正準変換の母関数
正準変換
正準変換とは､変換後もHamiltonの正準方程式が成り立つ座標変換である。
変換前 変換後
$${q_i(\{Q_j\},\{P_j\})}$$ $${Q_j(\{q_i\},\{p_i\})}$$
$${p_i(\{Q_j\},\{P_j\})}$$ $${\xrightarrow{正準変換}}$$ $${P_j(\{q_i\},\{p_i\})}$$
$${\dfrac{dq_i}{dt}=\dfrac{\partial\mathscr{H}}{\partial p_i}}$$ $${\dfrac{dQ_i}{dt}=\dfrac{\partial\overline{\mathscr{H}}}{\partial P_i}}$$
$${\dfrac{dp_i}{dt}=-\dfrac{\partial\mathscr{H}}{\partial q_i}}$$ $${\dfrac{dP_i}{dt}=-\dfrac{\partial\mathscr{\overline{H}}}{\partial P_i}}$$
$${\mathscr{H}(\{q_i\},\{p_i\})=\sum p_i\dot{q_i}-\mathscr{L}}$$ $${\mathscr{\overline{H}}(\{Q_i\},\{P_i\})=\sum P_i\dot{Q_i}-\mathscr{\overline{L}}}$$
※ $${q_i, p_i}$$は一般化座標と一般化運動量であるが､変換後の$${Q_i, P_i}$$は
Hamiltonの正準方程式を満たす正準変数であり、物理的な次元や
意味はない。
母関数の導入
Lagrangianの不定性より
$${\mathscr{L}(\{q_i\},\{\dot{q_i}\},t)=\mathscr{\overline{L}}(\{Q_i\},\{\dot{Q_i}\},t)+\dfrac{dW}{dt}}$$ (1)（$${W}$$は母関数）
Hamiltonian の定義より（Legendre 変換）
$${\mathscr{H}=\displaystyle\sum_{i=1}^fp_i\dot{q_i}-\mathscr{L}(\{q_i\},\{\dot{q_i}\},t)}$$, $${\mathscr{L}=\displaystyle\sum_{i=1}^fp_i\dot{q_i}-\mathscr{H}(\{q_i\},\{\dot{q_i}\},t)}$$ (2)
$${\mathscr{\overline{H}}=\displaystyle\sum_{i=1}^fP_i\dot{Q_i}-\mathscr{\overline{L}}(\{Q_i\},\{\dot{Q_i}\},t)}$$, $${\mathscr{\overline{L}}=\displaystyle\sum_{i=1}^fP_i\dot{Q_i}-\mathscr{\overline{H}}(\{q_i\},\{\dot{q_i}\},t)}$$
(2)を(1)に代入
$${\displaystyle\sum_{i=1}^fp_i\dot{q_i}-\mathscr{H}(\{q_i\},\{\dot{q_i}\},t)=\displaystyle\sum_{i=1}^fP_i\dot{Q_i}-\mathscr{\overline{H}}(\{Q_i\},\{\dot{Q_i}\},t)+\dfrac{dW}{dt}}$$ (3)
母関数の独立変数
ここで母関数$${W}$$の変数について考える。
$${\{q_i\},\{p_i\},\{Q_i\},\{P_i\}}$$は互いに独立ではないので､すべてを$${W}$$の変数とすることは出来ない。では､必要な変数はいくつか。
Lagrangianは２変数$${(\{q_i\},\{\dot{q_i}\})}$$であり､$${\mathscr{L}}$$をLegendre 変換して得られるHamiltonianも２変数$${(\{q_i\},\{p_i\})}$$である。(3)式での整合性を考えると$${W}$$も２変数が相応しい。（選ばれなかった２つの変数は選んだ２つの独立変数の関数になる）
$${\{q_i\},\{{p_i}\},\{Q_i\},\{{P_i}\}}$$のうち､２つの変数を選ぶ組み合わせは$${\{q_i\},\{{p_i}\}}$$と$${\{Q_i\},\{{P_i}\}}$$の組み合わせを除くと以下の４通りである。
i) $${\{q_i\},\{Q_i\}}$$が独立変数 つまり$${W=W_1\{q_i\}, \{Q_i\})}$$
残りの２変数は $${p_i(\{q_i\},\{Q_i\}),P_i(\{q_i\},\{Q_i\})}$$
ii) $${\{q_i\},\{P_i\}}$$が独立変数 つまり$${W=W_2\{q_i\}, \{P_i\})}$$
残りの２変数は $${p_i(\{q_i\},\{P_i\}),Q_i(\{q_i\},\{P_i\})}$$
iii) $${\{p_i\},\{Q_i\}}$$が独立変数 つまり$${W=W_3\{p_i\}, \{Q_i\})}$$
残りの２変数は $${q_i(\{p_i\},\{Q_i\}),P_i(\{p_i\},\{Q_i\})}$$
iv) $${\{p_i\},\{P_i\}}$$が独立変数 つまり$${W=W_4\{p_i\}, \{P_i\})}$$
残りの２変数は $${q_i(\{p_i\},\{P_i\}),Q_i(\{p_i\},\{p_i\})}$$
では､それぞれについて考える。
i) $${W=W_1(\{q_i\}, \{Q_i\}, t)}$$のとき
