# Huygens’ principle, 1.1

Why are there no backward secondary wavefronts?

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A secondary wave source is of different nature from a primary wave source. Consider an one dimensional transverse wave on a string:

Primary wave source is activated by the force from above, and then the wave propagates to both directions. Secondary wave source is activated by the particle on the left.

Primary wave source energy is from outside the string. Secondary wave source energy is from an adjacent string molecule.

When the secondary source S reaches its maximum, although it drags both P and Q, they are in different situations. At that moment, while the instantaneous velocity of Q is upward, that of P is downward.

Note that the red arrow at P is the force on P by S. It is not the net force on P.

— Me@2022-03-09 11:14:07 PM

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# 惜此際 5

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2022.03.29 Tuesday ACHK

# 伏線驅動程式 1.2

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我又是如何認識你們的呢？

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為何 HYC 會問我數學呢？

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為何 CKY 會找我教「附加數學」呢？

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為何我會教 CPM 物理呢？

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為何我在幾年前，會在這裡教書呢？

— Me@2022-03-08 12:06:18 PM

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# Ex 1.22 Driven pendulum, 3.1

Ex 1.24 Constraint forces, 1.1

Structure and Interpretation of Classical Mechanics

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~~~

[guess]

\displaystyle{ \begin{aligned} m \ddot{y} &= F \cos \theta - mg \\ m \ddot{x} &= - F \sin \theta \\ \end{aligned}}

\displaystyle{ \begin{aligned} m \ddot{y} &= F \frac{y_s - y}{l} - mg \\ m \ddot{x} &= - F \frac{x - x_s}{l} \\ \sqrt{(x_s - x)^2 + (y_s - y)^2} &= l \\ \end{aligned}}

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\displaystyle{ \begin{aligned} y_s &= l \\ x_s &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} m \ddot{y} &= F \frac{l - y}{l} - mg \\ m \ddot{x} &= - F \frac{x}{l} \\ \sqrt{x^2 + (l - y)^2} &= l \\ \end{aligned}}

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(define (KE-particle m v)
(* 1/2 m (square v)))

(define ((extract-particle pieces) local i)
(let* ((indices (apply up (iota pieces (* i pieces))))
(extract (lambda (tuple)
(vector-map (lambda (i)
(ref tuple i))
indices))))
(up (time local)
(extract (coordinate local))
(extract (velocity local)))))

(define (U-constraint q0 q1 F l)
(* (/ F (* 2 l))
(- (square (- q1 q0))
(square l))))

(define ((U-gravity g m) q)
(let* ((y (ref q 1)))
(* m g y)))

(define ((L-driven-free m l x_s y_s U) local)
(let* ((extract (extract-particle 2))

(p (extract local 0))
(q (coordinate p))
(qdot (velocity p))

(F (ref (coordinate local) 2)))

(- (KE-particle m qdot)
(U q)
(U-constraint (up (x_s (time local)) (y_s (time local)))
q
F
l))))

(let* ((U (U-gravity 'g 'm))
(x_s (literal-function 'x_s))
(y_s (literal-function 'y_s))
(L (L-driven-free 'm 'l x_s y_s U))
(q-rect (up (literal-function 'x)
(literal-function 'y)
(literal-function 'F))))
(show-expression
((compose L (Gamma q-rect)) 't)))



$\displaystyle{ L = \frac{1}{2} m \left[(Dx)^2 + (Dy)^2 \right] - mgy - \frac{F}{2l} \left[ (x-x_s)^2 + (y-y_s)^2 - l^2 \right] }$


(let* ((U (U-gravity 'g 'm))
(x_s (literal-function 'x_s))
(y_s (literal-function 'y_s))
(L (L-driven-free 'm 'l x_s y_s U))
(q-rect (up (literal-function 'x)
(literal-function 'y)
(literal-function 'F))))
(show-expression
(((Lagrange-equations L) q-rect) 't)))



