6. 벡터해석

6.1 서론

6.2 외적의 응용

두 개의 벡터 $\vec{A}, \vec{B}$에 대하여

내적 aka 스칼라곱

(1)
\begin{align} \vec{A} \cdot \vec{B} = \left\lvert \vec{A} \right\rvert \left\lvert \vec{B} \right\rvert \cos \theta = A_x B_x + A_y B_y + A_z B_z \end{align}
6-2-1.png

외적 aka 벡터곱

(2)
\begin{align} \left\lvert \vec{A} \times \vec{B} \right\rvert = \left\lvert \vec{A} \right\rvert \left\lvert \vec{B} \right\rvert \sin \theta \end{align}

6.3 삼중곱

삼중 내적

6-3-1.png
(3)
\begin{align} \left\lvert \vec{A} \cdot ( \vec{B} \times \vec{C} ) \right\rvert & = \left\lvert \vec{A} \right\rvert \left\lvert \vec{B} \times \vec{C} \right\rvert \cos \phi \\ & = \left\lvert \vec{B} \cdot ( \vec{C} \times \vec{A} ) \right\rvert \\ & = \left\lvert \vec{C} \cdot ( \vec{A} \times \vec{B} ) \right\rvert \end{align}

삼중 내적을 성분형태로 쓰려면 우선 $\vec{B} \times \vec{C}$를 행렬식 형태로 쓴다.

(4)
\begin{align} \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} \end{align}
(5)
\begin{align} \vec{A} \cdot ( \vec{B} \times \vec{C} ) = A_x ( \vec{B} \times \vec{C} )_x + A_y ( \vec{B} \times \vec{C} )_y + A_z ( \vec{B} \times \vec{C} )_z \end{align}
(6)
\begin{align} \vec{A} \cdot ( \vec{B} \times \vec{C} ) & = \begin{vmatrix} A_x & A_y & A_z \\ B_x & B_y & B_z \\ C_x & C_y & C_z \end{vmatrix} \\ & = A_x ( B_y C_z - B_z C_y ) - A_y ( B_x C_z - B_z C_x) + A_z ( B_x C_y - B_y C_x) \\ & = A_x (\vec{B} \times \vec{C} )_x + A_y ( \vec{B} \times \vec{C})_y + A_z ( \vec{B} \times \vec{C} )_z \end{align}

삼중 외적

6-3-4.png
(7)
\begin{align} \vec{B} & = B_x \hat{i} \\ \vec{C} & = C_x \hat{i} + C_y \hat{j} \\ \vec{A} & = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} \\ \vec{B} \times \vec{C} & = B_x \hat{i} \times \left[ C_x \hat{i} + C_y \hat{j} \right] = B_x C_y \hat{k} \end{align}
(8)
\begin{align} \vec{A} \times ( \vec{B} \times \vec{C} ) & = ( A_x \hat{i} + A_y \hat{j} + A_z \hat{k} ) \times ( B_x C_y \hat{k} ) \\ & = - A_x B_x C_y \hat{j} + A_y B_x C_y \hat{i} \\ & = a \vec{B} + b \vec{C} \\ & = a B_x \hat{i} + b ( C_x \hat{i} + C_y \hat{j} ) = \hat{i} ( a B_x + b C_x ) + \hat{j} ( b C_y ) \end{align}
(9)
\begin{align} b C_y & = - A_x B_x C_y \implies b = - A_x B_y = - \vec{A} \times \vec{B} \\ a B_x + b C_x & = A_y B_x C_y, \\ a B_x & = A_y B_x C_y + A_y B_x C_y \implies a = A_y C_y + A_x C_y = \vec{A} \times \vec{C} \end{align}
(10)
\begin{align} \therefore\ \vec{A} \times ( \vec{B} \times \vec{C} ) = ( \vec{A} \cdot \vec{C} ) \vec{B} - ( \vec{A} \cdot \vec{B} ) \vec{C} \end{align}

