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Recapitulation

- Vector space V (to axioms)

- basis

- inner product <\vec{v},\vec{u}>

- dot product VV->k

-cross product VV->V

\vec{B} \times \vec{C} = 
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}\\
B_{x} & B_{y} & B_{z} \\
C_{x} & C_{y} & C_{z} 
\end{vmatrix}
= \left | \vec{B} \right \vert \left | \vec{C} \right \vert 
\sin \theta_{k} \hat{n}

 dt
\begin{pmatrix}
a & b \\
c & d 
\end{pmatrix}
=ad-bc
dt
\begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i 
\end{pmatrix}
=aei+dhc+bfy-(ceg+bdi+hfa)

diagonal하게 생각하면, 기억하기 쉽다.

geometrical meaning of \vec{B} \times \vec{C}

벡터 B가 x축 위에 누워있고, 벡터 C가 xy평면의 1사분면 중 어느 점을 가리키고 있는 상황이다.
벡터 B와 벡터 C 사이의 각은 θBC이다.

\vec{B}=B \hat{e_{1}} \vec{C}=C \cos \theta_{BC} \hat{e_{1}} + 
C \sin \theta_{BC} \hat{e_{2}} \Rightarrow \vec{B} \times \vec{C} 
= BC \cos \theta_{BC} \hat{e_{1}} \times \hat{e_{1}} 
+ BC \sin \theta_{BC} \hat{e_{1}} \times \hat{e_{2}} =BC \sin \theta_{BC} \hat{e_{3}}

 \hat{e_{1}} \times \hat{e_{1}} = 
\begin{vmatrix}
\hat{e_{1}} & \hat{e_{2}} & \hat{e_{3}} \\
1 & 0 & 0 \\
1 & 0 & 0 
\end{vmatrix}
=0

\hat{e_{1}} \times \hat{e_{2}} = 
\begin{vmatrix}
\hat{e_{1}} & \hat{e_{2}} & \hat{e_{3}} \\
1 & 0 & 0 \\
0 & 1 & 0
\end{vmatrix}
= \hat{e_{3}}

\hat{e_{1}} \times \hat{e_{2}} = \hat{e_{3}}; \hat{e_{2}} \times \hat{e_{3}} = -\hat{e_{1}}; \hat{e_{3}} \times \hat{e_{1}} = \hat{e_{2}} \hat{e_{2}} \times \hat{e_{1}} = -\hat{e_{3}}; \hat{e_{3}} \times \hat{e_{2}} = -\hat{e_{1}}; \hat{e_{3}} \times \hat{e_{1}} = -\hat{e_{2}}

\vec{A} \times \vec{B} = - \left( \vec{B} \times \vec{A} \right) 

순서를 바꾸면 negation이 되는 이유

\begin{vmatrix}
\hat{e_{1}} & \hat{e_{2}} & \hat{e_{3}} \\
1 & 0 & 0 \\
0 & 1 & 0
\end{vmatrix} \begin{vmatrix}
\hat{e_{1}} & \hat{e_{2}} & \hat{e_{3}} \\
0 & 1 & 0 \\
1 & 0 & 0
\end{vmatrix}


triple scalar product

\vec{A} \cdot \left( \vec{B} \times \vec{C} \right)

\left | \vec{A} \right \vert \underbrace{\left | \vec{B} \right \vert \left | \vec{C} \right \vert \sin \theta_{BC}}_{area} \cos \beta

geometrical meaning: volume of pipe


triple cross product

\vec{A} \times \left( \vec{B} \times \vec{C} \right) \ne \left( \vec{A} \times \vec{B} \right) \times \vec{C} VVV->V \vec{A} \times \left( \vec{B} \times \vec{C} \right) = -\vec{C} \times \left( \vec{A} \times \vec{B} \right)
= \vec{C} \times \left( \vec{B} \times \vec{A} \right) \vec{A} \times \left( \vec{B} \times \vec{C} \right) = \vec{B} \left( \vec{A} \cdot \vec{C} \right)
-\vec{C} \left( \vec{A} - \vec{B} \right)

Levi-civita symbol

εijk

http://en.wikipedia.org/wiki/Levi-Civita_symbol

ε123 = ε231 = ε312 = 1 ε132 = ε213 = ε321 = − 1

나란히 있는 두 숫자의 순서가 바뀌면 1의 부호가 바뀐다라고 생각하면 쉬움. \epsilon_{122}=\epsilon_{233}=\epsilon_{113}= \cdots =0

같은 숫자가 두개 있으면 0이라고 생각하기.

