Spherical coordinates

In spatial polar coordinates one point is indicated by its distance from the origin and by two angles. If the distance from the origin is constant (on one Sphere = kugeloberflaeche), one needs only the two angles, in order to designate one point clearly, and speaks then of spherical coordinates. The term Spherical coordinates can be regarded as generic term for these two cases.

For Polar coordinates in the level (a distance, an angle) and Cylindrical coordinates (two distances, an angle) see the article .

1 Usual convention
2 Other conventions
3 Transformation of differentials
4 Transformation of vector fields and operators

Usual convention

The illustration shows one point P with the cartesian coordinates (x, y, z) and the spherical coordinates (r?, ?):

(those xAxle shows those in 0° -, yAxle in 90°-Richtung, those zAxle upward).

The transformation equations of cartesian in spherical coordinates read

<math>{r}=\sqrt{x^2+y^2+z^2}</math>;

<math>{\varphi}=\begin{cases}\arccos\frac{x}{\sqrt{x^2+y^2}} & \mathrm{f\ddot ur} \ y\geq0, \\[, 5em ] 2\pi-\arccos\frac x{\sqrt{x^2+y^2}} & \mathrm{f\ddot ur} \ y < 0;\end{cases} </math>

<math>{\theta}=\arccot\frac z{\sqrt{x^2+y^2}}</math>.

In that and their applications become spherical coordinates (in the general. all angles) always in indicated; therefore stands in the definition by cases for? not 360°, but 2?.

In order to explain the descriptive meaning of the spherical coordinates verbal, is r the radius vector of P (thus the vector, which connects the origin O with P) and r_XY the projection of r into those x-yLevel. Then the spherical coordinates have the following meaning of P:

r (radius) the distance of the point P from the origin O is, thus the length of the vector r;

? (polarwinkel) is the angle between the positive zAxle and r, counted from 0 to? (0° to 180°), and

? (azimuth angles) is the angle between the positive xAxle and r_XY, counted from 0 to 2? (0° to 360°) against the clockwise direction.

The inverse transform takes place after the equations

<math>x = r \sin \theta \cos \varphi</math>,

<math>y = r \sin \theta \sin \varphi</math> and

<math>z = r \cos \theta \quad</math>.

In spherical coordinates becomes r by 1 replaces and not as independent coordinate specified.

Other conventions

The above coordinate choice is international consent in theoretical physics. Sometimes do the indications become? and? straight in the reverse sense related, in particular in American literature. One should always pay attention therefore to it, an author follows which conventions.

The polarwinkel? is not those geographical latitude. This is rather as angle between that and the radius vector defines and takes values between -90° and 90°. Does it become also? designates, then is? = 90°? ?, ? = 90°? ?. However can one? easily with that geographical length east of Greenwich equate. See in addition the article geographical coordinates.

Furthermore the above construction is in certain respects inconsistent for the setting up of the even Polarkoordianten. For some problems it is more practical, the representation

<math>x = r \cos \theta \cos \varphi</math>,

<math>y = r \cos \theta \sin \varphi</math> and

<math>z = r \sin \theta \quad</math>.

to use.

Transformation of differentials

Jacobi matrix

The local characteristics of the koordinatentransformation become by those Jacobi matrix described. For the transformation of spherical coordinates into cartesian coordinates this reads

<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> with spherical polar coordinates (only?, ?) the first column is omitted.

The Jacobi matrix of the opposite transformation is defined only for spatial, not for spherical polar coordinates; one computes it most simply as Inverse one of J:

<math>

J^{-1} = \frac{\partial(r, \theta, \varphi)}{\partial(x, y, z)}

 = \begin{pmatrix}
    \sin\theta\cos\varphi & \sin\theta\sin\varphi & \cos\theta \ \
    \frac{1}{r}\cos\theta\cos\varphi & \frac{1}{r}\cos\theta\sin\varphi & -\frac{1}{r}\sin\theta \ \
    -\frac{1}{r}\frac{\sin\varphi}{\sin\theta} & \frac{1}{r}\frac{\cos\varphi}{\sin\theta} & 0
  \end{pmatrix}.

