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), andφ 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 ) 
