Betrachtet man Funktionen, die von mehreren Ortskoordinaten abhängen, so kann man sie nach jeder dieser Ortskoordinaten ableiten und das ggfs. auch mehrfach.
Einige Linearkombinationen solcher Ableitungen werden besonders häufig verwendet, z. B der Gradient, die Divergenz, der Laplace-Operator oder der Drehimpulsoperator.
Zusammenfassend bezeichnet man diese als Differentialoperatoren.
Man kann Ortskoordinaten in verschiedenen Koordinatensystemen angeben. Häufig verwendet werden kartesische Koordinaten, Zylinderkoordinaten und Kugelkoordinaten.
Betrachtet man dieselbe Funktion dargestellt in unterschiedlichen Koordinatensystemen, so sieht die Funktionsgleichung meist sehr unterschiedlich aus, je nachdem, in welchem Koordinatensystem man sie darstellt.
Genauso sehen die Differentialoperatoren in unterschiedlichen Koordinatensystem unterschiedlich aus. Im Folgenden geben wir an, wie einige Differentialoperatoren in verschiedenen Koordinatensystemen aussehen und anschließend rechnen wir die angegebenen Formeln nach.
Siehe Ing Mathematik: Differenziation im Rn und Partielle Ableitung .
Übungen:
Sei
f
(
x
,
y
)
=
x
2
+
y
3
{\displaystyle f(x,y)=x^{2}+y^{3}}
. Berechne
∂
f
∂
x
{\displaystyle {\frac {\partial f}{\partial x}}}
und
∂
f
∂
y
{\displaystyle {\frac {\partial f}{\partial y}}}
Sei
f
(
x
,
y
)
=
cos
(
x
3
y
4
)
{\displaystyle f(x,y)=\cos(x^{3}y^{4})}
. Berechne
∂
f
∂
x
{\displaystyle {\frac {\partial f}{\partial x}}}
,
∂
f
∂
y
{\displaystyle {\frac {\partial f}{\partial y}}}
und
∂
2
f
∂
x
2
{\displaystyle {\frac {\partial ^{2}f}{\partial x^{2}}}}
Satz von Schwarz:
Hermann Amandus Schwarz (1843-1921), deutscher Mathematiker
Der Satz von Schwarz besagt, dass bei mehrfach stetig differenzierbaren Funktionen mehrerer Variablen die Reihenfolge, in der die partiellen Differentiationen (Ableitungen) nach den einzelnen Variablen durchgeführt werden, nicht entscheidend für das Ergebnis ist (von Satz von Schwarz zitiert). D.h.
∂
2
f
∂
x
∂
y
=
∂
2
f
∂
y
∂
x
{\displaystyle {\frac {\partial ^{2}f}{\partial x\partial y}}={\frac {\partial ^{2}f}{\partial y\partial x}}}
. Ist die Funktion nicht stetig, so muss das nicht gelten (siehe z.B. Satz_von_Schwarz#Gegenbeispiel )!
Übungen:
Sei
f
(
x
,
y
)
=
x
3
sin
y
{\displaystyle f(x,y)=x^{3}\sin y}
. Berechne
∂
2
f
∂
x
∂
y
{\displaystyle {\frac {\partial ^{2}f}{\partial x\partial y}}}
und
∂
2
f
∂
y
∂
x
{\displaystyle {\frac {\partial ^{2}f}{\partial y\partial x}}}
.
Sei
f
(
x
,
y
)
=
log
|
x
|
y
{\displaystyle f(x,y)={\frac {\log |x|}{y}}}
. Berechne
∂
2
f
∂
x
∂
y
{\displaystyle {\frac {\partial ^{2}f}{\partial x\partial y}}}
und
∂
2
f
∂
y
∂
x
{\displaystyle {\frac {\partial ^{2}f}{\partial y\partial x}}}
.
In kartesischen Koordinaten ist der Gradient folgendermaßen definiert:
grad
f
=
(
∂
f
∂
x
∂
f
∂
y
∂
f
∂
z
)
{\displaystyle \mathbf {\text{grad}} \ f=\left({\begin{matrix}{\frac {\partial f}{\partial x}}\\{\frac {\partial f}{\partial y}}\\{\frac {\partial f}{\partial z}}\end{matrix}}\right)}
Übung : Sei
f
(
x
,
y
,
z
)
=
x
2
y
+
sin
z
{\displaystyle f(x,y,z)=x^{2}y+\sin z}
. Berechne
grad
f
{\displaystyle \mathbf {\text{grad}} \ f}
.
Das totale Differenzial lautet:
d
f
=
∂
f
∂
x
d
x
+
∂
f
∂
y
d
y
+
∂
f
∂
z
d
z
=
grad
f
d
r
{\displaystyle \mathrm {d} f={\frac {\partial f}{\partial x}}\mathrm {d} x+{\frac {\partial f}{\partial y}}\mathrm {d} y+{\frac {\partial f}{\partial z}}\mathrm {d} z=\mathbf {\text{grad}} \ f\;\mathrm {d} \mathbf {r} \quad }
mit
d
r
=
(
d
x
d
y
d
z
)
{\displaystyle \quad \mathrm {d} \mathbf {r} =\left({\begin{matrix}\mathrm {d} x\\\mathrm {d} y\\\mathrm {d} z\end{matrix}}\right)}
Der Nabla-Operator sei:
∇
=
(
∂
∂
x
∂
∂
y
∂
∂
z
)
{\displaystyle \nabla =\left({\begin{matrix}{\frac {\partial }{\partial x}}\\{\frac {\partial }{\partial y}}\\{\frac {\partial }{\partial z}}\end{matrix}}\right)}
Da der Nabla-Operator ein symbolischer Vektor ist, wird er manchmal auch fett oder mit einem Pfeil geschrieben:
∇
,
∇
→
{\displaystyle {\boldsymbol {\nabla }},\;{\vec {\nabla }}}
. Eingeführt wurde er von Hamilton.
