1
$\begingroup$

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 ?

$\endgroup$
  • 3
    $\begingroup$ Can you provide some basic code? The form of the function. Preferably in a way that would be easy to cope and paste it? $\endgroup$ – Darth_Bane May 12 '18 at 15:40
4
$\begingroup$

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
 ]

enter image description here

$\endgroup$
  • $\begingroup$ I am an amateur at using mathematica. Can you please tell me what the function V is in the code above? I couldn't spot any explicit definition for it $\endgroup$ – Lelouch May 12 '18 at 16:45
  • $\begingroup$ @Lelouch V = Function[{r, ϕ, z}, {r, 1, 3}]; $\endgroup$ – b3m2a1 May 12 '18 at 16:59
  • $\begingroup$ Again, sorry for asking , can you explicitly write the function and domain in mathematical form. I haven't used coordinate transform before. $\endgroup$ – Lelouch May 12 '18 at 17:38
2
$\begingroup$

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}]

enter image description here

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]

enter image description here

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

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.