Note: This page uses common physics notation for spherical coordinates, in which
θ
{\displaystyle \theta }
z axis and the radius vector connecting the origin to the point in question, while
ϕ
{\displaystyle \phi }
x-y plane and the x axis. Several other definitions are in use, and so care must be taken in comparing different sources.[1]
Spherical coordinates (r , θ , φ ) as commonly used in physics : radial distance r , polar angle θ (theta ), and azimuthal angle φ (phi ). The symbol ρ (rho ) is often used instead of r . Cylindrical coordinate system
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Vector fields
edit
Vectors are defined in cylindrical coordinates by (ρ , φ , z ), where
ρ is the length of the vector projected onto the xy -plane,φ is the angle between the projection of the vector onto the xy -plane (i.e. ρ ) and the positive x -axis (0 ≤ φ < 2π ),z is the regular z -coordinate.(ρ , φ , z ) is given in Cartesian coordinates by:
[
ρ
ϕ
z
]
=
[
x
2
+
y
2
arctan
(
y
/
x
)
z
]
,
0
≤
ϕ
<
2
π
,
{\displaystyle {\begin{bmatrix}\rho \\\phi \\z\end{bmatrix}}={\begin{bmatrix}{\sqrt {x^{2}+y^{2}}}\\\operatorname {arctan} (y/x)\\z\end{bmatrix}},\ \ \ 0\leq \phi <2\pi ,}
or inversely by:
[
x
y
z
]
=
[
ρ
cos
ϕ
ρ
sin
ϕ
z
]
.
{\displaystyle {\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}\rho \cos \phi \\\rho \sin \phi \\z\end{bmatrix}}.}
Any vector field can be written in terms of the unit vectors as:
A
=
A
x
x
^
+
A
y
y
^
+
A
z
z
^
=
A
ρ
ρ
^
+
A
ϕ
ϕ
^
+
A
z
z
^
{\displaystyle \mathbf {A} =A_{x}\mathbf {\hat {x}} +A_{y}\mathbf {\hat {y}} +A_{z}\mathbf {\hat {z}} =A_{\rho }\mathbf {\hat {\rho }} +A_{\phi }{\boldsymbol {\hat {\phi }}}+A_{z}\mathbf {\hat {z}} }
[
ρ
^
ϕ
^
z
^
]
=
[
cos
ϕ
sin
ϕ
0
−
sin
ϕ
cos
ϕ
0
0
0
1
]
[
x
^
y
^
z
^
]
{\displaystyle {\begin{bmatrix}\mathbf {\hat {\rho }} \\{\boldsymbol {\hat {\phi }}}\\\mathbf {\hat {z}} \end{bmatrix}}={\begin{bmatrix}\cos \phi &\sin \phi &0\\-\sin \phi &\cos \phi &0\\0&0&1\end{bmatrix}}{\begin{bmatrix}\mathbf {\hat {x}} \\\mathbf {\hat {y}} \\\mathbf {\hat {z}} \end{bmatrix}}}
Note: the matrix is an orthogonal matrix , that is, its inverse is simply its transpose .
Time derivative of a vector field
edit
To find out how the vector field A changes in time, the time derivatives should be calculated.
For this purpose Newton's notation will be used for the time derivative (
A
˙
{\displaystyle {\dot {\mathbf {A} }}}
A
˙
=
A
˙
x
x
^
+
A
˙
y
y
^
+
A
˙
z
z
^
{\displaystyle {\dot {\mathbf {A} }}={\dot {A}}_{x}{\hat {\mathbf {x} }}+{\dot {A}}_{y}{\hat {\mathbf {y} }}+{\dot {A}}_{z}{\hat {\mathbf {z} }}}
However, in cylindrical coordinates this becomes:
A
˙
=
A
˙
ρ
ρ
^
+
A
ρ
ρ
^
˙
+
A
˙
ϕ
ϕ
^
+
A
ϕ
ϕ
^
˙
+
A
˙
z
z
^
+
A
z
z
^
˙
{\displaystyle {\dot {\mathbf {A} }}={\dot {A}}_{\rho }{\hat {\boldsymbol {\rho }}}+A_{\rho }{\dot {\hat {\boldsymbol {\rho }}}}+{\dot {A}}_{\phi }{\hat {\boldsymbol {\phi }}}+A_{\phi }{\dot {\hat {\boldsymbol {\phi }}}}+{\dot {A}}_{z}{\hat {\boldsymbol {z}}}+A_{z}{\dot {\hat {\boldsymbol {z}}}}}
The time derivatives of the unit vectors are needed.