Sir William Rowan Hamilton (1805 - 1865) irischer Mathematiker und Physiker
Gradient und Nabla-Operator hängen somit folgendermaßen zusammen
grad
f
=
∇
f
{\displaystyle \mathbf {\text{grad}} \ f=\nabla f}
Pierre-Simon Laplace (1749-1827) französischer Mathematiker und Physiker
Der Laplace-Operator sei
Δ
=
∇
∇
=
∇
2
=
∂
2
∂
x
2
+
∂
2
∂
y
2
+
∂
2
∂
z
2
{\displaystyle \Delta =\nabla \nabla =\nabla ^{2}={\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}}
Wir betrachten eine Funktion
f
:
R
3
→
R
3
{\displaystyle \mathbf {f} \colon \,\mathbb {R} ^{3}\to \mathbb {R} ^{3}}
. In Komponentenschreibweise ist
f
{\displaystyle \mathbf {f} }
gegeben durch:
f
=
(
f
x
f
y
f
z
)
{\displaystyle \mathbf {f} =\left({\begin{matrix}f_{\mathrm {x} }\\f_{\mathrm {y} }\\f_{\mathrm {z} }\end{matrix}}\right)}
Es ist zu beachten, dass wir
f
{\displaystyle \mathbf {f} }
fett (bold) gesetzt haben, wobei wir
f
x
{\displaystyle f_{\mathrm {x} }}
normal (regular) gesetzt haben. Hierdurch drückt man in der Regel aus, dass
f
{\displaystyle \mathbf {f} }
in einen mehrdimensionalen Vektorraum abbildet, bzw, dass
f
{\displaystyle \mathbf {f} }
mehr als eine Komponente hat oder, anders ausgedrückt,
f
{\displaystyle \mathbf {f} }
eine vektorwertige Funktion ist. Hingegen wird
f
x
{\displaystyle f_{\mathrm {x} }}
normal gesetzt, weil es nur genau eine Komponente hat. Man beachte auch, dass hier mit
f
x
{\displaystyle f_{\mathrm {x} }}
nicht die partielle Ableitung gemeint ist. Diese wird auch oft als
∂
f
d
x
=
f
x
{\displaystyle {\frac {\partial f}{\mathrm {d} x}}=f_{x}}
geschrieben. Eine andere Möglichkeit, die Vektorwertigkeit einer Funktion auszudrücken, ist ein Pfeil über dem Funktionssymbol:
f
→
{\displaystyle {\vec {f}}}
Eine weitere Möglichkeit der Komponentenschreibweise ist:
f
=
(
f
1
f
2
f
3
)
{\displaystyle \mathbf {f} =\left({\begin{matrix}f_{1}\\f_{2}\\f_{3}\end{matrix}}\right)}
Abkürzend hierfür schreibt man auch:
f
i
{\displaystyle f_{i}}
Hierbei haben wir alle drei Komponenten der Funktion
f
{\displaystyle \mathbf {f} }
zusammenfassend durch das Symbol
f
i
{\displaystyle f_{i}}
ausgedrückt. Hierbei fällt auf, dass der Index
i
{\displaystyle i}
im Symbol
f
i
{\displaystyle f_{i}}
kursiv gesetzt wurde. Hingegen wurde der Index
z
{\displaystyle \mathrm {z} }
im Symbol
f
z
{\displaystyle f_{\mathbf {z} }}
normal gesetzt. Auch dies ist eine Konvention. Die Vektorwertigkeit von
f
i
{\displaystyle f_{i}}
wird durch die Kursivschrift des
i
{\displaystyle i}
ausgedrückt.
Die Divergenz in kartesischen Koordinaten ist definiert durch:
d
i
v
f
=
∇
⋅
f
=
∂
f
x
∂
x
+
∂
f
y
∂
y
+
∂
f
z
∂
z
{\displaystyle \mathrm {div} \mathbf {f} =\nabla \cdot \mathbf {f} ={\frac {\partial f_{\mathrm {x} }}{\partial \mathrm {x} }}+{\frac {\partial f_{\mathrm {y} }}{\partial \mathrm {y} }}+{\frac {\partial f_{\mathrm {z} }}{\partial \mathrm {z} }}}
Eine alternative Schreibweise ist:
∑
i
=
1
3
∂
f
i
∂
x
i
{\displaystyle \sum _{i=1}^{3}{\frac {\partial f_{i}}{\partial \mathrm {x} _{i}}}}
Diese kann man noch weiter verkürzen zu:
∂
i
f
i
{\displaystyle \partial _{i}f_{i}}
Übung : Sei
f
=
(
x
y
z
)
{\displaystyle \mathbf {f} =\left({\begin{matrix}x\\y\\z\end{matrix}}\right)}
. Berechne
d
i
v
f
{\displaystyle \mathrm {div} \mathbf {f} }
.
Umrechnung von Zylinderkoordinaten in kartesische Koordinaten [ Bearbeiten ]
Die Zylinderkoordinaten werden durch folgende Gleichungen definiert:
x
=
ρ
cos
(
ϕ
)
y
=
ρ
sin
(
ϕ
)
z
=
z
{\displaystyle {\begin{matrix}x&=&\rho \cos(\phi )\\y&=&\rho \sin(\phi )\\z&=&z\end{matrix}}}
Umrechnung von kartesischen Koordinaten in Zylinderkoordinaten [ Bearbeiten ]
Aus den Definitionsgleichungen erhält man:
ρ
=
x
2
+
y
2
=
(
x
2
+
y
2
)
1
2
ϕ
=
a
r
c
t
a
n
(
y
x
)
{\displaystyle {\begin{matrix}\rho ={\sqrt {x^{2}+y^{2}}}&=&\left(x^{2}+y^{2}\right)^{\frac {1}{2}}\\\phi &=&\mathrm {arctan} ({\frac {y}{x}})\end{matrix}}}
Übung: Leiten Sie die Formeln für
ρ
{\displaystyle \rho }
und
ϕ
{\displaystyle \phi }
her.
Ableitungen der Zylinderkoordinaten nach den kartesischen Koordinaten [ Bearbeiten ]
