# 2.10 A spacetime orbifold in two dimensions, 3

A First Course in String Theory

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(b) Draw a spacetime diagram, indicate the $\displaystyle{x^+}$ and $\displaystyle{x^-}$ axes, and sketch the family of curves $\displaystyle{x^+ x^- = a^2}$,

where $\displaystyle{a > 0}$ is a real constant that labels the various curves.

~~~ — Me@2021-08-31 08:42:40 PM

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# Visualizing higher dimensions, 2

Geometry is global.

Space is what we can see at once.

Dynamics is local.

Time is what we cannot see at once.

— Me@2017-02-07 10:11:34 PM

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If we could see, for example, several minutes at once, that several minutes would become a spatial dimension.

In other words, that dimension is visualized for us.

— Me@2017-02-03 07:31:25 AM

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# Broken

A: Before meeting you, my life is chaotic.

M: It is because you missed, me. With me, you are completed.

— Me@2021-06-07 06:16:29 PM

— Me@2021-08-22 05:40:53 PM

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# 上山尋寶, 2

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— Me@2021-08-16 04:31:19 PM

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# 1986, 2

— Me@2021-08-13 04:52:08 PM

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# Ex 1.21 The dumbbell, 3.2

Structure and Interpretation of Classical Mechanics

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c. Make a change of coordinates to a coordinate system with center of mass coordinates $\displaystyle{x_{cm}}$, $\displaystyle{y_{cm}}$, angle $\displaystyle{\theta}$, distance between the particles $\displaystyle{c}$, and tension force $\displaystyle{F}$. Write the Lagrangian in these coordinates, and write the Lagrange equations.

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[guess] \displaystyle{ \begin{aligned} m_0 \ddot y_0 &= F \sin \theta \\ m_0 \ddot x_0 &= F \cos \theta \\ m_1 \ddot y_1 &= -F \sin \theta \\ m_1 \ddot x_1 &= -F \cos \theta \\ \end{aligned}} \displaystyle{ \begin{aligned} y_{cm} &= \frac{m_0 y_0 + m_1 y_1}{m_0 + m_1} \\ x_{cm} &= \frac{m_0 x_0 + m_1 x_1}{m_0 + m_1} \\ \end{aligned}} \displaystyle{ \begin{aligned} \ddot y_{cm} &= \frac{F \sin \theta - F \sin \theta}{m_0 + m_1} = 0 \\ \ddot x_{cm} &= \frac{F \cos \theta - F \cos \theta}{m_0 + m_1} = 0 \\ \end{aligned}}

. \displaystyle{ \begin{aligned} y_0 &= y_{cm} - \frac{m_1}{M} c(t) \sin \theta \\ x_0 &= x_{cm} - \frac{m_1}{M} c(t) \cos \theta \\ y_1 &= y_{cm} + \frac{m_0}{M} c(t) \sin \theta \\ x_1 &= x_{cm} + \frac{m_0}{M} c(t) \cos \theta \\ \end{aligned}} \displaystyle{ \begin{aligned} x_1 - x_0 &= \frac{m_0}{M} c(t) \cos \theta + \frac{m_1}{M} c(t) \cos \theta \\ &= c(t) \cos \theta \\ \end{aligned}} \displaystyle{ \begin{aligned} \dot x_1 - \dot x_0 &= \dot c(t) \cos \theta - c(t) \dot \theta \sin \theta \\ \end{aligned}} \displaystyle{ \begin{aligned} \ddot x_1 - \ddot x_0 &= \ddot c(t) \cos \theta - \dot c(t) \dot \theta \sin \theta - \dot c(t) \dot \theta \sin \theta - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}} \displaystyle{ \begin{aligned} y_1 - y_0 &= c(t) \sin \theta \\ \dot y_1 - \dot y_0 &= \dot c(t) \sin \theta + c(t) \dot \theta \cos \theta \\ \ddot y_1 - \ddot y_0 &= \ddot c(t) \sin \theta + \dot c(t) \dot \theta \cos \theta + \dot c(t) \dot \theta \cos \theta + c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta \\ \end{aligned}}

. \displaystyle{ \begin{aligned} m_0 \ddot y_0 &= F \sin \theta \\ m_0 \ddot x_0 &= F \cos \theta \\ m_1 \ddot y_1 &= -F \sin \theta \\ m_1 \ddot x_1 &= -F \cos \theta \\ \end{aligned}}

When $\displaystyle{\dot c(t) = 0}$ and $\displaystyle{\ddot c(t) = 0}$, \displaystyle{ \begin{aligned} \ddot x_1 - \ddot x_0 &= - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta \\ - \left( \frac{1}{m_1} + \frac{1}{m_1} \right) F \cos \theta &= - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}} \displaystyle{ \begin{aligned} \ddot y_1 - \ddot y_0 &= ... \\ - \left( \frac{1}{m_1} + \frac{1}{m_0} \right) F \sin \theta &= c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta \\ \end{aligned}}

