Sistem koordinat bola: Perbedaan antara revisi
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Baris 49:
== Dalam Koordinat Kartesius ==
Koordinat bola dari suatu titik dalam konvensi ISO anda bisa melihat catatan dibawah ini, yaitu (khususnya untuk fisika):
* {{math|''r''}} adalah jari-jari.
* {{math|''a''}} adalah kemiringan.
Baris 60:
\theta &= \arccos\frac{z}{\sqrt{x^2 + y^2 + z^2}} = \arccos\frac{z}{r}=\arctan\frac{\sqrt{x^2+y^2}}{z}.
\end{align}</math>
Garis singgung iversi dilambangkan dengan nilai {{math|''φ'' {{=}} arctan {{sfrac|''y''|''x''}}}} harus didefinisikan dengan tepat cara mempertimbangkan kuadran yang benar dari nilai {{math|(''x'', ''y'')}}.
Sebaliknya, koordinat kartesius dapat diambil dari koordinat bola yaitu lihat catatan dibawah ini:
* {{mvar|r}} ''jari jari''.
* {{mvar|θ}} ''inklinasi''.
* {{mvar|φ}} ''azimut''.
darimana {{math|''r'' ∈ {{tutup-buka|0, ∞}}}}, {{math|''θ'' ∈ {{tutup-tutup|0, π}}}}, {{math|''φ'' ∈ {{tutup-buka|0, 2π}}}} adalah ...,oleh
: <math>\begin{align}
x &= r \sin\theta \, \cos\varphi, \\
y &= r \sin\theta \, \sin\varphi, \\
z &= r \cos\theta.
\end{align}</math>
== Sistem Koordinat Tabung ==
{{stub-matematika}}
{{Dalam perbaikan}}
{{Artikel|Sistem Koordinat Tabung}}
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: <math>\begin{align}
r &= \sqrt{\rho^2 + z^2}, \\
\theta &= \arctan\frac{\rho}{z} = \arccos\frac{z}{\sqrt{\rho^2 + z^2}}, \\
\varphi &= \varphi.
\end{align}</math>
----------------------------------
: <math>\begin{align}
\rho &= r \sin \theta, \\
\varphi &= \varphi, \\
z &= r \cos \theta.
\end{align}</math>
----------------------------------
== Koordinat bola yang dimodifikasi ==
Kemungkinan cara modifikasi pada elipsoid adalah dengan menggunakan versi koordinat bola yang dimodifikasi.
Misalkan P adalah ellipsoid yang ditentukan oleh nilai level
: <math>ax^2 + by^2 + cz^2 = d.</math>
Koordinat yang dimodifikasi oleh koordinat bola dari titik P saat konvensi ISO dapat diperoleh dari koordinat kartesius pada nilai {{math|(''x'', ''y'', ''z'')}} oleh karena itu rumusnya adalah:
: <math>\begin{align}
x &= \frac{1}{\sqrt{a}} r \sin\theta \, \cos\varphi, \\
y &= \frac{1}{\sqrt{b}} r \sin\theta \, \sin\varphi, \\
z &= \frac{1}{\sqrt{c}} r \cos\theta, \\
r &= ax^2 + by^2 + cz^2.
\end{align}</math>
Elemen volume yang sangat kecil diberikan oleh:
: <math>
\mathrm{d}V = \left|\frac{\partial(x, y, z)}{\partial(r, \theta, \varphi)}\right| =
\frac{1}{\sqrt{abc}} r^2 \sin \theta \,\mathrm{d}r \,\mathrm{d}\theta \,\mathrm{d}\varphi =
\frac{1}{\sqrt{abc}} r^2 \,\mathrm{d}r \,\mathrm{d}\Omega.
</math>
Faktor akar kuadrat yang berasal dari properti determinan yang memungkinkan sebuah konstanta ditarik oleh kolom:
: <math>
\begin{vmatrix}
ka & b & c \\
kd & e & f \\
kg & h & i
\end{vmatrix} =
k \begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{vmatrix}.
</math>
== Integrasi dan diferensiasi dalam koordinat bola ==
- Dalam pengembangan -
<!--[[File:Kugelkoord-lokale-Basis-s.svg|thumb|240px|right|Unit vectors in spherical coordinates]]
The following equations (Iyanaga 1977) assume that the colatitude {{mvar|θ}} is the inclination from the {{mvar|z}} (polar) axis (ambiguous since {{mvar|x}}, {{mvar|y}}, and {{mvar|z}} are mutually normal), as in the physics convention discussed.
