4時間で4問という構成の試験です。第1問は Aharonov–Casher のゼロモード数え上げを、演算子代数から正則関数の議論まで誘導つきで辿らせる問題で、最後は磁束量子数の整数部分という位相的な答えに落ちます。第2問と第3問は基礎的な統計力学と相対論ですが、第3問後半の4元運動量保存は微小量の扱いを丁寧にやらないと合いません。選択問題の3問はいずれも実験に近く、桁の評価とグラフの読み取りが得点を分けます。
問題 分野 主題 第1問 量子力学・複素解析 磁場中の2次元電子とゼロエネルギー束縛状態 第2問 統計力学 調和振動子系・1次元連成振動子系の内部エネルギーと比熱 第3問 力学・相対論 相対論的ロケットの質量減少 第4問 電磁気学・波動 プラズマ中の電磁波と干渉計による電子密度計測 第5問 原子核 高エネルギー電子弾性散乱と原子核の荷電半径 第6問 電気回路 フィルタ回路・整流回路とダイオードの非線形性
第1問から第3問が必答で、第4問から第6問のうち1問を選択する形式ですが、ここでは全6問の解答を載せます。
x y xy x y z z z R R R b > 0 b>0 b > 0 0 0 0
B z ( x , y ) = { b > 0 , ( r < R ) , 0 , ( r ≥ R ) , r = x 2 + y 2 B_z(x,y)=\begin{cases} b>0, & (r<R),\\ 0, & (r\ge R),\end{cases}\qquad r=\sqrt{x^2+y^2} B z ( x , y ) = { b > 0 , 0 , ( r < R ) , ( r ≥ R ) , r = x 2 + y 2 波動関数は 2 成分 Ψ = ( ψ ↑ , ψ ↓ ) T \Psi=(\psi_\uparrow,\psi_\downarrow)^{\mathrm T} Ψ = ( ψ ↑ , ψ ↓ ) T
H = p x 2 + p y 2 2 m 1 2 × 2 + ℏ e B z ( x , y ) 2 m σ z , p x = − i ℏ ∂ ∂ x + e A x , p y = − i ℏ ∂ ∂ y + e A y H=\frac{p_x^{\,2}+p_y^{\,2}}{2m}\mathbf{1}_{2\times2}+\frac{\hbar e B_z(x,y)}{2m}\sigma_z,\qquad
p_x=-i\hbar\frac{\partial}{\partial x}+eA_x,\quad p_y=-i\hbar\frac{\partial}{\partial y}+eA_y H = 2 m p x 2 + p y 2 1 2 × 2 + 2 m ℏ e B z ( x , y ) σ z , p x = − i ℏ ∂ x ∂ + e A x , p y = − i ℏ ∂ y ∂ + e A y です。− e < 0 -e<0 − e < 0 B z = ∂ x A y − ∂ y A x B_z=\partial_x A_y-\partial_y A_x B z = ∂ x A y − ∂ y A x E = 0 E=0 E = 0
パウリ行列を直接掛けると
σ x σ y = ( i 0 0 − i ) = i σ z , σ y σ x = − i σ z \sigma_x\sigma_y=\begin{pmatrix}i&0\\0&-i\end{pmatrix}=i\sigma_z,\qquad
\sigma_y\sigma_x=-i\sigma_z σ x σ y = ( i 0 0 − i ) = i σ z , σ y σ x = − i σ z なので [ σ x , σ y ] = 2 i σ z [\sigma_x,\sigma_y]=2i\sigma_z [ σ x , σ y ] = 2 i σ z
[ S x , S y ] = ℏ 2 4 [ σ x , σ y ] = ℏ 2 4 ⋅ 2 i σ z = i ℏ ⋅ ℏ 2 σ z = i ℏ S z . [S_x,S_y]=\frac{\hbar^2}{4}[\sigma_x,\sigma_y]=\frac{\hbar^2}{4}\cdot 2i\sigma_z=i\hbar\cdot\frac{\hbar}{2}\sigma_z=i\hbar S_z . [ S x , S y ] = 4 ℏ 2 [ σ x , σ y ] = 4 ℏ 2 ⋅ 2 i σ z = i ℏ ⋅ 2 ℏ σ z = i ℏ S z . [ H , S z ] [H,S_z] [ H , S z ] H H H σ z \sigma_z σ z B z ( x , y ) B_z(x,y) B z ( x , y ) σ z \sigma_z σ z σ z \sigma_z σ z p x , p y p_x,p_y p x , p y A x , A y A_x,A_y A x , A y
[ H , S z ] = 0. [H,S_z]=0 . [ H , S z ] = 0. 答えは [ S x , S y ] = i ℏ S z [S_x,S_y]=i\hbar S_z [ S x , S y ] = i ℏ S z [ H , S z ] = 0 [H,S_z]=0 [ H , S z ] = 0 S z S_z S z
D = p x σ x + p y σ y D=p_x\sigma_x+p_y\sigma_y D = p x σ x + p y σ y p x , p y p_x,p_y p x , p y σ z \sigma_z σ z σ x σ z = − σ z σ x \sigma_x\sigma_z=-\sigma_z\sigma_x σ x σ z = − σ z σ x σ y σ z = − σ z σ y \sigma_y\sigma_z=-\sigma_z\sigma_y σ y σ z = − σ z σ y
D σ z = p x σ x σ z + p y σ y σ z = − σ z ( p x σ x + p y σ y ) = − σ z D . D\sigma_z=p_x\sigma_x\sigma_z+p_y\sigma_y\sigma_z=-\sigma_z(p_x\sigma_x+p_y\sigma_y)=-\sigma_z D . D σ z = p x σ x σ z + p y σ y σ z = − σ z ( p x σ x + p y σ y ) = − σ z D . 次に D 2 D^2 D 2 σ x 2 = σ y 2 = 1 \sigma_x^2=\sigma_y^2=\mathbf{1} σ x 2 = σ y 2 = 1 σ x σ y = i σ z \sigma_x\sigma_y=i\sigma_z σ x σ y = i σ z σ y σ x = − i σ z \sigma_y\sigma_x=-i\sigma_z σ y σ x = − i σ z
D 2 = p x 2 σ x 2 + p y 2 σ y 2 + p x p y σ x σ y + p y p x σ y σ x = ( p x 2 + p y 2 ) 1 2 × 2 + i [ p x , p y ] σ z . \begin{aligned}
D^2&=p_x^{\,2}\sigma_x^2+p_y^{\,2}\sigma_y^2+p_xp_y\,\sigma_x\sigma_y+p_yp_x\,\sigma_y\sigma_x\\
&=(p_x^{\,2}+p_y^{\,2})\mathbf{1}_{2\times2}+i\,[p_x,p_y]\,\sigma_z .
\end{aligned} D 2 = p x 2 σ x 2 + p y 2 σ y 2 + p x p y σ x σ y + p y p x σ y σ x = ( p x 2 + p y 2 ) 1 2 × 2 + i [ p x , p y ] σ z . 交換子 は、任意の関数への作用を見れば
[ p x , p y ] = [ − i ℏ ∂ x , e A y ] + [ e A x , − i ℏ ∂ y ] = − i ℏ e ( ∂ A y ∂ x − ∂ A x ∂ y ) = − i ℏ e B z [p_x,p_y]=[-i\hbar\partial_x,\,eA_y]+[eA_x,\,-i\hbar\partial_y]
=-i\hbar e\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)=-i\hbar e B_z [ p x , p y ] = [ − i ℏ ∂ x , e A y ] + [ e A x , − i ℏ ∂ y ] = − i ℏ e ( ∂ x ∂ A y − ∂ y ∂ A x ) = − i ℏ e B z です。これを代入すると i [ p x , p y ] σ z = ℏ e B z σ z i[p_x,p_y]\sigma_z=\hbar e B_z\sigma_z i [ p x , p y ] σ z = ℏ e B z σ z
D 2 = ( p x 2 + p y 2 ) 1 2 × 2 + ℏ e B z σ z = 2 m H , D^2=(p_x^{\,2}+p_y^{\,2})\mathbf{1}_{2\times2}+\hbar e B_z\sigma_z=2mH, D 2 = ( p x 2 + p y 2 ) 1 2 × 2 + ℏ e B z σ z = 2 m H , すなわち H = D 2 / ( 2 m ) H=D^2/(2m) H = D 2 / ( 2 m )
∣ Φ n ⟩ = D ∣ Ψ n ⟩ |\Phi_n\rangle=D|\Psi_n\rangle ∣ Φ n ⟩ = D ∣ Ψ n ⟩ D D D
⟨ Φ n ∣ Φ n ⟩ = ⟨ Ψ n ∣ D † D ∣ Ψ n ⟩ = ⟨ Ψ n ∣ D 2 ∣ Ψ n ⟩ = 2 m ⟨ Ψ n ∣ H ∣ Ψ n ⟩ = 2 m E n ⟨ Ψ n ∣ Ψ n ⟩ . \langle\Phi_n|\Phi_n\rangle=\langle\Psi_n|D^\dagger D|\Psi_n\rangle=\langle\Psi_n|D^2|\Psi_n\rangle
=2m\langle\Psi_n|H|\Psi_n\rangle=2mE_n\langle\Psi_n|\Psi_n\rangle . ⟨ Φ n ∣ Φ n ⟩ = ⟨ Ψ n ∣ D † D ∣ Ψ n ⟩ = ⟨ Ψ n ∣ D 2 ∣ Ψ n ⟩ = 2 m ⟨ Ψ n ∣ H ∣ Ψ n ⟩ = 2 m E n ⟨ Ψ n ∣ Ψ n ⟩ . 3番目の等号で式 (9)、最後で H ∣ Ψ n ⟩ = E n ∣ Ψ n ⟩ H|\Psi_n\rangle=E_n|\Psi_n\rangle H ∣ Ψ n ⟩ = E n ∣ Ψ n ⟩
設問3の等式を E n E_n E n
E n = ⟨ Φ n ∣ Φ n ⟩ 2 m ⟨ Ψ n ∣ Ψ n ⟩ E_n=\frac{\langle\Phi_n|\Phi_n\rangle}{2m\langle\Psi_n|\Psi_n\rangle} E n = 2 m ⟨ Ψ n ∣ Ψ n ⟩ ⟨ Φ n ∣ Φ n ⟩ です。内積は正定値なので分子は ≥ 0 \ge 0 ≥ 0 > 0 >0 > 0 ∣ Ψ n ⟩ |\Psi_n\rangle ∣ Ψ n ⟩ E n ≥ 0 E_n\ge0 E n ≥ 0 E n ≠ 0 E_n\ne0 E n = 0 E n > 0 E_n>0 E n > 0
次に ∣ Ψ n ⟩ |\Psi_n\rangle ∣ Ψ n ⟩ H ∣ Ψ n ⟩ = E n ∣ Ψ n ⟩ H|\Psi_n\rangle=E_n|\Psi_n\rangle H ∣ Ψ n ⟩ = E n ∣ Ψ n ⟩ E n > 0 E_n>0 E n > 0 σ z ∣ Ψ n ⟩ = + ∣ Ψ n ⟩ \sigma_z|\Psi_n\rangle=+|\Psi_n\rangle σ z ∣ Ψ n ⟩ = + ∣ Ψ n ⟩ ∣ Φ n ⟩ = D ∣ Ψ n ⟩ |\Phi_n\rangle=D|\Psi_n\rangle ∣ Φ n ⟩ = D ∣ Ψ n ⟩ ⟨ Φ n ∣ Φ n ⟩ = 2 m E n ⟨ Ψ n ∣ Ψ n ⟩ > 0 \langle\Phi_n|\Phi_n\rangle=2mE_n\langle\Psi_n|\Psi_n\rangle>0 ⟨ Φ n ∣ Φ n ⟩ = 2 m E n ⟨ Ψ n ∣ Ψ n ⟩ > 0 H = D 2 / ( 2 m ) H=D^2/(2m) H = D 2 / ( 2 m ) D D D
H ∣ Φ n ⟩ = D 2 2 m D ∣ Ψ n ⟩ = D D 2 2 m ∣ Ψ n ⟩ = D H ∣ Ψ n ⟩ = E n ∣ Φ n ⟩ , H|\Phi_n\rangle=\frac{D^2}{2m}D|\Psi_n\rangle=D\,\frac{D^2}{2m}|\Psi_n\rangle=D\,H|\Psi_n\rangle=E_n|\Phi_n\rangle, H ∣ Φ n ⟩ = 2 m D 2 D ∣ Ψ n ⟩ = D 2 m D 2 ∣ Ψ n ⟩ = D H ∣ Ψ n ⟩ = E n ∣ Φ n ⟩ , エネルギーは変わりません。一方、式 (8) の反交換関係から
σ z ∣ Φ n ⟩ = σ z D ∣ Ψ n ⟩ = − D σ z ∣ Ψ n ⟩ = − D ∣ Ψ n ⟩ = − ∣ Φ n ⟩ \sigma_z|\Phi_n\rangle=\sigma_z D|\Psi_n\rangle=-D\sigma_z|\Psi_n\rangle=-D|\Psi_n\rangle=-|\Phi_n\rangle σ z ∣ Φ n ⟩ = σ z D ∣ Ψ n ⟩ = − D σ z ∣ Ψ n ⟩ = − D ∣ Ψ n ⟩ = − ∣ Φ n ⟩ となり、σ z \sigma_z σ z − 1 -1 − 1 ∣ Φ n ⟩ |\Phi_n\rangle ∣ Φ n ⟩ E n E_n E n σ z = − 1 \sigma_z=-1 σ z = − 1 E n > 0 E_n>0 E n > 0 D D D
E = 0 E=0 E = 0 ⟨ Φ ∣ Φ ⟩ = 0 \langle\Phi|\Phi\rangle=0 ⟨ Φ∣Φ ⟩ = 0
D ∣ Ψ ⟩ = 0. D|\Psi\rangle=0 . D ∣Ψ ⟩ = 0. D D D
D = p x σ x + p y σ y = ( 0 p x − i p y p x + i p y 0 ) D=p_x\sigma_x+p_y\sigma_y=\begin{pmatrix}0& p_x-ip_y\\ p_x+ip_y&0\end{pmatrix} D = p x σ x + p y σ y = ( 0 p x + i p y p x − i p y 0 ) なので、D Ψ = 0 D\Psi=0 D Ψ = 0
( p x + i p y ) ψ ↑ = 0 , ( p x − i p y ) ψ ↓ = 0 (p_x+ip_y)\psi_\uparrow=0,\qquad (p_x-ip_y)\psi_\downarrow=0 ( p x + i p y ) ψ ↑ = 0 , ( p x − i p y ) ψ ↓ = 0 になります。ここで A x = − ∂ y ρ A_x=-\partial_y\rho A x = − ∂ y ρ A y = + ∂ x ρ A_y=+\partial_x\rho A y = + ∂ x ρ B z = ∂ x A y − ∂ y A x = ∇ 2 ρ B_z=\partial_xA_y-\partial_yA_x=\nabla^2\rho B z = ∂ x A y − ∂ y A x = ∇ 2 ρ A x + i A y = i ( ∂ x + i ∂ y ) ρ A_x+iA_y=i(\partial_x+i\partial_y)\rho A x + i A y = i ( ∂ x + i ∂ y ) ρ A x − i A y = − i ( ∂ x − i ∂ y ) ρ A_x-iA_y=-i(\partial_x-i\partial_y)\rho A x − i A y = − i ( ∂ x − i ∂ y ) ρ
p x + i p y = − i ℏ ( ∂ x + i ∂ y ) + e ( A x + i A y ) = − i ℏ [ ( ∂ x + i ∂ y ) − e ℏ ( ∂ x + i ∂ y ) ρ ] , p x − i p y = − i ℏ ( ∂ x − i ∂ y ) + e ( A x − i A y ) = − i ℏ [ ( ∂ x − i ∂ y ) + e ℏ ( ∂ x − i ∂ y ) ρ ] . \begin{aligned}
p_x+ip_y&=-i\hbar(\partial_x+i\partial_y)+e(A_x+iA_y)=-i\hbar\left[(\partial_x+i\partial_y)-\frac{e}{\hbar}(\partial_x+i\partial_y)\rho\right],\\
p_x-ip_y&=-i\hbar(\partial_x-i\partial_y)+e(A_x-iA_y)=-i\hbar\left[(\partial_x-i\partial_y)+\frac{e}{\hbar}(\partial_x-i\partial_y)\rho\right].