</math> Some components of this matrix are breaks, at their denominators one the Uneindeutigkeit of polar coordinates r= 0 and with sin?=0 (thus?=0 or?) recognizes. The representation is more uncommon in cartesian coordinates:

<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>

Differentials, volume element, linear element

The Jacobi matrix makes it possible to write the conversion of differentials clearly as linear illustration:

<math>(dx, dy, dz)^T=J\cdot(dr, d\theta, d\varphi)^T</math>

and/or

<math>(dr, d\theta, d\varphi)^T=J^{-1}\cdot(dx, dy, dz)^T</math>.

The volume element <math>\mathrm{d}V=\mathrm{d}x\cdot\mathrm{d}y\cdot\mathrm{d}z</math> leaves itself particularly simple with the help of that Functional determinant

<math>|J|=r^2\sin\theta</math>

convert:

<math>\mathrm{d}V=r^2 \sin \theta \, \mathrm{d}\varphi \, \mathrm{d}\theta \, \mathrm{d}r </math>.

By deviation DV/dr one receives there for a two dimensional element on a sphere with radius r

<math>\mathrm{d}A=r^2 \sin \theta \, \mathrm{d}\varphi \, \mathrm{d}\theta </math>.

A linear element Ds one counts in accordance with

<math>\mathrm{d}s^2=\mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2

                   = \mathrm{d}r^2 + r^2\mathrm{d}\theta^2 +
                     r^2\siñ2\theta\mathrm{d}\varphi^2</math>

over.

Metrik and rotation matrix

In the absence of mixed members in the linear element Ds it is reflected again that that

<math>g=J^T J=\begin{pmatrix}1&0&0 \ \ 0&r^2&0 \ \ 0&0&r^2\siñ2\theta\end{pmatrix}</math>

also in spherical coordinates no Ausserdiagonalelemente has.

The metric tensor is obviously the square that Diagonal matrix

<math>h=diag(1, r, r\sin\theta)</math>.

With the help of this matrix the Jacobi matrix leaves itself as J=Sh writes, how S those

<math>

S = \begin{pmatrix}

    \sin\theta\cos\varphi&\cos\theta\cos\varphi& \sin\varphi \ \
    \sin\theta\sin\varphi&\cos\theta\sin\varphi&\cos\varphi \ \
    \cos\theta&-\sin\theta&0
  \end{pmatrix}

</math> is.

Transformation of vector fields and operators

In the following the transformation is to be deduced by vectors and differential operators exemplary. The results are written preferentially in compact form using transformation stencils.

Transformation of the vector space basis

That Basis vector e_? to the coordinate? indicates, in which direction itself one point P(r?, ?) moves, if the coordinate? around an infinitesimal amount D? one changes:

<math>\mathbf{e}_\varphi \sim \frac{\partial \mathrm{P}}{\partial\varphi}</math>.

From this one receives

<math>\mathbf{e}_\varphi \sim \frac{\partial \mathrm{P}}{\partial\varphi}

   = \frac{\partial x}{\partial \varphi}\frac{\partial \mathrm{P}}{x}
    +\frac{\partial y}{\partial \varphi}\frac{\partial \mathrm{P}}{y}
    +\frac{\partial z}{\partial \varphi}\frac{\partial \mathrm{P}}{z}
   = - r\sin\theta\sin\varphi\mathbf{e}_x
    +r\sin\theta\cos\varphi\mathbf{e}_y</math>.

In order to receive an orthonormal basis, must e_? still on the length 1 to be standardized:

<math>\mathbf{e}_\varphi = \sin\varphi\mathbf{e}_x + \cos\varphi\mathbf{e}_y</math>.

In similar way one receives the basis vectors e_r and e_?. In order to write the following transformations in compact form, we use the rotation matrix introduced above S. This matrix is orthogonal, i.e., S^-1=S^T. One can communicate the standardized basis vectors of the spherical coordinate system in summary in such a way then:

<math>(\mathbf{e}_r, \mathbf{e}_\theta, \mathbf{e}_\varphi)^T

    = S^T \cdot (\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z)^T</math>.

