Sferik koordinatalar sistemasi — uch oʻlchamli koordinatalar sistemasi boʻlib, fazodagi nuqtaning vaziyati uchta kattalik bilan (r,θ,φ{\displaystyle r, \theta, \varphi}) bilan aniqlanadi. Bu yerda r{\displaystyle {\displaystyle r}} — koordinatalar boshigacha boʻlgan masofa, θ{\displaystyle {\displaystyle \theta }} va φ{\displaystyle {\displaystyle \varphi }}— mos holda zenit va azimutal burchaklar.

Zenit va azimut tushunchalari astronomiyada keng qoʻllanadi. Zenit — ixtiyoriy tanlangan nuqta (kuzatish nuqtasi) dan vertikal yuqoriga yoʻnalgan boʻlib, fundamental tekislikda yotadi. Astronomiyada fundamental tekislik sifatida ekvator yotgan tekislik yoki ekliptika tekisligi olinadi. Azimut — fundamental tekislikdagi ixtiyoriy tanlangan nur bilan boshlangʻich kuzatish nuqtasi orasidagi burchak.

1-rasm. Ixtiyoriy nuqtaning fazodagi vaziyatini uchta dekart koordinatalari yoki uchta sferik koordinatalar orqali aniqlash mumkin.

Boshqa koordinata sistemalariga oʻtish

Dekart koordinatalar sistemasi

Agar nuqtaning sferik koordinatalari (r,θ,φ){\displaystyle (r,\;\theta,\;\varphi)} berilgan boʻlsa, dekart koordinatalariga oʻtish uchun quyidagi formulalardan foydalaniladi:

{x=rsin⁡θcos⁡φ,y=rsin⁡θsin⁡φ,z=rcos⁡θ.{\displaystyle \begin{cases} x=r\sin\theta\cos\varphi, \\ y=r\sin\theta\sin\varphi, \\ z=r\cos\theta. \end{cases}}

Dekart koordinatalaridan sferik koordinatalarga oʻtish uchun esa:

{r=x2+y2+z2,θ=arccos⁡zx2+y2+z2=arctgx2+y2z,φ=arctgyx.{\displaystyle \begin{cases} r=\sqrt{x^2+y^2+z^2}, \\ \theta=\arccos\dfrac{z}{\sqrt{x^2+y^2+z^2}}=\mathrm{arctg}\dfrac{\sqrt{x^2+y^2}}{z}, \\ \varphi=\mathrm{arctg}\dfrac{y}{x}. \end{cases}}

Sferik koordinatalarga oʻtish yakobiani:

Failed to parse (Unknown environment alignat): {\displaystyle \begin{alignat}{2}

J & =\frac{\partial(x,y,z)}{\partial (r,\theta,\varphi)}=\begin{vmatrix} \sin \theta \cos \varphi & r \cos \theta \cos \varphi & -r\sin\theta\sin\varphi\\ \sin \theta \sin \varphi & r \cos \theta \sin \varphi & r \sin \theta \cos \varphi \\ \cos \theta & - r\sin\theta & 0 \end{vmatrix} = \\ & =\cos \theta (r^2 \cos \varphi^2 \cos \theta \sin \theta + r^2\sin^2\varphi \cos\theta\sin\theta) + r\sin\theta(r\sin^2\theta\cos^2\varphi + r\sin^2\theta\sin^2\varphi) = \\ & = r^2 \cos^2\theta \sin \theta + r^2 \sin^2 \theta \sin\theta =\\ &= r^2\sin\theta. \end{alignat}}

Shunday qilib, dekart koordinatalaridan sferik koordinatalarga oʻtishdagi hajm elementi quyidagi koʻrinishga ega boʻladi:

dV=dxdydz=J(r,θ,φ)drdθdφ=r2sin⁡θdrdθdφ{\displaystyle \mathrm{d}V = \mathrm{d}x\, \mathrm{d}y\, \mathrm{d}z = J(r,\theta,\varphi) \,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\varphi = r^2 \sin \theta \, \,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\varphi}

Silindrik koordinatalar sistemasi

Agar nuqtaning silindrik koordinatalari berilgan boʻlsa, sferik koordinatalarga oʻtish uchun quyidagi formulalardan foydalaniladi:

{ρ=rsin⁡θφ=φz=rcos⁡θ{\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:

{r=ρ2+z2,θ=arctgρz,φ=φ.{\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 :

J=r{\displaystyle J = r}

Sferik koordinatalar sistemasida differensiallash va integrallash

(r,θ,φ){\displaystyle (r,\theta,\varphi)} nuqtadan (r+dr,θ+dθ,φ+dφ){\displaystyle (r+\mathrm{d}r, \,\theta+\mathrm{d}\theta, \, \varphi+\mathrm{d}\varphi)} nuqtaga oʻtkazilgan vektor dr{\displaystyle \mathrm{d}\mathbf{r}} ning uzunligi quyidagiga teng:

dr=drr^+rdθθ^+rsin⁡θdφφ^,{\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

r^=sin⁡θcos⁡φı^+sin⁡θsin⁡φȷ^+cos⁡θk^{\displaystyle \boldsymbol{\hat r} =\sin \theta \cos \varphi \boldsymbol{\hat{\imath}} + \sin \theta \sin \varphi \boldsymbol{\hat{\jmath}} + \cos \theta \boldsymbol{\hat{k}}}