삼중 내적의 응용 — 토크

6-3-5.png
(11)
\begin{align} \vec{\tau} & = \vec{r} \times \vec{F} \\ \hat{n} \cdot \vec{ \tau} & = \hat{n} \cdot ( \vec{r} \times \vec{F} ) \end{align}
(12)
\begin{align} \vec{r} = \vec{r}_\parallel + \vec{r}_\perp, \qquad \vec{F} = \vec{F}_\parallel + \vec{F}_\perp \end{align}
(13)
\begin{align} \vec{r} \times \vec{F} & = ( \vec{r}_\parallel +\vec{r}_\perp ) \times ( \vec{F}_\parallel + \vec{F}_\perp ) \\ & = \vec{r}_\parallel \times \vec{F}_\parallel + \vec{r}_\perp \times \vec{F}_\parallel + \vec{r}_\parallel \times \vec{F}_\perp + \vec{r}_\perp \times \vec{F}_\perp \\ \hat{n} \cdot ( \vec{r} \times \vec{F} ) & = \hat{n} \cdot ( \vec{r}_\perp \times \vec{F}_\perp ) \end{align}

예제 1:
$\vec{F} = (1, 3, -1)$이 점 $\vec{r} = (1, 1, 1)$에 작용할 때 직선 $\vec{r} = (3, 0, 2) + t(2, -2 , 1) \propto \hat{n} = 1/3 (2, -2, 1)$에 대한 토크 구하기

삼중 외적의 응용 — 각운동량

6-3-8.png
(14)
\begin{align} \vec{L} & = \vec{r} \times \vec{p} = \vec{r} \times ( m vec{v} ) = m \vec{r} \times ( \vec{ \omega } \times \vec{r} ) \\ \vec{v} & = \vec{\omega} \times \vec{r} \\ \left\lvert \vec{v} \right\rvert & = \left\lvert \vec{ \omega} \right\rvert \left\lvert \vec{r} \right\rvert \sin \theta \\ {{ds} \over {dt}} & = {d \over {dt}} \left[ \left\lvert \vec{r} \right\rvert \sin \theta\ d \phi \right] \end{align}

6.4 벡터의 미분

(15)
\begin{align} \vec{A} & = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} \\ {{d \vec{A} } \over {dt}} & = {{d A_x} \over {dt}} \hat{i} + {{d A_y} \over {dt}} \hat{j} + {{d A_z} \over {dt}} \hat{k} \end{align}
(16)
\begin{align} {d \over {dt}} ( a \vec{A} ) & = \left( {{da} \over {dt}} \right) \vec{A} + a {{d \vec{A} } \over {dt}} \\ {d \over {dt}} ( \vec{A} \cdot \vec{B} ) & = {{d \vec{A} } \over {dt}} \cdot \vec{B} + \vec{A} \cdot {{ d \vec{B} } \over {dt}} \\ {d \over {dt}} ( \vec{A} \times \vec{B} ) & = {{d \vec{A} } \over {dt}} \times \vec{B} + \vec{A} \times {{ d \vec{B} } \over {dt}} \end{align}

예제 2: 등속원운동의 가속도가 원의 중심을 향하고 크기가 $v^2 / r$임을 증명

(17)
\begin{align} r^2 = \vec{r} \cdot \vec{r} & = \mathrm{const.} \\ 2 \vec{v} \cdot {{d \vec{v} } \over {dt}} & = 0, \\ \vec{a} \cdot \vec{v} & = 0 \implies \vec{a} \perp \vec{v} \\ v^2 = \vec{v} \cdot \vec{v} & = \mathrm{const.} \\ 2 \vec{r} \cdot {{d \vec{r} } \over {dt}} & = 0, \\ \vec{v} \cdot \vec{r} & = 0 \implies \vec{r} \perp \vec{v} \end{align}
(18)
\begin{align} \vec{a} \cdot \vec{r} + \vec{v} \cdot \vec{v} & = 0 \\ \vec{a} \cdot \vec{r} = - v^2 & = ar \cos \theta \implies \theta = \pi \\ \therefore\ ar (-1) = - v^2, & \qquad a = {v^2 \over r} \end{align}

