我想用物理计算的输出来绘制输出,但使用python的matplotlib,但是我的程序生成的绘图对我来说没有意义。我想绘制的是三个等式,1/a1,1/a2和1/a3,它们的梯度在点mn处变化,但是现在我的代码会导致值以及梯度发生变化。Python/matplotlib绘图值在渐变不连续处跳跃
似乎a1,a2,a3的方程式在函数从ms定义切换到ms> = mn定义时变得更大,但是使用像a1(10e11,mn)这样的测试值尝试它们并不会没有显示任何大的变化。
import matplotlib.pyplot as plt
import numpy as np
mn = 10312054216.046213
mz = 91.1876
a1z = 98.330 #NOTE: these values are for inverses of aplha at mz, ie 1/a1, etc
a2z = 29.571
a3z = 8.396
b1, b2, b3 = -13/(4*pi), 19/(12*pi), 7/(2*pi)
b1p, b2p, b3p = -157/(36*pi), 11/(12*pi), 17/(6*pi) #above new physics scale
c1, c2, c3 = 3/5, 1, 1
ms = np.linspace(10e8, 10e11, num = 1000000, endpoint = True)
def a1(ms, mn):
if (ms < mn):
return (c1*(a1z + b1*np.log(ms/mz)))**(-1)
elif (ms >= mn):
a1n = (c1*(a1z + b1*log(mn/mz)))**(-1)
return (c1*(a1n + b1p*log(ms/mn)))**(-1)
def a2(ms, mn):
if (ms < mn):
return (c2*(a2z + b2*np.log(ms/mz)))**(-1)
elif (ms >= mn):
a2n = (c2*(a2z + b2*log(mn/mz)))**(-1)
return (c2*(a2n + b2p*log(ms/mn)))**(-1)
def a3(ms, mn):
if (ms < mn):
return (c3*(a3z + b3*np.log(ms/mz)))**(-1)
elif (ms >= mn):
a3n = (c3*(a3z + b3*log(mn/mz)))**(-1)
return (c3*(a3n + b3p*log(ms/mn)))**(-1)
plt.xscale('log')
plt.plot(ms, [1/a1(x, mn) for x in ms])
plt.plot(ms, [1/a2(x, mn) for x in ms])
plt.plot(ms, [1/a3(x, mn) for x in ms])
plt.show()
在这里的任何洞察将非常欢迎,谢谢。
我不明白你的问题是什么。你想让我们调试你的代码吗? – nicoguaro