& Any permutations of two indices will introduce "-" sign


Usefulness of εijk

\hat{e_{i}} \times \hat{e_{j}} = \sum_{k} \epsilon_{ijk} \hat{e_{k}} where i,j,k=1,2,3

eg.

\hat{e_{1}} \times \hat{e_{2}} = \sum_{k} \epsilon_{12k}\hat{e_{k}}=\underbrace{\epsilon_{123}}_{1}

Another useful symbol: Kronedeker delta


http://en.wikipedia.org/wiki/Kronecker_delta

http://mathworld.wolfram.com/KroneckerDelta.html

\delta_{ij} =
\begin{cases}
0, & \mbox{if }i \ne j \\
1, & \mbox{if }i=j
\end{cases}

eg. (m & n are dummy indices)

\vec{B} \times \vec{C} = \left( \sum_{m=1}^{3} B_{m} \hat{e_{m}} \right)
\times \left( \sum_{n=1}^{3} C_{n} \hat{e_{n}} \right)

 = \sum_{m, n} B_{m}C_{n} 
\underbrace{\left( \hat{e_{m}} \times \hat{e_{n}} \right)}_{\sum_{j} 
\epsilon_{mnj} \hat{e_{j}}}

=\sum_{m,n,j} B_{m}C_{n} \epsilon_{mnj} \hat{e_{j}}

jth component

BmCnεmnj
m,n
BmCnεmn1
m,n

= B2C3B3C2

\sum_{k=1}^{3} \epsilon_{mnk} \epsilon_{ijk} = 
\delta_{mi} \delta_{nj} - \delta_{mj} \delta_{ni}

εmjkεnjk = 2δmn
j,k

\sum_{i,j,k} \epsilon_{ijk}^{2}=6


eg 2. (Einstein convention - sum을 나타내는 sigma 기호 생략)

\vec{A} \times \left( \vec{B} \times \vec{C} \right)

 = \left \{ \left( A_{i} \hat{e_{i}} \right) \times 
\left \{ \vec{B} \times \vec{C} \right \}_{j} \hat{e_{j}} \right \}

= A_{i} \underbrace{\left( \vec{B} \times \vec{C} \right)_{j}}_{B_{m}C_{n} \epsilon_{mnj}} \underbrace{\hat{e_{i}} \times \hat{e_{j}}}_{\sum_{k} \underbrace{\epsilon_{ijk}}_{-\epsilon_{ikj}} \hat{e_{k}}}

 =-A_{i}B_{m}C_{n} \underbrace{\epsilon_{mnj} \epsilon_{ikj}}_{\delta_{mi} \delta_{nk} -  \delta_{mk} \delta_{ni}} \hat{e_{k}} 

=-A_{i}B_{m}C_{n} \delta_{mi} \delta_{nk} \hat{e_{k}} + A_{i}B_{m}C_{n} \delta_{mk} \delta_{ni} \hat{e_{k}} =-A_{m}B_{m}C_{n} \delta_{nk} \hat{e_{k}} + A_{i}B_{k}C_{n} \delta_{ni} \hat{e_{k}} =-A_{m}B_{m}C_{n} \hat{e_{n}} + A_{n}B_{k}C_{n} \hat{e_{k}} =-\left( \vec{A} \cdot \vec{B} \right) \vec{C} + \left( \vec{A} \cdot \vec{C} \right) \vec{B}


gradient operator

\nabla \phi \left( x,y \right)

=\frac{\partial \phi}{\partial x} \hat{e_{x}} + \frac{\partial \phi}{\partial y} \hat{e_{y}}

d \phi \left( x,y \right) = \frac{\partial \phi}{\partial x} dx + \frac{\partial \phi}{\partial y} dy = \nabla \phi \cdot d \vec{r} -(1)

여기서 d \vec{r} = dx \hat{e_{x}} + dy \hat{e_{y}}

식 (1)이 최대가 되려면 \cos \theta \left | \nabla \phi \right \vert \left | d \vec{r} \right \vert 에서 maximum at θ = 0

θd \vec{r} 벡터와 \nabla \phi 벡터 사이의 각

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