극좌표에서 단위벡터성분

6-4-1.png
(19)
\begin{align} \hat{e}_r & = \hat{i} \cos \theta + \hat{j} \sin \theta = ( \cos \theta , \sin \theta ) \\ \hat{e}_\theta & = \hat{i} \cos \left( \theta + {\pi \over 2} \right) + \hat{j} \sin \left( \theta + { \pi \over 2} \right) = ( - \sin \theta , \cos \theta ) \end{align}

이것을 $t$에 대해 미분하면

(20)
\begin{align} {{d \hat{e}_r } \over {dt}} & = ( - \sin \theta , \cos \theta ) {{d \theta } \over {dt}} = \hat{e}_\theta {{d \theta } \over {dt}} \\ {{ d \hat{e}_\theta } \over {dt}} & = ( - \cos \theta , - \sin \theta ) {{d \theta} \over {dt}} = - \hat{e}_r {{d \theta } \over {dt}} \end{align}

예제 3: $\vec{A} = A_r \hat{e}_r + A_\theta \hat{e}_\theta$일 때

(21)
\begin{align} { { d \vec{A} } \over {dt}} & = {{d A_r } \over {dt}} \hat{e}_r + A_r {{d \hat{e}_r } \over {dt}} + {{d A_\theta } \over {dt}} \hat{e}_\theta + A_\theta {{d \hat{e}_\theta} \over {dt}} \\ & = {{d A_r } \over {dt}} \hat{e}_r + A_r \hat{e}_\theta {{d \theta } \over {dt}} + {{d A_\theta} \over {dt}} \hat{e}_\theta - A_\theta \hat{e}_r {{d \theta } \over {dt}} \\ & = \left[ {{d A_r} \over {dt}} - A_\theta {{d \theta } \over {dt}} \right] \hat{e}_r + \left[ {{d A_\theta} \over {dt}} + A_r {{d \theta } \over {dt}} \right] \hat{e}_\theta \end{align}

6.5 장

6.6 방향도함수; 기울기 벡터

6-6-1.png
(22)
\begin{align} \hat{u} = (a, b, c) \end{align}
(23)
\begin{align} & \begin{cases} x & = x_0 + as = x(s) \\ y & = y_0 + bs = y(s) \\ z & = z_0 + cs = z(s) \end{cases} \\ \phi (x, y, z) & = \phi ( x(s), y(s), z(s) ) \\ {{d \phi} \over {ds}} & = {{ \partial \phi } \over {dx}} {{dx} \over {ds}} + {{\partial \phi} \over {dy}} {{dy} \over {ds}} + {{\partial \phi} \over {dz}} {{dz} \over {ds}} \\ & = {{ \partial \phi } \over {dx}} a + {{\partial \phi} \over {dy}} b + {{\partial \phi} \over {dz}} c \\ & = ( \vec{\nabla} \phi ) \cdot \hat{u} \end{align}
(24)
\begin{align} \vec{\nabla} \phi = = \operatorname{grad} \phi = \hat{i} {{\partial \phi} \over {\partial x}} +\hat{j} {{\partial \phi} \over {\partial y}} + \hat{k} {{\partial \phi} \over {\partial z}} \end{align}
6-6-2.png

예제 2:

6-6-3.png
(25)
\begin{align} \phi & = \mathrm{const.} \qquad \mathrm{i.e.} \\ \Delta \phi & = 0, \quad {{ \Delta \phi} \over {\Delta s}} = 0 \\ \lim_{\Delta s \to 0} {{ \Delta \phi } \over {\Delta s}} & = 0 = {{d \phi} \over {ds}} = ( \vec{\nabla} \phi ) \cdot \hat{u} \end{align}

내적이 수직인 즉 $\vec{\nabla} \phi \perp \hat{u}$,
$\vec{\nabla} \phi$$\phi = \mathrm{const.}$인 표면에 수직하다.
$\left\lvert \vec{\nabla} \phi \right\rvert$가 표면에 수직인 방향으로의 방향도함수값이므로 이를 수직도함수라고 부른다.

