#!/usr/bin/env python
# coding: utf-8

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# 
# # Code and Physical Units
# 
# ## GRHD Units in terms of EOS 
# 
# ## This notebook introduces an equation of state for general relativistic hydrodynamics (GRHD) and demonstrates the conversion of dimensional quantities to dimensionless units.
# 
# **Notebook Status:** <font color='red'><b> Not Validated </b></font>
# 
# 
# ## Introduction
# $\newcommand{\rhoCode}{{\tilde{\rho}}}$
# $\newcommand{\MCode}{{\tilde{M}}}$ $\newcommand{\rCode}{{\tilde{r}}}$ $\newcommand{\PCode}{{\tilde{P}}}$$\newcommand{\tCode}{{\tilde{t}}}$$\newcommand{\Mfid}{{M_{\rm fid}}}$$\newcommand{\MfidBar}{\bar{M}_{\rm fid}}$$\newcommand{\Mbar}{\bar{M}}$
# $\newcommand{\rBar}{\bar{r}}$$\newcommand{\tBar}{\bar{t}}$
# In GRHD, we can set an equation of state of the form
# \begin{equation}
# P = K\rho^{1+1/n}
# \end{equation}
# Taking $c_s^2 = \partial P/\partial \rho = (1+1/n) K\rho^{1/n}$.  This gives for some fiducial $\rho_0$
# \begin{equation}
# c_{s,0}^2 = \left(1 + \frac 1 n\right)K\rho_0^{1/n}.
# \end{equation}
# Selecting $c_s^2 = c^2\left(1 + 1/n\right)$, we have
# \begin{equation}
# \rho_0 = \left(\frac {c^2}{K}\right)^n
# \end{equation}
# This is equivalent to setting the isothermal sound speed to $c$.  With this definition of $\rho_0$, we can write
# \begin{equation}
# P = \rho_0c^2\left(\frac{\rho}{\rho_0}\right)^{1+1/n}
# \end{equation}
# which allows us to define the dimensionless density $\rhoCode = \rho/\rho_0$ and dimensionless pressure $\PCode = P/\rho_0 c^2$
# \begin{equation}
# \PCode = \rhoCode^{1+1/n},
# \end{equation}
# where we adopt code units where $c=1$.  These dimensionless pressure and density are in $G=c=1$ units and can be used in GRHD code including inclusion in the spacetime solver via $T_{\mu\nu}$.  Note that this sets $K=1$ in these units.
# 
# To find a dimensionless mass, $\MCode$, dimensionless distance, $\rCode$, and dimensionless time, $\tCode$, we note
# $GM/rc^2$ is dimensionless
# \begin{equation}
# \frac{GM}{rc^2} = \frac{G\rho_0 r^2}{c^2} = \frac{Gc^{2n-2}}{K^n}r^2 \rightarrow \rCode = \frac{\sqrt{G}c^{n-1}}{K^{n/2}} r = \frac r {r_0},
# \end{equation}
# where $r_0 = K^{n/2}/\sqrt{G}c^{n-1}$.  Then
# \begin{eqnarray}
# \tCode &=& \frac{t}{t_0} = \frac{t}{r_0/c} = \frac{\sqrt{G}c^n}{K^{n/2}} t \\
# \MCode &=& \frac{M}{M_0} = \frac{M}{\rho_0 r_0^3} = M\frac{K^n}{c^{2n}}\frac{G^{3/2}c^{3(n-1)}}{K^{3n/2}} = \frac{G^{3/2}c^{n-3}}{K^{n/2}} M,
# \end{eqnarray}
# Hence, we have 
# \begin{eqnarray}
# \rho_0 &=& \left(\frac{K}{c^2}\right)^n\\
# r_0 &=& \frac{c^{n+1}}{\sqrt{G}K^{n/2}}\\
# t_0 &=& \frac{c^{n}}{\sqrt{G}K^{n/2}}\\
# M_0 &=& \frac{c^{n+3}}{G^{3/2}K^{n/2}}
# \end{eqnarray}
# 
# ## Mapping to SENR or any NR code
# 
# So we will need a $\Mfid$ which is defined such that the (SENR) code units $\MfidBar = 1$ or in other words in SENR codes units: 
# \begin{equation}
# \Mbar = \frac{M}{\Mfid}
# \end{equation}
# In these units:
# \begin{eqnarray}
# \rBar &=& \frac{c^2}{G\Mfid} r\\
# \tBar &=& \frac{c^3}{G\Mfid} t
# \end{eqnarray}
# At some level $\Mfid$ is arbitrary, so we can select $M_0 = \Mfid$.  In this case, this means that $\rBar = \rCode$, $\tBar = \tCode$, and $\Mbar = \MCode$, which fixes all the quantities. This comes at a cost the $\bar{M}_{\rm ADM}$ is not something nice like 1 or 2, but the choice is consistent.

