-
Notifications
You must be signed in to change notification settings - Fork 0
/
Bulk.py
354 lines (248 loc) · 11 KB
/
Bulk.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
import numpy as np
import math as m
import matplotlib.pyplot as plt
from scipy.integrate import odeint
from scipy.integrate import simps
from scipy.signal import savgol_filter
import pandas as pd
from decimal import *
from scipy.integrate import solve_bvp
def Bulk_Scalar(H, V, r0, LamB=0.0, LamV=None, LamH=None, k: float = 1, m: float = 0.2, visualise: bool = True):
c=np.sqrt(4+(np.power(m,2)/np.power(k,2)))
E=np.sqrt(4+(np.power(m,2)/np.power(k,2)))-2
def dU_dx(U, x, r0, k):
# Here U is a vector such that y=U[0] and z=U[1]. This function should return [y', z']
return [U[1], 4*k*U[1] + m*m*U[0]+LamB*U[0]*U[0]*U[0]]
def func2(x, U):
# Here U is a vector such that phi=U[0] and phi'=U[1]. This function should return [phi', phi'']
return np.vstack((U[1], 4*k*U[1] + m*m*U[0]+LamB*U[0]*U[0]*U[0]))
def boundaries(phi, Lam, B):
return(2*Lam*phi*((phi*phi)-(B*B)))
def no_bc_bvp(Func, r0):
def bc(ya, yb):
#return np.array([ya[1]-(lamH*ya[0]*(ya[0]**2-H**2)), yb[1]-(lamH*yb[0]*(yb[0]**2-V**2))])
return np.array([ya[0]-H, yb[0]-V])
xs = np.linspace(0, r0, round((2**10)*r0*r0))
y_a = np.zeros((2, xs.size))
y_a[1]=-0.01396517
y_b = np.zeros((2, xs.size))
#xs2 = np.linspace(0, r0, (2**12)*r0)
res_a = solve_bvp(Func, bc, xs, y_a, tol=1e-9)
y0 = res_a.sol(xs)[0]
d0 = res_a.sol(xs)[1]
return([xs, y0], [xs, d0])
def bc_bvp(Func, BCs, r0):
#creates an array of 9999 points equally spaced from 0 to r0 (the position of the visible brane)
xs = np.linspace(r0, 0, 9999)
it=0 #counts number of interations in th shooting method
abserr=50 #sets the "initial" error so the the while loop can functiond
BH0=V+0.1 #estimate for the initial boundary that the while loop will iteratively improve
DBH0=BCs(BH0, LamV, V) #estimate for the initial gradient
shoot_J=V#sets the intial interval over which the shooting method varies
# the while loop will interate, improving upon the initial conditions until it has produced the
# boundary condition the the visible boundary to the error as described in the while loop condition
while(abserr>1e-14):
#print(BH0)
DBH0=-BCs(BH0, LamV, V)
U0 = [BH0, DBH0]
Us0 = odeint(dU_dx, U0, xs, args=(r0, k))
y0 = Us0[:,0]
d0 = Us0[:,1]
#print(y0)
trunc=0
for i in range(0, y0.size):
if(((y0[i]>2*H) or (y0[i]<0)) and trunc==0):
trunc=1
cont=y0[i]
cont2=d0[i]
if(trunc==1):
y0[i]=cont
d0[i]=cont2
BV1=y0[-1]#finds y value at r0
dsBC=BCs(BV1, LamH, H)
DBV1=d0[-1]
#for a given gradient, the following lines find the largest non imaginary solution for the boundaries
temp=[]
coeff=[-2*LamH, 0, 2*LamH*H*H, +DBV1]
solutions=np.roots(coeff)
for x in solutions:
if (np.iscomplex(x)==False):
temp=np.append(temp, x)
solution=np.real(np.amax(temp))
err=solution-BV1
abserr=abs(err)
if(err>0):
BH0=BH0+shoot_J
if(err<0):
BH0=BH0-shoot_J
if(it>=500):
break
shoot_J=shoot_J*0.5
it=it+1
#print("uncertainty on boundary: ", abserr)
return([np.flip(xs), np.flip(y0)], [np.flip(xs), np.flip(d0)])
if(LamV and LamH):
phi, d_phi = bc_bvp(dU_dx, boundaries, r0)
else:
phi, d_phi = no_bc_bvp(func2, r0)
ys=phi[1]; ds=d_phi[1]
xs=phi[0]
# BStiff =(H*(np.exp((2+c)*k*r0))-V)/(np.exp((2+c)*k*r0) - np.exp((2-c)*k*r0))
# AStiff=H-BStiff
AStiff = ((V * np.exp((c-2)*k*r0)) - H)/(np.exp(2*c*k*r0)-1)
BStiff = H-AStiff
def Phi(A0, B0, r):
return(np.exp(2*k*r)*((A0*np.exp(c*k*r))+(B0*np.exp(-c*k*r))))
if visualise:
fig = plt.figure(figsize=(15,8))
plt.plot(xs, ys, label = "With Coupling")
plt.plot(xs, Phi(AStiff, BStiff, xs), label = "Analytic Limit")
plt.xlabel("$y$", fontsize=13)
plt.ylabel("$\Phi(y)$", fontsize=13)
plt.legend()
plt.show()
return(phi, d_phi)
def Bulk_Potential(H: float, V: float, Ymin: float, Ymax: float, LamH=None, LamV=None, LamB=0.0, k: float = 1, m: float = 0.2, visualise: bool = True, value_returned=None):
c=np.sqrt(4+(np.power(m,2)/np.power(k,2)))
if(Ymin==0):
print("Please do not set Ymin to 0, as you can't intergrate over 0 distance!!!")