Leitet man die obigen Gleichungen ab, so erhält man:
∂
ρ
∂
x
=
1
2
(
x
2
+
y
2
)
−
1
2
⋅
2
x
=
1
x
2
+
y
2
x
=
x
ρ
=
cos
(
ϕ
)
{\displaystyle {\frac {\partial \rho }{\partial x}}={\frac {1}{2}}(x^{2}+y^{2})^{-{\frac {1}{2}}}\cdot 2x={\frac {1}{\sqrt {x^{2}+y^{2}}}}x={\frac {x}{\rho }}=\cos(\phi )}
∂
ρ
∂
y
=
1
2
(
x
2
+
y
2
)
−
1
2
⋅
2
y
=
y
ρ
=
sin
(
ϕ
)
{\displaystyle {\frac {\partial \rho }{\partial y}}={\frac {1}{2}}(x^{2}+y^{2})^{-{\frac {1}{2}}}\cdot 2y={\frac {y}{\rho }}=\sin(\phi )}
∂
ϕ
∂
x
=
1
1
+
(
y
x
)
2
⋅
−
y
x
2
=
−
y
ρ
2
=
−
sin
(
ϕ
)
ρ
{\displaystyle {\frac {\partial \phi }{\partial x}}={\frac {1}{1+\left({\frac {y}{x}}\right)^{2}}}\cdot {\frac {-y}{x^{2}}}=-{\frac {y}{\rho ^{2}}}=-{\frac {\sin(\phi )}{\rho }}}
∂
ϕ
∂
y
=
1
1
+
(
y
x
)
2
⋅
1
x
=
x
ρ
2
=
cos
(
ϕ
)
ρ
{\displaystyle {\frac {\partial \phi }{\partial y}}={\frac {1}{1+\left({\frac {y}{x}}\right)^{2}}}\cdot {\frac {1}{x}}={\frac {x}{\rho ^{2}}}={\frac {\cos(\phi )}{\rho }}}
Ableitung einer Funktion in Zylinderkoordinaten nach kartesischen Koordinaten [ Bearbeiten ]
Will man eine Funktion in Zylinderkoordinaten nach kartesischen Koordinaten ableiten, so muss man die (mehrdimensionale) Kettenregel berücksichtigen und erhält:
∂
∂
x
f
(
ρ
,
ϕ
,
z
)
=
∂
ρ
∂
x
∂
∂
ρ
f
(
ρ
,
ϕ
,
z
)
+
∂
ϕ
∂
x
∂
∂
ϕ
f
(
ρ
,
ϕ
,
z
)
=
(
cos
(
ϕ
)
∂
∂
ρ
−
sin
(
ϕ
)
ρ
∂
∂
ϕ
)
f
(
ρ
,
ϕ
,
z
)
{\displaystyle {\frac {\partial }{\partial x}}f(\rho ,\phi ,z)={\frac {\partial \rho }{\partial x}}{\frac {\partial }{\partial \rho }}f(\rho ,\phi ,z)+{\frac {\partial \phi }{\partial x}}{\frac {\partial }{\partial \phi }}f(\rho ,\phi ,z)=\left(\cos(\phi ){\frac {\partial }{\partial \rho }}-{\frac {\sin(\phi )}{\rho }}{\frac {\partial }{\partial \phi }}\right)f(\rho ,\phi ,z)}
∂
∂
y
f
(
ρ
,
ϕ
,
z
)
=
∂
ρ
∂
y
∂
∂
ρ
f
(
ρ
,
ϕ
,
z
)
+
∂
ϕ
∂
y
∂
∂
ϕ
f
(
ρ
,
ϕ
,
z
)
=
(
sin
(
ϕ
)
∂
∂
ρ
+
cos
(
ϕ
)
ρ
∂
∂
ϕ
)
f
(
ρ
,
ϕ
,
z
)
{\displaystyle {\frac {\partial }{\partial y}}f(\rho ,\phi ,z)={\frac {\partial \rho }{\partial y}}{\frac {\partial }{\partial \rho }}f(\rho ,\phi ,z)+{\frac {\partial \phi }{\partial y}}{\frac {\partial }{\partial \phi }}f(\rho ,\phi ,z)=\left(\sin(\phi ){\frac {\partial }{\partial \rho }}+{\frac {\cos(\phi )}{\rho }}{\frac {\partial }{\partial \phi }}\right)f(\rho ,\phi ,z)}
Ableitungen der kartesischen Koordinaten nach den Zylinderkoordinaten [ Bearbeiten ]
∂
x
∂
ρ
=
(
cos
(
ϕ
)
sin
(
ϕ
)
0
)
;
∂
x
∂
ϕ
=
ρ
(
−
sin
(
ϕ
)
cos
(
ϕ
)
0
)
;
∂
x
∂
z
=
ρ
(
0
0
1
)
{\displaystyle {\frac {\partial \mathbf {x} }{\partial \rho }}={\begin{pmatrix}\cos(\phi )\\\sin(\phi )\\0\end{pmatrix}};\;{\frac {\partial \mathbf {x} }{\partial \phi }}=\rho {\begin{pmatrix}-\sin(\phi )\\\cos(\phi )\\0\end{pmatrix}};\;{\frac {\partial \mathbf {x} }{\partial z}}=\rho {\begin{pmatrix}0\\0\\1\end{pmatrix}}}
Normiert:
ρ
^
=
∂
x
∂
ρ
|
∂
x
∂
ρ
|
=
(
cos
(
ϕ
)
sin
(
ϕ
)
0
)
{\displaystyle {\boldsymbol {\hat {\rho }}}={\frac {\frac {\partial \mathbf {x} }{\partial \rho }}{\left|{\frac {\partial \mathbf {x} }{\partial \rho }}\right|}}={\begin{pmatrix}\cos(\phi )\\\sin(\phi )\\0\end{pmatrix}}}
ϕ
^
=
∂
x
∂
ϕ
|
∂
x
∂
ϕ
|
=
(
−
sin
(
ϕ
)
cos
(
ϕ
)
0
)
{\displaystyle {\boldsymbol {\hat {\phi }}}={\frac {\frac {\partial \mathbf {x} }{\partial \phi }}{\left|{\frac {\partial \mathbf {x} }{\partial \phi }}\right|}}={\begin{pmatrix}-\sin(\phi )\\\cos(\phi )\\0\end{pmatrix}}}
z
^
=
∂
x
∂
z
|
∂
x
∂
z
|
=
(
0
0
1
)
{\displaystyle {\boldsymbol {\hat {z}}}={\frac {\frac {\partial \mathbf {x} }{\partial z}}{\left|{\frac {\partial \mathbf {x} }{\partial z}}\right|}}={\begin{pmatrix}0\\0\\1\end{pmatrix}}}
Gradient:
∇
f
(
r
,
ϕ
,
z
)
=
(
ρ
^
∂
∂
ρ
+
ϕ
^
1
ρ
∂
∂
ϕ
+
z
^
∂
∂
z
)
f
(
r
,
ϕ
,
z
)
{\displaystyle \nabla f(r,\phi ,z)=\left({\boldsymbol {\hat {\rho }}}{\frac {\partial }{\partial \rho }}+{\boldsymbol {\hat {\phi }}}{\frac {1}{\rho }}{\frac {\partial }{\partial \phi }}+{\boldsymbol {\hat {z}}}{\frac {\partial }{\partial z}}\right)f(r,\phi ,z)}
Divergenz:
A
:=
(
A
x
A
y
A
z
)
=
(
A
ρ
cos
(
ϕ
)
−
A