\displaystyle{ \begin{aligned} mD^2x(t) + \frac{F(t)}{l} \left[x(t) - x_s(t)\right] &= 0 \\ mg + m D^2y(t) + \frac{F(t)}{l} [y(t) - y_s(t)] &= 0 \\ -l^2 + [y(t)-y_s(t)]^2 + [x(t)-x_s(t)]^2 &= 0 \\ \end{aligned}}


(define ((F->C F) local)
(->local (time local)
(F local)
(+ (((partial 0) F) local)
(* (((partial 1) F) local)
(velocity local)))))

(define ((q->r x_s y_s l) local)
(let* ((q (coordinate local))
(t (time local))
(theta (ref q 0))
(F (ref q 1)))
(up (+ (x_s t) (* l (sin theta)))
(- (y_s t) (* l (cos theta)))
F)))

(let ((q (up (literal-function 'theta)
(literal-function 'F))))
(show-expression (q 't)))



(let* ((x_s (literal-function 'x_s))
(y_s (literal-function 'y_s))
(q (up (literal-function 'theta)
(literal-function 'F))))
(show-expression ((Gamma q) 't)))




(let* ((xs (literal-function 'x_s))
(ys (literal-function 'y_s))
(q (up (literal-function 'theta)
(literal-function 'F))))
(show-expression ((compose (q->r xs ys 'l) (Gamma q)) 't)))




(let* ((xs (literal-function 'x_s))
(ys (literal-function 'y_s))
(q (up (literal-function 'theta)
(literal-function 'F))))
(show-expression ((F->C (q->r xs ys 'l)) ((Gamma q) 't))))





(define (L-theta m l x_s y_s U)
(compose
(L-driven-free m l x_s y_s U) (F->C (q->r x_s y_s l))))

(let* ((U (U-gravity 'g 'm))
(xs (literal-function 'x_s))
(ys (literal-function 'y_s))
(q (up (literal-function 'theta)
(literal-function 'F))))
(show-expression ((Gamma
(compose (q->r xs ys 'l) (Gamma q)))
't)))





(let* ((U (U-gravity 'g 'm))
(xs (literal-function 'x_s))
(ys (literal-function 'y_s))
(q (up (literal-function 'theta)
(literal-function 'F))))
(show-expression (U ((compose (q->r xs ys 'l) (Gamma q)) 't))))




(let* ((U (U-gravity 'g 'm))
(xs (literal-function 'x_s))
(ys (literal-function 'y_s))
(q (up (literal-function 'theta)
(literal-function 'F))))
(show-expression ((L-driven-free 'm 'l xs ys U)
((Gamma (compose (q->r xs ys 'l) (Gamma q))) 't))))





(let* ((U (U-gravity 'g 'm))
(xs (literal-function 'x_s))
(ys (literal-function 'y_s))
(q (up (literal-function 'theta)
(literal-function 'F))))
(show-expression ((L-theta 'm 'l xs ys U) ((Gamma q) 't))))



\displaystyle{ \begin{aligned} L_\theta &= \frac{1}{2} m (D x_s(t))^2 + \frac{1}{2} m (D y_s(t))^2 - m g \left[ y_s(t) - l \cos \theta(t) \right] \\ &+ \frac{1}{2} m l^2 (D \theta(t))^2 + lm D \theta(t) \left[ D x_s(t) \cos \theta(t) + \sin \theta(t) D y_s(t) \right] \\ \end{aligned}}

[guess]

— Me@2022-03-24 04:38:10 PM

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Experiment C:

What if during Experiment A, the observer changes his mind to turn on the detector before the particle’s arrival?

Explanation X:

We should regard this whole process as an experiment-setup.

The probability patterns encoded in the quantum state is of this experiment-setup, which in this case is equivalent to Experiment B. With respect to this experiment-process/experiment-setup, the variable-to-be-measured is always in a mixed state.

