# 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, ),

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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