ϕ
sin
(
ϕ
)
A
ρ
sin
(
ϕ
)
+
A
ϕ
cos
(
ϕ
)
A
z
)
{\displaystyle \mathbf {A} :={\begin{pmatrix}A_{x}\\A_{y}\\A_{z}\end{pmatrix}}={\begin{pmatrix}A_{\rho }\cos(\phi )-A_{\phi }\sin(\phi )\\A_{\rho }\sin(\phi )+A_{\phi }\cos(\phi )\\A_{z}\end{pmatrix}}}
∂
∂
x
A
x
=
(
x
ρ
∂
∂
ρ
−
y
ρ
2
∂
∂
ϕ
)
(
A
ρ
cos
(
ϕ
)
−
A
ϕ
sin
(
ϕ
)
)
=
x
ρ
∂
∂
ρ
A
ρ
cos
(
ϕ
)
+
y
ρ
2
∂
∂
ϕ
A
ϕ
sin
(
ϕ
)
−
x
ρ
∂
∂
ρ
A
ϕ
sin
(
ϕ
)
−
y
ρ
2
∂
∂
ϕ
A
ρ
cos
(
ϕ
)
{\displaystyle {\frac {\partial }{\partial x}}A_{x}=\left({\frac {x}{\rho }}{\frac {\partial }{\partial \rho }}-{\frac {y}{\rho ^{2}}}{\frac {\partial }{\partial \phi }}\right)\left(A_{\rho }\cos(\phi )-A_{\phi }\sin(\phi )\right)={\frac {x}{\rho }}{\frac {\partial }{\partial \rho }}A_{\rho }\cos(\phi )+{\frac {y}{\rho ^{2}}}{\frac {\partial }{\partial \phi }}A_{\phi }\sin(\phi )-{\frac {x}{\rho }}{\frac {\partial }{\partial \rho }}A_{\phi }\sin(\phi )-{\frac {y}{\rho ^{2}}}{\frac {\partial }{\partial \phi }}A_{\rho }\cos(\phi )}
∂
∂
x
A
x
=
x
ρ
cos
(
ϕ
)
∂
A
ρ
∂
ρ
+
y
ρ
2
sin
(
ϕ
)
∂
A
ϕ
∂
ϕ
+
y
ρ
2
A
ϕ
cos
(
ϕ
)
−
x
ρ
sin
(
ϕ
)
∂
A
ϕ
∂
ρ
−
y
ρ
2
cos
(
ϕ
)
∂
A
ρ
∂
ϕ
+
y
ρ
2
A
ρ
sin
(
ϕ
)
{\displaystyle {\frac {\partial }{\partial x}}A_{x}={\frac {x}{\rho }}\cos(\phi ){\frac {\partial A_{\rho }}{\partial \rho }}+{\frac {y}{\rho ^{2}}}\sin(\phi ){\frac {\partial A_{\phi }}{\partial \phi }}+{\frac {y}{\rho ^{2}}}A_{\phi }\cos(\phi )-{\frac {x}{\rho }}\sin(\phi ){\frac {\partial A_{\phi }}{\partial \rho }}-{\frac {y}{\rho ^{2}}}\cos(\phi ){\frac {\partial A_{\rho }}{\partial \phi }}+{\frac {y}{\rho ^{2}}}A_{\rho }\sin(\phi )}
∂
∂
x
A
x
=
cos
(
ϕ
)
2
∂
A
ρ
∂
ρ
+
1
ρ
sin
(
ϕ
)
2
∂
A
ϕ
∂
ϕ
+
1
ρ
A
ϕ
cos
(
ϕ
)
sin
(
ϕ
)
−
cos
(
ϕ
)
sin
(
ϕ
)
∂
A
ϕ
∂
ρ
−
1
ρ
sin
(
ϕ
)
cos
(
ϕ
)
∂
A
ρ
∂
ϕ
+
1
ρ
A
ρ
sin
(
ϕ
)
2
{\displaystyle {\frac {\partial }{\partial x}}A_{x}=\cos(\phi )^{2}{\frac {\partial A_{\rho }}{\partial \rho }}+{\frac {1}{\rho }}\sin(\phi )^{2}{\frac {\partial A_{\phi }}{\partial \phi }}+{\frac {1}{\rho }}A_{\phi }\cos(\phi )\sin(\phi )-\cos(\phi )\sin(\phi ){\frac {\partial A_{\phi }}{\partial \rho }}-{\frac {1}{\rho }}\sin(\phi )\cos(\phi ){\frac {\partial A_{\rho }}{\partial \phi }}+{\frac {1}{\rho }}A_{\rho }\sin(\phi )^{2}}
∂
∂
y
A
y
=
(
y
ρ
∂
∂
ρ
+
x
ρ
2
∂
∂
ϕ
)
(
A
ρ
sin
(
ϕ
)
+
A
ϕ
cos
(
ϕ
)
)
=
y
ρ
∂
∂
ρ
A
ρ
sin
(
ϕ
)
+
x
ρ
2
∂
∂
ϕ
A
ϕ
cos
(
ϕ
)
+
y
ρ
∂
∂
ρ
A
ϕ
cos
(
ϕ
)
+
x
ρ
2
∂
∂
ϕ
A
ρ
sin
(
ϕ
)
{\displaystyle {\frac {\partial }{\partial y}}A_{y}=\left({\frac {y}{\rho }}{\frac {\partial }{\partial \rho }}+{\frac {x}{\rho ^{2}}}{\frac {\partial }{\partial \phi }}\right)\left(A_{\rho }\sin(\phi )+A_{\phi }\cos(\phi )\right)={\frac {y}{\rho }}{\frac {\partial }{\partial \rho }}A_{\rho }\sin(\phi )+{\frac {x}{\rho ^{2}}}{\frac {\partial }{\partial \phi }}A_{\phi }\cos(\phi )+{\frac {y}{\rho }}{\frac {\partial }{\partial \rho }}A_{\phi }\cos(\phi )+{\frac {x}{\rho ^{2}}}{\frac {\partial }{\partial \phi }}A_{\rho }\sin(\phi )}
∂
∂
y
A
y
=
y
ρ
sin
(
ϕ
)
∂
A
ρ
∂
ρ
−
x
ρ
2
A
ϕ
sin
(
ϕ
)
+
x
ρ
2
cos
(
ϕ
)
∂
A
ϕ
∂
ϕ
+
y
ρ
cos
(
ϕ
)
∂
A
ϕ
∂
ρ
+
x
ρ
2
sin
(
ϕ
)
∂
A
ρ
∂
ϕ
+
x
ρ
2
A
ρ
cos
(
ϕ
)
{\displaystyle {\frac {\partial }{\partial y}}A_{y}={\frac {y}{\rho }}\sin(\phi ){\frac {\partial A_{\rho }}{\partial \rho }}-{\frac {x}{\rho ^{2}}}A_{\phi }\sin(\phi )+{\frac {x}{\rho ^{2}}}\cos(\phi ){\frac {\partial A_{\phi }}{\partial \phi }}+{\frac {y}{\rho }}\cos(\phi ){\frac {\partial A_{\phi }}{\partial \rho }}+{\frac {x}{\rho ^{2}}}\sin(\phi ){\frac {\partial A_{\rho }}{\partial \phi }}+{\frac {x}{\rho ^{2}}}A_{\rho }\cos(\phi )}
∂
∂
y
A
y
=
sin
(
ϕ
)
2
∂
A
ρ
∂
ρ
−
1
ρ
A
ϕ
cos
(
ϕ
)
sin
(
ϕ
)
+
1
ρ
cos
(
ϕ
)
2
∂
A
ϕ
∂
ϕ
+
cos
(
ϕ
)
sin
(
ϕ
)
∂
A
ϕ
∂
ρ
+
1
ρ
cos
(
ϕ
)
sin
(
ϕ
)
∂
A
ρ
∂
ϕ
+
1
ρ
A
ρ
cos
(
ϕ
)
2
{\displaystyle {\frac {\partial }{\partial y}}A_{y}=\sin(\phi )^{2}{\frac {\partial A_{\rho }}{\partial \rho }}-{\frac {1}{\rho }}A_{\phi }\cos(\phi )\sin(\phi )+{\frac {1}{\rho }}\cos(\phi )^{2}{\frac {\partial A_{\phi }}{\partial \phi }}+\cos(\phi )\sin(\phi ){\frac {\partial A_{\phi }}{\partial \rho }}+{\frac {1}{\rho }}\cos(\phi )\sin(\phi ){\frac {\partial A_{\rho }}{\partial \phi }}+{\frac {1}{\rho }}A_{\rho }\cos(\phi )^{2}}
