Vector Analysis

$$ \vec{r} = \left( \begin{array}{c} r \cos { \varphi } \\ r \sin { \varphi } \\ z \end{array}\right) \tag{4}$$
$$\begin{eqnarray} x \left( {r,\varphi ,z} \right) &=& r \cos { \varphi } \\ y \left( {r,\varphi ,z} \right) &=& r \sin { \varphi } \\ z \left( {r,\varphi ,z} \right) &=& z\end{eqnarray}$$
$$\begin{eqnarray}\, dx &=& \frac{\, \partial x}{\, \partial r}\, dr + \frac{\, \partial x}{\, \partial \varphi }\, d\varphi + \frac{\, \partial x}{\, \partial z}\, dz \\ \, dy &=& \frac{\, \partial y}{\, \partial r}\, dr + \frac{\, \partial y}{\, \partial \varphi }\, d\varphi + \frac{\, \partial y}{\, \partial z}\, dz \\ \, dz &=& \frac{\, \partial z}{\, \partial r}\, dr + \frac{\, \partial z}{\, \partial \varphi }\, d\varphi + \frac{\, \partial z}{\, \partial z}\, dz\end{eqnarray}$$
$$\begin{eqnarray}\, dx &=& \cos { \varphi } \, dr - r \sin { \varphi } \, d\varphi \\ \, dy &=& \sin { \varphi } \, dr + r \cos { \varphi } \, d\varphi \\ \, dz &=&\, dz\end{eqnarray}$$
Das begleitende Dreibein in Zylinderkoordinaten ergibt sich wie folgt:

$$\begin{eqnarray} { \vec{e} }_{ r} &=& \frac{ \frac{ \partial \vec{r}}{\, \partial r}}{ \left| { \frac{ \partial \vec{r}}{\, \partial r}} \right|} \\ { \vec{e} }_{ \varphi } &=& \frac{ \frac{ \partial \vec{r}}{\, \partial \varphi }}{ \left| { \frac{ \partial \vec{r}}{\, \partial \varphi }} \right|} \\ { \vec{e} }_{ z} &=& \frac{ \frac{ \partial \vec{r}}{\, \partial z}}{ \left| { \frac{ \partial \vec{r}}{\, \partial z}} \right|}\end{eqnarray}$$

$$\begin{eqnarray} \frac{ \partial \vec{r}}{\, \partial r} &=& \left( \begin{array}{c} \cos { \varphi } \\ \sin { \varphi } \\ 0 \end{array}\right) \\ \frac{ \partial \vec{r}}{\, \partial \varphi } &=& \left( \begin{array}{c} - r \sin { \varphi } \\ r \cos { \varphi } \\ 0 \end{array}\right) \\ \frac{ \partial \vec{r}}{\, \partial z} &=& \left( \begin{array}{c} 0\\ 0\\ 1 \end{array}\right)\end{eqnarray}$$Wir erhalten daraus:

$$\begin{eqnarray} { \vec{e} }_{ r} &=& \left( \begin{array}{c} \cos { \varphi } \\ \sin { \varphi } \\ 0 \end{array}\right) \\ { \vec{e} }_{ \varphi } &=& \left( \begin{array}{c} - \sin { \varphi } \\ \cos { \varphi } \\ 0 \end{array}\right) \\ { \vec{e} }_{ z} &=& \left( \begin{array}{c} 0\\ 0\\ 1 \end{array}\right)\end{eqnarray}$$
$$\begin{eqnarray} \frac{ \partial \vec{r}}{\, \partial r} &=& { \vec{e} }_{ r} \\ \frac{ \partial \vec{r}}{\, \partial \varphi } &=& r { \vec{e} }_{ \varphi } \\ \frac{ \partial \vec{r}}{\, \partial z} &=& { \vec{e} }_{ z}\end{eqnarray}$$Wir erhalten daraus:

Wir können durch Bildung von Skalarprodukten leicht zeigen, dass diese Einheitsvektoren senkrecht aufeinander stehen.

Der Ortsvektor in Zylinderkoordinaten lautet

$$ \vec{r} = r { \vec{e} }_{ r} + z { \vec{e} }_{ z}$$

Der Ortsvektor kann sich nur solcher Einheitsvektoren des begleitenden Dreibeins bedienen, die nicht von sich selbst abhängen. Da $ { \vec{e} }_{ \varphi }$ von $ \varphi $ abhängt, lassen wir ihn weg. Jeder Punkt im Raum ist mit dem obigen Ortsvektor erreichbar. Der Ortsvektor ergibt sich auch durch Vergleich von Eq. 4 mit den berechneten Einheitsvektoren in Zylinderkoordinaten.

Geschwindigkeit und Beschleunigung in Zylinderkoordinaten

Bei der Berechnung der Geschwindigkeit ist die Zeitabhängigkeit der Einheitsvektoren zu berücksichtigen. $ { \vec{e} }_{ z}$ ist konstant, nicht so aber $ { \vec{e} }_{ r}$.

$$ \frac{d { \vec{r}} }{\, dt} = \frac{d { r} }{\, dt} { \vec{e} }_{ r} + r \frac{d\left( { { \vec{e} }_{ r}} \right)}{\, dt} + \frac{d { z} }{\, dt} { \vec{e} }_{ z}$$

$$\begin{eqnarray} \frac{ { \, \vec{de} }_{ r}}{\, dt} &=& \frac{ { \, \vec{de} }_{ r}}{\, dr} \frac{\, dr}{\, dt} + \frac{ { \, \vec{de} }_{ r}}{\, d\varphi } \frac{\, d\varphi }{\, dt} + \frac{ { \, \vec{de} }_{ r}}{\, dz} \frac{\, dz}{\, dt} \\ \frac{ { \, \vec{de} }_{ \varphi }}{\, dt} &=& \frac{ { \, \vec{de} }_{ \varphi }}{\, dr} \frac{\, dr}{\, dt} + \frac{ { \, \vec{de} }_{ \varphi }}{\, d\varphi } \frac{\, d\varphi }{\, dt} + \frac{ { \, \vec{de} }_{ \varphi }}{\, dz} \frac{\, dz}{\, dt} \\ \frac{ { \, \vec{de} }_{ z}}{\, dt} &=& \frac{ { \, \vec{de} }_{ z}}{\, dr} \frac{\, dr}{\, dt} + \frac{ { \, \vec{de} }_{ z}}{\, d\varphi } \frac{\, d\varphi }{\, dt} + \frac{ { \, \vec{de} }_{ z}}{\, dz} \frac{\, dz}{\, dt}\end{eqnarray}$$

$$\begin{eqnarray} \frac{ { \, \vec{de} }_{ r}}{\, dt} &=& \left( \begin{array}{c} - \sin { \varphi } \\ \cos { \varphi } \\ 0 \end{array}\right) \frac{\, d\varphi }{\, dt} \\ \frac{ { \, \vec{de} }_{ \varphi }}{\, dt} &=& \left( \begin{array}{c} - \cos { \varphi } \\ - \sin { \varphi } \\ 0 \end{array}\right) \frac{\, d\varphi }{\, dt} \\ \frac{ { \, \vec{de} }_{ z}}{\, dt} &=& 0\end{eqnarray}$$

Gradient in Zylinderkoordinaten

Es sei ein Skalarfeld $ \Phi $ gegeben. Das totale Differential lautet allgemein

$$\, d\Phi = \left( { \nabla \Phi } \right) \cdot \, \vec{dr} \tag{5}$$
und in Zylinderkoordinaten

$$\, d\Phi = \frac{\, \partial \Phi }{\, \partial r}\, dr + \frac{\, \partial \Phi }{\, \partial \varphi }\, d\varphi + \frac{\, \partial \Phi }{\, \partial z}\, dz \tag{6}$$

