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(R is \ the distance where the pressure p(r=R) = 0.) \nEnergy density and pressure, \ epsilon(r) and p(r), have units \nMeV/fm^3 or ergs/cm^3 or Msun*c^2/km^3. \ However, it is more\nconvenient to switch to dimensionless forms for these \ quantities. \n\n\tThe equation for p(r) is\n \n dp(r)/dr = \ -Rschw*epsilon(r)*curlyMbar(r)/r^2\n \ntimes three dimensionless \ correction factors from GR.\nHere Rschw = G * Msun/c^2 = 1.48 km is ", StyleBox["HALF", FontWeight->"Bold"], " the Schwartzschild radius \nof the Sun. In CGS units, to check that:" }], "Text"], Cell[CellGroupData[{ Cell["\<\ G = 6.67 10^(-8); (* dyne-cm^2/g^2 *) Msun = 1.989 10^33 ; (* gram *) c = 2.998 10^10; (* cm/sec *) Rschw = G Msun/c^2; (* cm *) Rschw = Rschw/10^5 (* km *)\ \>", "Input"], Cell[BoxData[ \(1.476037393841836`\)], "Output"] }, Open ]], Cell["\<\ \tWhat is the central pressure for a star of constant energy \ density, mass = Msun, and radius 10 km? From Aidan Parker's eq. (6),\ \>", "Text"], Cell[CellGroupData[{ Cell["\<\ bigR = 10.0 10^3 10^2; (* 10 km radius, in cm *) p0 = 3 G Msun^2/(8 Pi bigR^4); (* dynes/cm^2 *) p0 = p0 / 10.0; (* in pascals = kg m/s^2 /m^2 *) p0 = p0 / 10^9 (* in gigapascals *) \ \>", "Input"], Cell[BoxData[ \(3.14975515362981`*^24\)], "Output"] }, Open ]], Cell[TextData[{ "\tThe second equation needed to complete the solution is\n \n d \ curlyMbar(r)/dr = (4", StyleBox["\[Pi]", FontFamily->"Symbol"], "/Msun)*r^2*epsilon(r)\n \nwithout any further dimensionless factors. \ Epsilon must, at this \npoint, be in units of Msuns/km^3 for this equation to \ have overall \nunits of 1/km. We will make it dimensionless below.\n\n\tFor \ an Equation of State we take the fit to the Prakash EoS, which \nincludes the \ free Fermi gas contributions ", StyleBox["plus", FontWeight->"Bold"], " nuclear interactions \nthat fit, more or less, the BE per nucleon, the \ nuclear compressibility, \nand the symmetry energy. With dimensions, we \ found\n \n p(r) = A1*epsilon(r)^gam,\n \nwhere p and epsilon both \ have units of MeV/fm^3. Thus the coefficient \nA1 carries units of \ (MeV/fm^3)^(1-gam)\n\nDefine the ", StyleBox["dimensionless", FontWeight->"Bold"], " energy density and pressure, epsbar and pbar,\n \n epsilon(pbar) = \ eps0*epsbar(pbar)\n p = eps0*pbar\n eps0 = mneutron^4 c^8/(3 Pi^2 \ ", "hbarPlanck", "^3 c^3)\n \nFrom the non-causal and the causal formulations we found \ similar fits,\n\n p = A1*epsilon^gam -- in MeV/fm^3\n \ A1 = 0.000401195, gam = 2.0 -- causal, PrakashEoScausal.nb\n A1 = \ 0.000354842, gam = 2.1 -- non-causal, PrakashEoS.nb\n \nWe'll \ choose the gam = 2 case just to keep the dimensions of A1 simple, \ni.e., \ fm^3/MeV as it stands above. (It is also more physical.) We \nwill need to \ convert to ergs/cm^3 and then, later, to Msuns/km^3. ", StyleBox[" ", "Text"] }], "Text"], Cell[CellGroupData[{ Cell["\<\ {ergPerMeV = 1.602*10^(-6), (* ergs *) cmPerFm = 10.0^(-13), (* cm *) hbarPlanck = 1.055 10^(-27), (* erg-sec *) c = 2.99 10^10, (* cm/sec *) mneutron = 1.67 10^(-24)} (* grams *) eps0 = mneutron^4 c^8/(3 Pi^2 hbarPlanck^3 c^3) (* erg/cm^3 *) A1 = 0.000401195*cmPerFm^3/ergPerMeV (* now in (erg/cm^3)^(-1) *) \ \>", "Input"], Cell[BoxData[ \({1.602`*^-6, 1.