# 1980s

— DeepAI colorization

— Me@2020-11-25 04:39:20 PM

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# Problem 2.3b5

A First Course in String Theory

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2.3 Lorentz transformations, derivatives, and quantum operators.

(b) Show that the objects $\displaystyle{\frac{\partial}{\partial x^\mu}}$ transform under Lorentz transformations in the same way as the $\displaystyle{a_\mu}$ considered in (a) do. Thus, partial derivatives with respect to conventional upper-index coordinates $\displaystyle{x^\mu}$ behave as a four-vector with lower indices – as reflected by writing it as $\displaystyle{\partial_\mu}$.

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Denoting $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{L^{~\nu}_{\mu}}$ is misleading, because that presupposes that $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ is directly related to the matrix $\displaystyle{L}$.

To avoid this bug, instead, we denote $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{M ^\nu_{~\mu}}$. So

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ x^\mu x_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ \end{aligned}}

\displaystyle{ \begin{aligned} \nu \neq \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 0 \\ \nu = \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 1 \\ \end{aligned}}

Using the Kronecker Delta and Einstein summation notation, we have

\displaystyle{ \begin{aligned} L^\mu_{~\nu} M^{\beta}_{~\mu} &= M^{\beta}_{~\mu} L^\mu_{~\nu} \\ &= \delta^{\beta}_{~\nu} \\ \end{aligned}}

So

\displaystyle{ \begin{aligned} \sum_{\mu=0}^{4} L^\mu_{~\nu} M^{\beta}_{~\mu} &= \delta^{\beta}_{~\nu} \\ \end{aligned}}

\displaystyle{ \begin{aligned} M^{\beta}_{~\mu} &= [L^{-1}]^{\beta}_{~\mu} \\ \end{aligned}}

In other words,

\displaystyle{ \begin{aligned} \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma} &= [L^{-1}]^{\beta}_{~\mu} \\ \end{aligned}}

— Me@2020-11-23 04:27:13 PM

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One defines (as a matter of notation),

${\displaystyle {\Lambda _{\nu }}^{\mu }\equiv {\left(\Lambda ^{-1}\right)^{\mu }}_{\nu },}$

and may in this notation write

${\displaystyle {A'}_{\nu }={\Lambda _{\nu }}^{\mu }A_{\mu }.}$

Now for a subtlety. The implied summation on the right hand side of

${\displaystyle {A'}_{\nu }={\Lambda _{\nu }}^{\mu }A_{\mu }={\left(\Lambda ^{-1}\right)^{\mu }}_{\nu }A_{\mu }}$

is running over a row index of the matrix representing $\displaystyle{\Lambda^{-1}}$. Thus, in terms of matrices, this transformation should be thought of as the inverse transpose of $\displaystyle{\Lambda}$ acting on the column vector $\displaystyle{A_\mu}$. That is, in pure matrix notation,

${\displaystyle A'=\left(\Lambda ^{-1}\right)^{\mathrm {T} }A.}$

— Wikipedia on Lorentz transformation

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So

\displaystyle{ \begin{aligned} M^{\beta}_{~\mu} &= [L^{-1}]^{\beta}_{~\mu} \\ \end{aligned}}

In other words,

\displaystyle{ \begin{aligned} \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} &= [L^{-1}]^{\beta}_{~\mu} \\ \end{aligned}}

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Denote $\displaystyle{[L^{-1}]^{\beta}_{~\mu}}$ as

\displaystyle{ \begin{aligned} N^{~\beta}_{\mu} \\ \end{aligned}}

In other words,

\displaystyle{ \begin{aligned} N^{~\beta}_{\mu} &= M^{\beta}_{~\mu} \\ [N^T] &= [M] \\ \end{aligned}}

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The Lorentz transformation:

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')_\mu &= \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \\ \end{aligned}}

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\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')_\mu &= N^{~\nu}_{\mu} x_\nu \\ \end{aligned}}

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\displaystyle{ \begin{aligned} x^\mu &= [L^{-1}]^\mu_{~\nu} (x')^\nu \\ (x')_\mu &= M^{\nu}_{~\mu} x_\nu \\ \end{aligned}}

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\displaystyle{ \begin{aligned} x^\mu &= [L^{-1}]^\mu_{~\nu} (x')^\nu \\ (x')_\mu &= [L^{-1}]^{\nu}_{~\mu} x_\nu \\ \end{aligned}}

