京都大学 理学研究科 物理学・宇宙物理学専攻 2016年8月実施 問題II-1 Author 祭音Myyura (refined with GPT 6 Astra)
Description 真空中の Maxwell 方程式は、電場 E \mathbf E E B \mathbf B B ρ \rho ρ j \mathbf j j ϵ 0 \epsilon_0 ϵ 0 μ 0 \mu_0 μ 0
∇ × E + ∂ B ∂ t = 0 (A) \nabla\times\mathbf E+\frac{\partial\mathbf B}{\partial t}=0\tag{A} ∇ × E + ∂ t ∂ B = 0 ( A ) ∇ ⋅ B = 0 (B) \nabla\cdot\mathbf B=0\tag{B} ∇ ⋅ B = 0 ( B ) 1 μ 0 ∇ × B − ϵ 0 ∂ E ∂ t = j (C) \frac1{\mu_0}\nabla\times\mathbf B-\epsilon_0\frac{\partial\mathbf E}{\partial t}=\mathbf j\tag{C} μ 0 1 ∇ × B − ϵ 0 ∂ t ∂ E = j ( C ) ∇ ⋅ E = 1 ϵ 0 ρ (D) \nabla\cdot\mathbf E=\frac1{\epsilon_0}\rho\tag{D} ∇ ⋅ E = ϵ 0 1 ρ ( D ) (1) 物質中において、分極電荷密度を ρ d \rho_d ρ d P \mathbf P P ρ d = − ∇ ⋅ P \rho_d=-\nabla\cdot\mathbf P ρ d = − ∇ ⋅ P j d \mathbf j_d j d
j d = ∂ P ∂ t (E) \mathbf j_d=\frac{\partial\mathbf P}{\partial t}\tag{E} j d = ∂ t ∂ P ( E ) この式から分極電荷密度と分極電流密度において電荷保存則が成り立っていることを示せ。
(2) ここで、分極電荷密度以外の電荷密度を ρ e \rho_e ρ e j e \mathbf j_e j e M \mathbf M M j m \mathbf j_m j m j m = ∇ × M \mathbf j_m=\nabla\times\mathbf M j m = ∇ × M D \mathbf D D H \mathbf H H D = ϵ 0 E + P \mathbf D=\epsilon_0\mathbf E+\mathbf P D = ϵ 0 E + P B = μ 0 ( H + M ) \mathbf B=\mu_0(\mathbf H+\mathbf M) B = μ 0 ( H + M ) ρ e \rho_e ρ e j e \mathbf j_e j e
ここで、電荷 e e e − e -e − e N N N ω \omega ω E 0 E_0 E 0 c c c
(3) 電子の運動方程式は、質量を m m m x x x j j j
m d 2 x d t 2 = − e E 0 sin ( ω t ) (F) m\frac{d^2x}{dt^2}=-eE_0\sin(\omega t)\tag{F} m d t 2 d 2 x = − e E 0 sin ( ω t ) ( F ) (4) ここで、このような真空中の有限領域における電子とイオンの集団を、角振動数 ω \omega ω ϵ ( ω ) \epsilon(\omega) ϵ ( ω ) μ ( ω ) \mu(\omega) μ ( ω ) D = ϵ E \mathbf D=\epsilon\mathbf E D = ϵ E B = μ H \mathbf B=\mu\mathbf H B = μ H ϵ \epsilon ϵ
(5) 次に、電子に変位 x x x − m ω 0 2 x -m\omega_0^2x − m ω 0 2 x
(6) 問 (5) の場合の媒質の屈折率 n n n n 2 n^2 n 2 ω \omega ω ω ≠ ω 0 \omega\ne\omega_0 ω = ω 0 n n n ϵ μ / ( ϵ 0 μ 0 ) \sqrt{\epsilon\mu/(\epsilon_0\mu_0)} ϵ μ / ( ϵ 0 μ 0 )
题目描述
使用电场 E \mathbf E E B \mathbf B B ρ \rho ρ j \mathbf j j ϵ 0 \epsilon_0 ϵ 0 μ 0 \mu_0 μ 0
