ここでは、球面座標系 (r, θ, φ) から直交座標系 (x, y, z) への変換から球面座標系の線素の2乗を導出する。
球面座標 (r, θ, φ) から直交座標 (x, y, z) への変換は
{ x = r sin θ cos ϕ y = r sin θ sin ϕ z = r cos θ {\displaystyle {\begin{cases}x=r\sin \theta \,\cos \phi \\y=r\sin \theta \,\sin \phi \\z=r\cos \theta \end{cases}}}
で与えられる。これを微分すれば
{ d x = sin θ cos ϕ d r + r cos θ cos ϕ d θ − r sin θ sin ϕ d ϕ d y = sin θ sin ϕ d r + r cos θ sin ϕ d θ + r sin θ cos ϕ d ϕ d z = cos θ d r − r sin θ d θ {\displaystyle {\begin{cases}dx=\sin \theta \,\cos \phi \,dr+r\cos \theta \,\cos \phi \,d\theta -r\sin \theta \,\sin \phi \,d\phi \\dy=\sin \theta \,\sin \phi \,dr+r\cos \theta \,\sin \phi \,d\theta +r\sin \theta \,\cos \phi \,d\phi \\dz=\cos \theta \,dr-r\sin \theta \,d\theta \end{cases}}}
が得られる。したがって、線素の2乗は
d s 2 = d x 2 + d y 2 + d z 2 = sin 2 θ cos 2 ϕ d r 2 + r 2 cos 2 θ cos 2 ϕ d θ 2 + r 2 sin 2 θ sin 2 ϕ d ϕ 2 + sin 2 θ sin 2 ϕ d r 2 + r 2 cos 2 θ sin 2 ϕ d θ 2 + r 2 sin 2 θ cos 2 ϕ d ϕ 2 + cos 2 θ d r 2 + r 2 sin 2 θ d θ 2 − 2 r sin θ cos θ d r d θ + 2 r sin θ cos θ cos 2 ϕ d r d θ − 2 r 2 sin θ cos θ sin ϕ cos ϕ d θ d ϕ − 2 r sin 2 θ sin ϕ cos ϕ d r d ϕ + 2 r sin θ cos θ sin 2 ϕ d r d θ + 2 r 2 sin θ cos θ sin ϕ cos ϕ d θ d ϕ + 2 r sin 2 θ sin ϕ cos ϕ d r d ϕ = ( sin 2 θ cos 2 ϕ + sin 2 θ sin 2 ϕ + cos 2 θ ) d r 2 + ( r 2 cos 2 θ cos 2 ϕ + r 2 cos 2 θ sin 2 ϕ + r 2 sin 2 θ ) d θ 2 + ( r 2 sin 2 θ sin 2 ϕ + r 2 sin 2 θ cos 2 ϕ ) d ϕ 2 + ( 2 r sin θ cos θ sin 2 ϕ + 2 r sin θ cos θ cos 2 ϕ − 2 r sin θ cos θ ) d r d θ = { sin 2 θ ( cos 2 ϕ + sin 2 ϕ ) + cos 2 θ } d r 2 + { cos 2 θ ( cos 2 ϕ + sin 2 ϕ ) + sin 2 θ } r 2 d θ 2 + ( sin 2 ϕ + cos 2 ϕ ) r 2 sin 2 θ d ϕ 2 + ( sin 2 ϕ + cos 2 ϕ − 1 ) 2 r sin θ cos θ d r d θ = ( sin 2 θ + cos 2 θ ) d r 2 + ( cos 2 θ + sin 2 θ ) r 2 d θ 2 + r 2 sin 2 θ d ϕ 2 = d r 2 + r 2 d θ 2 + r 2 sin 2 θ d ϕ 2 {\displaystyle {\begin{aligned}ds^{2}&=dx^{2}+dy^{2}+dz^{2}\\&{\begin{aligned}=~&\sin ^{2}\theta \,\cos ^{2}\phi \,dr^{2}+r^{2}\cos ^{2}\theta \,\cos ^{2}\phi \,d\theta ^{2}+r^{2}\sin ^{2}\theta \,\sin ^{2}\phi \,d\phi ^{2}\\&+\sin ^{2}\theta \,\sin ^{2}\phi \,dr^{2}+r^{2}\cos ^{2}\theta \,\sin ^{2}\phi \,d\theta ^{2}+r^{2}\sin ^{2}\theta \,\cos ^{2}\phi \,d\phi ^{2}\\&+\cos ^{2}\theta \,dr^{2}+r^{2}\sin ^{2}\theta \,d\theta ^{2}-2r\sin \theta \,\cos \theta \,dr\,d\theta \\&+2r\sin \theta \,\cos \theta \,\cos ^{2}\phi \,dr\,d\theta -2r^{2}\sin \theta \,\cos \theta \,\sin \phi \,\cos \phi \,d\theta \,d\phi -2r\sin ^{2}\theta \,\sin \phi \,\cos \phi \,dr\,d\phi \\&+2r\sin \theta \,\cos \theta \,\sin ^{2}\phi \,dr\,d\theta +2r^{2}\sin \theta \,\cos \theta \,\sin \phi \,\cos \phi \,d\theta \,d\phi +2r\sin ^{2}\theta \,\sin \phi \,\cos \phi \,dr\,d\phi \end{aligned}}\\&{\begin{aligned}=~&(\sin ^{2}\theta \,\cos ^{2}\phi +\sin ^{2}\theta \,\sin ^{2}\phi +\cos ^{2}\theta )dr^{2}+(r^{2}\cos ^{2}\theta \,\cos ^{2}\phi +r^{2}\cos ^{2}\theta \,\sin ^{2}\phi +r^{2}\sin ^{2}\theta )d\theta ^{2}\\&+(r^{2}\sin ^{2}\theta \,\sin ^{2}\phi +r^{2}\sin ^{2}\theta \,\cos ^{2}\phi )d\phi ^{2}+(2r\sin \theta \,\cos \theta \,\sin ^{2}\phi +2r\sin \theta \,\cos \theta \,\cos ^{2}\phi -2r\sin \theta \,\cos \theta )dr\,d\theta \end{aligned}}\\&{\begin{aligned}=~&\{\sin ^{2}\theta \,(\cos ^{2}\phi +\sin ^{2}\phi )+\cos ^{2}\theta \}dr^{2}+\{\cos ^{2}\theta \,(\cos ^{2}\phi +\sin ^{2}\phi )+\sin ^{2}\theta \}r^{2}\,d\theta ^{2}\\&+(\sin ^{2}\phi +\cos ^{2}\phi )r^{2}\sin ^{2}\theta \,d\phi ^{2}+(\sin ^{2}\phi +\cos ^{2}\phi -1)2r\sin \theta \,\cos \theta \,dr\,d\theta \end{aligned}}\\&=(\sin ^{2}\theta +\cos ^{2}\theta )dr^{2}+(\cos ^{2}\theta +\sin ^{2}\theta )r^{2}\,d\theta ^{2}+r^{2}\sin ^{2}\theta \,d\phi ^{2}\\&=dr^{2}+r^{2}d\theta ^{2}+r^{2}\,\sin ^{2}\theta \,d\phi ^{2}\end{aligned}}}
となる。
