4時間で 4 問という構成の年度です。前半 3 問は量子力学・統計力学・力学の王道で、いずれも教科書の標準的な道具（生成消滅演算子、転送行列、ラグランジアンと保存量）を最後まで走らせる体力を問います。後半の選択 3 問は実験寄りで、第3問後半と第4問後半、第6問は数値評価まで求められるため、単位と桁の管理が得点を分けます。
問題 分野 主題 第1問 量子力学 生成消滅演算子と四次非調和項の摂動、準古典極限 第2問 統計力学 1次元2状態スピン鎖の分配関数・転送行列・相関長 第3問 解析力学・天体物理 重力2体問題と連星ブラックホールのインスパイラル 第4問 光学・フーリエ解析 マイケルソン干渉計によるフーリエ分光と CMB 測定 第5問 素粒子・原子核 反応断面積、不変質量、制動放射、運動量測定の誤差 第6問 電磁気学・電気回路 同軸ケーブルの伝送線路方程式と表皮効果
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。
1次元調和振動子
H 0 = 1 2 m p 2 + m ω 2 2 x 2 H_0 = \frac{1}{2m}p^2 + \frac{m\omega^2}{2}x^2 H 0 = 2 m 1 p 2 + 2 m ω 2 x 2 を出発点とし、[ x , p ] = i ℏ [x,p]=i\hbar [ x , p ] = i ℏ
a = m ω 2 ℏ ( x + i p m ω ) , a † = m ω 2 ℏ ( x − i p m ω ) , N = a † a a = \sqrt{\frac{m\omega}{2\hbar}}\left(x+\frac{ip}{m\omega}\right),\qquad
a^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(x-\frac{ip}{m\omega}\right),\qquad
N = a^\dagger a a = 2ℏ mω ( x + mω i p ) , a † = 2ℏ mω ( x − mω i p ) , N = a † a を導入します。後半では V = λ x 4 V=\lambda x^4 V = λ x 4 λ > 0 \lambda>0 λ > 0 H = H 0 + V H=H_0+V H = H 0 + V
定義式から x + i p m ω x+\dfrac{ip}{m\omega} x + mω i p x − i p m ω x-\dfrac{ip}{m\omega} x − mω i p
[ x + i p m ω , x − i p m ω ] = − i m ω [ x , p ] + i m ω [ p , x ] = − i m ω ( i ℏ ) + i m ω ( − i ℏ ) = 2 ℏ m ω \left[x+\frac{ip}{m\omega},\,x-\frac{ip}{m\omega}\right]
= -\frac{i}{m\omega}[x,p] + \frac{i}{m\omega}[p,x]
= -\frac{i}{m\omega}(i\hbar) + \frac{i}{m\omega}(-i\hbar) = \frac{2\hbar}{m\omega} [ x + mω i p , x − mω i p ] = − mω i [ x , p ] + mω i [ p , x ] = − mω i ( i ℏ ) + mω i ( − i ℏ ) = mω 2ℏ 前の因子 m ω / ( 2 ℏ ) m\omega/(2\hbar) mω / ( 2ℏ )
[ a , a † ] = 1. [a,a^\dagger] = 1 . [ a , a † ] = 1. これを使えば
[ N , a ] = [ a † a , a ] = a † [ a , a ] + [ a † , a ] a = − a , [ N , a † ] = a † [ a , a † ] + [ a † , a † ] a = a † . [N,a] = [a^\dagger a,\,a] = a^\dagger[a,a] + [a^\dagger,a]a = -a,\qquad
[N,a^\dagger] = a^\dagger[a,a^\dagger] + [a^\dagger,a^\dagger]a = a^\dagger . [ N , a ] = [ a † a , a ] = a † [ a , a ] + [ a † , a ] a = − a , [ N , a † ] = a † [ a , a † ] + [ a † , a † ] a = a † . つぎに N N N x , p x,p x , p
N = a † a = m ω 2 ℏ ( x − i p m ω ) ( x + i p m ω ) = m ω 2 ℏ ( x 2 + p 2 m 2 ω 2 + i m ω [ x , p ] ) = 1 ℏ ω ( p 2 2 m + m ω 2 2 x 2 ) − 1 2 N = a^\dagger a = \frac{m\omega}{2\hbar}\left(x-\frac{ip}{m\omega}\right)\left(x+\frac{ip}{m\omega}\right)
= \frac{m\omega}{2\hbar}\left(x^2+\frac{p^2}{m^2\omega^2}+\frac{i}{m\omega}[x,p]\right)
= \frac{1}{\hbar\omega}\left(\frac{p^2}{2m}+\frac{m\omega^2}{2}x^2\right)-\frac{1}{2} N = a † a = 2ℏ mω ( x − mω i p ) ( x + mω i p ) = 2ℏ mω ( x 2 + m 2 ω 2 p 2 + mω i [ x , p ] ) = ℏ ω 1 ( 2 m p 2 + 2 m ω 2 x 2 ) − 2 1 したがって H 0 = ℏ ω ( N + 1 2 ) H_0 = \hbar\omega\left(N+\tfrac12\right) H 0 = ℏ ω ( N + 2 1 ) α = 1 \alpha=1 α = 1 β = 1 2 \beta=\tfrac12 β = 2 1
N N N ν \nu ν ∣ ν ⟩ |\nu\rangle ∣ ν ⟩
(i) 非負性は内積の正定値性から直接出ます。
ν = ⟨ ν ∣ N ∣ ν ⟩ = ⟨ ν ∣ a † a ∣ ν ⟩ = ∥ a ∣ ν ⟩ ∥ 2 ≥ 0 \nu = \langle\nu|N|\nu\rangle = \langle\nu|a^\dagger a|\nu\rangle = \bigl\lVert a|\nu\rangle \bigr\rVert^2 \ge 0 ν = ⟨ ν ∣ N ∣ ν ⟩ = ⟨ ν ∣ a † a ∣ ν ⟩ = a ∣ ν ⟩ 2 ≥ 0 等号成立は a ∣ ν ⟩ = 0 a|\nu\rangle=0 a ∣ ν ⟩ = 0
(ii) [ N , a ] = − a [N,a]=-a [ N , a ] = − a N a ∣ ν ⟩ = ( a N − a ) ∣ ν ⟩ = ( ν − 1 ) a ∣ ν ⟩ Na|\nu\rangle = (aN-a)|\nu\rangle = (\nu-1)a|\nu\rangle N a ∣ ν ⟩ = ( a N − a ) ∣ ν ⟩ = ( ν − 1 ) a ∣ ν ⟩ a ∣ ν ⟩ a|\nu\rangle a ∣ ν ⟩ ν − 1 \nu-1 ν − 1 a k ∣ ν ⟩ a^k|\nu\rangle a k ∣ ν ⟩ ν − k \nu-k ν − k
∥ a k ∣ ν ⟩ ∥ 2 = ν ( ν − 1 ) ⋯ ( ν − k + 1 ) \bigl\lVert a^k|\nu\rangle \bigr\rVert^2 = \nu(\nu-1)\cdots(\nu-k+1) a k ∣ ν ⟩ 2 = ν ( ν − 1 ) ⋯ ( ν − k + 1 ) となります（k = 1 k=1 k = 1 ∥ a ∣ ν − k ⟩ ∥ 2 = ν − k \lVert a\,|\nu-k\rangle\rVert^2=\nu-k ∥ a ∣ ν − k ⟩ ∥ 2 = ν − k ν \nu ν ν \nu ν n 0 n_0 n 0 k ≤ n 0 + 1 k\le n_0+1 k ≤ n 0 + 1 ν − k + 1 ≥ ν − n 0 > 0 \nu-k+1 \ge \nu-n_0 > 0 ν − k + 1 ≥ ν − n 0 > 0 a n 0 + 1 ∣ ν ⟩ ≠ 0 a^{n_0+1}|\nu\rangle \neq 0 a n 0 + 1 ∣ ν ⟩ = 0 ν − n 0 − 1 \nu-n_0-1 ν − n 0 − 1 − 1 < ν − n 0 − 1 < 0 -1<\nu-n_0-1<0 − 1 < ν − n 0 − 1 < 0 ν \nu ν ν = n \nu=n ν = n ∥ a n + 1 ∣ n ⟩ ∥ 2 = n ( n − 1 ) ⋯ 1 ⋅ 0 = 0 \lVert a^{n+1}|n\rangle\rVert^2 = n(n-1)\cdots 1\cdot 0 = 0 ∥ a n + 1 ∣ n ⟩ ∥ 2 = n ( n − 1 ) ⋯ 1 ⋅ 0 = 0
N N N 0 , 1 , 2 , … 0,1,2,\dots 0 , 1 , 2 , … ∥ a † ∣ n ⟩ ∥ 2 = ⟨ n ∣ ( N + 1 ) ∣ n ⟩ = n + 1 > 0 \lVert a^\dagger|n\rangle\rVert^2 = \langle n|(N+1)|n\rangle = n+1 > 0 ∥ a † ∣ n ⟩ ∥ 2 = ⟨ n ∣ ( N + 1 ) ∣ n ⟩ = n + 1 > 0 a † a^\dagger a † { 0 , 1 , 2 , … } \{0,1,2,\dots\} { 0 , 1 , 2 , … } n n n n n n H 0 = ℏ ω ( N + 1 2 ) H_0=\hbar\omega(N+\tfrac12) H 0 = ℏ ω ( N + 2 1 )
E n = ℏ ω ( n + 1 2 ) . E_n = \hbar\omega\left(n+\frac{1}{2}\right). E n = ℏ ω ( n + 2 1 ) . 固有状態は a † ∣ n ⟩ a^\dagger|n\rangle a † ∣ n ⟩ n + 1 \sqrt{n+1} n + 1 ∣ n + 1 ⟩ = a † n + 1 ∣ n ⟩ |n+1\rangle = \dfrac{a^\dagger}{\sqrt{n+1}}|n\rangle ∣ n + 1 ⟩ = n + 1 a † ∣ n ⟩
∣ n ⟩ = 1 n ! ( a † ) n ∣ 0 ⟩ ( n ≥ 1 ) . |n\rangle = \frac{1}{\sqrt{n!}}\left(a^\dagger\right)^n |0\rangle \qquad (n\ge 1). ∣ n ⟩ = n ! 1 ( a † ) n ∣0 ⟩ ( n ≥ 1 ) . (i) H = H 0 + V H=H_0+V H = H 0 + V V V V H 0 ∣ k ⟩ = E k ( 0 ) ∣ k ⟩ H_0|k\rangle = E_k^{(0)}|k\rangle H 0 ∣ k ⟩ = E k ( 0 ) ∣ k ⟩ V k l = ⟨ k ∣ V ∣ l ⟩ V_{kl}=\langle k|V|l\rangle V k l = ⟨ k ∣ V ∣ l ⟩ ∣ n ⟩ + ∣ n ( 1 ) ⟩ + ⋯ |n\rangle + |n^{(1)}\rangle + \cdots ∣ n ⟩ + ∣ n ( 1 ) ⟩ + ⋯ E n ( 0 ) + E n ( 1 ) + ⋯ E_n^{(0)}+E_n^{(1)}+\cdots E n ( 0 ) + E n ( 1 ) + ⋯ ⟨ n ∣ n ( 1 ) ⟩ = 0 \langle n|n^{(1)}\rangle = 0 ⟨ n ∣ n ( 1 ) ⟩ = 0
H 0 ∣ n ( 1 ) ⟩ + V ∣ n ⟩ = E n ( 0 ) ∣ n ( 1 ) ⟩ + E n ( 1 ) ∣ n ⟩ . H_0|n^{(1)}\rangle + V|n\rangle = E_n^{(0)}|n^{(1)}\rangle + E_n^{(1)}|n\rangle . H 0 ∣ n ( 1 ) ⟩ + V ∣ n ⟩ = E n ( 0 ) ∣ n ( 1 ) ⟩ + E n ( 1 ) ∣ n ⟩ . 左から ⟨ n ∣ \langle n| ⟨ n ∣ ⟨ n ∣ H 0 = E n ( 0 ) ⟨ n ∣ \langle n|H_0 = E_n^{(0)}\langle n| ⟨ n ∣ H 0 = E n ( 0 ) ⟨ n ∣
E n ( 1 ) = V n n . E_n^{(1)} = V_{nn}. E n ( 1 ) = V nn . 左から ⟨ k ∣ \langle k| ⟨ k ∣ k ≠ n k\ne n k = n ∣ n ( 1 ) ⟩ = ∑ k ≠ n c k ( 1 ) ∣ k ⟩ |n^{(1)}\rangle = \sum_{k\ne n}c_k^{(1)}|k\rangle ∣ n ( 1 ) ⟩ = ∑ k = n c k ( 1 ) ∣ k ⟩
c k ( 1 ) = V k n E n ( 0 ) − E k ( 0 ) . c_k^{(1)} = \frac{V_{kn}}{E_n^{(0)}-E_k^{(0)}} . c k ( 1 ) = E n ( 0 ) − E k ( 0 ) V k n . 2次の項は
H 0 ∣ n ( 2 ) ⟩ + V ∣ n ( 1 ) ⟩ = E n ( 0 ) ∣ n ( 2 ) ⟩ + E n ( 1 ) ∣ n ( 1 ) ⟩ + E n ( 2 ) ∣ n ⟩ H_0|n^{(2)}\rangle + V|n^{(1)}\rangle = E_n^{(0)}|n^{(2)}\rangle + E_n^{(1)}|n^{(1)}\rangle + E_n^{(2)}|n\rangle H 0 ∣ n ( 2 ) ⟩ + V ∣ n ( 1 ) ⟩ = E n ( 0 ) ∣ n ( 2 ) ⟩ + E n ( 1 ) ∣ n ( 1 ) ⟩ + E n ( 2 ) ∣ n ⟩ で、左から ⟨ n ∣ \langle n| ⟨ n ∣ ⟨ n ∣ n ( 1 ) ⟩ = 0 \langle n|n^{(1)}\rangle=0 ⟨ n ∣ n ( 1 ) ⟩ = 0 E n ( 2 ) = ⟨ n ∣ V ∣ n ( 1 ) ⟩ E_n^{(2)} = \langle n|V|n^{(1)}\rangle E n ( 2 ) = ⟨ n ∣ V ∣ n ( 1 ) ⟩
E n ( 2 ) = ∑ k ≠ n V n k V k n E n ( 0 ) − E k ( 0 ) = ∑ k ≠ n ∣ V n k ∣ 2 E n ( 0 ) − E k ( 0 ) . E_n^{(2)} = \sum_{k\ne n}\frac{V_{nk}V_{kn}}{E_n^{(0)}-E_k^{(0)}} = \sum_{k\ne n}\frac{\lvert V_{nk}\rvert^2}{E_n^{(0)}-E_k^{(0)}} . E n ( 2 ) = k = n ∑ E n ( 0 ) − E k ( 0 ) V nk V k n = k = n ∑ E n ( 0 ) − E k ( 0 ) ∣ V nk ∣ 2 . 縮退がないので分母はどこでも 0 になりません。
(ii) x = ℓ ( a + a † ) x = \ell\,(a+a^\dagger) x = ℓ ( a + a † ) ℓ ≡ ℏ 2 m ω \ell \equiv \sqrt{\dfrac{\hbar}{2m\omega}} ℓ ≡ 2 mω ℏ ( a + a † ) 4 ∣ 0 ⟩ (a+a^\dagger)^4|0\rangle ( a + a † ) 4 ∣0 ⟩
( a + a † ) ∣ 0 ⟩ = ∣ 1 ⟩ , ( a + a † ) ∣ 1 ⟩ = ∣ 0 ⟩ + 2 ∣ 2 ⟩ , ( a + a † ) ( ∣ 0 ⟩ + 2 ∣ 2 ⟩ ) = 3 ∣ 1 ⟩ + 6 ∣ 3 ⟩ , ( a + a † ) ( 3 ∣ 1 ⟩ + 6 ∣ 3 ⟩ ) = 3 ∣ 0 ⟩ + 6 2 ∣ 2 ⟩ + 2 6 ∣ 4 ⟩ . \begin{aligned}
(a+a^\dagger)|0\rangle &= |1\rangle,\\
(a+a^\dagger)|1\rangle &= |0\rangle + \sqrt{2}\,|2\rangle,\\
(a+a^\dagger)\left(|0\rangle+\sqrt{2}\,|2\rangle\right) &= 3|1\rangle + \sqrt{6}\,|3\rangle,\\
(a+a^\dagger)\left(3|1\rangle+\sqrt{6}\,|3\rangle\right) &= 3|0\rangle + 6\sqrt{2}\,|2\rangle + 2\sqrt{6}\,|4\rangle .