`*^-13, 1.0549999999999998`*^-27, 2.9900000000000004`*^10, 1.67`*^-24}\)], "Output"], Cell[BoxData[ \(5.346178403625608`*^36\)], "Output"], Cell[BoxData[ \(2.5043383270911365`*^-37\)], "Output"] }, Open ]], Cell["\<\ To go to dimensionless epsbar and pbar, we have (after dividing out \ one factor of eps0) pbar = A1*eps0*epsbar^2\ \>", "Text"], Cell["\<\ Actually, we'll want epsbar as a function of pbar in the TOV \ equations. epsbar = 1/(A1*eps0)^(1/2) * pbar*(1/2) = capA * pbar*(1/2) It is interesting that this exponent for pbar is not dissimilar to the kF << Mnuc exponent of gamma = 3/5.\ \>", "Text"], Cell[CellGroupData[{ Cell["capA = 1/(A1*eps0)^(1/2) (* dimensionless *)", "Input"], Cell[BoxData[ \(0.8642348524266978`\)], "Output"] }, Open ]], Cell["\<\ The first differential equation, again after dividing out eps0 on both sides, then becomes d pbar(r)/dr = -Rschw*epsbar(pbar(r))*curlyMbar(r)/r^2 = -alpha * pbar(r)^(1/2) * curlyMbar(r)/r^2 alpha = Rschw*capA where alpha has, at this point, units of km. The second DEqn is d curlyMbar(r)/dr = 4*Pi*eps0/(Msun*c^2)*r^2*epsbar(pbar(r)) = myBeta * r^2 * pbar(r)^(1/2) where myBeta = 4*Pi*eps0*capA/(Msun*c^2) must have units of 1/km^3 so the equation itself has units 1/km. \ \>", \ "Text"], Cell[CellGroupData[{ Cell["alpha = Rschw*capA", "Input"], Cell[BoxData[ \(1.2756429592431868`\)], "Output"] }, Open ]], Cell["\<\ Convert the CGS value of eps0 (g/cm^3) into units of \ Msun/km^3:\ \>", "Text"], Cell[CellGroupData[{ Cell["\<\ {Msuncsqd = Msun*c^2, km3 = (10.0^5)^3} (* ergs and cm^3 *) eps0sunkm = eps0*km3/Msuncsqd (* units Msun/km^3 *)\ \>", "Input"], Cell[BoxData[ \({1.7781858900000004`*^54, 1.`*^15}\)], "Output"], Cell[BoxData[ \(0.0030065351624320937`\)], "Output"] }, Open ]], Cell["\tSo, what is beta for this value of eps0 in units of 1/km^3?", "Text"], Cell[CellGroupData[{ Cell["myBeta = 4*Pi*capA*eps0sunkm", "Input"], Cell[BoxData[ \(0.03265186015516843`\)], "Output"] }, Open ]], Cell["\<\ which is probably OK, beta being not too small. \ \>", "Text"], Cell["\<\ \tWe will integrate these two equations using a length scale unit of 1 km. \t d pbar(r)/dr = -alpha*capA * pbar(r)^2 * curlyMbar(r)/r^2 d curlyMbar(r)/dr = myBeta*capA * r^2 * pbar(r)^2 We may also need a starting value to avoid any singularity at r = 0. \ \>", \ "Text"], Cell[CellGroupData[{ Cell["{alpha, myBeta, rstart = 0.001}", "Input"], Cell[BoxData[ \({1.2756429592431868`, 0.03265186015516843`, 0.001`}\)], "Output"] }, Open ]], Cell["\<\ \tNow see if we can solve the coupled DE's given a value pzero for the energy density at the center of the star. We need a starting value of pbar(r=0) = pzero. If a neutron star has an energy density of \t = 3 Msun*c^2/(4 Pi Rnstar^3), Rnstar = 10 km, \t then\t \t = /eps0, \t \t = Kbar*^gamma, \t might be a good starting value.