Ex 1.9 Lagrange’s equations

Posted on September 30, 2020 by rpflee

Structure and Interpretation of Classical Mechanics

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.

~~~

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

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

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

Consistent histories, 8

Posted on September 27, 2020 by rpflee

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

2020.09.27 Sunday ACHK

Tenet

Posted on September 25, 2020 by rpflee

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

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

香港譯名：

1998 《Following》

2000 《凶心人》

取「空心人」同音。「心」，是指「記憶」。所以，電影譯名的意思是，沒有記憶的人。

2002 《白夜追兇》

2005 《俠影之謎》

2006 《死亡魔法》

此電影的主題為魔術，所以導演把電影本身，化成一個魔術。

2008 《黑夜之神》

2010 《潛行凶間》

此電影的主題為夢境，所以導演把電影本身，化成一個夢境。

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

相對論加量子力學

Posted on September 21, 2020 by rpflee

三一萬能俠, 2.2 | 太極滅世戰 2.3 | PhD, 4.2 | 財政自由 4.2

如果是最快樂的一個時代，我選預科的那兩年。

我暫時不能重造那個時代，主要因為還未有足夠的金錢儲備，令我可以，毋須做工維生，從而「全天候研究數學和物理」。

預科時代，需要為自己大學選擇主修。如果入到大學的話，我要選物理。但是，那時，我已經知道，中學的物理，其實不只對應於，大學的物理。其實，中學的物理有大半是，對應於大學的工程。

我想要全部，所以，如果成績許可的話，我會主修物理，副修工程。但是，一位回我中學講講座的師兄說：「工程無得（作為）副修架喎。」

工程屬於專業科目，只能作主修。所以，我立刻計劃倒轉，打算主修工程，副修物理；大學本科畢業後，到研究院時，才再研究物理。

預科後，有幸升到大學。我真的以工程為主修。

大學第二年時，可以開始有副修。但是，工程所需要讀的科目，多於其他主修一點。所以，我的時間表中，其實沒有足夠的空檔，去兼顧一個完整的副修。

我現在不記得副修物理，要修讀多少科物理科目。以下假設是八科。亦即是話，我要修夠八科，才可以於畢業成績表中，標籤物理為我的「副修」。

最終，我在第二年的上學期，和第三年的下學期，各自選修了一科。換句話說，我在大學本科時期，只修了兩科物理。

不幸中的大幸是，那兩科偏偏是，最重要的那兩科。

— Me@2020-09-16 04:01:32 PM

1990s, 4

Posted on September 14, 2020 by rpflee

— Me@2020-09-14 05:37:04 PM

Problem 2.3b3

Posted on September 13, 2020 by rpflee

…

Now we lower the indices, by expressing the upper-index coordinates (contravariant components) by lower-index coordinates (covariant components), in order to find the Lorentz transformation for the covariant components:

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

After raising the indices, we lower the indices again:

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

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

Prove that $\displaystyle{\eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} = \left[L^{-1}\right]^\sigma_{~\alpha}}$ .

By index renaming, $\displaystyle{ \begin{aligned} (x')_\mu &= \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma} x_\nu \\ \end{aligned}}$ , the question becomes

Prove that $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma} = \left[L^{-1}\right]^\nu_{~\mu}}$ .

Denote $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{L^{~\nu}_{\mu}}$ . Then the question is simplified to

Prove that $\displaystyle{ L^{~\nu}_{\mu} = \left[L^{-1}\right]^\nu_{~\mu}}$ .

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

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

$\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 (x')^\mu (x')_\mu &= \sum_{\mu = 0}^4 \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right) \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 (x')^\mu (x')_\mu &= \sum_{\mu = 0}^4 \left( \sum_{\nu = 0}^4 L^\mu_{~\nu} x^\nu \right) \left( \sum_{\beta = 0}^4 {L^{~\beta}_{\mu}} x_\beta \right) \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} &(x')^0 (x')_0 + (x')^1 (x')_1 + (x')^2 (x')_2 + (x')^3 (x')_3 \\ &= \sum_{\mu = 0}^4 \left( L^\mu_{~0} x^0 + L^\mu_{~1} x^1 + L^\mu_{~2} x^2 + L^\mu_{~3} x^3 \right) \left( L^{~0}_{\mu} x_0 + L^{~1}_{\mu} x_1 + L^{~2}_{\mu} x_2 + L^{~3}_{\mu} x_3 \right) \\ \end{aligned}}$

