Vector fields

Vector fields in cylindrical and spherical coordinates

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 X-Y-plane,
  • φ is the angle of the projected vector with the positive X-axis (0 ≤ φ < 2π),
  • z is the regular z-coordinate.

(ρ,φ,z) is given in cartesian coordinates by:

begin{bmatrix}rho phi z end{bmatrix} =
begin{bmatrix} sqrt{x^2 + y^2} operatorname{arctan}(y / x) z end{bmatrix}, 0 le phi < 2pi,

or inversely by:

begin{bmatrix} x y z end{bmatrix} =
begin{bmatrix} rhocosphi rhosinphi z end{bmatrix}.

Any vector field can be written in terms of the unit vectors as:

mathbf A = A_x mathbf{hat x} + A_y mathbf{hat y} + A_z mathbf{hat z}
= A_rho boldsymbol{hat rho} + A_phi boldsymbol{hat phi} + A_z boldsymbol{hat z} The cylindrical unit vectors are related to the cartesian unit vectors by:
begin{bmatrix}boldsymbol{hatrho} boldsymbol{hatphi} boldsymbol{hat z}end{bmatrix}
= begin{bmatrix} cosphi & sinphi & 0
                  -sinphi & cosphi & 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 (mathbf{dot A}). In cartesian coordinates this is simply:

mathbf{dot A} = dot A_x mathbf{hat x} + dot A_y mathbf{hat y} + dot A_z mathbf{hat z}

However, in cylindrical coordinates this becomes:

mathbf{dot A} = dot A_rho boldsymbol{hatrho} + A_rho boldsymbol{dot{hatrho}}
+ dot A_phi boldsymbol{hatphi} + A_phi boldsymbol{dot{hatphi}} + dot A_z boldsymbol{hat z} + A_z boldsymbol{dot{hat z}}

We need the time derivatives of the unit vectors. They are given by:

begin{align}
boldsymbol{dot{hatrho}} &= dotphi boldsymbol{hatphi} boldsymbol{dot{hatphi}} &= - dotphi boldsymbol{hatrho} boldsymbol{dot{hat z}} &= 0 end{align}

So the time derivative simplifies to:

mathbf{dot A} = boldsymbol{hatrho} (dot A_rho - A_phi dotphi)
+ boldsymbol{hatphi} (dot A_phi + A_rho dotphi) + boldsymbol{hat z} dot A_z

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 with the positive Z-axis (0 <= θ <= π),
  • φ is the angle with the X-Z-plane (0 <= φ < 2π).

(r,θ,φ) is given in cartesian coordinates by:

begin{align}
r &= sqrt{x^2 + y^2 + z^2}
   theta &= arccosleft(z / rright), & 0 le theta le pi 
phi &= operatorname{arctan}(y / x), & 0 le phi < 2pi, end{align}

or inversely by:

begin{align}
   x &= rsinthetacosphi 
   y &= rsinthetasinphi 
z &= rcostheta. end{align}

Any vector field can be written in terms of the unit vectors as:

mathbf A = A_xmathbf{hat x} + A_ymathbf{hat y} + A_zmathbf{hat z}
= A_rboldsymbol{hat r} + A_thetaboldsymbol{hat theta} + A_phiboldsymbol{hat phi} The spherical unit vectors are related to the cartesian unit vectors by:
begin{bmatrix}boldsymbol{hat r} boldsymbol{hattheta} boldsymbol{hatphi} end{bmatrix}
= begin{bmatrix} sinthetacosphi & sinthetasinphi & costheta
                   costhetacosphi & costhetasinphi & -sintheta 
-sinphi & cosphi & 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:

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:

mathbf{dot A} = dot A_r boldsymbol{hat r} + A_r boldsymbol{dot{hat r}}
+ dot A_theta boldsymbol{hattheta} + A_theta boldsymbol{dot{hattheta}} + dot A_phi boldsymbol{hatphi} + A_phi boldsymbol{dot{hatphi}}

We need the time derivatives of the unit vectors. They are given by:

begin{bmatrix}boldsymbol{dot{hat r}} boldsymbol{dot{hattheta}} boldsymbol{dot{hatphi}} end{bmatrix}
= begin{bmatrix} 0 & dottheta & dotphi sintheta
                   -dottheta & 0          & dotphi costheta 
-dotphi sintheta & -dotphi costheta & 0 end{bmatrix} begin{bmatrix} boldsymbol{hat r} boldsymbol{hattheta} boldsymbol{hatphi} end{bmatrix}

So the time derivative becomes:

mathbf{dot A} = boldsymbol{hat r} (dot A_r - A_theta dottheta - A_phi dotphi sintheta)
+ boldsymbol{hattheta} (dot A_theta + A_r dottheta - A_phi dotphi costheta) + boldsymbol{hatphi} (dot A_phi + A_r dotphi sintheta + A_theta dotphi costheta)

See also

Search another word or see Vector fieldson Dictionary | Thesaurus |Spanish
Copyright © 2014 Dictionary.com, LLC. All rights reserved.
  • Please Login or Sign Up to use the Recent Searches feature