\t \ \>", "Text"], Cell[CellGroupData[{ Cell["\<\ Rnstar = 10.0 (* km *) epsilonave = 3/(4 Pi Rnstar^3) (* Msun/km^3 *) epsbarave = epsilonave/eps0sunkm (* dimless *) pbarave = A1*eps0*epsbarave^2\ \>", "Input"], Cell[BoxData[ \(10.`\)], "Output"], Cell[BoxData[ \(0.000238732414637843`\)], "Output"], Cell[BoxData[ \(0.07940449778233222`\)], "Output"], Cell[BoxData[ \(0.008441636626763652`\)], "Output"] }, Open ]], Cell["\<\ That is, we should try values of pzero ~ 0.01 to get values of neutron star radii that are physically reasonable. \ \>", "Text"], Cell["\<\ First, we'll integrate the Newtonian case, i.e., without the three additional factors in the TOV equation which come from GR.\ \>", "Text"], Cell["\<\ Arhs[r_] := -alpha * pbar[r]^(1/2) * curlyMbar[r]/r^2 BRhs[r_] := myBeta * r^2 * pbar[r]^(1/2)\ \>", "Input"], Cell[CellGroupData[{ Cell["{alpha, myBeta}", "Input"], Cell[BoxData[ \({1.2756429592431868`, 0.03265186015516843`}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ RMlist = {}; Do[{pzero = 10.0^(-i), Print[\"pzero = \",pzero], Do[{ \ts1 = NDSolve[{pbar'[r] == Arhs[r], pbar[rstart] == pzero, \t\t\tcurlyMbar'[r] == BRhs[r], \t\t\tcurlyMbar[rstart] == 0.0}, \t\t\t{pbar, curlyMbar},{r,rstart,x}, \t\t\tMaxSteps->20000]; \ty = Re[pbar[x]/.s1]; \tz = Re[curlyMbar[x]/.s1]; \tIf[y[[1]] < 0, Break[], {capR = x, capM = z[[1]]}] \t}, {x,1.0,50.0,0.2}]; RMlist = Append[RMlist,{pzero, capR, capM}]; }, {i,2,2}] RMlist;\ \>", "Input"], Cell[BoxData[ InterpretationBox[\("pzero = "\[InvisibleSpace]0.01`\), SequenceForm[ "pzero = ", 0.01], Editable->False]], "Print"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Print[\" pzero R(km) M(Msun)\"] Print[\" \"] Do[Print[ScientificForm[RMlist[[i,1]],{12,2}],\" \", \tPaddedForm[RMlist[[i,2]],{8,1}], \tPaddedForm[RMlist[[i,3]],{10,4}]], \t{i,1}]\ \>", 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There are three factors,\nall functions of r:\n\n\t factorpbyeps = 1 + \ p/epsilon = 1 + pbar/epsbar\n\t \n\t factorpbyM = 1 + (4 Pi/c^2) r^3 \ p/curlyM\n\t = 1 + (4 Pi eps0/(Msun c^2)) r^3 pbar/curlyMbar\n\t\ \n\t factorgravity = 1 - 2 bigG curlyM/r\t= 1 - 2 Rschw curlyMbar/r\n\t \n\ This middle factor looks singular at r = 0, since curlyMbar(0) = 0.\nThe r^3 \ in the numerator, however, cancels the singularity since \ncurlyMbar(r) near \ r = 0 is approximately 4 Pi r^3 epsbar(0)/3.\nThus, we can handle this \ problem by setting\n\n\t factorpbyM = 1 + (3 eps0/(Msun c^2)) \ pbar(r)/epsbar(0)\n\t \nwhen, say, r < 0.1 km, and otherwise using \n\n\t \ factorpbyM = 1 + (4 Pi eps0/(Msun c^2)) r^3 pbar(r)/curlyMbar(r)\n\t \nfor \ all other values of r. We will do this using the If[...] function:\n\n\t \ If[r > 0.1, (1 + (4 Pi eps0/(Msun c^2)) r^3 pbar(r)/curlyMbar(r)),\n\t \n\t \ \t\t\t, (1 + (3 eps0/(Msun c^2)) pbar(r)/epsbar(0))]\n\t \nIt appears \ necessary to include all these factors ", StyleBox["within", FontWeight->"Bold"], " the NDSolve block.