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I'm trying to plot a vector field over a circle, Particularly the vector field in this plot:
VectorPlot[{(Sin[x])^3, 0}, {x, 0, 2 \[Pi]}, {y, -0.001, 0.001},
FrameTicks -> {{\[Pi]/2, \[Pi], 3 \[Pi]/2, 2 \[Pi]}, {-1, 1}},
Epilog ->
Plot[(Sin[x])^3, {x, 0, 2 \[Pi]}, PlotStyle -> Red,
Ticks -> None][[1]], PlotRange -> {{0, 2 \[Pi]}, {-1, 1}},
GridLines -> Automatic
]
That was my way to plot a vector field over the line. But I want to plot this one-dimensional over a circle.
Perhaps this?
With[{rad = 1, center = {0, 0}},
Show[
PolarPlot[{Sin[t]^2, 1}, {t, 0, 2 Pi},
PlotStyle -> Thickness[Medium], PlotRange -> 1.15 rad,
PolarGridLines -> Automatic],
VectorPlot[(Sin[ArcTan[x, y]])^3 {-y, x}/rad, {x, -1.1 rad,
1.1 rad}, {y, -1.1 rad, 1.1 rad},
VectorPoints -> CirclePoints[center, rad, 24]]
]
]
In your example, the horizontal lengths of the vectors are the y-values of the function for the given x-values; that is Sin[x]^3
Using that same approach, the horizontal lengths of the vectors should be the y-values of the points on the circle. For each x-value there are two points on the circle - for the one-dimensional vector field below I picked the positive ones.
Each point on the unit half-circle above the x-axis is of the form {Cos[theta], Sin[theta]} for some angle theta between 0 and pi. The vector field is defined in terms of x and y - so we need the theta for a given x-value: this can be computed using ArcCos[x].
The code below shows a red unit circle with the one-dimensional vector field similar to the one you used.
Show[Graphics[{Red, Circle[]}],
VectorPlot[{Sin[ArcCos[x]], 0}, {x, -1, 1}, {y, -0.001, 0.001}]]
To only show the positive half-circle you can use
Show[ParametricPlot[{Cos[theta], Sin[theta]}, {theta, 0, \[Pi]},
PlotStyle -> Red],
VectorPlot[{Sin[ArcCos[x]], 0}, {x, -1, 1}, {y, -0.001, 0.001}]]
For a circle which has a radius other than 1, you may use the code below (which uses radius = 3) which looks pretty much the same, except the scale, which is why the axes are shown.
radius = 3; Show[Graphics[{Red, Circle[{0, 0}, radius]}],
VectorPlot[{radius Sin[ArcCos[x/radius]], 0}, {x, -radius,
radius}, {y, -0.001, 0.001}], Axes -> True]
{Sin[x]^3, 0}? (Note: No parentheses needed aroundSin[x]. On exactly what circle? (The xy-rectangle in your main plot would only accommodate a very tiny circle, given the constraints onythere.) $\endgroup$ – murray Jun 23 '19 at 14:46