$${\dfrac{dW_1}{dt}=\displaystyle\sum_{i=i}^f\Big(\dfrac{\partial W_1}{\partial q_i}\dot{q_i}+\dfrac{\partial W_1}{\partial Q_i}\dot{Q_i}\Big)+\dfrac{\partial W_1}{\partial t}}$$
これを(3)に代入すると
$${\displaystyle\sum_{i=1}^f p_i\dot{q_i}-\mathscr{H}=\displaystyle\sum _{i=1}^fP_i\dot{Q_i}-\mathscr{\overline{H}}+\sum_{i=i}^f\Big(\dfrac{\partial W_1}{\partial q_i}\dot{q_i}+\dfrac{\partial W_1}{\partial Q_i}\dot{Q_i}\Big)+\dfrac{\partial W_1}{\partial t}}$$
$${\displaystyle\sum_{i=1}^f p_i\dot{q_i}-\mathscr{H}-\sum _{i=1}^fP_i\dot{Q_i}+\mathscr{\overline{H}}-\displaystyle\sum_{i=i}^f\Big(\dfrac{\partial W_1}{\partial q_i}\dot{q_i}+\dfrac{\partial W_1}{\partial Q_i}\dot{Q_i}\Big)-\dfrac{\partial W_1}{\partial t}=0}$$
$${\displaystyle\sum_{i=1}^f\Big( p_i-\dfrac{\partial W_1}{\partial q_i}\Big)\dot{q_i}-\sum _{i=1}^f\Big(P_i+\dfrac{\partial W_1}{\partial Q_i}\Big)\dot{Q_i}+\Big(\mathscr{\overline{H}}-\mathscr{H}-\dfrac{\partial W_1}{\partial t}\Big)=0}$$
両辺×$${dt}$$
$${\displaystyle\sum_{i=1}^f\Big( p_i-\dfrac{\partial W_1}{\partial q_i}\Big)dq_i-\sum _{i=1}^f\Big(P_i+\dfrac{\partial W_1}{\partial Q_i}\Big)dQ_i}$$
$${+\Big(\mathscr{\overline{H}}-\mathscr{H}-\dfrac{\partial W_1}{\partial t}\Big)dt=0}$$
各項を比較して
$${p_i=\dfrac{\partial W_1}{\partial q_i}}$$ (i-1) $${P_i=-\dfrac{\partial W_1}{\partial Q_i},}$$ (i-2)
$${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_1}{\partial t}}$$ (i-3)（$${W_1}$$に$${t}$$が陽に含まれなければ $${\mathscr{\overline{H}}=\mathscr{H}}$$）
$${p_i,P_i}$$は$${(\{q_i\},\{Q_i\})}$$を変数とする関数である。
$${p_i=p_i(\{q_i\},\{Q_i\},t)}$$を逆に辿ると
$${Q_i=p_i^{-1}(\{q_i\},\{p_i\},t)}$$ (i-4)
これを$${P_i=P_i(\{q_i\},\{Q_i\},t)}$$に代入すると
$${P_i=P_i(\{q_i\},\{Q_i\},t)=P_i(\{q_i\},\{p_i^{-1}(\{q_i\},\{p_i\},t)\},t)}$$ (i-5)
(i-4)(i-5)を使って$${\{q_i\},\{p_i\}}$$を$${\{Q_i\},\{P_i\}}$$に変換する式を作ればよい。
ii) $${W=W_2(\{q_i\},\{P_i\},t)}$$のとき
このまま$${\dfrac{dW_2}{dt}= …}$$といきたいところだが、これでは変数が合わない。
ここで小細工をする。
$${W_1(\{q_i\},\{Q_i\},t)}$$から$${W_2(\{q_i\},\{P_i\},t)}$$へのLugendre変換をおこなう。
$${\{q_i\},t}$$が共通で$${\{Q_i\}\rightarrow \{P_i\}}$$だから
$${W_2(\{q_i\},\{P_i\},t)=W_1(\{q_i\},\{Q_i\},t)-\displaystyle\sum_{i=1}^f\dfrac{\partial W_1}{\partial Q_i}Q_i}$$
← (i-2)$${\dfrac{\partial W_1}{\partial Q_i}=-P_i}$$
$${=W_1(\{q_i\},\{Q_i\},t)+\displaystyle\sum_{i=1}^fP_iQ_i}$$ (ii-1)
(ii-1)の両辺を$${q_i}$$で偏微分すると
$${\dfrac{\partial W_2(\{q_i\},\{P_i\},t)}{\partial q_i}=\dfrac{\partial W_1(\{\partial q_i\},\{Q_i\},t)}{\partial q_i}+\dfrac{\partial }{\partial q_i}\displaystyle\sum_{i=1}^fP_iQ_i}$$