Für $ \nabla \Phi $ schreiben wir mit noch unbekannten Koeefizienzen $ { G }_{ i}$

$$\fbox{$ \displaystyle \nabla \Phi = { G }_{ r} { \vec{e} }_{ r} + { G }_{ \varphi } { \vec{e} }_{ \varphi } + { G }_{ z} { \vec{e} }_{ z} $} \tag{7}$$

Wir können $\, \vec{dr}$ allgemein wie folgt darstellen:

$$\, \vec{dr} = \frac{\, \vec{\partial r}}{\, \partial r}\, dr + \frac{\, \vec{\partial r}}{\, \partial \varphi }\, d\varphi + \frac{\, \vec{\partial r}}{\, \partial z}\, dz$$
Wegen

$$ \vec{r} = \left( \begin{array}{c} r \cos { \varphi } \\ r \sin { \varphi } \\ z \end{array}\right)$$

$$\begin{eqnarray} \frac{\, \vec{dr}}{\, dr} &=& \left( \begin{array}{c} \cos { \varphi } \\ \sin { \varphi } \\ 0 \end{array}\right) \\ \frac{\, \vec{dr}}{\, d\varphi } &=& \left( \begin{array}{c} - r \sin { \varphi } \\ r \cos { \varphi } \\ 0 \end{array}\right) \\ \frac{\, \vec{dr}}{\, dz} &=& \left( \begin{array}{c} 0\\ 0\\ 1 \end{array}\right)\end{eqnarray}$$wird daraus

$$\, \vec{dr} = \left( \begin{array}{c} \cos { \varphi } \\ \sin { \varphi } \\ 0 \end{array}\right)\, dr + \left( \begin{array}{c} - r \sin { \varphi } \\ r \cos { \varphi } \\ 0 \end{array}\right)\, d\varphi + \left( \begin{array}{c} 0\\ 0\\ 1 \end{array}\right)\, dz$$

$$\fbox{$ \displaystyle \, \vec{dr} = { \vec{e} }_{ r}\, dr + r { \vec{e} }_{ \varphi }\, d\varphi + { \vec{e} }_{ z}\, dz $}$$

Wenn wir die beiden eingerahmten Gleichungen in Eq. 5 einsetzen, erhalten wir

$$\begin{eqnarray}\, d\Phi &=& \left( \begin{array}{c} { G }_{ r}\\ { G }_{ \varphi }\\ { G }_{ z} \end{array}\right) \cdot \left( \begin{array}{c} \, dr\\ r\, d\varphi \\\, dz \end{array}\right) \\ \, d\Phi &=& { G }_{ r}\, dr + { G }_{ \varphi } r\, d\varphi + { G }_{ z}\, dz\end{eqnarray}$$
und dann durch Koeffizientenvergleich mit Eq. 6
$$\, d\Phi = \frac{\, \partial \Phi }{\, \partial r}\, dr + \frac{\, \partial \Phi }{\, \partial \varphi }\, d\varphi + \frac{\, \partial \Phi }{\, \partial z}\, dz$$

$$\begin{eqnarray} { G }_{ r} &=& \frac{\, \partial \Phi }{\, \partial r} \\ { G }_{ \varphi } &=& \frac{ 1}{ r} \frac{\, \partial \Phi }{\, \partial \varphi } \\ { G }_{ z} &=& \frac{\, \partial \Phi }{\, \partial z}\end{eqnarray}$$Da eingesetzt in Eq. 7 ergibt

$$\fbox{$ \displaystyle \nabla \Phi = \frac{\, \partial \Phi }{\, \partial r} { \vec{e} }_{ r} + \frac{ 1}{ r} \frac{\, \partial \Phi }{\, \partial \varphi } { \vec{e} }_{ \varphi } + \frac{\, \partial \Phi }{\, \partial z} { \vec{e} }_{ z} $} \tag{8}$$

Divergenz in Zylinderkoordinaten

$$ \mbox{div}\, { \vec{A}} = \frac{ 1}{ r} \left( { \frac{ \partial }{\, \partial r} \left( { { A }_{ r} r} \right) + \frac{ \partial }{\, \partial \varphi } { A }_{ \varphi } + \frac{ \partial }{\, \partial z} \left( { { A }_{ z} r} \right)} \right) \tag{2}$$

Laplace in Zylinderkoordinaten

Mitunter muss $ { \nabla }^{ 2} \Phi $ in Zylinderkoordinaten gebildet werden. Der Gradient ist ein Vektor-Feld.

$$ \nabla \Phi = \left( \begin{array}{c} \frac{\, \partial \Phi }{\, \partial r}\\ \frac{ 1}{ r} \frac{\, \partial \Phi }{\, \partial \varphi }\\ \frac{\, \partial \Phi }{\, \partial z} \end{array}\right) = \vec{A}$$

Wenn wir mit dem Nabla-Operator auf ein Vektorfeld gehen, bilden wir eine Divergenz. Damit können wir Eq. 2 anwenden. Wenn wir $ { A }_{ r}$, $ { A }_{ \varphi }$ und $ { A }_{ z}$ dieser Gelichung entnehmen und in MARK einsetzen, erhalten wir

$$ { \nabla }^{ 2} \Phi = \mbox{div}\, { \vec{A}} = \frac{ 1}{ r} \left( { \frac{ \partial }{\, \partial r} \left( { \frac{\, \partial \Phi }{\, \partial r} r} \right) + \frac{ \partial }{\, \partial \varphi } \left( { \frac{ 1}{ r} \frac{\, \partial \Phi }{\, \partial \varphi }} \right) + \frac{ \partial }{\, \partial z} \left( { \frac{\, \partial \Phi }{\, \partial z} r} \right)} \right)$$

$$\fbox{$ \displaystyle { \nabla }^{ 2} \Phi = \frac{ 1}{ r} \frac{ \partial }{\, \partial r} \left( { r \frac{\, \partial \Phi }{\, \partial r}} \right) + \frac{ 1}{ { r }^{ 2}} \frac{\, \partial \Phi }{ { \, \partial \varphi }^{ 2}} + \frac{{\partial }^{2} { \Phi } }{ { \, \partial z }^{ 2}} $}$$

Rotation in Zylinderkoordinaten

$$ \mbox{rot}\, { \vec{A}} = \frac{ 1}{ r} \left| { \begin{array}{ccc}
{ \vec{e} }_{ r} & r { \vec{e} }_{ \varphi } & { \vec{e} }_{ z}\\ \frac{ \partial }{\, \partial r} & \frac{ \partial }{\, \partial \varphi } & \frac{ \partial }{\, \partial z}\\ { A }_{ r} & r { A }_{ \varphi } & { A }_{ z}\\
\end{array}} \right| \tag{9}$$

$$ \mbox{rot}\, { \vec{A}} = \left( { \frac{ \partial { A }_{ z}}{\, \partial \varphi } - \frac{ \partial \left( { r { A }_{ \varphi }} \right)}{\, \partial z}} \right) { \vec{e} }_{ r} - \left( { \frac{ \partial { A }_{ z}}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial z}} \right) r { \vec{e} }_{ \varphi } + \left( { \frac{ \partial \left( { r { A }_{ \varphi }} \right)}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial \varphi }} \right) { \vec{e} }_{ z} \tag{3}$$
$$\fbox{$ \displaystyle \mbox{rot}\, { \vec{A}} = \left( { \frac{ 1}{ r} \frac{ \partial { A }_{ z}}{\, \partial \varphi } - \frac{ \partial { A }_{ \varphi }}{\, \partial z}} \right) { \vec{e} }_{ r} + \left( { \frac{ \partial { A }_{ r}}{\, \partial z} - \frac{ \partial { A }_{ z}}{\, \partial r}} \right) { \vec{e} }_{ \varphi } + \frac{ 1}{ r} \left( { \frac{ \partial \left( { r { A }_{ \varphi }} \right)}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial \varphi }} \right) { \vec{e} }_{ z} $}$$