# <a id='toc'></a>
# 
# # Table of Contents
# $$\label{toc}$$
# 
# The module is organized as follows:
# 
# 1. [Step 1](#comments): Zach's comments
# 1. [Step 2](#tov_solver): TOV solver as illustration
#     1. [Step 2.a](#prescription): Another dimensionless prescription
#     1. [Step 2.b](#metric): Metric for TOV equation
# 1. [Step 3](#adm_conversion): Convert metric to be in terms of ADM quantities
# 1. [Step 4](#cartesian_conversion): Convert to Cartesian coordinates
# 1. [Step 5](#latex_pdf_output): Output this notebook to $\LaTeX$-formatted PDF file

# <a id='comments'></a>
# 
# # Step 1: Zach's comments \[Back to [top](#toc)\]
# $$\label{comments}$$
# 
# Sound speed $c_s$ is defined as
# 
# $$\frac{\partial P}{\partial \rho} = c_s^2,$$
# 
# so if we have a polytropic EOS, where
# 
# $$P = K \rho^{(1 + 1/n)},$$
# 
# then
# 
# \begin{align}
# \frac{\partial P}{\partial \rho} &= c_s^2 \\
# &= \left(1 + \frac{1}{n}\right) K \rho^{1/n}.
# \end{align}
# 
# Let's adopt the notation 
# 
# $$[\rho] = \text{"the units of $\rho$"}$$
# 
# Using this notation and the fact that $n$ is dimensionless, the above expression implies
# 
# \begin{align}
# \left[\rho^{1/n}\right] &= \left[\frac{c_s^2}{K}\right] \\
# \implies \left[\rho\right] &= \left[\frac{c_s^2}{K}\right]^n
# \end{align}
# 
# I think you found the inverse to be true.

# <a id='tov_solver'></a>
# 
# # Step 2: TOV Solver as illustration \[Back to [top](#toc)\]
# $$\label{tov_solver}$$
# 
# The TOV equations are 
# \begin{eqnarray}
# \frac{dP}{dr} &=& -\mu\frac{GM}{r^2}\left(1 + \frac P {\mu c^2}\right)\left(1 + \frac {4\pi r^3 P}{Mc^2}\right)\left(1 - \frac {2GM}{rc^2}\right)^{-1}\\
# \frac{dM}{dr} &=& 4\pi \mu r^2,
# \end{eqnarray}
# Here $M$ is the rest mass measured by a distant observer when we take $r\rightarrow \infty$.  Note this is different from the mass measured by integrating the density over the volume
# \begin{equation}
# M' = \int_0^{\infty} \frac{4\pi r^2\mu}{\sqrt{1 - \frac {2 GM}{rc^2}}} dr
# \end{equation}
# Additionally and annoyingly, $\mu = \rho h$ is the mass-energy density.  A lot of the literature uses $\rho$ for this, which is incredibly annoying. 
# 
# $\newcommand{\muCode}{{\tilde{\mu}}}$
# 
# In dimensionless units they are 
# \begin{eqnarray}
# \frac{d\PCode}{d\rCode} &=& -\frac {\left(\muCode + \PCode\right)\left(\MCode + 4\pi \rCode^3 \PCode\right)}{\rCode^2\left(1 - \frac {2\MCode}{\rCode}\right)}\\
# \frac{d\MCode}{d\rCode} &=& 4\pi \muCode\rCode^2
# \end{eqnarray}
# 
# At this point, we need to discuss how to numerically integrate these models. First, we pick a central baryonic mass density $\rhoCode_{0,c}$, then we compute a central pressure $\PCode_c$ and central mass-energy density $\muCode_c$.  At $\rCode=0$, we assume that $\muCode=\muCode_c$ is a constant and so 
# \begin{eqnarray}
# \frac{d\PCode}{d\rCode} &=& -\frac {\left(\muCode_c + \PCode_c\right)\left(\MCode(\rCode \ll 1) + 4\pi \rCode^3 \PCode_c\right)}{\rCode^2\left(1 - \frac {2\MCode(\rCode \ll 1)}{\rCode}\right)}\\
# \frac{d\MCode}{d\rCode} &=& 4\pi \muCode_c\rCode^2 \rightarrow \MCode(\rCode \ll 1) = \frac{4\pi}{3} \muCode_c \rCode^3 
# \end{eqnarray}