. \displaystyle{ \begin{aligned} \tan \theta &= \frac{ c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta }{- c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta} \\ &= \frac{ \ddot \theta - \dot \theta^2 \tan \theta }{- \ddot \theta \tan \theta - \dot \theta^2} \\ \end{aligned}} \displaystyle{ \begin{aligned} \tan \theta \left( - \ddot \theta \tan \theta - \dot \theta^2 \right) &= \ddot \theta - \dot \theta^2 \tan \theta \\ &... \\ 0 &= \ddot \theta (1 + \tan^2 \theta) \\ \ddot \theta &= 0 \\ \end{aligned}}

. \displaystyle{ \begin{aligned} - \left( \frac{1}{m_1} + \frac{1}{m_0} \right) F \sin \theta &= c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta \\ - \left( \frac{1}{m_1} + \frac{1}{m_1} \right) F \cos \theta &= - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}

Let $\displaystyle{\frac{1}{\mu} = \left( \frac{1}{m_1} + \frac{1}{m_1} \right)}$ and since $\displaystyle{\ddot \theta = 0}$, \displaystyle{ \begin{aligned} - \frac{1}{\mu} F \sin \theta &= - c(t) \dot \theta^2 \sin \theta \\ - \frac{1}{\mu} F \cos \theta &= - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}

Since $\displaystyle{\sin \theta}$ and $\displaystyle{\cos \theta}$ cannot be both zero at the same time, \displaystyle{ \begin{aligned} - \frac{1}{\mu} F &= - c(t) \dot \theta^2 \\ \end{aligned}}

Put $\displaystyle{c(t) = l}$, \displaystyle{ \begin{aligned} \frac{1}{\mu} F &= l \dot \theta^2 \\ \dot \theta^2 &= \frac{1}{l \mu} F \\ \end{aligned}}

[guess]

— Me@2021-08-08 05:41:21 PM

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# Logical arrow of time, 9.4

The second law of thermodynamics’ derivation (Ludwig Boltzmann’s H-theorem) is with respect to an observer.

How does an observer keep losing microscopic information about a system?

— Me@2017-02-12 07:37:54 PM

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This drew the objection from Loschmidt that it should not be possible to deduce an irreversible process from time-symmetric dynamics and a time-symmetric formalism: something must be wrong (Loschmidt’s paradox).

The resolution (1895) of this paradox is that the velocities of two particles after a collision are no longer truly uncorrelated. By asserting that it was acceptable to ignore these correlations in the population at times after the initial time, Boltzmann had introduced an element of time asymmetry through the formalism of his calculation.

— Wikipedia on Molecular chaos

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Physical entropy’s value is with respect to an observer.

— Me@2017-02-12 07:37:54 PM

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This “paradox” can be explained by carefully considering the definition of entropy. In particular, as concisely explained by Edwin Thompson Jaynes, definitions of entropy are arbitrary.

As a central example in Jaynes’ paper points out, one can develop a theory that treats two gases as similar even if those gases may in reality be distinguished through sufficiently detailed measurement. As long as we do not perform these detailed measurements, the theory will have no internal inconsistencies. (In other words, it does not matter that we call gases A and B by the same name if we have not yet discovered that they are distinct.) If our theory calls gases A and B the same, then entropy does not change when we mix them. If our theory calls gases A and B different, then entropy does increase when they are mixed. This insight suggests that the ideas of “thermodynamic state” and of “entropy” are somewhat subjective.

— Wikipedia on The mixing paradox

— Wikipedia on Gibbs paradox

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Turn my life into an ideal graduate school. Be my own professor.

— Me@2011.05.26

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Doing real research, doing real teaching.

— Me@2021-08-06 05:03:16 PM

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