\end{aligned} p x + i p y p x − i p y = − i ℏ ( ∂ x + i ∂ y ) + e ( A x + i A y ) = − i ℏ [ ( ∂ x + i ∂ y ) − ℏ e ( ∂ x + i ∂ y ) ρ ] , = − i ℏ ( ∂ x − i ∂ y ) + e ( A x − i A y ) = − i ℏ [ ( ∂ x − i ∂ y ) + ℏ e ( ∂ x − i ∂ y ) ρ ] . そこで ψ ↑ = f ↑ exp [ ( e / ℏ ) ρ ] \psi_\uparrow=f_\uparrow\exp[(e/\hbar)\rho] ψ ↑ = f ↑ exp [( e /ℏ ) ρ ] ψ ↓ = f ↓ exp [ − ( e / ℏ ) ρ ] \psi_\downarrow=f_\downarrow\exp[-(e/\hbar)\rho] ψ ↓ = f ↓ exp [ − ( e /ℏ ) ρ ] ρ \rho ρ
( p x + i p y ) ψ ↑ = − i ℏ [ ( ∂ x + i ∂ y ) f ↑ ] e e ℏ ρ , ( p x − i p y ) ψ ↓ = − i ℏ [ ( ∂ x − i ∂ y ) f ↓ ] e − e ℏ ρ (p_x+ip_y)\psi_\uparrow=-i\hbar\left[(\partial_x+i\partial_y)f_\uparrow\right]e^{\frac{e}{\hbar}\rho},\qquad
(p_x-ip_y)\psi_\downarrow=-i\hbar\left[(\partial_x-i\partial_y)f_\downarrow\right]e^{-\frac{e}{\hbar}\rho} ( p x + i p y ) ψ ↑ = − i ℏ [ ( ∂ x + i ∂ y ) f ↑ ] e ℏ e ρ , ( p x − i p y ) ψ ↓ = − i ℏ [ ( ∂ x − i ∂ y ) f ↓ ] e − ℏ e ρ となります。よって求める方程式は
( ∂ ∂ x + i ∂ ∂ y ) f ↑ ( x , y ) = 0 , ( ∂ ∂ x − i ∂ ∂ y ) f ↓ ( x , y ) = 0 \left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)f_\uparrow(x,y)=0,\qquad
\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)f_\downarrow(x,y)=0 ( ∂ x ∂ + i ∂ y ∂ ) f ↑ ( x , y ) = 0 , ( ∂ x ∂ − i ∂ y ∂ ) f ↓ ( x , y ) = 0 です。ゲージ場が指数因子に押し込められ、残るのは自由な微分方程式になります。
w = x + i y w=x+iy w = x + i y w ‾ = x − i y \overline w=x-iy w = x − i y
∂ ∂ w ‾ = 1 2 ( ∂ ∂ x + i ∂ ∂ y ) , ∂ ∂ w = 1 2 ( ∂ ∂ x − i ∂ ∂ y ) \frac{\partial}{\partial \overline w}=\frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right),\qquad
\frac{\partial}{\partial w}=\frac12\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right) ∂ w ∂ = 2 1 ( ∂ x ∂ + i ∂ y ∂ ) , ∂ w ∂ = 2 1 ( ∂ x ∂ − i ∂ y ∂ ) なので、上の 2 式はそれぞれ ∂ f ↑ / ∂ w ‾ = 0 \partial f_\uparrow/\partial\overline w=0 ∂ f ↑ / ∂ w = 0 ∂ f ↓ / ∂ w = 0 \partial f_\downarrow/\partial w=0 ∂ f ↓ / ∂ w = 0 f = u + i v f=u+iv f = u + i v ( ∂ x + i ∂ y ) f = 0 (\partial_x+i\partial_y)f=0 ( ∂ x + i ∂ y ) f = 0
∂ u ∂ x = ∂ v ∂ y , ∂ v ∂ x = − ∂ u ∂ y \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\qquad \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y} ∂ x ∂ u = ∂ y ∂ v , ∂ x ∂ v = − ∂ y ∂ u すなわち w = x + i y w=x+iy w = x + i y f ↑ f_\uparrow f ↑ w w w ( ∂ x − i ∂ y ) f ↓ = 0 (\partial_x-i\partial_y)f_\downarrow=0 ( ∂ x − i ∂ y ) f ↓ = 0 w ‾ = x − i y \overline w=x-iy w = x − i y f ↓ f_\downarrow f ↓ w ‾ \overline w w
r ≥ R r\ge R r ≥ R ρ = ( b R 2 / 2 ) ( 1 / 2 + log ( r / R ) ) \rho=(bR^2/2)\left(1/2+\log(r/R)\right) ρ = ( b R 2 /2 ) ( 1/2 + log ( r / R ) ) ν \nu ν
ν ≡ e b R 2 2 ℏ = e Φ 2 π ℏ = Φ h / e , Φ ≡ π R 2 b \nu\equiv\frac{ebR^2}{2\hbar}=\frac{e\Phi}{2\pi\hbar}=\frac{\Phi}{h/e},\qquad \Phi\equiv\pi R^2 b ν ≡ 2ℏ e b R 2 = 2 π ℏ e Φ = h / e Φ , Φ ≡ π R 2 b （Φ \Phi Φ h / e h/e h / e
e e ℏ ρ = e ν / 2 ( r R ) ν , e − e ℏ ρ = e − ν / 2 ( r R ) − ν ( r ≥ R ) e^{\frac{e}{\hbar}\rho}=e^{\nu/2}\left(\frac{r}{R}\right)^{\nu},\qquad
e^{-\frac{e}{\hbar}\rho}=e^{-\nu/2}\left(\frac{r}{R}\right)^{-\nu}\qquad (r\ge R) e ℏ e ρ = e ν /2 ( R r ) ν , e − ℏ e ρ = e − ν /2 ( R r ) − ν ( r ≥ R ) です。b > 0 b>0 b > 0 e > 0 e>0 e > 0 ν > 0 \nu>0 ν > 0 r → ∞ r\to\infty r → ∞ r ν r^{\nu} r ν r − ν r^{-\nu} r − ν
スピン上向きの場合を考えます。f ↑ f_\uparrow f ↑ w w w f ↑ = ∑ k ≥ 0 c k w k f_\uparrow=\sum_{k\ge0}c_k w^k f ↑ = ∑ k ≥ 0 c k w k w k = r k e i k θ w^k=r^ke^{ik\theta} w k = r k e ik θ k k k
∫ ∣ ψ ↑ ∣ 2 d 2 r = 2 π ∑ k ∣ c k ∣ 2 ∫ 0 ∞ r 2 k e 2 e ℏ ρ r d r \int |\psi_\uparrow|^2\,d^2r = 2\pi\sum_k |c_k|^2\int_0^\infty r^{2k}e^{\frac{2e}{\hbar}\rho}\,r\,dr ∫ ∣ ψ ↑ ∣ 2 d 2 r = 2 π k ∑ ∣ c k ∣ 2 ∫ 0 ∞ r 2 k e ℏ 2 e ρ r d r で、r → ∞ r\to\infty r → ∞ r 2 k + 2 ν + 1 r^{2k+2\nu+1} r 2 k + 2 ν + 1 k ≥ 0 k\ge0 k ≥ 0 ν > 0 \nu>0 ν > 0 c k c_k c k 0 0 0 f ↑ ≡ 0 f_\uparrow\equiv0 f ↑ ≡ 0
スピン下向きでは f ↓ = ∑ k ≥ 0 c k w ‾ k f_\downarrow=\sum_{k\ge0}c_k\overline w^{\,k} f ↓ = ∑ k ≥ 0 c k w k w ‾ k = r k e − i k θ \overline w^{\,k}=r^k e^{-ik\theta} w k = r k e − ik θ k k k
∫ ∣ ψ ↓ ∣ 2 d 2 r = 2 π ∑ k ∣ c k ∣ 2 ∫ 0 ∞ r 2 k e − 2 e ℏ ρ r d r \int|\psi_\downarrow|^2 d^2r=2\pi\sum_k |c_k|^2\int_0^\infty r^{2k}e^{-\frac{2e}{\hbar}\rho}\,r\,dr ∫ ∣ ψ ↓ ∣ 2 d 2 r = 2 π k ∑ ∣ c k ∣ 2 ∫ 0 ∞ r 2 k e − ℏ 2 e ρ r d r となり、各単項式が別々に規格化可能である必要があります。r → ∞ r\to\infty r → ∞ r 2 k − 2 ν + 1 r^{2k-2\nu+1} r 2 k − 2 ν + 1 2 k − 2 ν + 1 < − 1 2k-2\nu+1<-1 2 k − 2 ν + 1 < − 1
k < ν − 1 = e b R 2 2 ℏ − 1. k<\nu-1=\frac{ebR^2}{2\hbar}-1 . k < ν − 1 = 2ℏ e b R 2 − 1. （2 k − 2 ν + 1 = − 1 2k-2\nu+1=-1 2 k − 2 ν + 1 = − 1 ∫ d r / r \int dr/r ∫ d r / r e − e b r 2 / ( 2 ℏ ) e^{-ebr^2/(2\hbar)} e − e b r 2 / ( 2ℏ ) r 2 k + 1 r^{2k+1} r 2 k + 1 f ↓ = 1 , w ‾ , w ‾ 2 , … f_\downarrow=1,\overline w,\overline w^{\,2},\dots f ↓ = 1 , w , w 2 , … k < ν − 1 k<\nu-1 k < ν − 1
N 0 = ⌈ ν ⌉ − 1 = ⌈ e b R 2 2 ℏ ⌉ − 1 ( ν > 1 ) , N 0 = 0 ( ν ≤ 1 ) N_0=\lceil \nu\rceil-1=\left\lceil \frac{ebR^2}{2\hbar}\right\rceil-1\qquad(\nu>1),\qquad N_0=0\quad(\nu\le1) N 0 = ⌈ ν ⌉ − 1 = ⌈ 2ℏ e b R 2 ⌉ − 1 ( ν > 1 ) , N 0 = 0 ( ν ≤ 1 ) です。言い換えると、ν \nu ν N 0 = ⌊ ν ⌋ N_0=\lfloor\nu\rfloor N 0 = ⌊ ν ⌋ N 0 = ν − 1 N_0=\nu-1 N 0 = ν − 1 ν = Φ / ( h / e ) \nu=\Phi/(h/e) ν = Φ/ ( h / e ) h / e h/e h / e ρ \rho ρ b b b
自然長 a a a m ω 2 m\omega^2 m ω 2 m m m T T T N N N
H = ∑ j = 1 N ( p j 2 2 m + m ω 2 2 ( x j − a ) 2 ) H=\sum_{j=1}^{N}\left(\frac{p_j^2}{2m}+\frac{m\omega^2}{2}(x_j-a)^2\right) H = j = 1 ∑ N ( 2 m p j 2 + 2 m ω 2 ( x j − a ) 2 ) に従う場合、後半は N N N N + 1 N+1 N + 1 ( N + 1 ) a (N+1)a ( N + 1 ) a β = 1 / ( k B T ) \beta=1/(k_{\mathrm B}T) β = 1/ ( k B T )
1 質点の古典分配関数 は、u = x 1 − a u=x_1-a u = x 1 − a
z = 1 h ∫ d p e − β p 2 / 2 m ∫ d u e − β m ω 2 u 2 / 2 = 1 h 2 π m β 2 π β m ω 2 = 2 π h β ω z=\frac1h\int dp\,e^{-\beta p^2/2m}\int du\,e^{-\beta m\omega^2u^2/2}
=\frac1h\sqrt{\frac{2\pi m}{\beta}}\sqrt{\frac{2\pi}{\beta m\omega^2}}=\frac{2\pi}{h\beta\omega} z = h 1 ∫ d p e − β p 2 /2 m ∫ d u e − β m ω 2 u 2 /2 = h 1 β 2 π m β m ω 2 2 π = h β ω 2 π です。z ∝ β − 1 z\propto\beta^{-1} z ∝ β − 1
U = − ∂ ln z ∂ β = 1 β = k B T , C = ∂ U ∂ T = k B . U=-\frac{\partial \ln z}{\partial\beta}=\frac1\beta=k_{\mathrm B}T,\qquad
