朱莉娅的3D矢量绘图

Question

我正在尝试在Julia中绘制EM波（沿z方向传播）矢量场。我环顾四周，看起来像颤抖是我需要使用的，但我尝试了失败的结果。据我了解，（x，y，z）是向量的原点，（u，v，w）是向量本身在（x，y，z）点的原点。这是我到目前为止的内容，但这似乎无法产生正确的情节。我该如何工作？我也愿意尝试其他绘图库。预先感谢。

using Plots; gr()
t = 0; n = 100; k = 1; ω = 1; φ = π/4
x = y = w = zeros(n)
z = range(0, stop=10, length=n)
u = @. cos(k*z - ω*t)
v = @. sin(k*z - ω*t)
quiver(x, y, z, quiver=(u, v, w), projection="3d")

Answer 1

我不确定这是否是您想要的结果，但是我设法使您的代码在Julia v1.1中工作：

using PyPlot

pygui(true)

fig = figure()
ax = fig.gca(projection="3d")
t = 0; n = 100; k = 1; ω = 1; φ = π/4
x = y = w = zeros(n)
z = range(0, stop=10, length=n)
u = cos.(k*z .- ω*t)
v = sin.(k*z .- ω*t)
ax.quiver(x,y,z, u,v,w)

或者，带有颜色：

using PyPlot
using Random

function main()
    pygui(true)

    fig = figure()
    ax = fig.gca(projection="3d")
    t = 0; n = 100; k = 1; ω = 1; φ = π/4
    x = y = w = zeros(n)
    z = range(0, stop=10, length=n)
    u = cos.(k*z .- ω*t)
    v = sin.(k*z .- ω*t)
    a = ((u[1], 0.8, 0.5), (u[2], 0.8, 0.5))
    for i in 3:length(u)-2
        a = (a..., (abs(u[i]), 0.8, 0.5))
    end
    c = ((0.4, 0.5, 0.4), (0.4, 0.9, 0.4), (0.1, 0.1, 0.1))
    q = ax.quiver(x,y,z, u,v,w, color = a)
end
main()