$${\dfrac{\partial W_2}{\partial q_i}=\dfrac{\partial W_1}{\partial q_i}}$$ (i-1)$${p_i=\dfrac{\partial W_1}{\partial q_i}}$$とあわせて $${p_i=\dfrac{\partial W_2}{\partial q_i}}$$ (ii-2)
(ii-1)の両辺を$${P_i}$$で偏微分すると
$${\dfrac{\partial W_2(\{q_i\},\{P_i\},t)}{\partial P_i}=\dfrac{\partial W_1(\{\partial q_i\},\{Q_i\},t)}{\partial P_i}+\dfrac{\partial }{\partial P_i}\displaystyle\sum_{i=1}^fP_iQ_i}$$
$${\dfrac{\partial W_2}{\partial P_i}=Q_i}$$ よって $${Q_i=\dfrac{\partial W_2}{\partial P_i}}$$ (ii-3)
(ii-1)の両辺を$${Q_i}$$で偏微分すると
$${\dfrac{\partial W_2(\{q_i\},\{P_i\},t)}{\partial Q_i}=\dfrac{\partial W_1(\{\partial q_i\},\{Q_i\},t)}{\partial Q_i}+\dfrac{\partial }{\partial Q_i}\displaystyle\sum_{i=1}^fP_iQ_i}$$
$${0=\dfrac{\partial W_1}{\partial Q_i}+P_i, P_i=-\dfrac{\partial W_1}{\partial Q_i}}$$(i-2)と同じ
(ii-1)の両辺を$${p_i}$$で偏微分すると
$${\dfrac{\partial W_2(\{q_i\},\{P_i\},t)}{\partial p_i}=\dfrac{\partial W_1(\{\partial q_i\},\{Q_i\},t)}{\partial p_i}+\dfrac{\partial }{\partial p_i}\displaystyle\sum_{i=1}^fP_iQ_i, 0=0+0}$$
(ii-1)の両辺を$${t}$$で偏微分すると
$${\dfrac{\partial W_2(\{q_i\},\{P_i\},t)}{\partial t}=\dfrac{\partial W_1(\{\partial q_i\},\{Q_i\},t)}{\partial t}+\dfrac{\partial }{\partial t}\displaystyle\sum_{i=1}^fP_iQ_i}$$
$${\dfrac{\partial W_2}{\partial t}=\dfrac{\partial W_1}{\partial t}+0}$$ (i-3)$${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_1}{\partial t}}$$とあわせて
$${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_2}{\partial t}}$$ (ii-4)
まとめると
$${p_i=\dfrac{\partial W_2}{\partial q_i}}$$ (ii-2) $${Q_i=\dfrac{\partial W_2}{\partial P_i}}$$ (ii-3) $${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_2}{\partial t}}$$ (ii-4)
iii) $${W=W_3(\{p_i\},\{Q_i\},t)}$$のとき
$${W_1(\{q_i\},\{Q_i\},t)}$$から$${W_3(\{p_i\},\{Q_i\},t)}$$へのLugendre変換をおこなう。
$${\{Q_i\},t}$$が共通で$${\{q_i\}\rightarrow \{p_i\}}$$だから
$${W_3(\{p_i\},\{Q_i\},t)=W_1(\{q_i\},\{Q_i\},t)-\displaystyle\sum_{i=1}^f\dfrac{\partial W_1}{\partial q_i}q_i}$$
← (i-1)$${\dfrac{\partial W_1}{\partial q_i}=p_i}$$
$${=W_1(\{q_i\},\{Q_i\},t)-p_iq_i}$$ (iii-1)
(iii-1)の両辺を$${q_i}$$で偏微分
$${\dfrac{\partial W_3(\{p_i\},\{Q_i\},t)}{\partial q_i}=\dfrac{\partial W_1(\{q_i\},\{Q_i\},t)}{\partial q_i}-\dfrac{\partial}{\partial q_i}\displaystyle\sum_{i=1}^fp_iq_i}$$
$${0=\dfrac{\partial W_1}{\partial q_i}-p_1, p_1=\dfrac{\partial W_1}{\partial q_i} }$$(i-1)と同じ
(iii-1)の両辺を$${p_i}$$で偏微分
$${\dfrac{\partial W_3(\{p_i\},\{Q_i\},t)}{\partial p_i}=\dfrac{\partial W_1(\{q_i\},\{Q_i\},t)}{\partial p_i}-\dfrac{\partial}{\partial p_i}\displaystyle\sum_{i=1}^fp_iq_i}$$