They are given by:
ρ
^
˙
=
ϕ
˙
ϕ
^
ϕ
^
˙
=
−
ϕ
˙
ρ
^
z
^
˙
=
0
{\displaystyle {\begin{aligned}{\dot {\hat {\mathbf {\rho } }}}&={\dot {\phi }}{\hat {\boldsymbol {\phi }}}\\{\dot {\hat {\boldsymbol {\phi }}}}&=-{\dot {\phi }}{\hat {\mathbf {\rho } }}\\{\dot {\hat {\mathbf {z} }}}&=0\end{aligned}}}
So the time derivative simplifies to:
A
˙
=
ρ
^
(
A
˙
ρ
−
A
ϕ
ϕ
˙
)
+
ϕ
^
(
A
˙
ϕ
+
A
ρ
ϕ
˙
)
+
z
^
A
˙
z
{\displaystyle {\dot {\mathbf {A} }}={\hat {\boldsymbol {\rho }}}\left({\dot {A}}_{\rho }-A_{\phi }{\dot {\phi }}\right)+{\hat {\boldsymbol {\phi }}}\left({\dot {A}}_{\phi }+A_{\rho }{\dot {\phi }}\right)+{\hat {\mathbf {z} }}{\dot {A}}_{z}}
Second time derivative of a vector field
edit
The second time derivative is of interest in physics , as it is found in equations of motion for classical mechanical systems.
The second time derivative of a vector field in cylindrical coordinates is given by:
A
¨
=
ρ
^
(
A
¨
ρ
−
A
ϕ
ϕ
¨
−
2
A
˙
ϕ
ϕ
˙
−
A
ρ
ϕ
˙
2
)
+
ϕ
^
(
A
¨
ϕ
+
A
ρ
ϕ
¨
+
2
A
˙
ρ
ϕ
˙
−
A
ϕ
ϕ
˙
2
)
+
z
^
A
¨
z
{\displaystyle \mathbf {\ddot {A}} =\mathbf {\hat {\rho }} \left({\ddot {A}}_{\rho }-A_{\phi }{\ddot {\phi }}-2{\dot {A}}_{\phi }{\dot {\phi }}-A_{\rho }{\dot {\phi }}^{2}\right)+{\boldsymbol {\hat {\phi }}}\left({\ddot {A}}_{\phi }+A_{\rho }{\ddot {\phi }}+2{\dot {A}}_{\rho }{\dot {\phi }}-A_{\phi }{\dot {\phi }}^{2}\right)+\mathbf {\hat {z}} {\ddot {A}}_{z}}
To understand this expression, A is substituted for P , where P is the vector (ρ , φ , z ).
This means that
A
=
P
=
ρ
ρ
^
+
z
z
^
{\displaystyle \mathbf {A} =\mathbf {P} =\rho \mathbf {\hat {\rho }} +z\mathbf {\hat {z}} }
After substituting, the result is given:
P
¨
=
ρ
^
(
ρ
¨
−
ρ
ϕ
˙
2
)
+
ϕ
^
(
ρ
ϕ
¨
+
2
ρ
˙
ϕ
˙
)
+
z
^
z
¨
{\displaystyle {\ddot {\mathbf {P} }}=\mathbf {\hat {\rho }} \left({\ddot {\rho }}-\rho {\dot {\phi }}^{2}\right)+{\boldsymbol {\hat {\phi }}}\left(\rho {\ddot {\phi }}+2{\dot {\rho }}{\dot {\phi }}\right)+\mathbf {\hat {z}} {\ddot {z}}}
In mechanics, the terms of this expression are called:
ρ
¨
ρ
^
=
central outward acceleration
−
ρ
ϕ
˙
2
ρ
^
=
centripetal acceleration
ρ
ϕ
¨
ϕ
^
=
angular acceleration
2
ρ
˙
ϕ
˙
ϕ
^
=
Coriolis effect
z
¨
z
^
=
z-acceleration