\end{aligned} ( a + a † ) ∣0 ⟩ ( a + a † ) ∣1 ⟩ ( a + a † ) ( ∣0 ⟩ + 2 ∣2 ⟩ ) ( a + a † ) ( 3∣1 ⟩ + 6 ∣3 ⟩ ) = ∣1 ⟩ , = ∣0 ⟩ + 2 ∣2 ⟩ , = 3∣1 ⟩ + 6 ∣3 ⟩ , = 3∣0 ⟩ + 6 2 ∣2 ⟩ + 2 6 ∣4 ⟩ . よって x 4 ∣ 0 ⟩ = ℓ 4 ( 3 ∣ 0 ⟩ + 6 2 ∣ 2 ⟩ + 2 6 ∣ 4 ⟩ ) x^4|0\rangle = \ell^4\left(3|0\rangle+6\sqrt2\,|2\rangle+2\sqrt6\,|4\rangle\right) x 4 ∣0 ⟩ = ℓ 4 ( 3∣0 ⟩ + 6 2 ∣2 ⟩ + 2 6 ∣4 ⟩ ) V = λ x 4 V=\lambda x^4 V = λ x 4
V 00 = 3 λ ℓ 4 , V 20 = 6 2 λ ℓ 4 , V 40 = 2 6 λ ℓ 4 , V_{00} = 3\lambda\ell^4,\qquad V_{20} = 6\sqrt{2}\,\lambda\ell^4,\qquad V_{40} = 2\sqrt{6}\,\lambda\ell^4, V 00 = 3 λ ℓ 4 , V 20 = 6 2 λ ℓ 4 , V 40 = 2 6 λ ℓ 4 , それ以外の k k k V k 0 = 0 V_{k0}=0 V k 0 = 0 ℓ 4 = ℏ 2 / ( 4 m 2 ω 2 ) \ell^4 = \hbar^2/(4m^2\omega^2) ℓ 4 = ℏ 2 / ( 4 m 2 ω 2 ) x 4 x^4 x 4 k k k ( a + a † ) 4 (a+a^\dagger)^4 ( a + a † ) 4 k ≥ 5 k\ge 5 k ≥ 5 V 00 = 3 λ ℓ 4 V_{00}=3\lambda\ell^4 V 00 = 3 λ ℓ 4 ⟨ x 2 ⟩ 0 = ℓ 2 \langle x^2\rangle_0=\ell^2 ⟨ x 2 ⟩ 0 = ℓ 2 ⟨ x 4 ⟩ = 3 ⟨ x 2 ⟩ 2 \langle x^4\rangle = 3\langle x^2\rangle^2 ⟨ x 4 ⟩ = 3 ⟨ x 2 ⟩ 2
基底状態のエネルギーは E 0 ( 0 ) − E k ( 0 ) = − k ℏ ω E_0^{(0)}-E_k^{(0)} = -k\hbar\omega E 0 ( 0 ) − E k ( 0 ) = − k ℏ ω
E 0 = ℏ ω 2 + 3 λ ℓ 4 + ( 6 2 λ ℓ 4 ) 2 − 2 ℏ ω + ( 2 6 λ ℓ 4 ) 2 − 4 ℏ ω + O ( λ 3 ) = ℏ ω 2 + 3 λ ℓ 4 − λ 2 ℓ 8 ℏ ω ( 72 2 + 24 4 ) + O ( λ 3 ) = ℏ ω 2 + 3 ℏ 2 4 m 2 ω 2 λ − 21 ℏ 3 8 m 4 ω 5 λ 2 + O ( λ 3 ) . \begin{aligned}
E_0 &= \frac{\hbar\omega}{2} + 3\lambda\ell^4
+ \frac{(6\sqrt2\,\lambda\ell^4)^2}{-2\hbar\omega} + \frac{(2\sqrt6\,\lambda\ell^4)^2}{-4\hbar\omega} + O(\lambda^3)\\
&= \frac{\hbar\omega}{2} + 3\lambda\ell^4 - \frac{\lambda^2\ell^8}{\hbar\omega}\left(\frac{72}{2}+\frac{24}{4}\right) + O(\lambda^3)\\
&= \frac{\hbar\omega}{2} + \frac{3\hbar^2}{4m^2\omega^2}\,\lambda - \frac{21\hbar^3}{8m^4\omega^5}\,\lambda^2 + O(\lambda^3).
\end{aligned} E 0 = 2 ℏ ω + 3 λ ℓ 4 + − 2ℏ ω ( 6 2 λ ℓ 4 ) 2 + − 4ℏ ω ( 2 6 λ ℓ 4 ) 2 + O ( λ 3 ) = 2 ℏ ω + 3 λ ℓ 4 − ℏ ω λ 2 ℓ 8 ( 2 72 + 4 24 ) + O ( λ 3 ) = 2 ℏ ω + 4 m 2 ω 2 3 ℏ 2 λ − 8 m 4 ω 5 21 ℏ 3 λ 2 + O ( λ 3 ) . 答えは E 0 = ℏ ω 2 [ 1 + 3 2 ℏ λ m 2 ω 3 − 21 4 ( ℏ λ m 2 ω 3 ) 2 + ⋯ ] E_0 = \dfrac{\hbar\omega}{2}\left[1+\dfrac{3}{2}\dfrac{\hbar\lambda}{m^2\omega^3}-\dfrac{21}{4}\left(\dfrac{\hbar\lambda}{m^2\omega^3}\right)^2+\cdots\right] E 0 = 2 ℏ ω [ 1 + 2 3 m 2 ω 3 ℏ λ − 4 21 ( m 2 ω 3 ℏ λ ) 2 + ⋯ ] ℏ λ / ( m 2 ω 3 ) = λ ℓ 4 / ( ℏ ω ) × 4 \hbar\lambda/(m^2\omega^3) = \lambda\ell^4/(\hbar\omega)\times 4 ℏ λ / ( m 2 ω 3 ) = λ ℓ 4 / ( ℏ ω ) × 4 λ ℓ 4 ≪ ℏ ω \lambda\ell^4 \ll \hbar\omega λ ℓ 4 ≪ ℏ ω
n n n x 0 x_0 x 0 λ x 4 \lambda x^4 λ x 4 1 2 m ω 2 x 2 \tfrac12 m\omega^2x^2 2 1 m ω 2 x 2 2 λ x 0 2 / ( m ω 2 ) 2\lambda x_0^2/(m\omega^2) 2 λ x 0 2 / ( m ω 2 )
∮ p d x = 2 ∫ − x 0 x 0 2 m ( E − λ x 4 ) d x = 2 π ℏ ( n + 1 2 ) , x 0 = ( E λ ) 1 / 4 \oint p\,dx = 2\int_{-x_0}^{x_0}\sqrt{2m\left(E-\lambda x^4\right)}\,dx = 2\pi\hbar\left(n+\frac{1}{2}\right),
\qquad x_0=\left(\frac{E}{\lambda}\right)^{1/4} ∮ p d x = 2 ∫ − x 0 x 0 2 m ( E − λ x 4 ) d x = 2 π ℏ ( n + 2 1 ) , x 0 = ( λ E ) 1/4 において x = x 0 u x=x_0u x = x 0 u
∮ p d x = 2 2 m E x 0 ∫ − 1 1 1 − u 4 d u = c 0 m λ − 1 / 4 E 3 / 4 , c 0 = 2 2 ∫ − 1 1 1 − u 4 d u ≃ 4.9 \oint p\,dx = 2\sqrt{2mE}\,x_0\int_{-1}^{1}\sqrt{1-u^4}\,du
= c_0\sqrt{m}\,\lambda^{-1/4}E^{3/4},\qquad c_0 = 2\sqrt{2}\int_{-1}^{1}\sqrt{1-u^4}\,du \simeq 4.9 ∮ p d x = 2 2 m E x 0 ∫ − 1 1 1 − u 4 d u = c 0 m λ − 1/4 E 3/4 , c 0 = 2 2 ∫ − 1 1 1 − u 4 d u ≃ 4.9 となり、位相空間の面積は E 3 / 4 E^{3/4} E 3/4 2 π ℏ n 2\pi\hbar n 2 π ℏ n E 3 / 4 ∝ n E^{3/4}\propto n E 3/4 ∝ n
E n ∝ n 4 / 3 ( E n ∼ ℏ 4 / 3 λ 1 / 3 m − 2 / 3 n 4 / 3 ) . E_n \propto n^{4/3}\qquad\left(E_n \sim \hbar^{4/3}\lambda^{1/3}m^{-2/3}\,n^{4/3}\right). E n ∝ n 4/3 ( E n ∼ ℏ 4/3 λ 1/3 m − 2/3 n 4/3 ) . 答えは n n n 4 / 3 4/3 4/3 V ∝ ∣ x ∣ s V\propto |x|^s V ∝ ∣ x ∣ s E n ∝ n 2 s / ( s + 2 ) E_n\propto n^{2s/(s+2)} E n ∝ n 2 s / ( s + 2 ) s = 2 s=2 s = 2 n 1 n^1 n 1 s = 4 s=4 s = 4 n 4 / 3 n^{4/3} n 4/3 s → ∞ s\to\infty s → ∞ n 2 n^2 n 2 ℏ 4 / 3 λ 1 / 3 m − 2 / 3 \hbar^{4/3}\lambda^{1/3}m^{-2/3} ℏ 4/3 λ 1/3 m − 2/3 [ λ ] = J / m 4 [\lambda]=\mathrm{J/m^4} [ λ ] = J/ m 4
L L L S i \mathbf{S}_i S i x ^ = ( 1 , 0 , 0 ) \hat x=(1,0,0) x ^ = ( 1 , 0 , 0 ) y ^ = ( 0 , 1 , 0 ) \hat y=(0,1,0) y ^ = ( 0 , 1 , 0 )
E ( S 1 , … , S L ) = − J ∑ i = 1 L − 1 S i ⋅ S i + 1 , J > 0 E(\mathbf{S}_1,\dots,\mathbf{S}_L) = -J\sum_{i=1}^{L-1}\mathbf{S}_i\cdot\mathbf{S}_{i+1},\qquad J>0 E ( S 1 , … , S L ) = − J i = 1 ∑ L − 1 S i ⋅ S i + 1 , J > 0 で与えられる古典系です。x ^ ⋅ x ^ = y ^ ⋅ y ^ = 1 \hat x\cdot\hat x=\hat y\cdot\hat y=1 x ^ ⋅ x ^ = y ^ ⋅ y ^ = 1 x ^ ⋅ y ^ = 0 \hat x\cdot\hat y=0 x ^ ⋅ y ^ = 0 − J -J − J 0 0 0 β = 1 / ( k B T ) \beta=1/(k_\mathrm{B}T) β = 1/ ( k B T ) Z L ( β ) Z_L(\beta) Z L ( β )
L = 2 L=2 L = 2
( x ^ , x ^ ) : E = − J , ( y ^ , y ^ ) : E = − J , ( x ^ , y ^ ) : E = 0 , ( y ^ , x ^ ) : E = 0. \begin{aligned}
(\hat x,\hat x):&\ E=-J, & (\hat y,\hat y):&\ E=-J,\\
(\hat x,\hat y):&\ E=0, & (\hat y,\hat x):&\ E=0.
\end{aligned} ( x ^ , x ^ ) : ( x ^ , y ^ ) : E = − J , E = 0 , ( y ^ , y ^ ) : ( y ^ , x ^ ) : E = − J , E = 0. したがって
Z 2 ( β ) = 2 e β J + 2 = 2 ( e β J + 1 ) . Z_2(\beta) = 2e^{\beta J} + 2 = 2\left(e^{\beta J}+1\right). Z 2 ( β ) = 2 e β J + 2 = 2 ( e β J + 1 ) . L = 3 L=3 L = 3
( x ^ , x ^ , x ^ ) , ( y ^ , y ^ , y ^ ) : E = − 2 J ( x ^ , x ^ , y ^ ) , ( y ^ , x ^ , x ^ ) , ( x ^ , y ^ , y ^ ) , ( y ^ , y ^ , x ^ ) : E = − J ( x ^ , y ^ , x ^ ) , ( y ^ , x ^ , y ^ ) : E = 0 \begin{aligned}
(\hat x,\hat x,\hat x),\ (\hat y,\hat y,\hat y):&\quad E=-2J\\
(\hat x,\hat x,\hat y),\ (\hat y,\hat x,\hat x),\ (\hat x,\hat y,\hat y),\ (\hat y,\hat y,\hat x):&\quad E=-J\\
(\hat x,\hat y,\hat x),\ (\hat y,\hat x,\hat y):&\quad E=0
\end{aligned} ( x ^ , x ^ , x ^ ) , ( y ^ , y ^ , y ^ ) : ( x ^ , x ^ , y ^ ) , ( y ^ , x ^ , x ^ ) , ( x ^ , y ^ , y ^ ) , ( y ^ , y ^ , x ^ ) : ( x ^ , y ^ , x ^ ) , ( y ^ , x ^ , y ^ ) : E = − 2 J E = − J E = 0 内訳は E = − 2 J E=-2J E = − 2 J E = − J E=-J E = − J E = 0 E=0 E = 0
Z 3 ( β ) = 2 e 2 β J + 4 e β J + 2 = 2 ( e β J + 1 ) 2 . Z_3(\beta) = 2e^{2\beta J} + 4e^{\beta J} + 2 = 2\left(e^{\beta J}+1\right)^2 . Z 3 ( β ) = 2 e 2 β J + 4 e β J + 2 = 2 ( e β J + 1 ) 2 . L − 1 L-1 L − 1 S L \mathbf{S}_L S L S L = S L − 1 \mathbf{S}_L=\mathbf{S}_{L-1} S L = S L − 1 e β J e^{\beta J} e β J 1 1 1 e β J + 1 e^{\beta J}+1 e β J + 1 S L − 1 \mathbf{S}_{L-1} S L − 1
Z L ( β ) = ( e β J + 1 ) Z L − 1 ( β ) ( L ≥ 3 ) Z_L(\beta) = \left(e^{\beta J}+1\right)Z_{L-1}(\beta)\qquad (L\ge 3) Z L ( β ) = ( e β J + 1 ) Z L − 1 ( β ) ( L ≥ 3 ) が成り立ち、Z 2 = 2 ( e β J + 1 ) Z_2 = 2(e^{\beta J}+1) Z 2 = 2 ( e β J + 1 )
Z L ( β ) = 2 ( e β J + 1 ) L − 1 . Z_L(\beta) = 2\left(e^{\beta J}+1\right)^{L-1}. Z L ( β ) = 2 ( e β J + 1 ) L − 1 . L = 2 , 3 L=2,3 L = 2 , 3 L − 1 L-1 L − 1 e β J e^{\beta J} e β J 1 1 1
F L ( β ) = − 1 β ln Z L ( β ) = − 1 β [ ln 2 + ( L − 1 ) ln ( 1 + e β J ) ] . F_L(\beta) = -\frac{1}{\beta}\ln Z_L(\beta)
= -\frac{1}{\beta}\left[\ln 2 + (L-1)\ln\left(1+e^{\beta J}\right)\right]. F L ( β ) = − β 1 ln Z L ( β ) = − β 1 [ ln 2 + ( L − 1 ) ln ( 1 + e β J ) ] . 内部エネルギーは
U L = − ∂ ∂ β ln Z L = − ( L − 1 ) J e β J 1 + e β J U_L = -\frac{\partial}{\partial\beta}\ln Z_L = -(L-1)\frac{Je^{\beta J}}{1+e^{\beta J}} U L = − ∂ β ∂ ln Z L = − ( L − 1 ) 1 + e β J J e β J なので、L → ∞ L\to\infty L → ∞
u ( β ) = lim L → ∞ U L L = − J e β J 1 + e β J = − J 1 + e − β J . u(\beta) = \lim_{L\to\infty}\frac{U_L}{L} = -\frac{Je^{\beta J}}{1+e^{\beta J}} = -\frac{J}{1+e^{-\beta J}} . u ( β ) = L → ∞ lim L U L = − 1 + e β J J e β J = − 1 + e − β J J . 高温極限 β J → 0 \beta J\to 0 β J → 0 u → − J / 2 u\to -J/2 u → − J /2 β J → ∞ \beta J\to\infty β J → ∞ u → − J u\to -J u → − J
出現確率との関係は次のように読めます。L → ∞ L\to\infty L → ∞
p ∥ = e β J e β J + 1 = 1 1 + e − β J p_\parallel = \frac{e^{\beta J}}{e^{\beta J}+1} = \frac{1}{1+e^{-\beta J}} p ∥ = e β J + 1 e β J = 1 + e − β J 1 をもちます。平行なボンドのエネルギーは − J -J − J 0 0 0 u = − J p ∥ u = -J p_\parallel u = − J p ∥ p ∥ = 1 / 2 p_\parallel = 1/2 p ∥ = 1/2 u = − J / 2 u=-J/2 u = − J /2 e β J e^{\beta J} e β J p ∥ → 1 p_\parallel\to 1 p ∥ → 1 − J ( L − 1 ) -J(L-1) − J ( L − 1 ) u → − J u\to -J u → − J
S 1 = a ^ \mathbf{S}_1=\hat a S 1 = a ^ L − 1 L-1 L − 1 S L = b ^ \mathbf{S}_L=\hat b S L = b ^ w ( c ^ , b ^ ) = e β J w(\hat c,\hat b)=e^{\beta J} w ( c ^ , b ^ ) = e β J c ^ = b ^ \hat c=\hat b c ^ = b ^ w ( c ^ , b ^ ) = 1 w(\hat c,\hat b)=1 w ( c ^ , b ^ ) = 1 c ^ ≠ b ^ \hat c\ne\hat b c ^ = b ^
Z L P L ( a ^ , b ^ ) = ∑ c ^ [ Z L − 1 P L − 1 ( a ^ , c ^ ) ] w ( c ^ , b ^ ) Z_L\,P_L(\hat a,\hat b) = \sum_{\hat c}\left[Z_{L-1}P_{L-1}(\hat a,\hat c)\right]w(\hat c,\hat b) Z L P L ( a ^ , b ^ ) = c ^ ∑ [ Z L − 1 P L − 1 ( a ^ , c ^ ) ] w ( c ^ , b ^ ) であり、設問3 の Z L − 1 / Z L = 1 / ( e β J + 1 ) Z_{L-1}/Z_L = 1/(e^{\beta J}+1) Z L − 1 / Z L = 1/ ( e β J + 1 )
( Q ( x ^ , x ^ ) Q ( x ^ , y ^ ) Q ( y ^ , x ^ ) Q ( y ^ , y ^ ) ) = 1 e β J + 1 ( e β J 1 1 e β J ) , \begin{pmatrix} Q(\hat x,\hat x) & Q(\hat x,\hat y)\\ Q(\hat y,\hat x) & Q(\hat y,\hat y)\end{pmatrix}
= \frac{1}{e^{\beta J}+1}\begin{pmatrix} e^{\beta J} & 1\\ 1 & e^{\beta J}\end{pmatrix}, ( Q ( x ^ , x ^ ) Q ( y ^ , x ^ ) Q ( x ^ , y ^ ) Q ( y ^ , y ^ ) ) = e β J + 1 1 ( e β J 1 1 e β J ) , すなわち Q ( x ^ , x ^ ) = Q ( y ^ , y ^ ) = e β J e β J + 1 Q(\hat x,\hat x)=Q(\hat y,\hat y)=\dfrac{e^{\beta J}}{e^{\beta J}+1} Q ( x ^ , x ^ ) = Q ( y ^ , y ^ ) = e β J + 1 e β J Q ( x ^ , y ^ ) = Q ( y ^ , x ^ ) = 1 e β J + 1 Q(\hat x,\hat y)=Q(\hat y,\hat x)=\dfrac{1}{e^{\beta J}+1} Q ( x ^ , y ^ ) = Q ( y ^ , x ^ ) = e β J + 1 1 Q ( a ^ , b ^ ) Q(\hat a,\hat b) Q ( a ^ , b ^ ) S L − 1 = a ^ \mathbf{S}_{L-1}=\hat a S L − 1 = a ^ S L = b ^ \mathbf{S}_L=\hat b S L = b ^
設問6 の関係を繰り返すと、P L = P 2 Q L − 2 P_L = P_2\,Q^{L-2} P L = P 2 Q L − 2 L = 2 L=2 L = 2
P 2 = 1 2 ( e β J + 1 ) ( e β J 1 1 e β J ) = 1 2 Q P_2 = \frac{1}{2\left(e^{\beta J}+1\right)}\begin{pmatrix} e^{\beta J} & 1\\ 1 & e^{\beta J}\end{pmatrix} = \frac{1}{2}Q P 2 = 2 ( e β J + 1 ) 1 ( e β J 1 1 e β J ) = 2 1 Q なので P L = 1 2 Q L − 1 P_L = \tfrac12 Q^{L-1} P L = 2 1 Q L − 1 Q Q Q ( 1 , 1 ) (1,1) ( 1 , 1 ) 1 1 1 ( 1 , − 1 ) (1,-1) ( 1 , − 1 )
q ≡ e β J − 1 e β J + 1 = tanh β J 2 q \equiv \frac{e^{\beta J}-1}{e^{\beta J}+1} = \tanh\frac{\beta J}{2} q ≡ e β J + 1 e β J − 1 = tanh 2 β J なので、対応する射影演算子を使って
Q L − 1 = 1 2 ( 1 + q L − 1 1 − q L − 1 1 − q L − 1 1 + q L − 1 ) , P L ( x ^ , x ^ ) = P L ( y ^ , y ^ ) = 1 + q L − 1 4 . Q^{L-1} = \frac{1}{2}\begin{pmatrix} 1+q^{L-1} & 1-q^{L-1}\\ 1-q^{L-1} & 1+q^{L-1}\end{pmatrix},
\qquad
P_L(\hat x,\hat x) = P_L(\hat y,\hat y) = \frac{1+q^{L-1}}{4}. Q L − 1 = 2 1 ( 1 + q L − 1 1 − q L − 1 1 − q L − 1 1 + q L − 1 ) , P L ( x ^ , x ^ ) = P L ( y ^ , y ^ ) = 4 1 + q L − 1 . a ^ ⋅ b ^ \hat a\cdot\hat b a ^ ⋅ b ^
⟨ S 1 ⋅ S L ⟩ = P L ( x ^ , x ^ ) + P L ( y ^ , y ^ ) = 1 2 [ 1 + ( e β J − 1 e β J + 1 ) L − 1 ] = 1 2 + 1 2 tanh L − 1 β J 2 . \langle \mathbf{S}_1\cdot\mathbf{S}_L\rangle = P_L(\hat x,\hat x)+P_L(\hat y,\hat y)
= \frac{1}{2}\left[1+\left(\frac{e^{\beta J}-1}{e^{\beta J}+1}\right)^{L-1}\right]
= \frac{1}{2}+\frac{1}{2}\tanh^{L-1}\!\frac{\beta J}{2}. ⟨ S 1 ⋅ S L ⟩ = P L ( x ^ , x ^ ) + P L ( y ^ , y ^ ) = 2 1 [ 1 + ( e β J + 1 e β J − 1 ) L − 1 ] = 2 1 + 2 1 tanh L − 1 2 β J . L = 2 L=2 L = 2 e β J / ( e β J + 1 ) = p ∥ e^{\beta J}/(e^{\beta J}+1)=p_\parallel e β J / ( e β J + 1 ) = p ∥ β → 0 \beta\to 0 β → 0 1 / 2 1/2 1/2 β → ∞ \beta\to\infty β → ∞ 1 1 1
式(3)と比べれば A = B = 1 2 A=B=\tfrac12 A = B = 2 1 exp ( − 1 / ξ ) = q \exp(-1/\xi)=q exp ( − 1/ ξ ) = q
ξ = − 1 ln tanh ( β J / 2 ) = 1 ln coth ( β J / 2 ) . \xi = -\frac{1}{\ln\tanh(\beta J/2)} = \frac{1}{\ln\coth(\beta J/2)} . ξ = − ln tanh ( β J /2 ) 1 = ln coth ( β J /2 ) 1 . 高温極限 β J ≪ 1 \beta J\ll 1 β J ≪ 1 tanh ( β J / 2 ) ≃ β J / 2 \tanh(\beta J/2)\simeq \beta J/2 tanh ( β J /2 ) ≃ β J /2
