偏移与pcolormesh,轮廓(Python matplotlib)

时间:2017-05-26 11:55:06

标签: python matplotlib

我正在尝试使用pcolormesh绘制2D数组。这是一个黑洞磁层的二维轴对称球形图。问题是为什么在θ中存在偏移?...通常磁场线必须是垂直的并且由赤道处的黑洞稍微弯曲。地图似乎也被抵消了。 Black hole magnetosphere (64 points)这部分代码可能很有趣:

import numpy as np, matplotlib.pyplot as plt
import matplotlib.animation as animation
import scipy.integrate as integrate

##### Natural unities 
M = 1.0
G = 1.0
c = 1.0 

##### Gravitational radius
rg = (G * M)/(c*c)

##### spin
a = 0.0

##### Horizon radius
rh = rg + np.sqrt(rg*rg - a*a)

##### r, theta parameters
rmin = 0.9*rh

rmax = 5.0

thmin = 0.001*np.pi
thmax = 0.999*np.pi

Nr = 64
Nth = 64

r = np.logspace(np.log10(rmin),np.log10(rmax),Nr)

th = np.linspace(thmin,thmax,Nth)

r_grid, th_grid = np.meshgrid(r,th)


x = r_grid*np.cos(th_grid)
y = r_grid*np.sin(th_grid)

##### ergosphere

rerg = 1.0 + np.sqrt(1.0-a*a*np.cos(th)*np.cos(th))

xerg = rerg*np.cos(th)
yerg = rerg*np.sin(th)

##### Horizon
xrh = rh*np.cos(th)
yrh = rh*np.sin(th)


############################# schwarzschild's metric


Alpha_sch = 1.0/np.sqrt(1.0+(2.0/r_grid))

sqr_det_gamma_sch = np.sqrt((1.0+2.0/r_grid)*r_grid*r_grid*r_grid*r_grid*np.sin(th_grid)*np.sin(th_grid))


Brg = np.zeros((Nth,Nr))


############################# Flux function

Psy = np.zeros((Nth,Nr))

V = [2,4,6,8,10,12]



##### Wald's solution
Brg = Alpha_sch*np.cos(th_grid)


for i in range (0,Nr):
    for j in range(1,Nth):
        th3 = th[0:j]
        Psy[j,i]=integrate.simps(sqr_det_gamma_sch[0:j,i]*Brg[0:j,i],th3)



plt.figure(figsize=(8,8))
ax = plt.subplot(111)
plt.title('$B^r$')
circle1 = plt.Circle((0,0),rh,color = 'k')
ax.add_artist(circle1)
plt.contour(y,x,Psy,V,colors = 'k')
plt.pcolormesh(y,x,Brg,cmap='bwr',vmin=-1,vmax=1)
cbar = plt.colorbar()
cbar.set_label('Intensité', rotation=270)
plt.xlim(0,rmax)
plt.xlabel('$r_g$')
plt.ylabel('$r_g$')
plt.legend()
plt.show()

当然,如果我在r和theta中占据更多的点,那么偏移量会减少但仍然在Black hole magnetosphere (256 points)。如果你们中的一些人可以解释我为什么会很有帮助。提前谢谢你。

的Jérémy

1 个答案:

答案 0 :(得分:0)

您正在使用integrate.simps()进行集成,需要很多点才能获得精确度。

您可以使用integrate.quad()进行集成,以下是代码:

import numpy as np, matplotlib.pyplot as plt
import scipy.integrate as integrate
from math import sqrt, sin, cos

##### Natural unities 
M = 1.0
G = 1.0
c = 1.0 

##### Gravitational radius
rg = (G * M)/(c*c)

##### spin
a = 0.0

##### Horizon radius
rh = rg + np.sqrt(rg*rg - a*a)

##### r, theta parameters
rmin = 0.9*rh

rmax = 5.0

thmin = 0
thmax = np.pi

Nr = 64
Nth = 64

r = np.logspace(np.log10(rmin),np.log10(rmax),Nr)

th = np.linspace(thmin,thmax,Nth, endpoint=True)

r_grid, th_grid = np.meshgrid(r,th)

x = r_grid*np.cos(th_grid)
y = r_grid*np.sin(th_grid)

def f(th, r):
    th = float(th)
    r = float(r)
    Alpha_sch = 1.0/sqrt(1.0+(2.0/r))
    sqr_det_gamma_sch = sqrt((1.0+2.0/r)*r*r*r*r*sin(th)*sin(th))
    Brg = Alpha_sch*cos(th)
    return sqr_det_gamma_sch * Brg

Psy = np.zeros_like(x)
for i in range (0,Nr):
    for j in range(Nth):
        Psy[j, i] = integrate.quad(f, 0, th[j], args=(r[i],))[0]

fig, ax = plt.subplots(figsize=(8, 8))
ax.set_aspect("equal")
ax.pcolormesh(y, x, Psy)
ax.contour(y, x, Psy, colors="k")

和输出:

enter image description here