The [[line element]] for an infinitesimal displacement from {{math|(''r'', ''θ'', ''φ'')}} to {{math|(''r'' + d''r'', ''θ'' + d''θ'', ''φ'' + d''φ'')}} is
: <math>
\mathrm{d}\mathbf{r} = \mathrm{d}r\,\hat{\mathbf r} + r\,\mathrm{d}\theta \,\hat{\boldsymbol\theta } + r \sin{\theta} \, \mathrm{d}\varphi\,\mathbf{\hat{\boldsymbol\varphi}},</math>
where
:<math>\begin{align}
\hat{\mathbf r} &= \sin \theta \cos \varphi \,\hat{\mathbf x} +
\sin \theta \sin \varphi \,\hat{\mathbf y} + \cos \theta \,\hat{\mathbf z}, \\
\hat{\boldsymbol\theta} &= \cos \theta \cos \varphi \,\hat{\mathbf x} +
\cos \theta \sin \varphi \,\hat{\mathbf y} - \sin \theta \,\hat{\mathbf z}, \\
\hat{\boldsymbol\varphi} &= -\sin \varphi \,\hat{\mathbf x} +
\cos \varphi \,\hat{\mathbf y}
\end{align}</math>
are the local orthogonal [[unit vectors]] in the directions of increasing {{mvar|r}}, {{mvar|θ}}, and {{mvar|φ}}, respectively,
and {{math|'''x̂'''}}, {{math|'''ŷ'''}}, and {{math|'''ẑ'''}} are the unit vectors in Cartesian coordinates. The linear transformation to this right-handed coordinate triplet is a [[rotation matrix]],
:<math>
R =\begin{pmatrix}
\sin\theta\cos\varphi&\sin\theta\sin\varphi& \cos\theta\\
\cos\theta\cos\varphi&\cos\theta\sin\varphi&-\sin\theta\\
-\sin\varphi&\cos\varphi &0
\end{pmatrix}.
</math>
The general form of the formula to prove the differential line element, is<ref name="q74503">{{cite web |title=Line element (dl) in spherical coordinates derivation/diagram |date=October 21, 2011 |work=[[Stack Exchange]] |url=https://math.stackexchange.com/q/74503 }}</ref>
: <math>
\mathrm{d}\mathbf{r} =
\sum_i \frac{\partial \mathbf{r}}{\partial x_i} \,\mathrm{d}x_i =
\sum_i \left|\frac{\partial \mathbf{r}}{\partial x_i}\right|
\frac{\frac{\partial \mathbf{r}}{\partial x_i}}{\left|\frac{\partial \mathbf{r}}{\partial x_i}\right|} \, \mathrm{d}x_i =
\sum_i \left|\frac{\partial \mathbf{r}}{\partial x_i}\right| \,\mathrm{d}x_i \hat{\boldsymbol{x}}_i,
</math>
that is, the change in <math>\mathbf r</math> is decomposed into individual changes corresponding to changes in the individual coordinates.
To apply this to the present case, one needs to calculate how <math>\mathbf r</math> changes with each of the coordinates. In the conventions used,
: <math>
\mathbf{r} = \begin{bmatrix}
r \sin\theta \, \cos\varphi \\
r \sin\theta \, \sin\varphi \\
r \cos\theta
\end{bmatrix}.
</math>
Thus,
: <math>
\frac{\partial\mathbf r}{\partial r} = \begin{bmatrix}
\sin\theta \, \cos\varphi \\
\sin\theta \, \sin\varphi \\
\cos\theta
\end{bmatrix}, \quad
\frac{\partial\mathbf r}{\partial \theta} = \begin{bmatrix}
r \cos\theta \, \cos\varphi \\
r \cos\theta \, \sin\varphi \\
-r \sin\theta
\end{bmatrix}, \quad
\frac{\partial\mathbf r}{\partial \varphi} = \begin{bmatrix}
-r \sin\theta \, \sin\varphi \\
r \sin\theta \, \cos\varphi \\
0
\end{bmatrix}.
</math>
The desired coefficients are the magnitudes of these vectors:<ref name="q74503" />
: <math>
\left|\frac{\partial\mathbf r}{\partial r}\right| = 1, \quad
\left|\frac{\partial\mathbf r}{\partial \theta}\right| = r, \quad
\left|\frac{\partial\mathbf r}{\partial \varphi}\right| = r \sin\theta.
</math>
The [[Surface integral|surface element]] spanning from {{mvar|θ}} to {{math|''θ'' + d''θ''}} and {{mvar|φ}} to {{math|''φ'' + d''φ''}} on a spherical surface at (constant) radius {{mvar|r}} is then
: <math>
\mathrm{d}S_r =
\left\|\frac{\partial r \hat{\mathbf r}}{\partial \theta} \times \frac{\partial r \hat{\mathbf r}}{\partial \varphi}\right\| \mathrm{d}\theta \,\mathrm{d}\varphi =
r^2 \sin\theta \,\mathrm{d}\theta \,\mathrm{d}\varphi ~.
