Skip to main content

plotting - Color vectors according to a user-defined function


V[x_,z_]:=x/Sqrt[x^2+z^2];
Ex[x_,z_]=-D[V[x,z],x];
Ez[x_,z_]=-D[V[x,z],z];

The electric field components (Ex,Ez) are to be plotted as a vector field in the x-z plane. In addition, I would like to color the vectors according to |E|^2. The default option:


  VectorColorFunction -> "ThermometerColors"


does not suit here because it plots magnitude of the vectors, but my vectors only contain the real part of the field. Below is my attempt:


vectorplot = 
VectorPlot[{Re[Ex[x, z]], Re[Ez[x, z]]}, {x, -2*R, 2*R}, {z, -2*R,
2*R},
VectorColorFunction -> Function[{x, z, vx, vz, n}, ColorData["ThermometerColors"]
[Abs[Ex[x,z]]^2+Abs[Ez[x,z]]^2]],

VectorScale -> {0.03, Automatic, None},
VectorColorFunctionScaling -> True, VectorPoints -> 25,
PlotLegends -> BarLegend[Automatic,

LegendLabel -> HoldForm[Superscript["|E|", 2]],
LabelStyle -> {FontFamily -> "Helvetica", FontSize -> 18, Black}], AspectRatio -> 1];

Would be grateful for your suggestions.



Answer



So there was nothing wrong with your user-defined color function, the only problem is that you paired it with VectorColorFunctionScaling->True. This means that the x and z values fed to the color function were scaled to lie between 0 and 1. What you really want to do is to scale the field intensity, not the coordinates.


Since your field intensity has a singularity at the origin, you need to choose some maximum intensity value for the color scale, otherwise the arrow at the origin will be red and every other arrow will be blue. Here I choose a maximum intensity of 2,


V[x_, z_] := x/Sqrt[x^2 + z^2];
Ex[x_, z_] = -D[V[x, z], x];
Ez[x_, z_] = -D[V[x, z], z];

R = 1;

With[{vectorscale = {0, 2}},
Legended[
VectorPlot[Re@{Ex[x, z], Ez[x, z]}, {x, -2*R, 2*R}, {z, -2*R, 2*R},
VectorColorFunction ->
Function[{x, z, vx, vz, n},
ColorData[{"ThermometerColors", vectorscale}][
Abs[Ex[x, z]]^2 + Abs[Ez[x, z]]^2]],
VectorScale -> {0.03, Automatic, None},

VectorColorFunctionScaling -> False,
VectorPoints -> 25,
AspectRatio -> 1],
BarLegend[{ColorData[{"ThermometerColors", vectorscale}],
vectorscale}, LegendLabel -> HoldForm[Superscript["|E|", 2]],
LabelStyle -> {FontFamily -> "Helvetica", FontSize -> 18, Black}]
]
]

enter image description here



Also I moved your BarLegend outside because it seems that VectorPlot will not take a PlotLegends option. I would like to point out that you could write that a bit simpler since your field is real-valued, and therefore the intensity that you are coloring according to is simply the norm squared, and so you could replace Abs[Ex[x, z]]^2 + Abs[Ez[x, z]]^2 with n^2, but I left it like this to be most general, i.e. you could color the arrows according to any scalar field


Finally, you could also display this field using VectorDensityPlot


With[{vectorscale = {0, 5}},
VectorDensityPlot[{{Re[Ex[x, z]], Re[Ez[x, z]]},
Abs[Ex[x, z]]^2 + Abs[Ez[x, z]]^2}, {x, -2*R, 2*R}, {z, -2*R,
2*R},
VectorScale -> {0.03, Automatic, None},
VectorPoints -> 25,
VectorStyle -> White,
ColorFunction ->(*"ThermometerColors"*)

Function[{x, z, vx, vz, n},
ColorData["ThermometerColors"][Rescale[n^2, vectorscale]]],
ColorFunctionScaling -> False,
PlotLegends ->
BarLegend[{ColorData[{"ThermometerColors", vectorscale}],
vectorscale},
LegendLabel -> HoldForm[Superscript["|E|", 2]],
LabelStyle -> {FontFamily -> "Helvetica", FontSize -> 18, Black}]
]
]


enter image description here


Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.