The word “always” here means that a quantum state is not a physical state of a particle in physical spacetime during the experiment. Instead, it is a property (of a physical variable) of the experiment-setup (physical system) itself.

“Physical system” means the experimental-setup design, which includes not just objects and devices, but also operations.

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However, for an experiment like Experiment C, doesn't the violation of Bell's inequality have proved that the spin of the particle is still in a superposition before the activation of the detector?

Let’s translate this question to a double-slit experiment version. (The consistency of the left detector and the right detector is as strange as the EPR consistency.)

Experiment C:

What if in the double-slit experiment, the detector is off when the particle is emitted, but the observer changes his mind to turn on the detector before the particle’s passing through of the double-slit plate?

Is the particle in a superposition or not when the detector is still off?

Instead of a superposition pure state, we should regard the particle path state as a mixed state, even before the detector is activated.

The reason is given in Explanation X.

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However, for an experiment like Experiment C, doesn't the violation of Bell's inequality have proved that the path of the particle is still in a superposition before the activation of the detector?

The violation of Bell’s inequality should be interpreted in this way:

Experiment A:

If there is no detector activated throughout the experiment until the particle reaches the final screen, the resulting statistical pattern (aka the interference pattern) is possible only if the particle has no definite path. “An unknown but definite path exists” would not be able to create such statistics.

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Experiment B:

If there is a detector activated since the beginning of the experiment, the particle path is in a mixed state since that beginning, meaning that the path is definite although unknown before measurement.

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Experiment C:

What if in the double-slit experiment, the detector is off when the particle is emitted, but the observer changes his mind to turn on the detector before the particle’s passing through of the double-slit plate?

Is the particle in a superposition or not when the detector is still off?

After the detector is activated, the path is in a mixed state.

Before the detector is activated, regarding the path state as in a superposition or as in a mixed state makes no physical difference, because by definition, there is no measurement before the detector is activated.

Also, a superposition’s coefficients are always about the potential-activation of detectors.

A superposition state has a corresponding mixed state. The superposition coefficients can be modulus-squared to give the mixed state coefficients. That is the exact original meaning of the superposition coefficients. (This is the statistical interpretation given by Born rule.)

In other words, each coefficient in a superposition state by default encodes the probability of each potential measurement result.

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However, in Experiment C, regarding the state as a superposition state before the activation of detector creates conceptual paradoxes, such as:

If the particle path variable has no definite state before measurement, how come the left detector and right detector always give consistent results?

This is the double-slit experiment version of the EPR paradox.

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When the experimenter follows the plan of Experiment A, the particle path variable is always in a superposition, since the beginning of the experiment.

However, when he changes his mind to turn on the detector before the particle’s passing through of the double-slit plate, the particle path variable is always in a mixed state, since the beginning of the experiment.

Wouldn’t that create retro-causality?

It seems to be retro-causality, but it is not, because this description is a language shortcut only. The apparent violation of causality does not exist in the language longcut version, which is provided in Explanation X.

Also, due to the indistinguishability of identical particles, particle identities and thus particle trajectories are post hoc stories only.

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In short, the violation of Bell’s inequality means that:

If there is no path detector activated throughout the experiment until the particle reaches the final screen, the path variable never has a definite value, known or unknown, throughout that experiment.

The violation of Bell’s inequality should not be applied to Experiment C, because it is not the case of “having no detector throughout the experiment”.

Also, note that we do not need the violation of Bell’s inequality to prove that the path variable has no definite state, known or unknown.

Just the interference pattern’s existence is already enough for us to do so.

— Me@2022-03-08 11:56:00 PM

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Someone who causes people to do that which they ought to do but which they would not do in his absence.