C=\frac{\partial U}{\partial T}=k_{\mathrm B}. U = − ∂ β ∂ ln z = β 1 = k B T , C = ∂ T ∂ U = k B . 運動エネルギーと位置エネルギーがそれぞれ k B T / 2 k_{\mathrm B}T/2 k B T /2
位置の平均は、分布 ∝ e − β m ω 2 ( x 1 − a ) 2 / 2 \propto e^{-\beta m\omega^2(x_1-a)^2/2} ∝ e − β m ω 2 ( x 1 − a ) 2 /2 x 1 = a x_1=a x 1 = a
⟨ x 1 ⟩ = a . \langle x_1\rangle=a . ⟨ x 1 ⟩ = a . 分散は α = β m ω 2 / 2 \alpha=\beta m\omega^2/2 α = β m ω 2 /2
⟨ x 1 2 ⟩ − ⟨ x 1 ⟩ 2 = ⟨ ( x 1 − a ) 2 ⟩ = ∫ u 2 e − α u 2 d u ∫ e − α u 2 d u = 1 2 α = k B T m ω 2 . \langle x_1^2\rangle-\langle x_1\rangle^2=\langle (x_1-a)^2\rangle
=\frac{\int u^2e^{-\alpha u^2}du}{\int e^{-\alpha u^2}du}=\frac1{2\alpha}=\frac{k_{\mathrm B}T}{m\omega^2}. ⟨ x 1 2 ⟩ − ⟨ x 1 ⟩ 2 = ⟨( x 1 − a ) 2 ⟩ = ∫ e − α u 2 d u ∫ u 2 e − α u 2 d u = 2 α 1 = m ω 2 k B T . これも 1 2 m ω 2 ⟨ u 2 ⟩ = 1 2 k B T \frac12 m\omega^2\langle u^2\rangle=\frac12 k_{\mathrm B}T 2 1 m ω 2 ⟨ u 2 ⟩ = 2 1 k B T
量子論では 1 質点のエネルギー準位が ℏ ω ( n + 1 / 2 ) \hbar\omega(n+1/2) ℏ ω ( n + 1/2 )
z Q = ∑ n = 0 ∞ e − β ℏ ω ( n + 1 / 2 ) = e − β ℏ ω / 2 1 − e − β ℏ ω , z_{\mathrm Q}=\sum_{n=0}^{\infty}e^{-\beta\hbar\omega(n+1/2)}=\frac{e^{-\beta\hbar\omega/2}}{1-e^{-\beta\hbar\omega}}, z Q = n = 0 ∑ ∞ e − β ℏ ω ( n + 1/2 ) = 1 − e − β ℏ ω e − β ℏ ω /2 , U Q = − ∂ ln z Q ∂ β = ℏ ω 2 + ℏ ω e ℏ ω / k B T − 1 , U_{\mathrm Q}=-\frac{\partial\ln z_{\mathrm Q}}{\partial\beta}
=\frac{\hbar\omega}{2}+\frac{\hbar\omega}{e^{\hbar\omega/k_{\mathrm B}T}-1}, U Q = − ∂ β ∂ ln z Q = 2 ℏ ω + e ℏ ω / k B T − 1 ℏ ω , C Q = ∂ U Q ∂ T = k B ( ℏ ω k B T ) 2 e ℏ ω / k B T ( e ℏ ω / k B T − 1 ) 2 . C_{\mathrm Q}=\frac{\partial U_{\mathrm Q}}{\partial T}
=k_{\mathrm B}\left(\frac{\hbar\omega}{k_{\mathrm B}T}\right)^2
\frac{e^{\hbar\omega/k_{\mathrm B}T}}{\left(e^{\hbar\omega/k_{\mathrm B}T}-1\right)^2}. C Q = ∂ T ∂ U Q = k B ( k B T ℏ ω ) 2 ( e ℏ ω / k B T − 1 ) 2 e ℏ ω / k B T . 位置については X = x 1 − a X=x_1-a X = x 1 − a X X X X ∣ n ⟩ X|n\rangle X ∣ n ⟩ ∣ n ± 1 ⟩ |n\pm1\rangle ∣ n ± 1 ⟩ ∣ n ⟩ |n\rangle ∣ n ⟩ ⟨ n ∣ X ∣ n ⟩ = 0 \langle n|X|n\rangle=0 ⟨ n ∣ X ∣ n ⟩ = 0
⟨ x 1 ⟩ = a \langle x_1\rangle=a ⟨ x 1 ⟩ = a です。分散は式 (4) を 2 回使って ⟨ n ∣ X 2 ∣ n ⟩ = ℏ 2 m ω ( 2 n + 1 ) \langle n|X^2|n\rangle=\dfrac{\hbar}{2m\omega}(2n+1) ⟨ n ∣ X 2 ∣ n ⟩ = 2 mω ℏ ( 2 n + 1 ) ⟨ n ⟩ = 1 / ( e β ℏ ω − 1 ) \langle n\rangle=1/(e^{\beta\hbar\omega}-1) ⟨ n ⟩ = 1/ ( e β ℏ ω − 1 )
⟨ x 1 2 ⟩ − ⟨ x 1 ⟩ 2 = ℏ 2 m ω ( 2 ⟨ n ⟩ + 1 ) = ℏ 2 m ω coth ( ℏ ω 2 k B T ) . \langle x_1^2\rangle-\langle x_1\rangle^2=\frac{\hbar}{2m\omega}\left(2\langle n\rangle+1\right)
=\frac{\hbar}{2m\omega}\coth\!\left(\frac{\hbar\omega}{2k_{\mathrm B}T}\right). ⟨ x 1 2 ⟩ − ⟨ x 1 ⟩ 2 = 2 mω ℏ ( 2 ⟨ n ⟩ + 1 ) = 2 mω ℏ coth ( 2 k B T ℏ ω ) . 比較します。高温極限 k B T ≫ ℏ ω k_{\mathrm B}T\gg\hbar\omega k B T ≫ ℏ ω e β ℏ ω − 1 ≃ β ℏ ω e^{\beta\hbar\omega}-1\simeq\beta\hbar\omega e β ℏ ω − 1 ≃ β ℏ ω
C Q → k B ( β ℏ ω ) 2 1 ( β ℏ ω ) 2 = k B = C C_{\mathrm Q}\to k_{\mathrm B}(\beta\hbar\omega)^2\frac{1}{(\beta\hbar\omega)^2}=k_{\mathrm B}=C C Q → k B ( β ℏ ω ) 2 ( β ℏ ω ) 2 1 = k B = C で古典結果に一致します（分散も coth x ≃ 1 / x \coth x\simeq1/x coth x ≃ 1/ x k B T / ( m ω 2 ) k_{\mathrm B}T/(m\omega^2) k B T / ( m ω 2 ) k B T ≪ ℏ ω k_{\mathrm B}T\ll\hbar\omega k B T ≪ ℏ ω
C Q ≃ k B ( ℏ ω k B T ) 2 e − ℏ ω / k B T ⟶ 0 C_{\mathrm Q}\simeq k_{\mathrm B}\left(\frac{\hbar\omega}{k_{\mathrm B}T}\right)^2e^{-\hbar\omega/k_{\mathrm B}T}\longrightarrow 0 C Q ≃ k B ( k B T ℏ ω ) 2 e − ℏ ω / k B T ⟶ 0 と指数関数的に消え、C = k B C=k_{\mathrm B} C = k B ℏ ω \hbar\omega ℏ ω ℏ / ( 2 m ω ) \hbar/(2m\omega) ℏ/ ( 2 mω ) T → 0 T\to0 T → 0 0 0 0
平衡位置からのずれ u j ≡ x j − j a u_j\equiv x_j-ja u j ≡ x j − j a j = 1 , … , N j=1,\dots,N j = 1 , … , N u 0 ≡ 0 u_0\equiv0 u 0 ≡ 0 u N + 1 ≡ 0 u_{N+1}\equiv0 u N + 1 ≡ 0 x 1 − a = u 1 x_1-a=u_1 x 1 − a = u 1 x j − x j − 1 − a = u j − u j − 1 x_j-x_{j-1}-a=u_j-u_{j-1} x j − x j − 1 − a = u j − u j − 1 ( N + 1 ) a − x N − a = − u N (N+1)a-x_N-a=-u_N ( N + 1 ) a − x N − a = − u N
H = ∑ j = 1 N p j 2 2 m + m ω 2 2 ∑ j = 1 N + 1 ( u j − u j − 1 ) 2 H=\sum_{j=1}^{N}\frac{p_j^2}{2m}+\frac{m\omega^2}{2}\sum_{j=1}^{N+1}(u_j-u_{j-1})^2 H = j = 1 ∑ N 2 m p j 2 + 2 m ω 2 j = 1 ∑ N + 1 ( u j − u j − 1 ) 2 と書けます。ポテンシャルは m ω 2 2 ∑ i , j u i K i j u j \dfrac{m\omega^2}{2}\sum_{i,j}u_iK_{ij}u_j 2 m ω 2 ∑ i , j u i K ij u j K K K 2 2 2 − 1 -1 − 1
K K K v j ( ℓ ) = 2 N + 1 sin ( j ℓ π N + 1 ) v_j^{(\ell)}=\sqrt{\dfrac{2}{N+1}}\sin\!\left(\dfrac{j\ell\pi}{N+1}\right) v j ( ℓ ) = N + 1 2 sin ( N + 1 j ℓ π ) θ ℓ = ℓ π / ( N + 1 ) \theta_\ell=\ell\pi/(N+1) θ ℓ = ℓ π / ( N + 1 )
− v j − 1 ( ℓ ) + 2 v j ( ℓ ) − v j + 1 ( ℓ ) = ( 2 − 2 cos θ ℓ ) v j ( ℓ ) -v_{j-1}^{(\ell)}+2v_j^{(\ell)}-v_{j+1}^{(\ell)}=(2-2\cos\theta_\ell)\,v_j^{(\ell)} − v j − 1 ( ℓ ) + 2 v j ( ℓ ) − v j + 1 ( ℓ ) = ( 2 − 2 cos θ ℓ ) v j ( ℓ ) が成り立ちます（sin ( ( j ± 1 ) θ ) \sin((j\pm1)\theta) sin (( j ± 1 ) θ ) v 0 ( ℓ ) = 0 v_0^{(\ell)}=0 v 0 ( ℓ ) = 0 v N + 1 ( ℓ ) = sin ( ( N + 1 ) θ ℓ ) = 0 v_{N+1}^{(\ell)}=\sin((N+1)\theta_\ell)=0 v N + 1 ( ℓ ) = sin (( N + 1 ) θ ℓ ) = 0 θ ℓ = ℓ π / ( N + 1 ) \theta_\ell=\ell\pi/(N+1) θ ℓ = ℓ π / ( N + 1 ) ℓ = 1 , 2 , … , N \ell=1,2,\dots,N ℓ = 1 , 2 , … , N N N N
∑ j = 1 N v j ( ℓ ) v j ( ℓ ′ ) = δ ℓ ℓ ′ \sum_{j=1}^{N}v_j^{(\ell)}v_j^{(\ell')}=\delta_{\ell\ell'} j = 1 ∑ N v j ( ℓ ) v j ( ℓ ′ ) = δ ℓ ℓ ′ なので、{ v ( ℓ ) } \{v^{(\ell)}\} { v ( ℓ ) }
そこで直交変換 u j = ∑ ℓ v j ( ℓ ) Q ℓ u_j=\sum_\ell v_j^{(\ell)}Q_\ell u j = ∑ ℓ v j ( ℓ ) Q ℓ p j = ∑ ℓ v j ( ℓ ) P ℓ p_j=\sum_\ell v_j^{(\ell)}P_\ell p j = ∑ ℓ v j ( ℓ ) P ℓ ∑ j p j 2 = ∑ ℓ P ℓ 2 \sum_j p_j^2=\sum_\ell P_\ell^2 ∑ j p j 2 = ∑ ℓ P ℓ 2
H = ∑ ℓ = 1 N ( P ℓ 2 2 m + m ω ℓ 2 2 Q ℓ 2 ) , ω ℓ = ω 2 − 2 cos ( ℓ N + 1 π ) H=\sum_{\ell=1}^{N}\left(\frac{P_\ell^2}{2m}+\frac{m\omega_\ell^2}{2}Q_\ell^2\right),\qquad
\omega_\ell=\omega\sqrt{2-2\cos\left(\frac{\ell}{N+1}\pi\right)} H = ℓ = 1 ∑ N ( 2 m P ℓ 2 + 2 m ω ℓ 2 Q ℓ 2 ) , ω ℓ = ω 2 − 2 cos ( N + 1 ℓ π ) となります。これで系が N N N ω ℓ = 2 ω ∣ sin ℓ π 2 ( N + 1 ) ∣ \omega_\ell=2\omega\left|\sin\dfrac{\ell\pi}{2(N+1)}\right| ω ℓ = 2 ω sin 2 ( N + 1 ) ℓ π 0 < ω ℓ < 2 ω 0<\omega_\ell<2\omega 0 < ω ℓ < 2 ω