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\displaystyle{ \begin{aligned} \frac{\partial}{\partial (x')^\mu} &= \frac{\partial x^\nu}{\partial (x')^\mu} \frac{\partial}{\partial x^\nu} \\ &= \frac{\partial x^0}{\partial (x')^\mu} \frac{\partial}{\partial x^0} + \frac{\partial x^1}{\partial (x')^\mu} \frac{\partial}{\partial x^1} + \frac{\partial x^2}{\partial (x')^\mu} \frac{\partial}{\partial x^2} + \frac{\partial x^3}{\partial (x')^\mu} \frac{\partial}{\partial x^3} \\ \end{aligned}}

Now we consider $\displaystyle{f}$ as a function of $\displaystyle{x^{\mu}}$‘s:

$\displaystyle{f(x^0, x^1, x^2, x^3)}$

Since $\displaystyle{x^{\mu}}$‘s and $\displaystyle{(x')^{\mu}}$‘s are related by Lorentz transform, $\displaystyle{f}$ is also a function of $\displaystyle{(x')^{\mu}}$‘s, although indirectly.

$\displaystyle{f(x^0((x')^0, (x')^1, (x')^2, (x')^3), x^1((x')^0, ...), x^2((x')^0, ...), x^3((x')^0, ...))}$

For notational simplicity, we write $\displaystyle{f}$ as

$\displaystyle{f(x^\alpha((x')^\beta))}$

Since $\displaystyle{f}$ is a function of $\displaystyle{(x')^{\mu}}$‘s, we can differentiate it with respect to $\displaystyle{(x')^{\mu}}$‘s.

\displaystyle{ \begin{aligned} \frac{\partial}{\partial (x')^\mu} f(x^\alpha((x')^\beta))) &= \sum_{\nu = 0}^4 \frac{\partial x^\nu}{\partial (x')^\mu} \frac{\partial}{\partial x^\nu} f(x^\alpha) \\ \end{aligned}}

Since

\displaystyle{ \begin{aligned} x^\nu &= [L^{-1}]^\nu_{~\beta} (x')^\beta \\ \end{aligned}},

\displaystyle{ \begin{aligned} \frac{\partial f}{\partial (x')^\mu} &= \sum_{\nu = 0}^4 \frac{\partial}{\partial (x')^\mu} \left[ \sum_{\beta = 0}^4 [L^{-1}]^\nu_{~\beta} (x')^\beta \right] \frac{\partial f}{\partial x^\nu} \\ &= \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 [L^{-1}]^\nu_{~\beta} \frac{\partial (x')^\beta}{\partial (x')^\mu} \frac{\partial f}{\partial x^\nu} \\ &= \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 [L^{-1}]^\nu_{~\beta} \delta^\beta_\mu \frac{\partial f}{\partial x^\nu} \\ &= \sum_{\nu = 0}^4 [L^{-1}]^\nu_{~\mu} \frac{\partial f}{\partial x^\nu} \\ &= [L^{-1}]^\nu_{~\mu} \frac{\partial f}{\partial x^\nu} \\ \end{aligned}}

Therefore,

\displaystyle{ \begin{aligned} \frac{\partial}{\partial (x')^\mu} &= [L^{-1}]^\nu_{~\mu} \frac{\partial}{\partial x^\nu} \\ \end{aligned}}

It is the same as the Lorentz transform for covariant vectors:

\displaystyle{ \begin{aligned} (x')_\mu &= [L^{-1}]^{\nu}_{~\mu} x_\nu \\ \end{aligned}}

— Me@2020-11-23 04:27:13 PM

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# Global symmetry, 2

In physics, a global symmetry is a symmetry that holds at all points in the spacetime under consideration, as opposed to a local symmetry which varies from point to point.

Global symmetries require conservation laws, but not forces, in physics.

— Wikipedia on Global symmetry

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2020.11.22 Sunday ACHK

# Light, 3

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The one in the mirror is your Light.

— Me@2011.06.24

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Thou shalt have no other gods before Me.