Kugelkoordinaten

Zylinder-Koordinaten sind durch folgende Transformationsgleichungen definiert.

$$ \vec{r} = x { \vec{e} }_{ x} + y { \vec{e} }_{ y} + z { \vec{e} }_{ z}$$

$$ \vec{r} = \left( \begin{array}{c} r \cos { \alpha } \sin { \beta } \\ r \sin { \alpha } \sin { \beta } \\ r \cos { \beta } \end{array}\right) \tag{10}$$
$$\begin{eqnarray} x \left( {r,\alpha ,\beta } \right) &=& r \cos { \alpha } \sin { \beta } \\ y \left( {r,\alpha ,\beta } \right) &=& r \sin { \alpha } \sin { \beta } \\ z \left( {r,\alpha ,\beta } \right) &=& r \cos { \beta } \end{eqnarray}$$
$$\begin{eqnarray}\, dx &=& \frac{\, \partial x}{\, \partial r}\, dr + \frac{\, \partial x}{\, \partial \alpha }\, d\alpha + \frac{\, \partial x}{\, \partial \beta }\, d\beta \\ \, dy &=& \frac{\, \partial y}{\, \partial r}\, dr + \frac{\, \partial y}{\, \partial \alpha }\, d\alpha + \frac{\, \partial y}{\, \partial \beta }\, d\beta \\ \, dz &=& \frac{\, \partial z}{\, \partial r}\, dr + \frac{\, \partial z}{\, \partial \alpha }\, d\alpha + \frac{\, \partial z}{\, \partial \beta }\, d\beta \end{eqnarray}$$
$$\begin{eqnarray}\, dx &=& \cos { \alpha } \sin { \beta } \, dr - r \sin { \beta } \sin { \alpha } \, d\alpha + r \cos { \alpha } \cos { \beta } \\ \, dy &=& \sin { \alpha } \sin { \beta } \, dr + r \sin { \beta } \cos { \alpha } \, d\alpha + r \sin { \alpha } \cos { \beta } \\ \, dz &=& \cos { \beta } \, dr - r \sin { \beta } \, d\beta \end{eqnarray}$$

Das begleitende Dreibein in Zylinderkoordinaten ergibt sich wie folgt:

$$\begin{eqnarray} { \vec{e} }_{ r} &=& \frac{ \frac{ \partial \vec{r}}{\, \partial r}}{ \left| { \frac{ \partial \vec{r}}{\, \partial r}} \right|} \\ { \vec{e} }_{ \alpha } &=& \frac{ \frac{ \partial \vec{r}}{\, \partial \alpha }}{ \left| { \frac{ \partial \vec{r}}{\, \partial \alpha }} \right|} \\ { \vec{a} }_{ \beta } &=& \frac{ \frac{ \partial \vec{r}}{\, \partial \beta }}{ \left| { \frac{ \partial \vec{r}}{\, \partial \beta }} \right|}\end{eqnarray}$$
$$\begin{eqnarray} \frac{ \partial \vec{r}}{\, \partial r} &=& \left( \begin{array}{c} \cos { \alpha } \sin { \beta } \\ \sin { \alpha } \sin { \beta } \\ \cos { \beta } \end{array}\right) \\ \frac{ \partial \vec{r}}{\, \partial \alpha } &=& \left( \begin{array}{c} - r \sin { \alpha } \sin { \beta } \\ r \cos { \alpha } \sin { \beta } \\ 0 \end{array}\right) \\ \frac{ \partial \vec{r}}{\, \partial \beta } &=& \left( \begin{array}{c} r \cos { \alpha } \cos { \beta } \\ r \sin { \alpha } \cos { \beta } \\ - r \sin { \beta } \end{array}\right)\end{eqnarray}$$
Wir erhalten daraus:

$$\begin{eqnarray} { \vec{e} }_{ r} &=& \left( \begin{array}{c} \cos { \alpha } \sin { \beta } \\ \sin { \alpha } \sin { \beta } \\ \cos { \beta } \end{array}\right) \\ { \vec{e} }_{ \alpha } &=& \left( \begin{array}{c} - \sin { \alpha } \\ \cos { \alpha } \\ 0 \end{array}\right) \\ { \vec{e} }_{ \beta } &=& \left( \begin{array}{c} \cos { \alpha } \cos { \beta } \\ \sin { \alpha } \cos { \beta } \\ - \sin { \beta } \end{array}\right)\end{eqnarray}$$

$$\begin{eqnarray} \frac{ \partial \vec{r}}{\, \partial r} &=& { \vec{e} }_{ r} \\ \frac{ \partial \vec{r}}{\, \partial \alpha } &=& r \sin { \beta } { \vec{e} }_{ a} \\ \frac{ \partial \vec{r}}{\, \partial \beta } &=& r { \vec{e} }_{ \beta }\end{eqnarray}$$

Wir können durch Bildung von Skalarprodukten leicht zeigen, dass diese Einheitsvektoren senkrecht aufeinander stehen.

Der Ortsvektor in Kugelkoordinaten lautet

$$ \vec{r} = r { \vec{e} }_{ r}$$

Der Ortsvektor kann sich nur solcher Einheitsvektoren des begleitenden Dreibeins bedienen, die nicht von sich selbst abhängen. Da $ { \vec{e} }_{ a}$ von $ \alpha $ und $ { \vec{e} }_{ \beta }$ von $ \beta $ abhängen, lassen wir beide weg. Jeder Punkt im Raum ist mit dem obigen Ortsvektor erreichbar. Der Ortsvektor ergibt sich auch durch Vergleich von Eq. 10 mit den berechneten Einheitsvektoren in Kugelkoordinaten.

Inverse Transformationsgleichungen

Wir schreiben die Einheitsvektoren wie folgt auf

$$\begin{eqnarray} { \vec{e} }_{ r} &=& \cos { \alpha } \sin { \beta } { \vec{e} }_{ x} + \sin { \alpha } \sin { \beta } { \vec{e} }_{ y} + \cos { \beta } { \vec{e} }_{ z} \\ { \vec{e} }_{ \alpha } &=& - \sin { \alpha } { \vec{e} }_{ x} + \cos { \alpha } { \vec{e} }_{ y} + 0 { \vec{e} }_{ z} \\ { \vec{e} }_{ \beta } &=& \cos { \alpha } \cos { \beta } { \vec{e} }_{ x} + \sin { \alpha } \cos { \beta } { \vec{e} }_{ y} - \sin { \beta } { \vec{e} }_{ z}\end{eqnarray}$$
und lösen das Gleichungssystem mit Hilfe der Cramerschen Regel:

$$ { \vec{e} }_{ x} = \frac{ \left| { \begin{array}{ccc}
{ \vec{e} }_{ r} & \sin { \alpha } \sin { \beta } & \cos { \beta } \\ { \vec{e} }_{ \alpha } & \cos { \alpha } & 0\\ { \vec{e} }_{ \beta } & \sin { \alpha } \cos { \beta } & - \sin { \beta } \\
\end{array}} \right|}{ \left| { \begin{array}{ccc}
\cos { \alpha } \sin { \beta } & \sin { \alpha } \sin { \beta } & \cos { \beta } \\ - \sin { \alpha } & \cos { \alpha } & 0\\ \cos { \alpha } \cos { \beta } & \sin { \alpha } \cos { \beta } & - \sin { \beta } \\
\end{array}} \right|}$$

