"Lagrangian formulation of electromagetism"의 두 판 사이의 차이

2013년 3월 24일 (일) 17:22 판

Lagrangian for a particle

Lagrangian for a charged particle in an electromagnetic field $L=T-V$

\[L(q,\dot{q})=m||\dot{q}||-e\phi+eA_{i}\dot{q}^{i}\]

Euler-Lagrange equations

\[p_{i}=\frac{\partial{L}}{\partial{\dot{q}^{i}}}=m\frac{\dot{q}_{i}}{||\dot{q}_{i}||}+eA_{i}=mv_{i}+eA_{i}\] $$ F_{i}=\frac{\partial{L}}{\partial{q^{i}}}=\frac{\partial}{\partial{{q}^{i}}}(eA_{j}\dot{q}^{j})=e\frac{\partial{A_{j}}}{\partial{q}^{i}}\dot{q}^{j} $$

equation of motion

\[\dot{p}=F,\] or \[m\frac{dv_{i}}{dt}=eF_{ij}\dot{q}^{j}.\]

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

$$ \partial_\mu F^{\mu\nu}=0 $$

see http://www.damtp.cam.ac.uk/user/tong/qft/six.pdf

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

http://dexterstory.tistory.com/888

related items

expositions

blogs

Higgs mechanism

questions

http://physics.stackexchange.com/questions/3005/derivation-of-maxwells-equations-from-field-tensor-lagrangian?rq=1

@@ 7번째 줄: / 7번째 줄: @@
 F_{i}=\frac{\partial{L}}{\partial{q^{i}}}=\frac{\partial}{\partial{{q}^{i}}}(eA_{j}\dot{q}^{j})=e\frac{\partial{A_{j}}}{\partial{q}^{i}}\dot{q}^{j}
 $$
-*  equation of motion<math>\dot{p}=F</math> Therefore we get
+*  equation of motion
-:<math>m\frac{dv_{i}}{dt}=eF_{ij}\dot{q}^{j}</math>. This is what we call the Lorentz force law.
+:<math>\dot{p}=F,</math> or
+:<math>m\frac{dv_{i}}{dt}=eF_{ij}\dot{q}^{j}.</math>
+* 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>
+* http://en.wikipedia.org/wiki/Lorentz_force