$${\dfrac{\partial W_3}{\partial p_i}=0-q_i, q_i=-\dfrac{\partial W_3}{\partial p_i}}$$ (iii-2)
(iii-1)の両辺を$${P_i}$$で偏微分
$${\dfrac{\partial W_3(\{p_i\},\{Q_i\},t)}{\partial P_i}=\dfrac{\partial W_1(\{q_i\},\{Q_i\},t)}{\partial P_i}-\dfrac{\partial}{\partial P_i}\displaystyle\sum_{i=1}^fp_iq_i}$$ $${0=0+0}$$
(iii-1)の両辺を$${Q_i}$$で偏微分
$${\dfrac{\partial W_3(\{p_i\},\{Q_i\},t)}{\partial Q_i}=\dfrac{\partial W_1(\{q_i\},\{Q_i\},t)}{\partial Q_i}-\dfrac{\partial}{\partial Q_i}\displaystyle\sum_{i=1}^fp_iq_i}$$
$${\dfrac{\partial W_3}{\partial Q_i}=\dfrac{\partial W_1}{\partial Q_i}}$$ (i-2)$${P_i=-\dfrac{\partial W_1}{\partial Q_i}}$$とあわせて $${P_i=-\dfrac{\partial W_3}{\partial Q_i}}$$ (iii-3)
(iii-1)の両辺を$${t}$$で偏微分
$${\dfrac{\partial W_3(\{p_i\},\{Q_i\},t)}{\partial t}=\dfrac{\partial W_1(\{q_i\},\{Q_i\},t)}{\partial t}-\dfrac{\partial}{\partial t}\displaystyle\sum_{i=1}^fp_iq_i}$$
$${\dfrac{\partial W_3}{\partial t}=\dfrac{\partial W_1}{\partial t}+0}$$ (i-3)$${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_1}{\partial t}}$$とあわせて
$${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_3}{\partial t}}$$ (iii-4)
まとめると
$${q_i=-\dfrac{\partial W_3}{\partial p_i}}$$ (iii-2) $${P_i=-\dfrac{\partial W_3}{\partial Q_i}}$$ (iii-3) $${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_3}{\partial t}}$$ (iii-4)
iv) $${W=W_4(\{p_i\},\{P_i\},t)}$$のとき
$${W_2(\{q_i\},\{P_i\},t)}$$から$${W_4(\{p_i\},\{P_i\},t)}$$へのLugendre変換をおこなう。
$${\{P_i\},t}$$が共通で$${\{q_i\}\rightarrow \{p_i\}}$$だから
$${W_4(\{p_i\},\{P_i\},t)=W_2(\{q_i\},\{P_i\},t)-\displaystyle\sum_{i=1}^f\dfrac{\partial W_2}{\partial q_i}q_i}$$
← (ii-2)$${p_i=\dfrac{\partial W_2}{\partial q_i}}$$
$${=W_2(\{q_i\},\{P_i\},t)-\displaystyle\sum_{i=1}^fp_iq_i}$$ (iv-1)
(iv-1)の両辺を$${q_i}$$で偏微分
$${\dfrac{\partial W_4(\{p_i\},\{P_i\},t)}{\partial q_i}=\dfrac{\partial W_2(\{q_i\},\{P_i\},t)}{\partial q_i}-\dfrac{\partial}{\partial q_i}\displaystyle\sum_{i=1}^fp_iq_i}$$
$${0=\dfrac{\partial W_2}{\partial q_i}-p_i, p_i=\dfrac{\partial W_2}{\partial q_i}}$$ (ii-2)と同じ
(iv-1)の両辺を$${p_i}$$で偏微分
$${\dfrac{\partial W_4(\{p_i\},\{P_i\},t)}{\partial p_i}=\dfrac{\partial W_2(\{q_i\},\{P_i\},t)}{\partial p_i}-\dfrac{\partial}{\partial p_i}\displaystyle\sum_{i=1}^fp_iq_i}$$
$${\dfrac{\partial W_4}{\partial p_i}=0-q_i, q_i=-\dfrac{\partial W_4}{\partial p_i}}$$ (iv-2)
(iv-1)の両辺を$${Q_i}$$で偏微分