— one of the Ten Commandments

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— Me@the Last Century

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# 一萬個小時 2.4

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— Me@2020-11-16 04:55:18 PM

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# 2006, 2

— Me@2020-11-14 09:27:04 PM

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# Possion’s Lagrange Equation

Structure and Interpretation of Classical Mechanics

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Ex 1.10 Higher-derivative Lagrangians

Derive Lagrange’s equations for Lagrangians that depend on accelerations. In particular, show that the Lagrange equations for Lagrangians of the form $\displaystyle{L(t, q, \dot q, \ddot q)}$ with $\displaystyle{\ddot{q}}$ terms are

$\displaystyle{D^2(\partial_3L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + \partial_1 L \circ \Gamma[q] = 0}$

In general, these equations, first derived by Poisson, will involve the fourth derivative of $\displaystyle{q}$. Note that the derivation is completely analogous to the derivation of the Lagrange equations without accelerations; it is just longer. What restrictions must we place on the variations so that the critical path satisfies a differential equation?

Varying the action

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

\displaystyle{ \begin{aligned} \delta_\eta S[q] (t_1, t_2) &= 0 \\ \end{aligned}}

\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}}

\displaystyle{ \begin{aligned} \delta_\eta I [q] &= \eta \\ \delta_\eta g[q] &= D \eta~~~\text{with}~~~g[q] = Dq \\ \end{aligned}}

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Let $\displaystyle{h[q] = D^2 q}$.

\displaystyle{ \begin{aligned} \delta_\eta h[q] &= \lim_{\epsilon \to 0} \frac{h[q+\epsilon \eta] - h[q]}{\epsilon} \\ &= \lim_{\epsilon \to 0} \frac{D^2 (q+\epsilon \eta) - D^2 q}{\epsilon} \\ &= \lim_{\epsilon \to 0} \frac{D^2 q + D^2 \epsilon \eta - D^2 q}{\epsilon} \\ &= \lim_{\epsilon \to 0} \frac{D^2 \epsilon \eta}{\epsilon} \\ &= D^2 \eta \\ \end{aligned}}

\displaystyle{ \begin{aligned} \Gamma [q] (t) &= (t, q(t), D q(t), D^2 q(t)) \\ \delta_\eta \Gamma [q] (t) &= (0, \eta (t), D \eta (t), D^2 \eta (t)) \\ \end{aligned}}

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Chain rule of functional variation

\displaystyle{ \begin{aligned} &\delta_\eta F[g[q]] \\ &= \delta_\eta (F \circ g)[q] \\ &= \delta_{ \left( \delta_\eta g[q] \right)} F[g] \\ \end{aligned}}

Since variation commutes with integration,

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

By the chain rule of functional variation:

\displaystyle{ \begin{aligned} \delta_\eta L \circ \Gamma [q] = \delta_{ \left( \delta_\eta \Gamma[q] \right)} L[\Gamma[q]] \\ \end{aligned}}

If $\displaystyle{L}$ is path-independent,

\displaystyle{ \begin{aligned} \delta_\eta \left( L \circ \Gamma [q] \right) = \left( DL \circ \Gamma[q] \right) \delta_\eta \Gamma[q] \\ \end{aligned}}

But is $\displaystyle{L}$ path-independent?

The $\displaystyle{L \circ \Gamma [.]}$ is path-dependent. Its input is a path $\displaystyle{q}$, not just $\displaystyle{q(t)}$, the value of $\displaystyle{q}$ at the time $\displaystyle{t}$. However, $\displaystyle{L(.)}$ itself is a path-independent function, because its input is not a path $\displaystyle{q}$, but a quadruple of values $\displaystyle{(t, q(t), Dq(t), D^2 q(t))}$.

\displaystyle{ \begin{aligned} L \circ \Gamma [q] = L(t, q(t), Dq(t), D^2 q(t)) \\ \end{aligned}}

Since $\displaystyle{L}$ is path-independent,

\displaystyle{ \begin{aligned} \delta_\eta \left( L \circ \Gamma [q] \right) = \left( DL \circ \Gamma[q] \right) \delta_\eta \Gamma[q] \\ \end{aligned}}

