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 .
NOTE: This page uses common physics notation for spherical coordinates, in which
θ
{\displaystyle \theta }
is the angle between the z axis and the radius vector connecting the origin to the point in question, while
ϕ
{\displaystyle \phi }
is the angle between the projection of the radius vector onto the x-y plane and the x axis. Several other definitions are in use, and so care must be taken in comparing different sources.[ 1]
Cylindrical coordinate system
Vector fields
Vectors are defined in cylindrical coordinates by (r , θ, z ), where
r is the length of the vector projected onto the xy -plane,
θ is the angle between the projection of the vector onto the xy -plane (i.e. r ) and the positive x -axis (0 ≤ θ < 2π),
z is the regular z -coordinate.
(r , θ, z ) is given in cartesian coordinates by:
[
r
θ
z
]
=
[
x
2
+
y
2
arctan
(
y
/
x
)
z
]
,
0
≤
θ
<
2
π
,
{\displaystyle {\begin{bmatrix}r\\\theta \\z\end{bmatrix}}={\begin{bmatrix}{\sqrt {x^{2}+y^{2}}}\\\operatorname {arctan} (y/x)\\z\end{bmatrix}},\ \ \ 0\leq \theta <2\pi ,}
or inversely by:
[
x
y
z
]
=
[
r
cos
θ
r
sin
θ
z
]
.
{\displaystyle {\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}r\cos \theta \\r\sin \theta \\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
r
r
^
+
A
θ
θ
^
+
A
z
z
^
{\displaystyle \mathbf {A} =A_{x}\mathbf {\hat {x}} +A_{y}\mathbf {\hat {y}} +A_{z}\mathbf {\hat {z}} =A_{r}\mathbf {\hat {r}} +A_{\theta }{\boldsymbol {\hat {\theta }}}+A_{z}\mathbf {\hat {z}} }
The cylindrical unit vectors are related to the cartesian unit vectors by:
[
r
^
θ
^
z
^
]
=
[
cos
θ
sin
θ
0
−
sin
θ
cos
θ
0
0
0
1
]
[
x
^
y
^
z
^
]
{\displaystyle {\begin{bmatrix}\mathbf {\hat {r}} \\{\boldsymbol {\hat {\theta }}}\\\mathbf {\hat {z}} \end{bmatrix}}={\begin{bmatrix}\cos \theta &\sin \theta &0\\-\sin \theta &\cos \theta &0\\0&0&1\end{bmatrix}}{\begin{bmatrix}\mathbf {\hat {x}} \\\mathbf {\hat {y}} \\\mathbf {\hat {z}} \end{bmatrix}}}
Time derivative of a vector field
To find out how the vector field A changes in time we calculate the time derivatives.
For this purpose we use Newton's notation for the time derivative (
A
˙
{\displaystyle {\dot {\mathbf {A} }}}
).
In cartesian coordinates this is simply:
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
˙
r
r
^
+
A
r
r
^
˙
+
A
˙
θ
θ
^
+
A
θ
θ
^
˙
+
A
˙
z
z
^
+
A
z
z
^
˙
{\displaystyle {\dot {\mathbf {A} }}={\dot {A}}_{r}{\hat {\boldsymbol {r}}}+A_{r}{\dot {\hat {\boldsymbol {r}}}}+{\dot {A}}_{\theta }{\hat {\boldsymbol {\theta }}}+A_{\theta }{\dot {\hat {\boldsymbol {\theta }}}}+{\dot {A}}_{z}{\hat {\boldsymbol {z}}}+A_{z}{\dot {\hat {\boldsymbol {z}}}}}
We need the time derivatives of the unit vectors.