ξ ≃ 1 ln ( 2 / β J ) , \xi \simeq \frac{1}{\ln\left(2/\beta J\right)} , ξ ≃ ln ( 2/ β J ) 1 , これは 1 サイト分より短く、隣同士でしか相関が残らないことを表します。低温極限 β J ≫ 1 \beta J\gg 1 β J ≫ 1 tanh ( β J / 2 ) = 1 − e − β J 1 + e − β J ≃ 1 − 2 e − β J \tanh(\beta J/2) = \dfrac{1-e^{-\beta J}}{1+e^{-\beta J}} \simeq 1-2e^{-\beta J} tanh ( β J /2 ) = 1 + e − β J 1 − e − β J ≃ 1 − 2 e − β J ln tanh ( β J / 2 ) ≃ − 2 e − β J \ln\tanh(\beta J/2)\simeq -2e^{-\beta J} ln tanh ( β J /2 ) ≃ − 2 e − β J
ξ ≃ 1 2 e β J . \xi \simeq \frac{1}{2}e^{\beta J} . ξ ≃ 2 1 e β J . 相関長は T → 0 T\to 0 T → 0 J J J ∼ e β J \sim e^{\beta J} ∼ e β J ξ \xi ξ
質量 m 1 , m 2 m_1,m_2 m 1 , m 2 r = r 2 − r 1 \mathbf{r}=\mathbf{r}_2-\mathbf{r}_1 r = r 2 − r 1 ( r , φ ) (r,\varphi) ( r , φ ) M = m 1 + m 2 M=m_1+m_2 M = m 1 + m 2 μ = m 1 m 2 / ( m 1 + m 2 ) \mu=m_1m_2/(m_1+m_2) μ = m 1 m 2 / ( m 1 + m 2 )
L = 1 2 μ ( d r d t ) 2 + 1 2 μ r 2 ( d φ d t ) 2 + G μ M r \mathcal{L} = \frac{1}{2}\mu\left(\frac{dr}{dt}\right)^2 + \frac{1}{2}\mu r^2\left(\frac{d\varphi}{dt}\right)^2 + \frac{G\mu M}{r} L = 2 1 μ ( d t d r ) 2 + 2 1 μ r 2 ( d t d φ ) 2 + r G μ M が与えられています。後半では重力波放射によるエネルギー損失を、ニュートン力学の枠内で半径 a ( t ) a(t) a ( t ) E = − G μ M / ( 2 a ) E=-G\mu M/(2a) E = − G μ M / ( 2 a )
L G W = 32 G 4 5 c 5 μ 2 M 3 a 5 L_{\mathrm{GW}} = \frac{32G^4}{5c^5}\frac{\mu^2M^3}{a^5} L GW = 5 c 5 32 G 4 a 5 μ 2 M 3 を使います。μ , M \mu,M μ , M
∂ L ∂ r ˙ = μ r ˙ \dfrac{\partial\mathcal L}{\partial \dot r} = \mu\dot r ∂ r ˙ ∂ L = μ r ˙ ∂ L ∂ r = μ r φ ˙ 2 − G μ M r 2 \dfrac{\partial \mathcal L}{\partial r} = \mu r\dot\varphi^2 - \dfrac{G\mu M}{r^2} ∂ r ∂ L = μ r φ ˙ 2 − r 2 G μ M
μ d 2 r d t 2 = μ r ( d φ d t ) 2 − G μ M r 2 . \mu\frac{d^2r}{dt^2} = \mu r\left(\frac{d\varphi}{dt}\right)^2 - \frac{G\mu M}{r^2}. μ d t 2 d 2 r = μ r ( d t d φ ) 2 − r 2 G μ M . φ \varphi φ L \mathcal L L
d d t ( μ r 2 d φ d t ) = 0. \frac{d}{dt}\left(\mu r^2\frac{d\varphi}{dt}\right) = 0 . d t d ( μ r 2 d t d φ ) = 0. L \mathcal L L t t t E = r ˙ ∂ L / ∂ r ˙ + φ ˙ ∂ L / ∂ φ ˙ − L E=\dot r\,\partial\mathcal L/\partial\dot r + \dot\varphi\,\partial\mathcal L/\partial\dot\varphi - \mathcal L E = r ˙ ∂ L / ∂ r ˙ + φ ˙ ∂ L / ∂ φ ˙ − L
E = 1 2 μ ( d r d t ) 2 + 1 2 μ r 2 ( d φ d t ) 2 − G μ M r , J = ∂ L ∂ φ ˙ = μ r 2 d φ d t . E = \frac{1}{2}\mu\left(\frac{dr}{dt}\right)^2 + \frac{1}{2}\mu r^2\left(\frac{d\varphi}{dt}\right)^2 - \frac{G\mu M}{r},
\qquad
J = \frac{\partial\mathcal L}{\partial\dot\varphi} = \mu r^2\frac{d\varphi}{dt} . E = 2 1 μ ( d t d r ) 2 + 2 1 μ r 2 ( d t d φ ) 2 − r G μ M , J = ∂ φ ˙ ∂ L = μ r 2 d t d φ . J J J L \mathcal L L φ \varphi φ
d J d t = d d t ∂ L ∂ φ ˙ = ∂ L ∂ φ = 0. \frac{dJ}{dt} = \frac{d}{dt}\frac{\partial\mathcal L}{\partial\dot\varphi} = \frac{\partial\mathcal L}{\partial\varphi} = 0 . d t dJ = d t d ∂ φ ˙ ∂ L = ∂ φ ∂ L = 0. E E E
d E d t = μ r ˙ r ¨ + μ r r ˙ φ ˙ 2 + μ r 2 φ ˙ φ ¨ + G μ M r 2 r ˙ \frac{dE}{dt} = \mu\dot r\ddot r + \mu r\dot r\dot\varphi^2 + \mu r^2\dot\varphi\ddot\varphi + \frac{G\mu M}{r^2}\dot r d t d E = μ r ˙ r ¨ + μ r r ˙ φ ˙ 2 + μ r 2 φ ˙ φ ¨ + r 2 G μ M r ˙ に、設問1 の μ r ¨ = μ r φ ˙ 2 − G μ M / r 2 \mu\ddot r = \mu r\dot\varphi^2 - G\mu M/r^2 μ r ¨ = μ r φ ˙ 2 − G μ M / r 2 d d t ( μ r 2 φ ˙ ) = 0 \dfrac{d}{dt}(\mu r^2\dot\varphi)=0 d t d ( μ r 2 φ ˙ ) = 0 μ r 2 φ ¨ = − 2 μ r r ˙ φ ˙ \mu r^2\ddot\varphi = -2\mu r\dot r\dot\varphi μ r 2 φ ¨ = − 2 μ r r ˙ φ ˙
d E d t = ( μ r r ˙ φ ˙ 2 − G μ M r ˙ r 2 ) + μ r r ˙ φ ˙ 2 − 2 μ r r ˙ φ ˙ 2 + G μ M r ˙ r 2 = 0. \frac{dE}{dt} = \left(\mu r\dot r\dot\varphi^2 - \frac{G\mu M\dot r}{r^2}\right) + \mu r\dot r\dot\varphi^2 - 2\mu r\dot r\dot\varphi^2 + \frac{G\mu M\dot r}{r^2} = 0 . d t d E = ( μ r r ˙ φ ˙ 2 − r 2 G μ M r ˙ ) + μ r r ˙ φ ˙ 2 − 2 μ r r ˙ φ ˙ 2 + r 2 G μ M r ˙ = 0. よって E E E J J J L \mathcal L L
E = − G μ M 2 a E=-\dfrac{G\mu M}{2a} E = − 2 a G μ M d E d t = G μ M 2 a 2 d a d t \dfrac{dE}{dt} = \dfrac{G\mu M}{2a^2}\dfrac{da}{dt} d t d E = 2 a 2 G μ M d t d a d E / d t = − L G W dE/dt=-L_{\mathrm{GW}} d E / d t = − L GW
G μ M 2 a 2 d a d t = − 32 G 4 5 c 5 μ 2 M 3 a 5 ⟹ d a d t = − 64 G 3 5 c 5 μ M 2 a 3 . \frac{G\mu M}{2a^2}\frac{da}{dt} = -\frac{32G^4}{5c^5}\frac{\mu^2M^3}{a^5}
\quad\Longrightarrow\quad
\frac{da}{dt} = -\frac{64G^3}{5c^5}\frac{\mu M^2}{a^3}. 2 a 2 G μ M d t d a = − 5 c 5 32 G 4 a 5 μ 2 M 3 ⟹ d t d a = − 5 c 5 64 G 3 a 3 μ M 2 . 右辺が負なので軌道は縮み続けます。a 3 a^3 a 3 a → 0 a\to 0 a → 0
d P d t = − A ( P c P ) 5 / 3 \dfrac{dP}{dt} = -A\left(\dfrac{P_\mathrm{c}}{P}\right)^{5/3} d t d P = − A ( P P c ) 5/3
P 5 / 3 d P = − A P c 5 / 3 d t ⟹ 3 8 ( P 8 / 3 − P 0 8 / 3 ) = − A P c 5 / 3 t . P^{5/3}\,dP = -A P_\mathrm{c}^{5/3}\,dt
\quad\Longrightarrow\quad
\frac{3}{8}\left(P^{8/3}-P_0^{8/3}\right) = -AP_\mathrm{c}^{5/3}\,t . P 5/3 d P = − A P c 5/3 d t ⟹ 8 3 ( P 8/3 − P 0 8/3 ) = − A P c 5/3 t . P = 0 P=0 P = 0
τ G W = 3 P 0 8 / 3 8 A P c 5 / 3 = 3 8 A P c ( P 0 P c ) 8 / 3 . \tau_{\mathrm{GW}} = \frac{3\,P_0^{8/3}}{8A\,P_\mathrm{c}^{5/3}} = \frac{3}{8A}P_\mathrm{c}\left(\frac{P_0}{P_\mathrm{c}}\right)^{8/3}. τ GW = 8 A P c 5/3 3 P 0 8/3 = 8 A 3 P c ( P c P 0 ) 8/3 . A A A P c P_\mathrm{c} P c
ケプラーの方程式 G M P 2 = 4 π 2 a 3 GMP^2=4\pi^2a^3 GM P 2 = 4 π 2 a 3 2 G M P P ˙ = 12 π 2 a 2 a ˙ 2GMP\,\dot P = 12\pi^2a^2\dot a 2 GM P P ˙ = 12 π 2 a 2 a ˙
d P d t = 6 π 2 a 2 G M P d a d t . \frac{dP}{dt} = \frac{6\pi^2a^2}{GMP}\frac{da}{dt} . d t d P = GM P 6 π 2 a 2 d t d a . 設問4 の結果を入れると
d P d t = − 6 π 2 a 2 G M P ⋅ 64 G 3 μ M 2 5 c 5 a 3 = − 384 π 2 G 2 μ M 5 c 5 P a , \frac{dP}{dt} = -\frac{6\pi^2a^2}{GMP}\cdot\frac{64G^3\mu M^2}{5c^5a^3}
= -\frac{384\pi^2G^2\mu M}{5c^5\,P\,a}, d t d P = − GM P 6 π 2 a 2 ⋅ 5 c 5 a 3 64 G 3 μ M 2 = − 5 c 5 P a 384 π 2 G 2 μ M , さらに a = ( G M 4 π 2 ) 1 / 3 P 2 / 3 a = \left(\dfrac{GM}{4\pi^2}\right)^{1/3}P^{2/3} a = ( 4 π 2 GM ) 1/3 P 2/3
d P d t = − 384 π 2 ( 4 π 2 ) 1 / 3 5 G 5 / 3 μ M 2 / 3 c 5 P − 5 / 3 = − 96 5 ( 2 π ) 8 / 3 ( G μ 3 / 5 M 2 / 5 c 3 ) 5 / 3 P − 5 / 3 . \frac{dP}{dt} = -\frac{384\pi^2\left(4\pi^2\right)^{1/3}}{5}\,\frac{G^{5/3}\mu M^{2/3}}{c^5}\,P^{-5/3}
= -\frac{96}{5}(2\pi)^{8/3}\left(\frac{G\mu^{3/5}M^{2/5}}{c^3}\right)^{5/3}P^{-5/3}. d t d P = − 5 384 π 2 ( 4 π 2 ) 1/3 c 5 G 5/3 μ M 2/3 P − 5/3 = − 5 96 ( 2 π ) 8/3 ( c 3 G μ 3/5 M 2/5 ) 5/3 P − 5/3 . これが式(6)です。比較すれば
P c = G μ 3 / 5 M 2 / 5 c 3 , α = 3 5 , β = 2 5 , γ = − 3 , P_\mathrm{c} = \frac{G\mu^{3/5}M^{2/5}}{c^3},\qquad