\t\n \nFirst, just include the factorpbyeps and \ factorgravity factors:" }], "Text"], Cell["epsbar[r_] := capA * pbar[r]^(1/2)", "Input"], Cell["\<\ RMlistGR = {}; Do[{pzero = 10.0^(-i), epsbarzero = capA*pzero^(1/2), Do[{ \tsGR = NDSolve[{pbar'[r] == Arhs[r]* \t\t\t\t(1 + pbar[r]/epsbar[r]) / \t\t\t\t(1 - 2 Rschw curlyMbar[r]/r), \t\t\tcurlyMbar'[r] == BRhs[r], \t\t\tpbar[rstart] == pzero, curlyMbar[rstart] == 0.0}, \t\t\t{pbar, curlyMbar},{r,rstart,x}, \t\t\tMaxSteps->20000]; \ty = Re[pbar[x]/.sGR]; \tz = Re[curlyMbar[x]/.sGR]; \tIf[y[[1]] < 0, Break[], {capR = x, capM = z[[1]]}] \t}, {x,1.0,50.0,0.25}]; RMlistGR = Append[RMlistGR,{pzero, capR, capM}]; }, {i,1,4}] RMlistGR;\ \>", "Input"], Cell[CellGroupData[{ Cell["\<\ Print[\" pzero R(km) M(Msun)\"] Print[\" \"] Do[Print[ScientificForm[RMlistGR[[i,1]],{12,2}],\" \", \tPaddedForm[RMlistGR[[i,2]],{8,1}], \tPaddedForm[RMlistGR[[i,3]],{10,4}]],{i,1,4}]\ \>", "Input"], Cell[BoxData[ \(" pzero R(km) M(Msun)"\)], "Print"], Cell[BoxData[ \(" "\)], "Print"], Cell[BoxData[ InterpretationBox[ RowBox[{ TagBox[ InterpretationBox[\("1."\[Times]10\^"-1"\), 0.10000000000000001, AutoDelete->True], (ScientificForm[ #, {12, 2}]&)], "\[InvisibleSpace]", "\<\" \"\>", "\[InvisibleSpace]", TagBox[ InterpretationBox["\<\" 12.5\"\>", 12.5, AutoDelete->True], (PaddedForm[ #, {8, 1}]&)], "\[InvisibleSpace]", TagBox[ InterpretationBox["\<\" 2.7907\"\>", 2.7906667581951781, AutoDelete->True], (PaddedForm[ #, {10, 4}]&)]}], SequenceForm[ ScientificForm[ 0.10000000000000001, {12, 2}], " ", PaddedForm[ 12.5, {8, 1}], PaddedForm[ 2.7906667581951781, {10, 4}]], Editable->False]], "Print"], Cell[BoxData[ InterpretationBox[ RowBox[{ TagBox[ InterpretationBox[\("1."\[Times]10\^"-2"\), 0.01, AutoDelete->True], (ScientificForm[ #, {12, 2}]&)], "\[InvisibleSpace]", "\<\" \"\>", "\[InvisibleSpace]", TagBox[ InterpretationBox["\<\" 17.5\"\>", 17.5, AutoDelete->True], (PaddedForm[ #, {8, 1}]&)], "\[InvisibleSpace]", TagBox[ InterpretationBox["\<\" 2.0209\"\>", 2.0209112262350741, AutoDelete->True], (PaddedForm[ #, {10, 4}]&)]}], SequenceForm[ ScientificForm[ 0.01, {12, 2}], " ", PaddedForm[ 17.5, {8, 1}], PaddedForm[ 2.0209112262350741, {10, 4}]], Editable->False]], "Print"], Cell[BoxData[ InterpretationBox[ RowBox[{ TagBox[ InterpretationBox[\("1."\[Times]10\^"-3"\), 0.001, AutoDelete->True], (ScientificForm[ #, {12, 2}]&)], "\[InvisibleSpace]", "\<\" \"\>", "\[InvisibleSpace]", TagBox[ InterpretationBox["\<\" 20.3\"\>", 20.25, AutoDelete->True], (PaddedForm[ #, {8, 1}]&)], "\[InvisibleSpace]", TagBox[ InterpretationBox["\<\" 0.9028\"\>", 0.90284137389114805, AutoDelete->True], (PaddedForm[ #, {10, 4}]&)]}], SequenceForm[ ScientificForm[ 0.001, {12, 2}], " ", PaddedForm[ 20.25, {8, 1}], PaddedForm[ 0.90284137389114805, {10, 4}]], Editable->False]], "Print"], Cell[BoxData[ InterpretationBox[ RowBox[{ TagBox[ InterpretationBox[\("1."