta_\eta g[q]}$

~~~

$\displaystyle{RHS = \lim_{\Delta t \to 0} \left( \frac{F \circ g[q](t+\Delta t) - F \circ g[q](t)}{\Delta t} \right) \lim_{\epsilon \to 0} \left( \frac{g[q+\epsilon \eta]-g[q]}{\epsilon} \right)}$

\displaystyle{ \begin{aligned} LHS &= \delta_\eta F \circ g[q] \\ &= \lim_{\epsilon \to 0} \left( \frac{F \circ g[q+\epsilon \eta]-F \circ g[q]}{\epsilon} \right) \\ &= \lim_{\epsilon \to 0} \left( \frac{F \left[ g[q+\epsilon \eta] \right] - F \left[ g[q] \right]}{\epsilon} \right) \\ \end{aligned}}

Since $\displaystyle{F}$ is path-independent,

\displaystyle{ \begin{aligned} LHS &= \lim_{\epsilon \to 0} \left( \frac{F \left( g[q+\epsilon \eta ] \right) - F \left( g[q] \right)}{\epsilon} \right) \\ \end{aligned}}

Let $\displaystyle{ g[q+\epsilon \eta] = g + \Delta g}$.

\displaystyle{ \begin{aligned} LHS &= \lim_{\epsilon \to 0} \left( \frac{F \left( g[q] + \Delta g[q]] \right) - F \left( g[q] \right)}{\epsilon} \right) \\ &= \lim_{\epsilon \to 0} \left( \frac{F \left( g[q] + \Delta g[q]] \right) - F \left( g[q] \right)}{\Delta g[q]}\frac{\Delta g[q]}{\epsilon} \right) \\ \end{aligned}}

When $\displaystyle{ \epsilon \to 0}$, $\displaystyle{ \Delta g \to 0 }$.

\displaystyle{ \begin{aligned} LHS &= \lim_{\substack{\epsilon \to 0 \\ \Delta g \to 0}} \left( \frac{F \left( g[q] + \Delta g[q]] \right) - F \left( g[q] \right)}{\Delta g[q]}\frac{\Delta g[q]}{\epsilon} \right) \\ &= \lim_{\Delta g \to 0} \left( \frac{F \left( g[q] + \Delta g[q]] \right) - F \left( g[q] \right)}{\Delta g[q]} \lim_{\epsilon \to 0} \frac{g[q + \epsilon \eta] - g[q]}{\epsilon} \right) \\ &= DF \left( g[q] \right) \delta_\eta g[q] \\ &= RHS \\ \end{aligned}}

— Me@2019-06-24 10:55:28 PM

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From classical to quantum

From this viewpoint, the move from a classical to a quantum mechanical system is not a move from a comutative to a non-commutative algebra $\displaystyle{\mathcal{A}}$ of a real-valued observables, but, instead, a move from a commutative algebra to a partial commutative algebra of observables.

Of course, every non-commutative algebra determines an underlying partial commutative algebra and also its diagram of commutative subalgebras.

That fact that assuming the structure of a non-commutative algebra is the wrong assumption has already been observed in the literature (see, for example, [19]),

but it is often replaced by another wrong assumption, namely that of assuming the structure of a Jordan algebra.

These differing assumptions on the structure of $\displaystyle{\mathcal A}$ affect the size of its automorphisum group and, hence, of the allowable symmetries of the system (the weaker the assumed structure on $\displaystyle{\mathcal A}$, the larger is its automorphism group).

— The Mathematical Foundations of Quantum Mechanics

— David A. Edwards

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2019.06.18 Tuesday ACHK

Free will, 6

The factors are so many and so complex that no single observer can ever predict my actions with 100% accuracy, even in principle.

$\displaystyle{\equiv}$

I have free will.

— Me@2018-05-06 12:21:36 AM

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（反對）開夜車 4.1

（問：但是，在現今社會，無論是上班，或是讀書，完全不「開夜車」，又好像不切實際。）

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（問：那樣說有意思嗎？我正正是問你，在時間管理失宜，需要「開夜車」時，該如何自處？）

• 只可以間中，不可以經常。

• 日間中途要有小睡。

• 平均而言，你仍必須要有，充足的睡眠。亦即是話，某一天睡少了，必須於在當個星期，還回「睡債」。

• 例如，如果你的充足睡眠是，每天七小時，而你在某一天只睡了六小時的話，你就有義務，在當個星期的另一天，睡多一小時。

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— Me@2019-06-06 08:23:56 PM

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Physical laws are low-energy approximations to reality, 1.4

$\displaystyle{\vdots}$

$\displaystyle{\uparrow}$

quark

$\displaystyle{\uparrow}$

plasma

$\displaystyle{\uparrow}$

vapour

$\displaystyle{\uparrow}$

water

$\displaystyle{\uparrow}$

ice

$\displaystyle{\downarrow}$

f-magnetism

$\displaystyle{\downarrow}$

QCD

$\displaystyle{\vdots}$

— Me@2019-06-04 08:42:39 PM

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