"Lagrangian formulation of electromagetism"의 두 판 사이의 차이
imported>Pythagoras0 |
imported>Pythagoras0 |
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1번째 줄: | 1번째 줄: | ||
==Lagrangian for a particle== | ==Lagrangian for a particle== | ||
− | * | + | * Lagrangian for a charged particle in an electromagnetic field <math>L=T-V</math> |
− | :<math>L(q,\dot{q})=m||\dot{q}||-e\phi+eA_{i}\dot{q}^{i}</math> | + | :<math>L(q,\dot{q})=\frac{m||\dot{q}||^2}{2}-e\phi+eA_{i}\dot{q}^{i}</math> |
− | * | + | * Euler-Lagrange equations |
− | :<math>p_{i}=\frac{\partial{L}}{\partial{\dot{q}^{i}}}=m | + | :<math>p_{i}=\frac{\partial{L}}{\partial{\dot{q}^{i}}}=m \dot{q}_{i}+eA_{i}=mv_{i}+eA_{i}</math> |
$$ | $$ | ||
− | + | \dot{p}_{i}=m\frac{dv_{i}}{dt}+e\frac{\partial{A_{i}}}{\partial{q}^{j}}\dot{q}^{j} | |
$$ | $$ | ||
− | * | + | $$ |
− | + | F_{i}=\frac{\partial{L}}{\partial{q^{i}}}=\frac{\partial}{\partial{{q}^{i}}}(-e\phi+eA_{j}\dot{q}^{j})=-e\frac{\partial{\phi}}{\partial{q}^{i}} +e\frac{\partial{A_{j}}}{\partial{q}^{i}}\dot{q}^{j} | |
+ | $$ | ||
+ | * equation of motion <math>\dot{p}=F,</math> implies | ||
:<math>m\frac{dv_{i}}{dt}=eF_{ij}\dot{q}^{j}.</math> | :<math>m\frac{dv_{i}}{dt}=eF_{ij}\dot{q}^{j}.</math> | ||
* This is what we call the Lorentz force law. | * This is what we call the Lorentz force law. | ||
* force on a particle is same as <math>e\mathbf{E}+e\mathbf{v}\times \mathbf{B}</math> | * force on a particle is same as <math>e\mathbf{E}+e\mathbf{v}\times \mathbf{B}</math> | ||
* http://en.wikipedia.org/wiki/Lorentz_force | * http://en.wikipedia.org/wiki/Lorentz_force | ||
− | |||
==Lagrangian for electromagnetic tensor== | ==Lagrangian for electromagnetic tensor== |
2013년 3월 24일 (일) 17:42 판
Lagrangian for a particle
- Lagrangian for a charged particle in an electromagnetic field \(L=T-V\)
\[L(q,\dot{q})=\frac{m||\dot{q}||^2}{2}-e\phi+eA_{i}\dot{q}^{i}\]
- Euler-Lagrange equations
\[p_{i}=\frac{\partial{L}}{\partial{\dot{q}^{i}}}=m \dot{q}_{i}+eA_{i}=mv_{i}+eA_{i}\] $$ \dot{p}_{i}=m\frac{dv_{i}}{dt}+e\frac{\partial{A_{i}}}{\partial{q}^{j}}\dot{q}^{j} $$ $$ F_{i}=\frac{\partial{L}}{\partial{q^{i}}}=\frac{\partial}{\partial{{q}^{i}}}(-e\phi+eA_{j}\dot{q}^{j})=-e\frac{\partial{\phi}}{\partial{q}^{i}} +e\frac{\partial{A_{j}}}{\partial{q}^{i}}\dot{q}^{j} $$
- equation of motion \(\dot{p}=F,\) implies
\[m\frac{dv_{i}}{dt}=eF_{ij}\dot{q}^{j}.\]
- This is what we call the Lorentz force law.
- force on a particle is same as \(e\mathbf{E}+e\mathbf{v}\times \mathbf{B}\)
- http://en.wikipedia.org/wiki/Lorentz_force
Lagrangian for electromagnetic tensor
free
- 상호작용이 없는 전자기장의 라그랑지안은 다음과 같다
$$\mathcal{L}_{\text{EM}}= - \frac{1}{4}F_{\mu\nu}F^{\mu\nu}=\frac{1}{2}(\mathbf{E}^2-\mathbf{B}^2)$$ 이 때 \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \,\!\)는 전자기텐서, $A=(A_{\mu})$는 전자기 포텐셜
- action
\[S=-\frac{1}{4}\int F^{\alpha\beta}F_{\alpha\beta}\,d^{4}x\]
- 라그랑지안은 전자기 포텐셜의 다음과 같은 변환에 대하여 불변이다
\[A_{\mu}(x) \to A_{\mu}(x)-\partial_{\mu}\Lambda(x)\] 여기서 $\Lambda(x)$는 임의의 스칼라장
- equation of motion
$$ \partial_\mu F^{\mu\nu}=0 $$
in the presence of $j$ and $\rho$
- Lagrangian
$$L=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-ej_\mu A^\mu$$
- action
$$S[\phi,A]=\int_{t_1}^{t_2}\int_{\mathbb{R}^3}-\rho\phi+j\cdot A+\frac{\epsilon_0}{2}E^2-\frac{1}{2\mu_0}B^2\,dV\,dt$$
- w.r.t $\phi$
$$\nabla\cdot E=\frac{\rho}{\epsilon_0}$$
- w.r.t $A$
$$\nabla\times B=\mu_0j+\epsilon_0\mu_0\frac{\partial E}{\partial t}$$
memo
expositions
- THOMAS YU LAGRANGIAN FORMULATION OF THE ELECTROMAGNETIC FIELD
- The field Lagrangian
- http://www.lecture-notes.co.uk/susskind/classical-mechanics/lecture-8/the-electromagnetic-lagrangian/
blogs
- Higgs mechanism
- http://unapologetic.wordpress.com/2012/07/16/the-higgs-mechanism-part-1-lagrangians/
- http://unapologetic.wordpress.com/2012/07/17/the-higgs-mechanism-part-2-examples-of-lagrangian-field-equations/
- http://unapologetic.wordpress.com/2012/07/18/the-higgs-mechanism-part-3-gauge-symmetries/
- http://unapologetic.wordpress.com/2012/07/19/the-higgs-mechanism-part-4-symmetry-breaking/