HW5.nb HW #5. Boltzmann equation << Graphics`Graphics` We start with the Boltzmann equation d n + H n = -Xs v\ Hn 2 - n 2 d t eq L. We defined the yield Y = n ê s and rewrite the equation in terms of x = m ê T instead of time. Because the entropy density scales as R -, it satisfies the equation d s ÅÅÅÅÅÅ + H s = 0 d t and hence the H term cancels for the yield, d Y = -Xs v\ shy 2 - Y 2 d t eq L. To rewrite the time variable by x, we need the relationship between time and temperature in radiation dominated universe. The expansion rate is H 2 = 8 p G N r R = 8 p G N g p2 * ÅÅÅÅÅÅ 0 T4. Using H = ÅÅÅÅ Rÿ R = ÅÅÅ d R R d t, and noting T R- and hence d R R = - d T, we find d t = - ÅÅÅÅÅ d T T2. Because HHTL = HHmL, T H T m 2 d t = - ÅÅÅÅ m2 d T. Then the Boltzmann equation becomes HHmL T d Y d T = ÅÅÅÅ m2 ÅÅÅÅÅÅ Xs v\ shy 2 - Y 2 HHmL T eq L Using shtl = shml T, it becomes m d Y d T = ÅÅÅÅ m2 Å shml Xs v\ HY 2 - Y 2 HHmL m eq L = ÅÅÅÅ shml ÅÅÅÅÅ HHmL m Xs v\ HY 2 - Y 2 eq L. Finally, d T = d ÅÅÅÅÅ m x = -m ÅÅÅÅÅÅ d x x, and d Y 2 d x = - ÅÅÅÅÅ shml ÅÅÅÅ x 2 HHmL Xs v\ HY 2 - Y 2 eq L ê2 = -0.68 g * ÅÅÅÅÅ m M x 2 Pl Xs v\ HY 2 - Y 2 eq L Here, M Pl = H8 p G N L -ê2 = 2.44 µ 0 8 GeV. NA i j ÅÅÅÅ k p 2 0 0.67787 y -ê2 Zeta@D z ÅÅ ÅÅÅÅÅ E { p 2 To integrate the Boltzmann equation, we need to have an expression for the equilibrium yield. Once the particle is non-relativistic, the difference in statistics is not important. The number density is n eq = Ÿ d p ÅÅÅ Å e - b E = H2 pl Ÿ d p ÅÅÅ Å e - b Hm+p2 ê2 ml = e - b m H Å m T H2 pl 2 p Lê2. The yield is Y eq = n eq s = ÅÅÅÅÅ p 2 g * Å zhl ÅÅÅÅÅÅ T e - b m H Å m T p 2 NA ÅÅ Zeta@D H2 pl E ê2 0.522 2 p Lê2 = p2 g * zhl e-x H x 2 p Lê2 - = 0.52 g * x ê2 e -x.
HW5.nb 2 The boundary condition is such that Y = Y eq at x = 0. Unfortunately for x = 0 (or T p m), the non-relativistic approximation we made to work out Y eq above is no longer good. Because the result is not too sensitive to the initial condition, we set Y = Y eq at x =. We will verify later that indeed the result is insensitive to this choice. We use GeV unit for everything. S-wave Mathematica unfortunately seems to have trouble dealing with big numbers such as M Pl. We can help it by solving for y = ÅÅÅÅ shml Xs v\ Y HHmL d y d x = - ÅÅÅÅÅ x Hy 2 - y 2 2 eq L, y eq = ÅÅÅÅ shml HHmL Xs v\ ÅÅÅÅ s e- b m H Å m T 2 p Lê2 = I ÅÅÅÅ p2 ÅÅÅÅÅÅ 0 M-ê2 m M Pl ÅÅÅÅÅÅ T Xs v\ e - b m H Å m T 2 p Lê2 = 0.92 M Pl m Xs v\ x ê2 e -x NA i j ÅÅÅÅ k p 2 0 0.975 y z { -ê2 i j y Å z k 2 p { ê2 E Mathematica still balks when I let it integrate all the way from x = to 00. I break it up into < x < 50 and the rest. solution = x 2 Iy@xD2 - H0.92 M Pl m s x ê2 E -x L 2 M, y@d == 0.92 M Pl m s ê2 E - = ê. 8m Ø 0, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, y, 8x,, 50<E 88y Ø InterpolatingFunction@88., 50.<<, <>D<< Evaluate@y@50D ê. solutiond 842.44< solution2 = x 2 Iy@xD2 - H0.92 M Pl m s x ê2 E -x L 2 M, y@50d ã 42.44290272824`= ê. 8m Ø 0, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, y, 8x, 50, 00<E 88y Ø InterpolatingFunction@8850., 00.<<, <>D<<
HW5.nb LogLogPlot@Evaluate@y@xD ê. solutiond, 8x,, 50<, PlotRange Ø 88, 00<, 8, 0 <<D 00 0 0 00 LogLogPlot@Evaluate@y@xD ê. solution2d, 8x, 50, 00<, PlotRange Ø 88, 00<, 8, 0 <<D 00 0 0 00 Show@%, %%D 00 0 0 00 Compared to the equlibrium values,