</math>
Thus the differential [[solid angle]] is
: <math>\mathrm{d}\Omega = \frac{\mathrm{d}S_r}{r^2} = \sin\theta \,\mathrm{d}\theta \,\mathrm{d}\varphi.</math>
The surface element in a surface of polar angle {{mvar|θ}} constant (a cone with vertex the origin) is
: <math>\mathrm{d}S_\theta = r \sin\theta \,\mathrm{d}\varphi \,\mathrm{d}r.</math>
The surface element in a surface of azimuth {{mvar|φ}} constant (a vertical half-plane) is
: <math>\mathrm{d}S_\varphi = r \,\mathrm{d}r \,\mathrm{d}\theta.</math>
The [[volume element]] spanning from {{mvar|r}} to {{math|''r'' + d''r''}}, {{mvar|θ}} to {{math|''θ'' + d''θ''}}, and {{mvar|φ}} to {{math|''φ'' + d''φ''}} is specified by the [[determinant]] of the [[Jacobian matrix]] of [[partial derivative]]s,
:<math>
J =\frac{\partial(x,y,z)}{\partial(r,\theta,\varphi)}
=\begin{pmatrix}
\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{pmatrix},
</math>
namely
: <math>
\mathrm{d}V = \left|\frac{\partial(x, y, z)}{\partial(r, \theta, \varphi)}\right| =
r^2 \sin\theta \,\mathrm{d}r \,\mathrm{d}\theta \,\mathrm{d}\varphi =
r^2 \,\mathrm{d}r \,\mathrm{d}\Omega ~.
</math>
Thus, for example, a function {{math|''f''(''r'', ''θ'', ''φ'')}} can be integrated over every point in ℝ<sup>3</sup> by the [[Multiple integral#Spherical coordinates|triple integral]]
: <math>\int\limits_0^{2\pi} \int\limits_0^\pi \int\limits_0^\infty f(r, \theta, \varphi) r^2 \sin\theta \,\mathrm{d}r \,\mathrm{d}\theta \,\mathrm{d}\varphi ~.</math>
The [[del]] operator in this system leads to the following expressions for [[gradient]], [[divergence]], [[curl (mathematics)|curl]] and [[Laplacian]],
: <math>
\begin{align}
\nabla f = {} &{\partial f \over \partial r}\hat{\mathbf r}
+ {1 \over r}{\partial f \over \partial \theta}\hat{\boldsymbol\theta}
+ {1 \over r\sin\theta}{\partial f \over \partial \varphi}\hat{\boldsymbol\varphi}, \\[8pt]
\nabla\cdot \mathbf{A} = {} & \frac{1}{r^2}{\partial \over \partial r}\left( r^2 A_r \right) + \frac{1}{r \sin\theta}{\partial \over \partial\theta} \left( \sin\theta A_\theta \right) + \frac{1}{r \sin \theta} {\partial A_\varphi \over \partial \varphi}, \\[8pt]
\nabla \times \mathbf{A} = {} & \frac{1}{r\sin\theta}\left({\partial \over \partial \theta} \left( A_\varphi\sin\theta \right)
- {\partial A_\theta \over \partial \varphi}\right) \hat{\mathbf r} \\[8pt]
& {} + \frac 1 r \left({1 \over \sin\theta}{\partial A_r \over \partial \varphi}
- {\partial \over \partial r} \left( r A_\varphi \right) \right) \hat{\boldsymbol\theta} \\[8pt]
& {} + \frac 1 r \left({\partial \over \partial r} \left( r A_\theta \right)
- {\partial A_r \over \partial \theta}\right) \hat{\boldsymbol\varphi}, \\[8pt]
\nabla^2 f = {} & {1 \over r^2}{\partial \over \partial r} \left(r^2 {\partial f \over \partial r}\right) + {1 \over r^2 \sin\theta}{\partial \over \partial \theta} \left(\sin\theta {\partial f \over \partial \theta}\right)
+ {1 \over r^2 \sin^2\theta}{\partial^2 f \over \partial \varphi^2} \\[8pt]
= {} & \left(\frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r}\right)f + {1 \over r^2 \sin\theta}{\partial \over \partial \theta} \left(\sin\theta \frac{\partial}{\partial \theta}\right)f + \frac{1}{r^2 \sin^2\theta}\frac{\partial^2}{\partial \varphi^2}f ~.
\end{align}
</math>
Further, the inverse Jacobian in Cartesian coordinates is
:<math>
J^{-1}
=\begin{pmatrix}
\frac{x}{r}&\frac{y}{r}&\frac{z}{r}\\\\
\frac{xz}{r^2\sqrt{x^2+y^2}}&\frac{yz}{r^2\sqrt{x^2+y^2}}&\frac{-(x^2+y^2)}{r^2\sqrt{x^2+y^2}}\\\\
\frac{-y}{x^2+y^2}&\frac{x}{x^2+y^2}&0
\end{pmatrix}.
</math>
The [[metric tensor]] in the spherical coordinate system is <math>g=J^T J </math>.-->
== Jarak dalam Koordinat Bulat ==
- Dalam pengembangan -
== Kinematika ==
- Dalam pengembangan -
== Referensi ==
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