$$\begin{eqnarray} A &=& - \cos { \alpha } \sin { \beta } \cos { \alpha } \sin { \beta } - \sin { \alpha } \sin { \beta } \sin { \alpha } \sin { \beta } + \cos { \beta } \left( { - \sin { \alpha } \sin { \alpha } \cos { \beta } - \cos { \alpha } \cos { \alpha } \cos { \beta } } \right) \\ A &=& - { \cos { \alpha } }^{ 2} { \sin { \beta } }^{ 2} - { \sin { \alpha } }^{ 2} { \sin { \beta } }^{ 2} - \sin { \alpha } \sin { \alpha } \cos { \beta } \cos { \beta } - \cos { \alpha } \cos { \alpha } \cos { \beta } \cos { \beta } \\ A &=& - { \cos { \alpha } }^{ 2} { \sin { \beta } }^{ 2} - { \sin { \alpha } }^{ 2} { \sin { \beta } }^{ 2} - { \sin { \alpha } }^{ 2} { \cos { \beta } }^{ 2} - { \cos { \alpha } }^{ 2} { \cos { \beta } }^{ 2} \\ A &=& - { \sin { \beta } }^{ 2} \left( { { \cos { \alpha } }^{ 2} + { \sin { \alpha } }^{ 2}} \right) - { \cos { \beta } }^{ 2} \left( { { \sin { \alpha } }^{ 2} + { \cos { \alpha } }^{ 2}} \right) \\ A &=& - { \sin { \beta } }^{ 2} - { \cos { \beta } }^{ 2} \\ A &=& - 1\end{eqnarray}$$
$$\begin{eqnarray} { \vec{e} }_{ x} &=& \frac{ - \cos { \alpha } \sin { \beta } { \vec{e} }_{ r} + \sin { \alpha } { \sin { \beta } }^{ 2} { \vec{e} }_{ \alpha } + \cos { \beta } \left( { \sin { \alpha } \cos { \beta } { \vec{e} }_{ \alpha } - \cos { \alpha } { \vec{e} }_{ \beta }} \right)}{ - 1} \\ { \vec{e} }_{ x} &=& \frac{ - \cos { \alpha } \sin { \beta } { \vec{e} }_{ r} + \sin { \alpha } { \sin { \beta } }^{ 2} { \vec{e} }_{ \alpha } + \sin { \alpha } { \cos { \beta } }^{ 2} { \vec{e} }_{ \alpha } - \cos { \alpha } \cos { \beta } { \vec{e} }_{ \beta }}{ - 1} \\ { \vec{e} }_{ x} &=& \frac{ - \cos { \alpha } \sin { \beta } { \vec{e} }_{ r} + \sin { \alpha } { \vec{e} }_{ \alpha } - \cos { \alpha } \cos { \beta } { \vec{e} }_{ \beta }}{ - 1} \\ { \vec{e} }_{ x} &=& \cos { \alpha } \sin { \beta } { \vec{e} }_{ r} - \sin { \alpha } { \vec{e} }_{ \alpha } + \cos { \alpha } \cos { \beta } { \vec{e} }_{ \beta }\end{eqnarray}$$

$$\begin{eqnarray} { \vec{e} }_{ y} &=& \frac{ \left| { \begin{array}{ccc}
\cos { \alpha } \sin { \beta } & { \vec{e} }_{ r} & \cos { \beta } \\ - \sin { \alpha } & { \vec{e} }_{ a} & 0\\ \cos { \alpha } \cos { \beta } & { \vec{e} }_{ \beta } & - \sin { \beta } \\
\end{array}} \right|}{ \left| { \begin{array}{ccc}
\cos { \alpha } \sin { \beta } & \sin { \alpha } \sin { \beta } & \cos { \beta } \\ - \sin { \alpha } & \cos { \alpha } & 0\\ \cos { \alpha } \cos { \beta } & \sin { \alpha } \cos { \beta } & - \sin { \beta } \\
\end{array}} \right|} \\ { \vec{e} }_{ y} &=& \frac{ - \cos { \alpha } { \sin { \beta } }^{ 2} { \vec{e} }_{ \alpha } - \left( { \sin { \alpha } \sin { \beta } } \right) { \vec{e} }_{ r} + \cos { \beta } \left( { - \sin { \alpha } { \vec{e} }_{ \beta } - \cos { \alpha } \cos { \beta } { \vec{e} }_{ \alpha }} \right)}{ - 1} \\ { \vec{e} }_{ y} &=& \frac{ - \cos { \alpha } { \sin { \beta } }^{ 2} { \vec{e} }_{ \alpha } - \sin { \alpha } \sin { \beta } { \vec{e} }_{ r} - \sin { \alpha } \cos { \beta } { \vec{e} }_{ \beta } - \cos { \alpha } { \cos { \beta } }^{ 2} { \vec{e} }_{ \alpha }}{ - 1} \\ { \vec{e} }_{ y} &=& \frac{ - \cos { \alpha } { \sin { \beta } }^{ 2} { \vec{e} }_{ \alpha } - \sin { \alpha } \sin { \beta } { \vec{e} }_{ r} - \sin { \alpha } \cos { \beta } { \vec{e} }_{ \beta } - \cos { \alpha } { \cos { \beta } }^{ 2} { \vec{e} }_{ \alpha }}{ -1} \\ { \vec{e} }_{ y} &=& \cos { \alpha } { \sin { \beta } }^{ 2} { \vec{e} }_{ \alpha } + \sin { \alpha } \sin { \beta } { \vec{e} }_{ r} + \sin { \alpha } \cos { \beta } { \vec{e} }_{ \beta } + \cos { \alpha } { \cos { \beta } }^{ 2} { \vec{e} }_{ \alpha } \\ { \vec{e} }_{ y} &=& \cos { \alpha } { \vec{e} }_{ \alpha } + \sin { \alpha } \sin { \beta } { \vec{e} }_{ r} + \sin { \alpha } \cos { \beta } { \vec{e} }_{ \beta } \\ { \vec{e} }_{ y} &=& \sin { \alpha } \sin { \beta } { \vec{e} }_{ r} + \cos { \alpha } { \vec{e} }_{ \alpha } + \sin { \alpha } \cos { \beta } { \vec{e} }_{ \beta }\end{eqnarray}$$

$$ { \vec{e} }_{ z} = \frac{ \left| { \begin{array}{ccc}
\cos { \alpha } \sin { \beta } & \sin { \alpha } \sin { \beta } & { \vec{e} }_{ r}\\ - \sin { \alpha } & \cos { \alpha } & { \vec{e} }_{ \alpha }\\ \cos { \alpha } \cos { \beta } & \sin { \alpha } \cos { \beta } & { \vec{e} }_{ \beta }\\
\end{array}} \right|}{ \left| { \begin{array}{ccc}
\cos { \alpha } \sin { \beta } & \sin { \alpha } \sin { \beta } & \cos { \beta } \\ - \sin { \alpha } & \cos { \alpha } & 0\\ \cos { \alpha } \cos { \beta } & \sin { \alpha } \cos { \beta } & - \sin { \beta } \\
\end{array}} \right|}$$