They are given by:
r
^
˙
=
θ
˙
θ
^
θ
^
˙
=
−
θ
˙
r
^
z
^
˙
=
0
{\displaystyle {\begin{aligned}{\dot {\hat {\mathbf {r} }}}&={\dot {\theta }}{\hat {\boldsymbol {\theta }}}\\{\dot {\hat {\boldsymbol {\theta }}}}&=-{\dot {\theta }}{\hat {\mathbf {r} }}\\{\dot {\hat {\mathbf {z} }}}&=0\end{aligned}}}
So the time derivative simplifies to:
A
˙
=
r
^
(
A
˙
r
−
A
θ
θ
˙
)
+
θ
^
(
A
˙
θ
+
A
r
θ
˙
)
+
z
^
A
˙
z
{\displaystyle {\dot {\mathbf {A} }}={\hat {\boldsymbol {r}}}({\dot {A}}_{r}-A_{\theta }{\dot {\theta }})+{\hat {\boldsymbol {\theta }}}({\dot {A}}_{\theta }+A_{r}{\dot {\theta }})+{\hat {\mathbf {z} }}{\dot {A}}_{z}}
Second time derivative of a vector field
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
¨
=
r
^
(
A
¨
r
−
A
θ
θ
¨
−
2
A
˙
θ
θ
˙
−
A
r
θ
˙
2
)
+
θ
^
(
A
¨
θ
+
A
r
θ
¨
+
2
A
˙
r
θ
˙
−
A
θ
θ
˙
2
)
+
z
^
A
¨
z
{\displaystyle \mathbf {\ddot {A}} =\mathbf {\hat {r}} ({\ddot {A}}_{r}-A_{\theta }{\ddot {\theta }}-2{\dot {A}}_{\theta }{\dot {\theta }}-A_{r}{\dot {\theta }}^{2})+{\boldsymbol {\hat {\theta }}}({\ddot {A}}_{\theta }+A_{r}{\ddot {\theta }}+2{\dot {A}}_{r}{\dot {\theta }}-A_{\theta }{\dot {\theta }}^{2})+\mathbf {\hat {z}} {\ddot {A}}_{z}}
To understand this expression, we substitute A = P, where p is the vector (r, θ, z).
This means that
A
=
P
=
r
r
^
+
z
z
^
{\displaystyle \mathbf {A} =\mathbf {P} =r\mathbf {\hat {r}} +z\mathbf {\hat {z}} }
.
After substituting we get:
P
¨
=
r
^
(
r
¨
−
r
θ
˙
2
)
+
θ
^
(
r
θ
¨
+
2
r
˙
θ
˙
)
+
z
^
z
¨
{\displaystyle {\ddot {\mathbf {P} }}=\mathbf {\hat {r}} ({\ddot {r}}-r{\dot {\theta }}^{2})+{\boldsymbol {\hat {\theta }}}(r{\ddot {\theta }}+2{\dot {r}}{\dot {\theta }})+\mathbf {\hat {z}} {\ddot {z}}}
In mechanics, the terms of this expression are called:
r
¨
r
^
=
central outward acceleration
−
r
θ
˙
2
r
^
=
centripetal acceleration
r
θ
¨
θ
^
=
angular acceleration
2
r
˙
θ
˙
θ
^
=
Coriolis effect
z
¨
z
^
=
z-acceleration
{\displaystyle {\begin{aligned}{\ddot {r}}\mathbf {\hat {r}} &={\mbox{central outward acceleration}}\\-r{\dot {\theta }}^{2}\mathbf {\hat {r}} &={\mbox{centripetal acceleration}}\\r{\ddot {\theta }}{\boldsymbol {\hat {\theta }}}&={\mbox{angular acceleration}}\\2{\dot {r}}{\dot {\theta }}{\boldsymbol {\hat {\theta }}}&={\mbox{Coriolis effect}}\\{\ddot {z}}\mathbf {\hat {z}} &={\mbox{z-acceleration}}\end{aligned}}}
See also: Centripetal force , Angular acceleration , Coriolis effect .
Spherical coordinate system
Vector fields
Vectors are defined in spherical coordinates by (ρ,θ,φ), where
ρ is the length of the vector,
θ is the angle between the positive Z-axis and vector in question (0 ≤ θ ≤ π)
φ is the angle between the projection of the vector onto the X-Y-plane and the positive X-axis (0 ≤ φ < 2π),
(ρ,θ,φ) is given in cartesian coordinates by:
[
ρ
θ
ϕ
]
=
[
x
2
+
y
2
+
z
2
arccos
(
z
/
ρ
)
arctan
(
y
/
x
)
]
,
0
≤
θ
≤
π
,
0
≤
ϕ
<
2
π
,
{\displaystyle {\begin{bmatrix}\rho \\\theta \\\phi \end{bmatrix}}={\begin{bmatrix}{\sqrt {x^{2}+y^{2}+z^{2}}}\\\arccos(z/\rho )\\\arctan(y/x)\end{bmatrix}},\ \ \ 0\leq \theta \leq \pi ,\ \ \ 0\leq \phi <2\pi ,}
or inversely by:
[
x
y
z
]
=
[
ρ
sin
θ
cos
ϕ
ρ
sin
θ
sin
ϕ
ρ
cos
θ
]
.