HW5.nb 4 LogLogPlot@0.92 M Pl m s x ê2 E -x ê. 8m Ø 0, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, 8x,, 0<, PlotRange Ø 88, 00<, 8, 0 <<, PlotStyle Ø RGBColor@, 0, 0DD 00 0 0 00 Show@%, %%D 00 0 0 00 the solution is basically right on the equilibrium until x ~ 20, and then becomes approximately constant afterwards. This is why we don't expect the result to be sensitive to the initial condition as long as it starts on the equilibrium for x < 0 or so. Verify this point by taking the ratio
HW5.nb 5 LogLinearPlot@0.92 M Pl m s x ê2 E -x ê Evaluate@y@xD ê. solution@@ddd ê. 8m Ø 0, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, 8x,, 50<, PlotStyle Ø RGBColor@, 0, 0DD 0.8 0.6 0.4 0.2 0 2 5 0 20 50 LogLinearPlot@Evaluate@y@xD ê. solution2d, 8x, 50, 00<, PlotRange Ø 820, 0<D 0 28 26 24 22 200 500 0 2000 5000 00 Therefore, using the notation in the problem, YH L = Å HHmL shml s 0 yh L, and yh L is about when the abundance starts to deviate significantly from the equilibrium value. Because this behavior is mainly due to the exponential dropoff of Y eq, it is expected to be rather insensitive to the chioce of m and s 0. Indeed, by varying m, we find
HW5.nb 6 TableA90 t, Clear@yint, yint2, solution, solution2, solutiond; solution = x 2 Iy@xD2 - H0.92 M Pl m s x ê2 E -x L 2 M, y@d == 0.92 M Pl m s ê2 E - = ê. 8m Ø 0 t, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, y, 8x,, 20<E; yint = Evaluate@y@20D ê. solution@@ddd; solution2 = x 2 Iy@xD2 - H0.92 M Pl m s x ê2 E -x L 2 M, y@20d ã yint= ê. 8m Ø 0 t, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, y, 8x, 20, 0<E; yint2 = Evaluate@y@0D ê. solution2@@ddd; solution = x 2 Iy@xD2 - H0.92 M Pl m s x ê2 E -x L 2 M, y@0d ã yint2= ê. 8m Ø 0 t, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, y, 8x, 0, 00<E; Evaluate@y@00D ê. solution@@ddd=, 8t, 0.5,.5, 0.<E Part::partd : Part specification solutionpt is longer than depth of object. More ReplaceAll::reps : 8solutionPT< is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing. More 88.6228, 7.2766<, 8.9807, 7.2766<, 85.087, 7.506<, 86.0957, 7.7265<, 87.9428, 7.956<, 80., 8.768<, 82.589, 8.402<, 85.8489, 8.627<, 89.9526, 8.8527<, 825.89, 9.0782<, 8.6228, 9.07<, 89.807, 9.529<, 850.87, 9.755<, 86.0957, 9.9807<, 879.428, 20.2065<, 8., 20.424<, 825.89, 20.658<, 858.489, 20.884<, 899.526, 2.04<, 825.89, 2.65<, 86.228, 2.5627<, 898.07, 2.7889<, 850.87, 22.052<, 860.957, 22.246<, 8794.28, 22.468<, 80., 22.6945<, 8258.9, 22.92<, 8584.89, 2.476<, 8995.26, 2.742<, 825.89, 2.6009<, 862.28, 2.8277<< Only a mild variation for a very wide range of m. (caution: the version of Mathematica I have seems to have a strange memory effect that always gives the first entry wrong. I get the reasonable results by quitting the kernel once and re-run the command.) P-wave Going back to d Y d x = - ÅÅÅÅÅ shml ÅÅÅÅ x 2 HHmL Xs v\ HY 2 - Y 2 eq L ê2 = -0.68 g * ÅÅÅÅÅ m M x 2 Pl Xs v\ HY 2 - Y 2 eq L we now substitute Xs v\ = s 0 x -. d Y d x = - ÅÅÅÅÅ shml ÅÅÅÅ x HHmL s 0HY 2 - Y 2 eq L Mathematica again seems to have trouble dealing with big numbers such as M Pl. We can help it by solving for y = ÅÅÅÅ shml HHmL s 0 Y d y d x = - ÅÅÅÅÅ Hy 2 - y 2 x eq L, y eq = ÅÅÅÅ shml HHmL Xs v\ ÅÅÅÅ s e- b m H Å m T 2 p Lê2 = I ÅÅÅÅ p2 ÅÅÅÅÅÅ 0 M-ê2 m M Pl ÅÅÅÅÅÅ T Xs v\ e - b m H Å m T 2 p Lê2 = 0.92 M Pl m Xs v\ x ê2 e -x NA i j ÅÅÅÅ k p 2 0 0.975 y z { -ê2 i j y Å z k 2 p { ê2 E
