# 2004, 2

— Me@2021-01-22 07:24:57 PM

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# Ex 1.15 Central force motion

Structure and Interpretation of Classical Mechanics

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Find Lagrangians for central force motion in three dimensions in rectangular coordinates and in spherical coordinates. First, find the Lagrangians analytically, then check the results with the computer by generalizing the programs that we have presented.

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The Lagrangians in rectangular coordinates:

\displaystyle{ \begin{aligned} &L (t; x, y, x; v_x, v_y, v_z) \\ &= \frac{1}{2} m \left( v_x^2 + v_y^2 + v_z^2 \right) - U \left( \sqrt{x^2 + y^2 + z^2} \right) \\ \end{aligned}}

The Lagrangians in spherical coordinates:

\displaystyle{ \begin{aligned} &L (t; r, \theta, \phi; \dot r, \dot \theta, \dot \phi) \\ &= \frac{1}{2} m r^2 \dot \phi^2 (\sin \theta)^2 + \frac{1}{2} m \dot \theta^2 r^2 + \frac{1}{2} m \dot r^2 - U (r) \\ &= \frac{1}{2} m \left( \dot r^2 + r^2 \dot \theta^2 + r^2 (\sin^2 \theta) \dot \phi^2 \right) - U (r) \\ \end{aligned}}

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

(define (p->r local)
(let ((polar-tuple (coordinate local)))
(let ((r (ref polar-tuple 0))
(theta (ref polar-tuple 1))
(phi (ref polar-tuple 2)))
(let ((x (* r (sin theta) (cos phi)))
(y (* r (sin theta) (sin phi)))
(z (* r (cos theta))))
(up x y z)))))

(define ((L-central-rectangular m U) local)
(let ((q (coordinate local))
(v (velocity local)))
(- (* 1/2 m (square v))
(U (sqrt (square q))))))

(define (L-central-polar m U)
(compose (L-central-rectangular m U) (F->C p->r)))

(show-expression
((L-central-polar 'm (literal-function 'U))
(->local 't
(up 'r 'theta 'phi)



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— Me@2021-01-22 02:59:11 PM

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# Quantum Computing, 2.2

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2021.01.21 Thursday ACHK

# Fallacy of infinity thinking, prequel

Why should you NOT murder one innocent person in order to save millions of people?

(Note: This question is NOT the same as “Could we give up one innocent person’s life in order to save millions of people’s?”)

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Could we murder one innocent person in order to save millions of people?

— Me@2017-06-20 01:04:56 PM

— Me@2021-01-19 06:15:54 PM

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NO.

If we can murder one innocent person in order to save all other people, then no one is safe after all, because anyone could be THAT innocent person, being sacrificed at any time.

(In the situation that we cannot save all the people at the same time, which person or which group has higher or lower priorities depends on context. There is no universal answer.)

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Instead, if we protect each person’s life, then all the people’s lives are protected.

— Me@2021-01-19 06:01:24 PM

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If you start with protecting each one, then every one person and thus the whole society will be protected.

If you start with protecting the whole society at all costs, then no one will be safe, because anyone could be that cost; any innocent person could be sacrificed at any time.

— Me@2021-01-20 6:48 AM

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# 鐵達尼極限 2.2

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Enjoy everything, need nothing, …

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… especially for human relationships.

– Conversations with God

（那某些原因，其實就是「基因」。「基因」只顧自己生命的延續，企圖不斷複製自己，即使犧牲「基因載體」的利益、幸福，甚至生命，也在所不惜。「基因載體」者，人或其他生物也。）

— Me@2021-01-03 04:18:55 PM

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# 1985

— Photomyne colorization

— Me@2021-01-18 10:14:30 PM

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# Problem 2.5b1

A First Course in String Theory

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2.5 Constructing simple orbifolds

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The wikipedia page “Fundamental polygon”, specifically the subsection entitled “group generators”, has a serious mathematical error. You cannot derive a presentation for the fundamental group from the fundamental polygon using the side labels in the manner described on that page (and which you have copied), unless all of the vertices of the polygon are identified to the same point. In the picture you provided and which can be seen on that page, one opposite pair of vertices of the square is identified to one point on the sphere, the other opposite pair of vertices is identified to a different point on the sphere.

