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王月昊-第7天-2403-计算材料学实战
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推荐镜像 :DeePMD-kit:3.0.0a0-cuda12.1
推荐机型 :c2_m4_cpu
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组队共读活动打卡Day7（3月15日，周五）
▶ 今日学习内容：LAMMPS力学性质计算实战2-体模量（对应视频 11 部分）。
▶ 课后任务：
第6个 LAMMPS程序：不断调整晶格常数大小，用lammps计算出在平衡距离附近时，Cu晶体能量与其体积的关系，在此之上拟合出体弹性模量数据。分别尝试使用二次、三次多项式和BM方程进行拟合。分别考虑 LJ 势函数和 EAM 势函数的情况。尝试撰写Notebook，分析哪种结构最稳定，并在报告中提供LAMMPS脚本及分析方法。
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Cu体积模量计算脚本
LJ势
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[1]
%%writefile in.CuLJ
units metal
boundary p p p
atom_style atomic
#---------- setup loop -----------------
variable i loop 40
variable x equal 3.40+0.01*$i
lattice fcc $x
region box block 0 1 0 1 0 1
create_box 1 box
create_atoms 1 box
mass 1 64
#---------- use Cu LJ parameter----------
pair_style lj/cut 10.0
pair_coeff 1 1 0.40933 2.338
variable v equal ($x)^3
variable n equal count(all)
variable P equal pe
#-------------- run ----------------------
run 0
print "Cohesive Energy of Cu v = $v x= $x E = $P "
clear
next i
jump SELF
Writing in.CuLJ
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[2]
%%capture
!OMP_NUM_THREADS=2 lmp -i in.CuLJ
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[3]
!grep 'Cohesive' log.lammps
print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 39.651821 x= 3.41 E = -11.6743276227989 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 40.001688 x= 3.42 E = -11.9405476050061 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 40.353607 x= 3.43 E = -12.1595640975458 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 40.707584 x= 3.44 E = -12.3842526979336 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 41.063625 x= 3.45 E = -12.5895372366665 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 41.421736 x= 3.46 E = -12.7764125325894 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 41.781923 x= 3.47 E = -12.9458260014581 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 42.144192 x= 3.48 E = -13.0986799313477 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 42.508549 x= 3.49 E = -13.2358336452587 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 42.875 x= 3.5 E = -13.3581055567528 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 43.243551 x= 3.51 E = -13.4662751241234 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 43.614208 x= 3.52 E = -13.5610847083218 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 43.986977 x= 3.53 E = -13.6432413395693 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 44.361864 x= 3.54 E = -13.7070496808943 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 44.738875 x= 3.55 E = -13.7659953596674 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 45.118016 x= 3.56 E = -13.8142114984977 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 45.499293 x= 3.57 E = -13.852279849293 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 45.882712 x= 3.58 E = -13.8807544504762 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 46.268279 x= 3.59 E = -13.9001629355792 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 46.656 x= 3.6 E = -13.9110077784364 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 47.045881 x= 3.61 E = -13.9137674781707 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 47.437928 x= 3.62 E = -13.9088976869966 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 47.832147 x= 3.63 E = -13.8968322836988 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 48.228544 x= 3.64 E = -13.8779843955076 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 48.627125 x= 3.65 E = -13.8527473709409 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 49.027896 x= 3.66 E = -13.7961834075309 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 49.430863 x= 3.67 E = -13.759684579032 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 49.836032 x= 3.68 E = -13.7178592687278 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 50.243409 x= 3.69 E = -13.6710307078911 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 50.653 x= 3.7 E = -13.6195065548619 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 51.064811 x= 3.71 E = -13.5635796208348 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 51.478848 x= 3.72 E = -13.5035285612586 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 51.895117 x= 3.73 E = -13.4396185345301 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 52.313624 x= 3.74 E = -13.3721018295858 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 52.734375 x= 3.75 E = -13.3012184639042 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 53.157376 x= 3.76 E = -13.2271967533648 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 53.582633 x= 3.77 E = -13.1502538553272 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 54.010152 x= 3.78 E = -13.0705962862321 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 54.439939 x= 3.79 E = -12.9884204149546 print "Cohesive Energy of Cu v = $v x= $x E = $P " Cohesive Energy of Cu v = 54.872 x= 3.8 E = -12.9039129330838
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[4]
# Read the file and search for lines containing the pattern "Cohesive Energy of Cu v = $v x= $x E = $P"
import re
# Initialize list to hold the matching lines
matching_lines = []
# Define the regex pattern for capturing v and x and E values
pattern = re.compile(r"Cohesive Energy of Cu v = ([\d\.-]+) x= ([\d\.-]+) E = ([\d\.-]+)")
# Read the file and search for the pattern
with open('./log.lammps', 'r') as file:
for line in file:
match = pattern.search(line)
if match:
v_value = float(match.group(1))
x_value = float(match.group(2))
E_value = float(match.group(3))
matching_lines.append((v_value,x_value,E_value))
# Show some of the matching lines
matching_lines[:40]
[(39.651821, 3.41, -11.6743276227989), (40.001688, 3.42, -11.9405476050061), (40.353607, 3.43, -12.1595640975458), (40.707584, 3.44, -12.3842526979336), (41.063625, 3.45, -12.5895372366665), (41.421736, 3.46, -12.7764125325894), (41.781923, 3.47, -12.9458260014581), (42.144192, 3.48, -13.0986799313477), (42.508549, 3.49, -13.2358336452587), (42.875, 3.5, -13.3581055567528), (43.243551, 3.51, -13.4662751241234), (43.614208, 3.52, -13.5610847083218), (43.986977, 3.53, -13.6432413395693), (44.361864, 3.54, -13.7070496808943), (44.738875, 3.55, -13.7659953596674), (45.118016, 3.56, -13.8142114984977), (45.499293, 3.57, -13.852279849293), (45.882712, 3.58, -13.8807544504762), (46.268279, 3.59, -13.9001629355792), (46.656, 3.6, -13.9110077784364), (47.045881, 3.61, -13.9137674781707), (47.437928, 3.62, -13.9088976869966), (47.832147, 3.63, -13.8968322836988), (48.228544, 3.64, -13.8779843955076), (48.627125, 3.65, -13.8527473709409), (49.027896, 3.66, -13.7961834075309), (49.430863, 3.67, -13.759684579032), (49.836032, 3.68, -13.7178592687278), (50.243409, 3.69, -13.6710307078911), (50.653, 3.7, -13.6195065548619), (51.064811, 3.71, -13.5635796208348), (51.478848, 3.72, -13.5035285612586), (51.895117, 3.73, -13.4396185345301), (52.313624, 3.74, -13.3721018295858), (52.734375, 3.75, -13.3012184639042), (53.157376, 3.76, -13.2271967533648), (53.582633, 3.77, -13.1502538553272), (54.010152, 3.78, -13.0705962862321), (54.439939, 3.79, -12.9884204149546), (54.872, 3.8, -12.9039129330838)]
