AKA Einstein summation convention??? 같은건가 sub인가?
2022-02-18 summation convention 이 더 나은 pagename?

Chasnov https://youtu.be/CWIaPrwLyjM?si=T8or3V_5-92PhnLY
Ex.
$\sum_{i=1}^3\delta_{ii}\equiv\delta_{ii}=3$
$\epsilon_{ijk}\epsilon_{ijk}=\epsilon_{123}\epsilon_{123}+\cdots=6$
$\epsilon_{ijk}\epsilon_{lmn}=\begin{vmatrix}\delta_{il}&\delta_{im}&\delta_{in}\\\delta_{jl}&\delta_{jm}&\delta_{jn}\\\delta_{kl}&\delta_{km}&\delta_{kn}\end{vmatrix}$
$\vec{A}\cdot\vec{B}=A_iB_i$
$(\vec{A}\times\vec{B})_i = \epsilon_{ijk} A_j B_k$
$\begin{align}(\vec{A}\times\vec{B})_1 &= \epsilon_{1jk} A_j B_k \\ &= \epsilon_{123}A_2B_3 + \epsilon_{132}A_3B_2 \\ &= A_2B_3-A_3B_2\end{align}$
MKL 크로네커_델타,Kronecker_delta 레비치비타_기호,Levi-Civita_symbol

MKL
합,sum (curr goto 덧셈,addition)

einstein.notation

einstein.convention

[edit]

LCY ¶

ex.
일단 전제는

$\vec{A}=iA_x+jA_y+kA_z$
$\vec{A}=(A_x,A_y,A_z)$
$\partial_x\equiv\frac{\partial}{\partial x}$
$\vec{\nabla}=\hat{i}\partial_x+\hat{j}\partial_y+\hat{k}\partial_z$
$\vec{\nabla}=(\partial_x,\partial_y,\partial_z)$
$\vec{A}{}_j$ 는 A의 j번째 성분????

이하 파란색이 아인슈타인 표기법으로 표기된 것

내적,inner_product
$\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y+A_zB_z=\sum_{i=1}^3 A_iB_i$

$\color{blue}=A_i B_i$

기울기,gradient관련하여
$\vec{\nabla}\phi(x,y,z)=\hat{i}\partial_x\phi+\hat{j}\partial_y\phi+\hat{k}\partial_z\phi$

$\blue{=\hat{e_i}\partial_i \phi}$

e-hat 표기법은 모르겠는데... 레비치비타_기호,Levi-Civita_symbol의 "벡터곱,vector_product,cross_product에 사용된다는데" 에서 유추해야

발산,divergence관련하여
$\vec{\nabla}\cdot\vec{A}=\partial_xA_x+\partial_yA_y+\partial_zA_z$

$\blue =\partial_i A_i$

회전,curl
$\vec{\nabla}\times\vec{A}=\begin{vmatrix}i&j&k\\ \partial_x&\partial_y&\partial_z \\ A_x&A_y&A_z \end{vmatrix}$

$\blue =\hat{e_i}\epsilon_{ijk}\partial_j A_k$

삼중곱,triple_product
$\nabla\cdot(\vec{A}\times\vec{B})=\vec{B}\cdot(\nabla\times\vec{A})-\vec{A}\cdot(\nabla\times\vec{B})$

$=\partial_i(\vec{A}\times\vec{B})_i$
$=\partial_i(\epsilon_{ijk}A_jB_k)$
$=\epsilon_{ijk}\partial_i(A_jB_k)$
$=\epsilon_{ijk}\left[(\partial_iA_j)B_k+A_j(\partial_iB_k)\right]$
$=(\epsilon_{ijk} \partial_i A_j)B_k+A_j(\epsilon_{ijk} \partial_i B_k)$
$=(\nabla\times\vec{A})_k + A_j (-\epsilon_{jik}\partial_i B_k)$
$=(\nabla\times\vec{A})_k + A_j(-(\nabla\times\vec{B})_j)$
$=(\nabla\times\vec{A})\cdot\vec{B}-\vec{A}\cdot(\nabla\times\vec{B})$