解析力学から様々な理論分野を組み立てる 第2部 第6章 §2.6.4 ★★★ ∇ ⋅ B = 0 \nabla\cdot\bm{B}=0 ∇ ⋅ B = 0 F = ∂ A − ∂ A F=\partial A-\partial A F = ∂ A − ∂ A
Bianchi 恒等式
F μ ν = ∂ μ A ν − ∂ ν A μ F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu F μν = ∂ μ A ν − ∂ ν A μ
∂ λ F μ ν + ∂ μ F ν λ + ∂ ν F λ μ = 0 (2.6.17) \partial_\lambda F_{\mu\nu}+\partial_\mu F_{\nu\lambda}+\partial_\nu F_{\lambda\mu}=0\tag{2.6.17} ∂ λ F μν + ∂ μ F ν λ + ∂ ν F λ μ = 0 ( 2.6.17 )
が、A μ A_\mu A μ A A A ∂ λ ∂ μ A ν − ∂ λ ∂ ν A μ + ∂ μ ∂ ν A λ − ∂ μ ∂ λ A ν + ∂ ν ∂ λ A μ − ∂ ν ∂ μ A λ \partial_\lambda\partial_\mu A_\nu-\partial_\lambda\partial_\nu A_\mu+\partial_\mu\partial_\nu A_\lambda-\partial_\mu\partial_\lambda A_\nu+\partial_\nu\partial_\lambda A_\mu-\partial_\nu\partial_\mu A_\lambda ∂ λ ∂ μ A ν − ∂ λ ∂ ν A μ + ∂ μ ∂ ν A λ − ∂ μ ∂ λ A ν + ∂ ν ∂ λ A μ − ∂ ν ∂ μ A λ λ , μ , ν \lambda,\mu,\nu λ , μ , ν F F F 0 0 0 { 0 , 1 , 2 , 3 } \{0,1,2,3\} { 0 , 1 , 2 , 3 }
( λ , μ , ν ) = ( 1 , 2 , 3 ) (\lambda,\mu,\nu)=(1,2,3) ( λ , μ , ν ) = ( 1 , 2 , 3 ) F 23 = − B x F_{23}=-B_x F 23 = − B x F 31 = − B y F_{31}=-B_y F 31 = − B y F 12 = − B z F_{12}=-B_z F 12 = − B z
∂ x F 23 + ∂ y F 31 + ∂ z F 12 = − ∇ ⋅ B = 0. (2.6.18) \partial_xF_{23}+\partial_yF_{31}+\partial_zF_{12}=-\nabla\cdot\bm{B}=0 .\tag{2.6.18} ∂ x F 23 + ∂ y F 31 + ∂ z F 12 = − ∇ ⋅ B = 0. ( 2.6.18 )
( λ , μ , ν ) = ( 0 , 1 , 2 ) (\lambda,\mu,\nu)=(0,1,2) ( λ , μ , ν ) = ( 0 , 1 , 2 ) F 12 = − B z F_{12}=-B_z F 12 = − B z F 20 = − E y / c F_{20}=-E_y/c F 20 = − E y / c F 01 = E x / c F_{01}=E_x/c F 01 = E x / c
1 c ∂ F 12 ∂ t + ∂ x F 20 + ∂ y F 01 = − 1 c ( ∂ B z ∂ t + ∂ x E y − ∂ y E x ) = − 1 c ( ∂ B ∂ t + ∇ × E ) z = 0. (2.6.19) \frac1c\frac{\partial F_{12}}{\partial t}+\partial_xF_{20}+\partial_yF_{01}=-\frac1c\Bigl(\frac{\partial B_z}{\partial t}+\partial_xE_y-\partial_yE_x\Bigr)=-\frac1c\Bigl(\frac{\partial \bm{B}}{\partial t}+\nabla\times\bm{E}\Bigr)_z=0 .\tag{2.6.19} c 1 ∂ t ∂ F 12 + ∂ x F 20 + ∂ y F 01 = − c 1 ( ∂ t ∂ B z + ∂ x E y − ∂ y E x ) = − c 1 ( ∂ t ∂ B + ∇ × E ) z = 0. ( 2.6.19 )
( 0 , 2 , 3 ) (0,2,3) ( 0 , 2 , 3 ) ( 0 , 3 , 1 ) (0,3,1) ( 0 , 3 , 1 ) x x x y y y
∇ ⋅ B = 0 , ∇ × E + ∂ B ∂ t = 0 \nabla\cdot\bm{B}=0,\qquad\nabla\times\bm{E}+\frac{\partial \bm{B}}{\partial t}=0 ∇ ⋅ B = 0 , ∇ × E + ∂ t ∂ B = 0
が出る。磁荷が無いことと Faraday の電磁誘導の法則である。
方程式と恒等式の違い
前の節の2本と、この節の2本は、出どころが違う。前の節の2本は作用 S [ A ] S[A] S [ A ] A A A E , B \bm{E},\bm{B} E , B (2.6.2) で作れば、作用の停留と無関係に、どんな A A A