예제 3: 곡면좌표계에서의 기울기

(26)
\begin{align} \vec{\nabla} f & = {{\partial f} \over {\partial r}} \hat{e}_r +{1 \over r} {{\partial f} \over {\partial \theta}} \hat{e}_\theta + {{\partial f} \over {\partial z}} \hat{e}_z \qquad \qquad (원통좌표계) \\ \vec{\nabla} f & = {{\partial f} \over {\partial r}} \hat{e}_r + {1 \over 2} {{\partial f} \over {\partial \theta }} \hat{e}_\theta + {1 \over {r \sin \theta }} {{\partial f} \over {\partial \phi}} \hat{e}_\phi \qquad (구면좌표계) \end{align}

6.7 $\vec{\nabla}$을 포함하는 다른 표헌들

(27)
\begin{align} \vec{\nabla} = \hat{i} {{\partial } \over {\partial x}} + \hat{j} {{\partial } \over {\partial y}} +\hat{k} {{\partial } \over {\partial z}} \end{align}
(28)
\begin{align} 발산: \vec{\nabla} \cdot \vec{V} & = {{\partial V_x} \over {\partial x}} + {{\partial V_y} \over {\partial y}} +{{\partial V_z} \over {\partial z}} \\ 회전: \vec{\nabla} \times \vec{V} & = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ {\partial \over {\partial x}} & { \partial \over {\partial y}} & { \partial \over {\partial z}} \\ V_x & V_y & V_z \end{vmatrix} \end{align}
(29)
\begin{align} \vec{\nabla} \cdot ( \vec{\nabla} \phi ) & = \left( {{\partial^2 } \over {\partial x^2}} + {{\partial^2} \over {\partial y^2}} + {{\partial^2} \over {\partial z^2}} \right) \phi = \nabla^2 \phi \end{align}

$\nabla^2$ = 라플라스 연산자

(30)
\begin{align} \nabla^2 \phi & = 0 & 라플라스\ 방정식 \\ \nabla^2 \phi & = {1 \over a^2} {{\partial^2 \phi} \over {\partial t^2}} & 파동방정식 \\ \nabla^2 \phi & = {1 \over a^2} {{\partial \phi} \over {\partial t}} & 확산, 열전도, 슈뢰딩거\ 방정식 \end{align}

예제 1:

(31)
\begin{align} \vec{\nabla} \times ( \vec{\nabla} \times \vec{V} ) & = ( \vec{\nabla} \cdot \vec{V} ) \vec{\nabla} - ( \vec{\nabla} \cdot \vec{\nabla} ) \vec{V} \\ & = \vec{\nabla} ( \vec{\nabla} \cdot \vec{V} ) - \nabla^2 \vec{V} \\ & = \vec{\nabla} \left( {{\partial \vec{V}} \over {\partial x}} + {{\partial \vec{V} } \over {\partial y}} +{{\partial \vec{V}} \over {\partial z}} \right) \\ & = \hat{i} \nabla^2 V_x + \hat{j} \nabla^2 V_y +\hat{k} \nabla^2 V_z \end{align}

곡면좌표계에서의 발산과 라플라시안
원통좌표계 — ($z$항을 무시하면 극좌표계)

(32)
\begin{align} \vec{\nabla} \cdot \vec{V} & = {1 \over r} {\partial \over {\partial r}} (r V_r) +{1 \over r} {\partial \over {\partial \theta }} V_\theta + {\partial \over {\partial z}} V_z \\ \nabla^2 f & = {1 \over r} {{\partial} \over {\partial r}} \left( r {{\partial f} \over {\partial r}} \right) +{1 \over r^2 } {{\partial^2} \over {\partial \theta^2}} f + {\partial^2 \over {\partial z^2}} f \end{align}