∂
∂
x
A
x
+
∂
∂
y
A
y
=
∂
A
ρ
∂
ρ
+
1
ρ
∂
A
ϕ
∂
ϕ
+
1
ρ
A
ρ
=
1
ρ
∂
∂
ρ
ρ
A
ρ
+
1
ρ
∂
A
ϕ
∂
ϕ
{\displaystyle {\frac {\partial }{\partial x}}A_{x}+{\frac {\partial }{\partial y}}A_{y}={\frac {\partial A_{\rho }}{\partial \rho }}+{\frac {1}{\rho }}{\frac {\partial A_{\phi }}{\partial \phi }}+{\frac {1}{\rho }}A_{\rho }={\frac {1}{\rho }}{\frac {\partial }{\partial \rho }}\rho A_{\rho }+{\frac {1}{\rho }}{\frac {\partial A_{\phi }}{\partial \phi }}}
∇
⋅
A
=
1
ρ
∂
∂
ρ
ρ
A
ρ
+
1
ρ
∂
A
ϕ
∂
ϕ
+
∂
A
z
∂
z
{\displaystyle \nabla \cdot \mathbf {A} ={\frac {1}{\rho }}{\frac {\partial }{\partial \rho }}\rho A_{\rho }+{\frac {1}{\rho }}{\frac {\partial A_{\phi }}{\partial \phi }}+{\frac {\partial A_{z}}{\partial z}}}
x
=
r
sin
(
Θ
)
cos
(
ϕ
)
y
=
r
sin
(
Θ
)
sin
(
ϕ
)
z
=
r
cos
(
Θ
)
{\displaystyle {\begin{matrix}x&=&r\sin(\Theta )\cos(\phi )\\y&=&r\sin(\Theta )\sin(\phi )\\z&=&r\cos(\Theta )\end{matrix}}}
r
=
x
2
+
y
2
+
z
2
=
(
x
2
+
y
2
+
z
2
)
1
2
ϕ
=
a
r
c
t
a
n
(
y
x
)
Θ
=
a
r
c
t
a
n
(
x
2
+
y
2
z
)
{\displaystyle {\begin{matrix}r&=&{\sqrt {x^{2}+y^{2}+z^{2}}}=\left(x^{2}+y^{2}+z^{2}\right)^{\frac {1}{2}}\\\phi &=&\mathrm {arctan} ({\frac {y}{x}})\\\Theta &=&\mathrm {arctan} ({\frac {\sqrt {x^{2}+y^{2}}}{z}})\end{matrix}}}
∂
r
∂
x
=
1
2
(
x
2
+
y
2
+
z
2
)
−
1
2
⋅
2
x
=
x
r
=
sin
(
Θ
)
cos
(
ϕ
)
{\displaystyle {\frac {\partial r}{\partial x}}={\frac {1}{2}}(x^{2}+y^{2}+z^{2})^{-{\frac {1}{2}}}\cdot 2x={\frac {x}{r}}=\sin(\Theta )\cos(\phi )}
∂
r
∂
y
=
y
r
=
sin
(
Θ
)
sin
(
ϕ
)
{\displaystyle {\frac {\partial r}{\partial y}}={\frac {y}{r}}=\sin(\Theta )\sin(\phi )}
∂
r
∂
z
=
z
r
=
cos
(
Θ
)
{\displaystyle {\frac {\partial r}{\partial z}}={\frac {z}{r}}=\cos(\Theta )}
ρ
:=
x
2
+
y
2
=
r
sin
(
Θ
)
{\displaystyle \rho :={\sqrt {x^{2}+y^{2}}}=r\sin(\Theta )}
∂
ϕ
∂
x
=
1
1
+
(
y
x
)
2
⋅
−
y
x
2
=
−
y
ρ
2
=
−
sin
(
ϕ
)
ρ
=
−
sin
(
ϕ
)
r
sin
(
Θ
)
{\displaystyle {\frac {\partial \phi }{\partial x}}={\frac {1}{1+\left({\frac {y}{x}}\right)^{2}}}\cdot -{\frac {y}{x^{2}}}=-{\frac {y}{\rho ^{2}}}=-{\frac {\sin(\phi )}{\rho }}=-{\frac {\sin(\phi )}{r\sin(\Theta )}}}
∂
ϕ
∂
y
=
1
1
+
(
y
x
)
2
⋅
1
x
=
x
ρ
2
=
cos
(
ϕ
)
ρ
=
cos
(
ϕ
)
r
sin
(
Θ
)
{\displaystyle {\frac {\partial \phi }{\partial y}}={\frac {1}{1+\left({\frac {y}{x}}\right)^{2}}}\cdot {\frac {1}{x}}={\frac {x}{\rho ^{2}}}={\frac {\cos(\phi )}{\rho }}={\frac {\cos(\phi )}{r\sin(\Theta )}}}
∂
ϕ
∂
z
=
0
{\displaystyle {\frac {\partial \phi }{\partial z}}=0}
∂
Θ
∂
x
=
1
1
+
x
2
+
y
2
z
2
⋅
2
x
2
z
x
2
+
y
2
=
z
r
2
⋅
cos
(
ϕ
)
=
1
r
⋅
cos
(
Θ
)
cos
(
ϕ
)
{\displaystyle {\frac {\partial \Theta }{\partial x}}={\frac {1}{1+{\frac {x^{2}+y^{2}}{z^{2}}}}}\cdot {\frac {2x}{2z{\sqrt {x^{2}+y^{2}}}}}={\frac {z}{r^{2}}}\cdot \cos(\phi )={\frac {1}{r}}\cdot \cos(\Theta )\cos(\phi )}
∂
Θ
∂
y
=
1
1
+
x
2
+
y
2
z
2
⋅
2
y
2
z
x
2
+
y
2
=
z
r
2
⋅
sin
(
ϕ
)
=
1
r
⋅
cos
(
Θ
)
sin
(
ϕ
)
{\displaystyle {\frac {\partial \Theta }{\partial y}}={\frac {1}{1+{\frac {x^{2}+y^{2}}{z^{2}}}}}\cdot {\frac {2y}{2z{\sqrt {x^{2}+y^{2}}}}}={\frac {z}{r^{2}}}\cdot \sin(\phi )={\frac {1}{r}}\cdot \cos(\Theta )\sin(\phi )}
∂
Θ
∂
z
=
1
1
+
x
2
+
y
2
z
2
⋅
−
x
2
+
y
2
z
2
=
−
ρ
r
2
=
−
sin
(
Θ
)
r
{\displaystyle {\frac {\partial \Theta }{\partial z}}={\frac {1}{1+{\frac {x^{2}+y^{2}}{z^{2}}}}}\cdot {\frac {-{\sqrt {x^{2}+y^{2}}}}{z^{2}}}=-{\frac {\rho }{r^{2}}}=-{\frac {\sin(\Theta )}{r}}}
∂
x
∂
r
=
(
sin
(
Θ
)
cos
(
ϕ
)
sin
(
Θ
)
sin
(
ϕ
)
cos
(
Θ
)
)
{\displaystyle {\frac {\partial \mathbf {x} }{\partial r}}={\begin{pmatrix}\sin(\Theta )\cos(\phi )\\\sin(\Theta )\sin(\phi )\\\cos(\Theta )\end{pmatrix}}}
∂
x
∂
ϕ
=
(
−
r
sin
(
Θ
)
sin
(
ϕ
)
r
sin
(
Θ
)
cos
(
ϕ
)
0
)
{\displaystyle {\frac {\partial \mathbf {x} }{\partial \phi }}={\begin{pmatrix}-r\sin(\Theta )\sin(\phi )\\r\sin(\Theta )\cos(\phi )\\0\end{pmatrix}}}
∂
x
∂
Θ
=
(
r
cos
(
Θ
)
cos
(
ϕ
)
r
cos
(
Θ
)
sin
(
ϕ
)
−
r
sin
(
Θ
)
)
