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]}

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\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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2019.06.25 Tuesday (c) All rights reserved by ACHK

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

(反對)開夜車 4.1

本文章並(!)不(!)可作為醫學建議。如需醫學意見,請諮詢專業人士。

(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。)

其實,主要是講讀書時代。如果在工作時代,你的職位需要,你時常「開夜車」的話,你根本就應該另謀高就。

試問,世間上,有什麼工作,竟然值得你冒生命危險,去時常「開夜車」呢?

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言歸正傳,讀書時代,如果時間管理得宜,需要「開夜車」的情況,其實是很少。

(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?)

一定「開夜車」的話,你至少要做到以下幾點,去保障自己的安全:

  • 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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另外,有時,只要跳出框框,破格思考,或者,只要你的時間表稍改一點,就根本毋須「開夜車」。

本文章並(!)不(!)可作為醫學建議。如需醫學意見,請諮詢專業人士。

— Me@2019-06-06 08:23:56 PM

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2019.06.08 Saturday (c) All rights reserved by ACHK