# <a id='prescription'></a>
# 
# ## Step 2.a: Another dimensionless prescription \[Back to [top](#toc)\]
# $$\label{prescription}$$
# 
# Let consider an alternative formulation where rather than setting $K=1$, we set the characteristic mass $\Mfid = M_0$.  In this case,
# \begin{eqnarray}
# r_0 &=& \frac{GM_0}{c^2} \\
# t_0 &=& \frac{GM_0}{c^3} \\
# \rho_0 &=& \frac{M_0}{r_0^3} = \frac{c^6}{G^3 M_0^2} = 6.17\times 10^{17}\left(\frac {M_0} {1 M_{\odot}}\right)^{-2}
# \end{eqnarray}
# In this case we can define $\rhoCode = \rho/\rho_0$, $\rCode = r/r_0$, $t_0 = t/t_0$. The only remaining thing to do is to define $\PCode$.  Lets define $P_0'$ to be the pressure in dimensionful units at $\rho_0$ (density in units of $1/M_0^2$):
# \begin{equation}
# P = P_0'\rhoCode^{1+1/n} \rightarrow P_0' = K\rho_0^{1+1/n},
# \end{equation}
# So defining $P_0 = \rho_0 c^2$, we have
# \begin{equation}
# \PCode = \frac{P}{P_0} = \frac{K\rho_0^{1/n}}{c^2}\rhoCode^{1+1/n} = \PCode_0\rhoCode^{1+1/n}
# \end{equation}
# If we take $K=1$ and define $\rho_0$ such that the $\PCode_0 = 1$, we recover the results above.
# Finally for $\muCode = \rhoCode + \PCode/n$

# <a id='metric'></a>
# 
# ## Step 2.b: Metric for TOV equation \[Back to [top](#toc)\]
# $$\label{metric}$$
# 
# The metric for the TOV equation (taken) from Wikipedia is
# \begin{equation}
# ds^2 = - c^2 e^\nu dt^2 + \left(1 - \frac{2GM}{rc^2}\right)^{-1} dr^2 + r^2 d\Omega^2
# \end{equation}
# where $M$ is defined as above, the mass as measured by a distant observer.  The equation for $\nu$ is
# \begin{equation}
# \frac{d\nu}{dr} = -\left(\frac {2}{P +\mu}\right)\frac{dP}{dr}
# \end{equation}
# with the boundary condition
# \begin{equation}
# \exp(\nu) = \left(1-\frac {2Gm(R)}{Rc^2}\right)
# \end{equation}
# 
# Let's write this in dimensionless units:
# \begin{equation}
# ds^2 = \exp(\nu) d\tCode^2 - \left(1 - \frac{2\MCode}{\rCode}\right)^{-1} d\rCode^2 + \rCode^2 d\Omega^2
# \end{equation}
# \begin{equation}
# \frac{d\nu}{d\rCode} = -\left(\frac {2}{\PCode +\muCode}\right)\frac{d\PCode}{d\rCode}
# \end{equation}
# and BC:
# \begin{equation}
# \exp(\nu) = \left(1-\frac {2\MCode}{\rCode}\right)
# \end{equation}

# In[1]:


import sys
import numpy as np
import scipy.integrate as si
import math
import matplotlib.pyplot as pl
n = 1.
rho_central = 0.129285

