Referat I 2022 yil reja: Silindrik va sferik koordinatakar sistemasi. Chiziqlarning parametric tenglamasi


Silindrik koordinatalar sistemasi



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Silindrik koordinatalar sistemasi
Agar nuqtaning silindrik koordinatalari berilgan boʻlsa, sferik koordinatalarga oʻtish uchun quyidagi formulalardan foydalaniladi:
{\displaystyle {\begin{cases}\rho =r\sin \theta \\\varphi =\varphi \\z=r\cos \theta \end{cases}}}
Yoki aksincha, sferik koordinatalardan silindrik koordinatalarga oʻtish uchun quyidagi formulalardan foydalaniladi:
{\displaystyle {\begin{cases}r={\sqrt {\rho ^{2}+z^{2}}},\\\theta =\mathrm {arctg} {\dfrac {\rho }{z}},\\\varphi =\varphi .\end{cases}}}
Silindrik koordinatalardan sferik koordinatalarga oʻtish yakobiani :
{\displaystyle J=r}
Sferik koordinatalar sistemasida differensiallash va integrallash
{\displaystyle (r,\theta ,\varphi )}nuqtadan {\displaystyle (r+\mathrm {d} r,\,\theta +\mathrm {d} \theta ,\,\varphi +\mathrm {d} \varphi )}  nuqtaga oʻtkazilgan vektor {\displaystyle \mathrm {d} \mathbf {r} }  ning uzunligi quyidagiga teng:
{\displaystyle \mathrm {d} \mathbf {r} =\mathrm {d} r\,{\boldsymbol {\hat {r}}}+r\,\mathrm {d} \theta \,{\boldsymbol {\hat {\theta }}}+r\sin {\theta }\,\mathrm {d} \varphi \,\mathbf {\boldsymbol {\hat {\varphi }}} ,}bu yerda
{\displaystyle {\boldsymbol {\hat {r}}}=\sin \theta \cos \varphi {\boldsymbol {\hat {\imath }}}+\sin \theta \sin \varphi {\boldsymbol {\hat {\jmath }}}+\cos \theta {\boldsymbol {\hat {k}}}}{\displaystyle {\boldsymbol {\hat {\theta }}}=\cos \theta \cos \varphi {\boldsymbol {\hat {\imath }}}+\cos \theta \sin \varphi {\boldsymbol {\hat {\jmath }}}-\sin \theta {\boldsymbol {\hat {k}}}}{\displaystyle {\boldsymbol {\hat {\varphi }}}=-\sin \varphi {\boldsymbol {\hat {\imath }}}+\cos \varphi {\boldsymbol {\hat {\jmath }}}}
Sferik koordinatalar ortogonal hisoblanadi. Shu sababli ularning metrik tenzori diagonal koʻrinishda boʻladi:
{\displaystyle g_{ij}={\begin{pmatrix}1&0&0\\0&r^{2}&0\\0&0&r^{2}\sin ^{2}\theta \end{pmatrix}},\quad g^{ij}={\begin{pmatrix}1&0&0\\0&{\dfrac {1}{r^{2}}}&0\\0&0&{\dfrac {1}{r^{2}\sin ^{2}\theta }}\end{pmatrix}}}

  • {\displaystyle \det(g_{ij})=r^{4}\sin ^{2}\theta .\ }

  • Yoy uzunligi differensialining kvadrati:

{\displaystyle ds^{2}=dr^{2}+r^{2}\,d\theta ^{2}+r^{2}\sin ^{2}\theta \,d\varphi ^{2}.}

{\displaystyle H_{r}=1,\quad H_{\theta }=r,\quad H_{\varphi }=r\sin \theta .}

  • Kristoffel belgilari {\displaystyle \{r,\;\theta ,\;\varphi \}} :

{\displaystyle \Gamma _{22}^{1}=-r,\quad \Gamma _{33}^{1}=-r\sin ^{2}\theta ,}
{\displaystyle \Gamma _{21}^{2}=\Gamma _{12}^{2}=\Gamma _{13}^{3}=\Gamma _{31}^{3}={\frac {1}{r}},}
{\displaystyle \Gamma _{33}^{2}=-\cos \theta \sin \theta ,\quad \Gamma _{23}^{3}=\Gamma _{32}^{3}=\mathrm {ctg} \,\theta .}

Sferik koordinatalar sistemasida birlik vektorlar
Sferik koordinatalar sistemasida masofa
Fazodagi vaziyati sferik koordinatalar sistemasida berilgan ikki nuqtaning joylashuvi quyidagicha boʻlsin:
{\displaystyle {\begin{aligned}{\mathbf {r} }&=(r,\theta ,\varphi ),\\{\mathbf {r} '}&=(r',\theta ',\varphi ')\end{aligned}}}
U holda ushbu nuqtalar orasidagi masofani quyidagi formula orqali hisoblash mumkin:
{\displaystyle {\begin{aligned}{\mathbf {D} }&={\sqrt {r^{2}+r'^{2}-2rr'(\sin {\theta }\sin {\theta '}\cos {(\varphi -\varphi ')}+\cos {\theta }\cos {\theta '})}}\end{aligned}}}


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