$$\begin{eqnarray} { \vec{e} }_{ z} &=& \frac{ \cos { \alpha } \sin { \beta } \left( { \cos { \alpha } { \vec{e} }_{ \beta } - \sin { \alpha } \cos { \beta } { \vec{e} }_{ \alpha }} \right) - \sin { \alpha } \sin { \beta } \left( { - \sin { \alpha } { \vec{e} }_{ \beta } - \cos { \alpha } \cos { \beta } { \vec{e} }_{ \alpha }} \right) + \left( { - { \sin { \alpha } }^{ 2} \cos { \beta } - { \cos { \alpha } }^{ 2} \cos { \beta } } \right) { \vec{e} }_{ r}}{ - 1} \\ { \vec{e} }_{ z} &=& \frac{ { \cos { \alpha } }^{ 2} \sin { \beta } { \vec{e} }_{ \beta } - \sin { \alpha } \cos { \beta } \cos { \alpha } \sin { \beta } { \vec{e} }_{ \alpha } - \left( { - { \sin { \alpha } }^{ 2} \sin { \beta } { \vec{e} }_{ \beta } - \sin { \alpha } \sin { \beta } \cos { \alpha } \cos { \beta } { \vec{e} }_{ \alpha }} \right) - \left( { { \sin { \alpha } }^{ 2} \cos { \beta } + { \cos { \alpha } }^{ 2} \cos { \beta } } \right) { \vec{e} }_{ r}}{ - 1} \\ { \vec{e} }_{ z} &=& \frac{ { \cos { \alpha } }^{ 2} \sin { \beta } { \vec{e} }_{ \beta } - \sin { \alpha } \cos { \beta } \cos { \alpha } \sin { \beta } { \vec{e} }_{ \alpha } + { \sin { \alpha } }^{ 2} \sin { \beta } { \vec{e} }_{ \beta } + \sin { \alpha } \sin { \beta } \cos { \alpha } \cos { \beta } { \vec{e} }_{ \alpha } - \left( { { \sin { \alpha } }^{ 2} + { \cos { \alpha } }^{ 2}} \right) \cos { \beta } { \vec{e} }_{ r}}{ -1} \\ { \vec{e} }_{ z} &=& \frac{ \sin { \beta } { \vec{e} }_{ \beta } - \left( { { \sin { \alpha } }^{ 2} + { \cos { \alpha } }^{ 2}} \right) \cos { \beta } { \vec{e} }_{ r}}{ -1} \\ { \vec{e} }_{ z} &=& \frac{ \sin { \beta } { \vec{e} }_{ \beta } - \cos { \beta } { \vec{e} }_{ r}}{ -1} \\ { \vec{e} }_{ z} &=& \cos { \beta } { \vec{e} }_{ r} - \sin { \beta } { \vec{e} }_{ \beta }\end{eqnarray}$$
Zusammengefasst lauten die invsersen Transformationsgleichungen also

$$\begin{eqnarray} { \vec{e} }_{ x} &=& \cos { \alpha } \sin { \beta } { \vec{e} }_{ r} - \sin { \alpha } { \vec{e} }_{ \alpha } + \cos { \alpha } \cos { \beta } { \vec{e} }_{ \beta } \\ { \vec{e} }_{ y} &=& \sin { \alpha } \sin { \beta } { \vec{e} }_{ r} + \cos { \alpha } { \vec{e} }_{ \alpha } + \sin { \alpha } \cos { \beta } { \vec{e} }_{ \beta } \\ { \vec{e} }_{ z} &=& \cos { \beta } { \vec{e} }_{ r} - \sin { \beta } { \vec{e} }_{ \beta }\end{eqnarray}$$

Geschwindigkeit und Beschleunigung in Kugelkoordinaten

Für die Berechnung der Geschwindigkeit und Beschleunigung in Kugelkoordianten benötigen wir die zeitlichen Ableitungen der Einheitsvektoren.

$$ \frac{\, \vec{dr}}{\, dt} = \frac{\, dr}{\, dt} { \vec{e} }_{ r} + r \frac{ { \, \vec{de} }_{ r}}{\, dt}$$

$$\begin{eqnarray} \frac{ { \, \vec{de} }_{ r}}{\, dt} &=& \frac{ { \, \vec{de} }_{ r}}{\, dr} \frac{\, dr}{\, dt} + \frac{ { \, \vec{de} }_{ r}}{\, d\alpha } \frac{\, d\alpha }{\, dt} + \frac{ { \, \vec{de} }_{ r}}{\, d\beta } \frac{\, d\beta }{\, dt} \\ \frac{ { \, \vec{de} }_{ r}}{\, dt} &=& \left( \begin{array}{c} - \sin { \alpha } \sin { \beta } \\ \cos { \alpha } \sin { \beta } \\ 0 \end{array}\right) \frac{\, d\alpha }{\, dt} + \left( \begin{array}{c} \cos { \alpha } \cos { \beta } \\ \sin { \alpha } \cos { \beta } \\ - \sin { \beta } \end{array}\right) \frac{\, d\beta }{\, dt} \\ \frac{ { \, \vec{de} }_{ r}}{\, dt} &=& \sin { \beta } \frac{\, d\alpha }{\, dt} { \vec{e} }_{ \alpha } + \frac{\, d\beta }{\, dt} { \vec{e} }_{ \beta }\end{eqnarray}$$

$$\begin{eqnarray} \frac{ { \, \vec{de} }_{ \alpha }}{\, dt} &=& \frac{ { \, \vec{de} }_{ \alpha }}{\, dr} \frac{\, dr}{\, dt} + \frac{ { \, \vec{de} }_{ \alpha }}{\, d\alpha } \frac{\, d\alpha }{\, dt} + \frac{ { \, \vec{de} }_{ \alpha }}{\, d\beta } \frac{\, d\beta }{\, dt} \\ \frac{ { \, \vec{de} }_{ \alpha }}{\, dt} &=& \left( \begin{array}{c} - \cos { \alpha } \\ - \sin { \alpha } \\ 0 \end{array}\right) \frac{\, d\alpha }{\, dt} \\ \frac{ { \, \vec{de} }_{ \alpha }}{\, dt} &=& - \left( { \sin { \beta } { \vec{e} }_{ r} + \cos { \beta } { \vec{e} }_{ \beta }} \right) \frac{\, d\alpha }{\, dt} \\ \frac{ { \, \vec{de} }_{ \alpha }}{\, dt} &=& - \sin { \beta } \frac{\, d\alpha }{\, dt} { \vec{e} }_{ r} - \cos { \beta } \frac{\, d\alpha }{\, dt} { \vec{e} }_{ \beta }\end{eqnarray}$$

$$\begin{eqnarray} \frac{ { \, \vec{de} }_{ \beta }}{\, dt} &=& \frac{ { \, \vec{de} }_{ \beta }}{\, dr} \frac{\, dr}{\, dt} + \frac{ { \, \vec{de} }_{ \beta }}{\, d\alpha } \frac{\, d\alpha }{\, dt} + \frac{ { \, \vec{de} }_{ \beta }}{\, d\beta } \frac{\, d\beta }{\, dt} \\ \frac{ { \, \vec{de} }_{ \beta }}{\, dt} &=& \left( \begin{array}{c} - \sin { \alpha } \cos { \beta } \\ \cos { \alpha } \cos { \beta } \\ 0 \end{array}\right) \frac{\, d\alpha }{\, dt} + \left( \begin{array}{c} - \cos { \alpha } \sin { \beta } \\ - \sin { \alpha } \sin { \beta } \\ - \cos { \beta } \end{array}\right) \frac{\, d\beta }{\, dt} \\ \frac{ { \, \vec{de} }_{ \beta }}{\, dt} &=& \cos { \beta } \frac{\, d\alpha }{\, dt} { \vec{e} }_{ \alpha } - \frac{\, d\beta }{\, dt} { \vec{e} }_{ r} \\ \frac{ { \, \vec{de} }_{ \beta }}{\, dt} &=& - \frac{\, d\beta }{\, dt} { \vec{e} }_{ r} + \cos { \beta } \frac{\, d\alpha }{\, dt} { \vec{e} }_{ \alpha }\end{eqnarray}$$
Zusammengfasst lauten die Ableitungen der Einheitsvektoren in Kugelkoordinaten:

$$\begin{eqnarray} \frac{ { \, \vec{de} }_{ r}}{\, dt} &=& \sin { \beta } \frac{\, d\alpha }{\, dt} { \vec{e} }_{ \alpha } + \frac{\, d\beta }{\, dt} { \vec{e} }_{ \beta } \\ \frac{ { \, \vec{de} }_{ \alpha }}{\, dt} &=& - \sin { \beta } \frac{\, d\alpha }{\, dt} { \vec{e} }_{ r} - \cos { \beta } \frac{\, d\alpha }{\, dt} { \vec{e} }_{ \beta } \\ \frac{ { \, \vec{de} }_{ \beta }}{\, dt} &=& - \frac{\, d\beta }{\, dt} { \vec{e} }_{ r} + \cos { \beta } \frac{\, d\alpha }{\, dt} { \vec{e} }_{ \alpha }\end{eqnarray}$$

Damit lassen sich die Geschwindigkeit und Beschleuinigung in Kugelkoordinaten leicht angeben:

$$\begin{eqnarray} \frac{\, \vec{dr}}{\, dt} &=& \frac{\, dr}{\, dt} { \vec{e} }_{ r} + r \frac{ { \, \vec{de} }_{ r}}{\, dt} \\ \frac{\, \vec{dr}}{\, dt} &=& \frac{\, dr}{\, dt} { \vec{e} }_{ r} + r \left( { \sin { \beta } \frac{\, d\alpha }{\, dt} { \vec{e} }_{ \alpha } + \frac{\, d\beta }{\, dt} { \vec{e} }_{ \beta }} \right) \\ \frac{\, \vec{dr}}{\, dt} &=& \frac{\, dr}{\, dt} { \vec{e} }_{ r} + r \sin { \beta } \frac{\, d\alpha }{\, dt} { \vec{e} }_{ \alpha } + r \frac{\, d\beta }{\, dt} { \vec{e} }_{ \beta }\end{eqnarray}$$

Gradient in Kugelkoordinaten

Es sei ein Skalarfeld $ \Phi $ gegeben. Der Gradient in Kugelkoordinaten ergibt sich wie folgt. Das totale Differential lautet allgemein

$$\, d\Phi = \left( { \nabla \Phi } \right) \cdot \, \vec{dr}$$
und in Kugelkoordinaten

$$\, d\Phi = \frac{\, \partial \Phi }{\, \partial r}\, dr + \frac{\, \partial \Phi }{\, \partial \alpha }\, d\alpha + \frac{\, \partial \Phi }{\, \partial \beta }\, d\beta $$

Wir können $\, \vec{dr}$ wie folgt darstellen:

$$\begin{eqnarray}\, \vec{dr} &=& \frac{\, \vec{\partial r}}{\, \partial r}\, dr + \frac{\, \vec{\partial r}}{\, \partial \alpha }\, d\alpha + \frac{\, \vec{\partial r}}{\, \partial \beta }\, d\beta \\ \, \vec{dr} &=& { \vec{e} }_{ r}\, dr + r \sin { \beta } { \vec{e} }_{ \alpha }\, d\alpha + r { \vec{e} }_{ \beta }\, d\beta \end{eqnarray}$$
Durch Koeffizientenvergleich ergibt sich dann

$$ \nabla \Phi = { G }_{ r} { \vec{e} }_{ r} + { G }_{ \alpha } { \vec{e} }_{ \alpha } + { G }_{ \beta } { \vec{e} }_{ \beta }$$

$$ \frac{\, \partial \Phi }{\, \partial r}\, dr + \frac{\, \partial \Phi }{\, \partial \alpha }\, d\alpha + \frac{\, \partial \Phi }{\, \partial \beta }\, d\beta = { G }_{ r}\, dr + { G }_{ \alpha } r \sin { \beta } \, d\alpha + { G }_{ \beta } r\, d\beta $$

$$\begin{eqnarray} { G }_{ r} &=& \frac{\, \partial \Phi }{\, \partial r} \\ { G }_{ a} &=& \frac{ 1}{ r \sin { \beta } } \frac{\, \partial \Phi }{\, \partial \alpha } \\ { G }_{ \beta } &=& \frac{ 1}{ r} \frac{\, \partial \Phi }{\, \partial \beta }\end{eqnarray}$$
$$ \nabla \Phi = \frac{\, \partial \Phi }{\, \partial r} { \vec{e} }_{ r} + \frac{ 1}{ r \sin { \beta } } \frac{\, \partial \Phi }{\, \partial \alpha } { \vec{e} }_{ \alpha } + \frac{ 1}{ r} \frac{\, \partial \Phi }{\, \partial \beta } { \vec{e} }_{ \beta } \tag{1}$$

Divergenz in Kugelkoordinaten

Die Divergenz an einem gegebenen Punkt ist definiert durch

$$ \mbox{div}\, { \vec{A}} = \frac{ 1}{ \Delta V} \int \vec{A} \cdot \, \vec{dS}$$

Das inifinitesimale Volumenelement in Kugelkoordinaten ist gegeben durch

$$\, dV = { r }^{ 2} \sin { \beta } \, d\beta \, d\alpha \, dr$$

Der Fluss in Richtung $ { \vec{e} }_{ r}$ ist gegeben durch

$$\begin{eqnarray} { F }_{ r} &=& { A }_{ r} { r }^{ 2} \sin { \beta } \, d\beta \, d\alpha \\ { F }_{ r'} &=& { A }_{ r} { r }^{ 2} \sin { \beta } \, d\beta \, d\alpha + \frac{ \partial \left( { { A }_{ r} { r }^{ 2} \sin { \beta } \, d\beta \, d\alpha } \right)}{\, \partial r}\, dr\end{eqnarray}$$
$$ F = { F }_{ r'} - { F }_{ r} = \sin { \beta } \frac{\, \partial \left( { { A }_{ r} { r }^{ 2}} \right)}{\, \partial r}\, d\beta \, d\alpha \, dr$$

Der Fluss in Richtung $ { \vec{e} }_{ \beta }$ is gegeben durch

$$\begin{eqnarray} { F }_{ \beta } &=& { A }_{ \beta } r\, d\alpha \sin { \beta } \, dr \\ { F }_{ \beta '} &=& { A }_{ \beta } r\, d\alpha \sin { \beta } \, dr + \frac{ \partial \left( { { A }_{ \beta } r\, d\alpha \sin { \beta } \, dr} \right)}{\, \partial \beta }\, d\beta \end{eqnarray}$$
$$ F = { F }_{ \beta '} - { F }_{ \beta } = r \frac{ \partial \left( { { A }_{ \beta } \sin { \beta } } \right)}{\, \partial \beta }\, d\beta \, d\alpha \, dr$$

Der Fluss in Richtung $ { \vec{e} }_{ \alpha }$ ist gegeben durch

$$\begin{eqnarray} { F }_{ \alpha } &=& { A }_{ \alpha } r\, d\beta \, dr \\ { F }_{ \alpha '} &=& { A }_{ \alpha } r\, d\beta \, dr + \frac{ \partial \left( { { A }_{ \alpha } r\, d\beta \, dr} \right)}{\, \partial \alpha }\, d\alpha \end{eqnarray}$$
$$ F = { F }_{ \alpha '} - { F }_{ \alpha } = r \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha }\, d\beta \, d\alpha \, dr$$