Accordingly the transformation reads into the opposite direction

<math>(\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z)^T

    = S\cdot (\mathbf{e}_r, \mathbf{e}_\theta, \mathbf{e}_\varphi)^T</math>.

Transformation of a vector field

Vector, as a geometrical object, must be independent of the coordinate system:

<math>A_x\mathbf{e}_x + A_y\mathbf{e}_y + A_z\mathbf{e}_z = \mathbf{A}

     = A_r\mathbf{e}_r + A_\theta\mathbf{e}_\theta + A_\varphi\mathbf{e}_\varphi.</math>

This condition fulfilled through

<math>(A_r, A_\theta, A_\varphi)^T

    = S^T \cdot (A_x, A_y, A_z)^T</math>

and/or

<math>(A_x, A_y, A_z)^T

    = S \cdot (A_r, A_\theta, A_\varphi)^T</math>.

Transformation of the partial derivatives

The partial derivatives transform like the basis vectors, but without standardisation. One can to be counted exactly as above, only one leaves the point P in the counter away (that becomes actual in the modern formulation the unit vectors of the Tangential area and equated) and the Jacobi matrix uses the partial derivatives J=Sh in place of the rotation matrix S. The transformation reads thus:

<math>

(\frac{\partial}{\partial r}, \frac{\partial}{\partial\theta}, \frac{\partial}{\partial\varphi})^T
 = J^T \cdot 
(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})^T
</math>,

and into the opposite direction

<math>

(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})^T
 = {\left(J^{-1}\right)}^T \cdot 
(\frac{\partial}{\partial r}, \frac{\partial}{\partial\theta}, \frac{\partial}{\partial\varphi})^T
</math>.

Transformation of the Del

That ∇ the simple form has only in cartesian coordinates

<math>\mathbf{\nabla}

 = \frac{\partial}{\partial x}\mathbf{e}_x
  +\frac{\partial}{\partial y}\mathbf{e}_y
  +\frac{\partial}{\partial z}\mathbf{e}_z</math>.

Both one must transform the partial derivatives and the unit vectors in the way deduced above. One finds:

<math>\mathbf{\nabla}

 = \frac{\partial}{\partial r}\mathbf{e}_r 
  + \frac{1}{r}\frac{\partial}{\partial\theta}\mathbf{e}_\theta
  + \frac{1}{r\sin\theta}\frac{\partial}{\partial\varphi}\mathbf{e}_\varphi</math>.

In this form the transformed Del can be used directly, around that a scalar field given in spherical coordinates.

Around those a vector field given in spherical coordinates A to compute, it is to be considered however that ∇ not only on the coefficients A_r, ... works, but also on in A implicitly contained basis vectors e_r, ... According to longer calculation one finds

<math>\mathbf{\nabla}\mathbf{A} = \frac{1}{r^2}\frac{\partial}{\partial r}r^2 A_r

 + \frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\sin\theta A_\theta
 +\frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}A_\phi</math>.

Transformation of the Laplace operator

If one in the divergence formula as vector field A the gradient operator ∇ , finds one uses that

<math>\mathbf{\Delta}=\mathbf{\nabla}^2 =

   \frac{1}{r^2}\frac{\partial}{\partial r}r^2 \frac{\partial}{\partial r}
 + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\sin\theta \frac{\partial}{\partial\theta}
 +\frac{1}{r^2\siñ2\theta}\frac{\partial^2}{\partial\phi^2} </math>.

Spherical coordinates

Table of contents

Usual convention

Other conventions

Transformation of differentials

Jacobi matrix

Differentials, volume element, linear element

Metrik and rotation matrix

Transformation of vector fields and operators

Transformation of the vector space basis

Transformation of a vector field

Transformation of the partial derivatives

Transformation of the Del

Transformation of the Laplace operator

> German to English > de.wikipedia.org (Machine translated into English)

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