P0 = 1. # ZACH NOTES: CHANGED FROM 100.
gamma = 1. + 1./n
gam1 = gamma - 1.

def pressure( rho) :
    return P0*rho**gamma

def rhs( r, y) :
# In \tilde units
#
    P = y[0]
    m   = y[1]
    nu  = y[2]
    rbar = y[3]

    rho = (P/P0)**(1./gamma)
    mu = rho + P/gam1
    dPdr = 0.
    drbardr = 0.
    if( r < 1e-4 or m <= 0.) :
        m = 4*math.pi/3. * mu*r**3
        dPdr = -(mu + P)*(4.*math.pi/3.*r*mu + 4.*math.pi*r*P)/(1.-8.*math.pi*mu*r*r)
        drbardr = 1./(1. - 8.*math.pi*mu*r*r)**0.5
    else :
        dPdr = -(mu + P)*(m + 4.*math.pi*r**3*P)/(r*r*(1.-2.*m/r))
        drbardr = 1./(1. - 2.*m/r)**0.5*rbar/r

    dmdr = 4.*math.pi*r*r*mu
    dnudr = -2./(P + mu)*dPdr
    return [dPdr, dmdr, dnudr, drbardr]

def integrateStar( P, showPlot = False, dumpData = False, compareFile="TOV/output_EinsteinToolkitTOVSolver.txt") :
    integrator = si.ode(rhs).set_integrator('dop853')
    y0 = [P, 0., 0., 0.]
    integrator.set_initial_value(y0,0.)
    dr = 1e-5
    P = y0[0]

    PArr = []
    rArr = []
    mArr = []
    nuArr = []
    rbarArr = []
    r = 0.

    while integrator.successful() and P > 1e-9*y0[0] :
        P, m, nu, rbar = integrator.integrate(r + dr)
        r = integrator.t

        dPdr, dmdr, dnudr, drbardr = rhs( r+dr, [P,m,nu,rbar])
        dr = 0.1*min(abs(P/dPdr), abs(m/dmdr))
        dr = min(dr, 1e-2)
        PArr.append(P)
        rArr.append(r)
        mArr.append(m)
        nuArr.append(nu)
        rbarArr.append( rbar)

    M = mArr[-1]
    R = rArr[-1]

    nuArr_np = np.array(nuArr)
    # Rescale solution to nu so that it satisfies BC: exp(nu(R))=exp(nutilde-nu(r=R)) * (1 - 2m(R)/R)
    #   Thus, nu(R) = (nutilde - nu(r=R)) + log(1 - 2*m(R)/R)
    nuArr_np = nuArr_np - nuArr_np[-1] + math.log(1.-2.*mArr[-1]/rArr[-1])

    rArrExtend_np = 10.**(np.arange(0.01,5.0,0.01))*rArr[-1]

    rArr.extend(rArrExtend_np)
    mArr.extend(rArrExtend_np*0. + M)
    PArr.extend(rArrExtend_np*0.)
    phiArr_np = np.append( np.exp(nuArr_np), 1. - 2.*M/rArrExtend_np)
    rbarArr.extend( 0.5*(np.sqrt(rArrExtend_np*rArrExtend_np-2.*M*rArrExtend_np) + rArrExtend_np - M))
    # Appending a Python array does what one would reasonably expect.
    #   Appending a numpy array allocates space for a new array with size+1,
    #   then copies the data over... over and over... super inefficient.
    mArr_np       = np.array(mArr)
    rArr_np       = np.array(rArr)
    PArr_np       = np.array(PArr)
    rbarArr_np    = np.array(rbarArr)
    rhoArr_np     = (PArr_np/P0)**(1./gamma)
    confFactor_np = rArr_np/rbarArr_np
    #confFactor_np = (1.0 / 12.0) * np.log(1.0/(1.0 - 2.0*mArr_np/rArr_np))
    Grr_np        = 1.0/(1.0 - 2.0*mArr_np/rArr_np)
    Gtt_np        = phiArr_np
    if( showPlot) :
        r,rbar,rprop,rho,m,phi = np.loadtxt( compareFile, usecols=[0,1,2,3,4,5],unpack=True)
        pl.plot(rArr_np[rArr_np < r[-1]], rbarArr_np[rArr_np < r[-1]],lw=2,color="black")
        #pl.plot(r, rbar, lw=2,color="red")

        pl.show()


        if( dumpData) :
            np.savetxt( "output.txt", np.transpose([rArr_np,rhoArr_np,PArr_np,mArr_np,phiArr_np,confFactor_np,rbarArr_np]), fmt="%.15e")
            np.savetxt( "metric.txt", np.transpose([rArr_np, Grr_np, Gtt_np]),fmt="%.15e")
        #        np.savetxt( "output.txt", zip(rArr,rhoArr,mArr,phiArr), fmt="%12.7e")