{\displaystyle {\frac {\partial \mathbf {x} }{\partial \Theta }}={\begin{pmatrix}r\cos(\Theta )\cos(\phi )\\r\cos(\Theta )\sin(\phi )\\-r\sin(\Theta )\end{pmatrix}}}
r
^
=
∂
x
∂
r
|
∂
x
∂
r
|
=
(
sin
(
Θ
)
cos
(
ϕ
)
sin
(
Θ
)
sin
(
ϕ
)
cos
(
Θ
)
)
{\displaystyle {\boldsymbol {\hat {r}}}={\frac {\frac {\partial \mathbf {x} }{\partial r}}{|{\frac {\partial \mathbf {x} }{\partial r}}|}}={\begin{pmatrix}\sin(\Theta )\cos(\phi )\\\sin(\Theta )\sin(\phi )\\\cos(\Theta )\end{pmatrix}}}
ϕ
^
=
∂
x
∂
ϕ
|
∂
x
∂
ϕ
|
=
(
−
sin
(
ϕ
)
cos
(
ϕ
)
0
)
{\displaystyle {\boldsymbol {\hat {\phi }}}={\frac {\frac {\partial \mathbf {x} }{\partial \phi }}{|{\frac {\partial \mathbf {x} }{\partial \phi }}|}}={\begin{pmatrix}-\sin(\phi )\\\cos(\phi )\\0\end{pmatrix}}}
Θ
^
=
∂
x
∂
Θ
|
∂
x
∂
Θ
|
=
(
cos
(
Θ
)
cos
(
ϕ
)
cos
(
Θ
)
sin
(
ϕ
)
−
sin
(
Θ
)
)
{\displaystyle {\boldsymbol {\hat {\Theta }}}={\frac {\frac {\partial \mathbf {x} }{\partial \Theta }}{|{\frac {\partial \mathbf {x} }{\partial \Theta }}|}}={\begin{pmatrix}\cos(\Theta )\cos(\phi )\\\cos(\Theta )\sin(\phi )\\-\sin(\Theta )\end{pmatrix}}}
∂
∂
x
f
(
r
,
Θ
,
ϕ
)
=
(
∂
r
∂
x
∂
∂
r
+
∂
ϕ
∂
x
∂
∂
ϕ
+
∂
Θ
∂
x
∂
∂
Θ
)
f
(
r
,
Θ
,
ϕ
)
=
(
sin
(
Θ
)
cos
(
ϕ
)
∂
∂
r
−
sin
(
ϕ
)
r
sin
(
Θ
)
∂
∂
ϕ
+
1
r
⋅
cos
(
Θ
)
cos
(
ϕ
)
∂
∂
Θ
)
f
(
r
,
Θ
,
ϕ
)
{\displaystyle {\frac {\partial }{\partial x}}f(r,\Theta ,\phi )=\left({\frac {\partial r}{\partial x}}{\frac {\partial }{\partial r}}+{\frac {\partial \phi }{\partial x}}{\frac {\partial }{\partial \phi }}+{\frac {\partial \Theta }{\partial x}}{\frac {\partial }{\partial \Theta }}\right)f(r,\Theta ,\phi )=\left(\sin(\Theta )\cos(\phi ){\frac {\partial }{\partial r}}-{\frac {\sin(\phi )}{r\sin(\Theta )}}{\frac {\partial }{\partial \phi }}+{\frac {1}{r}}\cdot \cos(\Theta )\cos(\phi ){\frac {\partial }{\partial \Theta }}\right)f(r,\Theta ,\phi )}
∂
∂
y
f
(
r
,
Θ
,
ϕ
)
=
(
∂
r
∂
y
∂
∂
r
+
∂
ϕ
∂
y
∂
∂
ϕ
+
∂
Θ
∂
y
∂
∂
Θ
)
f
(
r
,
Θ
,
ϕ
)
=
(
sin
(
Θ
)
sin
(
ϕ
)
∂
∂
r
+
cos
(
ϕ
)
r
sin
(
Θ
)
∂
∂
ϕ
+
1
r
⋅
cos
(
Θ
)
sin
(
ϕ
)
∂
∂
Θ
)
f
(
r
,
Θ
,
ϕ
)
{\displaystyle {\frac {\partial }{\partial y}}f(r,\Theta ,\phi )=\left({\frac {\partial r}{\partial y}}{\frac {\partial }{\partial r}}+{\frac {\partial \phi }{\partial y}}{\frac {\partial }{\partial \phi }}+{\frac {\partial \Theta }{\partial y}}{\frac {\partial }{\partial \Theta }}\right)f(r,\Theta ,\phi )=\left(\sin(\Theta )\sin(\phi ){\frac {\partial }{\partial r}}+{\frac {\cos(\phi )}{r\sin(\Theta )}}{\frac {\partial }{\partial \phi }}+{\frac {1}{r}}\cdot \cos(\Theta )\sin(\phi ){\frac {\partial }{\partial \Theta }}\right)f(r,\Theta ,\phi )}
∂
∂
z
f
(
r
,
Θ
,
ϕ
)
=
(
∂
r
∂
z
∂
∂
r
+
∂
ϕ
∂
z
∂
∂
ϕ
+
∂
Θ
∂
z
∂
∂
Θ
)
f
(
r
,
Θ
,
ϕ
)
=
(
cos
(
Θ
)
∂
∂
r
−
1
r
⋅
sin
(
Θ
)
∂
∂
Θ
)
f
(
r
,
Θ
,
ϕ
)
{\displaystyle {\frac {\partial }{\partial z}}f(r,\Theta ,\phi )=\left({\frac {\partial r}{\partial z}}{\frac {\partial }{\partial r}}+{\frac {\partial \phi }{\partial z}}{\frac {\partial }{\partial \phi }}+{\frac {\partial \Theta }{\partial z}}{\frac {\partial }{\partial \Theta }}\right)f(r,\Theta ,\phi )=\left(\cos(\Theta ){\frac {\partial }{\partial r}}-{\frac {1}{r}}\cdot \sin(\Theta ){\frac {\partial }{\partial \Theta }}\right)f(r,\Theta ,\phi )}
∇
f
(
r
,
Θ
,
ϕ
)
=
r
^
∂
∂
r
+
1
r
sin
(
Θ
)
ϕ
^
∂
∂
ϕ
+
1
r
Θ
^
∂
∂
Θ
{\displaystyle \nabla f(r,\Theta ,\phi )={\boldsymbol {\hat {r}}}{\frac {\partial }{\partial r}}+{\frac {1}{r\sin(\Theta )}}{\boldsymbol {\hat {\phi }}}{\frac {\partial }{\partial \phi }}+{\frac {1}{r}}{\boldsymbol {\hat {\Theta }}}{\frac {\partial }{\partial \Theta }}}
A
(
r
,
Θ
,
ϕ
)
=
(
A
x
A
y
A
z
)
=
A
r
r
^
+
A
Θ
Θ
^
+
A
ϕ
ϕ
^
=
(
A
r
sin
(
Θ
)
cos
(
ϕ
)
+
A
Θ
cos
(
Θ
)
cos
(
ϕ
)
−
A
ϕ
sin
(
ϕ
)
A
r
sin
(
Θ
)
sin
(
ϕ
)
+
A
Θ
cos
(
Θ
)
sin
(
ϕ
)
+
A
ϕ
cos
(
ϕ
)
A
r
cos
(
Θ
)
−
A
Θ
sin
(
Θ
)
)
{\displaystyle \mathbf {A} (r,\Theta ,\phi )={\begin{pmatrix}A_{x}\\A_{y}\\A_{z}\end{pmatrix}}=A_{r}{\boldsymbol {\hat {r}}}+A_{\Theta }{\boldsymbol {\hat {\Theta }}}+A_{\phi }{\boldsymbol {\hat {\phi }}}={\begin{pmatrix}A_{r}\sin(\Theta )\cos(\phi )+A_{\Theta }\cos(\Theta )\cos(\phi )-A_{\phi }\sin(\phi )\\A_{r}\sin(\Theta )\sin(\phi )+A_{\Theta }\cos(\Theta )\sin(\phi )+A_{\phi }\cos(\phi )\\A_{r}\cos(\Theta )-A_{\Theta }\sin(\Theta )\end{pmatrix}}}
