# 寧化飛灰，不作浮塵

And what of that truth which more than anything else gives me confidence in Hong Kong? The truth is this. The qualities, the beliefs, the ideals that have made Hong Kong’s present will still be here to shape Hong Kong’s future.

Hong Kong, it seems to me, has always lived by the author, Jack London’s credo:

“I would rather be ashes than dust,
I would rather my spark should burn out in a brilliant blaze,
Than it should be stifled in dry rot.
I would rather be a superb meteor,
With every atom of me in magnificent glow,
Than a sleepy and permanent planet.”

Whatever the challenges ahead, nothing should bring this meteor crashing to earth, nothing should snuff out its glow. I hope that Hong Kong will take tomorrow by storm. And when it does, History will stand and cheer.

「寧化飛灰，不作浮塵。

— Christopher Francis Patten

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2019.10.13 Sunday ACHK

# Varying the action, 2.1

Equation (1.28):

$\displaystyle{S[q](t_1, t_2) = \int_{t_1}^{t_2} L \circ \Gamma[q]}$

Equation (1.30):

$\displaystyle{h[q] = L \circ \Gamma[q]}$

$\displaystyle{\delta_\eta S[q](t_1, t_2) = \int_{t_1}^{t_2} \delta_\eta h[q]}$

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

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

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

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— 1.5.1 Varying a path

— Structure and Interpretation of Classical Mechanics

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\displaystyle{ \begin{aligned} &\delta_\eta S[q] (t_1, t_2) \\ &= \int_{t_1}^{t_2} \delta_\eta \left( L \circ \Gamma[q] \right) \\ \end{aligned}}

Assume that $\displaystyle{L}$ is a path-independent function, so that we can use Eq. 1.26:

\displaystyle{ \begin{aligned} &= \int_{t_1}^{t_2} (D L \circ \Gamma[q]) \delta_\eta \Gamma[q] \\ \end{aligned}}

\displaystyle{ \begin{aligned} &= \int_{t_1}^{t_2} (D L \circ \Gamma[q]) (0, \eta(t), D\eta(t)) \\ &= \int_{t_1}^{t_2} (D L \left[ \Gamma[q] \right]) (0, \eta(t), D\eta(t)) \\ \end{aligned}}

Assume that $\displaystyle{L}$ is a path-independent function, so that any value of $\displaystyle{L}$ depends on the value of $\displaystyle{\Gamma}$ at that moment only, instead of depending on the whole path $\displaystyle{\Gamma}$:

\displaystyle{ \begin{aligned} &= \int_{t_1}^{t_2} (D L (\Gamma[q])) (0, \eta(t), D\eta(t)) \\ &= \int_{t_1}^{t_2} (D L (t, q, v)) (0, \eta(t), D\eta(t)) \\ &= \int_{t_1}^{t_2} [\partial_0 L (t, q, v), \partial_1 L (t, q, v), \partial_2 L (t, q, v)] (0, \eta(t), D\eta(t)) \\ \end{aligned}}

What kind of product is it here? Is it just a dot product? Probably not.

\displaystyle{ \begin{aligned} &= \int_{t_1}^{t_2} [\partial_1 L (t, q, v) \eta(t) + \partial_2 L (t, q, v) D\eta(t)] \\ \end{aligned}}

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— Me@2019-10-12 03:42:01 PM

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# Introduction to Differential Equations

llamaz 1 hour ago [-]

I think the calculus of variations might be a better approach to introducing ODEs in first year.

You can show that by generalizing calculus so the values are functions rather than real numbers, then trying to find a max/min using the functional version of $\displaystyle{\frac{dy}{dx} = 0}$, you end up with an ODE (viz. the Euler-Lagrange equation).

This also motivates Lagrange multipliers which are usually taught around the same time as ODEs. They are similar to the Hamiltonian, which is a synonym for energy and is derived from the Euler-Lagrange equations of a system.

Of course you would brush over most of this mechanics stuff in a single lecture (60 min). But now you’ve motivated ODEs and given the students a reason to solve ODEs with constant coefficients.

— Hacker News

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2019.10.02 Wednesday ACHK

# 少補習

— Me@2017-11-09 01:26:59 PM

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