∇ × E + ∂ B ∂ t = 0 , (A) \nabla\times\mathbf E+\frac{\partial\mathbf B}{\partial t}=0,\tag{A} ∇ × E + ∂ t ∂ B = 0 , ( A ) ∇ ⋅ B = 0 , (B) \nabla\cdot\mathbf B=0,\tag{B} ∇ ⋅ B = 0 , ( B ) 1 μ 0 ∇ × B − ϵ 0 ∂ E ∂ t = j , (C) \frac1{\mu_0}\nabla\times\mathbf B-\epsilon_0\frac{\partial\mathbf E}{\partial t}=\mathbf j,\tag{C} μ 0 1 ∇ × B − ϵ 0 ∂ t ∂ E = j , ( C ) ∇ ⋅ E = 1 ϵ 0 ρ . (D) \nabla\cdot\mathbf E=\frac1{\epsilon_0}\rho.\tag{D} ∇ ⋅ E = ϵ 0 1 ρ . ( D ) (1) 在物质中，设极化电荷密度为 ρ d \rho_d ρ d P \mathbf P P ρ d = − ∇ ⋅ P \rho_d=-\nabla\cdot\mathbf P ρ d = − ∇ ⋅ P j d \mathbf j_d j d
j d = ∂ P ∂ t . (E) \mathbf j_d=\frac{\partial\mathbf P}{\partial t}.\tag{E} j d = ∂ t ∂ P . ( E ) 由此证明极化电荷密度与极化电流密度满足电荷守恒定律。
(2) 设极化电荷以外的电荷密度为 ρ e \rho_e ρ e j e \mathbf j_e j e M \mathbf M M j m \mathbf j_m j m j m = ∇ × M \mathbf j_m=\nabla\times\mathbf M j m = ∇ × M D \mathbf D D H \mathbf H H D = ϵ 0 E + P \mathbf D=\epsilon_0\mathbf E+\mathbf P D = ϵ 0 E + P B = μ 0 ( H + M ) \mathbf B=\mu_0(\mathbf H+\mathbf M) B = μ 0 ( H + M ) ρ e \rho_e ρ e j e \mathbf j_e j e
考虑一个有限区域，其中每单位体积内分别存在 N N N e e e N N N − e -e − e ω \omega ω E 0 E_0 E 0 c c c
(3) 电子质量为 m m m x x x
m d 2 x d t 2 = − e E 0 sin ( ω t ) . (F) m\frac{d^2x}{dt^2}=-eE_0\sin(\omega t).\tag{F} m d t 2 d 2 x = − e E 0 sin ( ω t ) . ( F ) 求电流密度 j j j
(4) 将真空中此类有限区域内的电子和离子群体看作介电常数为 ϵ ( ω ) \epsilon(\omega) ϵ ( ω ) μ ( ω ) \mu(\omega) μ ( ω ) ω \omega ω D = ϵ E \mathbf D=\epsilon\mathbf E D = ϵ E B = μ H \mathbf B=\mu\mathbf H B = μ H ϵ \epsilon ϵ
(5) 接下来，电子还受到与位移 x x x − m ω 0 2 x -m\omega_0^2x − m ω 0 2 x
(6) 求第 (5) 问中介质的折射率 n n n n 2 n^2 n 2 ω \omega ω ω ≠ ω 0 \omega\ne\omega_0 ω = ω 0 n = ϵ μ / ( ϵ 0 μ 0 ) n=\sqrt{\epsilon\mu/(\epsilon_0\mu_0)} n = ϵ μ / ( ϵ 0 μ 0 )
Kai (1)
∂ t ρ d + ∇ ⋅ j d = − ∂ t ∇ ⋅ P + ∇ ⋅ ∂ t P = 0. \partial_t\rho_d+\nabla\cdot\mathbf j_d
=-\partial_t\nabla\cdot\mathbf P+\nabla\cdot\partial_t\mathbf P=0. ∂ t ρ d + ∇ ⋅ j d = − ∂ t ∇ ⋅ P + ∇ ⋅ ∂ t P = 0. (2)
ρ = ρ e + ρ d \rho=\rho_e+\rho_d ρ = ρ e + ρ d j = j e + ∂ t P + ∇ × M \mathbf j=\mathbf j_e+\partial_t\mathbf P+\nabla\times\mathbf M j = j e + ∂ t P + ∇ × M
∇ × E = − ∂ t B , ∇ ⋅ B = 0 , ∇ × H = j e + ∂ t D , ∇ ⋅ D = ρ e . \boxed{\begin{aligned}
\nabla\times\mathbf E&=-\partial_t\mathbf B,\\
\nabla\cdot\mathbf B&=0,\\
\nabla\times\mathbf H&=\mathbf j_e+\partial_t\mathbf D,\\
\nabla\cdot\mathbf D&=\rho_e.