∂
A
x
∂
x
=
(
sin
(
Θ
)
cos
(
ϕ
)
∂
∂
r
−
sin
(
ϕ
)
r
sin
(
Θ
)
∂
∂
ϕ
+
1
r
⋅
cos
(
Θ
)
cos
(
ϕ
)
∂
∂
Θ
)
(
A
r
sin
(
Θ
)
cos
(
ϕ
)
+
A
Θ
cos
(
Θ
)
cos
(
ϕ
)
−
A
ϕ
sin
(
ϕ
)
)
=
sin
2
(
Θ
)
cos
2
(
ϕ
)
∂
A
r
∂
r
+
sin
(
Θ
)
cos
(
Θ
)
cos
2
(
ϕ
)
∂
A
Θ
∂
r
−
sin
(
Θ
)
cos
(
ϕ
)
sin
(
ϕ
)
∂
A
ϕ
∂
r
−
sin
(
ϕ
)
cos
(
ϕ
)
r
∂
A
r
∂
ϕ
+
A
r
sin
2
(
ϕ
)
r
−
sin
(
ϕ
)
cos
(
ϕ
)
cos
(
Θ
)
r
sin
(
Θ
)
∂
A
Θ
∂
ϕ
+
A
Θ
sin
2
(
ϕ
)
cos
(
Θ
)
r
sin
(
Θ
)
+
sin
2
(
ϕ
)
r
sin
(
Θ
)
∂
A
ϕ
∂
ϕ
+
A
ϕ
sin
(
ϕ
)
cos
(
ϕ
)
r
sin
(
Θ
)
+
1
r
cos
(
Θ
)
sin
(
Θ
)
cos
2
(
ϕ
)
∂
A
r
∂
Θ
+
1
r
A
r
cos
2
(
Θ
)
cos
2
(
ϕ
)
+
1
r
cos
2
(
Θ
)
cos
2
(
ϕ
)
∂
A
Θ
∂
Θ
−
1
r
A
Θ
cos
(
Θ
)
sin
(
Θ
)
cos
2
(
ϕ
)
−
1
r
cos
(
Θ
)
cos
(
ϕ
)
sin
(
ϕ
)
∂
A
ϕ
∂
Θ
{\displaystyle {\begin{matrix}{\frac {\partial A_{x}}{\partial x}}&=&\left(\sin(\Theta )\cos(\phi ){\frac {\partial }{\partial r}}-{\frac {\sin(\phi )}{r\sin(\Theta )}}{\frac {\partial }{\partial \phi }}+{\frac {1}{r}}\cdot \cos(\Theta )\cos(\phi ){\frac {\partial }{\partial \Theta }}\right)\left(A_{r}\sin(\Theta )\cos(\phi )+A_{\Theta }\cos(\Theta )\cos(\phi )-A_{\phi }\sin(\phi )\right)\\&=&\sin ^{2}(\Theta )\cos ^{2}(\phi ){\frac {\partial A_{r}}{\partial r}}+\sin(\Theta )\cos(\Theta )\cos ^{2}(\phi ){\frac {\partial A_{\Theta }}{\partial r}}-\sin(\Theta )\cos(\phi )\sin(\phi ){\frac {\partial A_{\phi }}{\partial r}}\\&&-{\frac {\sin(\phi )\cos(\phi )}{r}}{\frac {\partial A_{r}}{\partial \phi }}+A_{r}{\frac {\sin ^{2}(\phi )}{r}}-{\frac {\sin(\phi )\cos(\phi )\cos(\Theta )}{r\sin(\Theta )}}{\frac {\partial A_{\Theta }}{\partial \phi }}+A_{\Theta }{\frac {\sin ^{2}(\phi )\cos(\Theta )}{r\sin(\Theta )}}+{\frac {\sin ^{2}(\phi )}{r\sin(\Theta )}}{\frac {\partial A_{\phi }}{\partial \phi }}+A_{\phi }{\frac {\sin(\phi )\cos(\phi )}{r\sin(\Theta )}}\\&&+{\frac {1}{r}}\cos(\Theta )\sin(\Theta )\cos ^{2}(\phi ){\frac {\partial A_{r}}{\partial \Theta }}+{\frac {1}{r}}A_{r}\cos ^{2}(\Theta )\cos ^{2}(\phi )\\&&+{\frac {1}{r}}\cos ^{2}(\Theta )\cos ^{2}(\phi ){\frac {\partial A_{\Theta }}{\partial \Theta }}-{\frac {1}{r}}A_{\Theta }\cos(\Theta )\sin(\Theta )\cos ^{2}(\phi )-{\frac {1}{r}}\cos(\Theta )\cos(\phi )\sin(\phi ){\frac {\partial A_{\phi }}{\partial \Theta }}\end{matrix}}}
∂
A
y
∂
y
=
(
sin
(
Θ
)
sin
(
ϕ
)
∂
∂
r
+
cos
(
ϕ
)
r
sin
(
Θ
)
∂
∂
ϕ
+
1
r
⋅
cos
(
Θ
)
sin
(
ϕ
)
∂
∂
Θ
)
(
A
r
sin
(
Θ
)
sin
(
ϕ
)
+
A
Θ
cos
(
Θ
)
sin
(
ϕ
)
+
A
ϕ
cos
(
ϕ
)
)
=
sin
2
(
Θ
)
sin
2
(
ϕ
)
∂
A
r
∂
r
+
sin
(
Θ
)
cos
(
Θ
)
sin
2
(
ϕ
)
∂
A
Θ
∂
r
+
sin
(
Θ
)
cos
(
ϕ
)
sin
(
ϕ
)
∂
A
ϕ
∂
r
+
sin
(
ϕ
)
cos
(
ϕ
)
r
∂
A
r
∂
ϕ
+
A
r
cos
2
(
ϕ
)
r
+
sin
(
ϕ
)
cos
(
ϕ
)
cos
(
Θ
)
r
sin
(
Θ
)
∂
A
Θ
∂
ϕ
+
A
Θ
cos
2
(
ϕ
)
cos
(
Θ
)
r
sin
(
Θ
)
+
cos
2
(
ϕ
)
r
sin
(
Θ
)
∂
A
ϕ
∂
ϕ
−
A
ϕ
sin
(
ϕ
)
cos
(
ϕ
)
r
sin
(
Θ
)
+
1
r
cos
(
Θ
)
sin
(
Θ
)
sin
2
(
ϕ
)
∂
A
r
∂
Θ
+
1
r
A
r
cos
2
(
Θ
)
sin
2
(
ϕ
)
+
1
r
cos
2
(
Θ
)
sin
2
(
ϕ
)
∂
A
Θ
∂
Θ
−
1
r
A
Θ
cos
(
Θ
)
sin
(
Θ
)
sin
2
(
ϕ
)
+
1
r
cos
(
Θ
)
sin
(
ϕ
)
cos
(
ϕ
)
∂
A
ϕ
∂
Θ
{\displaystyle {\begin{matrix}{\frac {\partial A_{y}}{\partial y}}&=&\left(\sin(\Theta )\sin(\phi ){\frac {\partial }{\partial r}}+{\frac {\cos(\phi )}{r\sin(\Theta )}}{\frac {\partial }{\partial \phi }}+{\frac {1}{r}}\cdot \cos(\Theta )\sin(\phi ){\frac {\partial }{\partial \Theta }}\right)\left(A_{r}\sin(\Theta )\sin(\phi )+A_{\Theta }\cos(\Theta )\sin(\phi )+A_{\phi }\cos(\phi )\right)\\&=&\sin ^{2}(\Theta )\sin ^{2}(\phi ){\frac {\partial A_{r}}{\partial r}}+\sin(\Theta )\cos(\Theta )\sin ^{2}(\phi ){\frac {\partial A_{\Theta }}{\partial r}}+\sin(\Theta )\cos(\phi )\sin(\phi ){\frac {\partial A_{\phi }}{\partial r}}\\&&+{\frac {\sin(\phi )\cos(\phi )}{r}}{\frac {\partial A_{r}}{\partial \phi }}+A_{r}{\frac {\cos ^{2}(\phi )}{r}}+{\frac {\sin(\phi )\cos(\phi )\cos(\Theta )}{r\sin(\Theta )}}{\frac {\partial