\displaystyle{ \begin{aligned} &\delta_\eta S[q] (t_1, t_2) \\ &= \int_{t_1}^{t_2} \delta_\eta L \circ \Gamma [q] \\ &= \int_{t_1}^{t_2} \left( D \left( L \circ \Gamma[q] \right) \right) \delta_\eta \Gamma[q] \\ &= \int_{t_1}^{t_2} \left( D \left( L(t, q, D q, D^2 q) \right) \right) (0, \eta (t), D \eta (t), D^2 \eta (t)) \\ &= \int_{t_1}^{t_2} \left[ \partial_0 L \circ \Gamma[q], \partial_1 L \circ \Gamma[q], \partial_2 L \circ \Gamma[q], \partial_3 L \circ \Gamma[q] \right] (0, \eta (t), D \eta (t), D^2 \eta (t)) \\ &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta + (\partial_2 L \circ \Gamma[q]) D \eta + (\partial_3 L \circ \Gamma[q]) D^2 \eta \\ &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta + \left[ \left. (\partial_2 L \circ \Gamma[q]) \eta \right|_{t_1}^{t_2} - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q]) \eta \right] + \int_{t_1}^{t_2} (\partial_3 L \circ \Gamma[q]) D^2 \eta \\ \end{aligned}}

Since $\displaystyle{\eta(t_1) = 0}$ and $\displaystyle{\eta(t_2) = 0}$,

\displaystyle{ \begin{aligned} \delta_\eta S[q] (t_1, t_2) &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q]) \eta + \int_{t_1}^{t_2} (\partial_3 L \circ \Gamma[q]) D^2 \eta \\ \end{aligned}}

Here is a trick for integration by parts:

As long as the boundary term $\displaystyle{\left. u(t)v(t) \right|_{t_1}^{t_2} = 0}$,

$\displaystyle{\int_{t_1}^{t_2} u(t) dv(t) = - \int_{t_1}^{t_2} v(t) du(t)}$

So if $\displaystyle{D \eta(t_1) = 0}$ and $\displaystyle{D \eta(t_2) = 0}$,

\displaystyle{ \begin{aligned} \delta_\eta S[q] (t_1, t_2) &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q]) \eta - \int_{t_1}^{t_2} D(\partial_3 L \circ \Gamma[q]) D \eta \\ \end{aligned}}

Since $\displaystyle{\eta(t_1) = 0}$ and $\displaystyle{\eta(t_2) = 0}$,

\displaystyle{ \begin{aligned} \delta_\eta S[q] (t_1, t_2) &= \int_{t_1}^{t_2} (\partial_1 L \circ \Gamma[q]) \eta - \int_{t_1}^{t_2} D(\partial_2 L \circ \Gamma[q]) \eta + \int_{t_1}^{t_2} D^2 (\partial_3 L \circ \Gamma[q]) \eta \\ \end{aligned}}

\displaystyle{ \begin{aligned} \delta_\eta S[q] (t_1, t_2) &= \int_{t_1}^{t_2} \left[ (\partial_1 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + D^2 (\partial_3 L \circ \Gamma[q]) \right] \eta \\ \end{aligned}}

By the principle of stationary action, $\displaystyle{ \delta_\eta S[q] (t_1, t_2) = 0}$. So

\displaystyle{ \begin{aligned} 0 &= \int_{t_1}^{t_2} \left[ (\partial_1 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + D^2 (\partial_3 L \circ \Gamma[q]) \right] \eta \\ \end{aligned}}

Since this is true for any function $\displaystyle{\eta(t)}$ that satisfies $\displaystyle{\eta(t_1) = \eta(t_2) = 0}$ and $\displaystyle{D\eta(t_1) = D\eta(t_2) = 0}$,

\displaystyle{ \begin{aligned} (\partial_1 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + D^2 (\partial_3 L \circ \Gamma[q]) &= 0 \\ D^2 (\partial_3 L \circ \Gamma[q]) - D(\partial_2 L \circ \Gamma[q]) + \partial_1 L \circ \Gamma[q] &= 0 \\ \end{aligned}}

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Note:

The notation of the path function $\displaystyle{\Gamma}$ is $\displaystyle{\Gamma[q](t)}$, not $\displaystyle{\Gamma[q(t)]}$.

The notation $\displaystyle{\Gamma[q](t)}$ means that $\displaystyle{\Gamma}$ takes a path $\displaystyle{q}$ as input. And then returns a path-independent function $\displaystyle{\Gamma[q]}$, which takes time $\displaystyle{t}$ as input, returns a value $\displaystyle{\Gamma[q](t)}$.