古典論では、系は N N N k B T k_{\mathrm B}T k B T N k B T Nk_{\mathrm B}T N k B T
C 1 = k B C_1=k_{\mathrm B} C 1 = k B です。振動数 ω ℓ \omega_\ell ω ℓ C C C
質点 1 個当たりの量子比熱は、設問2の 1 振動子の結果を基準振動について足したものです。x ℓ = β ℏ ω ℓ x_\ell=\beta\hbar\omega_\ell x ℓ = β ℏ ω ℓ
C 1 Q = k B N ∑ ℓ = 1 N x ℓ 2 e x ℓ ( e x ℓ − 1 ) 2 . C_{1\mathrm Q}=\frac{k_{\mathrm B}}{N}\sum_{\ell=1}^{N}\frac{x_\ell^2 e^{x_\ell}}{(e^{x_\ell}-1)^2}. C 1 Q = N k B ℓ = 1 ∑ N ( e x ℓ − 1 ) 2 x ℓ 2 e x ℓ . 被和関数 x 2 e x / ( e x − 1 ) 2 x^2e^x/(e^x-1)^2 x 2 e x / ( e x − 1 ) 2 x ≫ 1 x\gg1 x ≫ 1 k B T ≪ ℏ ω k_{\mathrm B}T\ll\hbar\omega k B T ≪ ℏ ω ℏ ω ℓ ≲ k B T \hbar\omega_\ell\lesssim k_{\mathrm B}T ℏ ω ℓ ≲ k B T ω ℓ ≪ ω \omega_\ell\ll\omega ω ℓ ≪ ω ℓ ≪ N \ell\ll N ℓ ≪ N
ω ℓ = 2 ω sin ℓ π 2 ( N + 1 ) ≃ π ω ℓ N + 1 \omega_\ell=2\omega\sin\frac{\ell\pi}{2(N+1)}\simeq\frac{\pi\omega\,\ell}{N+1} ω ℓ = 2 ω sin 2 ( N + 1 ) ℓ π ≃ N + 1 π ω ℓ と線形分散になります。N N N x ℓ x_\ell x ℓ Δ = π β ℏ ω / ( N + 1 ) \Delta=\pi\beta\hbar\omega/(N+1) Δ = π β ℏ ω / ( N + 1 ) 1 1 1 π ℏ ω / ( N + 1 ) ≪ k B T ≪ ℏ ω \pi\hbar\omega/(N+1)\ll k_{\mathrm B}T\ll\hbar\omega π ℏ ω / ( N + 1 ) ≪ k B T ≪ ℏ ω
C 1 Q ≃ k B N ⋅ 1 Δ ∫ 0 ∞ x 2 e x ( e x − 1 ) 2 d x = k B N ⋅ N + 1 π β ℏ ω ⋅ π 2 3 ≃ π k B 3 β ℏ ω . C_{1\mathrm Q}\simeq\frac{k_{\mathrm B}}{N}\cdot\frac{1}{\Delta}\int_0^{\infty}\frac{x^2e^x}{(e^x-1)^2}dx
=\frac{k_{\mathrm B}}{N}\cdot\frac{N+1}{\pi\beta\hbar\omega}\cdot\frac{\pi^2}{3}
\simeq\frac{\pi k_{\mathrm B}}{3\beta\hbar\omega}. C 1 Q ≃ N k B ⋅ Δ 1 ∫ 0 ∞ ( e x − 1 ) 2 x 2 e x d x = N k B ⋅ π β ℏ ω N + 1 ⋅ 3 π 2 ≃ 3 β ℏ ω π k B . 与えられた公式から ∫ 0 ∞ x 2 e x ( e x − 1 ) 2 d x = 2 ⋅ π 2 6 = π 2 3 \displaystyle\int_0^\infty\frac{x^2e^x}{(e^x-1)^2}dx=2\cdot\frac{\pi^2}{6}=\frac{\pi^2}{3} ∫ 0 ∞ ( e x − 1 ) 2 x 2 e x d x = 2 ⋅ 6 π 2 = 3 π 2 1 / β = k B T 1/\beta=k_{\mathrm B}T 1/ β = k B T
C 1 Q ≃ π k B 2 3 ℏ ω T , C_{1\mathrm Q}\simeq\frac{\pi k_{\mathrm B}^2}{3\hbar\omega}\,T, C 1 Q ≃ 3ℏ ω π k B 2 T , すなわち
b = 1 , A = π k B 2 3 ℏ ω b=1,\qquad A=\frac{\pi k_{\mathrm B}^2}{3\hbar\omega} b = 1 , A = 3ℏ ω π k B 2 です。1 次元デバイ模型の T 1 T^1 T 1 [ k B 2 T / ( ℏ ω ) ] = ( J / K ) 2 ⋅ K / J = J / K [k_{\mathrm B}^2T/(\hbar\omega)]=(\mathrm{J/K})^2\cdot\mathrm{K}/\mathrm{J}=\mathrm{J/K} [ k B 2 T / ( ℏ ω )] = ( J/K ) 2 ⋅ K / J = J/K k B k_{\mathrm B} k B k B T ≪ ℏ ω k_{\mathrm B}T\ll\hbar\omega k B T ≪ ℏ ω C 1 Q ≪ C 1 C_{1\mathrm Q}\ll C_1 C 1 Q ≪ C 1 ω ℓ → 0 \omega_\ell\to0 ω ℓ → 0
外力を受けず x x x − v g a s -v_{\mathrm{gas}} − v gas v g a s > 0 v_{\mathrm{gas}}>0 v gas > 0 α \alpha α 慣性系 （この系での速度を対地速度と呼ぶ）で議論します。前半は非相対論、後半は特殊相対論です。γ ( v ) = 1 / 1 − v 2 / c 2 \gamma(v)=1/\sqrt{1-v^2/c^2} γ ( v ) = 1/ 1 − v 2 / c 2
時刻 t t t t + d t t+dt t + d t m → m + d m m\to m+dm m → m + d m d m < 0 dm<0 d m < 0 v → v + d v v\to v+dv v → v + d v − d m > 0 -dm>0 − d m > 0 v − v g a s v-v_{\mathrm{gas}} v − v gas
m v = ( m + d m ) ( v + d v ) + ( − d m ) ( v − v g a s ) mv=(m+dm)(v+dv)+(-dm)(v-v_{\mathrm{gas}}) m v = ( m + d m ) ( v + d v ) + ( − d m ) ( v − v gas ) で、微小量の 2 次を落として整理すると
0 = m d v + v g a s d m . 0=m\,dv+v_{\mathrm{gas}}\,dm . 0 = m d v + v gas d m . d v / d t = α dv/dt=\alpha d v / d t = α
d m d t = − α v g a s m ( t ) ( < 0 ) . \frac{dm}{dt}=-\frac{\alpha}{v_{\mathrm{gas}}}\,m(t)\;(<0). d t d m = − v gas α m ( t ) ( < 0 ) . 上の式は d ln m / d t = − α / v g a s d\ln m/dt=-\alpha/v_{\mathrm{gas}} d ln m / d t = − α / v gas m ( 0 ) = m 0 m(0)=m_0 m ( 0 ) = m 0
m ( t ) = m 0 exp ( − α t v g a s ) . m(t)=m_0\exp\left(-\frac{\alpha t}{v_{\mathrm{gas}}}\right). m ( t ) = m 0 exp ( − v gas α t ) . v ( t ) = α t v(t)=\alpha t v ( t ) = α t v = v g a s ln ( m 0 / m ) v=v_{\mathrm{gas}}\ln(m_0/m) v = v gas ln ( m 0 / m )
イ について。d s 2 = − c 2 d t 2 + d x 2 ds^2=-c^2dt^2+dx^2 d s 2 = − c 2 d t 2 + d x 2 d τ = − d s 2 / c = d t 2 − d x 2 / c 2 d\tau=\sqrt{-ds^2}/c=\sqrt{dt^2-dx^2/c^2} d τ = − d s 2 / c = d t 2 − d x 2 / c 2
1 − v 2 c 2 ( = 1 γ ( v ) ) . \sqrt{1-\frac{v^2}{c^2}}\;\left(=\frac{1}{\gamma(v)}\right). 1 − c 2 v 2 ( = γ ( v ) 1 ) . ロ について。u μ ⊙ u μ = − c 2 d t 2 + d x 2 d τ 2 = d s 2 d τ 2 u^\mu\odot u^\mu=\dfrac{-c^2dt^2+dx^2}{d\tau^2}=\dfrac{ds^2}{d\tau^2} u μ ⊙ u μ = d τ 2 − c 2 d t 2 + d x 2 = d τ 2 d s 2 d τ 2 = − d s 2 / c 2 d\tau^2=-ds^2/c^2 d τ 2 = − d s 2 / c 2 − c 2 -c^2 − c 2
ハ について。ロの式を τ \tau τ 2 u μ ⊙ a μ = 0 2\,u^\mu\odot a^\mu=0 2 u μ ⊙ a μ = 0 0 0 0
ニ と ホ について。ハより − u 0 a 0 + u 1 a 1 = 0 -u^0a^0+u^1a^1=0 − u 0 a 0 + u 1 a 1 = 0 a 0 = u 1 a 1 / u 0 a^0=u^1a^1/u^0 a 0 = u 1 a 1 / u 0 − ( a 0 ) 2 + ( a 1 ) 2 = α 2 -(a^0)^2+(a^1)^2=\alpha^2 − ( a 0 ) 2 + ( a 1 ) 2 = α 2
( a 1 ) 2 ( u 0 ) 2 − ( u 1 ) 2 ( u 0 ) 2 = α 2 . (a^1)^2\frac{(u^0)^2-(u^1)^2}{(u^0)^2}=\alpha^2 . ( a 1 ) 2 ( u 0 ) 2 ( u 0 ) 2 − ( u 1 ) 2 = α 2 . ロより ( u 0 ) 2 − ( u 1 ) 2 = c 2 (u^0)^2-(u^1)^2=c^2 ( u 0 ) 2 − ( u 1 ) 2 = c 2 a 1 > 0 a^1>0 a 1 > 0
a 0 = α u 1 c , a 1 = α u 0 c . a^0=\frac{\alpha\,u^1}{c},\qquad a^1=\frac{\alpha\,u^0}{c}. a 0 = c α u 1 , a 1 = c α u 0 . 瞬間静止系では u μ = ( c , 0 ) u^\mu=(c,0) u μ = ( c , 0 ) a μ = ( 0 , α ) a^\mu=(0,\alpha) a μ = ( 0 , α ) α \alpha α
設問3の結果は
d u 0 d τ = α c u 1 , d u 1 d τ = α c u 0 \frac{du^0}{d\tau}=\frac{\alpha}{c}u^1,\qquad \frac{du^1}{d\tau}=\frac{\alpha}{c}u^0 d τ d u 0 = c α u 1 , d τ d u 1 = c α u 0 です。和と差をとると d d τ ( u 0 ± u 1 ) = ± α c ( u 0 ± u 1 ) \dfrac{d}{d\tau}(u^0\pm u^1)=\pm\dfrac{\alpha}{c}(u^0\pm u^1) d τ d ( u 0 ± u 1 ) = ± c α ( u 0 ± u 1 ) τ = 0 \tau=0 τ = 0 v = 0 v=0 v = 0 u 0 = c , u 1 = 0 u^0=c,\ u^1=0 u 0 = c , u 1 = 0
u 0 ± u 1 = c e ± α τ / c , u^0\pm u^1=c\,e^{\pm\alpha\tau/c}, u 0 ± u 1 = c e ± α τ / c , すなわち
u 0 = c cosh α τ c , u 1 = c sinh α τ c . u^0=c\cosh\frac{\alpha\tau}{c},\qquad u^1=c\sinh\frac{\alpha\tau}{c}. u 0 = c cosh c α τ , u 1 = c sinh c α τ . 対地速度は v = d x d t = d x / d τ d t / d τ = c u 1 u 0 v=\dfrac{dx}{dt}=\dfrac{dx/d\tau}{dt/d\tau}=\dfrac{c\,u^1}{u^0} v = d t d x = d t / d τ d x / d τ = u 0 c u 1