θ^=cos⁡θcos⁡φı^+cos⁡θsin⁡φȷ^−sin⁡θk^{\displaystyle \boldsymbol{\hat\theta } =\cos \theta \cos \varphi \boldsymbol{\hat{\imath}} + \cos \theta \sin \varphi \boldsymbol{\hat{\jmath}} -\sin \theta \boldsymbol{\hat{k}}}

φ^=−sin⁡φı^+cos⁡φȷ^{\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[1]:

gij=(1000r2000r2sin⁡2θ),gij=(10001r20001r2sin⁡2θ){\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}}

  • Failed to parse (trailing backslash): {\displaystyle \det(g_{ij})=r^4\sin^2\theta.\}
  • Yoy uzunligi differensialining kvadrati:
ds2=dr2+r2dθ2+r2sin⁡2θdφ2.{\displaystyle ds^2=dr^2+r^2\,d\theta^2+r^2\sin^2\theta\,d\varphi^2.}
Hr=1,Hθ=r,Hφ=rsin⁡θ.{\displaystyle H_r=1,\quad H_\theta=r,\quad H_\varphi=r\sin\theta.}
Γ221=−r,Γ331=−rsin⁡2θ,{\displaystyle \Gamma^1_{22}=-r,\quad \Gamma^1_{33}=-r\sin^2\theta,}
Γ212=Γ122=Γ133=Γ313=1r,{\displaystyle \Gamma^2_{21}=\Gamma^2_{12}=\Gamma^3_{13}=\Gamma^3_{31}=\frac{1}{r},}
Γ332=−cos⁡θsin⁡θ,Γ233=Γ323=ctgθ.{\displaystyle \Gamma^2_{33}=-\cos\theta\sin\theta,\quad \Gamma^3_{23}=\Gamma^3_{32}=\mathrm{ctg}\,\theta.}
Sferik koordinatalar sistemasida birlik vektorlar

Sferik koordinatalar sistemasida masofa

Fazodagi vaziyati sferik koordinatalar sistemasida berilgan ikki nuqtaning joylashuvi quyidagicha boʻlsin:

r=(r,θ,φ),r′=(r′,θ′,φ′){\displaystyle \begin{align}

{\mathbf r} &= (r,\theta,\varphi), \\
{\mathbf r'} &= (r',\theta',\varphi')

\end{align}} U holda ushbu nuqtalar orasidagi masofani quyidagi formula orqali hisoblash mumkin:

D=r2+r′2−2rr′(sin⁡θsin⁡θ′cos⁡(φ−φ′)+cos⁡θcos⁡θ′){\displaystyle \begin{align}

{\mathbf D} &= \sqrt{r^2+r'^2-2rr'(\sin{\theta}\sin{\theta'}\cos{(\varphi-\varphi')} + \cos{\theta}\cos{\theta'})}

\end{align}}

Harakat tenglamasi

Nuqtaning vaziyati sferik koordinatalarda quyidagi koʻrinishda berilgan boʻlsin:

r=rr^.{\displaystyle \mathbf{r} = r \mathbf{\hat r}.}

U holda uning tezligi:

v=r˙r^+rθ˙θ^+rφ˙sin⁡θφ^,{\displaystyle \mathbf{v} = \dot{r} \mathbf{\hat r} + r\,\dot\theta\,\hat{\boldsymbol\theta } + r\,\dot\varphi \sin\theta\,\mathbf{\hat{\boldsymbol\varphi}},}

hamda tezlanishi:

a=(r¨−rθ˙2−rφ˙2sin⁡2θ)r^+(rθ¨+2r˙θ˙−rφ˙2sin⁡θcos⁡θ)θ^+(rφ¨sin⁡θ+2r˙φ˙sin⁡θ+2rθ˙φ˙cos⁡θ)φ^.{\displaystyle \begin{align} \mathbf{a} = {} & \left( \ddot{r} - r\,\dot\theta^2 - r\,\dot\varphi^2\sin^2\theta \right)\mathbf{\hat r} \\ & {} + \left( r\,\ddot\theta + 2\dot{r}\,\dot\theta - r\,\dot\varphi^2\sin\theta\cos\theta \right) \hat{\boldsymbol\theta } \\ & {} + \left( r\ddot\varphi\,\sin\theta + 2\dot{r}\,\dot\varphi\,\sin\theta + 2 r\,\dot\theta\,\dot\varphi\,\cos\theta \right) \hat{\boldsymbol\varphi}. \end{align}}

ga teng boʻladi.

Burchak momenti:

L=mr×v=mr2(θ˙φ^−φ˙sin⁡θθ^).{\displaystyle \mathbf{L} = m \mathbf{r} \times \mathbf{v} = m r^2 (\dot\theta\,\hat{\boldsymbol\varphi } - \dot\varphi \sin\theta\,\mathbf{\hat{\boldsymbol\theta}} ).}

φ{\displaystyle \varphi} oʻzgarmas boʻlganda yoki θ=π2{\displaystyle \theta=\frac{\pi}{2}} boʻlganda, moddiy nuqtaning harakat tenglamasi qutb koordinatalar sistemasiga oʻtadi.

L=−iℏr×∇=iℏ(θ^sin⁡(θ)∂∂ϕ−ϕ^∂∂θ).{\displaystyle \mathbf{L }= -i\hbar ~\mathbf{r} \times \nabla =i \hbar \left(\frac{\hat{\boldsymbol{\theta}}}{\sin(\theta)} \frac{\partial}{\partial\phi} - \hat{\boldsymbol{\phi}} \frac{\partial}{\partial\theta}\right) .}

Yana qarang

Manbalar