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[5]
# Write the extracted v, x, and E values to a CSV file
with open('lj_Cu.csv', 'w') as f:
f.write("v,x,E\n") # Write header
for v, x, E in matching_lines:
f.write(f"{v},{x},{E}\n")
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1. 二次多项式拟合
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[6]
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
# 读取数据（之前存储的csv文件）：
filename = "lj_Cu.csv"
DataFrame = pd.read_csv(filename)
lat = np.array(DataFrame['v'])
ene = np.array(DataFrame['E'])
# 用二次函数拟合：
a,b,c = np.polyfit(lat,ene,2)
mesh = np.linspace(np.min(lat),np.max(lat),1000)
opt_lat = -b/(2*a)
Emin = np.polyval([a,b,c], opt_lat)
print("a,b,c =",a,b,c)
print("Emin =",Emin,"opt_lat =",opt_lat)
# 数据可视化：
plt.scatter(lat, ene, 10)
plt.plot(mesh, a*mesh**2+b*mesh+c)
plt.xlabel('Volume')
plt.ylabel('Energy')
plt.show()
a,b,c = 0.027859136511323872 -2.677008042125913 50.35779796262389 Emin = -13.951195019749292 opt_lat = 48.04542382420561
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2. 三次多项式拟合
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[7]
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
# 读取数据（之前存储的csv文件）：
filename = "lj_Cu.csv"
DataFrame = pd.read_csv(filename)
lat = np.array(DataFrame['v'])
ene = np.array(DataFrame['E'])
# 用三次函数拟合：
a, b, c, d = np.polyfit(lat, ene, 3)
mesh = np.linspace(np.min(lat), np.max(lat), 1000)
opt_lat = -b / (3 * a) # 三次函数极小值
Emin = np.polyval([a, b, c, d], opt_lat)
print("a, b, c, d =", a, b, c, d)
print("Emin =", Emin, "opt_lat =", opt_lat)
# 数据可视化：
plt.scatter(lat, ene, 10)
plt.plot(mesh, a * mesh**3 + b * mesh**2 + c * mesh + d)
plt.xlabel('Volume')
plt.ylabel('Energy')
plt.show()
a, b, c, d = -0.0017381601310075745 0.27402339262618536 -14.234388038073991 230.2287757280661 Emin = -13.3084620063795 opt_lat = 52.55046945671606
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3. 用Birch-Murnaghan方程拟合
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import pandas as pd
import numpy as np
# 读取数据（之前存储的csv文件）：
df = pd.read_csv('lj_Cu.csv') # 修正文件名为正确的 'lj_Cu.csv'
v = np.array(df['v'])
x = np.array(df['x'])
E = np.array(df['E'])
# 先做一次多项式拟合，根据以上多项式拟合的结果设置初始猜测解X0：
a, b, c = np.polyfit(v, E, 2)
v0 = -b / (2 * a)
e0 = a * v0 ** 2 + b * v0 + c
b0 = 2 * a * v0
bP = 3.5
x0 = [e0, b0, bP, v0]
def Murnaghan(parameters, vol):
E0 = parameters[0]
B0 = parameters[1]
BP = parameters[2]
V0 = parameters[3]
E = E0 + B0 * vol / BP * (((V0 / vol) ** BP) / (BP - 1) + 1) - V0 * B0 / (BP - 1.)
return E
# 定义误差（优化目标）
def residual(pars, y, x):
err = y - Murnaghan(pars, x)
return err
# 进行拟合：
from scipy.optimize import leastsq
murnpars, ier = leastsq(residual, x0, args=(E, v))
# 输出结果
print('Bulk Modulus:', murnpars[1])
print('Lattice constant:', murnpars[3] ** (1 / 3))
# 拟合效果可视化：
from matplotlib import pyplot as plt
v_mesh = np.linspace(np.min(v), np.max(v), 1000)
plt.scatter(v, E, 40)
plt.plot(v_mesh, Murnaghan(murnpars, v_mesh))
plt.ylabel('Energy')
plt.xlabel('Volume')
plt.show()
Bulk Modulus: 2.3208178189964985 Lattice constant: 3.6078449224812617