\[Times]10\^"-4"\), 0.0001, AutoDelete->True], (ScientificForm[ #, {12, 2}]&)], "\[InvisibleSpace]", "\<\" \"\>", "\[InvisibleSpace]", TagBox[ InterpretationBox["\<\" 21.3\"\>", 21.25, AutoDelete->True], (PaddedForm[ #, {8, 1}]&)], "\[InvisibleSpace]", TagBox[ InterpretationBox["\<\" 0.3221\"\>", 0.32210253566871588, AutoDelete->True], (PaddedForm[ #, {10, 4}]&)]}], SequenceForm[ ScientificForm[ 0.0001, {12, 2}], " ", PaddedForm[ 21.25, {8, 1}], PaddedForm[ 0.32210253566871588, {10, 4}]], Editable->False]], "Print"] }, Open ]], Cell["\<\ Now include the factorpbyM as well, using the If[] discussed \ above.. First do the pzero = 0.01 case, to compare the Newtonian and GR \ solutions.\ \>", "Text"], Cell[CellGroupData[{ Cell["\<\ RMlistGR = {}; const1 = 4 Pi eps0sunkm const2 = 3 eps0sunkm Do[{pzero = 10.0^(-i), epsbarzero = capA*pzero^(1/2), Do[{ \tsGR = NDSolve[{pbar'[r] == Arhs[r]* \t\t\t\t(1 + pbar[r]/epsbar[r]) * \t\t\t\tIf[r > 0.5, (1 + const1 r^3 pbar[r]/curlyMbar[r]), \t\t\t\t (1 + const2 pbar[r]/epsbarzero)] / \t\t\t\t(1 - 2 Rschw curlyMbar[r]/r), \t\t\tcurlyMbar'[r] == BRhs[r], \t\t\tpbar[rstart] == pzero, curlyMbar[rstart] == 0.0}, \t\t\t{pbar, curlyMbar},{r,rstart,x}, \t\t\tMaxSteps->20000]; \ty = Re[pbar[x]/.sGR]; \tz = Re[curlyMbar[x]/.sGR]; \tIf[y[[1]] < 0, Break[], {capR = x, capM = z[[1]]}] \t}, {x,1.0,50.0,0.1}]; RMlistGR = Append[RMlistGR,{pzero, capR, capM}]; }, {i,2,2}] RMlistGR;\ \>", "Input"], Cell[BoxData[ \(0.03778123511622424`\)], "Output"], Cell[BoxData[ \(0.009019605487296281`\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Print[\" pzero R(km) M(Msun)\"] Print[\" \"] Do[Print[ScientificForm[RMlistGR[[i,1]],{12,2}],\" \", \tPaddedForm[RMlistGR[[i,2]],{8,2}], \tPaddedForm[RMlistGR[[i,3]],{10,4}]], \t{i,1}]\ \>", "Input"], Cell[BoxData[ \(" pzero R(km) M(Msun)"\)], "Print"], Cell[BoxData[ \(" "\)], "Print"], Cell[BoxData[ InterpretationBox[ RowBox[{ TagBox[ InterpretationBox[\("1."\[Times]10\^"-2"\), 0.01, AutoDelete->True], (ScientificForm[ #, {12, 2}]&)], "\[InvisibleSpace]", "\<\" \"\>", "\[InvisibleSpace]", TagBox[ InterpretationBox["\<\" 17.50\"\>", 17.5, AutoDelete->True], (PaddedForm[ #, {8, 2}]&)], "\[InvisibleSpace]", TagBox[ InterpretationBox["\<\" 1.7647\"\>", 1.7647033697995298, AutoDelete->True], (PaddedForm[ #, {10, 4}]&)]}], SequenceForm[ ScientificForm[ 0.01, {12, 2}], " ", PaddedForm[ 17.5, {8, 2}], PaddedForm[ 1.7647033697995298, {10, 4}]], Editable->False]], "Print"] }, Open ]], Cell["to be compared with the Newtonian solution,", "Text"], Cell[CellGroupData[{ Cell["\<\ Print[\" pzero R(km) M(Msun)\"] Print[\" \"] 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