The other notation $\displaystyle{\Gamma[q(t)]}$ makes no sense, because $\displaystyle{\Gamma[.]}$ takes a path $\displaystyle{q}$, not a value $\displaystyle{q(t)}$, as input.

— Me@2020-11-11 05:37:13 PM

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# Memory as past microstate information encoded in present devices

Logical arrow of time, 4.2

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Memory is of the past.

The main point of memories or records is that without them, most of the past microstate information would be lost for a macroscopic observer forever.

For example, if a mixture has already reached an equilibrium state, we cannot deduce which previous microstate it is from, unless we have the memory of it.

This work is free and may be used by anyone for any purpose. Wikimedia Foundation has received an e-mail confirming that the copyright holder has approved publication under the terms mentioned on this page.

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memory/record

~ some of the past microstate and macrostate information encoded in present macroscopic devices, such as paper, electronic devices, etc.

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How come macroscopic time is cumulative?

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Quantum states are unitary.

A quantum state in the present is evolved from one and only one quantum state at any particular time point in the past.

Also, that quantum state in the present will evolve to one and only one quantum state at any particular time point in the future.

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Let

$\displaystyle{t_1}$ = a past time point

$\displaystyle{t_2}$ = now

$\displaystyle{t_3}$ = a future time point

Also, let state $\displaystyle{S_1}$ at time $\displaystyle{t_1}$ evolve to state $\displaystyle{S_2}$ at time $\displaystyle{t_2}$. And then state $\displaystyle{S_2}$ evolves to state $\displaystyle{S_3}$ at time $\displaystyle{t_3}$.

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State $\displaystyle{S_2}$ has one-one correspondence to its past state $\displaystyle{S_1}$. So for the state $\displaystyle{S_2}$, it does not need memory to store any information of state $\displaystyle{S_1}$.

Instead, just by knowing that $\displaystyle{t_2}$ microstate is $\displaystyle{S_2}$, we already can deduce that it is evolved from state $\displaystyle{S_1}$ at time $\displaystyle{t_1}$.

In other words, microstate does not require memory.

— Me@2020-10-28 10:26 AM

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# Action | Uncertainty, 3

taking action ~ teleporting yourself into the future

— Me@2020-10-28 6:46 PM

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# 尋覓

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（CPK: 可不可以講呀？可以？

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（CSY: 是。）

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— Me@2020-10-28 10:26:23 PM

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# 1990s, 5

— DeepAI colorization

— Me@2020-10-24 04:59:43 PM

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# Kronecker delta in tensor component form

Problem 2.3b4

A First Course in String Theory

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Continue the previous calculation:

\displaystyle{ \begin{aligned} \nu \neq \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 0 \\ \nu = \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 1 \\ \end{aligned}}

The two cases can be grouped into one, by replacing the right hand sides with the Kronecker delta. However, there are 4 possible forms and I am not sure which one should be used.

$\displaystyle{\delta^i_{~j}}$
$\displaystyle{\delta_i^{~j}}$
$\displaystyle{\delta^{ij}}$
$\displaystyle{\delta_{ij}}$

So I do a little research on Kronecker delta in this post.

— Me@2020-10-21 03:40:36 PM

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The inverse Lorentz transformation should satisfy $\displaystyle{\left( \Lambda^{-1} \right)^\beta_{~\mu} \Lambda^\mu_{~\nu} = \delta^\beta_{~\nu}}$, where $\displaystyle{\delta^\beta_{~\nu} \equiv \text{diag}(1,1,1,1)}$ is the Kronecker delta. Then, multiply by the inverse on both sides of Eq. 4 to find

\displaystyle{ \begin{aligned} \left( \Lambda^{-1} \right)^\beta_{~\mu} \left( \Delta x' \right)^\mu &= \delta^\beta_{~\nu} \Delta x^\nu \\ &= \Delta x^\beta \\ \end{aligned}}

(6)

The inverse $\displaystyle{\left( \Lambda^{-1} \right)^\beta_{~\mu}}$ is also written as $\displaystyle{\Lambda_\mu^{~\beta}}$. The notation is as follows: the left index denotes a row while the right index denotes a column, while the top index denotes the frame we’re transforming to and the bottom index denotes the frame we’re transforming from. Then, the operation $\displaystyle{\Lambda_\mu^{~\beta} \Lambda^\mu_{~\nu}}$ means sum over the index $\displaystyle{\mu}$ which lives in the primed frame, leaving unprimed indices $\displaystyle{\beta}$ and $\displaystyle{\nu}$ (so that the RHS of Eq. 6 is unprimed as it should be), where the sum is over a row of $\displaystyle{\Lambda_\mu^{~\beta}}$ and a column of $\displaystyle{\Lambda_{~\nu}^\mu}$ which is precisely the operation of matrix multiplication.