\alpha = \frac{3}{5},\quad \beta = \frac{2}{5},\quad \gamma = -3, P c = c 3 G μ 3/5 M 2/5 , α = 5 3 , β = 5 2 , γ = − 3 , A = 384 ⋅ 2 2 / 3 π 8 / 3 5 = 96 5 ( 2 π ) 8 / 3 ≃ 2.6 × 10 3 . A = \frac{384\cdot 2^{2/3}\pi^{8/3}}{5} = \frac{96}{5}(2\pi)^{8/3} \simeq 2.6\times 10^{3}. A = 5 384 ⋅ 2 2/3 π 8/3 = 5 96 ( 2 π ) 8/3 ≃ 2.6 × 1 0 3 . P c P_\mathrm{c} P c [ G ] = m 3 k g − 1 s − 2 [G]=\mathrm{m^3\,kg^{-1}\,s^{-2}} [ G ] = m 3 k g − 1 s − 2 m 3 − 3 k g − 1 + 3 / 5 + 2 / 5 s − 2 + 3 = s \mathrm{m^{3-3}\,kg^{-1+3/5+2/5}\,s^{-2+3}}=\mathrm{s} m 3 − 3 k g − 1 + 3/5 + 2/5 s − 2 + 3 = s α + β = 1 \alpha+\beta=1 α + β = 1 γ = − 3 \gamma=-3 γ = − 3 μ 3 / 5 M 2 / 5 \mu^{3/5}M^{2/5} μ 3/5 M 2/5
設問5 の結果を P c P_\mathrm{c} P c
P c = ( 3 8 A τ G W ) 3 / 5 P 0 8 / 5 P_\mathrm{c} = \left(\frac{3}{8A\tau_{\mathrm{GW}}}\right)^{3/5}P_0^{8/5} P c = ( 8 A τ GW 3 ) 3/5 P 0 8/5 等質量 m m m M = 2 m M=2m M = 2 m μ = m / 2 \mu=m/2 μ = m /2 μ 3 / 5 M 2 / 5 = ( m / 2 ) 3 / 5 ( 2 m ) 2 / 5 = 2 − 1 / 5 m \mu^{3/5}M^{2/5} = (m/2)^{3/5}(2m)^{2/5}=2^{-1/5}m μ 3/5 M 2/5 = ( m /2 ) 3/5 ( 2 m ) 2/5 = 2 − 1/5 m
P c = G m 2 1 / 5 c 3 ⟹ m M ⊙ = 2 1 / 5 P c c 3 G M ⊙ . P_\mathrm{c} = \frac{Gm}{2^{1/5}c^3}
\quad\Longrightarrow\quad
\frac{m}{M_\odot} = 2^{1/5}\,P_\mathrm{c}\,\frac{c^3}{GM_\odot} . P c = 2 1/5 c 3 G m ⟹ M ⊙ m = 2 1/5 P c G M ⊙ c 3 . 数値を入れます。3 8 A τ G W = 3 8 × 2500 × 0.15 = 1.0 × 10 − 3 \dfrac{3}{8A\tau_{\mathrm{GW}}} = \dfrac{3}{8\times 2500\times 0.15} = 1.0\times 10^{-3} 8 A τ GW 3 = 8 × 2500 × 0.15 3 = 1.0 × 1 0 − 3 3 / 5 3/5 3/5 10 − 1.8 10^{-1.8} 1 0 − 1.8 P 0 8 / 5 = ( 6 × 10 − 2 ) 8 / 5 = 6 ⋅ 6 3 / 5 × 10 − 3.2 ≃ 6 × 2.9 × 10 − 3.2 = 17.4 × 10 − 3.2 P_0^{8/5} = (6\times 10^{-2})^{8/5} = 6\cdot 6^{3/5}\times 10^{-3.2} \simeq 6\times 2.9\times 10^{-3.2}=17.4\times 10^{-3.2} P 0 8/5 = ( 6 × 1 0 − 2 ) 8/5 = 6 ⋅ 6 3/5 × 1 0 − 3.2 ≃ 6 × 2.9 × 1 0 − 3.2 = 17.4 × 1 0 − 3.2
P c ≃ 17.4 × 10 − 1.8 − 3.2 = 1.7 × 10 − 4 s . P_\mathrm{c} \simeq 17.4\times 10^{-1.8-3.2} = 1.7\times 10^{-4}\,\mathrm{s}. P c ≃ 17.4 × 1 0 − 1.8 − 3.2 = 1.7 × 1 0 − 4 s . 2 G M ⊙ / c 3 ≈ 10 − 5 s 2GM_\odot/c^3\approx 10^{-5}\,\mathrm{s} 2 G M ⊙ / c 3 ≈ 1 0 − 5 s G M ⊙ / c 3 ≈ 5 × 10 − 6 s GM_\odot/c^3 \approx 5\times 10^{-6}\,\mathrm{s} G M ⊙ / c 3 ≈ 5 × 1 0 − 6 s
m M ⊙ ≃ 1.1 × 1.74 × 10 − 4 5 × 10 − 6 ≃ 38. \frac{m}{M_\odot} \simeq \frac{1.1\times 1.74\times 10^{-4}}{5\times 10^{-6}} \simeq 38 . M ⊙ m ≃ 5 × 1 0 − 6 1.1 × 1.74 × 1 0 − 4 ≃ 38. 答えは m ≈ 4 × 10 1 M ⊙ m \approx 4\times 10^{1}\,M_\odot m ≈ 4 × 1 0 1 M ⊙
μ , M \mu,M μ , M a 0 a_0 a 0 a min a_{\min} a m i n E r a d = E ( a 0 ) − E ( a min ) E_{\mathrm{rad}} = E(a_0)-E(a_{\min}) E rad = E ( a 0 ) − E ( a m i n )
E r a d = G μ M 2 a min − G μ M 2 a 0 ≃ G μ M 2 a min E_{\mathrm{rad}} = \frac{G\mu M}{2a_{\min}} - \frac{G\mu M}{2a_0} \simeq \frac{G\mu M}{2a_{\min}} E rad = 2 a m i n G μ M − 2 a 0 G μ M ≃ 2 a m i n G μ M です（a 0 ≫ a min a_0\gg a_{\min} a 0 ≫ a m i n a a a a min a_{\min} a m i n m m m 0.1 m c 2 0.1\,mc^2 0.1 m c 2 μ M = m 2 ⋅ 2 m = m 2 \mu M = \dfrac{m}{2}\cdot 2m = m^2 μ M = 2 m ⋅ 2 m = m 2
G m 2 2 a min = m c 2 10 ⟹ a min = 5 G m c 2 . \frac{Gm^2}{2a_{\min}} = \frac{mc^2}{10}
\quad\Longrightarrow\quad
a_{\min} = \frac{5Gm}{c^2}. 2 a m i n G m 2 = 10 m c 2 ⟹ a m i n = c 2 5 G m . これが求める関係式です。1 つのブラックホールのシュワルツシルト半径 r s = 2 G m / c 2 r_\mathrm{s}=2Gm/c^2 r s = 2 G m / c 2 a min = 2.5 r s a_{\min}=2.5\,r_\mathrm{s} a m i n = 2.5 r s 2 r s 2r_\mathrm{s} 2 r s m ≈ 38 M ⊙ m\approx 38M_\odot m ≈ 38 M ⊙ G M ⊙ / c 2 = c ⋅ ( G M ⊙ / c 3 ) ≈ 1.5 k m GM_\odot/c^2 = c\cdot(GM_\odot/c^3)\approx 1.5\,\mathrm{km} G M ⊙ / c 2 = c ⋅ ( G M ⊙ / c 3 ) ≈ 1.5 km a min ≈ 5 × 38 × 1.5 k m ≈ 3 × 10 2 k m a_{\min}\approx 5\times 38\times 1.5\,\mathrm{km} \approx 3\times 10^{2}\,\mathrm{km} a m i n ≈ 5 × 38 × 1.5 km ≈ 3 × 1 0 2 km
光源からの光がビームスプリッタ（BS）で分割され、反射光は鏡 A、透過光は鏡 B へ向かい、それぞれ戻って BS で再び合流します。検出器へ向かう光は「A から戻って BS を透過した成分」と「B から戻って BS で反射された成分」の和、光源へ戻る光はその逆の組み合わせです。複素電場を E e − i 2 π ν t Ee^{-i2\pi\nu t} E e − i 2 π ν t Δ t \Delta t Δ t e + i 2 π ν Δ t e^{+i2\pi\nu\Delta t} e + i 2 π ν Δ t p , r p,r p , r ϕ p \phi_\mathrm{p} ϕ p ϕ r \phi_\mathrm{r} ϕ r r = p = 1 / 2 r=p=1/\sqrt2 r = p = 1/ 2 ϕ r = π \phi_\mathrm{r}=\pi ϕ r = π ϕ p = π / 2 \phi_\mathrm{p}=\pi/2 ϕ p = π /2 L + l L+l L + l L − l L-l L − l h = 6.6 × 10 − 34 J s h=6.6\times10^{-34}\,\mathrm{J\,s} h = 6.6 × 1 0 − 34 J s c = 3.0 × 10 8 m / s c=3.0\times10^{8}\,\mathrm{m/s} c = 3.0 × 1 0 8 m/s k B = 1.4 × 10 − 23 J / K k_\mathrm{B}=1.4\times10^{-23}\,\mathrm{J/K} k B = 1.4 × 1 0 − 23 J/K
鏡 A の側へ行く光は、BS で 1 回反射（係数 r r r ϕ r \phi_\mathrm{r} ϕ r ϕ r \phi_\mathrm{r} ϕ r e i 2 π ν Δ t A e^{i2\pi\nu\Delta t_\mathrm{A}} e i 2 π ν Δ t A
E A = r E i e i ( 2 ϕ r + 2 π ν Δ t A ) . E_\mathrm{A} = r\,E_\mathrm{i}\,e^{i\left(2\phi_\mathrm{r}+2\pi\nu\Delta t_\mathrm{A}\right)} . E A = r E i e i ( 2 ϕ r + 2 π ν Δ t A ) . 与えられた E B = p E i e i ( ϕ p + ϕ r + 2 π ν Δ t B ) E_\mathrm{B}=pE_\mathrm{i}e^{i(\phi_\mathrm{p}+\phi_\mathrm{r}+2\pi\nu\Delta t_\mathrm{B})} E B = p E i e i ( ϕ p + ϕ r + 2 π ν Δ t B ) p ↔ r p\leftrightarrow r p ↔ r ϕ p ↔ ϕ r \phi_\mathrm{p}\leftrightarrow\phi_\mathrm{r} ϕ p ↔ ϕ r
検出器へ向かう成分は、A からの光が BS を透過（p p p ϕ p \phi_\mathrm{p} ϕ p r r r ϕ r \phi_\mathrm{r} ϕ r
E o = p e i ϕ p E A + r e i ϕ r E B = p r E i e i ( ϕ p + 2 ϕ r ) [ e i 2 π ν Δ t A + e i 2 π ν Δ t B ] . E_\mathrm{o} = p\,e^{i\phi_\mathrm{p}}E_\mathrm{A} + r\,e^{i\phi_\mathrm{r}}E_\mathrm{B}
= p\,r\,E_\mathrm{i}\,e^{i\left(\phi_\mathrm{p}+2\phi_\mathrm{r}\right)}\left[e^{i2\pi\nu\Delta t_\mathrm{A}} + e^{i2\pi\nu\Delta t_\mathrm{B}}\right]. E o = p e i ϕ p E A + r e i ϕ r E B = p r E i e i ( ϕ p + 2 ϕ r ) [ e i 2 π ν Δ t A + e i 2 π ν Δ t B ] . 光源へ戻る成分は逆の組み合わせで、A からの光が BS で反射、B からの光が BS を透過したものの和です。
E s = r e i ϕ r E A + p e i ϕ p E B = E i e i ϕ r [ r 2 e i ( 2 ϕ r + 2 π ν Δ t A ) + p 2 e i ( 2 ϕ p + 2 π ν Δ t B ) ] . E_\mathrm{s} = r\,e^{i\phi_\mathrm{r}}E_\mathrm{A} + p\,e^{i\phi_\mathrm{p}}E_\mathrm{B}
= E_\mathrm{i}\,e^{i\phi_\mathrm{r}}\left[r^2 e^{i\left(2\phi_\mathrm{r}+2\pi\nu\Delta t_\mathrm{A}\right)} + p^2 e^{i\left(2\phi_\mathrm{p}+2\pi\nu\Delta t_\mathrm{B}\right)}\right]. E s = r e i ϕ r E A + p e i ϕ p E B = E i e i ϕ r [ r 2 e i ( 2 ϕ r + 2 π ν Δ t A ) + p 2 e i ( 2 ϕ p + 2 π ν Δ t B ) ] . E o E_\mathrm{o} E o p r pr p r E s E_\mathrm{s} E s r 2 r^2 r 2 p 2 p^2 p 2
往復距離から Δ t A = 2 ( L + l ) c \Delta t_\mathrm{A} = \dfrac{2(L+l)}{c} Δ t A = c 2 ( L + l ) Δ t B = 2 ( L − l ) c \Delta t_\mathrm{B} = \dfrac{2(L-l)}{c} Δ t B = c 2 ( L − l )
2 π ν ( Δ t A − Δ t B ) = 2 π ν ⋅ 4 l c = 8 π ν l c . 2\pi\nu\left(\Delta t_\mathrm{A}-\Delta t_\mathrm{B}\right) = 2\pi\nu\cdot\frac{4l}{c} = \frac{8\pi\nu l}{c}. 2 π ν ( Δ t A − Δ t B ) = 2 π ν ⋅ c 4 l = c 8 π ν l . E o E_\mathrm{o} E o ϕ p , ϕ r \phi_\mathrm{p},\phi_\mathrm{r} ϕ p , ϕ r ∣ e i θ 1 + e i θ 2 ∣ 2 = 4 cos 2 θ 1 − θ 2 2 \lvert e^{i\theta_1}+e^{i\theta_2}\rvert^2 = 4\cos^2\dfrac{\theta_1-\theta_2}{2} ∣ e i θ 1 + e i θ 2 ∣ 2 = 4 cos 2 2 θ 1 − θ 2 p r = 1 / 2 pr=1/2 p r = 1/2
