"Classical field theory and classical mechanics"의 두 판 사이의 차이

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<h5>Lagrangian formalism</h5>
 
<h5>Lagrangian formalism</h5>
  
* Lagrangian f
+
* [[Lagrangian formalism]]
  
 
 
 
 
 
 
 
 
 
 
 
<h5 style="margin: 0px; line-height: 2em;">action</h5>
 
 
*  functional which takes a trajectory(history or path) to a number<br>
 
*  integral of Lagrangian<br><math>\mathcal{S} = \int L\, \mathrm{d}t</math><br>
 
*  this describes the 'total amount that happend' from one moment to another as a particle traces out a path<br>
 
*  applying Hamilton's action principle gives rise to a equation of motion<br><math>{\partial L\over\partial q} - {\mathrm{d}\over \mathrm{d}t }{\partial L\over\partial \dot{q}} = 0</math><br>
 
*  mass particle<br><math>L(q,\dot{q})=T-V=\frac{1}{2}m{\dot{q}}^2-V(q)</math><br><math>{\partial L\over\partial q} - {\mathrm{d}\over \mathrm{d}t }{\partial L\over\partial \dot{q}} = 0</math> becomes <br><math>\mathcal{S} = \int_{t_0}^{t_1} L(q,\dot{q}) \,dt</math><br>
 
*  in [[quantum mechanics]], <math>e^{iS/\hbar}</math> will describe the 'change in phase' of a quantum system as it traces out a path<br>
 
 
 
 
 
 
 
 
<h5>Euler-Lagrange equation</h5>
 
 
*  if field satisfies the equation of motion, EL is satisfied<br><math>\partial_\mu \left( \frac{\partial \mathcal{L}}{\partial ( \partial_\mu \psi )} \right) - \frac{\partial \mathcal{L}}{\partial \psi} = 0.</math><br>
 
 
 
 
 
 
 
 
<h5>equation of continuity</h5>
 
 
*  current density <math>J_{\mu}=(J_0,J_1,J_2,J_3)</math> satisfies<br><math>\partial^{\mu} J_{\mu}=0</math><br>
 
*  we get a conserved quantity<br><math>G=\int_V J_0(x) \,d^3 x</math><br>
 
* Lagrangian can be used to express the current density explicity
 
 
 
 
 
 
 
 
<h5>currents</h5>
 
 
* quantum analogues of the conserved densities arising by Noether's theorem
 
* due to the close relation to observable quantities, they behave similarly to free fields forming the current algebra
 
 
 
 
 
 
 
 
<h5 style="margin: 0px; line-height: 2em;">Lagrangian mechanics</h5>
 
 
From Lagrangian we obtain the conjugate momentum variable
 
  
 
 
 
 

2010년 9월 22일 (수) 17:32 판

introduction
  • can be formulated using classical fields and lagrangian density
  • change the coordinates and fields accordingly
  • require the invariance of action integral over arbitrary region
  • this invariance consists of two parts : Euler-Lagrange equation and the equation of continuity
  • three important conserved quantity
    • energy
    • momentum
    • angular momentum

 

 

notation
  • \(T\) kinetic energy
  • \(V\) potential energy
  • We have Lagrangian \(L=T-V\)
  • Define the Hamiltonian
  • \(H =p\dot q-L\)
  • \(p\dot q\) is twice of kinetic energy
  • Thus the Hamiltonian represents \(H=T+V\) the total energy of the system

 

 

Lagrangian formalism

 

 

 

Hamiltonian mechanics

conjugate variables are on the equal footing

 

 

 

Poisson bracket

For \(f(p_i,q_i,t), g(p_i,q_i,t)\) , we define the Poisson bracket

\(\{f,g\} = \sum_{i=1}^{N} \left[ \frac{\partial f}{\partial q_{i}} \frac{\partial g}{\partial p_{i}} - \frac{\partial f}{\partial p_{i}} \frac{\partial g}{\partial q_{i}} \right]\)

In quantization we have correspondence

\(\{f,g\} = \frac{1}{i}[u,v]\)

 

 

phase space

 

 

canonically conjugate momentum

 

 

 

links and webpages

 

 

history

 

 

related items

 

 

encyclopedia

 

 

books

 

 

articles

 

 

 

question and answers(Math Overflow)

 

 

blogs

 

 

experts on the field

 

 

links