#    return rArr[-1], mArr[-1], phiArr[-1]
    return R, M

mass = []
radius = []

R_TOV,M_TOV = integrateStar(pressure(rho_central), showPlot=True, dumpData=True)
print("Just generated a TOV star with r= "+str(R_TOV)+" , m = "+str(M_TOV)+" , m/r = "+str(M_TOV/R_TOV)+" .")

#for rho0 in np.arange(0.01, 1., 0.01):
#    r,m = integrateStar(pressure(rho0))
#    mass.append(m)
#    radius.append(r)

#print(mass, radius)
#pl.clf()
#pl.plot(radius,mass)
#pl.show()


# In[2]:


# Generate the Sedov Problem

rArr_np = np.arange(0.01,5.,0.01)
rbarArr_np = rArr_np
rhoArr_np = np.ones(rArr_np.size)*0.1
mArr_np = 4.*np.pi/3.*rArr_np**3*rhoArr_np
PArr_np = rhoArr_np*1e-6
PArr_np[rArr_np < 0.5] = 1e-2
phiArr_np = np.ones(rArr_np.size)
confFactor_np = rArr_np/rbarArr_np
if sys.version_info[0] < 3:
    np.savetxt( "sedov.txt", zip(rArr_np,rhoArr_np,PArr_np,mArr_np,phiArr_np,confFactor_np,rbarArr_np), fmt="%.15e")
else:
    np.savetxt( "sedov.txt", list(zip(rArr_np,rhoArr_np,PArr_np,mArr_np,phiArr_np,confFactor_np,rbarArr_np)), fmt="%.15e")

pl.semilogx(rArr_np, rhoArr_np)
pl.show()


# <a id='adm_conversion'></a>
# 
# # Step 3: Convert metric to be in terms of ADM quantities \[Back to [top](#toc)\]
# $$\label{adm_conversion}$$
# 
# Above, the line element was written:
# $$
# ds^2 = - c^2 e^\nu dt^2 + \left(1 - \frac{2GM}{rc^2}\right)^{-1} dr^2 + r^2 d\Omega^2.
# $$
# 
# In terms of $G=c=1$ units adopted by NRPy+, this becomes:
# $$
# ds^2 = - e^\nu dt^2 + \left(1 - \frac{2M}{r}\right)^{-1} dr^2 + r^2 d\Omega^2.
# $$
# 
# The ADM 3+1 line element for this diagonal metric in spherical coordinates is given by:
# $$
# ds^2 = (-\alpha^2 + \beta_k \beta^k) dt^2 + \gamma_{rr} dr^2 + \gamma_{\theta\theta} d\theta^2+ \gamma_{\phi\phi} d\phi^2,
# $$
# 
# from which we can immediately read off the ADM quantities:
# \begin{align}
# \alpha &= e^{\nu/2} \\
# \beta^k &= 0 \\
# \gamma_{rr} &= \left(1 - \frac{2M}{r}\right)^{-1}\\
# \gamma_{\theta\theta} &= r^2 \\
# \gamma_{\phi\phi} &= r^2 \sin^2 \theta \\
# \end{align}

# <a id='cartesian_conversion'></a>
# 
# # Step 4: Convert to Cartesian coordinates \[Back to [top](#toc)\]
# $$\label{cartesian_conversion}$$
# 
# The above metric is given in spherical coordinates and we need everything in Cartesian coordinates.  Given this the 
# transformation to Cartesian coordinates is 
# \begin{equation}
# g_{\mu\nu} = \Lambda^{\mu'}_{\mu} \Lambda^{\nu'}_{\nu} g_{\mu'\nu'},
# \end{equation}
# where $\Lambda^{\mu'}_{\mu}$ is the Jacobian defined as
# \begin{equation}
# \Lambda^{\mu'}_{\mu} = \frac{\partial x'^{\mu'}}{\partial x^{\mu}}
# \end{equation}
# In this particular case $x'$ is in spherical coordinates and $x$ is in Cartesian coordinates.  