∣ E o ∣ 2 = ( p r ) 2 ∣ E i ∣ 2 ⋅ 4 cos 2 ( 4 π ν l c ) = ∣ E i ∣ 2 cos 2 ( 4 π ν l c ) = 1 2 ∣ E i ∣ 2 [ cos ( 8 π ν l c ) + 1 ] , \lvert E_\mathrm{o}\rvert^2 = (pr)^2\lvert E_\mathrm{i}\rvert^2\cdot 4\cos^2\left(\frac{4\pi\nu l}{c}\right)
= \lvert E_\mathrm{i}\rvert^2\cos^2\left(\frac{4\pi\nu l}{c}\right)
= \frac{1}{2}\lvert E_\mathrm{i}\rvert^2\left[\cos\left(\frac{8\pi\nu l}{c}\right)+1\right], ∣ E o ∣ 2 = ( p r ) 2 ∣ E i ∣ 2 ⋅ 4 cos 2 ( c 4 π ν l ) = ∣ E i ∣ 2 cos 2 ( c 4 π ν l ) = 2 1 ∣ E i ∣ 2 [ cos ( c 8 π ν l ) + 1 ] , 最後は cos 2 θ = ( 1 + cos 2 θ ) / 2 \cos^2\theta = (1+\cos2\theta)/2 cos 2 θ = ( 1 + cos 2 θ ) /2
E s E_\mathrm{s} E s r 2 = p 2 = 1 / 2 r^2=p^2=1/2 r 2 = p 2 = 1/2 2 ϕ r = 2 π 2\phi_\mathrm{r}=2\pi 2 ϕ r = 2 π 2 ϕ p = π 2\phi_\mathrm{p}=\pi 2 ϕ p = π
E s = 1 2 E i e i ϕ r [ e i 2 π ν Δ t A − e i 2 π ν Δ t B ] , ∣ E s ∣ 2 = 1 4 ∣ E i ∣ 2 ⋅ 4 sin 2 ( 4 π ν l c ) = ∣ E i ∣ 2 sin 2 ( 4 π ν l c ) . E_\mathrm{s} = \frac{1}{2}E_\mathrm{i}e^{i\phi_\mathrm{r}}\left[e^{i2\pi\nu\Delta t_\mathrm{A}} - e^{i2\pi\nu\Delta t_\mathrm{B}}\right],
\qquad
\lvert E_\mathrm{s}\rvert^2 = \frac{1}{4}\lvert E_\mathrm{i}\rvert^2\cdot 4\sin^2\left(\frac{4\pi\nu l}{c}\right)
= \lvert E_\mathrm{i}\rvert^2\sin^2\left(\frac{4\pi\nu l}{c}\right). E s = 2 1 E i e i ϕ r [ e i 2 π ν Δ t A − e i 2 π ν Δ t B ] , ∣ E s ∣ 2 = 4 1 ∣ E i ∣ 2 ⋅ 4 sin 2 ( c 4 π ν l ) = ∣ E i ∣ 2 sin 2 ( c 4 π ν l ) . したがって
∣ E o ∣ 2 + ∣ E s ∣ 2 = ∣ E i ∣ 2 [ cos 2 ( 4 π ν l c ) + sin 2 ( 4 π ν l c ) ] = ∣ E i ∣ 2 \lvert E_\mathrm{o}\rvert^2 + \lvert E_\mathrm{s}\rvert^2 = \lvert E_\mathrm{i}\rvert^2\left[\cos^2\left(\frac{4\pi\nu l}{c}\right)+\sin^2\left(\frac{4\pi\nu l}{c}\right)\right] = \lvert E_\mathrm{i}\rvert^2 ∣ E o ∣ 2 + ∣ E s ∣ 2 = ∣ E i ∣ 2 [ cos 2 ( c 4 π ν l ) + sin 2 ( c 4 π ν l ) ] = ∣ E i ∣ 2 が成り立ちます。検出器側で暗くなる分はちょうど光源側へ戻り、エネルギーは保存しています。この相補性は ϕ r \phi_\mathrm{r} ϕ r ϕ p \phi_\mathrm{p} ϕ p π / 2 \pi/2 π /2
I ( ν ) = I 1 δ ( ν − ν 1 ) + I 2 δ ( ν − ν 2 ) I(\nu)=I_1\delta(\nu-\nu_1)+I_2\delta(\nu-\nu_2) I ( ν ) = I 1 δ ( ν − ν 1 ) + I 2 δ ( ν − ν 2 )
P o k = 1 2 I 1 [ cos ( 2 π ν 1 k Δ l c ) + 1 ] + 1 2 I 2 [ cos ( 2 π ν 2 k Δ l c ) + 1 ] . P_\mathrm{o}^k = \frac{1}{2}I_1\left[\cos\left(\frac{2\pi\nu_1 k\Delta l}{c}\right)+1\right]
+ \frac{1}{2}I_2\left[\cos\left(\frac{2\pi\nu_2 k\Delta l}{c}\right)+1\right]. P o k = 2 1 I 1 [ cos ( c 2 π ν 1 k Δ l ) + 1 ] + 2 1 I 2 [ cos ( c 2 π ν 2 k Δ l ) + 1 ] . α ≡ 2 π Δ l c \alpha \equiv \dfrac{2\pi\Delta l}{c} α ≡ c 2 π Δ l ( I 1 , I 2 ) = ( I 0 , 0 ) (I_1,I_2)=(I_0,0) ( I 1 , I 2 ) = ( I 0 , 0 ) P o k = I 0 2 [ cos ( k α ( ν ˉ + Δ ν ) ) + 1 ] P_\mathrm{o}^k=\dfrac{I_0}{2}\left[\cos\left(k\alpha(\bar\nu+\Delta\nu)\right)+1\right] P o k = 2 I 0 [ cos ( k α ( ν ˉ + Δ ν ) ) + 1 ] ( 0 , I 0 ) (0,I_0) ( 0 , I 0 ) ν ˉ − Δ ν \bar\nu-\Delta\nu ν ˉ − Δ ν
cos ( k α ( ν ˉ + Δ ν ) ) − cos ( k α ( ν ˉ − Δ ν ) ) = − 2 sin ( k α ν ˉ ) sin ( k α Δ ν ) \cos\left(k\alpha(\bar\nu+\Delta\nu)\right) - \cos\left(k\alpha(\bar\nu-\Delta\nu)\right)
= -2\sin\left(k\alpha\bar\nu\right)\sin\left(k\alpha\Delta\nu\right) cos ( k α ( ν ˉ + Δ ν ) ) − cos ( k α ( ν ˉ − Δ ν ) ) = − 2 sin ( k α ν ˉ ) sin ( k α Δ ν ) となります。これが任意の Δ ν \Delta\nu Δ ν k = 0 , 1 , 2 , … k=0,1,2,\dots k = 0 , 1 , 2 , … sin ( α ν ˉ ) = 0 \sin(\alpha\bar\nu)=0 sin ( α ν ˉ ) = 0
2 π Δ l c ν ˉ = π N ( N = 1 , 2 , … ) ⟺ ν ˉ = N c 2 Δ l \frac{2\pi\Delta l}{c}\bar\nu = \pi N\quad (N=1,2,\dots)
\qquad\Longleftrightarrow\qquad
\bar\nu = \frac{Nc}{2\Delta l} c 2 π Δ l ν ˉ = π N ( N = 1 , 2 , … ) ⟺ ν ˉ = 2Δ l N c です（このとき sin ( k α ν ˉ ) = sin ( k π N ) = 0 \sin(k\alpha\bar\nu)=\sin(k\pi N)=0 sin ( k α ν ˉ ) = sin ( k π N ) = 0 k k k
ν n = c 2 Δ l . \nu_\mathrm{n} = \frac{c}{2\Delta l}. ν n = 2Δ l c . 式(3)の位相 2 π ν k Δ l / c 2\pi\nu k\Delta l/c 2 π ν k Δ l / c Δ l / c \Delta l/c Δ l / c Δ l / 4 \Delta l/4 Δ l /4 Δ l \Delta l Δ l
x ≡ h ν / ( k B T ) x\equiv h\nu/(k_\mathrm{B}T) x ≡ h ν / ( k B T ) h ν ≪ k B T h\nu\ll k_\mathrm{B}T h ν ≪ k B T e x − 1 ≃ x e^{x}-1\simeq x e x − 1 ≃ x
I ( ν ; T ) ≃ 2 h ν 3 c 2 ⋅ k B T h ν = 2 ν 2 k B T c 2 I(\nu;T) \simeq \frac{2h\nu^3}{c^2}\cdot\frac{k_\mathrm{B}T}{h\nu} = \frac{2\nu^2 k_\mathrm{B}T}{c^2} I ( ν ; T ) ≃ c 2 2 h ν 3 ⋅ h ν k B T = c 2 2 ν 2 k B T がレイリー・ジーンズの放射法則です。h ν ≫ k B T h\nu\gg k_\mathrm{B}T h ν ≫ k B T e x − 1 ≃ e x e^{x}-1\simeq e^{x} e x − 1 ≃ e x
I ( ν ; T ) ≃ 2 h ν 3 c 2 exp ( − h ν k B T ) I(\nu;T) \simeq \frac{2h\nu^3}{c^2}\exp\left(-\frac{h\nu}{k_\mathrm{B}T}\right) I ( ν ; T ) ≃ c 2 2 h ν 3 exp ( − k B T h ν ) がウィーンの放射法則です。前者に h h h
I ( ν ; T ) I(\nu;T) I ( ν ; T ) x = h ν / ( k B T ) x=h\nu/(k_\mathrm{B}T) x = h ν / ( k B T ) f ( x ) = x 3 / ( e x − 1 ) f(x)=x^3/(e^x-1) f ( x ) = x 3 / ( e x − 1 ) x ≈ 2.8 x\approx 2.8 x ≈ 2.8
ν max = 2.8 k B T h = 2.8 × 1.4 × 10 − 23 × 2.7 6.6 × 10 − 34 = 1.06 × 10 − 22 6.6 × 10 − 34 = 1.6 × 10 11 H z \nu_{\max} = \frac{2.8\,k_\mathrm{B}T}{h} = \frac{2.8\times 1.4\times 10^{-23}\times 2.7}{6.6\times 10^{-34}}
= \frac{1.06\times 10^{-22}}{6.6\times 10^{-34}} = 1.6\times 10^{11}\,\mathrm{Hz} ν m a x = h 2.8 k B T = 6.6 × 1 0 − 34 2.8 × 1.4 × 1 0 − 23 × 2.7 = 6.6 × 1 0 − 34 1.06 × 1 0 − 22 = 1.6 × 1 0 11 Hz すなわち約 160 G H z 160\,\mathrm{GHz} 160 GHz c / ν max ≈ 1.9 m m c/\nu_{\max}\approx 1.9\,\mathrm{mm} c / ν m a x ≈ 1.9 mm ν n = 10 ν max = 1.6 × 10 12 H z \nu_\mathrm{n}=10\nu_{\max}=1.6\times 10^{12}\,\mathrm{Hz} ν n = 10 ν m a x = 1.6 × 1 0 12 Hz Δ l \Delta l Δ l ν n = c / ( 2 Δ l ) \nu_\mathrm{n}=c/(2\Delta l) ν n = c / ( 2Δ l )
Δ l = c 2 ν n = 3.0 × 10 8 2 × 1.6 × 10 12 = 9.4 × 10 − 5 m \Delta l = \frac{c}{2\nu_\mathrm{n}} = \frac{3.0\times 10^{8}}{2\times 1.6\times 10^{12}} = 9.4\times 10^{-5}\,\mathrm{m} Δ l = 2 ν n c = 2 × 1.6 × 1 0 12 3.0 × 1 0 8 = 9.4 × 1 0 − 5 m です。答えは Δ l ≈ 9.4 × 10 − 5 m \Delta l \approx 9.4\times 10^{-5}\,\mathrm{m} Δ l ≈ 9.4 × 1 0 − 5 m 94 μ m 94\,\mu\mathrm{m} 94 μ m Δ l / 4 \Delta l/4 Δ l /4 23 μ m 23\,\mu\mathrm{m} 23 μ m Δ l = λ max / 20 \Delta l = \lambda_{\max}/20 Δ l = λ m a x /20
中性子と原子核の反応断面積、中性 π \pi π c c c
中性子は 940 M e V / c 2 940\,\mathrm{MeV}/c^2 940 MeV / c 2 938 M e V / c 2 938\,\mathrm{MeV}/c^2 938 MeV / c 2 π \pi π 135 M e V / c 2 135\,\mathrm{MeV}/c^2 135 MeV / c 2 0.511 0.511 0.511 K K K 494 494 494 Δ \Delta Δ 1232 1232 1232
n → p + e − + ν ˉ e n \to p + e^- + \bar\nu_e n → p + e − + ν ˉ e 生じる粒子は陽子、電子、反電子ニュートリノの 3 つです。
Δ I = − n σ I Δ x \Delta I = -n\sigma I\Delta x Δ I = − nσ I Δ x d I d x = − n σ I \dfrac{dI}{dx} = -n\sigma I d x d I = − nσ I n , σ n,\sigma n , σ
I ( x ) = I 0 e − n σ x , I 1 = I ( d ) = I 0 e − n σ d . I(x) = I_0 e^{-n\sigma x},\qquad I_1 = I(d) = I_0 e^{-n\sigma d}. I ( x ) = I 0 e − nσ x , I 1 = I ( d ) = I 0 e − nσ d . これを σ \sigma σ