구면좌표계 —

(33)
\begin{align} \vec{\nabla} \cdot \vec{V} & = {1 \over r^2 } {\partial \over {\partial r}} ( r^2 V_r) +{1 \over {r \sin \theta}} {\partial \over {\partial \theta}} ( V_\theta \sin \theta ) + {1 \over {r \sin \theta }} {\partial \over {\partial \phi}} V_\phi \\ \nabla^2 f & = {1 \over r^2} {\partial \over {\partial r}} \left( r^2 {{\partial f} \over {\partial r}} \right) +{1 \over {r^2 \sin \theta}} {\partial \over {\partial \theta }} \left( \sin \theta {{\partial} \over {\partial \theta }} f \right) + {1 \over {r^2 \sin^2 \theta }} {\partial^2 \over {\partial \phi^2}} f \end{align}

6.8 선적분

6-8-1.png

물체에 힘 $\vec{F}$가 작용하여 미소변위 $d \vec{r}$이 발생했을 때 일은

(34)
\begin{align} dW & = \vec{F} \cdot d\vec{r} \end{align}

전체 일은 그 경로를 따라 적분하여

(35)
\begin{align} W = \int_\mathrm{path} \vec{F} ( \vec{r})\ d \vec{r} \end{align}

6.9 평면에서의 그린 정리

6.10 발산과 발산정리

6-10-1.png
(36)
\begin{align} A' & = A \cos \theta \\ \rho\ vt\ A' & = \rho t \vec{V} \cdot \hat{n} A \end{align}
6-10-2.png
(37)
\begin{align} (\rho \vec{V} \cdot \hat{n} ) dx dz \\ \rho \vec{V} ( \vec{r}) \hat{n} (\vec{r} ) dx dz \end{align}
(38)
\begin{align} \rho \left[ \vec{V} ( y + dy) \cdot \hat{y} - \vec{V} ( y) \cdot g \right] dx dz & = \rho {{\partial V_y} \over {\partial y}} dy dx dz \\ \left( dV_y = {{\partial V_y} \over {\partial y}} dy \right) & \\ & \implies \rho \left( {{\partial V_x} \over {\partial x}} + {{\partial V_y} \over {\partial y}} + {{\partial V_z} \over {\partial z}} \right) dx dy dz = \vec{\nabla} \cdot \vec{V} dx dy dz \end{align}
(39)
\begin{align} {{\partial \rho } \over {\partial t}} dx dy dz = - \rho \vec{\nabla} \cdot \vec{V} dx dy dz + \psi\ dx dy dz \end{align}
(40)
\begin{align} {{\partial \rho } \over {\partial t}} + \rho \vec{\nabla} \cdot \vec{V} = \psi = 0 \end{align}
(41)
\begin{align} \vec{\nabla} \cdot \vec{V} = \psi, \\ \vec{\nabla} \cdot \vec{D} = \rho \end{align}
(42)
\begin{align} \rho {{\partial V_y} \over {\partial y}} dx dy dz = \rho \left[ \vec{V} (y + dy ) \cdot \hat{n} ( y + dy) + \vec{V} (y) \cdot \hat{n} (y) \right] dx dz \end{align}
(43)
\begin{align} \rho ( \vec{\nabla} \cdot \vec{V} ) dx dy dz = \rho ( \vec{\nabla} \cdot \vec{V} ) d \tau_1 & = \iint \rho \vec{V} \cdot \hat{n} d \sigma \quad \longrightarrow \sum_i \iint_{\partial \tau_1 } \rho \vec{V} \cdot \hat{n} d \sigma_i \\ \sum_i \rho (\vec{\nabla} \cdot \vec{V} ) d \tau_1 & = \iiint \rho ( \vec{\nabla} \cdot \vec{V} ) d \tau \end{align}

발산정리의 예제: $\vec{V} = \hat{i} x + \hat{j} y + \hat{k} z$에 대하여

6-10-5.png

위 그림의 원통 모양 폐곡면에 대해 $\oint \hat{V} \cdot \hat{n}\ d \sigma$ 계산하기

(1) 윗면

(44)
\begin{align} \hat{n} & = \hat{k}, \\ \vec{V} \cdot \hat{n} & = z = h \oint h\ d \sigma & = h \pi a^2 \end{align}

