Classical Mechanics: Difference between revisions

no edit summary
mNo edit summary
No edit summary
Line 19: Line 19:


For the Lagrangian given, at initial condition <math> (1, 0, 0) </math> the solution is <math> cos(\sqrt{\frac{k}{m}}t) </math> which is the familiar oscillator of mass m and spring constant k. For higher dimensional and multi-particle systems, the generalization is from considering one EL equation to one for each dimension of position for each particle. For N particles in three dimensions, this gives 3N equations. Often, the Lagrangian is divided into two terms <math> L=T(\dot{q}_1, \dot{q}_2, \cdots, \dot{q}_{3N})-U(q_1, q_2, \cdots, q_{3N}) </math>, with the term T depending only on the velocities is called the kinetic energy and U the potential energy. Now, we have an opportunity to analyze these terms and equations more generally. The derivatives <math> \frac{\partial L}{\partial \dot{q}_i} = \frac{\partial T}{\partial \dot{q}_i}, \quad i = 1, 2, \cdots, 3N</math> assembled as a vector are known as the momentum of the system. This another reason Lagrangians are powerful. Letting the <math> q_i </math> coordinates be coordinates other than Cartesian, e.g. spherical coordinates for each particle, allows us to discuss linear and angular momentum on the same footing or momentum in any convenient coordinate system. Given <math> m_{1+((i-1)-(i-1)\,mod 3)/3} = m_k </math> as the mass of the k'th particle in order, in cartesian coordinates <math> T=\sum_{i=1}^{3N} \frac{1}{2}m_k \dot{q}^2_i </math> gives <math> \frac{\partial T}{\partial \dot{q}_i}=m_k \dot{q}_i=m_k v_i = p_i</math>. Similarly, the coordinate derivative gives <math> \frac{\partial L}{\partial q_i}=-\frac{\partial U}{\partial q_i} = F_i </math> which has the interpretation of force on the i'th coordinate. Then the i'th EL equation is expressed as <math> m_k \frac{d}{dt} v_i = m_k a_i= F_i </math> which is just Newton's second law. Indeed, all of Newton's laws can be derived from this formulation of motion, but to do so fully we need an equation using finite properties of the Lagrangian and not just an infinitesimal condition.
For the Lagrangian given, at initial condition <math> (1, 0, 0) </math> the solution is <math> cos(\sqrt{\frac{k}{m}}t) </math> which is the familiar oscillator of mass m and spring constant k. For higher dimensional and multi-particle systems, the generalization is from considering one EL equation to one for each dimension of position for each particle. For N particles in three dimensions, this gives 3N equations. Often, the Lagrangian is divided into two terms <math> L=T(\dot{q}_1, \dot{q}_2, \cdots, \dot{q}_{3N})-U(q_1, q_2, \cdots, q_{3N}) </math>, with the term T depending only on the velocities is called the kinetic energy and U the potential energy. Now, we have an opportunity to analyze these terms and equations more generally. The derivatives <math> \frac{\partial L}{\partial \dot{q}_i} = \frac{\partial T}{\partial \dot{q}_i}, \quad i = 1, 2, \cdots, 3N</math> assembled as a vector are known as the momentum of the system. This another reason Lagrangians are powerful. Letting the <math> q_i </math> coordinates be coordinates other than Cartesian, e.g. spherical coordinates for each particle, allows us to discuss linear and angular momentum on the same footing or momentum in any convenient coordinate system. Given <math> m_{1+((i-1)-(i-1)\,mod 3)/3} = m_k </math> as the mass of the k'th particle in order, in cartesian coordinates <math> T=\sum_{i=1}^{3N} \frac{1}{2}m_k \dot{q}^2_i </math> gives <math> \frac{\partial T}{\partial \dot{q}_i}=m_k \dot{q}_i=m_k v_i = p_i</math>. Similarly, the coordinate derivative gives <math> \frac{\partial L}{\partial q_i}=-\frac{\partial U}{\partial q_i} = F_i </math> which has the interpretation of force on the i'th coordinate. Then the i'th EL equation is expressed as <math> m_k \frac{d}{dt} v_i = m_k a_i= F_i </math> which is just Newton's second law. Indeed, all of Newton's laws can be derived from this formulation of motion, but to do so fully we need an equation using finite properties of the Lagrangian and not just an infinitesimal condition.
<br>
<br>
Supposing <math> q_i(t) </math> solves the EL equations, we can analyze properties of the integral across finite time <math> S = \int_{t_1}^{t_2} L(q_1(t), \cdots, q_{3N}(t), \dot{q}_1(t), \cdots, \dot{q}_{3N}(t), t) dt </math>, since substituting the trajectory gives a strict function of time.