There is still a way to derive a presentation for the fundamental group from a fundamental polygon, but it is not the way described on the wikipedia page. In the sphere example of your question, you have to ignore one of the two letters $\displaystyle{A}$, $\displaystyle{B}$, keeping only the other letter. For example, ignoring $\displaystyle{A}$ and keeping $\displaystyle{B}$, you get a presentation $\displaystyle{ \langle B \mid B B^{-1} = 1 \rangle }$, which is a presentation of the trivial group. The way you tell which to ignore and which to keep is by taking the quotient of the boundary of the polygon which is a graph with vertices and edges, choosing a maximal tree in that graph, ignoring all edge labels in the maximal tree, and keeping all edge labels not in the maximal tree.

On that wikipedia page, the Klein bottle and the torus examples are correct and you do not have to ignore any edge labels: all vertices are identified to a single point and the maximal tree is just a point. The sphere and the projective plane examples are incorrect: the four vertices are identified to two separate points, the maximal tree has one edge, and you have to ignore one edge label. The example of a hexagon fundamental domain for the torus is also incorrect: the six vertices are identified to two separate points, the maximal tree has one edge, and you have to ignore one edge label.

edited Jul 23 ’14 at 17:17

answered Jul 23 ’14 at 17:11

Lee Mosher

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yes, i thought that the fundamental polygon is this quotient space. – user159356 Jul 23 ’14 at 17:28

That’s backward: in your example, the sphere is the quotient space of the fundamental polygon, not the other way around. – Lee Mosher Jul 23 ’14 at 17:30

— Mathematics Stack Exchange

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2021.01.18 Monday ACHK

# Summing over histories, 2

The square root of the probability, 5

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If there is more than one way to achieve the present state, present == sum over all possible pasts, with weightings.

— Me@2011.06.26

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This is false for a physical state. This is only true for wave functions, which are NOT probabilities.

Wave functions are used for calculating probabilities; but they are not themselves probabilities.

Wave functions are quantum states, but not physical states.

Wave functions are logical and mathematical, but not physical.

A physical state is something observable, something can be measured, at least in principle.

A physical state is something that exists in spacetime, a wave function is not.

— Me@2021-01-16 06:12:08 PM

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# Zooming out, 2

In this morning’s dream, I could hold my lucid dream without feeling a headache. I felt comfortable as awake.

I was in a city and listening to an Inception-style music.

I tried to fold the city as in Inception so that I could have walked on a vertical land.

However, it did not work. The city got folded together, not as an L shape, [?]but as an L with a horizontal line head shape.[?]

Then I got myself waking up by flying-falling downwards.

— Me@2011.08.03

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# 機遇創生論, 2.1

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「緣份攻略」都不行，因為感覺有點怪。

（安：那就不如叫做「緣份理論」。）

「理論」很空泛。不應把「理論」，視為名字的一部分。

（安：不如叫做「超級種子理論」，或者「廣義種子理論」？）

（安：無錯，那相當「拗口」。)

（安：我剛才都是想到「大種子論」。）

（安：但「種子大論」比較奇怪。）

— Me@2021-01-13 04:52:02 PM

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# 2006, 4

— Me@2021-01-10 06:11:40 PM

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# Ex 1.14 Lagrange equations for L’, 2

Structure and Interpretation of Classical Mechanics

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Show by direct calculation that the Lagrange equations for $\displaystyle{L'}$ are satisfied if the Lagrange equations for $\displaystyle{L}$ are satisfied.

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$\displaystyle{x = F(t, x')}$

$\displaystyle{x' = G(t, x)}$

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\displaystyle{ \begin{aligned} &\partial_2 L \circ \Gamma[q] \\ &= \frac{\partial}{\partial v} L(t, x, v) \\ &= \frac{\partial}{\partial v} L'(t, x', v') \\ &= \frac{\partial}{\partial x'} L'(t, x', v') \frac{\partial x'}{\partial v} + \frac{\partial}{\partial v'} L'(t, x', v') \frac{\partial v'}{\partial v} \\ &= \frac{\partial}{\partial v'} L'(t, x', v') \frac{\partial v'}{\partial v} \\ \end{aligned}}

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$\displaystyle{v' = \partial_0 G(t, x) + \partial_1 G(t,x) v}$

\displaystyle{ \begin{aligned} \frac{\partial v'}{\partial v} &= \partial_1 G(t,x) = \frac{\partial x'}{\partial x} \\ \end{aligned}}