逆に、∇ ⋅ B = 0 \nabla\cdot\bm{B}=0 ∇ ⋅ B = 0 ∇ × E + ∂ t B = 0 \nabla\times\bm{E}+\partial_t\bm{B}=0 ∇ × E + ∂ t B = 0 E , B \bm{E},\bm{B} E , B (2.6.2) の形に書ける。∇ ⋅ B = 0 \nabla\cdot\bm{B}=0 ∇ ⋅ B = 0 B = ∇ × A \bm{B}=\nabla\times\bm{A} B = ∇ × A ∇ × ( E + ∂ t A ) = 0 \nabla\times(\bm{E}+\partial_t\bm{A})=0 ∇ × ( E + ∂ t A ) = 0 E + ∂ t A = − ∇ ϕ \bm{E}+\partial_t\bm{A}=-\nabla\phi E + ∂ t A = − ∇ ϕ A μ A_\mu A μ ρ m \rho_{\mathrm m} ρ m ∇ ⋅ B ≠ 0 \nabla\cdot\bm{B}\neq0 ∇ ⋅ B = 0 B = ∇ × A \bm{B}=\nabla\times\bm{A} B = ∇ × A
図 2.6.2 Maxwell 方程式の4本の出どころ。左の列は A μ A_\mu A μ
例題 2.6.4 一様に強くなる磁場と誘導電場 ★★
オリジナル ϕ = 0 \phi=0 ϕ = 0 A = B ( t ) 2 ( − y , x , 0 ) \bm{A}=\frac{B(t)}{2}(-y,\,x,\,0) A = 2 B ( t ) ( − y , x , 0 ) B ( t ) = B 0 + B ˙ t B(t)=B_0+\dot Bt B ( t ) = B 0 + B ˙ t B ˙ \dot B B ˙
(1) B \bm{B} B E \bm{E} E (2) ∇ × E + ∂ t B = 0 \nabla\times\bm{E}+\partial_t\bm{B}=0 ∇ × E + ∂ t B = 0 A \bm{A} A (3) B ˙ = 0.1 T s − 1 \dot B=0.1\,\mathrm{T\,s^{-1}} B ˙ = 0.1 T s − 1 z z z r = 5 c m r=5\,\mathrm{cm} r = 5 cm
解答 (1) B = ∇ × A \bm{B}=\nabla\times\bm{A} B = ∇ × A z z z ∂ x A y − ∂ y A x = B 2 + B 2 = B ( t ) \partial_xA_y-\partial_yA_x=\frac B2+\frac B2=B(t) ∂ x A y − ∂ y A x = 2 B + 2 B = B ( t ) x , y x,y x , y A A A z z z A z = 0 A_z=0 A z = 0 0 0 0 B = B ( t ) e z \bm{B}=B(t)\,\bm{e}_z B = B ( t ) e z E = − ∂ t A = − B ˙ 2 ( − y , x , 0 ) \bm{E}=-\partial_t\bm{A}=-\frac{\dot B}{2}(-y,\,x,\,0) E = − ∂ t A = − 2 B ˙ ( − y , x , 0 ) ∣ E ∣ = B ˙ r / 2 \lvert \bm{E}\rvert =\dot B\,r/2 ∣ E ∣ = B ˙ r /2 r = x 2 + y 2 r=\sqrt{x^2+y^2} r = x 2 + y 2 z z z A \bm{A} A
(2) ( ∇ × E ) z = ∂ x E y − ∂ y E x = − B ˙ 2 − B ˙ 2 = − B ˙ (\nabla\times\bm{E})_z=\partial_xE_y-\partial_yE_x=-\frac{\dot B}{2}-\frac{\dot B}{2}=-\dot B ( ∇ × E ) z = ∂ x E y − ∂ y E x = − 2 B ˙ − 2 B ˙ = − B ˙ x , y x,y x , y 0 0 0 ∂ t B = B ˙ e z \partial_t\bm{B}=\dot B\,\bm{e}_z ∂ t B = B ˙ e z 0 0 0 E = − ∇ ϕ − ∂ t A \bm{E}=-\nabla\phi-\partial_t\bm{A} E = − ∇ ϕ − ∂ t A B = ∇ × A \bm{B}=\nabla\times\bm{A} B = ∇ × A ∇ × E = − ∂ t ( ∇ × A ) = − ∂ t B \nabla\times\bm{E}=-\partial_t(\nabla\times\bm{A})=-\partial_t\bm{B} ∇ × E = − ∂ t ( ∇ × A ) = − ∂ t B ∇ × ∇ ϕ = 0 \nabla\times\nabla\phi=0 ∇ × ∇ ϕ = 0 A \bm{A} A A \bm{A} A
(3) ∣ E ∣ = B ˙ r / 2 = 0.1 × 0.05 / 2 = 2.5 × 10 − 3 V m − 1 \lvert \bm{E}\rvert =\dot B r/2=0.1\times0.05/2=2.5\times10^{-3}\,\mathrm{V\,m^{-1}} ∣ E ∣ = B ˙ r /2 = 0.1 × 0.05/2 = 2.5 × 1 0 − 3 V m − 1 ∮ E ⋅ d l \oint\bm{E}\cdot\mathrm{d}\bm{l} ∮ E ⋅ d l 2 π r ⋅ B ˙ r / 2 = π r 2 B ˙ = 7.85 × 10 − 4 V 2\pi r\cdot\dot Br/2=\pi r^2\dot B=7.85\times10^{-4}\,\mathrm V 2 π r ⋅ B ˙ r /2 = π r 2 B ˙ = 7.85 × 1 0 − 4 V π r 2 B \pi r^2B π r 2 B B \bm{B} B
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