# In[3]:


import sympy as sp                  # SymPy: The Python computer algebra package upon which NRPy+ depends
import NRPy_param_funcs as par      # NRPy+: parameter interface
from outputC import outputC         # NRPy+: Basic C code output functionality
import indexedexp as ixp            # NRPy+: Symbolic indexed expression (e.g., tensors, vectors, etc.) support
import reference_metric as rfm      # NRPy+: Reference metric support

# The ADM & BSSN formalisms only work in 3D; they are 3+1 decompositions of Einstein's equations.
#    To implement axisymmetry or spherical symmetry, simply set all spatial derivatives in
#    the relevant angular directions to zero; DO NOT SET DIM TO ANYTHING BUT 3.
# Step 0: Set spatial dimension (must be 3 for BSSN)
DIM = 3

# Set the desired *output* coordinate system to Cylindrical:
par.set_parval_from_str("reference_metric::CoordSystem","Cartesian")
rfm.reference_metric()

CoordType_in = "Spherical"

r_th_ph_or_Cart_xyz_of_xx = []
if CoordType_in == "Spherical":
    r_th_ph_or_Cart_xyz_of_xx = rfm.xxSph
elif CoordType_in == "Cartesian":
    r_th_ph_or_Cart_xyz_of_xx = rfm.xx_to_Cart

Jac_dUSphorCart_dDrfmUD = ixp.zerorank2()
for i in range(DIM):
    for j in range(DIM):
        Jac_dUSphorCart_dDrfmUD[i][j] = sp.diff(r_th_ph_or_Cart_xyz_of_xx[i],rfm.xx[j])

Jac_dUrfm_dDSphorCartUD, dummyDET = ixp.generic_matrix_inverter3x3(Jac_dUSphorCart_dDrfmUD)

betaU   = ixp.zerorank1()
gammaDD = ixp.zerorank2()
gammaSphDD = ixp.zerorank2()
grr, gthth, gphph = sp.symbols("grr gthth gphph")
gammaSphDD[0][0] = grr
gammaSphDD[1][1] = gthth
gammaSphDD[2][2] = gphph
betaSphU = ixp.zerorank1()
for i in range(DIM):
    for j in range(DIM):
        betaU[i] += Jac_dUrfm_dDSphorCartUD[i][j] * betaSphU[j]
        for k in range(DIM):
            for l in range(DIM):
                gammaDD[i][j] += Jac_dUSphorCart_dDrfmUD[k][i]*Jac_dUSphorCart_dDrfmUD[l][j] * gammaSphDD[k][l]

outputC([gammaDD[0][0], gammaDD[0][1], gammaDD[0][2], gammaDD[1][1], gammaDD[1][2], gammaDD[2][2]],
              ["mi.gamDDxx", "mi.gamDDxy", "mi.gamDDxz", "mi.gamDDyy", "mi.gamDDyz","mi.gamDDzz"], filename="NRPY+spherical_to_Cartesian_metric.h")


# <a id='latex_pdf_output'></a>
# 
# # Step 5: Output this notebook to $\LaTeX$-formatted PDF file \[Back to [top](#toc)\]
# $$\label{latex_pdf_output}$$
# 
# The following code cell converts this Jupyter notebook into a proper, clickable $\LaTeX$-formatted PDF file. After the cell is successfully run, the generated PDF may be found in the root NRPy+ tutorial directory, with filename
# [Tutorial-GRHD_UnitConversion.pdf](Tutorial-GRHD_UnitConversion.pdf). (Note that clicking on this link may not work; you may need to open the PDF file through another means.)

# In[2]:


import cmdline_helper as cmd    # NRPy+: Multi-platform Python command-line interface
cmd.output_Jupyter_notebook_to_LaTeXed_PDF("Tutorial-GRHD_UnitConversion")