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\displaystyle{ \begin{aligned} &\partial_2 L \circ \Gamma[q] \\ &= \frac{\partial L'}{\partial v'} \frac{\partial x'}{\partial x} \\ \end{aligned}}

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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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\displaystyle{ \begin{aligned} &\partial_1 L \circ \Gamma[q] \\ &= \partial_1 L' \circ \Gamma[q'] \\ &= \frac{\partial}{\partial x} L' (t, x', v') \\ &= \frac{\partial L'}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} \\ \end{aligned}}

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\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ D \left( \frac{\partial L'}{\partial v'} \frac{\partial x'}{\partial x} \right) - \left( \frac{\partial L'}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} \right) &= 0 \\ D \left( \frac{\partial L'}{\partial v'} \right) \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} D \left( \frac{\partial x'}{\partial x} \right) - \left( \frac{\partial L'}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} \right) &= 0 \\ \end{aligned}}

\displaystyle{ \begin{aligned} \left[ D \left( \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} \right] \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} D \left( \frac{\partial x'}{\partial x} \right) - \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} &= 0 \\ \end{aligned}}

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\displaystyle{ \begin{aligned} &D \left( \frac{\partial x'}{\partial x} \right) \\ &= \frac{d}{dt} \left( \frac{\partial x'}{\partial x} \right) \\ &= \frac{\partial}{\partial t} \left( \frac{\partial x'}{\partial x} \right) \frac{\partial t}{\partial t} + \frac{\partial}{\partial x'} \left( \frac{\partial x'}{\partial x} \right) \frac{\partial x'}{\partial t} + \frac{\partial}{\partial v'} \left( \frac{\partial x'}{\partial x} \right) \frac{\partial v'}{\partial t} \\ &= \frac{\partial}{\partial t} \left( \frac{\partial x'}{\partial x} \right) + \frac{\partial}{\partial x} \left( \frac{\partial x'}{\partial x'} \right) \frac{\partial x'}{\partial t} + \frac{\partial}{\partial x} \left( \frac{\partial x'}{\partial v'} \right) \frac{\partial v'}{\partial t} \\ &= \frac{\partial}{\partial t} \left( \frac{\partial x'}{\partial x} \right) + \frac{\partial}{\partial x} \left( 1 \right) \frac{\partial x'}{\partial t} + \frac{\partial}{\partial x} \left( 0 \right) \frac{\partial v'}{\partial t} \\ &= \frac{\partial}{\partial x} \left( \frac{\partial x'}{\partial t} \right) + 0 + 0 \\ &= \frac{\partial v'}{\partial x} \\ \end{aligned}}

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\displaystyle{ \begin{aligned} \left[ D \left( \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} \right] \frac{\partial x'}{\partial x} + \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} - \frac{\partial L'}{\partial v'} \frac{\partial v'}{\partial x} &= 0 \\ \left[ D \left( \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} \right] \frac{\partial x'}{\partial x} &= 0 \\ D \left( \frac{\partial L'}{\partial v'} \right) - \frac{\partial L'}{\partial x'} &= 0 \\ \end{aligned}}

— Me@2021-01-06 08:10:46 PM

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# The square root of the probability, 4.3

Eigenstates 3.4.3

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The indistinguishability of cases is where the quantum probability comes from.

— Me@2020-12-25 06:21:48 PM

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In the double slit experiment, there are 4 cases:

1. only the left slit is open

2. only the right slit is open

3. both slits are open and a measuring device is installed somewhere in the experiment setup so that we can know which slit each photon passes through

4. both slits are open but no measuring device is installed; so for each photon, we have no which-way information

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For simplicity, we rephrase the case-3 and case-4:

1. only the left slit is open

2. only the right slit is open

3. both slits are open, with which-way information

4. both slits are open, without which-way information

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Case-3 can be regarded as a classical overlapping of case-1 and case-2, because if you check the result of case-3, you will find that it is just an overlapping of result case-1 and result case-2.

However, case-4 cannot be regarded as a classical overlapping of case-1 and case-2. Instead, case-4 is a quantum superposition. A quantum superposition canNOT be regarded as a classical overlapping of possibilities/probabilities/worlds/universes.

Experimentally, no classical overlapping can explain the interference pattern, especially the destruction interference part. An addition of two non-zero probability values can never result in a zero.

Logically, case-4 is a quantum superposition of go-left and go-right. Case-4 is neither AND nor OR of the case-1 and case-2.