— John T. Reed

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2022.03.21 Monday ACHK

# 十萬七千里 3

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（安：另外，他提的另一個，有關學習數學的要點是，即使假設你在大學中，學到的數學，在日常生活中沒有用，單單是為獲取，那些嶄新的元素概念本身，就已經能夠令你有超能力；令你有一些，常人沒有的思考工具、比喻語言。）

$\displaystyle{\lim_{x \to \infty} \frac{1}{x} = 0}$

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— Me@2022-03-14 05:11:29 PM

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1990s, 17

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# 2.11 Extra dimension and statistical mechanics, 3

A First Course in String Theory

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(b) Assume that $\displaystyle{R \ll a}$ in such a way that …

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[guess]

$\displaystyle{\frac{\hbar^2}{ma^2} \ll k T \ll \frac{\hbar^2}{mR^2}}$

\displaystyle{ \begin{aligned} Z &= Z(a) \tilde{Z}(R) \\ \end{aligned}}

\displaystyle{ \begin{aligned} Z(a) &= \sum_{k=1}^\infty \exp\left[- \beta \left( \frac{\hbar^2}{2m} \right) \left(\frac{k \pi}{a} \right)^2 \right] \\ \tilde Z (R) &= 1 + 2 Z(R \pi) \\ \end{aligned}}

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$\displaystyle{\beta = \frac{1}{kT} \ll \frac{ma^2}{\hbar^2}}$

\displaystyle{ \begin{aligned} Z(a) &\approx \frac{1}{\beta} \int_{x = 0}^\infty e^{\left[- \frac{1}{\beta} \left( \frac{\hbar^2}{2m} \right) \left(\frac{x \pi}{a} \right)^2 \right]} dx = \sqrt{\frac{m a^2}{2 \beta \pi \hbar^2}} \\ \end{aligned}}

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$\displaystyle{\beta \gg \frac{mR^2}{\hbar^2}}$

\displaystyle{ \begin{aligned} Z(R\pi) &= \sum_{k=1}^\infty \exp\left[- \beta \left( \frac{\hbar^2}{2m} \right) \left(\frac{k}{R} \right)^2 \right] \\ &= \exp\left[- \beta \left( \frac{\hbar^2}{2mR^2} \right) \right] + ... \\ &= 1 - \beta \left( \frac{\hbar^2}{2mR^2} \right) + ... \\ \end{aligned}}

\displaystyle{ \begin{aligned} \tilde Z (R) &= 1 + 2 Z(R \pi) \\ &= 1 + 2 \sum_{k=1}^\infty \exp\left[- \beta \left( \frac{\hbar^2}{2m} \right) \left(\frac{k}{R} \right)^2 \right]\\ &= 1 + 2 \left\{ \exp\left[- \beta \left( \frac{\hbar^2}{2mR^2} \right) \right] + ... \right\} \\ \end{aligned}}

[guess]

— Me@2022-03-10 10:56:10 AM

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A wave function (for a particular variable) is an intrinsic property of a physical system.

“Physical system” means the experimental-setup design, which includes not just objects and devices, but also operations.

In other words, “where and when an observer should do what during the experiment” is actually part of your experimental-setup design, defining what probability distribution (for any particular variable) you (the observer) will get.

If the experimenter does not follow the original experiment design, such as not turning on the detector at the pre-defined time, then he is actually doing another experiment, which will have a completely different probability distribution (for any particular variable).

— Me@2022-02-18 07:40:14 AM

— Me@2022-02-22 07:01:40 PM

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Note that different observers see different “physical systems”. They see different quantum states, because a quantum state is actually not a “state”, but a (statistical) property of a system, encoded in the superposition coefficients.

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Experiment A:

In an EPR experiment, if it is by design that an observer will not turn on the detector, then the story should be that this experiment-setup is always in a superposition state (with respect to the variable-not-to-be-measured).

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Experiment B:

In an EPR experiment, if it is by design that an observer will turn on the detector before a particle’s arrival, then the story should be that this experiment-setup is always in a mixed state (with respect to the variable-to-be-measured).

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No concept of “wave function collapse” is needed.

