ランデン変換の導出
出典: フリー百科事典『ウィキペディア（Wikipedia）』 (2018/09/19 10:13 UTC 版)
ランデン変換は sin ϕ = 2 1 + k sin θ cos θ 1 − 4 k ( 1 + k ) 2 sin 2 θ {\displaystyle \sin \phi ={\frac {{\frac {2}{1+k}}\sin \theta \cos \theta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}} cos ϕ d ϕ = 2 1 + k ( cos 2 θ − sin 2 θ ) 1 − 4 k ( 1 + k ) 2 sin 2 θ d θ + 2 1 + k ( 4 k ( 1 + k ) 2 sin 2 θ cos 2 θ ) ( 1 − 4 k ( 1 + k ) 2 sin 2 θ ) 3 d θ = 2 1 + k ( 1 − 2 1 + k sin 2 θ ) ( 1 − 2 k 1 + k sin 2 θ ) ( 1 − 4 k ( 1 + k ) 2 sin 2 θ ) 3 d θ {\displaystyle {\begin{aligned}\cos \phi {d\phi }&={\frac {{\frac {2}{1+k}}\left(\cos ^{2}\theta -\sin ^{2}\theta \right)}{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}{d\theta }+{\frac {{\frac {2}{1+k}}\left({\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta \cos ^{2}\theta \right)}{\left({\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)^{3}}}{d\theta }\\&={\frac {{\frac {2}{1+k}}\left(1-{\frac {2}{1+k}}\sin ^{2}\theta \right)\left(1-{\frac {2k}{1+k}}\sin ^{2}\theta \right)}{\left({\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)^{3}}}{d\theta }\end{aligned}}} の置換により導かれる。 F ( sin α , k ) = ∫ ϕ = 0 α d ϕ 1 − k 2 sin 2 ϕ = ∫ ϕ = 0 α cos ϕ d ϕ 1 − sin 2 ϕ 1 − k 2 sin 2 ϕ = ∫ θ = 0 β 2 1 + k ( 1 − 2 1 + k sin 2 θ ) ( 1 − 2 k 1 + k sin 2 θ ) ( 1 − 4 k ( 1 + k ) 2 sin 2 θ ) 3 1 − 4 ( 1 + k ) 2 sin 2 θ cos 2 θ 1 − 4 k ( 1 + k ) 2 sin 2 θ 1 − k 2 4 ( 1 + k ) 2 sin 2 θ cos 2 θ 1 − 4 k ( 1 + k ) 2 sin 2 θ d θ = ∫ θ = 0 β 2 1 + k ( 1 − 2 1 + k sin 2 θ ) ( 1 − 2 k 1 + k sin 2 θ ) ( 1 − 4 k ( 1 + k ) 2 sin 2 θ ) 3 1 − 2 1 + k sin 2 θ 1 − 4 k ( 1 + k ) 2 sin 2 θ 1 − 2 k 1 + k sin 2 θ 1 − 4 k ( 1 + k ) 2 sin 2 θ d θ = 2 1 + k ∫ θ = 0 β d θ 1 − 4 k ( 1 + k ) 2 sin 2 θ = 2 1 + k F ( sin β , 2 k 1 + k ) {\displaystyle {\begin{aligned}F\left(\sin \alpha ,k\right)&=\int _{\phi =0}^{\alpha }{\frac {d\phi }{\sqrt {1-k^{2}\sin ^{2}\phi }}}\\&=\int _{\phi =0}^{\alpha }{\frac {\cos \phi {d\phi }}{{\sqrt {1-\sin ^{2}\phi }}{\sqrt {1-k^{2}\sin ^{2}\phi }}}}\\&=\int _{\theta =0}^{\beta }{\frac {\frac {{\frac {2}{1+k}}\left(1-{\frac {2}{1+k}}\sin ^{2}\theta \right)\left(1-{\frac {2k}{1+k}}\sin ^{2}\theta \right)}{\left({\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)^{3}}}{{\sqrt {1-{\frac {{\frac {4}{(1+k)^{2}}}\sin ^{2}\theta \cos ^{2}\theta }{1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}{\sqrt {1-k^{2}{\frac {{\frac {4}{(1+k)^{2}}}\sin ^{2}\theta \cos ^{2}\theta }{1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}}}{d\theta }\\&=\int _{\theta =0}^{\beta }{\frac {\frac {{\frac {2}{1+k}}\left(1-{\frac {2}{1+k}}\sin ^{2}\theta \right)\left(1-{\frac {2k}{1+k}}\sin ^{2}\theta \right)}{\left({\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}\right)^{3}}}{{\frac {1-{\frac {2}{1+k}}\sin ^{2}\theta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}\;{\frac {1-{\frac {2k}{1+k}}\sin ^{2}\theta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}}}{d\theta }\\&={\frac {2}{1+k}}\int _{\theta =0}^{\beta }{\frac {d\theta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\theta }}}\\&={\frac {2}{1+k}}F\left(\sin \beta ,{\frac {2{\sqrt {k}}}{1+k}}\right)\end{aligned}}} sin β {\displaystyle \sin \beta } を陽にすると sin α = 2 1 + k sin β cos β 1 − 4 k ( 1 + k ) 2 sin 2 β {\displaystyle \sin \alpha ={\frac {{\frac {2}{1+k}}\sin \beta \cos \beta }{\sqrt {1-{\frac {4k}{(1+k)^{2}}}\sin ^{2}\beta }}}} sin 2 α = 4 sin 2 β cos 2 β ( 1 + k ) 2 − 4 k sin 2 β = 1 − cos 2 ( 2 β ) 1 + k 2 + 2 k cos ( 2 β ) {\displaystyle \sin ^{2}\alpha ={\frac {4\sin ^{2}\beta \cos ^{2}\beta }{(1+k)^{2}-4k\sin ^{2}\beta }}={\frac {1-\cos ^{2}(2\beta )}{1+k^{2}+2k\cos(2\beta )}}} cos 2 ( 2 β ) + 2 k sin 2 α cos ( 2 β ) + k 2 sin 2 α − 1 + sin 2 α = 0 {\displaystyle \cos ^{2}(2\beta )+2k\sin ^{2}\alpha \cos(2\beta )+k^{2}\sin ^{2}\alpha -1+\sin ^{2}\alpha =0} cos ( 2 β ) = − k sin 2 α + k 2 sin 4 α − k 2 sin 2 α + 1 − sin 2 α {\displaystyle \cos(2\beta )=-k\sin ^{2}\alpha +{\sqrt {k^{2}\sin ^{4}\alpha -k^{2}\sin ^{2}\alpha +1-\sin ^{2}\alpha }}} sin β = 1 − cos ( 2 β ) 2 = 1 2 ( 1 + k ) 2 sin 2 α + ( 1 − k 2 sin 2 ϕ − 1 − sin 2 ϕ ) 2 {\displaystyle \sin \beta ={\frac {1-\cos(2\beta )}{2}}={\frac {1}{2}}{\sqrt {\left(1+k\right)^{2}\sin ^{2}\alpha +\left({\sqrt {1-k^{2}\sin ^{2}\phi }}-{\sqrt {1-\sin ^{2}\phi }}\right)^{2}}}} である。
※この「ランデン変換の導出」の解説は、「ランデン変換」の解説の一部です。
「ランデン変換の導出」を含む「ランデン変換」の記事については、「ランデン変換」の概要を参照ください。
- ランデン変換の導出のページへのリンク
