이것은 Analytical Mechanics 강의를 듣고 적은 필기입니다.
정리가 안 되어 있고, 개인적인 생각과 풀이가 섞여 있을 수도 있습니다.
지난 강의
AM lecture note - Coordinate transformation invariance of Lagrangian equation
오늘의 핵심
어떻게 constraint system을 general하게 풀 수 있을까?
라그랑주 multiplier 를 도입하여, 라그랑지안을 수정한다:
L’(x^a, \dot{x}^a, \lambda_\alpha) = L(x^a, \dot{x}^a) + \lambda_\alpha f_\alpha(x^a, t)
f_\alpha(x^a, t) = 0
x^2 + y^2 = l^2 \implies \begin{cases} x = l\sin\theta \ y = l\cos\theta \end{cases}
\begin{cases} m\ddot{x} = -T\dfrac{x}{l} \ m\ddot{y} = mg - T\dfrac{y}{l} \end{cases}
\begin{cases} \ddot{\theta} = -\dfrac{g}{l}\sin\theta \ T = ml\dot{\theta}^2 + mg\cos\theta \end{cases}
L’(x^a, \dot{x}^a, \lambda_\alpha) = L(x^a, \dot{x}^a) + \lambda_\alpha f_\alpha(x^a, t)
\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}^a}\right) - \frac{\partial L}{\partial x^a} = 0
\quad \Longleftrightarrow \quad
\frac{d}{dt}\left(\frac{\partial L’}{\partial \dot{x}^a}\right) - \frac{\partial L’}{\partial x^a} = 0
\frac{\partial L’}{\partial \lambda_\alpha} = f_\alpha(x^a, t) = 0
L’ = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) + mgy + \frac{1}{2}\lambda(x^2 + y^2 - l^2)
\begin{cases} m\ddot{x} = \lambda x \ m\ddot{y} = mg + \lambda y \end{cases}
\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}i}\right) - \frac{\partial L}{\partial q_i} = \lambda\alpha \frac{\partial f_\alpha}{\partial q_i}
\frac{\partial f_\alpha}{\partial q_i} = 0
\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right) - \frac{\partial L}{\partial q_i} = 0
\begin{cases}
x = l\sin\theta \
y = l\cos\theta \
\dot{x} = \dot{\theta}, l\cos\theta \
\dot{y} = -l\dot{\theta}\sin\theta \
x^2 + y^2 = l^2 \
\dot{x}^2 + \dot{y}^2 = (l\dot{\theta})^2
\end{cases}
L’(\theta, \dot{\theta}) = \frac{1}{2}m(l\dot{\theta})^2 + mgl\cos\theta
\frac{\partial L’}{\partial \theta} = -mgl\sin\theta
\frac{\partial L’}{\partial \dot{\theta}} = ml^2\dot{\theta}
\frac{d}{dt}\left(\frac{\partial L’}{\partial \dot{\theta}}\right) - \frac{\partial L’}{\partial \theta} = ml^2\ddot{\theta} + mgl\sin\theta = 0
\therefore\quad \ddot{\theta} = -\frac{g}{l}\sin\theta
\langle x_f, y_f | e^{\frac{i}{\hbar}H(t_f - t_i)} | x_i, y_i \rangle = \int_{(x_i,y_i)}^{(x_f,y_f)} \mathcal{D}x,\mathcal{D}y; e^{\frac{i}{\hbar}\int_{t_i}^{t_f} dt, L}
- \int_{-\infty}^{\infty} \mathcal{D}x; e^{ix, f_\alpha(x,t)}