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What if during Experiment A, the observer changes his mind to turn on the detector before the particle's arrival?

He has actually replaced experiment-setup A with experiment-setup B.

Then is the experiment always in a superposition state? Or always in a mixed state?

As a language shortcut, you can say that the superposition wave function collapses at the moment of system replacement.

However, it is a language shortcut, a story only.

A story is not reality.

A story is post hoc.

physical definition

~ define unobservable events in terms of observable events

Any story would be fine as long as it is compatible with reality, aka measurement results.

But some stories are better than others.

Here, only the longcut version can avoid common meaningless questions.

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Experiment C:

What if during Experiment A, the observer changes his mind to turn on the detector before the particle’s arrival?

We should regard this whole process as an experiment-setup.

The probability patterns encoded in the quantum state is of this experiment-setup, which in this case is equivalent to Experiment B. With respect to this experiment-process/experiment-setup, the variable-to-be-measured is always in a mixed state.

The word “always” here means that a quantum state is not a physical state of a particle in physical spacetime during the experiment. Instead, it is a property (of a physical variable) of the experiment-setup (physical system) itself.

“Physical system” means the experimental-setup design, which includes not just objects and devices, but also operations.

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However, for an experiment like Experiment C, doesn't the violation of Bell's inequality have proved that the spin of the particle is still in a superposition before the activation of the detector?

Let’s translate this question to a double-slit experiment version.

— Me@2022-03-08 11:56:00 PM

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— Me@2011.10.08

— Me@2022-03-08

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# 伏線驅動程式 1.1

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— Me@2022-03-08 12:06:18 PM

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# 2014

— Me@2022-03-07 01:12:19 PM

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# Ex 1.23 Fill in the details

Structure and Interpretation of Classical Mechanics

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Show that the Lagrange equations for Lagrangian (1.97) are the same as the Lagrange equations for Lagrangian (1.95) with the substitution $\displaystyle{c(t) = l}$, $\displaystyle{Dc(t) = D^2 c(t) = 0}$.

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[guess]

The Lagrange equation:

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

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Eq. 1.97:

$\displaystyle{ L''(t,q,\dot q)}$

$\displaystyle{ = \sum_\alpha \frac{1}{2} m_\alpha \left( \partial_0 f_\alpha (t, q) + \partial_1 f_\alpha (t, q) \dot q \right)^2 - V(t, f(t,q,l)) }$

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Eq. 1.95:

$\displaystyle{ L'(t;q,c,F; \dot q, \dot c, \dot F)}$

$\displaystyle{ = \sum_\alpha \frac{1}{2} m_\alpha \left( \partial_0 f_\alpha (t, q, c) + \partial_1 f_\alpha (t, q,c) \dot q + \partial_2 f_\alpha (t, q, c) \dot c \right)^2 }$

$\displaystyle{ - V(t, f(t,q,c)) - \sum_{\{ \alpha, \beta | \alpha < \beta, \alpha \leftrightarrow \beta \}} \frac{F_{\alpha \beta}}{2 l_{\alpha \beta}} \left[ c_{\alpha \beta}^2 - l_{\alpha \beta}^2 \right] }$

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\displaystyle{ \begin{aligned} &L'(t;q,c,F; \dot q, \dot c, \dot F) \\ &= L'(t;\{q_{\alpha i}\}, \{c_{\alpha \beta}\}, \{F_{\alpha \beta}\}; \{\dot q_{\alpha i}\}, \{\dot c_{\alpha \beta}\}, \{\dot F_{\alpha \beta}\}) \\ \end{aligned}}

