from getBoundary import getBoundary

import numpy as np

import pylab as pl

from fig import fig, saveFig, printText, getPlotY

from printStatus import printRatio

from pickle import load, dump

hpsis=np.logspace(-6,1.5,300)

a2=0.0

plotImag=False

from solveBC import sweep, findrho

oscN=6

try:

bbs,sols=load(open('cache/sols_oscN='+str(oscN)+'_a2='+str(a2)))

except IOError:

print('Solve for sweep of different horizon BCs...')

bbs,sols=sweep(lambda x,y: getBoundary(x,y,[0,a2]), hpsis, oscN=oscN, status=True)

dump((bbs,sols),open('cache/sols_oscN='+str(oscN)+'_a2='+str(a2),'w'))

rho=-np.array(bbs)[:,:,1,1].real

mu=np.array(bbs)[:,:,1,0].real

O=np.array(bbs)[:,:,0,1].real

#const=mu

const=np.sqrt(rho)

cConst=min([min(r) for r in const])#rho, in units used at Tc

zh=1.#choice of length units makes zh numerically constant

T=3./(zh*4.*np.pi)#temp of solution in units used, this is constant by choice of units

for osci in range(len(mu)):

sign=-1 if O[osci][0]<0 else 1

Tc=T/cConst*const[osci]

fig(1)

pl.plot(cConst/const[osci],np.sqrt(sign*O[osci]*np.sqrt(2))/Tc,ls='-' if sign==1 else '--',c='k')

fig(2)

pl.plot(cConst/const[osci],mu[osci]/Tc,ls='-' if sign==1 else '--',c='k')

print O[osci]

fig(1)

pl.plot([0,1.2],[0,0],lw=1)

pl.ylim([-1,9])

#pl.legend(loc='upper right')

pl.xlabel(r'$\frac{T}{T_c}$')

pl.ylabel(r'$\frac{\sqrt{\sqrt{2}|\langle\mathcal{O}\rangle|}}{T_c}$')

saveFig('O_constRho_a2_'+str(a2))

fig(2)

Trs=np.linspace(0,1.2)[1:]

#pl.plot(Trs,rho*zh/Tc,lw=1)

pl.plot(Trs,min([min(r) for r in rho])*(zh/Trs)/T,lw=1)

#pl.plot(Trs,Trs*7.71285,ls=':')

pl.ylim([0,50])

#pl.legend(loc='upper right')

pl.xlabel(r'$\frac{T}{T_c}$')

pl.ylabel(r'$\frac{\mu}{T_c}$')

saveFig('mu_constRho_a2_'+str(a2))

pl.show()