— Lorentz tensor redux

— Emily Nardoni

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This one is WRONG:

$\displaystyle{(\Lambda^T)^{\mu}{}_{\nu} = \Lambda_{\nu}{}^{\mu}}$

This one is RIGHT:

$\displaystyle{(\Lambda^T)_{\nu}{}^{\mu} ~:=~ \Lambda^{\mu}{}_{\nu}}$

— Me@2020-10-23 06:30:57 PM

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1. $\displaystyle{(\Lambda^T)_{\nu}{}^{\mu} ~:=~\Lambda^{\mu}{}_{\nu}}$

2. [Kronecker delta] is invariant in all coordinate systems, and hence it is an isotropic tensor.

3. Covariant, contravariant and mixed type of this tensor are the same, that is

$\displaystyle{\delta^i_{~j} = \delta_i^{~j} = \delta^{ij} = \delta_{ij}}$

— Introduction to Tensor Calculus

— Taha Sochi

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Raising and then lowering the same index (or conversely) are inverse operations, which is reflected in the covariant and contravariant metric tensors being inverse to each other:

${\displaystyle g^{ij}g_{jk}=g_{kj}g^{ji}={\delta ^{i}}_{k}={\delta _{k}}^{i}}$

where $\displaystyle{\delta^i_{~k}}$ is the Kronecker delta or identity matrix. Since there are different choices of metric with different metric signatures (signs along the diagonal elements, i.e. tensor components with equal indices), the name and signature is usually indicated to prevent confusion.

— Wikipedia on Raising and lowering indices

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So

${\displaystyle g^{ij}g_{jk}={\delta ^{i}}_{k}}$

and

${\displaystyle g_{kj}g^{ji}={\delta _{k}}^{i}}$

— Me@2020-10-19 05:21:49 PM

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$\displaystyle{ T_{i}^{\; j} = \boldsymbol{T}(\boldsymbol{e}_i,\boldsymbol{e}^j) }$ and $\displaystyle{T_{j}^{\; i} = \boldsymbol{T}(\boldsymbol{e}_j,\boldsymbol{e}^i) }$ are both 1-covariant 2-contravariant coordinates of T. The only difference between them is the notation used for sub- and superscripts;

$\displaystyle{ T^{i}_{\; j} = \boldsymbol{T}(\boldsymbol{e}^i,\boldsymbol{e}_j) }$ and $\displaystyle{ T^{j}_{\; i} = \boldsymbol{T}(\boldsymbol{e}^j,\boldsymbol{e}_i) }$ are both 1-contravariant 2-covariant coordinates of T. The only difference between them is the notation used for sub- and superscripts.

— edited Oct 11 ’17 at 14:14

— answered Oct 11 ’17 at 10:58

— EditPiAf

— Tensor Notation Upper and Lower Indices

— Physics StackExchange

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Rather, the dual basis one-forms are defined by imposing the following
16 requirements at each spacetime point:

$\displaystyle{\langle \tilde{e}^\mu \mathbf{x}, \vec e_\nu \mathbf{x} \rangle = \delta^{\mu}_{~\nu}}$

is the Kronecker delta, $\displaystyle{\delta^{\mu}_{~\nu} = 1}$ if $\displaystyle{\mu = \nu}$ and $\displaystyle{\delta^{\mu}_{~\nu} = 0}$ otherwise, with the same values for each spacetime point. (We must always distinguish subscripts from superscripts; the Kronecker delta always has one of each.)