\displaystyle{ \begin{aligned} &= \sum_\alpha \frac{1}{2} m_\alpha \left( \partial_0 f_\alpha (t;\{q_{\alpha i}\}, \{c_{\alpha \beta}\}) + \partial_1 f_\alpha (t;\{q_{\alpha i}\}, \{c_{\alpha \beta}\}) \dot q + \partial_2 f_\alpha (t;\{q_{\alpha i}\}, \{c_{\alpha \beta}\}) \dot c \right)^2 \\ &- V(t, f(t;\{q_{\alpha i}\}, \{c_{\alpha \beta}\})) - \sum_{\{ \alpha, \beta | \alpha < \beta, \alpha \leftrightarrow \beta \}} \frac{F_{\alpha \beta}}{2 l_{\alpha \beta}} \left[ c_{\alpha \beta}^2 - l_{\alpha \beta}^2 \right] \\ \end{aligned}}

\displaystyle{ \begin{aligned} &= \sum_\alpha \frac{1}{2} m_\alpha \left( \frac{\partial f_\alpha}{\partial t} + \sum_i \frac{\partial f_\alpha}{\partial q_{\alpha i}} \dot q_{\alpha i} + \sum_\beta \frac{\partial f_\alpha}{\partial c_{\alpha \beta}} \dot c_{\alpha \beta} \right)^2 \\ &- V(t, f(t;\{q_{\alpha i}\}, \{c_{\alpha \beta}\})) - \sum_{\{ \alpha, \beta | \alpha < \beta, \alpha \leftrightarrow \beta \}} \frac{F_{\alpha \beta}}{2 l_{\alpha \beta}} \left[ c_{\alpha \beta}^2 - l_{\alpha \beta}^2 \right] \\ \end{aligned}}

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Consider the mass $\displaystyle{m_\alpha}$:

\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot q_{\alpha i}} \right) - \frac{\partial L}{\partial q_{\alpha i}} &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot c_{\alpha \beta}} \right) - \frac{\partial L}{\partial c_{\alpha \beta}} &= 0 \\ \end{aligned}}

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\displaystyle{ \begin{aligned} &\frac{\partial L}{\partial q_{\alpha i}} \\ &= \frac{\partial}{\partial q_{\alpha i}} \left[ \sum_\alpha \frac{1}{2} m_\alpha \left( \frac{\partial f_\alpha}{\partial t} + \sum_j \frac{\partial f_\alpha}{\partial q_{\alpha j}} \dot q_{\alpha j} + \sum_\beta \frac{\partial f_\alpha}{\partial c_{\alpha \beta}} \dot c_{\alpha \beta} \right)^2 - V(t, f) \right] - 0 \\ &= \sum_\alpha \frac{1}{2} m_\alpha \frac{\partial}{\partial q_{\alpha i}} \left( G \right)^2 - \frac{\partial}{\partial q_{\alpha i}} V(t, f(t;\{q_{\alpha j}\}, \{c_{\alpha \beta}\})) \\ &= \sum_\alpha m_\alpha G \frac{\partial G}{\partial q_{\alpha i}} - \frac{\partial V}{\partial q_{\alpha i}} \\ \end{aligned}}

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\displaystyle{ \begin{aligned} G &= \left( \frac{\partial f_\alpha}{\partial t} + \sum_j \frac{\partial f_\alpha}{\partial q_{\alpha j}} \dot q_{\alpha j} + \sum_\beta \frac{\partial f_\alpha}{\partial c_{\alpha \beta}} \dot c_{\alpha \beta} \right) \\ \frac{\partial G}{\partial q_{\alpha i}} &= \frac{\partial}{\partial q_{\alpha i}} \left( \frac{\partial f_\alpha}{\partial t} + \sum_j \frac{\partial f_\alpha}{\partial q_{\alpha j}} \dot q_{\alpha j} + \sum_\beta \frac{\partial f_\alpha}{\partial c_{\alpha \beta}} \dot c_{\alpha \beta} \right) \\ &= \left( \frac{\partial}{\partial q_{\alpha i}} \frac{\partial f_\alpha}{\partial t} + \sum_j \frac{\partial}{\partial q_{\alpha i}} \frac{\partial f_\alpha}{\partial q_{\alpha j}} \dot q_{\alpha j} + \sum_\beta \frac{\partial}{\partial q_{\alpha i}} \frac{\partial f_\alpha}{\partial c_{\alpha \beta}} \dot c_{\alpha \beta} \right) \\ \end{aligned}}