$${\dfrac{\partial W_4(\{p_i\},\{P_i\},t)}{\partial Q_i}=\dfrac{\partial W_2(\{q_i\},\{P_i\},t)}{\partial Q_i}-\dfrac{\partial}{\partial Q_i}\displaystyle\sum_{i=1}^fp_iq_i,}$$ $${0=0-0}$$
(iv-1)の両辺を$${P_i}$$で偏微分
$${\dfrac{\partial W_4(\{p_i\},\{P_i\},t)}{\partial P_i}=\dfrac{\partial W_2(\{q_i\},\{P_i\},t)}{\partial P_i}-\dfrac{\partial}{\partial P_i}\displaystyle\sum_{i=1}^fp_iq_i}$$
$${\dfrac{\partial W_4}{\partial P_i}=\dfrac{\partial W_2}{\partial P_i}-0}$$ (ii-3)$${Q_i=\dfrac{\partial W_2}{\partial P_i}}$$とあわせて $${Q_i=\dfrac{\partial W_4}{\partial P_i}}$$ ((iv-3)
(iv-1)の両辺を$${t}$$で偏微分
$${\dfrac{\partial W_4(\{p_i\},\{P_i\},t)}{\partial t}=\dfrac{\partial W_2(\{q_i\},\{P_i\},t)}{\partial t}-\dfrac{\partial}{\partial t}\displaystyle\sum_{i=1}^fp_iq_i}$$
$${\dfrac{\partial W_4}{\partial t}=\dfrac{\partial W_2}{\partial t}-0}$$ (ii-4)$${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_2}{\partial t}}$$とあわせて
$${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_4}{\partial t}}$$ (iv-4)
まとめると
$${q_i=-\dfrac{\partial W_4}{\partial p_i}}$$ (iv-2) $${Q_i=\dfrac{\partial W_4}{\partial P_i}}$$ ((iv-3) $${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_4}{\partial t}}$$ (iv-4)
最後にすべてをまとめる
i) $${W=W_1(\{q_i\}, \{Q_i\}, t)}$$のとき
$${p_i=\dfrac{\partial W_1}{\partial q_i}}$$(i-1) $${P_i=-\dfrac{\partial W_1}{\partial Q_i},}$$(i-2) $${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_1}{\partial t}}$$(i-3)
ii) $${W=W_2(\{q_i\},\{P_i\},t)}$$のとき
$${p_i=\dfrac{\partial W_2}{\partial q_i}}$$(ii-2) $${Q_i=\dfrac{\partial W_2}{\partial P_i}}$$(ii-3) $${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_2}{\partial t}}$$ (ii-4)
iii) $${W=W_3(\{p_i\},\{Q_i\},t)}$$のとき
$${q_i=-\dfrac{\partial W_3}{\partial p_i}}$$(iii-2) $${P_i=-\dfrac{\partial W_3}{\partial Q_i}}$$(iii-3) $${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_3}{\partial t}}$$ (iii-4)
iv) $${W=W_4(\{p_i\},\{P_i\},t)}$$のとき
$${q_i=-\dfrac{\partial W_4}{\partial p_i}}$$(iv-2) $${Q_i=\dfrac{\partial W_4}{\partial P_i}}$$(iv-3) $${\mathscr{\overline{H}}=\mathscr{H}+\dfrac{\partial W_4}{\partial t}}$$ (iv-4)
（$${W_1,W_2,W_3,W_4}$$に$${t}$$が陽に含まれなければ $${\mathscr{\overline{H}}=\mathscr{H}}$$）
はじめにも書いたが、変換後の各変数は残りの２変数の関数であり物理的な次元や意味はない。（例えば (i-4)(i-5) を見よ）
$${\mathscr{L}}$$から$${\mathscr{H}}$$への Legendre 変換では､位置$${q}$$と加速度$${\dot{q}}$$から位置$${q}$$と運動量$${ p}$$とはっきり意味づけられているのに対して、ただ正準方程式を満たす変換だということである。
〇 〇 〇
$${W}$$の変数の場合分けのところから異常に手間取った。たくさんの本やネットの資料をあたっても､本当にさらっとしか書いていない。わたしには難しすぎる。もっと丁寧に書いてほしい。Legendre 変換の理解をもっと深めて､いつかもう一度整理したい。今はこんなもんにしといたろ。やれやれ。