{\displaystyle {\begin{aligned}{\ddot {\rho }}\mathbf {\hat {\rho }} &={\text{central outward acceleration}}\\-\rho {\dot {\phi }}^{2}\mathbf {\hat {\rho }} &={\text{centripetal acceleration}}\\\rho {\ddot {\phi }}{\boldsymbol {\hat {\phi }}}&={\text{angular acceleration}}\\2{\dot {\rho }}{\dot {\phi }}{\boldsymbol {\hat {\phi }}}&={\text{Coriolis effect}}\\{\ddot {z}}\mathbf {\hat {z}} &={\text{z-acceleration}}\end{aligned}}}
Spherical coordinate system
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Vector fields
edit
Vectors are defined in spherical coordinates by (r , θ , φ ), where
r is the length of the vector,θ is the angle between the positive Z-axis and the vector in question (0 ≤ θ ≤ π ), andφ is the angle between the projection of the vector onto the xy -plane and the positive X-axis (0 ≤ φ < 2π ).(r , θ , φ ) is given in Cartesian coordinates by:
[
r
θ
ϕ
]
=
[
x
2
+
y
2
+
z
2
arccos
(
z
/
x
2
+
y
2
+
z
2
)
arctan
(
y
/
x
)
]
,
0
≤
θ
≤
π
,
0
≤
ϕ
<
2
π
,
{\displaystyle {\begin{bmatrix}r\\\theta \\\phi \end{bmatrix}}={\begin{bmatrix}{\sqrt {x^{2}+y^{2}+z^{2}}}\\\arccos(z/{\sqrt {x^{2}+y^{2}+z^{2}}})\\\arctan(y/x)\end{bmatrix}},\ \ \ 0\leq \theta \leq \pi ,\ \ \ 0\leq \phi <2\pi ,}
[
x
y
z
]
=
[
r
sin
θ
cos
ϕ
r
sin
θ
sin
ϕ
r
cos
θ
]
.
{\displaystyle {\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}r\sin \theta \cos \phi \\r\sin \theta \sin \phi \\r\cos \theta \end{bmatrix}}.}
Any vector field can be written in terms of the unit vectors as:
A
=
A
x
x
^
+
A
y
y
^
+
A
z
z
^
=
A
r
r
^
+
A
θ
θ
^
+
A
ϕ
ϕ
^
{\displaystyle \mathbf {A} =A_{x}\mathbf {\hat {x}} +A_{y}\mathbf {\hat {y}} +A_{z}\mathbf {\hat {z}} =A_{r}{\boldsymbol {\hat {r}}}+A_{\theta }{\boldsymbol {\hat {\theta }}}+A_{\phi }{\boldsymbol {\hat {\phi }}}}
[
r
^
θ
^
ϕ
^
]
=
[
sin
θ
cos
ϕ
sin
θ
sin
ϕ
cos
θ
cos
θ
cos
ϕ
cos
θ
sin
ϕ
−
sin
θ
−
sin
ϕ
cos
ϕ
0
]
[
x
^
y
^
z
^
]
{\displaystyle {\begin{bmatrix}{\boldsymbol {\hat {r}}}\\{\boldsymbol {\hat {\theta }}}\\{\boldsymbol {\hat {\phi }}}\end{bmatrix}}={\begin{bmatrix}\sin \theta \cos \phi &\sin \theta \sin \phi &\cos \theta \\\cos \theta \cos \phi &\cos \theta \sin \phi &-\sin \theta \\-\sin \phi &\cos \phi &0\end{bmatrix}}{\begin{bmatrix}\mathbf {\hat {x}} \\\mathbf {\hat {y}} \\\mathbf {\hat {z}} \end{bmatrix}}}
Note: the matrix is an orthogonal matrix , that is, its inverse is simply its transpose .
