# 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.

~~~ \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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# Problem 2.3b3

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}}

So \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}}

So \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}}

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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}}

— Me@2020-09-12 09:33:00 PM

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# Space and Causality

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

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# 機遇創生論 1.6.4

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— Me@2020-09-01 03:44:39 PM

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