A_{\Theta }}{\partial \phi }}+A_{\Theta }{\frac {\cos ^{2}(\phi )\cos(\Theta )}{r\sin(\Theta )}}+{\frac {\cos ^{2}(\phi )}{r\sin(\Theta )}}{\frac {\partial A_{\phi }}{\partial \phi }}-A_{\phi }{\frac {\sin(\phi )\cos(\phi )}{r\sin(\Theta )}}\\&&+{\frac {1}{r}}\cos(\Theta )\sin(\Theta )\sin ^{2}(\phi ){\frac {\partial A_{r}}{\partial \Theta }}+{\frac {1}{r}}A_{r}\cos ^{2}(\Theta )\sin ^{2}(\phi )\\&&+{\frac {1}{r}}\cos ^{2}(\Theta )\sin ^{2}(\phi ){\frac {\partial A_{\Theta }}{\partial \Theta }}-{\frac {1}{r}}A_{\Theta }\cos(\Theta )\sin(\Theta )\sin ^{2}(\phi )+{\frac {1}{r}}\cos(\Theta )\sin(\phi )\cos(\phi ){\frac {\partial A_{\phi }}{\partial \Theta }}\end{matrix}}}
∂
A
x
∂
x
+
∂
A
y
∂
y
=
sin
2
(
Θ
)
∂
A
r
∂
r
+
sin
(
Θ
)
cos
(
Θ
)
∂
A
Θ
∂
r
+
A
r
r
+
A
Θ
cos
(
Θ
)
r
sin
(
Θ
)
+
1
r
sin
(
Θ
)
∂
A
ϕ
∂
ϕ
+
1
r
sin
(
Θ
)
cos
(
Θ
)
∂
A
r
∂
Θ
+
1
r
cos
2
(
Θ
)
A
r
+
1
r
cos
2
(
Θ
)
∂
A
Θ
∂
Θ
−
1
r
sin
(
Θ
)
cos
(
Θ
)
A
Θ
{\displaystyle {\begin{matrix}{\frac {\partial A_{x}}{\partial x}}+{\frac {\partial A_{y}}{\partial y}}&=&\sin ^{2}(\Theta ){\frac {\partial A_{r}}{\partial r}}+\sin(\Theta )\cos(\Theta ){\frac {\partial A_{\Theta }}{\partial r}}+{\frac {A_{r}}{r}}+{\frac {A_{\Theta }\cos(\Theta )}{r\sin(\Theta )}}+{\frac {1}{r\sin(\Theta )}}{\frac {\partial A_{\phi }}{\partial \phi }}\\&&+{\frac {1}{r}}\sin(\Theta )\cos(\Theta ){\frac {\partial A_{r}}{\partial \Theta }}+{\frac {1}{r}}\cos ^{2}(\Theta )A_{r}+{\frac {1}{r}}\cos ^{2}(\Theta ){\frac {\partial A_{\Theta }}{\partial \Theta }}-{\frac {1}{r}}\sin(\Theta )\cos(\Theta )A_{\Theta }\end{matrix}}}
∂
A
z
∂
z
=
(
cos
(
Θ
)
∂
∂
r
−
1
r
⋅
sin
(
Θ
)
∂
∂
Θ
)
(
A
r
cos
(
Θ
)
−
A
Θ
sin
(
Θ
)
)
=
cos
2
Θ
∂
A
r
∂
r
−
sin
(
Θ
)
cos
(
Θ
)
∂
A
Θ
∂
r
−
1
r
sin
(
Θ
)
cos
(
Θ
)
∂
A
r
∂
Θ
+
1
r
A
r
sin
2
(
Θ
)
+
1
r
A
Θ
sin
(
Θ
)
cos
(
Θ
)
+
1
r
sin
2
(
Θ
)
∂
A
Θ
∂
Θ
{\displaystyle {\begin{matrix}{\frac {\partial A_{z}}{\partial z}}&=&\left(\cos(\Theta ){\frac {\partial }{\partial r}}-{\frac {1}{r}}\cdot \sin(\Theta ){\frac {\partial }{\partial \Theta }}\right)\left(A_{r}\cos(\Theta )-A_{\Theta }\sin(\Theta )\right)\\&=&\cos ^{2}{\Theta }{\frac {\partial A_{r}}{\partial r}}-\sin(\Theta )\cos(\Theta ){\frac {\partial A_{\Theta }}{\partial r}}-{\frac {1}{r}}\sin(\Theta )\cos(\Theta ){\frac {\partial A_{r}}{\partial \Theta }}+{\frac {1}{r}}A_{r}\sin ^{2}(\Theta )+{\frac {1}{r}}A_{\Theta }\sin(\Theta )\cos(\Theta )+{\frac {1}{r}}\sin ^{2}(\Theta ){\frac {\partial A_{\Theta }}{\partial \Theta }}\end{matrix}}}
∇
⋅
A
=
∂
A
r
∂
r
+
A
r
r
+
A
Θ
cos
(
Θ
)
r
sin
(
Θ
)
+
1
r
sin
(
Θ
)
∂
A
ϕ
∂
ϕ
+
A
r
r
+
1
r
∂
A
Θ
∂
Θ
=
1
r
2
∂
r
2
A
r
∂
r
+
1
r
sin
(
Θ
)
∂
∂
Θ
A
Θ
sin
(
Θ
)
+
1
r
sin
(
Θ
)
∂
A
ϕ
∂
ϕ
{\displaystyle \nabla \cdot \mathbf {A} ={\frac {\partial A_{r}}{\partial r}}+{\frac {A_{r}}{r}}+{\frac {A_{\Theta }\cos(\Theta )}{r\sin(\Theta )}}+{\frac {1}{r\sin(\Theta )}}{\frac {\partial A_{\phi }}{\partial \phi }}+{\frac {A_{r}}{r}}+{\frac {1}{r}}{\frac {\partial A_{\Theta }}{\partial \Theta }}={\frac {1}{r^{2}}}{\frac {\partial r^{2}A_{r}}{\partial r}}+{\frac {1}{r\sin(\Theta )}}{\frac {\partial }{\partial \Theta }}A_{\Theta }\sin(\Theta )+{\frac {1}{r\sin(\Theta )}}{\frac {\partial A_{\phi }}{\partial \phi }}}
Δ
f
(
r
,
Θ
,
ϕ
)
=
∇
⋅
(
∇
f
(
r
,
Θ
,
ϕ
)
)
=
∇
⋅
(
r
^
∂
∂
r
+
1
r
sin
(
Θ
)
ϕ
^
∂
∂
ϕ
+
1
r
Θ
^
∂
∂
Θ
)
f
(
r
,
Θ
,
ϕ
)
{\displaystyle \Delta f(r,\Theta ,\phi )=\nabla \cdot (\nabla f(r,\Theta ,\phi ))=\nabla \cdot \left({\boldsymbol {\hat {r}}}{\frac {\partial }{\partial r}}+{\frac {1}{r\sin(\Theta )}}{\boldsymbol {\hat {\phi }}}{\frac {\partial }{\partial \phi }}+{\frac {1}{r}}{\boldsymbol {\hat {\Theta }}}{\frac {\partial }{\partial \Theta }}\right)f(r,\Theta ,\phi )}
Δ
f
(
r
,
Θ
,
ϕ
)
=
(
1
r
2
∂
∂
r
r
2
∂
∂
r
+
1
r
2
sin
(
Θ
)
∂
∂
Θ
sin
(
Θ
)
∂
∂
Θ
+
1
r
2
sin
(
Θ
)
2
∂
2
∂
ϕ
2
)
f
(
r
,
Θ
,
ϕ
)
{\displaystyle \Delta f(r,\Theta ,\phi )=\left({\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}\sin(\Theta )^{2}}}{\frac {\partial ^{2}}{\partial \phi ^{2}}}\right)f(r,\Theta ,\phi )}