σ = 1 n d ln I 0 I 1 . \sigma = \frac{1}{nd}\ln\frac{I_0}{I_1}. σ = n d 1 ln I 1 I 0 . [ n ] = m − 3 [n]=\mathrm{m^{-3}} [ n ] = m − 3 [ d ] = m [d]=\mathrm{m} [ d ] = m σ \sigma σ
π 0 → γ γ \pi^0\to\gamma\gamma π 0 → γ γ ∣ p ⃗ i ∣ = E i / c \lvert \vec p_i\rvert = E_i/c ∣ p i ∣ = E i / c
m 2 c 4 = ( E 1 + E 2 ) 2 − ∣ p ⃗ 1 + p ⃗ 2 ∣ 2 c 2 = ( E 1 + E 2 ) 2 − ( E 1 2 + E 2 2 + 2 E 1 E 2 cos ϕ ) = 2 E 1 E 2 ( 1 − cos ϕ ) . m^2c^4 = \left(E_1+E_2\right)^2 - \left\lvert \vec p_1+\vec p_2\right\rvert^2c^2
= \left(E_1+E_2\right)^2 - \left(E_1^2+E_2^2+2E_1E_2\cos\phi\right)
= 2E_1E_2\left(1-\cos\phi\right). m 2 c 4 = ( E 1 + E 2 ) 2 − ∣ p 1 + p 2 ∣ 2 c 2 = ( E 1 + E 2 ) 2 − ( E 1 2 + E 2 2 + 2 E 1 E 2 cos ϕ ) = 2 E 1 E 2 ( 1 − cos ϕ ) . 1 − cos ϕ = 2 sin 2 ( ϕ / 2 ) 1-\cos\phi = 2\sin^2(\phi/2) 1 − cos ϕ = 2 sin 2 ( ϕ /2 )
m = 2 E 1 E 2 ( 1 − cos ϕ ) c 2 = 2 E 1 E 2 c 2 sin ϕ 2 . m = \frac{\sqrt{2E_1E_2\left(1-\cos\phi\right)}}{c^2} = \frac{2\sqrt{E_1E_2}}{c^2}\sin\frac{\phi}{2}. m = c 2 2 E 1 E 2 ( 1 − cos ϕ ) = c 2 2 E 1 E 2 sin 2 ϕ . ϕ → 0 \phi\to 0 ϕ → 0 m → 0 m\to 0 m → 0 ϕ = π \phi=\pi ϕ = π m = 2 E / c 2 m=2E/c^2 m = 2 E / c 2
MKSA 単位系で、質量 M \mathsf{M} M L \mathsf{L} L T \mathsf{T} T I \mathsf{I} I
[ P ] = M L 2 T − 3 , [ a ] = L T − 2 , [ e ] = I T , [ ε 0 ] = M − 1 L − 3 T 4 I 2 , [ c ] = L T − 1 [P] = \mathsf{M}\mathsf{L}^2\mathsf{T}^{-3},\quad
[a] = \mathsf{L}\mathsf{T}^{-2},\quad
[e] = \mathsf{I}\mathsf{T},\quad
[\varepsilon_0] = \mathsf{M}^{-1}\mathsf{L}^{-3}\mathsf{T}^{4}\mathsf{I}^{2},\quad
[c] = \mathsf{L}\mathsf{T}^{-1} [ P ] = M L 2 T − 3 , [ a ] = L T − 2 , [ e ] = IT , [ ε 0 ] = M − 1 L − 3 T 4 I 2 , [ c ] = L T − 1 （[ ε 0 ] [\varepsilon_0] [ ε 0 ] F = q 1 q 2 / ( 4 π ε 0 r 2 ) F=q_1q_2/(4\pi\varepsilon_0r^2) F = q 1 q 2 / ( 4 π ε 0 r 2 ) P = A a f e g ε 0 h c i P=Aa^fe^g\varepsilon_0^hc^i P = A a f e g ε 0 h c i
M : 1 = − h I : 0 = g + 2 h L : 2 = f − 3 h + i T : − 3 = − 2 f + g + 4 h − i \begin{aligned}
\mathsf{M}:&\quad 1 = -h\\
\mathsf{I}:&\quad 0 = g+2h\\
\mathsf{L}:&\quad 2 = f-3h+i\\
\mathsf{T}:&\quad -3 = -2f+g+4h-i
\end{aligned} M : I : L : T : 1 = − h 0 = g + 2 h 2 = f − 3 h + i − 3 = − 2 f + g + 4 h − i 第1式から h = − 1 h=-1 h = − 1 g = 2 g=2 g = 2 f + i = − 1 f+i=-1 f + i = − 1 − 2 f − i = − 1 -2f-i=-1 − 2 f − i = − 1 f = 2 f=2 f = 2 i = − 3 i=-3 i = − 3
f = 2 , g = 2 , h = − 1 , i = − 3 , P = A e 2 a 2 ε 0 c 3 . f=2,\quad g=2,\quad h=-1,\quad i=-3,\qquad P = A\,\frac{e^2a^2}{\varepsilon_0c^3}. f = 2 , g = 2 , h = − 1 , i = − 3 , P = A ε 0 c 3 e 2 a 2 . これは A = 1 / ( 6 π ) A=1/(6\pi) A = 1/ ( 6 π )
アルミニウムと鉛の比較では、鉛のほうが P P P P ∝ a 2 P\propto a^2 P ∝ a 2 a a a Z e Ze Z e r r r a = Z e 2 4 π ε 0 m e r 2 a = \dfrac{Ze^2}{4\pi\varepsilon_0 m_e r^2} a = 4 π ε 0 m e r 2 Z e 2 Z Z Z P ∝ Z 2 P\propto Z^2 P ∝ Z 2 Z A l = 13 Z_{\mathrm{Al}}=13 Z Al = 13 Z P b = 82 Z_{\mathrm{Pb}}=82 Z Pb = 82 ( 82 / 13 ) 2 ≈ 40 (82/13)^2\approx 40 ( 82/13 ) 2 ≈ 40 Z Z Z Z 2 Z^2 Z 2
磁場中では磁気力が仕事をしないので ∣ v ⃗ ∣ \lvert\vec v\rvert ∣ v ∣ ∣ p ⃗ ∣ \lvert\vec p\rvert ∣ p ∣
d p ⃗ d t = q v ⃗ × B ⃗ , ∣ d p ⃗ d t ∣ = q v B \frac{d\vec p}{dt} = q\,\vec v\times\vec B,\qquad \left\lvert\frac{d\vec p}{dt}\right\rvert = qvB d t d p = q v × B , d t d p = q v B で、半径 R R R v / R v/R v / R ∣ d p ⃗ / d t ∣ = p v / R \left\lvert d\vec p/dt\right\rvert = p\,v/R ∣ d p / d t ∣ = p v / R v v v
p v R = q v B ⟹ p = q B R . \frac{pv}{R} = qvB \qquad\Longrightarrow\qquad p = qBR . R p v = q v B ⟹ p = q B R . つぎに幾何です。軌道の接線が x x x θ \theta θ s s s d x d s = cos θ \dfrac{dx}{ds}=\cos\theta d s d x = cos θ d θ d s = 1 R \dfrac{d\theta}{ds}=\dfrac{1}{R} d s d θ = R 1 B B B q > 0 q>0 q > 0 + x +x + x q v ⃗ × B ⃗ q\vec v\times\vec B q v × B θ \theta θ s s s d x = R cos θ d θ dx = R\cos\theta\,d\theta d x = R cos θ d θ
L = ∫ θ 1 θ 2 R cos θ d θ = R ( sin θ 2 − sin θ 1 ) ⟹ R = L sin θ 2 − sin θ 1 . L = \int_{\theta_1}^{\theta_2}R\cos\theta\,d\theta = R\left(\sin\theta_2-\sin\theta_1\right)
\qquad\Longrightarrow\qquad
R = \frac{L}{\sin\theta_2-\sin\theta_1}. L = ∫ θ 1 θ 2 R cos θ d θ = R ( sin θ 2 − sin θ 1 ) ⟹ R = sin θ 2 − sin θ 1 L . 磁場領域の x x x R R R
κ = sin θ 2 − sin θ 1 L ≃ θ 2 − θ 1 L \kappa = \dfrac{\sin\theta_2-\sin\theta_1}{L} \simeq \dfrac{\theta_2-\theta_1}{L} κ = L sin θ 2 − sin θ 1 ≃ L θ 2 − θ 1 θ 1 \theta_1 θ 1 θ 2 \theta_2 θ 2 L L L
( Δ κ ) 2 = ( ∂ κ ∂ θ 1 ) 2 ( Δ θ 1 ) 2 + ( ∂ κ ∂ θ 2 ) 2 ( Δ θ 2 ) 2 = ( Δ θ 1 ) 2 + ( Δ θ 2 ) 2 L 2 \left(\Delta\kappa\right)^2 = \left(\frac{\partial\kappa}{\partial\theta_1}\right)^2\left(\Delta\theta_1\right)^2 + \left(\frac{\partial\kappa}{\partial\theta_2}\right)^2\left(\Delta\theta_2\right)^2
= \frac{\left(\Delta\theta_1\right)^2+\left(\Delta\theta_2\right)^2}{L^2} ( Δ κ ) 2 = ( ∂ θ 1 ∂ κ ) 2 ( Δ θ 1 ) 2 + ( ∂ θ 2 ∂ κ ) 2 ( Δ θ 2 ) 2 = L 2 ( Δ θ 1 ) 2 + ( Δ θ 2 ) 2 したがって
Δ κ = ( Δ θ 1 ) 2 + ( Δ θ 2 ) 2 L . \Delta\kappa = \frac{\sqrt{\left(\Delta\theta_1\right)^2+\left(\Delta\theta_2\right)^2}}{L}. Δ κ = L ( Δ θ 1 ) 2 + ( Δ θ 2 ) 2 . p = q B / κ p=qB/\kappa p = q B / κ Δ p p = Δ κ κ = ( Δ θ 1 ) 2 + ( Δ θ 2 ) 2 θ 2 − θ 1 \dfrac{\Delta p}{p} = \dfrac{\Delta\kappa}{\kappa} = \dfrac{\sqrt{(\Delta\theta_1)^2+(\Delta\theta_2)^2}}{\theta_2-\theta_1} p Δ p = κ Δ κ = θ 2 − θ 1 ( Δ θ 1 ) 2 + ( Δ θ 2 ) 2
同軸ケーブルの中心軸を x x x C C C L L L I ( x , t ) I(x,t) I ( x , t ) V ( x , t ) V(x,t) V ( x , t )
∂ I ∂ x = − C ∂ V ∂ t , ∂ V ∂ x = − L ∂ I ∂ t \frac{\partial I}{\partial x} = -C\frac{\partial V}{\partial t},\qquad
\frac{\partial V}{\partial x} = -L\frac{\partial I}{\partial t} ∂ x ∂ I = − C ∂ t ∂ V , ∂ x ∂ V = − L ∂ t ∂ I に従います。内部導体半径 a a a b b b ε \varepsilon ε μ 0 \mu_0 μ 0 C = 2 π ε ln ( b / a ) C=\dfrac{2\pi\varepsilon}{\ln(b/a)} C = ln ( b / a ) 2 π ε L = μ 0 ln ( b / a ) 2 π L=\dfrac{\mu_0\ln(b/a)}{2\pi} L = 2 π μ 0 ln ( b / a ) c = 3.0 × 10 8 m / s c=3.0\times10^{8}\,\mathrm{m/s} c = 3.0 × 1 0 8 m/s μ 0 = 4 π × 10 − 7 H / m \mu_0=4\pi\times10^{-7}\,\mathrm{H/m} μ 0 = 4 π × 1 0 − 7 H/m
第2式を x x x
∂ 2 V ∂ x 2 = − L ∂ ∂ t ∂ I ∂ x = − L ∂ ∂ t ( − C ∂ V ∂ t ) = L C ∂ 2 V ∂ t 2 \frac{\partial^2V}{\partial x^2} = -L\frac{\partial}{\partial t}\frac{\partial I}{\partial x}
= -L\frac{\partial}{\partial t}\left(-C\frac{\partial V}{\partial t}\right)
= LC\frac{\partial^2V}{\partial t^2} ∂ x 2 ∂ 2 V = − L ∂ t ∂ ∂ x ∂ I = − L ∂ t ∂ ( − C ∂ t ∂ V ) = L C ∂ t 2 ∂ 2 V すなわち ∂ 2 V ∂ x 2 = L C ∂ 2 V ∂ t 2 \dfrac{\partial^2V}{\partial x^2} = LC\dfrac{\partial^2 V}{\partial t^2} ∂ x 2 ∂ 2 V = L C ∂ t 2 ∂ 2 V I I I ∂ 2 V ∂ x 2 = 1 v 2 ∂ 2 V ∂ t 2 \dfrac{\partial^2 V}{\partial x^2}=\dfrac{1}{v^2}\dfrac{\partial^2V}{\partial t^2} ∂ x 2 ∂ 2 V = v 2 1 ∂ t 2 ∂ 2 V
v = 1 L C . v = \frac{1}{\sqrt{LC}} . v = L C 1 . 与えられた C C C L L L ln ( b / a ) \ln(b/a) ln ( b / a )