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Birch-Murnaghan方程因为更接近于物理实际，所以得到的结果与解析解最为接 近；体模量的拟合，应优先使用Birch-Murnaghan方程。
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EAM势
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下载EAM势
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%%bash
## 这行命令可以下载 lammps安装包中，potentials文件夹下的所有势函数，用于后续eam势函数的导入
wget https://bohrium-api.dp.tech/ds-dl/lammps-potentials-8snt-v1.zip
unzip -o ./lammps-potentials-8snt-v1.zip
--2024-03-17 23:49:38-- https://bohrium-api.dp.tech/ds-dl/lammps-potentials-8snt-v1.zip Resolving ga.dp.tech (ga.dp.tech)... 10.255.254.7, 10.255.254.37, 10.255.254.18 Connecting to ga.dp.tech (ga.dp.tech)|10.255.254.7|:8118... connected. Proxy request sent, awaiting response... 307 Temporary Redirect Location: https://dataset-bohr-storage.dp.tech/lbg%2Fdataset%2Fzip%2Fdataset_tiefblue_bohr_11303_lammps-potentials-8snt_v105264.zip?Expires=1710726578&OSSAccessKeyId=LTAI5t8SinfkteepM4WuAEqc&Signature=62TpLNEYxmLTvp%2FDbVwbk%2F8TINY%3D [following] --2024-03-17 23:49:38-- https://dataset-bohr-storage.dp.tech/lbg%2Fdataset%2Fzip%2Fdataset_tiefblue_bohr_11303_lammps-potentials-8snt_v105264.zip?Expires=1710726578&OSSAccessKeyId=LTAI5t8SinfkteepM4WuAEqc&Signature=62TpLNEYxmLTvp%2FDbVwbk%2F8TINY%3D Connecting to ga.dp.tech (ga.dp.tech)|10.255.254.7|:8118... connected. Proxy request sent, awaiting response... 200 OK Length: 15066136 (14M) [application/zip] Saving to: ‘lammps-potentials-8snt-v1.zip’ 0K .......... .......... .......... .......... .......... 0% 25.8M 1s 50K .......... .......... .......... .......... .......... 0% 31.9M 0s 100K .......... .......... .......... .......... .......... 1% 50.9M 0s 150K .......... .......... .......... .......... .......... 1% 35.6M 0s 200K .......... .......... .......... .......... .......... 1% 45.7M 0s 250K .......... .......... .......... .......... .......... 2% 45.6M 0s 300K .......... .......... .......... .......... .......... 2% 8.40M 1s 350K .......... .......... .......... .......... .......... 2% 138M 1s 400K .......... .......... .......... .......... .......... 3% 204M 0s 450K .......... .......... .......... .......... .......... 3% 131M 0s 500K .......... .......... .......... .......... .......... 3% 173M 0s 550K .......... .......... .......... .......... .......... 4% 143M 0s 600K 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potentials/Ag_u3.eam inflating: potentials/AlCu.adp inflating: potentials/AlCu.bop.table inflating: potentials/AlCu.eam.alloy inflating: potentials/AlCuH.bop.table inflating: potentials/AlFe_mm.eam.fs inflating: potentials/AlO.eam.alloy inflating: potentials/AlO.streitz inflating: potentials/AlSiMgCuFe.meam inflating: potentials/Al_Batra_2019.agni inflating: potentials/Al_jnp.eam inflating: potentials/Al_jpc.agni inflating: potentials/Al_mm.eam.fs inflating: potentials/Al_prb.agni inflating: potentials/Al_zhou.eam.alloy inflating: potentials/Au_u3.eam inflating: potentials/BN.extep inflating: potentials/BNC.tersoff inflating: potentials/BNCH-old.ILP inflating: potentials/BNCH.ILP inflating: potentials/BNC_MBD_bulk.ILP inflating: potentials/BNC_TS_bulk.ILP inflating: potentials/Bi.meam inflating: potentials/C.drip inflating: potentials/C.lcbop inflating: potentials/CC.KC inflating: potentials/CC.KC-full inflating: potentials/CC.Lebedeva inflating: potentials/CCu_v2.bop.table inflating: 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[28]
%%writefile in.CuEAM
units metal