E=np.sqrt(m**2)/(4*k**2)
#E=np.sqrt(4+(np.power(m,2)/np.power(k,2)))-2
Potential=[]
Y = []
y = Ymin
while y <= Ymax:
phi, d_phi = Bulk_Scalar(H, V, y, LamB=LamB, LamH=LamH, LamV=LamV, k=k, m=m, visualise=False)
ys=phi[1]; ds=d_phi[1]
xs=phi[0]
#the values of the field and its derivative for a given r0 are plugged into the Lagrangian
e=np.exp(-4*k*xs)
L=e*((ds*ds)+(m*m*ys*ys)+(LamB*ys*ys*ys)) #our lagrangean
value = simps(L, xs)
if(LamV and LamH):
value = value + LamH*((ys[0]*ys[0]-H*H)*(ys[0]*ys[0]-H*H))+LamV*(np.exp(-4*k*y)*((ys[-1]*ys[-1]-V*V)*(ys[-1]*ys[-1]-V*V)))
Potential=np.append(Potential, value)
Y=np.append(Y, y)
y=y+0.01
print(y, end="\r")
A = ((V * np.exp((c-2)*k*Y)) - H)/(np.exp(2*c*k*Y)-1)
B = H-A
Pot=(A*A*k*(2+c)*(np.exp(2*c*k*Y)-1) + k*(c-2)*B*B*(1-np.exp(-2*c*k*Y)))
min_position_numerical=Potential.argmin()
second_deriv_num=np.gradient(np.gradient(Potential, Y), Y)
ym = Y[Potential.argmin()]
yc = Y[Pot.argmin()]
mass = second_deriv_num[min_position_numerical]*(np.exp(2*k*(yc+ym))/(8*k*k*k*E*E*V*V))
if visualise:
fig= plt.figure(figsize=(15,8))
plt.plot(Y, Potential, label = "With Coupling")
plt.plot(Y[1:], Pot[1:], label = "Analytic Limit")
#print()
plt.ylim(0, 2*Potential[0])
#plt.plot(xs, Phi(AStiff, BStiff, xs), label = "Analytic Limit")
plt.ylabel("$V_{GW}$", fontsize=13)
plt.xlabel("Brane Sepration $y_0$", fontsize=13)
plt.legend()
plt.show()
#mass1= d2Npot_dY2[NPotMinPos]*(np.exp(2*k*(Y[NPotMinPos]+Smpos))/(8*k*k*k*E*E*V*V))
print("mr/m0: ", np.sqrt(mass))
print("Minima, ", "y: ", ym, " V: ", Potential.min())
print("Analytical Min, y:", yc, "V:", Pot.min())
print("V(0): ", Potential[0], " Analytical Value: ", ((LamH*LamV)/(LamH+LamV))*(V*V - H*H)*(V*V - H*H) )
if(value_returned == "mass"):
return mass
if(value_returned == "depth"):
return Potential[-1] - Potential.min()
else:
return [Y, Potential]
# c=np.sqrt(4+(np.power(m,2)/np.power(k,2)))
# E=np.sqrt(4+(np.power(m,2)/np.power(k,2)))-2
# #the EL equation of motion that must be solved (Numerically)
# def dU_dx(U, x):
# # Here U is a vector such that phi=U[0] and phi'=U[1]. This function should return [phi', phi'']
# return [U[1], 4*k*U[1] + m*m*U[0]+LamB*U[0]*U[0]*U[0]]
# def Phi(A0, B0, r):
# return(np.exp(2*k*r)*((A0*np.exp(c*k*r))+(B0*np.exp(-c*k*r))))
# def V_lim(rc, V, H):
# A=np.float64((V-H*np.exp((2-c)*k*rc))/(np.exp((2+c)*k*rc)-np.exp((2-c)*k*rc)))
# B=np.float64(H-A)
# V=np.float64((A*A*k*(2+c)*(np.exp(2*c*k*rc)-1) + k*(c-2)*B*B*(1-np.exp(-2*c*k*rc))))
# return V
# def solve_bvp_HM(Func, r0):
# BStiff=((V*(np.exp(-(2+c)*k*r0))-H)/((np.exp(-2*k*c*r0))-1))
# AStiff=H-BStiff
# #creates an array of 9999 points equally spaced