v ( τ ) = c tanh ( α τ c ) . v(\tau)=c\tanh\left(\frac{\alpha\tau}{c}\right). v ( τ ) = c tanh ( c α τ ) . α τ ≪ c \alpha\tau\ll c α τ ≪ c v ≃ α τ v\simeq\alpha\tau v ≃ α τ τ → ∞ \tau\to\infty τ → ∞ v → c v\to c v → c
4元運動量 は p μ = m u μ = ( m γ c , m γ v ) p^\mu=m u^\mu=(m\gamma c,\ m\gamma v) p μ = m u μ = ( mγ c , mγ v ) ( m , v ) (m,v) ( m , v ) ( m + d m , v + d v ) (m+dm,\ v+dv) ( m + d m , v + d v ) d m ′ dm' d m ′ v ′ v' v ′
m u μ ( v ) = ( m + d m ) u μ ( v + d v ) + d m ′ u μ ( v ′ ) m\,u^\mu(v)=(m+dm)\,u^\mu(v+dv)+dm'\,u^\mu(v') m u μ ( v ) = ( m + d m ) u μ ( v + d v ) + d m ′ u μ ( v ′ ) です。d γ d v = γ 3 v c 2 \dfrac{d\gamma}{dv}=\gamma^3\dfrac{v}{c^2} d v d γ = γ 3 c 2 v γ 3 v 2 c 2 + γ = γ 3 \gamma^3\dfrac{v^2}{c^2}+\gamma=\gamma^3 γ 3 c 2 v 2 + γ = γ 3
γ d m + m γ 3 v c 2 d v + γ ′ d m ′ = 0 , \gamma\,dm+m\gamma^3\frac{v}{c^2}\,dv+\gamma'\,dm'=0, γ d m + m γ 3 c 2 v d v + γ ′ d m ′ = 0 , γ v d m + m γ 3 d v + γ ′ v ′ d m ′ = 0 \gamma v\,dm+m\gamma^3\,dv+\gamma' v'\,dm'=0 γ v d m + m γ 3 d v + γ ′ v ′ d m ′ = 0 となります。ここで γ = γ ( v ) \gamma=\gamma(v) γ = γ ( v ) γ ′ = γ ( v ′ ) \gamma'=\gamma(v') γ ′ = γ ( v ′ )
上の 2 式から d m ′ dm' d m ′ v ′ v' v ′
γ ( v ′ − v ) d m + m γ 3 ( v v ′ c 2 − 1 ) d v = 0 ⟹ d m m = γ 2 1 − v v ′ / c 2 v ′ − v d v . \gamma(v'-v)\,dm+m\gamma^3\left(\frac{vv'}{c^2}-1\right)dv=0
\;\Longrightarrow\;
\frac{dm}{m}=\gamma^2\,\frac{1-vv'/c^2}{v'-v}\,dv . γ ( v ′ − v ) d m + m γ 3 ( c 2 v v ′ − 1 ) d v = 0 ⟹ m d m = γ 2 v ′ − v 1 − v v ′ / c 2 d v . 速度変換則 − v g a s = v ′ − v 1 − v v ′ / c 2 -v_{\mathrm{gas}}=\dfrac{v'-v}{1-vv'/c^2} − v gas = 1 − v v ′ / c 2 v ′ − v 1 − v v ′ / c 2 v ′ − v = − 1 v g a s \dfrac{1-vv'/c^2}{v'-v}=-\dfrac{1}{v_{\mathrm{gas}}} v ′ − v 1 − v v ′ / c 2 = − v gas 1
d m m = − γ 2 d v v g a s . \frac{dm}{m}=-\frac{\gamma^2\,dv}{v_{\mathrm{gas}}}. m d m = − v gas γ 2 d v . ここに設問4の v = c tanh ( α τ / c ) v=c\tanh(\alpha\tau/c) v = c tanh ( α τ / c ) χ ≡ α τ / c \chi\equiv\alpha\tau/c χ ≡ α τ / c γ = cosh χ \gamma=\cosh\chi γ = cosh χ d v = c s e c h 2 χ d χ dv=c\,\mathrm{sech}^2\chi\,d\chi d v = c sech 2 χ d χ γ 2 d v = c d χ = α d τ \gamma^2dv=c\,d\chi=\alpha\,d\tau γ 2 d v = c d χ = α d τ
d m m = − α v g a s d τ ⟹ m ( τ ) = m 0 exp ( − α τ v g a s ) . \frac{dm}{m}=-\frac{\alpha}{v_{\mathrm{gas}}}\,d\tau
\;\Longrightarrow\;
m(\tau)=m_0\exp\left(-\frac{\alpha\tau}{v_{\mathrm{gas}}}\right). m d m = − v gas α d τ ⟹ m ( τ ) = m 0 exp ( − v gas α τ ) . 設問2の答え m = m 0 e − α t / v g a s m=m_0e^{-\alpha t/v_{\mathrm{gas}}} m = m 0 e − α t / v gas t t t 固有時 間 τ \tau τ α \alpha α v g a s v_{\mathrm{gas}} v gas d τ d\tau d τ d τ d\tau d τ t → τ t\to\tau t → τ
物理的な帰結を 2 つ挙げます。第一に、α τ / c = χ = a r t a n h ( v / c ) \alpha\tau/c=\chi=\mathrm{artanh}(v/c) α τ / c = χ = artanh ( v / c )
m 0 m = exp ( c v g a s a r t a n h v c ) = ( 1 + v / c 1 − v / c ) c / ( 2 v g a s ) \frac{m_0}{m}=\exp\left(\frac{c}{v_{\mathrm{gas}}}\,\mathrm{artanh}\frac{v}{c}\right)
=\left(\frac{1+v/c}{1-v/c}\right)^{c/(2v_{\mathrm{gas}})} m m 0 = exp ( v gas c artanh c v ) = ( 1 − v / c 1 + v / c ) c / ( 2 v gas ) で、非相対論の e v / v g a s e^{v/v_{\mathrm{gas}}} e v / v gas v → c v\to c v → c τ < t \tau<t τ < t t t t α t \alpha t α t d m ′ dm' d m ′ d m ′ = − d m 1 − v g a s 2 / c 2 < − d m dm'=-dm\sqrt{1-v_{\mathrm{gas}}^2/c^2}<-dm d m ′ = − d m 1 − v gas 2 / c 2 < − d m
磁場のないプラズマ中で、電子密度 n e n_{\mathrm e} n e n c = m e ε 0 ω 2 / e 2 n_{\mathrm c}=m_{\mathrm e}\varepsilon_0\omega^2/e^2 n c = m e ε 0 ω 2 / e 2 α = 1 − n e / n c \alpha=\sqrt{1-n_{\mathrm e}/n_{\mathrm c}} α = 1 − n e / n c k = α ω / c k=\alpha\omega/c k = α ω / c E ( x ⃗ , t ) = E 0 exp ( i k ⃗ ⋅ x ⃗ − i ω t ) E(\vec x,t)=E_0\exp(i\vec k\cdot\vec x-i\omega t) E ( x , t ) = E 0 exp ( i k ⋅ x − iω t ) n e / n c = ω p e 2 / ω 2 n_{\mathrm e}/n_{\mathrm c}=\omega_{\mathrm{pe}}^2/\omega^2 n e / n c = ω pe 2 / ω 2 ω p e = n e e 2 / ( ε 0 m e ) \omega_{\mathrm{pe}}=\sqrt{n_{\mathrm e}e^2/(\varepsilon_0m_{\mathrm e})} ω pe = n e e 2 / ( ε 0 m e ) m e = 9.1 × 10 − 31 k g m_{\mathrm e}=9.1\times10^{-31}\,\mathrm{kg} m e = 9.1 × 1 0 − 31 kg e = 1.6 × 10 − 19 C e=1.6\times10^{-19}\,\mathrm{C} e = 1.6 × 1 0 − 19 C ε 0 = 8.9 × 10 − 12 F / m \varepsilon_0=8.9\times10^{-12}\,\mathrm{F/m} ε 0 = 8.9 × 1 0 − 12 F/m c = 3.0 × 10 8 m / s c=3.0\times10^{8}\,\mathrm{m/s} c = 3.0 × 1 0 8 m/s
z ≤ 0 z\le0 z ≤ 0 z > 0 z>0 z > 0 n e n_{\mathrm e} n e z z z z > 0 z>0 z > 0 E = E 0 e i k z − i ω t E=E_0e^{ikz-i\omega t} E = E 0 e ik z − iω t k = ( ω / c ) 1 − n e / n c k=(\omega/c)\sqrt{1-n_{\mathrm e}/n_{\mathrm c}} k = ( ω / c ) 1 − n e / n c
n e < n c n_{\mathrm e}<n_{\mathrm c} n e < n c α \alpha α 0 < α < 1 0<\alpha<1 0 < α < 1 k k k z z z
λ = 2 π k = 2 π c ω 1 − n e / n c = λ v a c 1 − n e / n c \lambda=\frac{2\pi}{k}=\frac{2\pi c}{\omega\sqrt{1-n_{\mathrm e}/n_{\mathrm c}}}=\frac{\lambda_{\mathrm{vac}}}{\sqrt{1-n_{\mathrm e}/n_{\mathrm c}}} λ = k 2 π = ω 1 − n e / n c 2 π c = 1 − n e / n c λ vac で、真空中の波長 λ v a c = 2 π c / ω \lambda_{\mathrm{vac}}=2\pi c/\omega λ vac = 2 π c / ω 位相速度 ω / k = c / α \omega/k=c/\alpha ω / k = c / α c c c 群速度 α c \alpha c α c c c c
n e > n c n_{\mathrm e}>n_{\mathrm c} n e > n c α \alpha α α = i n e / n c − 1 \alpha=i\sqrt{n_{\mathrm e}/n_{\mathrm c}-1} α = i n e / n c − 1
E = E 0 exp ( − z δ ) e − i ω t , δ = c ω n e / n c − 1 = c ω p e 2 − ω 2 E=E_0\exp\left(-\frac{z}{\delta}\right)e^{-i\omega t},\qquad
\delta=\frac{c}{\omega\sqrt{n_{\mathrm e}/n_{\mathrm c}-1}}=\frac{c}{\sqrt{\omega_{\mathrm{pe}}^2-\omega^2}} E = E 0 exp ( − δ z ) e − iω t , δ = ω n e / n c − 1 c = ω pe 2 − ω 2 c となり、電場は位相を進めずに表面から長さ δ \delta δ ω ≪ ω p e \omega\ll\omega_{\mathrm{pe}} ω ≪ ω pe δ → c / ω p e \delta\to c/\omega_{\mathrm{pe}} δ → c / ω pe
半径 a a a z z z a a a r = ∣ z ∣ r=|z| r = ∣ z ∣
ϕ p = ∫ − a a k d z = ω c ∫ − a a 1 − n e ( r ) n c d z . \phi_{\mathrm p}=\int_{-a}^{a}k\,dz=\frac{\omega}{c}\int_{-a}^{a}\sqrt{1-\frac{n_{\mathrm e}(r)}{n_{\mathrm c}}}\,dz . ϕ p = ∫ − a a k d z = c ω ∫ − a a 1 − n c n e ( r ) d z . n e ( r ) ≪ n c n_{\mathrm e}(r)\ll n_{\mathrm c} n e ( r ) ≪ n c
