Electromagnetism with classical mechanics

09/08/2011

Lagrangianは次のように与えられる。

$\mathcal{L}=-\frac{1}{4}\left(F_{\mu\nu}\right)^{2}$

そしてEuler-Lagrange equationは次のようである。

$\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}A_{\nu}\right)}\right)-\frac{\partial\mathcal{L}}{\partial A_{\nu}}=0$

主要な微分計算

$\frac{\partial}{\partial\left(\partial_{\mu}A_{\nu}\right)}\left(\partial_{\rho}A_{\sigma}-\partial_{\sigma}A_{\rho}\right)={\delta^{\mu}}_{\rho}{\delta^{\nu}}_{\sigma}-{\delta^{\mu}}_{\sigma}{\delta^{\nu}}_{\rho}$

$\frac{\partial}{\partial\left(\partial_{\mu}A_{\nu}\right)}\left(\partial^{\rho}A^{\sigma}-\partial^{\sigma}A^{\rho}\right)=g^{\mu\rho}g^{\nu\sigma}-g^{\mu\sigma}g^{\nu\rho}$

$\frac{\partial}{\partial\left(\partial_{\mu}A_{\nu}\right)}\left\{ -\frac{1}{4}\left(\partial_{\rho}A_{\sigma}-\partial_{\sigma}A_{\rho}\right)\left(\partial^{\rho}A^{\sigma}-\partial^{\sigma}A^{\rho}\right)\right\} =-\left(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}\right)=-F^{\mu\nu}$

$\frac{\partial}{\partial A_{\nu}}\left\{ -\frac{1}{4}\left(\partial_{\rho}A_{\sigma}-\partial_{\sigma}A_{\rho}\right)\left(\partial^{\rho}A^{\sigma}-\partial^{\sigma}A^{\rho}\right)\right\} =0$

これらの結果を集めると、次がわかる。

$\partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}A_{\nu}\right)}\right)-\frac{\partial\mathcal{L}}{\partial A_{\nu}}=-\partial_{\mu}F^{\mu\nu}=0$

これを3次元ベクトルの形に直してみよう。

$E^{i}=\left(-\nabla\phi-\frac{\partial\mathbf{A}}{\partial t}\right)^{i}=-\frac{\partial\phi}{\partial x^{i}}-\frac{\partial A^{i}}{\partial t}=\partial^{i}A^{0}-\partial^{0}A^{i}=-F^{0i}$

$\varepsilon^{ijk}B^{k}=\varepsilon^{ijk}\left(\nabla\times\mathbf{A}\right)^{k}=\varepsilon^{ijk}\varepsilon^{klm}\frac{\partial}{\partial x^{l}}A^{m}=\left(\delta^{il}\delta^{jm}-\delta^{im}\delta^{jl}\right)\frac{\partial}{\partial x^{l}}A^{m}$
$=-\partial^{i}A^{j}+\partial^{j}A^{i}=-F^{ij}$