Der Gesamtfluss ergibt sich dann zu

$$ { F }_{ ges} = \left( { \sin { \beta } \frac{\, \partial { A }_{ r} { r }^{ 2}}{\, \partial r} + r \frac{ \partial { A }_{ \beta } \sin { \beta } }{\, \partial \beta } + r \frac{ \partial { A }_{ \alpha }}{\, \partial \alpha }} \right)\, d\beta \, d\alpha \, dr$$
und damit die Divergenz zu

$$ \mbox{div}\, { \vec{A}} = \frac{ { F }_{ ges}}{ { r }^{ 2} \sin { \beta } \, d\beta \, d\alpha \, dr}$$

$$ \mbox{div}\, { \vec{A}} = \frac{ 1}{ { r }^{ 2}} \frac{\partial \left( { { A }_{ r} { r }^{ 2}} \right)}{\, \partial r} + \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \beta } \sin { \beta } } \right)}{\, \partial \beta } + \frac{ 1}{ r \sin { \beta } } \frac{\partial \left( { { A }_{ \alpha }} \right)}{\, \partial \alpha } \tag{11}$$

Rotation in Kugelkoordinaten

Ihre geometrische Definition führt die Rotation auf ein Linienintegral zurück.

$$ \vec{n} \cdot \mbox{rot}\, { \vec{A}} = \frac{ \int \vec{A} \cdot \, \vec{dl}}{ \Delta F}$$

Krummlinige Koordinatensysteme

Jeder Ortsvektor läßt sich in beliebigen Koordinaten $ { q }_{ i}$ (z.B. $ { q }_{ 1} = r$, $ { q }_{ 2} = \alpha $, $ { q }_{ 3} = \beta $) mit Hilfe von im Allgemeinen vom Ortspunkt abhängigen Einheitsvektoren $ { \vec{e} }_{ i}$ wie folgt darstellen:

$$ \vec{r} = \sum { a }_{ i} { \vec{e} }_{ i}$$

Die Einheitsvektoren an einem gegebenen Ortspunkt ergeben sich aus

$$ { \vec{e} }_{ i} = \frac{ \frac{ \partial \vec{r}}{ { \, \partial q }_{ i}}}{ \left| { \frac{ \partial \vec{r}}{ { \, \partial q }_{ i}}} \right|}$$
oder

$$ \frac{ \partial \vec{r}}{ { \, \partial q }_{ i}} = { h }_{ i} { \vec{e} }_{ i}$$

mit

$$ { h }_{ i} = \left| { \frac{ \partial \vec{r}}{ { \, \partial q }_{ i}}} \right|$$

Die $ { h }_{ i}$ nennt man Skalenfaktoren. Die Einheitsvektoren $ { \vec{e} }_{ i}$ zeigen in die Richtung von wachsendem $ { q }_{ i}$ entlang der $ { q }_{ i}$-Koordinatenlinie.

Bogenlänge

Die Länge eines Bogens in krummlinigen Koordinaten ergibt sich mit

$$\, \vec{dr} = \sum \frac{ \partial \vec{r}}{ { \, \partial q }_{ i}} { \, dq }_{ i} = \sum { h }_{ i} { \, dq }_{ i} { \vec{e} }_{ i}$$

und

$$ { \, ds }^{ 2} =\, \vec{dr} \cdot \, \vec{dr}$$
wir folgt:

$$ { \, ds }^{ 2} = \sum_{ i = 1}^{ N} { h }_{ i} { \, dq }_{ i} { \vec{e} }_{ i} \cdot \sum_{ j = 1}^{ N} { h }_{ j} { \, dq }_{ j} { \vec{e} }_{ j}$$

Für Koordinatensysteme mit orthogonalen Einheitsvektoren ($ { \delta }_{ ij} = 0$ für $ i \ne j$) führt dies zu ...

Gradient

$$ \nabla = { \vec{e} }_{ q1} \left( { \frac{ 1}{ { h }_{ 1}} \frac{ \partial }{ { \, \partial q }_{ 1}}} \right) + { \vec{e} }_{ q2} \left( { \frac{ 1}{ { h }_{ 2}} \frac{ \partial }{ { \, \partial q }_{ 2}}} \right) + { \vec{e} }_{ q3} \left( { \frac{ 1}{ { h }_{ 3}} \frac{ \partial }{ { \, \partial q }_{ 3}}} \right)$$
Beispiel: Zylinderkordinaten

$$ \nabla = { \vec{e} }_{ r} \left( { \frac{ 1}{ { h }_{ r}} \frac{ \partial }{\, \partial r}} \right) + { \vec{e} }_{ \varphi } \left( { \frac{ 1}{ { h }_{ \varphi }} \frac{ \partial }{\, \partial \varphi }} \right) + { \vec{e} }_{ z} \left( { \frac{ 1}{ { h }_{ z}} \frac{ \partial }{\, \partial z}} \right)$$

$$ \nabla = { \vec{e} }_{ r} \frac{ \partial }{\, \partial r} + { \vec{e} }_{ \varphi } \left( { \frac{ 1}{ r} \frac{ \partial }{\, \partial \varphi }} \right) + { \vec{e} }_{ z} \frac{ \partial }{\, \partial z}$$

Divergenz

$$ \mbox{div}\, { \vec{A}} = \frac{ 1}{ { h }_{ 1} { h }_{ 2} { h }_{ 3}} \left( { \frac{ \partial }{ { \, \partial q }_{ 1}} \left( { { A }_{ 1} { h }_{ 2} { h }_{ 3}} \right) + \frac{ \partial }{ { \, \partial q }_{ 2}} \left( { { A }_{ 2} { h }_{ 1} { h }_{ 3}} \right) + \frac{ \partial }{ { \, \partial q }_{ 3}} \left( { { A }_{ 3} { h }_{ 1} { h }_{ 2}} \right)} \right)$$

Beispiel: Zylinderkoordinaten

$$\begin{eqnarray} \mbox{div}\, { \vec{A}} &=& \frac{ 1}{ { h }_{ r} { h }_{ \varphi } { h }_{ z}} \left( { \frac{ \partial }{\, \partial r} \left( { { A }_{ r} { h }_{ \varphi } { h }_{ z}} \right) + \frac{ \partial }{\, \partial \varphi } \left( { { A }_{ \varphi } { h }_{ r} { h }_{ z}} \right) + \frac{ \partial }{\, \partial z} \left( { { A }_{ z} { h }_{ r} { h }_{ \varphi }} \right)} \right) \\ \mbox{div}\, { \vec{A}} &=& \frac{ 1}{ r} \left( { \frac{ \partial }{\, \partial r} \left( { { A }_{ r} r} \right) + \frac{ \partial }{\, \partial \varphi } { A }_{ \varphi } + \frac{ \partial }{\, \partial z} \left( { { A }_{ z} r} \right)} \right)\end{eqnarray}$$

Rotation

$$ \mbox{rot}\, { \vec{A}} = \frac{ 1}{ { h }_{ 1} { h }_{ 2} { h }_{ 3}} \left| { \begin{array}{ccc}
{ h }_{ 1} { \vec{e} }_{ q1} & { h }_{ 2} { \vec{e} }_{ 2} & { h }_{ 3} { \vec{e} }_{ 3}\\ \frac{ \partial }{ { \, \partial q }_{ 1}} & \frac{ \partial }{ { \, \partial q }_{ 2}} & \frac{ \partial }{ { \, \partial q }_{ 3}}\\ { A }_{ 1} { h }_{ 1} & { A }_{ 2} { h }_{ 2} & { A }_{ 3} { h }_{ 3}\\
\end{array}} \right|$$