ϕ p ≃ ω c ∫ − a a ( 1 − n e ( r ) 2 n c ) d z = ω c ⋅ 2 a − ω 2 c n c ∫ − a a n e ( r ) d z . \phi_{\mathrm p}\simeq\frac{\omega}{c}\int_{-a}^{a}\left(1-\frac{n_{\mathrm e}(r)}{2n_{\mathrm c}}\right)dz
=\frac{\omega}{c}\cdot2a-\frac{\omega}{2cn_{\mathrm c}}\int_{-a}^{a}n_{\mathrm e}(r)\,dz . ϕ p ≃ c ω ∫ − a a ( 1 − 2 n c n e ( r ) ) d z = c ω ⋅ 2 a − 2 c n c ω ∫ − a a n e ( r ) d z . 真空の場合は ϕ v = ( ω / c ) ⋅ 2 a \phi_{\mathrm v}=(\omega/c)\cdot2a ϕ v = ( ω / c ) ⋅ 2 a
Δ ϕ = ϕ v − ϕ p = ω 2 c n c ∫ − a a n e ( r ) d z = e 2 2 ε 0 m e c ω ∫ − a a n e ( r ) d z . \Delta\phi=\phi_{\mathrm v}-\phi_{\mathrm p}=\frac{\omega}{2cn_{\mathrm c}}\int_{-a}^{a}n_{\mathrm e}(r)\,dz
=\frac{e^2}{2\varepsilon_0 m_{\mathrm e}c\,\omega}\int_{-a}^{a}n_{\mathrm e}(r)\,dz . Δ ϕ = ϕ v − ϕ p = 2 c n c ω ∫ − a a n e ( r ) d z = 2 ε 0 m e c ω e 2 ∫ − a a n e ( r ) d z . n c = m e ε 0 ω 2 / e 2 n_{\mathrm c}=m_{\mathrm e}\varepsilon_0\omega^2/e^2 n c = m e ε 0 ω 2 / e 2 Δ ϕ > 0 \Delta\phi>0 Δ ϕ > 0 Δ ϕ \Delta\phi Δ ϕ Δ ϕ ∝ 1 / ω ∝ λ \Delta\phi\propto1/\omega\propto\lambda Δ ϕ ∝ 1/ ω ∝ λ
n e ( r ) = n e 0 [ 1 − ( r / a ) 2 ] n_{\mathrm e}(r)=n_{\mathrm e0}[1-(r/a)^2] n e ( r ) = n e 0 [ 1 − ( r / a ) 2 ] a = 0.1 m a=0.1\,\mathrm{m} a = 0.1 m n e 0 = 1 × 10 19 m − 3 n_{\mathrm e0}=1\times10^{19}\,\mathrm{m^{-3}} n e 0 = 1 × 1 0 19 m − 3
周波数の下限を決めるのはカットオフです。電磁波が経路上のどこでも伝搬できなければ透過測定にならないので、最も密度が高い中心 r = 0 r=0 r = 0 n e 0 < n c ( ω ) n_{\mathrm e0}<n_{\mathrm c}(\omega) n e 0 < n c ( ω )
ω > ω p e ( n e 0 ) = n e 0 e 2 ε 0 m e \omega>\omega_{\mathrm{pe}}(n_{\mathrm e0})=\sqrt{\frac{n_{\mathrm e0}e^2}{\varepsilon_0 m_{\mathrm e}}} ω > ω pe ( n e 0 ) = ε 0 m e n e 0 e 2 が必要です。これより低い周波数では中心部で α \alpha α Δ ϕ \Delta\phi Δ ϕ n e 0 ≪ n c n_{\mathrm e0}\ll n_{\mathrm c} n e 0 ≪ n c
ω p e = 10 19 × ( 1.6 × 10 − 19 ) 2 8.9 × 10 − 12 × 9.1 × 10 − 31 = 1.8 × 10 11 r a d / s , f min = ω p e 2 π ≃ 3 × 10 10 H z \omega_{\mathrm{pe}}=\sqrt{\frac{10^{19}\times(1.6\times10^{-19})^2}{8.9\times10^{-12}\times9.1\times10^{-31}}}
=1.8\times10^{11}\,\mathrm{rad/s},\qquad
f_{\min}=\frac{\omega_{\mathrm{pe}}}{2\pi}\simeq3\times10^{10}\,\mathrm{Hz} ω pe = 8.9 × 1 0 − 12 × 9.1 × 1 0 − 31 1 0 19 × ( 1.6 × 1 0 − 19 ) 2 = 1.8 × 1 0 11 rad/s , f m i n = 2 π ω pe ≃ 3 × 1 0 10 Hz で、オーダーは 10 10 H z 10^{10}\,\mathrm{Hz} 1 0 10 Hz
この周波数 f ≃ 3 × 10 10 H z f\simeq3\times10^{10}\,\mathrm{Hz} f ≃ 3 × 1 0 10 Hz ω = 1.9 × 10 11 r a d / s \omega=1.9\times10^{11}\,\mathrm{rad/s} ω = 1.9 × 1 0 11 rad/s n c = 1.1 × 10 19 m − 3 n_{\mathrm c}=1.1\times10^{19}\,\mathrm{m^{-3}} n c = 1.1 × 1 0 19 m − 3 Δ ϕ \Delta\phi Δ ϕ r = ∣ z ∣ r=|z| r = ∣ z ∣
∫ − a a n e d z = n e 0 ∫ − a a ( 1 − z 2 a 2 ) d z = 4 3 n e 0 a = 4 3 × 10 19 × 0.1 = 1.3 × 10 18 m − 2 \int_{-a}^{a}n_{\mathrm e}\,dz=n_{\mathrm e0}\int_{-a}^{a}\left(1-\frac{z^2}{a^2}\right)dz=\frac43 n_{\mathrm e0}a
=\frac43\times10^{19}\times0.1=1.3\times10^{18}\,\mathrm{m^{-2}} ∫ − a a n e d z = n e 0 ∫ − a a ( 1 − a 2 z 2 ) d z = 3 4 n e 0 a = 3 4 × 1 0 19 × 0.1 = 1.3 × 1 0 18 m − 2 であり
Δ ϕ = 2 e 2 n e 0 a 3 ε 0 m e c ω = 2 × ( 1.6 × 10 − 19 ) 2 × 10 19 × 0.1 3 × 8.9 × 10 − 12 × 9.1 × 10 − 31 × 3.0 × 10 8 × 1.9 × 10 11 ≃ 4 × 10 1 r a d . \Delta\phi=\frac{2e^2n_{\mathrm e0}a}{3\varepsilon_0m_{\mathrm e}c\,\omega}
=\frac{2\times(1.6\times10^{-19})^2\times10^{19}\times0.1}
{3\times8.9\times10^{-12}\times9.1\times10^{-31}\times3.0\times10^{8}\times1.9\times10^{11}}
\simeq4\times10^{1}\,\mathrm{rad}. Δ ϕ = 3 ε 0 m e c ω 2 e 2 n e 0 a = 3 × 8.9 × 1 0 − 12 × 9.1 × 1 0 − 31 × 3.0 × 1 0 8 × 1.9 × 1 0 11 2 × ( 1.6 × 1 0 − 19 ) 2 × 1 0 19 × 0.1 ≃ 4 × 1 0 1 rad . 答えは Δ ϕ ≈ 4 × 10 1 r a d \Delta\phi\approx4\times10^1\,\mathrm{rad} Δ ϕ ≈ 4 × 1 0 1 rad 2 π 2\pi 2 π
参照波とプローブ波の重ね合わせは E t = E 1 e − i ω t + E 2 e − i ω t + i Δ ϕ E_{\mathrm t}=E_1e^{-i\omega t}+E_2e^{-i\omega t+i\Delta\phi} E t = E 1 e − iω t + E 2 e − iω t + i Δ ϕ
∣ E t ∣ 2 = ∣ E 1 + E 2 e i Δ ϕ ∣ 2 = E 1 2 + E 2 2 + 2 E 1 E 2 cos Δ ϕ . |E_{\mathrm t}|^2=\left|E_1+E_2e^{i\Delta\phi}\right|^2=E_1^2+E_2^2+2E_1E_2\cos\Delta\phi . ∣ E t ∣ 2 = E 1 + E 2 e i Δ ϕ 2 = E 1 2 + E 2 2 + 2 E 1 E 2 cos Δ ϕ . ここで E 1 , E 2 E_1,E_2 E 1 , E 2
参照波を ω 1 \omega_1 ω 1 ω 2 \omega_2 ω 2 E t = E 1 e − i ω 1 t + E 2 e − i ω 2 t + i Δ ϕ ( t ) E_{\mathrm t}=E_1e^{-i\omega_1t}+E_2e^{-i\omega_2t+i\Delta\phi(t)} E t = E 1 e − i ω 1 t + E 2 e − i ω 2 t + i Δ ϕ ( t )
∣ E t ∣ 2 = E 1 2 + E 2 2 + 2 E 1 E 2 cos [ Δ ω t − Δ ϕ ( t ) ] , Δ ω = ω 2 − ω 1 |E_{\mathrm t}|^2=E_1^2+E_2^2+2E_1E_2\cos\left[\Delta\omega\,t-\Delta\phi(t)\right],\qquad \Delta\omega=\omega_2-\omega_1 ∣ E t ∣ 2 = E 1 2 + E 2 2 + 2 E 1 E 2 cos [ Δ ω t − Δ ϕ ( t ) ] , Δ ω = ω 2 − ω 1 となります。検出器は Δ ω \Delta\omega Δ ω Δ ω \Delta\omega Δ ω
(a) が解決できる理由。位相情報はビート成分の位相にのみ現れ、E 1 , E 2 E_1,E_2 E 1 , E 2 Δ ω \Delta\omega Δ ω Δ ω \Delta\omega Δ ω Δ ϕ ( t ) \Delta\phi(t) Δ ϕ ( t ) Δ ω = 0 \Delta\omega=0 Δ ω = 0 Δ ϕ \Delta\phi Δ ϕ
(b) が解決できる理由。Δ ω = 0 \Delta\omega=0 Δ ω = 0 cos Δ ϕ \cos\Delta\phi cos Δ ϕ Δ ϕ → − Δ ϕ \Delta\phi\to-\Delta\phi Δ ϕ → − Δ ϕ Δ ω t − Δ ϕ ( t ) \Delta\omega t-\Delta\phi(t) Δ ω t − Δ ϕ ( t ) Δ ω − d ( Δ ϕ ) / d t \Delta\omega-d(\Delta\phi)/dt Δ ω − d ( Δ ϕ ) / d t Δ ϕ \Delta\phi Δ ϕ 2 π 2\pi 2 π
必要な条件は
Δ ω ≫ ∣ d ( Δ ϕ ) d t ∣ \Delta\omega\gg\left|\frac{d(\Delta\phi)}{dt}\right| Δ ω ≫ d t d ( Δ ϕ ) です。これが破れると瞬時周波数 Δ ω − d ( Δ ϕ ) / d t \Delta\omega-d(\Delta\phi)/dt Δ ω − d ( Δ ϕ ) / d t Δ ϕ \Delta\phi Δ ϕ
核子や原子核の荷電半径を高エネルギー電子の弾性散乱で測る問題です。c = 3 × 10 8 m / s c=3\times10^{8}\,\mathrm{m/s} c = 3 × 1 0 8 m/s ℏ c = 0.2 G e V ⋅ f m \hbar c=0.2\,\mathrm{GeV\cdot fm} ℏ c = 0.2 GeV ⋅ fm M c 2 = 1 G e V Mc^2=1\,\mathrm{GeV} M c 2 = 1 GeV
(a) 静電加速器は適しません。電極間の静電圧で得られるエネルギーは端子電圧そのままなので、数百 MeV には数百 MV の直流電圧が必要です。実際には絶縁破壊（放電）のため端子電圧は十数 MV 程度が限界で、タンデム型でもその 2 倍程度にしかなりません。
(b) サイクロトロンは適しません。電子の静止エネルギーは 0.511 M e V 0.511\,\mathrm{MeV} 0.511 MeV γ ≫ 1 \gamma\gg1 γ ≫ 1 T = 2 π γ m e / ( e B ) T=2\pi\gamma m_{\mathrm e}/(eB) T = 2 π γ m e / ( e B ) γ \gamma γ γ ∼ 600 \gamma\sim600 γ ∼ 600
(c) 線形加速器は適します。電子は 1 MeV 程度で既に v ≃ c v\simeq c v ≃ c c c c
入射電子数は N e = I t / e N_{\mathrm e}=It/e N e = I t / e ρ N A / A \rho N_{\mathrm A}/A ρ N A / A ℓ \ell ℓ ρ ℓ N A / A \rho\ell N_{\mathrm A}/A ρ ℓ N A / A ω \omega ω