The right hand side has 64 terms.

Since the spacetime interval is Lorentz-invariant, $\displaystyle{ (x')^\mu (x')_\mu = x^\mu x_\mu }$ . So the left hand side can be replaced by $\displaystyle{ x^\mu x_\mu }$ .

$\displaystyle{ \begin{aligned} &x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 \\ &= \sum_{\mu = 0}^4 \left( L^\mu_{~0} x^0 + L^\mu_{~1} x^1 + L^\mu_{~2} x^2 + L^\mu_{~3} x^3 \right) \left( L^{~0}_{\mu} x_0 + L^{~1}_{\mu} x_1 + L^{~2}_{\mu} x_2 + L^{~3}_{\mu} x_3 \right) \\ \end{aligned}}$

Note that the 4 terms on the left side also appear on the right hand side.

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

$\displaystyle{ \begin{aligned} x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 &= \sum_{\mu = 0}^4 \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 L^\mu_{~\nu} L^{~\beta}_{\mu} x^\nu x_\beta \\ \end{aligned}}$

Since this equation is true for any coordinates, it is an identity. By comparing coefficients, we have:

1. For any terms with $\displaystyle{\nu \ne \beta}$ , such as $\displaystyle{\nu = 0}$ and $\displaystyle{\beta=1}$ ,

$\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 L^\mu_{~0} L^{~1}_{\mu} x^0 x_1 &\equiv 0 \\ \left( \sum_{\mu = 0}^4 L^\mu_{~0} L^{~1}_{\mu} \right) x^0 x_1 &\equiv 0\\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 L^\mu_{~0} L^{~1}_{\mu} &= 0 \\ \end{aligned}}$

2. For any terms with $\displaystyle{\nu = \beta}$ .

$\displaystyle{ \begin{aligned} x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 &\equiv \sum_{\mu = 0}^4 \sum_{\nu = 0}^4 \sum_{\beta = \nu} L^\mu_{~\nu} L^{~\beta}_{\mu} x^\nu x_\beta \\ x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 &\equiv \sum_{\mu = 0}^4 \left( L^\mu_{~0} L^{~0}_{\mu} x^0 x_0 + L^\mu_{~1} L^{~1}_{\mu} x^1 x_1 + L^\mu_{~2} L^{~2}_{\mu} x^2 x_2 + L^\mu_{~3} L^{~3}_{\mu} x^3 x_3 \right) \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 L^\mu_{~0} L^{~0}_{\mu} &= 1 \\ \sum_{\mu = 0}^4 L^\mu_{~1} L^{~1}_{\mu} &= 1 \\ \sum_{\mu = 0}^4 L^\mu_{~2} L^{~2}_{\mu} &= 1 \\ \sum_{\mu = 0}^4 L^\mu_{~3} L^{~3}_{\mu} &= 1 \\ \end{aligned}}$

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

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

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

— Me@2020-09-12 09:33:00 PM

Observer, 0

Posted on September 6, 2020 by rpflee

What/Who is an/the observer?

Who am I?

— Me@2013-01-01 08:15:20 PM

Space and Causality

Posted on September 6, 2020 by rpflee

When we say that A and B are at different places at the same time, that implies that A and B have no causal relation[, at least for that moment of time].

— Me@2011.08.25

— Me@2020-09-06

機遇創生論 1.6.4

Posted on September 5, 2020 by rpflee

一萬個小時 2.2 | 十年 3.2

這就是「專業知識」和「通用知識」，結構相似的地方——都是由最初的「學海無涯，唯勤是岸」，發展到「學海無涯，回頭是岸」。

而結構不相似的地方就是，對於「專業知識」而言，由「唯勤是岸」到「回頭是岸」的過程，需要長很多的時間（十年一萬小時）。二來，對於「專業知識」而言，「回頭是岸」（純熟到可以用來維生）只是里程碑而已，而不是終點。

另一個「專業知識」和「通用知識」，結構不相似的地方是，精簡來說，「專業知識」並不「通用」，不可「轉化」。例如，一位醫生無論多聰明，都不可能在，未經詳細訓練和研習的情況下，竟然做到律師。

詳細而言，任何一門專業的知識，都有一部分，可以於其他行業中，循環再用，簡稱「可轉化部分」，或者「通用部分」；同時，亦有另一部分知識，不可以於其他行業中，循環再用，簡稱「不可轉化部分」，或者「專業部分」。例如，剛才的那位醫生，如果已下定了決心要，轉行做律行的話，他原本的部分才能是，可以循環再用的，例如良好的英文和細密的心思。

不記得從哪裡看到的文章，講述有研究員探討，「智力遊戲」可否提高智力。亦即是問，「益知遊戲」會否益智？

該文的結論是，「智力遊戲」可以提升，有關該個智力遊戲的智力；至於其他方面的智力，則沒有大幫忙。我猜想，那個「智力遊戲」甚至連，其他智力遊戲中，所需的智力，也未必能提升。

該文的結論，我不知真假。不過，我覺得那結論可信。

試想想，如果你不斷練習足球，你足球的技巧當然會提升。但是，你籃球的技巧則不會。

為什麼會這樣呢？

— Me@2020-09-01 03:44:39 PM

Physics Town

continues

Monthly Archives: September 2020