Vector fields in cylindrical and spherical coordinates

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:

:egin{bmatrix} ho \ phi \ z end{bmatrix} = egin{bmatrix}sqrt{x^2 + y^2} \ operatorname{arctan}(y / x) \ zend{bmatrix}, 0 le phi < 2pi,

or inversely by:

:egin{bmatrix} x \ y \ z end{bmatrix} =egin{bmatrix} hocosphi \ hosinphi \ 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_ ho oldsymbol{hat ho} + A_phi oldsymbol{hat phi} + A_z oldsymbol{hat z}The cylindrical unit vectors are related to the cartesian unit vectors by::egin{bmatrix}oldsymbol{hat ho} \ oldsymbol{hatphi} \ oldsymbol{hat z}end{bmatrix} = egin{bmatrix} cosphi & sinphi & 0 \ -sinphi & cosphi & 0 \ 0 & 0 & 1 end{bmatrix} egin{bmatrix} mathbf{hat x} \ mathbf{hat y} \ mathbf{hat z} end{bmatrix}
* 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 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_ ho oldsymbol{hat ho} + A_ ho oldsymbol{dot{hat ho + dot A_phi oldsymbol{hatphi} + A_phi oldsymbol{dot{hatphi + dot A_z oldsymbol{hat z} + A_z oldsymbol{dot{hat z

We need the time derivatives of the unit vectors. They are given by::egin{align} oldsymbol{dot{hat ho &= dotphi oldsymbol{hatphi} \ oldsymbol{dot{hatphi &= - dotphi oldsymbol{hat ho} \ oldsymbol{dot{hat z &= 0 end{align}

So the time derivative simplifies to::mathbf{dot A} = oldsymbol{hat ho} (dot A_ ho - A_phi dotphi) + oldsymbol{hatphi} (dot A_phi + A_ ho dotphi) + oldsymbol{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:

:egin{align} r &= sqrt{x^2 + y^2 + z^2} \ heta &= arccosleft( z / r ight), & 0 le heta le pi \ phi &= operatorname{arctan}(y / x), & 0 le phi < 2pi, end{align}

or inversely by:

:egin{align} x &= rsin hetacosphi \ y &= rsin hetasinphi \ z &= rcos heta. 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_roldsymbol{hat r} + A_ hetaoldsymbol{hat heta} + A_phioldsymbol{hat phi}The spherical unit vectors are related to the cartesian unit vectors by::egin{bmatrix}oldsymbol{hat r} \ oldsymbol{hat heta} \ oldsymbol{hatphi} end{bmatrix} = egin{bmatrix} sin hetacosphi & sin hetasinphi & cos heta \ cos hetacosphi & cos hetasinphi & -sin heta \ -sinphi & cosphi & 0 end{bmatrix} egin{bmatrix} mathbf{hat x} \ mathbf{hat y} \ mathbf{hat z} end{bmatrix}
* 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 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 oldsymbol{hat r} + A_r oldsymbol{dot{hat r + dot A_ heta oldsymbol{hat heta} + A_ heta oldsymbol{dot{hat heta + dot A_phi oldsymbol{hatphi} + A_phi oldsymbol{dot{hatphi

We need the time derivatives of the unit vectors. They are given by::egin{bmatrix}oldsymbol{dot{hat r \ oldsymbol{dot{hat heta \ oldsymbol{dot{hatphi end{bmatrix} = egin{bmatrix} 0 & dot heta & dotphi sin heta \ -dot heta & 0 & dotphi cos heta \ -dotphi sin heta & -dotphi cos heta & 0 end{bmatrix} egin{bmatrix} oldsymbol{hat r} \ oldsymbol{hat heta} \ oldsymbol{hatphi} end{bmatrix}

So the time derivative becomes::mathbf{dot A} = oldsymbol{hat r} (dot A_r - A_ heta dot heta - A_phi dotphi sin heta) + oldsymbol{hat heta} (dot A_ heta + A_r dot heta - A_phi dotphi cos heta) + oldsymbol{hatphi} (dot A_phi + A_r dotphi sin heta + A_ heta dotphi cos heta)

See also

* Del in cylindrical and spherical coordinates for the specification of gradient, divergence, curl, and laplacian in various coordinate systems.


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