HW8 Answer

1 textbook

拉格朗日方程是

\underset{- J + \nabla \times \frac{B}{2 μ_{0}}}{\underset{⏟}{\frac{\partial L}{\partial B}}} - \underset{ϵ_{0} \partial_{t} E}{\underset{⏟}{\frac{\partial}{\partial t} (\frac{\partial L}{\partial (\partial_{t} B)})}} - \underset{- \frac{\nabla \times B}{2 μ_{0}}}{\underset{⏟}{\partial_{i} (\frac{\partial L}{\partial (\partial_{i} B)})}} = 0

\underset{ϵ_{0} \partial_{t} B + \frac{ϵ_{0}}{2} \nabla \times E}{\underset{⏟}{\frac{\partial L}{\partial E}}} - \underset{0}{\underset{⏟}{\frac{\partial}{\partial t} (\frac{\partial L}{\partial (\partial_{t} E)})}} - \underset{- \frac{ϵ_{0}}{2} \nabla \times E}{\underset{⏟}{\partial_{i} (\frac{\partial L}{\partial (\partial_{i} E)})}} = 0

得到两个旋度方程

\nabla \times E = - \frac{\partial B}{\partial t}

\nabla \times B = μ_{0} J + \frac{1}{c^{2}} \frac{\partial E}{\partial t}

参见 chapter 4-E 节的 ppt，拉格朗日方程是

\underset{- \frac{1}{2} κ^{2} φ}{\underset{⏟}{\frac{\partial L}{\partial φ^{*}}}} - \underset{- \frac{1}{2} \partial^{μ} φ}{\underset{⏟}{\partial_{μ} (\frac{\partial L}{\partial (\partial_{μ} φ^{*})})}} = 0, ϕ = φ, φ^{*}

取 $ϕ = φ^{*}$ 得到 $φ$ 的 Klein-Gordon 方程

(◻ - κ^{2}) φ = 0

代入正则能动张量密度

T^{μ ν} = g^{μ ν} L - \partial^{μ} φ \underset{- \frac{1}{2} \partial^{ν} φ^{*}}{\underset{⏟}{\frac{\partial L}{\partial (\partial_{ν} φ)}}} - \partial^{μ} φ^{*} \underset{- \frac{1}{2} \partial^{ν} φ}{\underset{⏟}{\frac{\partial L}{\partial (\partial_{ν} φ^{*})}}}

得到

T^{μ ν} = - \frac{1}{2} g^{μ ν} (\partial_{ρ} φ^{*} \partial^{ρ} φ + κ^{2} φ^{*} φ) + \frac{1}{2} (\partial^{μ} φ \partial^{ν} φ^{*} + \partial^{μ} φ^{*} \partial^{ν} φ)