Plotting vector field in polar coordinates

Question

Is there an easy way(other than painfully decomposing into Cartesian components) to plot a vector field in polar coordinates? Basically I have a vector function $\vec V (r,\phi,z)$. How do I plot it ?

Answer 1

You can use CoordinateTransform to transform between different coordinate systems. Beware that you might have to use the differential of certain coordinate transformation in order to transform the vector components of the vector field appropriately

Φ = {x, y, z} \[Function] Evaluate[CoordinateTransform["Cartesian" -> "Cylindrical", {x, y, z}]];
Ψ = {r, ϕ, z} \[Function] Evaluate[CoordinateTransform["Cylindrical" -> "Cartesian", {r, ϕ, z}]];
DΨ = {r, ϕ, z} \[Function] Evaluate[D[Ψ[r, ϕ, z], {{r, ϕ, z}, 1}]];

Mathematically, this is

$$ \varPhi \colon \mathbb{R}^3 \to {[0,\infty[} \times {]-\pi,\pi[} \times \mathbb{R}, \quad \varPhi(x,y,z) = \left(\sqrt{x^2+y^2},\arctan(x,y),z \right)$$

and

$$ \varPsi \colon {[0,\infty[} \times {]-\pi,\pi[} \times \mathbb{R} \to \mathbb{R}^3, \quad \varPsi(r ,\varphi,z) = \left(r \,\cos(\varphi),r \, \sin(\varphi),z \right)$$

Here is a vector field V in cylindrical coordinates and its transformation into cartesian coordinates.

V = {r, ϕ, z} \[Function] {r, 1, 3};
F = {x, y, z} \[Function] Evaluate[(DΨ @@ Φ[x, y, z]).V @@ Φ[x, y, z]];

In polar coordinates, V has to be interpreted as a mapping with values in the tangent bundle of ${[0,\infty[} \times {]-\pi,\pi[} \times \mathbb{R}$:

$$V \colon {[0,\infty[} \times {]-\pi,\pi[} \times \mathbb{R} \to T({[0,\infty[} \times {]-\pi,\pi[} \times \mathbb{R}), \quad V(r,\varphi,z) = \left(r \tfrac{\partial}{\partial r} + \tfrac{\partial}{\partial \varphi} + 3\, \tfrac{\partial}{\partial z} \right) \Big|_{(r,\varphi,z)}.$$

It has to be transformed like this:

$$F(x,y,z) = D\varPsi(\varPhi(x,y,z)) \cdot V(\varPhi(x,y,z)).$$

And this is the plot:

R = Pi;
VectorPlot3D[
 F[x, y, z], {x, -R, R}, {y, -R, R}, {z, -R, R},
 VectorStyle -> Arrowheads[0.02],
 VectorScale -> 0.15,
 VectorPoints -> 6
 ]

Answer 2

Mathematica needs a coordinate-system option in VectorPlot.

For now a nice way to plot non-Cartesian fields is to use TransformedField which handles scalar, vector and even tensor fields.

Consider the cylindrical field used above.

{r, 1, 3}

we can very cleanly and without any thinking turn this into its Cartesian representation:

TransformedField["Cylindrical" -> "Cartesian", 

                {r, 1, 3}

              , {r, \[Phi], \[Zeta]} -> {x, y, z}]

and then plot it (like above)

VectorPlot3D[{x - y/Sqrt[x^2 + y^2], y + x/Sqrt[x^2 + y^2], 3}, {x, -\[Pi], \[Pi]}, {y, -\[Pi], \[Pi]}, {z, -\[Pi], \[Pi]}, VectorStyle -> Arrowheads[0.02], VectorScale -> 0.15, VectorPoints -> 6]

The coordinate-system option in Div, Grad, Laplacian etc. is excellent and it would be nice if this was incorporated into VectorPlot: