NOTE: This page uses common physics notation for spherical coordinates, in which
θ
is the angle between the z axis and the radius vector connecting the origin to the point in question, while
ϕ
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.
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
,
or inversely by:
[
x
y
z
]
=
[
r
cos
θ
r
sin
θ
z
]
.
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
^
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
^
]
Note: the matrix is an orthogonal matrix, that is, its inverse is simply its transpose.
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
˙
). In cartesian coordinates this is simply:
A
˙
=
A
˙
x
x
^
+
A
˙
y
y
^
+
A
˙
z
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
^
˙
We need the time derivatives of the unit vectors. They are given by:
r
^
˙
=
θ
˙
θ
^
θ
^
˙
=
−
θ
˙
r
^
z
^
˙
=
0
So the time derivative simplifies to:
A
˙
=
r
^
(
A
˙
r
−
A
θ
θ
˙
)
+
θ
^
(
A
˙
θ
+
A
r
θ
˙
)
+
z
^
A
˙
z
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
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
^
.
After substituting we get:
P
¨
=
r
^
(
r
¨
−
r
θ
˙
2
)
+
θ
^
(
r
θ
¨
+
2
r
˙
θ
˙
)
+
z
^
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
See also: Centripetal force, Angular acceleration, Coriolis effect.
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 the vector in question (0 ≤ θ ≤ π), and
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
,
or inversely by:
[
x
y
z
]
=
[
ρ
sin
θ
cos
ϕ
ρ
sin
θ
sin
ϕ
ρ
cos
θ
]
.
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
^
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
^
]
Note: the matrix is an orthogonal matrix, that is, its inverse is simply its transpose.
So the cartesian unit vectors are 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
]
[
ρ
^
θ
^
ϕ
^
]
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
^
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
^
˙
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
^
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
)