\end{aligned}} ∇ × E ∇ ⋅ B ∇ × H ∇ ⋅ D = − ∂ t B , = 0 , = j e + ∂ t D , = ρ e . (3)
並進成分を除く解は x = e E 0 sin ω t / ( m ω 2 ) x=eE_0\sin\omega t/(m\omega^2) x = e E 0 sin ω t / ( m ω 2 )
j = − N e x ˙ = − N e 2 E 0 m ω cos ω t . \boxed{j=-Ne\dot x=-\frac{Ne^2E_0}{m\omega}\cos\omega t.} j = − N e x ˙ = − mω N e 2 E 0 cos ω t . (4)
イオンは静止しているので P = − N e x P=-Nex P = − N e x
D = ϵ 0 E + P = ( ϵ 0 − N e 2 m ω 2 ) E , ϵ ( ω ) = ϵ 0 − N e 2 m ω 2 . D=\epsilon_0E+P=\left(\epsilon_0-\frac{Ne^2}{m\omega^2}\right)E,
\qquad
\boxed{\epsilon(\omega)=\epsilon_0-\frac{Ne^2}{m\omega^2}.} D = ϵ 0 E + P = ( ϵ 0 − m ω 2 N e 2 ) E , ϵ ( ω ) = ϵ 0 − m ω 2 N e 2 . (5)
x ¨ + ω 0 2 x = − e E 0 m sin ω t . \boxed{\ddot x+\omega_0^2x=-\frac{eE_0}{m}\sin\omega t.} x ¨ + ω 0 2 x = − m e E 0 sin ω t . (6)
周波数 ω \omega ω
x = e E 0 m ( ω 2 − ω 0 2 ) sin ω t , ϵ = ϵ 0 − N e 2 m ( ω 2 − ω 0 2 ) . x=\frac{eE_0}{m(\omega^2-\omega_0^2)}\sin\omega t,
\qquad \epsilon=\epsilon_0-\frac{Ne^2}{m(\omega^2-\omega_0^2)}. x = m ( ω 2 − ω 0 2 ) e E 0 sin ω t , ϵ = ϵ 0 − m ( ω 2 − ω 0 2 ) N e 2 . ω p 2 = N e 2 / ( m ϵ 0 ) \omega_p^2=Ne^2/(m\epsilon_0) ω p 2 = N e 2 / ( m ϵ 0 )
n 2 = μ ( ω ) μ 0 ( 1 − ω p 2 ω 2 − ω 0 2 ) . \boxed{n^2=\frac{\mu(\omega)}{\mu_0}\left(1-\frac{\omega_p^2}{\omega^2-\omega_0^2}\right).} n 2 = μ 0 μ ( ω ) ( 1 − ω 2 − ω 0 2 ω p 2 ) . 本問の電気双極子近似では磁気応答を無視して μ ≃ μ 0 \mu\simeq\mu_0 μ ≃ μ 0 n 2 ( 0 ) = 1 + ω p 2 / ω 0 2 n^2(0)=1+\omega_p^2/\omega_0^2 n 2 ( 0 ) = 1 + ω p 2 / ω 0 2 ω → ω 0 − 0 \omega\to\omega_0-0 ω → ω 0 − 0 + ∞ +\infty + ∞ ω → ω 0 + 0 \omega\to\omega_0+0 ω → ω 0 + 0 − ∞ -\infty − ∞ ω = ω 0 2 + ω p 2 \omega=\sqrt{\omega_0^2+\omega_p^2} ω = ω 0 2 + ω p 2 ω → ∞ \omega\to\infty ω → ∞ 1 1 1 n 2 < 0 n^2<0 n 2 < 0
Reference 