"구면(sphere)"의 두 판 사이의 차이

2012년 8월 15일 (수) 17:05 판

3차원상의 반지름이 R인 구면 \( x^2+y^2+z^2 = R^2\)
매개화
\(X(u,v)=R(\cos u \sin v, \sin u \sin v, \cos v)\)
\(0<u<2\pi\), \(0<v<\pi\)
\(X_u=R(- \sin u \sin v , \cos u \sin v ,0)\)
\(X_v=R( \cos u \cos v , \sin u \cos v ,-\sin v)\)
\(N=(-\cos u \sin v, -\sin u \sin v, -\cos v)\)
\(X_{uu}=R(-\cos u \sin v , -\sin u \sin v ,0)\)
\(X_{uv}=R(-\cos v \sin u , \cos u \cos v , 0)\)
\(X_{vv}=R(- \cos u \sin v , - \sin u \sin v , - \cos v )\)

리만 곡률 텐서
\(\begin{array}{ll} \begin{array}{ll} R_{111}^1 & 0 \\ R_{112}^1 & 0 \end{array} & \begin{array}{ll} R_{121}^1 & 0 \\ R_{122}^1 & 0 \end{array} \\ \begin{array}{ll} R_{211}^1 & 0 \\ R_{212}^1 & 1 \end{array} & \begin{array}{ll} R_{221}^1 & -1 \\ R_{222}^1 & 0 \end{array} \\ \begin{array}{ll} R_{111}^2 & 0 \\ R_{112}^2 & -\sin ^2(v) \end{array} & \begin{array}{ll} R_{121}^2 & \sin ^2(v) \\ R_{122}^2 & 0 \end{array} \\ \begin{array}{ll} R_{211}^2 & 0 \\ R_{212}^2 & 0 \end{array} & \begin{array}{ll} R_{221}^2 & 0 \\ R_{222}^2 & 0 \end{array} \end{array}\)
covariant tensor
\(\begin{array}{ll} \begin{array}{ll} R_{1111} & 0 \\ R_{1112} & 0 \end{array} & \begin{array}{ll} R_{1121} & 0 \\ R_{1122} & 0 \end{array} \\ \begin{array}{ll} R_{1211} & 0 \\ R_{1212} & R^2 \sin ^2(v) \end{array} & \begin{array}{ll} R_{1221} & -R^2 \sin ^2(v) \\ R_{1222} & 0 \end{array} \\ \begin{array}{ll} R_{2111} & 0 \\ R_{2112} & -R^2 \sin ^2(v) \end{array} & \begin{array}{ll} R_{2121} & R^2 \sin ^2(v) \\ R_{2122} & 0 \end{array} \\ \begin{array}{ll} R_{2211} & 0 \\ R_{2212} & 0 \end{array} & \begin{array}{ll} R_{2221} & 0 \\ R_{2222} & 0 \end{array} \end{array}\)

측지선 이 만족시키는 미분방정식
\(\frac{d^2\alpha_k }{dt^2} + \Gamma^{k}_{~i j }\frac{d\alpha_i }{dt}\frac{d\alpha_j }{dt} = 0\)
풀어쓰면,
\(\frac{d^2 u}{dt^2} + 2\Gamma^{1}_{~1 2 }\frac{du }{dt}\frac{dv }{dt} = 0\)
\(\frac{d^2 v}{dt^2} + \Gamma^{2}_{~1 1 }\frac{du }{dt}\frac{du }{dt} = 0\)

가우스곡률 항목 참조
\(K = -\frac{1}{2\sqrt{EG}}\left(\frac{\partial}{\partial u}\frac{G_u}{\sqrt{EG}} + \frac{\partial}{\partial v}\frac{E_v}{\sqrt{EG}}\right)\)
반지름 R인 구면의 가우스곡률
\(K=\frac{1}{R^2}\)

위의 좌표계에서 \(u=\phi,v=\theta\) 로 생각하자.
라플라시안
\(\Delta f = {1 \over r^2 \sin \theta} {\partial \over \partial \theta} \left( \sin \theta {\partial f \over \partial \theta} \right) + {1 \over r^2 \sin^2 \theta} {\partial^2 f \over \partial \phi^2}={1 \over r^2 }({\partial^2 f \over \partial \theta^2} +\cot\theta {\partial f \over \partial \theta} + \frac{1}{ \sin^2 \theta} {\partial^2 f \over \partial \phi^2})\)

@@ 25번째 줄: / 25번째 줄: @@
-<h5>제1기본형식</h5>
+<h5>제1기본형식 (메트릭 텐서)</h5>
 * <math>E=R^2\sin^2 v</math>
@@ 104번째 줄: / 104번째 줄: @@
 * [[수학과 지도학|지도와 수학]]
 * [[n차원 공의 부피]]
-* [[#]]
+* [[구면좌표계]]