from 0 to r0 (the position of the visible brane)
# xs = np.linspace(r0, 0, 9999)
# it=0 #counts number of interations in th shooting method
# abserr=50 #sets the "initial" error so the the while loop can functiond
# DBH0=-10 #estimate for the initial gradient that the while loop will iteratively improve
# shoot_J=DBH0#sets the intial interval over which the shooting method varies
# BH0=H#sets the intial condition of phi (unchanged by shooting method)
# # the while loop will interate, improving upon the initial conditions until it has produced the
# # boundary condition the the visible boundary to the error as described in the while loop condition
# while(abserr>1e-17):
# U0 = [V, DBH0]
# Us0 = odeint(Func, U0, xs)
# y0 = Us0[:,0]
# d0 = Us0[:,1]
# trunc=0
# for i in range(0, y0.size):
# if(((y0[i]>2*H) or (y0[i]<0)) and trunc==0):
# trunc=1
# cont=y0[i]
# if(trunc==1):
# y0[i]=cont
# #print(y0[0], y0[-1])
# BV1=y0[-1]#finds y value at 0
# #finds the difference between the visible boundary condition, and the value of this iterations
# #phi at the visible boundary. the initial conditions are then adjusted accordingly, by bisecting
# #the interval until it the error on the visible boundary is small enough
# err=BV1-H
# abserr=abs(err)
# if(err>0):
# DBH0=DBH0-shoot_J
# if(err<0):
# DBH0=DBH0+shoot_J
# if(it>=500):
# break
# shoot_J=shoot_J*0.5
# it=it+1
# print(DBH0, err)
# return([xs, y0], [xs, d0])
# #the EL equation of motion that must be solved (Numerically)
# def func2(x, U):
# global LamB
# # Here U is a vector such that phi=U[0] and phi'=U[1]. This function should return [phi', phi'']
# return np.vstack((U[1], 4*k*U[1] + m*m*U[0]+LamB*U[0]*U[0]*U[0]))
# from scipy.integrate import solve_bvp
# def New_solve_bvp(Func, r0):
# def bc(ya, yb):
# return np.array([ya[0]-H, yb[0]-V])
# xs = np.linspace(0, r0, round((2**10)*r0*r0))
# y_a = np.zeros((2, xs.size))
# y_a[1]=1
# y_b = np.zeros((2, xs.size))
# #xs2 = np.linspace(0, r0, (2**12)*r0)
# res_a = solve_bvp(Func, bc, xs, y_a, tol=1e-8)
# y0 = res_a.sol(xs)[0]
# d0 = res_a.sol(xs)[1]
# return([xs, y0], [xs, d0])
# def Phi(A0, B0, r):
# return(np.exp(2*k*r)*((A0*np.exp(c*k*r))+(B0*np.exp(-c*k*r))))
# def NumV(Func, Ymin, Ylim):
# global LamB
# Potential=[]
# Y=[]
# while(Ymin < Ylim):
# phi, d_phi = New_solve_bvp(Func, Ymin)
# ys=phi[1]; ds=d_phi[1]
# xs=phi[0]
# print(Ymin)
# #the values of the field and its derivative for a given r0 are plugged into the Lagrangian
# e=np.exp(-4*k*xs)
# L=e*((ds*ds)+(m*m*ys*ys)+(LamB*ys*ys*ys)) #our lagrangean
# V=simps(L, xs)
# Potential=np.append(Potential, V)
# Y=np.append(Y, Ymin)
# Ymin=Ymin+0.05
# return [Y, Potential]