boundary p p p
atom_style atomic
#---------- setup loop -----------------
variable i loop 40
variable x equal 3.40+0.01*$i
lattice fcc $x
region box block 0 1 0 1 0 1
create_box 1 box
create_atoms 1 box
mass 1 64
#---------- use Cu EAM parameter----------
pair_style eam
pair_coeff * * ./potentials/Cu_u3.eam Cu
variable v equal ($x)^3
variable n equal count(all)
variable P equal pe
#-------------- run ----------------------
run 0
print "Cohesive Energy of Cu v = $v x= $x E = $P "
clear
next i
jump SELF
Overwriting in.CuEAM
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[29]
%%capture
!OMP_NUM_THREADS=2 lmp -i in.CuEAM
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[30]
!grep 'Cohesive' log.lammps
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# Read the file and search for lines containing the pattern "Cohesive Energy of Cu v = $v x= $x E = $P"
import re
# Initialize list to hold the matching lines
matching_lines = []
# Define the regex pattern for capturing v and x and E values
pattern = re.compile(r"Cohesive Energy of Cu v = ([\d\.-]+) x= ([\d\.-]+) E = ([\d\.-]+)")
# Read the file and search for the pattern
with open('./log.lammps', 'r') as file:
for line in file:
match = pattern.search(line)
if match:
v_value = float(match.group(1))
x_value = float(match.group(2))
E_value = float(match.group(3))
matching_lines.append((v_value,x_value,E_value))
# Show some of the matching lines
matching_lines[:40]
[]
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[17]
# Write the extracted v, x, and E values to a CSV file
with open('lj_EAM.csv', 'w') as f:
f.write("v,x,E\n") # Write header
for v, x, E in matching_lines:
f.write(f"{v},{x},{E}\n")
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[18]
import pandas as pd
import numpy as np
# 读取数据（之前存储的csv文件）：
df = pd.read_csv('lj_EAM.csv')
v = np.array(df['v'])
x = np.array(df['x'])
E = np.array(df['E'])
# 先做一次多项式拟合，根据以上多项式拟合的结果设置初始猜测解X0：
a, b, c = np.polyfit(v, E, 2)
v0 = -b / (2 * a)
e0 = a * v0 ** 2 + b * v0 + c
b0 = 2 * a * v0
bP = 3.5
x0 = [e0, b0, bP, v0]
def Murnaghan(parameters, vol):
E0 = parameters[0]
B0 = parameters[1]
BP = parameters[2]
V0 = parameters[3]
E = E0 + B0 * vol / BP * (((V0 / vol) ** BP) / (BP - 1) + 1) - V0 * B0 / (BP - 1.)
return E
# 定义误差（优化目标）
def residual(pars, y, x):
err = y - Murnaghan(pars, x)
return err
# 进行拟合：
from scipy.optimize import leastsq
murnpars, ier = leastsq(residual, x0, args=(E, v))
# 输出结果
print('Bulk Modulus:', murnpars[1])
print('Lattice constant:', murnpars[3] ** (1 / 3))
# 拟合效果可视化：
from matplotlib import pyplot as plt
v_mesh = np.linspace(np.min(v), np.max(v), 1000)
plt.scatter(v, E, 40)
plt.plot(v_mesh, Murnaghan(murnpars, v_mesh))
plt.ylabel('Energy')
plt.xlabel('Volume')
plt.show()
--------------------------------------------------------------------------- TypeError Traceback (most recent call last) Cell In[18], line 11 8 E = np.array(df['E']) 10 # 先做一次多项式拟合，根据以上多项式拟合的结果设置初始猜测解X0： ---> 11 a, b, c = np.polyfit(v, E, 2) 12 v0 = -b / (2 * a) 13 e0 = a * v0 ** 2 + b * v0 + c File /opt/deepmd-kit-3.0.0/lib/python3.10/site-packages/numpy/lib/polynomial.py:639, in polyfit(x, y, deg, rcond, full, w, cov) 637 raise TypeError("expected 1D vector for x") 638 if x.size == 0: --> 639 raise TypeError("expected non-empty vector for x") 640 if y.ndim < 1 or y.ndim > 2: 641 raise TypeError("expected 1D or 2D array for y") TypeError: expected non-empty vector for x
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