使用Python为三体问题创建蛙跳算法时出现的问题

Question

我正试图用蛙跳算法为三体问题编写一个代码。我用的是Piet Hut的《移动的星星》；牧野君作为向导。

指南中的代码是用C编写的，但在尝试使用Python之前，我正试图遵循确切的工作流程

以下是我尝试遵循第5.1节中的代码。

import numpy as np
N = 3       #number of bodies
m = 1       #mass
dt = 0.01   #timestep
t_end = 10  #duration
r = [] 
v = []
a = [[0, 0, 0] for i in range(N)]
for i in range(N):
phi = i * 2 * np.pi/3
r.append([np.cos(phi), np.sin(phi), 0])
for i in range(N):
for j in range(i+1, N):
rji = []
for k in range(3):
rji.append(r[j][k] - r[i][k])
r2 = 0
for k in range(3):
r2 += rji[k]**2
r3 = r2 * np.sqrt(r2)
for k in range(3):
a[i][k] += m * rji[k] / r3
a[j][k] -= m * rji[k] / r3

v_abs = np.sqrt(-a[0][0])
for i in range(N):
phi = i * 2 * np.pi/3
v.append([-v_abs * np.sin(phi),
v_abs * np.cos(phi), 0])

ekin = 0
epot = 0
for i in range(N):
for j in range(i+1, N):
rji = [0, 0, 0]
for k in range(3):
rji[k] = r[j][k] - r[i][k]
r2 = 0
for k in range(3):
r2 += rji[k]**2
d = np.sqrt(r2)
epot -= m**2 / d
for k in range(3):
ekin += 0.5 * m * v[i][k]**2
e_in = ekin + epot
print('Initial total energy E_in = ', e_in)

dt_out = 0.01
t_out = dt_out
for t in np.arange(0, t_end, dt):
for i in range(N):
for k in range(3):
v[i][k] += a[i][k] * dt/2
for k in range(3):
r[i][k] += v[i][k] * dt
for i in range(N):
for k in range(3):
a[i][k] = 0
for i in range(N):
for j in range(i+1, N):
rji = []
for k in range(3):
rji.append(r[j][k] - r[i][k])
r2 = 0
for k in range(3):
r2 += rji[k]**2
r3 = r2 * np.sqrt(r2)
for k in range(3):
a[i][k] += m * rji[k] / r3
a[j][k] -= m * rji[k] / r3
for i in range(N):
for k in range(3):
v[i][k] += a[i][k] * dt/2
'''
if t >= t_out:
for i in range(N):
print(r[i][k], ' ')
for k in range(N):
print(v[i][k], ' ')
'''
t_out += dt_out

epot = 0
ekin = 0
for i in range(N):
for j in range(i+1, N):
rji = [0, 0, 0]
for k in range(3):
rji[k] = r[j][k] - r[i][k]
r2 = 0
for k in range(3):
r2 += rji[k]**2
d = np.sqrt(r2)
epot -= m**2 / d
for k in range(3):
ekin += 0.5 * m * v[i][k]**2
e_out = ekin + epot
print('Final total energy E_out = ', e_out)
print('absolute energy error: E_out - E_in = ', e_out - e_in)
print('relative energy error: (E_out - E_in)/E_in = ', (e_out - e_in)/e_in)

我已经定义了时间步长CCD_ 1和持续时间CCD_。在第5.4节中，结果应为：

|gravity> g++ -o leapfrog2 leapfrog2.C
|gravity> leapfrog2 > leapfrog2_0.01_10.out
Please provide a value for the time step
0.01
and for the duration of the run
10
Initial total energy E_in = -0.866025
Final total energy E_out = -0.866025
absolute energy error: E_out - E_in = 2.72254e-10
relative energy error: (E_out - E_in) / E_in = -3.14372e-10

以及圆形图。然而，我的代码产生了分歧：

Initial total energy E_in =  -0.8660254037844386
Final total energy E_out =  -0.39922101519288833
absolute energy error: E_out - E_in =  0.46680438859155027
relative energy error: (E_out - E_in)/E_in =  -0.5390192788244604

当然，在我画出我的结果后，它们不会循环。

我想知道我在翻译代码时是否犯了错误。任何帮助都将不胜感激！

Answer 1

欢迎使用堆栈溢出！

首先，这个bug是一个典型的python问题：代码的一部分缩进不正确。具体而言：

for i in range(N):
for j in range(i+1, N):
rji = []
for k in range(3):
rji.append(r[j][k] - r[i][k])
r2 = 0
for k in range(3):
r2 += rji[k]**2
r3 = r2 * np.sqrt(r2)
for k in range(3):
a[i][k] += m * rji[k] / r3
a[j][k] -= m * rji[k] / r3

应该是：

for i in range(N):
for j in range(i+1, N):
rji = []
for k in range(3):
rji.append(r[j][k] - r[i][k])
r2 = 0
for k in range(3):
r2 += rji[k]**2
r3 = r2 * np.sqrt(r2)
for k in range(3):
a[i][k] += m * rji[k] / r3
a[j][k] -= m * rji[k] / r3

让我给你一个建议：如果你想在阅读这本书的同时学习python，试着编写一个能完成任务的版本(就像我们在这里讨论的那个(，然后努力使它更地道。通过使用numpy，您可以删除空间维度上的大多数(如果不是全部的话(循环(至少！(。