L C = μ 0 ln ( b / a ) 2 π ⋅ 2 π ε ln ( b / a ) = μ 0 ε , v = 1 μ 0 ε = c ε r LC = \frac{\mu_0\ln(b/a)}{2\pi}\cdot\frac{2\pi\varepsilon}{\ln(b/a)} = \mu_0\varepsilon,
\qquad
v = \frac{1}{\sqrt{\mu_0\varepsilon}} = \frac{c}{\sqrt{\varepsilon_\mathrm{r}}} L C = 2 π μ 0 ln ( b / a ) ⋅ ln ( b / a ) 2 π ε = μ 0 ε , v = μ 0 ε 1 = ε r c となります（ε = ε r ε 0 \varepsilon=\varepsilon_\mathrm{r}\varepsilon_0 ε = ε r ε 0 c = 1 / μ 0 ε 0 c=1/\sqrt{\mu_0\varepsilon_0} c = 1/ μ 0 ε 0 ln ( b / a ) = 1.8 \ln(b/a)=1.8 ln ( b / a ) = 1.8 ε r = 2.2 \varepsilon_\mathrm{r}=2.2 ε r = 2.2 2.2 = 1.48 \sqrt{2.2}=1.48 2.2 = 1.48
v = 3.0 × 10 8 1.48 = 2.0 × 10 8 m / s , λ = v f = 2.0 × 10 8 5.0 × 10 8 ≈ 0.4 m . v = \frac{3.0\times 10^{8}}{1.48} = 2.0\times 10^{8}\,\mathrm{m/s},
\qquad
\lambda = \frac{v}{f} = \frac{2.0\times 10^{8}}{5.0\times 10^{8}} \approx 0.4\,\mathrm{m}. v = 1.48 3.0 × 1 0 8 = 2.0 × 1 0 8 m/s , λ = f v = 5.0 × 1 0 8 2.0 × 1 0 8 ≈ 0.4 m . 答えは λ ≈ 0.4 m \lambda \approx 0.4\,\mathrm{m} λ ≈ 0.4 m
x x x V = V ( x − v t ) V=V(x-vt) V = V ( x − v t ) I = I ( x − v t ) I=I(x-vt) I = I ( x − v t ) ∂ V ∂ x = − L ∂ I ∂ t \dfrac{\partial V}{\partial x}=-L\dfrac{\partial I}{\partial t} ∂ x ∂ V = − L ∂ t ∂ I V ′ = L v I ′ V' = Lv\,I' V ′ = Lv I ′ V = L v I V = Lv\,I V = Lv I
Z 0 = L v = L L C = L C . Z_0 = Lv = \frac{L}{\sqrt{LC}} = \sqrt{\frac{L}{C}} . Z 0 = Lv = L C L = C L . 与えられた C , L C,L C , L
Z 0 = μ 0 ln ( b / a ) / 2 π 2 π ε / ln ( b / a ) = ln ( b / a ) 2 π μ 0 ε = ln ( b / a ) 2 π ⋅ μ 0 c ε r . Z_0 = \sqrt{\frac{\mu_0\ln(b/a)/2\pi}{2\pi\varepsilon/\ln(b/a)}}
= \frac{\ln(b/a)}{2\pi}\sqrt{\frac{\mu_0}{\varepsilon}}
= \frac{\ln(b/a)}{2\pi}\cdot\frac{\mu_0c}{\sqrt{\varepsilon_\mathrm{r}}} . Z 0 = 2 π ε / ln ( b / a ) μ 0 ln ( b / a ) /2 π = 2 π ln ( b / a ) ε μ 0 = 2 π ln ( b / a ) ⋅ ε r μ 0 c . μ 0 c = 4 π × 10 − 7 × 3.0 × 10 8 = 120 π ≈ 3.8 × 10 2 Ω \mu_0c = 4\pi\times10^{-7}\times 3.0\times10^{8} = 120\pi \approx 3.8\times10^{2}\,\Omega μ 0 c = 4 π × 1 0 − 7 × 3.0 × 1 0 8 = 120 π ≈ 3.8 × 1 0 2 Ω
Z 0 = 1.8 2 π ⋅ 120 π 1.48 = 1.8 × 120 2 × 1.48 ≈ 7 × 10 1 Ω Z_0 = \frac{1.8}{2\pi}\cdot\frac{120\pi}{1.48} = \frac{1.8\times 120}{2\times 1.48} \approx 7\times 10^{1}\,\Omega Z 0 = 2 π 1.8 ⋅ 1.48 120 π = 2 × 1.48 1.8 × 120 ≈ 7 × 1 0 1 Ω （詳しくは 73 Ω 73\,\Omega 73 Ω 75 Ω 75\,\Omega 75 Ω
入射パルスと反射パルスの重ね合わせが終端で抵抗器のオームの法則を満たす、という条件で決まります。+ x +x + x V = Z 0 I V=Z_0I V = Z 0 I − x -x − x V = − Z 0 I V=-Z_0I V = − Z 0 I
V 1 + V 2 , V 1 Z 0 − V 2 Z 0 V_1+V_2,\qquad \frac{V_1}{Z_0}-\frac{V_2}{Z_0} V 1 + V 2 , Z 0 V 1 − Z 0 V 2 です。これらが V = r I V=rI V = r I
V 1 + V 2 = r V 1 − V 2 Z 0 ⟹ V 2 = r − Z 0 r + Z 0 V 1 . V_1+V_2 = r\,\frac{V_1-V_2}{Z_0}
\qquad\Longrightarrow\qquad
V_2 = \frac{r-Z_0}{r+Z_0}V_1 . V 1 + V 2 = r Z 0 V 1 − V 2 ⟹ V 2 = r + Z 0 r − Z 0 V 1 . r = Z 0 r=Z_0 r = Z 0 V 2 = 0 V_2=0 V 2 = 0 r → ∞ r\to\infty r → ∞ V 2 = V 1 V_2=V_1 V 2 = V 1 r = 0 r=0 r = 0 V 2 = − V 1 V_2=-V_1 V 2 = − V 1
V , I ∝ e i ( k x − ω t ) V,I \propto e^{i(kx-\omega t)} V , I ∝ e i ( k x − ω t ) k k k ∂ V ∂ x = − L ∂ I ∂ t − R I \dfrac{\partial V}{\partial x}=-L\dfrac{\partial I}{\partial t}-RI ∂ x ∂ V = − L ∂ t ∂ I − R I ∂ I ∂ x = − C ∂ V ∂ t \dfrac{\partial I}{\partial x}=-C\dfrac{\partial V}{\partial t} ∂ x ∂ I = − C ∂ t ∂ V
i k V ~ = ( i ω L − R ) I ~ , i k I ~ = i ω C V ~ ik\tilde V = \left(i\omega L - R\right)\tilde I,\qquad
ik\tilde I = i\omega C\tilde V ik V ~ = ( iω L − R ) I ~ , ik I ~ = iω C V ~ から I ~ \tilde I I ~
k 2 = ( i ω L − R ) ω C i = ω 2 L C ( 1 + i R ω L ) . k^2 = \frac{\left(i\omega L-R\right)\omega C}{i} = \omega^2LC\left(1+\frac{iR}{\omega L}\right). k 2 = i ( iω L − R ) ω C = ω 2 L C ( 1 + ω L i R ) . R ≪ L ω R\ll L\omega R ≪ L ω
k ≃ ω L C ( 1 + i R 2 ω L ) = ω v + i 2 R C L = ω v + i R 2 Z 0 . k \simeq \omega\sqrt{LC}\left(1+\frac{iR}{2\omega L}\right) = \frac{\omega}{v} + \frac{i}{2}R\sqrt{\frac{C}{L}} = \frac{\omega}{v}+\frac{iR}{2Z_0}. k ≃ ω L C ( 1 + 2 ω L i R ) = v ω + 2 i R L C = v ω + 2 Z 0 i R . 虚部が正なので e i k x e^{ikx} e ik x + x +x + x Z 0 Z_0 Z 0 d d d
V 2 = V 1 exp ( − R d 2 Z 0 ) = V 1 exp ( − R d 2 C L ) . V_2 = V_1\exp\left(-\frac{Rd}{2Z_0}\right) = V_1\exp\left(-\frac{Rd}{2}\sqrt{\frac{C}{L}}\right). V 2 = V 1 exp ( − 2 Z 0 R d ) = V 1 exp ( − 2 R d L C ) . この近似では減衰量が ω \omega ω R R R ω \omega ω
軸方向の電流密度 j = J ( r ) e − i ω t j=J(r)e^{-i\omega t} j = J ( r ) e − iω t B φ B_\varphi B φ E E E
1 r ∂ ( r B φ ) ∂ r = μ j , ∂ E ∂ r = ∂ B φ ∂ t \frac{1}{r}\frac{\partial\left(rB_\varphi\right)}{\partial r} = \mu j,
\qquad
\frac{\partial E}{\partial r} = \frac{\partial B_\varphi}{\partial t} r 1 ∂ r ∂ ( r B φ ) = μ j , ∂ r ∂ E = ∂ t ∂ B φ です（後者は ∇ × E ⃗ = − ∂ B ⃗ / ∂ t \nabla\times\vec E = -\partial\vec B/\partial t ∇ × E = − ∂ B / ∂ t φ \varphi φ j = σ E j=\sigma E j = σ E e − i ω t e^{-i\omega t} e − iω t B φ = B ( r ) e − i ω t B_\varphi = \mathcal B(r)e^{-i\omega t} B φ = B ( r ) e − iω t
1 σ d J d r = − i ω B ⟹ i ω σ 1 r d d r ( r d J d r ) = μ J . \frac{1}{\sigma}\frac{dJ}{dr} = -i\omega\mathcal B
\qquad\Longrightarrow\qquad
\frac{i}{\omega\sigma}\frac{1}{r}\frac{d}{dr}\left(r\frac{dJ}{dr}\right) = \mu J . σ 1 d r dJ = − iω B ⟹ ω σ i r 1 d r d ( r d r dJ ) = μ J . δ ≪ a \delta\ll a δ ≪ a J J J δ \delta δ 1 r d J d r \dfrac{1}{r}\dfrac{dJ}{dr} r 1 d r dJ d 2 J d r 2 \dfrac{d^2J}{dr^2} d r 2 d 2 J δ / a \delta/a δ / a
d 2 J d r 2 = − i μ ω σ J . \frac{d^2J}{dr^2} = -i\mu\omega\sigma\,J . d r 2 d 2 J = − i μ ω σ J . J ∝ e κ r J\propto e^{\kappa r} J ∝ e κ r κ 2 = − i μ ω σ \kappa^2 = -i\mu\omega\sigma κ 2 = − i μ ω σ κ = e − i π / 4 μ ω σ = 1 − i 2 μ ω σ \kappa = e^{-i\pi/4}\sqrt{\mu\omega\sigma} = \dfrac{1-i}{\sqrt2}\sqrt{\mu\omega\sigma} κ = e − iπ /4 μ ω σ = 2 1 − i μ ω σ R e κ > 0 \mathrm{Re}\,\kappa>0 Re κ > 0
∣ J ( r ) ∣ = ∣ J ( a ) ∣ exp [ ( R e κ ) ( r − a ) ] , 1 δ = R e κ = μ ω σ 2 \lvert J(r)\rvert = \lvert J(a)\rvert \exp\left[\left(\mathrm{Re}\,\kappa\right)\left(r-a\right)\right],
\qquad
\frac{1}{\delta} = \mathrm{Re}\,\kappa = \sqrt{\frac{\mu\omega\sigma}{2}} ∣ J ( r )∣ = ∣ J ( a )∣ exp [ ( Re κ ) ( r − a ) ] , δ 1 = Re κ = 2 μ ω σ すなわち
δ = 2 μ ω σ . \delta = \sqrt{\frac{2}{\mu\omega\sigma}} . δ = μ ω σ 2 . ω \omega ω 2 π a δ 2\pi a\delta 2 π a δ R ∝ 1 / δ ∝ ω R\propto 1/\delta \propto \sqrt{\omega} R ∝ 1/ δ ∝ ω exp ( − c o n s t ⋅ ω d ) \exp\left(-\mathrm{const}\cdot\sqrt{\omega}\,d\right) exp ( − const ⋅ ω d ) [ μ ω σ ] = H m − 1 ⋅ s − 1 ⋅ Ω − 1 m − 1 = m − 2 [\mu\omega\sigma] = \mathrm{H\,m^{-1}\cdot s^{-1}\cdot \Omega^{-1}m^{-1}} = \mathrm{m^{-2}} [ μ ω σ ] = H m − 1 ⋅ s − 1 ⋅ Ω − 1 m − 1 = m − 2 δ \delta δ
出典: 東京大学大学院理学系研究科 物理学専攻 平成31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。
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