n = N e × ρ ℓ N A A × d σ d Ω ( θ ) × ω n=N_{\mathrm e}\times\frac{\rho\ell N_{\mathrm A}}{A}\times\frac{d\sigma}{d\Omega}(\theta)\times\omega n = N e × A ρ ℓ N A × d Ω d σ ( θ ) × ω です。これを解いて
d σ d Ω = n e A I t ρ ℓ N A ω . \frac{d\sigma}{d\Omega}=\frac{n\,e\,A}{I\,t\,\rho\,\ell\,N_{\mathrm A}\,\omega}. d Ω d σ = I t ρ ℓ N A ω n e A . θ \theta θ Z Z Z c c c ρ \rho ρ g / c m 3 \mathrm{g/cm^3} g/c m 3 A A A g / m o l \mathrm{g/mol} g/mol N A N_{\mathrm A} N A m o l − 1 \mathrm{mol^{-1}} mo l − 1 c m 2 / s r \mathrm{cm^2/sr} c m 2 /sr
一様磁場中の円運動では p = e B r p=eBr p = e B r B = p / ( e r ) B=p/(er) B = p / ( er ) p = 300 M e V / c p=300\,\mathrm{MeV}/c p = 300 MeV / c
p = 300 × 10 6 × 1.6 × 10 − 19 3 × 10 8 = 1.6 × 10 − 19 k g ⋅ m / s , p=\frac{300\times10^{6}\times1.6\times10^{-19}}{3\times10^{8}}=1.6\times10^{-19}\,\mathrm{kg\cdot m/s}, p = 3 × 1 0 8 300 × 1 0 6 × 1.6 × 1 0 − 19 = 1.6 × 1 0 − 19 kg ⋅ m/s , B = 1.6 × 10 − 19 1.6 × 10 − 19 × 1.0 ≃ 1 T . B=\frac{1.6\times10^{-19}}{1.6\times10^{-19}\times1.0}\simeq1\,\mathrm{T}. B = 1.6 × 1 0 − 19 × 1.0 1.6 × 1 0 − 19 ≃ 1 T . 答えは B ≈ 1 T B\approx1\,\mathrm{T} B ≈ 1 T B [ T ] = p [ G e V / c ] / ( 0.3 r [ m ] ) B[\mathrm T]=p[\mathrm{GeV}/c]/(0.3\,r[\mathrm m]) B [ T ] = p [ GeV / c ] / ( 0.3 r [ m ])
電子質量を無視すると電子の 4 元運動量は光子と同じく p 2 = 0 p^2=0 p 2 = 0 ∣ p ⃗ ∣ = E / c |\vec p|=E/c ∣ p ∣ = E / c P = ( M c , 0 ⃗ ) P=(Mc,\vec 0) P = ( M c , 0 ) P ′ = P + p i − p f P'=P+p_{\mathrm i}-p_{\mathrm f} P ′ = P + p i − p f P ′ 2 = P 2 P'^2=P^2 P ′2 = P 2
p i ⋅ p f = P ⋅ ( p i − p f ) ⟹ E i E f c 2 ( 1 − cos θ ) = M ( E i − E f ) p_{\mathrm i}\cdot p_{\mathrm f}=P\cdot(p_{\mathrm i}-p_{\mathrm f})
\;\Longrightarrow\;
\frac{E_{\mathrm i}E_{\mathrm f}}{c^2}(1-\cos\theta)=M(E_{\mathrm i}-E_{\mathrm f}) p i ⋅ p f = P ⋅ ( p i − p f ) ⟹ c 2 E i E f ( 1 − cos θ ) = M ( E i − E f ) が得られます。両辺を M E i E f ME_{\mathrm i}E_{\mathrm f} M E i E f
1 E f − 1 E i = 1 − cos θ M c 2 ⟺ E f = E i 1 + E i M c 2 ( 1 − cos θ ) \frac{1}{E_{\mathrm f}}-\frac{1}{E_{\mathrm i}}=\frac{1-\cos\theta}{Mc^2}
\qquad\Longleftrightarrow\qquad
E_{\mathrm f}=\frac{E_{\mathrm i}}{1+\dfrac{E_{\mathrm i}}{Mc^2}(1-\cos\theta)} E f 1 − E i 1 = M c 2 1 − cos θ ⟺ E f = 1 + M c 2 E i ( 1 − cos θ ) E i になります。これに E i = 300 M e V E_{\mathrm i}=300\,\mathrm{MeV} E i = 300 MeV M c 2 = 1 G e V Mc^2=1\,\mathrm{GeV} M c 2 = 1 GeV θ = 90 ∘ \theta=90^\circ θ = 9 0 ∘ cos θ = 0 \cos\theta=0 cos θ = 0
E f = 300 1 + 0.3 ≃ 2.3 × 10 2 M e V . E_{\mathrm f}=\frac{300}{1+0.3}\simeq2.3\times10^{2}\,\mathrm{MeV}. E f = 1 + 0.3 300 ≃ 2.3 × 1 0 2 MeV . 答えは E f ≈ 230 M e V E_{\mathrm f}\approx230\,\mathrm{MeV} E f ≈ 230 MeV 70 M e V 70\,\mathrm{MeV} 70 MeV
ρ ( r ) \rho(r) ρ ( r ) ∫ ρ d 3 r = 1 \int\rho\,d^3r=1 ∫ ρ d 3 r = 1 q q q
F ( q 2 ) = ∫ ρ ( r ) [ 1 + i q ⃗ ⋅ r ⃗ ℏ − ( q ⃗ ⋅ r ⃗ ) 2 2 ℏ 2 + ⋯ ] d 3 r . F(q^2)=\int\rho(r)\left[1+i\frac{\vec q\cdot\vec r}{\hbar}-\frac{(\vec q\cdot\vec r)^2}{2\hbar^2}+\cdots\right]d^3r . F ( q 2 ) = ∫ ρ ( r ) [ 1 + i ℏ q ⋅ r − 2 ℏ 2 ( q ⋅ r ) 2 + ⋯ ] d 3 r . 第1項は規格化条件から 1 1 1 ρ \rho ρ r ⃗ \vec r r 0 0 0 q ⃗ \vec q q
∫ ρ ( q ⃗ ⋅ r ⃗ ) 2 d 3 r = q 2 ∫ ρ r 2 cos 2 θ d 3 r = q 2 3 ∫ r 2 ρ ( r ) d 3 r = q 2 ⟨ r 2 ⟩ 3 \int\rho\,(\vec q\cdot\vec r)^2d^3r=q^2\int\rho\,r^2\cos^2\theta\,d^3r=\frac{q^2}{3}\int r^2\rho(r)\,d^3r=\frac{q^2\langle r^2\rangle}{3} ∫ ρ ( q ⋅ r ) 2 d 3 r = q 2 ∫ ρ r 2 cos 2 θ d 3 r = 3 q 2 ∫ r 2 ρ ( r ) d 3 r = 3 q 2 ⟨ r 2 ⟩ です（⟨ cos 2 θ ⟩ = 1 / 3 \langle\cos^2\theta\rangle=1/3 ⟨ cos 2 θ ⟩ = 1/3
F ( q 2 ) ≃ 1 − q 2 ⟨ r 2 ⟩ 6 ℏ 2 , ∣ F ( q 2 ) ∣ 2 ≃ 1 − q 2 ⟨ r 2 ⟩ 3 ℏ 2 F(q^2)\simeq1-\frac{q^2\langle r^2\rangle}{6\hbar^2},\qquad
\left|F(q^2)\right|^2\simeq1-\frac{q^2\langle r^2\rangle}{3\hbar^2} F ( q 2 ) ≃ 1 − 6 ℏ 2 q 2 ⟨ r 2 ⟩ , F ( q 2 ) 2 ≃ 1 − 3 ℏ 2 q 2 ⟨ r 2 ⟩ となり、∣ F ∣ 2 |F|^2 ∣ F ∣ 2 q 2 q^2 q 2
s ≡ d ∣ F ( q 2 ) ∣ 2 d ( q 2 ) = − ⟨ r 2 ⟩ 3 ℏ 2 s\equiv\frac{d|F(q^2)|^2}{d(q^2)}=-\frac{\langle r^2\rangle}{3\hbar^2} s ≡ d ( q 2 ) d ∣ F ( q 2 ) ∣ 2 = − 3 ℏ 2 ⟨ r 2 ⟩ と読み取れば
⟨ r 2 ⟩ = ℏ 3 ∣ s ∣ \sqrt{\langle r^2\rangle}=\hbar\sqrt{3\,\lvert s\rvert} ⟨ r 2 ⟩ = ℏ 3 ∣ s ∣ として荷電半径が求まります。q 2 → 0 q^2\to0 q 2 → 0 ∣ F ∣ 2 → 1 |F|^2\to1 ∣ F ∣ 2 → 1
図1 の q 2 q^2 q 2 ( q 2 , ∣ F ∣ 2 ) = ( 0 , 1.00 ) , ( 0.003 , 0.87 ) , ( 0.008 , 0.75 ) , ( 0.012 , 0.62 ) (q^2,|F|^2)=(0,1.00),(0.003,0.87),(0.008,0.75),(0.012,0.62) ( q 2 , ∣ F ∣ 2 ) = ( 0 , 1.00 ) , ( 0.003 , 0.87 ) , ( 0.008 , 0.75 ) , ( 0.012 , 0.62 ) q 2 q^2 q 2 ( G e V / c ) 2 (\mathrm{GeV}/c)^2 ( GeV / c ) 2
∣ s ∣ ≃ 1.00 − 0.62 0.012 ≃ 3 × 10 1 ( G e V / c ) − 2 \lvert s\rvert\simeq\frac{1.00-0.62}{0.012}\simeq3\times10^{1}\,(\mathrm{GeV}/c)^{-2} ∣ s ∣ ≃ 0.012 1.00 − 0.62 ≃ 3 × 1 0 1 ( GeV / c ) − 2 です。ℏ 2 \hbar^2 ℏ 2 ( G e V / c ) − 2 (\mathrm{GeV}/c)^{-2} ( GeV / c ) − 2 ℏ 2 / ( G e V / c ) 2 = ( ℏ c ) 2 / G e V 2 = ( 0.2 G e V ⋅ f m ) 2 / G e V 2 = 0.04 f m 2 \hbar^2/(\mathrm{GeV}/c)^2=(\hbar c)^2/\mathrm{GeV}^2=(0.2\,\mathrm{GeV\cdot fm})^2/\mathrm{GeV}^2=0.04\,\mathrm{fm^2} ℏ 2 / ( GeV / c ) 2 = ( ℏ c ) 2 / GeV 2 = ( 0.2 GeV ⋅ fm ) 2 / GeV 2 = 0.04 f m 2
⟨ r 2 ⟩ = 3 ℏ 2 ∣ s ∣ ≃ 3 × 30 × 0.04 = 3.6 f m 2 , ⟨ r 2 ⟩ ≃ 1.9 f m ≈ 2 f m . \langle r^2\rangle=3\hbar^2\lvert s\rvert\simeq3\times30\times0.04=3.6\,\mathrm{fm^2},
\qquad
\sqrt{\langle r^2\rangle}\simeq1.9\,\mathrm{fm}\approx2\,\mathrm{fm}. ⟨ r 2 ⟩ = 3 ℏ 2 ∣ s ∣ ≃ 3 × 30 × 0.04 = 3.6 f m 2 , ⟨ r 2 ⟩ ≃ 1.9 fm ≈ 2 fm . 答えは重陽子の荷電半径 ⟨ r 2 ⟩ ≈ 2 f m \sqrt{\langle r^2\rangle}\approx2\,\mathrm{fm} ⟨ r 2 ⟩ ≈ 2 fm 0.8 f m 0.8\,\mathrm{fm} 0.8 fm
交流回路の設問です。α \alpha α V 1 V_1 V 1 V 2 V_2 V 2 ω = 2 π f \omega=2\pi f ω = 2 π f
図1 は R R R C C C C C C
α = ∣ 1 / ( i ω C ) R + 1 / ( i ω C ) ∣ = ∣ 1 1 + i ω R C ∣ = 1 1 + ( ω R C ) 2 . \alpha=\left|\frac{1/(i\omega C)}{R+1/(i\omega C)}\right|=\left|\frac{1}{1+i\omega RC}\right|
=\frac{1}{\sqrt{1+(\omega RC)^2}} . α = R + 1/ ( iω C ) 1/ ( iω C ) = 1 + iω R C 1 = 1 + ( ω R C ) 2 1 . 漸近線は ω R C ≪ 1 \omega RC\ll1 ω R C ≪ 1 α = 1 \alpha=1 α = 1 ω R C ≫ 1 \omega RC\gg1 ω R C ≫ 1 α = 1 / ( ω R C ) \alpha=1/(\omega RC) α = 1/ ( ω R C ) ω R C = 1 \omega RC=1 ω R C = 1