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We can discuss AND or OR only when there are really 2 distinguishable cases. Since there are not any kinds of measuring devices (for getting which-way information) installed anywhere in the case-4, go-left and go-right are actually indistinguishable cases. In other words, by defining case-4 as a no-measuring-device case, we have indirectly defined that go-left and go-right are actually indistinguishable cases, even in principle.

Note that saying “they are actually indistinguishable cases, even in principle” is equivalent to saying that “they are logically indistinguishable cases” or “they are logically the same case“. So discussing whether a photon has gone left or gone right is meaningless.

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If 2 cases are actually indistinguishable even in principle, then in a sense, there is actually only 1 case, the case of “both slits are open but without measuring device installed anywhere” (case-4). Mathematically, this case is expressed as the quantum superposition of go-left and go-right.

Since it is only 1 case, it is meaningless to discuss AND or OR. It is neither “go-left AND go-right” nor “go-left OR go-right“, because the phrases “go-left” and “go-right” are themselves meaningless in this case.

— Me@2020-12-19 10:38 AM

— Me@2020-12-26 11:02 AM

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It is a quantum superposition of go-left and go-right.

Quantum superposition is NOT an overlapping of worlds.

Quantum superposition is neither AND nor OR.

— Me@2020-12-26 09:07:22 AM

When the final states are distinguishable you add probabilities:

$\displaystyle{P_{dis} = P_1 + P_2 = \psi_1^*\psi_1 + \psi_2^*\psi_2}$

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When the final state are indistinguishable,[^2] you add amplitudes:

$\displaystyle{\Psi_{1,2} = \psi_1 + \psi_2}$

and

$\displaystyle{P_{ind} = \Psi_{1,2}^*\Psi_{1,2} = \psi_1^*\psi_1 + \psi_1^*\psi_2 + \psi_2^*\psi_1 + \psi_2^*\psi_2}$

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[^2]: This is not precise, the states need to be “coherent”, but you don’t want to hear about that today.

edited Mar 21 ’13 at 17:04
answered Mar 21 ’13 at 16:58

dmckee

— Physics Stack Exchange

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$\displaystyle{ P_{ind} = P_1 + P_2 + \psi_2^*\psi_1 + \psi_2^*\psi_2 }$

$\displaystyle{ P_{\text{indistinguishable}} = P_{\text{distinguishable}} + \text{interference terms} }$

— Me@2020-12-26 09:07:46 AM

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interference terms ~ indistinguishability effect

— Me@2020-12-26 01:22:36 PM

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# WHY NOT

You have only ONE life in this life.

WHY NOT do something extraordinary in order to help people to help others?

— Me@~2008

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# 鐵達尼極限 2

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— 黃子華

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Enjoy everything, need nothing, …

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… especially for human relationships.

– Conversations with God

— Me@2021-01-03 04:18:55 PM

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# 1954

— Photomyne colorization

— Me@2021-01-02 05:20:52 PM

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# Problem 2.5a

A First Course in String Theory

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2.5 Constructing simple orbifolds

(a) Consider a circle $\displaystyle{S^1}$, presented as the real line with the identification $\displaystyle{x \sim x + 2}$. Choose $\displaystyle{-1 < x \le 1}$ as the fundamental domain. The circle is the space $\displaystyle{-1 < x \le 1}$ with points $\displaystyle{x = \pm 1}$ identified. The orbifold $\displaystyle{S^1/\mathbb{Z}_2}$ is defined by imposing the (so-called) $\displaystyle{\mathbb{Z}_2}$ identification $\displaystyle{x \sim -x}$. Describe the action of this identification on the circle. Show that there are two points on the circle that are left fixed by the $\displaystyle{\mathbb{Z}_2}$ action. Find a fundamental domain for the two identifications. Describe the orbifold $\displaystyle{S^1/\mathbb{Z}_2}$ in simple terms.

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Put point $\displaystyle{x=0}$ and point $\displaystyle{x=1}$ on the positions that they can form a horizontal diameter.

Then the action is a reflection of the lower semi-circle through the horizontal diameter to the upper semi-circle.

Point $\displaystyle{x=0}$ and point $\displaystyle{x=1}$ are the two fixed points.

A possible fundamental domain is $\displaystyle{0 \le x \le 1}$.

If a variable point $\displaystyle{x}$ moves from 0 to 1 and then keeps going, that point will actually go back and forth between 0 and 1.

— Me@2020-12-31 04:43:07 PM

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