Vector fields in cylindrical and spherical coordinates

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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 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.[1]

Cylindrical coordinate system

Vector fields

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:

[ρϕz]=[x2+y2arctan(y/x)z],   0ϕ<2π,

or inversely by: [xyz]=[ρcosϕρsinϕz].

Any vector field can be written in terms of the unit vectors as: 𝐀=Ax𝐱^+Ay𝐲^+Az𝐳^=Aρρ^+Aϕϕ^+Az𝐳^ The cylindrical unit vectors are related to the Cartesian unit vectors by: [ρ^ϕ^𝐳^]=[cosϕsinϕ0sinϕcosϕ0001][𝐱^𝐲^𝐳^]

Note: the matrix is an orthogonal matrix, that is, its inverse is simply its transpose.

Time derivative of a vector field

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˙). In Cartesian coordinates this is simply: 𝐀˙=A˙x𝐱^+A˙y𝐲^+A˙z𝐳^

However, in cylindrical coordinates this becomes: 𝐀˙=A˙ρρ^+Aρρ^˙+A˙ϕϕ^+Aϕϕ^˙+A˙z𝒛^+Az𝒛^˙

The time derivatives of the unit vectors are needed. They are given by: ρ^˙=ϕ˙ϕ^ϕ^˙=ϕ˙ρ^𝐳^˙=0

So the time derivative simplifies to: 𝐀˙=ρ^(A˙ρAϕϕ˙)+ϕ^(A˙ϕ+Aρϕ˙)+𝐳^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¨=ρ^(A¨ρAϕϕ¨2A˙ϕϕ˙Aρϕ˙2)+ϕ^(A¨ϕ+Aρϕ¨+2A˙ρϕ˙Aϕϕ˙2)+𝐳^A¨z

To understand this expression, A is substituted for P, where P is the vector (ρ, φ, z).

This means that 𝐀=𝐏=ρρ^+z𝐳^.

After substituting, the result is given: 𝐏¨=ρ^(ρ¨ρϕ˙2)+ϕ^(ρϕ¨+2ρ˙ϕ˙)+𝐳^z¨

In mechanics, the terms of this expression are called:

ρ¨ρ^ central outward acceleration
ρϕ˙2ρ^ centripetal acceleration
ρϕ¨ϕ^ angular acceleration
2ρ˙ϕ˙ϕ^ Coriolis effect
z¨𝐳^ Template:Mvar-acceleration

Template:See also

Spherical coordinate system

Vector fields

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 ≤ θπ), and
  • φ 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: [rθϕ]=[x2+y2+z2arccos(z/x2+y2+z2)arctan(y/x)],   0θπ,   0ϕ<2π, or inversely by: [xyz]=[rsinθcosϕrsinθsinϕrcosθ].

Any vector field can be written in terms of the unit vectors as: 𝐀=Ax𝐱^+Ay𝐲^+Az𝐳^=Ar𝒓^+Aθθ^+Aϕϕ^

The spherical basis vectors are related to the Cartesian basis vectors by the Jacobian matrix:

[𝒓^θ^ϕ^]=[xryrzrxθyθzθxϕyϕzϕ][𝐱^𝐲^𝐳^]

Normalizing the Jacobian matrix so that the spherical basis vectors have unit length we get:

[𝒓^θ^ϕ^]=[sinθcosϕsinθsinϕcosθcosθcosϕcosθsinϕsinθsinϕcosϕ0][𝐱^𝐲^𝐳^]

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^]=[sinθcosϕcosθcosϕsinϕsinθsinϕcosθsinϕcosϕcosθsinθ0][𝒓^θ^ϕ^]

Time derivative of a vector field

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˙x𝐱^+A˙y𝐲^+A˙z𝐳^ However, in spherical coordinates this becomes: 𝐀˙=A˙r𝒓^+Ar𝒓^˙+A˙θθ^+Aθθ^˙+A˙ϕϕ^+Aϕϕ^˙ The time derivatives of the unit vectors are needed. They are given by: 𝒓^˙=θ˙θ^+ϕ˙sinθϕ^θ^˙=θ˙𝒓^+ϕ˙cosθϕ^ϕ^˙=ϕ˙sinθ𝒓^ϕ˙cosθθ^ Thus the time derivative becomes: 𝐀˙=𝒓^(A˙rAθθ˙Aϕϕ˙sinθ)+θ^(A˙θ+Arθ˙Aϕϕ˙cosθ)+ϕ^(A˙ϕ+Arϕ˙sinθ+Aθϕ˙cosθ)

See also

References