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# 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:

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

or inversely by:

$begin\left\{bmatrix\right\} x y z end\left\{bmatrix\right\} =$
begin{bmatrix} rhocosphi rhosinphi z end{bmatrix}.

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

$mathbf A = A_x mathbf\left\{hat x\right\} + A_y mathbf\left\{hat y\right\} + A_z mathbf\left\{hat z\right\}$
= 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:
$begin\left\{bmatrix\right\}boldsymbol\left\{hatrho\right\} boldsymbol\left\{hatphi\right\} boldsymbol\left\{hat z\right\}end\left\{bmatrix\right\}$
= 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 ($mathbf\left\{dot A\right\}$). In cartesian coordinates this is simply:

$mathbf\left\{dot A\right\} = dot A_x mathbf\left\{hat x\right\} + dot A_y mathbf\left\{hat y\right\} + dot A_z mathbf\left\{hat z\right\}$

However, in cylindrical coordinates this becomes:

$mathbf\left\{dot A\right\} = dot A_rho boldsymbol\left\{hatrho\right\} + A_rho boldsymbol\left\{dot\left\{hatrho\right\}\right\}$
+ 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\left\{align\right\}
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:

$mathbf\left\{dot A\right\} = boldsymbol\left\{hatrho\right\} \left(dot A_rho - A_phi dotphi\right)$
+ 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\left\{align\right\}
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\left\{align\right\}
`   x &= rsinthetacosphi `
`   y &= rsinthetasinphi `
z &= rcostheta. end{align}

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

$mathbf A = A_xmathbf\left\{hat x\right\} + A_ymathbf\left\{hat y\right\} + A_zmathbf\left\{hat z\right\}$
= A_rboldsymbol{hat r} + A_thetaboldsymbol{hat theta} + A_phiboldsymbol{hat phi} The spherical unit vectors are related to the cartesian unit vectors by:
$begin\left\{bmatrix\right\}boldsymbol\left\{hat r\right\} boldsymbol\left\{hattheta\right\} boldsymbol\left\{hatphi\right\} end\left\{bmatrix\right\}$
= 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:

$mathbf\left\{dot A\right\} = dot A_x mathbf\left\{hat x\right\} + dot A_y mathbf\left\{hat y\right\} + dot A_z mathbf\left\{hat z\right\}$

However, in spherical coordinates this becomes:

$mathbf\left\{dot A\right\} = dot A_r boldsymbol\left\{hat r\right\} + A_r boldsymbol\left\{dot\left\{hat r\right\}\right\}$
+ 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:

$begin\left\{bmatrix\right\}boldsymbol\left\{dot\left\{hat r\right\}\right\} boldsymbol\left\{dot\left\{hattheta\right\}\right\} boldsymbol\left\{dot\left\{hatphi\right\}\right\} end\left\{bmatrix\right\}$
= 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:

$mathbf\left\{dot A\right\} = boldsymbol\left\{hat r\right\} \left(dot A_r - A_theta dottheta - A_phi dotphi sintheta\right)$
+ 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)