{\displaystyle {\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}\rho \sin \theta \cos \phi \\\rho \sin \theta \sin \phi \\\rho \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
ρ
ρ
^
+
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_{\rho }{\boldsymbol {\hat {\rho }}}+A_{\theta }{\boldsymbol {\hat {\theta }}}+A_{\phi }{\boldsymbol {\hat {\phi }}}}
The spherical unit vectors are related to the cartesian unit vectors by:
[
ρ
^
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 {\rho }}}\\{\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}}}
Time derivative of a vector field
To find out how the vector field A changes in time we calculate the time derivatives.
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
˙
ρ
ρ
^
+
A
ρ
ρ
^
˙
+
A
˙
θ
θ
^
+
A
θ
θ
^
˙
+
A
˙
ϕ
ϕ
^
+
A
ϕ
ϕ
^
˙
{\displaystyle \mathbf {\dot {A}} ={\dot {A}}_{\rho }{\boldsymbol {\hat {\rho }}}+A_{\rho }{\boldsymbol {\dot {\hat {\rho }}}}+{\dot {A}}_{\theta }{\boldsymbol {\hat {\theta }}}+A_{\theta }{\boldsymbol {\dot {\hat {\theta }}}}+{\dot {A}}_{\phi }{\boldsymbol {\hat {\phi }}}+A_{\phi }{\boldsymbol {\dot {\hat {\phi }}}}}
We need the time derivatives of the unit vectors.
They are given by:
ρ
^
˙
=
θ
˙
θ
^
+
ϕ
˙
sin
θ
ϕ
^
θ
^
˙
=
−
θ
˙
ρ
^
+
ϕ
˙
cos
θ
ϕ
^
ϕ
^
˙
=
−
ϕ
˙
sin
θ
ρ
^
−
ϕ
˙
cos
θ
θ
^
{\displaystyle {\begin{aligned}{\boldsymbol {\dot {\hat {\rho }}}}&={\dot {\theta }}{\boldsymbol {\hat {\theta }}}+{\dot {\phi }}\sin \theta {\boldsymbol {\hat {\phi }}}\\{\boldsymbol {\dot {\hat {\theta }}}}&=-{\dot {\theta }}{\boldsymbol {\hat {\rho }}}+{\dot {\phi }}\cos \theta {\boldsymbol {\hat {\phi }}}\\{\boldsymbol {\dot {\hat {\phi }}}}&=-{\dot {\phi }}\sin \theta {\boldsymbol {\hat {\rho }}}-{\dot {\phi }}\cos \theta {\boldsymbol {\hat {\theta }}}\end{aligned}}}
So the time derivative becomes:
A
˙
=
ρ
^
(
A
˙
ρ
−
A
θ
θ
˙
−
A
ϕ
ϕ
˙
sin
θ
)
+
θ
^
(
A
˙
θ
+
A
ρ
θ
˙
−
A
ϕ
ϕ
˙
cos
θ
)
+
ϕ
^
(
A
˙
ϕ
+
A
ρ
ϕ
˙
sin
θ
+
A
θ
ϕ
˙
cos
θ
)
{\displaystyle \mathbf {\dot {A}} ={\boldsymbol {\hat {\rho }}}({\dot {A}}_{\rho }-A_{\theta }{\dot {\theta }}-A_{\phi }{\dot {\phi }}\sin \theta )+{\boldsymbol {\hat {\theta }}}({\dot {A}}_{\theta }+A_{\rho }{\dot {\theta }}-A_{\phi }{\dot {\phi }}\cos \theta )+{\boldsymbol {\hat {\phi }}}({\dot {A}}_{\phi }+A_{\rho }{\dot {\phi }}\sin \theta +A_{\theta }{\dot {\phi }}\cos \theta )}
See also
References