[guess]

— Me@2022-03-06 05:24:18 PM

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Measuring spin and polarization

In de Broglie–Bohm theory, the results of a spin experiment cannot be analyzed without some knowledge of the experimental setup. It is possible to modify the setup so that the trajectory of the particle is unaffected, but that the particle with one setup registers as spin-up, while in the other setup it registers as spin-down. Thus, for the de Broglie–Bohm theory, the particle’s spin is not an intrinsic property of the particle; instead spin is, so to speak, in the wavefunction of the particle in relation to the particular device being used to measure the spin. This is an illustration of what is sometimes referred to as contextuality and is related to naive realism about operators. Interpretationally, measurement results are a deterministic property of the system and its environment, which includes information about the experimental setup including the context of co-measured observables; in no sense does the system itself possess the property being measured, as would have been the case in classical physics.

— Wikipedia on De Broglie–Bohm theory

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2022.03.06 Sunday ACHK

# 預防離婚

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— Me@2022-03-05 07:13:14 PM

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「好婚姻」勝過「無婚姻」；

— Me@2022-03-06 10:33:05 AM

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# 同一屋簷下

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— Me@2022-03-05 07:13:14 PM

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# 1990s, 16

— Me@2022-03-05 11:41:03 AM

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# 2.11 Extra dimension and statistical mechanics

A First Course in String Theory

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(a) Explicitly calculate $\displaystyle{Z(a, R)}$ in the very high temperature limit $\displaystyle{\left( \beta = \frac{1}{kT} \to 0\right)}$.

~~~

\displaystyle{ \begin{aligned} Z &= Z(a) \tilde{Z}(R) \\ \end{aligned}}

\displaystyle{ \begin{aligned} Z(a) &= \sum_{k=1}^\infty \exp\left[- \beta \left( \frac{\hbar^2}{2m} \right) \left(\frac{k \pi}{a} \right)^2 \right] \\ \tilde Z (R) &= 1 + 2 Z(R \pi) \\ \end{aligned}}

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\displaystyle{ \begin{aligned} \Delta k &= 1 \\ Z(a) &= \frac{1}{\beta} \sum_{\beta k = \beta, 2\beta, ...} \exp\left[- \frac{1}{\beta} \left( \frac{\hbar^2}{2m} \right) \left(\frac{\beta k \pi}{a} \right)^2 \right] \Delta (\beta k) \\ \end{aligned}}

\displaystyle{ \begin{aligned} \int_{0}^{\infty } e^{-ax^{2}}\,dx &= \frac{1}{2} {\sqrt {\frac {\pi }{a}}} \\ \end{aligned}}

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When $\displaystyle{ \beta \to 0 }$,

\displaystyle{ \begin{aligned} Z(a) &\approx \frac{1}{\beta} \int_{x = 0}^\infty e^{\left[- \frac{1}{\beta} \left( \frac{\hbar^2}{2m} \right) \left(\frac{x \pi}{a} \right)^2 \right]} dx = \sqrt{\frac{m a^2}{2 \beta \pi \hbar^2}} \\ \end{aligned}}

\displaystyle{ \begin{aligned} \tilde Z (R) &\approx 1 + \sqrt{\frac{2 m R^2 \pi}{\beta \hbar^2}} \approx \sqrt{\frac{2 m R^2 \pi}{\beta \hbar^2}} = Z(2 \pi R) \\\\ \end{aligned}}

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\displaystyle{ \begin{aligned} Z &\approx Z(a) Z(2 \pi R) \\ \end{aligned}}

— Me@2022-03-04 07:12:49 PM

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