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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2020.10.21 Wednesday (c) All rights reserved by ACHK

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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2020.10.13 Tuesday (c) All rights reserved by ACHK

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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2020.10.11 Sunday (c) All rights reserved by ACHK

一萬個小時 2.3

機遇創生論 1.6.5 | 十年 3.3

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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。

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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,「益智遊戲」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?

「專業」,是由「通用」發展出來的。

「專業」,就是「通用」的分支。

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通用的極致,就是專業。而通用發展成極致的方法,有很多種。每種極致,就自成一門專業。通用是樹幹;專業是樹支。一支樹支的強壯與否,並不能保證,另一支樹支的健康。

那是否就代表,學生時代以後,就毋須再發展,通用知識或通用技能呢?

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不是。

學生時代以後,仍然需要發展通用知識,以防原有的通用技能退化,因為不進則退。但是,發展通用知識的策略上,一定和學生時代時,有所不同。大人需要維生,沒有學生時代,那麼多的學習時間。

學生時代追求通用知識時,應有的態度,是「窮盡」——可以學的都學,從而發掘自己當時的興趣,尋找自己將來的專業。

工作時代追求通用知識時,應有的角度,是「選擇」——每一個時期,選擇一樣新的學問,去研究和實踐,從而維持自己的智能和體能。只選一兩樣,原因是工作時代的工餘時間有限。

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情形就好像一個人在單身時,應該盡量識多一些,志同道合的朋友,從而提高找到另一半的機會率。但是,有了另一半後,是否就毋須再見,任何其他朋友呢?

當然不是,因為,尋找另一半,並不是交朋友的唯一目的。有了另一半後,見朋友再不是為了尋找另一半,而是純粹為了,見那些朋友。

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但是,因為有了另一半後,你自然會把,工餘時間中的大部分,花在他/她身上。所以,你可以見到朋友的機會,自然少了很多。

但是,一定要有。

一個人只有愛情,沒有友情,大慨不會快樂,反之亦然。

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學生時代,每天上學,都會學到新事物。工作時代,每天上班,大概很少會學到新想法。

學生時代,每天上學,都會見到朋友。工作時代,每天上班,大概不會見到朋友。

那是否代表,永久生存於學生時代,是好事呢?

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嬰兒的最可愛之處,在於他會長大,有著無限個未來;人們見到他時,會充滿希望。如果一個嬰兒不會長大,你再不會覺得可愛,而是覺得可惜和可悲。

學生時代的最精采之處,在於可以期望,在不久的將來,可以脫離學生時代。工作時代的恐怖之處,在於不能期望,在短時間內,可以跳出工作時代。

如果學生時代是永久的話,它就會有如,工作時代那麼恐怖。

— Me@2020-10-10 07:52:39 PM

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2020.10.10 Saturday (c) All rights reserved by ACHK