# 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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