Beispiel: Zylinderkoordinaten

Beispiel: Kugelkoordinaten

$$ \frac{ \partial \vec{r}}{\, \partial r} = \left( \begin{array}{c} \cos { \alpha } \sin { \beta } \\ \sin { \alpha } \sin { \beta } \\ \cos { \beta } \end{array}\right)$$

$$ { h }_{ 1} = \left| { \frac{ \partial \vec{r}}{\, \partial r}} \right| = { \cos { \alpha } }^{ 2} { \sin { \beta } }^{ 2} + { \sin { \alpha } }^{ 2} { \sin { \beta } }^{ 2} + { \cos { \beta } }^{ 2} = 1$$

$$ \frac{ \partial \vec{r}}{\, \partial \alpha } = \left( \begin{array}{c} - r \sin { \alpha } \sin { \beta } \\ r \cos { \alpha } \sin { \beta } \\ 0 \end{array}\right)$$

$$ { h }_{ 2} = \left| { \frac{ \partial \vec{r}}{\, \partial \alpha }} \right| = r \sin { \beta } $$

$$ \frac{ \partial \vec{r}}{\, \partial \beta } = \left( \begin{array}{c} r \cos { \alpha } \cos { \beta } \\ r \sin { \alpha } \cos { \beta } \\ - r \sin { \beta } \end{array}\right)$$

$$ { h }_{ 3} = \left| { \frac{ \partial \vec{r}}{\, \partial \beta }} \right| = \sqrt{ { r }^{ 2} { \cos { \beta } }^{ 2} + { r }^{ 2} { \sin { \beta } }^{ 2}} = r$$

$$ \mbox{rot}\, { \vec{A}} = \frac{ 1}{ { r }^{ 2} \sin { \beta } } \left| { \begin{array}{ccc}
{ \vec{e} }_{ r} & r \sin { \beta } { \vec{e} }_{ \alpha } & r { \vec{e} }_{ \beta }\\ \frac{ \partial }{\, \partial r} & \frac{ \partial }{\, \partial \alpha } & \frac{ \partial }{\, \partial \beta }\\ { A }_{ r} & r \sin { \beta } { A }_{ \alpha } & r { A }_{ \beta }\\
\end{array}} \right|$$

$$\begin{eqnarray} \mbox{rot}\, { \vec{A}} &=& \frac{ 1}{ { r }^{ 2} \sin { \beta } } \left( { { \vec{e} }_{ r} \left( { \frac{ \partial \left( { r { A }_{ \beta }} \right)}{\, \partial \alpha } - \frac{ \partial \left( { r \sin { \beta } { A }_{ \alpha }} \right)}{\, \partial \beta }} \right) - r \sin { \beta } { \vec{e} }_{ \alpha } \left( { \frac{ \partial \left( { r { A }_{ \beta }} \right)}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial \beta }} \right) + r { \vec{e} }_{ \beta } \left( { \frac{ \partial \left( { r \sin { \beta } { A }_{ \alpha }} \right)}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial \alpha }} \right)} \right) \\ \mbox{rot}\, { \vec{A}} &=& { \vec{e} }_{ r} \frac{ 1}{ { r }^{ 2} \sin { \beta } } \left( { \frac{ \partial \left( { r { A }_{ \beta }} \right)}{\, \partial \alpha } - \frac{ \partial \left( { r \sin { \beta } { A }_{ \alpha }} \right)}{\, \partial \beta }} \right) - \frac{ 1}{ r} { \vec{e} }_{ \alpha } \left( { \frac{ \partial \left( { r { A }_{ \beta }} \right)}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial \beta }} \right) + \frac{ 1}{ r \sin { \beta } } { \vec{e} }_{ \beta } \left( { \frac{ \partial \left( { r \sin { \beta } { A }_{ \alpha }} \right)}{\, \partial r} - \frac{ \partial { A }_{ r}}{\, \partial \alpha }} \right) \\ \mbox{rot}\, { \vec{A}} &=& \frac{ 1}{ r \sin { \beta } } \left( { \frac{ \partial { A }_{ \beta }}{\, \partial \alpha } - \frac{ \partial \sin { \beta } { A }_{ \alpha }}{\, \partial \beta }} \right) { \vec{e} }_{ r} + \frac{ 1}{ r} \left( { \frac{ \partial { A }_{ r}}{\, \partial \beta } - \frac{ \partial r { A }_{ \beta }}{\, \partial r}} \right) { \vec{e} }_{ \alpha } + \frac{ 1}{ r} \left( { \frac{ \partial r { A }_{ \alpha }}{\, \partial r} - \frac{ 1}{ \sin { \beta } } \frac{ \partial { A }_{ r}}{\, \partial \alpha }} \right) { \vec{e} }_{ \beta }\end{eqnarray}$$

Vektoridentitäten

Quelle: http://wwwex.physik.uni-ulm.de/lehre/krm-2008-2009/node67.html

$$ \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}$$

$$\begin{eqnarray} \vec{a} \times \left( { \vec{b} \times \vec{c}} \right) &=& \left( { \vec{a} \cdot \vec{c}} \right) \vec{b} - \left( { \vec{a} \cdot \vec{b}} \right) \vec{c} \\ \vec{a} \times \left( { \vec{b} \times \vec{c}} \right) &=& \left( { \vec{a} \cdot \vec{c}} \right) \vec{b} - \left( { \vec{a} \cdot \vec{b}} \right) \vec{c}\end{eqnarray}$$
$$ \left( { \vec{a} \times \vec{b}} \right) \cdot \vec{c} = \left( { \vec{b} \times \vec{c}} \right) \cdot \vec{a}$$

$$\begin{eqnarray} \mbox{div}\, \left( { \vec{B} \times \vec{C}} \right) &=& \vec{C} \cdot \mbox{rot}\, { \vec{B}} - \vec{B} \cdot \mbox{rot}\, { \vec{C}} \\ \mbox{div}\, \left( { \vec{A} \cdot \vec{B}} \right) &=& \frac{ \vec{A} \cdot \vec{B}}{ \left| { \vec{A}} \right|} \\ \mbox{rot}\, \left( { \varphi \vec{a}} \right) &=& \mbox{grad}\, { \varphi } \times \vec{a} + \varphi \mbox{rot}\, { \vec{a}} \\ \mbox{div}\, \left( { \varphi \vec{a}} \right) &=& \mbox{grad}\, { \varphi } \vec{a} + \varphi \mbox{div}\, { \vec{a}} \end{eqnarray}$$
$$ \mbox{rot}\, \left( { \vec{a} \times \vec{b}} \right) = \left( { \vec{b} \cdot \nabla } \right) \vec{a} - \left( { \vec{a} \cdot \nabla } \right) \vec{b} + a \mbox{div}\, { \vec{b}} - b \mbox{div}\, { \vec{a}} $$

$$\begin{eqnarray} \mbox{rot}\, \left( { \mbox{rot}\, { \vec{a}} } \right) &=& \mbox{grad}\, \left( { \mbox{div}\, { \vec{a}} } \right) - \mbox{div}\, \left( { \mbox{grad}\, { \vec{a}} } \right) \\ \nabla \times \left( { \nabla \times \vec{B}} \right) &=& \nabla \left( { \nabla \cdot \vec{B}} \right) - \vec{B} \\ \nabla \times \left( { \nabla \times \vec{A}} \right) &=& \nabla \left( { \nabla \cdot \vec{A}} \right) - { \nabla }^{ 2} \vec{A}\end{eqnarray}$$