The Cartesian unit vectors are thus related to the spherical unit vectors by:
[
x
^
y
^
z
^
]
=
[
sin
θ
cos
ϕ
cos
θ
cos
ϕ
−
sin
ϕ
sin
θ
sin
ϕ
cos
θ
sin
ϕ
cos
ϕ
cos
θ
−
sin
θ
0
]
[
r
^
θ
^
ϕ
^
]
{\displaystyle {\begin{bmatrix}\mathbf {\hat {x}} \\\mathbf {\hat {y}} \\\mathbf {\hat {z}} \end{bmatrix}}={\begin{bmatrix}\sin \theta \cos \phi &\cos \theta \cos \phi &-\sin \phi \\\sin \theta \sin \phi &\cos \theta \sin \phi &\cos \phi \\\cos \theta &-\sin \theta &0\end{bmatrix}}{\begin{bmatrix}{\boldsymbol {\hat {r}}}\\{\boldsymbol {\hat {\theta }}}\\{\boldsymbol {\hat {\phi }}}\end{bmatrix}}}
Time derivative of a vector field
edit
To find out how the vector field A changes in time, the time derivatives should be calculated.
In Cartesian coordinates this is simply:
A
˙
=
A
˙
x
x
^
+
A
˙
y
y
^
+
A
˙
z
z
^
{\displaystyle \mathbf {\dot {A}} ={\dot {A}}_{x}\mathbf {\hat {x}} +{\dot {A}}_{y}\mathbf {\hat {y}} +{\dot {A}}_{z}\mathbf {\hat {z}} }
However, in spherical coordinates this becomes:
A
˙
=
A
˙
r
r
^
+
A
r
r
^
˙
+
A
˙
θ
θ
^
+
A
θ
θ
^
˙
+
A
˙
ϕ
ϕ
^
+
A
ϕ
ϕ
^
˙
{\displaystyle \mathbf {\dot {A}} ={\dot {A}}_{r}{\boldsymbol {\hat {r}}}+A_{r}{\boldsymbol {\dot {\hat {r}}}}+{\dot {A}}_{\theta }{\boldsymbol {\hat {\theta }}}+A_{\theta }{\boldsymbol {\dot {\hat {\theta }}}}+{\dot {A}}_{\phi }{\boldsymbol {\hat {\phi }}}+A_{\phi }{\boldsymbol {\dot {\hat {\phi }}}}}
The time derivatives of the unit vectors are needed. They are given by:
r
^
˙
=
θ
˙
θ
^
+
ϕ
˙
sin
θ
ϕ
^
θ
^
˙
=
−
θ
˙
r
^
+
ϕ
˙
cos
θ
ϕ
^
ϕ
^
˙
=
−
ϕ
˙
sin
θ
r
^
−
ϕ
˙
cos
θ
θ
^
{\displaystyle {\begin{aligned}{\boldsymbol {\dot {\hat {r}}}}&={\dot {\theta }}{\boldsymbol {\hat {\theta }}}+{\dot {\phi }}\sin \theta {\boldsymbol {\hat {\phi }}}\\{\boldsymbol {\dot {\hat {\theta }}}}&=-{\dot {\theta }}{\boldsymbol {\hat {r}}}+{\dot {\phi }}\cos \theta {\boldsymbol {\hat {\phi }}}\\{\boldsymbol {\dot {\hat {\phi }}}}&=-{\dot {\phi }}\sin \theta {\boldsymbol {\hat {r}}}-{\dot {\phi }}\cos \theta {\boldsymbol {\hat {\theta }}}\end{aligned}}}
Thus the time derivative becomes:
A
˙
=
r
^
(
A
˙
r
−
A
θ
θ
˙
−
A
ϕ
ϕ
˙
sin
θ
)
+
θ
^
(
A
˙
θ
+
A
r
θ
˙
−
A
ϕ
ϕ
˙
cos
θ
)
+
ϕ
^
(
A
˙
ϕ
+
A
r
ϕ
˙
sin
θ
+
A
θ
ϕ
˙
cos
θ
)
{\displaystyle \mathbf {\dot {A}} ={\boldsymbol {\hat {r}}}\left({\dot {A}}_{r}-A_{\theta }{\dot {\theta }}-A_{\phi }{\dot {\phi }}\sin \theta \right)+{\boldsymbol {\hat {\theta }}}\left({\dot {A}}_{\theta }+A_{r}{\dot {\theta }}-A_{\phi }{\dot {\phi }}\cos \theta \right)+{\boldsymbol {\hat {\phi }}}\left({\dot {A}}_{\phi }+A_{r}{\dot {\phi }}\sin \theta +A_{\theta }{\dot {\phi }}\cos \theta \right)}
See also
edit
References
edit
^ Wolfram Mathworld, spherical coordinates