f c = 1 2 π R C f_{\mathrm c}=\frac{1}{2\pi RC} f c = 2 π R C 1 で、そこでの実測値は α = 1 / 2 ≃ 0.7 \alpha=1/\sqrt2\simeq0.7 α = 1/ 2 ≃ 0.7 f ≃ 2 × 10 3 H z f\simeq2\times10^{3}\,\mathrm{Hz} f ≃ 2 × 1 0 3 Hz f c ≃ 2 × 10 3 H z f_{\mathrm c}\simeq2\times10^{3}\,\mathrm{Hz} f c ≃ 2 × 1 0 3 Hz R = 1 k Ω R=1\,\mathrm{k}\Omega R = 1 k Ω
C = 1 2 π f c R = 1 2 π × 2 × 10 3 × 10 3 ≃ 8 × 10 − 8 F C=\frac{1}{2\pi f_{\mathrm c}R}=\frac{1}{2\pi\times2\times10^{3}\times10^{3}}\simeq8\times10^{-8}\,\mathrm{F} C = 2 π f c R 1 = 2 π × 2 × 1 0 3 × 1 0 3 1 ≃ 8 × 1 0 − 8 F です。答えは C ≈ 8 × 10 − 8 F C\approx8\times10^{-8}\,\mathrm{F} C ≈ 8 × 1 0 − 8 F 0.08 μ F 0.08\,\mu\mathrm F 0.08 μ F
抵抗 1 個とコイル 1 個で同じ低周波透過特性を作るには、直列側にコイル、並列（出力）側に抵抗を置きます。すなわち、交流源の一方の端子からコイル L L L R R R R R R V 2 V_2 V 2 R R R C C C
α = ∣ R R + i ω L ∣ = 1 1 + ( ω L / R ) 2 \alpha=\left|\frac{R}{R+i\omega L}\right|=\frac{1}{\sqrt{1+(\omega L/R)^2}} α = R + iω L R = 1 + ( ω L / R ) 2 1 となり、ω R C \omega RC ω R C ω L / R \omega L/R ω L / R ω L / R = 1 \omega L/R=1 ω L / R = 1 f c = R / ( 2 π L ) f_{\mathrm c}=R/(2\pi L) f c = R / ( 2 π L ) f c ≃ 2 × 10 3 H z f_{\mathrm c}\simeq2\times10^{3}\,\mathrm{Hz} f c ≃ 2 × 1 0 3 Hz R = 100 Ω R=100\,\Omega R = 100 Ω
L = R 2 π f c = 100 2 π × 2 × 10 3 ≃ 8 × 10 − 3 H L=\frac{R}{2\pi f_{\mathrm c}}=\frac{100}{2\pi\times2\times10^{3}}\simeq8\times10^{-3}\,\mathrm{H} L = 2 π f c R = 2 π × 2 × 1 0 3 100 ≃ 8 × 1 0 − 3 H です。答えは L ≈ 8 × 10 − 3 H L\approx8\times10^{-3}\,\mathrm{H} L ≈ 8 × 1 0 − 3 H
図3 は、R R R L L L C C C V 2 V_2 V 2
Z = ( 1 i ω L + i ω C ) − 1 = i ω L 1 − ω 2 L C Z=\left(\frac{1}{i\omega L}+i\omega C\right)^{-1}=\frac{i\omega L}{1-\omega^2LC} Z = ( iω L 1 + iω C ) − 1 = 1 − ω 2 L C iω L なので
α = ∣ Z R + Z ∣ = ω L R 2 ( 1 − ω 2 L C ) 2 + ω 2 L 2 . \alpha=\left|\frac{Z}{R+Z}\right|
=\frac{\omega L}{\sqrt{R^2\left(1-\omega^2LC\right)^2+\omega^2L^2}} . α = R + Z Z = R 2 ( 1 − ω 2 L C ) 2 + ω 2 L 2 ω L . ω 0 = 1 / L C \omega_0=1/\sqrt{LC} ω 0 = 1/ L C α = 1 \alpha=1 α = 1 ω ≪ ω 0 \omega\ll\omega_0 ω ≪ ω 0 α ≃ ω L / R \alpha\simeq\omega L/R α ≃ ω L / R + 1 +1 + 1 ω ≫ ω 0 \omega\gg\omega_0 ω ≫ ω 0 α ≃ 1 / ( ω R C ) \alpha\simeq1/(\omega RC) α ≃ 1/ ( ω R C ) − 1 -1 − 1 ω = ω 0 \omega=\omega_0 ω = ω 0
α a s y m = ω 0 L R = 1 R L C \alpha_{\mathrm{asym}}=\frac{\omega_0L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}} α asym = R ω 0 L = R 1 C L です。図4 からピーク位置 f 0 ≃ 1 × 10 5 H z f_0\simeq1\times10^{5}\,\mathrm{Hz} f 0 ≃ 1 × 1 0 5 Hz α a s y m ≃ 0.2 \alpha_{\mathrm{asym}}\simeq0.2 α asym ≃ 0.2 C = 10 n F = 1 × 10 − 8 F C=10\,\mathrm{nF}=1\times10^{-8}\,\mathrm{F} C = 10 nF = 1 × 1 0 − 8 F
L = 1 ( 2 π f 0 ) 2 C = 1 ( 6.28 × 10 5 ) 2 × 10 − 8 ≃ 2.5 × 10 − 4 H , L=\frac{1}{(2\pi f_0)^2C}=\frac{1}{(6.28\times10^{5})^2\times10^{-8}}\simeq2.5\times10^{-4}\,\mathrm{H}, L = ( 2 π f 0 ) 2 C 1 = ( 6.28 × 1 0 5 ) 2 × 1 0 − 8 1 ≃ 2.5 × 1 0 − 4 H , R = 1 α a s y m L C = 1 0.2 2.5 × 10 − 4 10 − 8 ≃ 8 × 10 2 Ω . R=\frac{1}{\alpha_{\mathrm{asym}}}\sqrt{\frac{L}{C}}=\frac{1}{0.2}\sqrt{\frac{2.5\times10^{-4}}{10^{-8}}}\simeq8\times10^{2}\,\Omega . R = α asym 1 C L = 0.2 1 1 0 − 8 2.5 × 1 0 − 4 ≃ 8 × 1 0 2 Ω. 答えは L ≈ 2.5 × 10 − 4 H L\approx2.5\times10^{-4}\,\mathrm{H} L ≈ 2.5 × 1 0 − 4 H R ≈ 8 × 10 2 Ω R\approx8\times10^{2}\,\Omega R ≈ 8 × 1 0 2 Ω f = 10 4 H z f=10^{4}\,\mathrm{Hz} f = 1 0 4 Hz α ≃ 0.02 = 0.2 / 10 \alpha\simeq0.02=0.2/10 α ≃ 0.02 = 0.2/10 α ∝ ω \alpha\propto\omega α ∝ ω
図5 では、交流源の 2 つの端子が B と D に出ており、出力側は A が V 2 V_2 V 2 V 1 = V B − V D V_1=V_{\mathrm B}-V_{\mathrm D} V 1 = V B − V D V 2 = V A − V C V_2=V_{\mathrm A}-V_{\mathrm C} V 2 = V A − V C V 2 = ∣ V 1 ∣ V_2=|V_1| V 2 = ∣ V 1 ∣
動作を確認します。V 1 > 0 V_1>0 V 1 > 0 R R R V A = V B V_{\mathrm A}=V_{\mathrm B} V A = V B V C = V D V_{\mathrm C}=V_{\mathrm D} V C = V D V 2 = V B − V D = V 1 > 0 V_2=V_{\mathrm B}-V_{\mathrm D}=V_1>0 V 2 = V B − V D = V 1 > 0 V 1 < 0 V_1<0 V 1 < 0 R R R V 2 = V D − V B = − V 1 > 0 V_2=V_{\mathrm D}-V_{\mathrm B}=-V_1>0 V 2 = V D − V B = − V 1 > 0 V 2 = ∣ V 1 ∣ V_2=|V_1| V 2 = ∣ V 1 ∣
A–C 間にコンデンサを入れると、V 2 V_2 V 2 V max = 12 V V_{\max}=12\,\mathrm{V} V m a x = 12 V R R R
T 2 = 1 2 f = 1 2 × 50 = 1 × 10 − 2 s \frac{T}{2}=\frac{1}{2f}=\frac{1}{2\times50}=1\times10^{-2}\,\mathrm{s} 2 T = 2 f 1 = 2 × 50 1 = 1 × 1 0 − 2 s です。時定数は R C = 500 × 3 × 10 − 3 = 1.5 s RC=500\times3\times10^{-3}=1.5\,\mathrm{s} R C = 500 × 3 × 1 0 − 3 = 1.5 s
Δ V = V max ( 1 − e − T / 2 R C ) ≃ V max T 2 R C = 12 × 1 × 10 − 2 1.5 ≃ 8 × 10 − 2 V . \Delta V=V_{\max}\left(1-e^{-\frac{T/2}{RC}}\right)\simeq V_{\max}\frac{T}{2RC}
=12\times\frac{1\times10^{-2}}{1.5}\simeq8\times10^{-2}\,\mathrm{V}. Δ V = V m a x ( 1 − e − R C T /2 ) ≃ V m a x 2 R C T = 12 × 1.5 1 × 1 0 − 2 ≃ 8 × 1 0 − 2 V . 答えは Δ V ≈ 0.08 V \Delta V\approx0.08\,\mathrm{V} Δ V ≈ 0.08 V 12 V 12\,\mathrm V 12 V
図8 は交流源、ダイオード、コンデンサの直列閉回路で、コンデンサの両端が V 2 V_2 V 2 V 2 V_2 V 2 0 0 0 0 0 0
ダイオードに加わる電圧は V = V 1 − V 2 V=V_1-V_2 V = V 1 − V 2 V 1 = V 0 sin ω t V_1=V_0\sin\omega t V 1 = V 0 sin ω t V 0 = 10 m V V_0=10\,\mathrm{mV} V 0 = 10 mV I = g V ( 1 + V 2 V T ) I=gV\left(1+\dfrac{V}{2V_{\mathrm T}}\right) I = g V ( 1 + 2 V T V ) ⟨ V ⟩ = − V 2 \langle V\rangle=-V_2 ⟨ V ⟩ = − V 2 ⟨ V 2 ⟩ = V 0 2 2 + V 2 2 \langle V^2\rangle=\dfrac{V_0^2}{2}+V_2^2 ⟨ V 2 ⟩ = 2 V 0 2 + V 2 2
⟨ I ⟩ = g [ ⟨ V ⟩ + ⟨ V 2 ⟩ 2 V T ] = g [ − V 2 + V 0 2 / 2 + V 2 2 2 V T ] = 0. \langle I\rangle=g\left[\langle V\rangle+\frac{\langle V^2\rangle}{2V_{\mathrm T}}\right]
=g\left[-V_2+\frac{V_0^2/2+V_2^2}{2V_{\mathrm T}}\right]=0 . ⟨ I ⟩ = g [ ⟨ V ⟩ + 2 V T ⟨ V 2 ⟩ ] = g [ − V 2 + 2 V T V 0 2 /2 + V 2 2 ] = 0. V 2 V_2 V 2 V 0 V_0 V 0 V 2 2 V_2^2 V 2 2
V 2 ≃ V 0 2 4 V T = ( 10 m V ) 2 4 × 25 m V = 1 m V . V_2\simeq\frac{V_0^2}{4V_{\mathrm T}}=\frac{(10\,\mathrm{mV})^2}{4\times25\,\mathrm{mV}}=1\,\mathrm{mV}. V 2 ≃ 4 V T V 0 2 = 4 × 25 mV ( 10 mV ) 2 = 1 mV . 答えは V 2 ≈ 1 m V V_2\approx1\,\mathrm{mV} V 2 ≈ 1 mV V 2 2 V_2^2 V 2 2 1.02 m V 1.02\,\mathrm{mV} 1.02 mV g g g
出典: 東京大学大学院理学系研究科 物理学専攻 平成28年度 修士課程 入学試験問題 物理学。問題文は要約して引用しています。