Lagrangian formulation of electromagetism
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
\[p_{i}=\frac{\partial{L}}{\partial{\dot{q}^{i}}}=m \dot{q}_{i}+eA_{i}=mv_{i}+eA_{i}\] $$ \dot{p}_{i}=m\frac{dv_{i}}{dt}+e\frac{\partial{A_{i}}}{\partial t}+e\frac{\partial{A_{i}}}{\partial{q}^{j}}\dot{q}^{j} $$ $$ F_{i}=\frac{\partial{L}}{\partial{q^{i}}}=\frac{\partial}{\partial{{q}^{i}}}(-e\phi+eA_{j}\dot{q}^{j})=-e\frac{\partial{\phi}}{\partial{q}^{i}} +e\frac{\partial{A_{j}}}{\partial{q}^{i}}\dot{q}^{j} $$
- equation of motion \(\dot{p}=F,\) implies
\[m\frac{dv_{i}}{dt}=eE_{i}+eF_{ij}\dot{q}^{j}\] where $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \,\!$
- for example, if $i=1$
$$ ma_1=eE_1+e(F_{11}\dot{q}^{1}+F_{12}\dot{q}^{2}+F_{13}\dot{q}^{3})=eE_1+e(F_{12}\dot{q}^{2}-F_{31}\dot{q}^{3})=eE_1+e(\mathbf{v}\times \mathbf{B})_{1} $$ where $F_{12}=-B_{3}$ and $F_{31}=-B_{2}$
- 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
- 틀:수학노트
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
$$S[\phi,A]=\int_{t_1}^{t_2}\int_{\mathbb{R}^3}\left(-\rho\phi+j\cdot A+\frac{\epsilon_0}{2}E^2-\frac{1}{2\mu_0}B^2\right)\,dV\,dt$$
- 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/