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The paper consists a no of plots which have color bars as axis but I think all of them followed the same procedure, I want to plot such good looking plots (like fig 1 or 2 or any)can anybody guide me how to do them, since I am very new to Mathematica

arxiv.org/abs/1910.00234

I think the problem is very similar like the below mentioned example

v = 246
   E1 = 9000

  PT = ((Subscript[E1, T]*v)/(
 Subscript[v, n]*E1))/((Subscript[v, n]/Subscript[T, n])^-1*(v/
  E1)) - 7*(Log[Subscript[v, n]]/Subscript[T, n]) + 
   Log[Subscript[T, n]/100]
  Subscript[v, n]/Subscript[T, n] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
   Subscript[T, n] = {300, 500, 469, 650, 546, 389, 456, 411, 523, 
    700};
    ((Subscript[E1, T]*v)/(Subscript[v, n]*E1)) = 
     RandomReal[{1, 3.1}, {10}]

Now I have to randomly vary the last ratio to some range and make it the color bar axis and the values for PT and v_n/T_n for each ratios can be plotted PT vs v_n/T_n plot making the ratio as color axis

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2
  • $\begingroup$ Something is wrong with your idea. If you take 1 value of x you get a curve F[x,y] vs. y, not just one point! Please, make a more realistic example. $\endgroup$
    – yarchik
    Commented Oct 3, 2020 at 12:58
  • $\begingroup$ Do you want the contour line(s) for y == Sqrt[Tan[x]] colored according to the value of F? $\endgroup$
    – Coolwater
    Commented Oct 3, 2020 at 13:16

3 Answers 3

Reset to default
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Clear["Global`*"]

F[x_, y_] := Sqrt[(5*10^-6)*x^2*(5*10^-9)^y]

y = Sqrt[Tan[x]];

Use ParametricPlot

ParametricPlot[
 {y, F[x, y]}, {x, 10^-5, 1},
 Frame -> True,
 FrameLabel -> (Style[#, 12, Bold] & /@ {"y", "F"}),
 ColorFunction -> Function[{y, F, x}, ColorData["Rainbow"][x]],
 AspectRatio -> 1,
 PlotLegends -> Placed[
   BarLegend[{"Rainbow", {10^-5, 1}},
    LegendLabel -> Style["x", 12, Bold],
    LegendMarkerSize -> 200],
   {.75, .5}]]

enter image description here

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Using Bob Hanlon's method in alternative way -- plotting two curves with ParametricPlot and making the second look like a bar legend:

ClearAll[F, x, y]
F[x_, y_] := Sqrt[(5*10^-6)*x^2*(5*10^-9)^y]
y = Sqrt[Tan[x]];


ParametricPlot[{{y, F[x, y]}, {y, -10^-6}}, {x, 10^-5, 1}, 
 Frame -> True, Axes -> False, 
 FrameLabel -> (Style[#, 12, Bold] & /@ {"y", "F"}), 
 PlotStyle -> {Automatic, Directive[Opacity[1], AbsoluteThickness[20], CapForm["Butt"]]},
 ColorFunction -> Function[{y, F, x}, ColorData["Rainbow"][x]], 
 AspectRatio -> 1, PlotRange -> All, ImageSize -> Large]

enter image description here

Same approach with an alternative color function:

ParametricPlot[{{y, F[x, y]}, {1.3, F[x, y]}}, {x, 10^-5, 1}, 
 Frame -> True, 
 FrameTicks -> {{Automatic, All}, {Automatic, Automatic}}, 
 Axes -> False, FrameLabel -> (Style[#, 12, Bold] & /@ {"y", "F"}), 
 PlotStyle -> {Automatic, Directive[Opacity[1], AbsoluteThickness[20], CapForm["Butt"]]},
 ColorFunction -> Function[{y, F, x}, ColorData["Rainbow"][F]], 
 AspectRatio -> 1, PlotRange -> All, 
 PlotRangePadding -> {{Automatic, .04}, {Automatic, Automatic}}, 
 ImageSize -> Large, PlotRangeClipping -> False]

enter image description here

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I assume you're asking how to generate a bar plot with your given function, and not how to solve some sort of thing.

This can easily be done with BarLegend

Plot[Sqrt[x], {x, 0, 1}, ColorFunction -> "Rainbow", PlotLegends -> BarLegend["Rainbow"]]

enter image description here

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3
  • $\begingroup$ It is quite far from the appearance in the OP. $\endgroup$
    – yarchik
    Commented Oct 3, 2020 at 12:59
  • $\begingroup$ The op needs to be more specific if they want an exact replicable of their drawing, or how to add a barlegend...at the moment the only question i see, is how to add a barlegend. Their equation model also doesn‘t represent their drawing. Which is a different problem $\endgroup$
    – DrMrstheMonarch
    Commented Oct 3, 2020 at 13:01
  • 2
    $\begingroup$ Yes, fully agree, I wrote a comment above. It is better to wait for clarifications. $\endgroup$
    – yarchik
    Commented Oct 3, 2020 at 13:02

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