— Introduction to Tensor Calculus for General Relativity

— Edmund Bertschinger

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However, since $\displaystyle{\delta_{~b}^a}$ is a tensor, we can raise or lower its indices using the metric tensor in the usual way. That is, we can get a version of $\displaystyle{\delta}$ with both indices raised or lowered, as follows:

[$\displaystyle{\delta^{ab} = \delta^a_{~c} g^{cb} = g^{ab}}$]

$\displaystyle{\delta_{ab} = g_{ac} \delta^c_{~b} = g_{ab}}$

In this sense, $\displaystyle{\delta^{ab}}$ and $\displaystyle{\delta_{ab}}$ are the upper and lower versions of the metric tensor. However, they can’t really be considered versions of the Kronecker delta any more, as they don’t necessarily satisfy [0 when $i \ne j$ and 1 when $i = j$]. In other words, the only version of $\delta$ that is both a Kronecker delta and a tensor is the version with one upper and one lower index: $\delta^a_{~b}$ [or $\delta^{~a}_{b}$].

— Kronecker Delta as a tensor

— physicspages

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Continue the calculation for the Problem 2.3b:

Denoting $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{L^{~\nu}_{\mu}}$ is misleading, because that presupposes that $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ is directly related to the matrix $\displaystyle{L}$.

To avoid this bug, instead, we denote $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{M ^\nu_{~\mu}}$. So

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ x^\mu x_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ \end{aligned}}

\displaystyle{ \begin{aligned} \nu \neq \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 0 \\ \nu = \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 1 \\ \end{aligned}}

Using the Kronecker Delta and Einstein summation notation, we have

\displaystyle{ \begin{aligned} L^\mu_{~\nu} M^{\beta}_{~\mu} &= M^{\beta}_{~\mu} L^\mu_{~\nu} \\ &= \delta^{\beta}_{~\nu} \\ \end{aligned}}

Note: After tensor contraction, the remaining left index should be kept on the left and the remaining right on the right.

— Me@2020-10-20 03:49:09 PM

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# Tenet, 2

T-symmetry 6.2 | Loschmidt’s paradox 4

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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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If an observer insists to monitor all the microstate information of the observed and the environment, i.e. without leaving any microstate information, that observer would see a time symmetric universe, in the sense that the second law of thermodynamics would not be there anymore.

It would then be meaningless to label any of the two directions of time as “past” or “future”.

— Me@2020-10-12 08:10:27 PM

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So in this sense, as long as an observer wants to save some mental power by ignoring some micro-information, the past and future distinction is created, in the sense that there will be the second law of thermodynamics.

— Me@2020-10-12 08:12:25 PM

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Time’s arrow is due to approximation. Time’s arrow is due to the coarse-grained description of reality. In other words, you use an inaccurate macroscopic description on an actually microscopic reality.

— Me@2020-10-12 10:41:48 PM

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# 信心動搖

Kyle：我都遇過類似的事情，大大動搖了我對自己的信心。

Me：你已經比我好。至起碼，你還有信心可以動搖。

— Me@2003

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# 一萬個小時 2.3

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「專業」，是由「通用」發展出來的。

「專業」，就是「通用」的分支。

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— Me@2020-10-10 07:52:39 PM

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# 星宿 3

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— Me@2020-10-02 04:41:36 PM

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# Ex 1.9 Lagrange’s equations

Structure and Interpretation of Classical Mechanics

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Derive the Lagrange equations for the following systems, showing all of the intermediate steps as in the harmonic oscillator and orbital motion examples.

b. An ideal planar pendulum consists of a bob of mass $\displaystyle{m}$ connected to a pivot by a massless rod of length $\displaystyle{l}$ subject to uniform gravitational acceleration $\displaystyle{g}$. A Lagrangian is $\displaystyle{L(t, \theta, \dot \theta) = \frac{1}{2} m l^2 \dot \theta^2 + mgl \cos \theta}$. The formal parameters of $\displaystyle{L}$ are $\displaystyle{t}$, $\displaystyle{\theta}$, and $\displaystyle{\dot \theta}$; $\displaystyle{\theta}$ measures the angle of the pendulum rod to a plumb line and $\displaystyle{\dot \theta}$ is the angular velocity of the rod.

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\displaystyle{ \begin{aligned} L (t, \xi, \eta) &= \frac{1}{2} m l^2 \eta^2 + m g l \cos \xi \\ \end{aligned}}

\displaystyle{ \begin{aligned} \partial_1 L (t, \xi, \eta) &= - m g l \sin \xi \\ \partial_2 L (t, \xi, \eta) &= m l^2 \eta \\ \end{aligned}}

Put $\displaystyle{q = \theta}$,

\displaystyle{ \begin{aligned} \Gamma[q](t) &= (t; \theta(t); D\theta(t)) \\ \end{aligned}}

\displaystyle{ \begin{aligned} \partial_1 L \circ \Gamma[q] (t) &= - m g l \sin \theta \\ \partial_2 L \circ \Gamma[q] (t) &= m l^2 D \theta \\ \end{aligned}}

The Lagrange equation:

\displaystyle{ \begin{aligned} D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\ D ( m l^2 D \theta ) - ( - m g l \sin \theta ) &= 0 \\ D^2 \theta + \frac{g}{l} \sin \theta &= 0 \\ \end{aligned}}

— Me@2020-09-28 05:40:42 PM

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# Consistent histories, 8

Relationship with other interpretations

The only group of interpretations of quantum mechanics with which RQM is almost completely incompatible is that of hidden variables theories. RQM shares some deep similarities with other views, but differs from them all to the extent to which the other interpretations do not accord with the “relational world” put forward by RQM.

Copenhagen interpretation

RQM is, in essence, quite similar to the Copenhagen interpretation, but with an important difference. In the Copenhagen interpretation, the macroscopic world is assumed to be intrinsically classical in nature, and wave function collapse occurs when a quantum system interacts with macroscopic apparatus. In RQM, any interaction, be it micro or macroscopic, causes the linearity of Schrödinger evolution to break down. RQM could recover a Copenhagen-like view of the world by assigning a privileged status (not dissimilar to a preferred frame in relativity) to the classical world. However, by doing this one would lose sight of the key features that RQM brings to our view of the quantum world.

Hidden variables theories

Bohm’s interpretation of QM does not sit well with RQM. One of the explicit hypotheses in the construction of RQM is that quantum mechanics is a complete theory, that is it provides a full account of the world. Moreover, the Bohmian view seems to imply an underlying, “absolute” set of states of all systems, which is also ruled out as a consequence of RQM.

We find a similar incompatibility between RQM and suggestions such as that of Penrose, which postulate that some processes (in Penrose’s case, gravitational effects) violate the linear evolution of the Schrödinger equation for the system.

Relative-state formulation

The many-worlds family of interpretations (MWI) shares an important feature with RQM, that is, the relational nature of all value assignments (that is, properties). Everett, however, maintains that the universal wavefunction gives a complete description of the entire universe, while Rovelli argues that this is problematic, both because this description is not tied to a specific observer (and hence is “meaningless” in RQM), and because RQM maintains that there is no single, absolute description of the universe as a whole, but rather a net of inter-related partial descriptions.

Consistent histories approach

In the consistent histories approach to QM, instead of assigning probabilities to single values for a given system, the emphasis is given to sequences of values, in such a way as to exclude (as physically impossible) all value assignments which result in inconsistent probabilities being attributed to observed states of the system. This is done by means of ascribing values to “frameworks”, and all values are hence framework-dependent.

RQM accords perfectly well with this view. However, the consistent histories approach does not give a full description of the physical meaning of framework-dependent value (that is it does not account for how there can be “facts” if the value of any property depends on the framework chosen). By incorporating the relational view into this approach, the problem is solved: RQM provides the means by which the observer-independent, framework-dependent probabilities of various histories are reconciled with observer-dependent descriptions of the world.

— Wikipedia on Relational quantum mechanics

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2020.09.27 Sunday ACHK

# Tenet

Christopher Nolan, 2 | 時空幻境 4 | Braid 4

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1998 Following
2000 Memento
2002 Insomnia
2005 Batman Begins
2006 The Prestige
2008 The Dark Knight

2010 Inception
2012 The Dark Knight Rises
2014 Interstellar
2017 Dunkirk
2020 Tenet

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1998 《Following》

2000 《凶心人》

2002 《白夜追兇》

2005 《俠影之謎》

2006 《死亡魔法》

2008 《黑夜之神》

2010 《潛行凶間》

2012 《夜神起義》

2014 《星際啓示錄》

「啓示」，即是「來自未來的訊息」。

2017 《鄧寇克大行動》

2020 《天能》

A lot of Nolan’s movies are about some kinds of time travel.

For those movies, each has a unique time logic. Each is like a stage of the computer game Braid.

In Braid, there are 6 stages. Each stage has a unique time mechanics.

— Me@2020-09-20 10:36:54 AM

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# 相對論加量子力學

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— Me@2020-09-16 04:01:32 PM

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