(2) 아랫면

(45)
\begin{align} \hat{n} & = - \hat{k}, \\ \vec{V} \cdot \hat{n} & = - z = 0 \end{align}

(3) 옆면

(46)
\begin{align} \vec{r} & = x \hat{i}+y \hat{j} \\ \hat{n} & = {{ x \hat{i} + y \hat{j} } \over \sqrt{ x^2 + y^2 } } = {{x \hat{i} + y \hat{j} } \over a} \\ \vec{V} \cdot \hat{n} & = {{ x^2 + y^2 } \over a} = a \\ \oint \vec{V} \cdot \hat{n}\ d \tau & = a\ 2 \pi a h = 2 \pi a^2 h \\ & = \int \vec{\nabla} \cdot \vec{V}\ d \tau = 3 \pi a^2 h \end{align}

가우스 법칙

(47)
\begin{align} \vec{E} & = {{ q} \over {4 \pi \epsilon_0 r^2 } } \hat{e}_r \qquad (쿨롱\ 법칙) \end{align}
(48)
\begin{align} \vec{D} & = \epsilon_0 \vec{E} = {q \over {4 \pi r^2 }} \hat{e}_r \end{align}

점전하 $q$가 폐곡면 내부에 있을 때

(49)
\begin{align} \oint \vec{D} \cdot \hat{n}\ d \sigma = \oint \left( {q \over {4 \pi r^2}} \right) \hat{e}_r ( r^2 \sin \theta\ d \theta d \phi ) = q \end{align}

점전하 $q$가 폐곡면 바깥에 있을 때

(50)
\begin{align} \oint \vec{D} \cdot \hat{n}\ d \tau = 0 \end{align}

폐곡면 안에 여러 개의 전하 $q_i$가 있을 때

(51)
\begin{align} \sum_i \oint \vec{D}_i \cdot \hat{n}\ d \tau & = \oint \vec{D} \cdot \hat{n}\ d \tau = \int \vec{\nabla} \cdot \vec{D}\ d \tau \qquad (발산\ 정리) \\ & = \sum_i {q_i} = \int \rho \ d \tau \\ & \implies \vec{\nabla} \cdot \vec{D} = \rho \end{align}

6.11 회전연산자와 스토크스 정리

6-3-8.png
(52)
\begin{align} \vec{V} & = \vec{\omega} \times \vec{r} \\ \vec{\nabal} \times \vec{V} & = \vec{\nabla} \times ( \vec{\omega} \times \vec{r} ) \\ & = \vec{\omega} ( \vec{\nabla} \cdot \vec{r} ) - ( \vec{\omega} \cdot \vec{\nabla} ) \vec{r} = 2 \vec{\omega} \\ & \qquad \left( \omega_x {{\partial } \over {\partial x}} + \omega_y {{\partial} \over {\partial y}} + \omega_z { \partial \over {\partial z}} \right) \vec{r} = \omega_x \hat{i} +\omega_y \hat{j} + \omega_z \hat{k} = \vec{\omega} \\ & = 3 \vec{\omega} - \vec{\omega} = 2 \vec{\omega} \end{align}
6-11-2.png
(53)
\begin{align} \iint_A ( \vec{ \nabla} \times \vec{V} ) \cdot \hat{n}\ d \sigma = \iint_A ( \operatorname{ curl} \vec{V} ) \cdot \hat{k}\ dx dy = \oint_{dA} \vec{V} \cdot d \vec{r} \end{align}

스토크스 정리

6-11-5.png
(54)
\begin{align} \sum_i \iint_+{ \sigma_i } ( \vec{\nabla} \times \vec{V} ) \hat{n}\ d \sigma_i = \oint_{ \partial \sigma_i } \vec{V} \cdot d \vec{r} \end{align}