HW5.nb 7 Mathematica still balks when I let it integrate all the way from x = to 00. I break it up into < x < 50 and the rest. solution = x Iy@xD2 - H0.92 M Pl m s x ê2 E -x L 2 M, y@d == 0.92 M Pl m s ê2 E - = ê. 8m Ø 0, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, y, 8x,, 50<E 88y Ø InterpolatingFunction@88., 50.<<, <>D<< Evaluate@y@50D ê. solutiond 894.006< solution2 = x Iy@xD2 - H0.92 M Pl m s x ê2 E -x L 2 M, y@50d ã 94.0060809`= ê. 8m Ø 0, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, y, 8x, 50, 00<E 88y Ø InterpolatingFunction@8850., 00.<<, <>D<< LogLogPlot@Evaluate@y@xD ê. solutiond, 8x,, 50<, PlotRange Ø 88, 00<, 8, 0 <<D 00 0 0 00 LogLogPlot@Evaluate@y@xD ê. solution2d, 8x, 50, 00<, PlotRange Ø 88, 00<, 8, 0 <<D 00 0 0 00
HW5.nb 8 Show@%, %%D 00 0 0 00 Compared to the equlibrium values, LogLogPlot@0.92 M Pl m s x ê2 E -x ê. 8m Ø 0, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, 8x,, 0<, PlotRange Ø 88, 00<, 8, 0 <<, PlotStyle Ø RGBColor@, 0, 0DD 00 0 0 00 Show@%, %%D 00 0 0 00
HW5.nb 9 the solution is basically right on the equilibrium until x ~ 20, and then becomes approximately constant afterwards. This is why we don't expect the result to be sensitive to the initial condition as long as it starts on the equilibrium for x < 0 or so. Verify this point by taking the ratio LogLinearPlot@0.92 M Pl m s x ê2 E -x ê Evaluate@y@xD ê. solution@@ddd ê. 8m Ø 0, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, 8x,, 50<, PlotStyle Ø RGBColor@, 0, 0DD 0.8 0.6 0.4 0.2 0 2 5 0 20 50 LogLinearPlot@Evaluate@y@xD ê. solution2d, 8x, 50, 00<, PlotRange Ø 8700, 900<D 900 875 850 825 800 775 750 725 200 500 0 2000 5000 00 Therefore, using the notation in the problem, YH L = Å HHmL 2 shml s 0 yh L, and yh L is about 2 x f where x f is when the abundance starts to deviate significantly from the equilibrium value. Because this behavior is mainly due to the exponential dropoff of Y eq, it is expected to be rather insensitive to the chioce of m and s 0. Indeed, by varying m, we find
HW5.nb 0 TableA90 t, Clear@yint, yint2, solution, solution2, solutiond; solution = x Iy@xD2 - H0.92 M Pl m s x ê2 E -x L 2 M, y@d == 0.92 M Pl m s ê2 E - = ê. 8m Ø 0 t, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, y, 8x,, 20<E; yint = Evaluate@y@20D ê. solution@@ddd; solution2 = x Iy@xD2 - H0.92 M Pl m s x ê2 E -x L 2 M, y@20d ã yint= ê. 8m Ø 0 t, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, y, 8x, 20, 0<E; yint2 = Evaluate@y@0D ê. solution2@@ddd; solution = x Iy@xD2 - H0.92 M Pl m s x ê2 E -x L 2 M, y@0d ã yint2= ê. 8m Ø 0 t, g Ø, s Ø 0-0, M Pl Ø 2.44 0 8 <, y, 8x, 0, 00<E; Evaluate@y@00D ê. solution@@ddd=, 8t, 0.5,.5, 0.<E NDSolve::ndcf : Repeated convergence test failure at x == 2.655257957994`; unable to continue. More InterpolatingFunction::dmval : Input value 820< lies outside the range of data in the interpolating function. Extrapolation will be used. More 88.6228, 24.0545<, 8.9807, 425.54<, 85.087, 47.854<, 86.0957, 450.85<, 87.9428, 46.08<, 80., 476.022<, 82.589, 489.29<, 85.8489, 502.428<, 89.9526, 55.99<, 825.89, 529.604<, 8.6228, 54.482<, 89.807, 557.55<, 850.87, 57.87<, 86.0957, 586.275<, 879.428, 600.928<, 8., 65.774<, 825.89, 60.85<, 858.489, 646.05<, 899.526, 66.48<, 825.89, 677.07<, 86.228, 692.928<, 898.07, 708.944<, 850.87, 725.55<, 860.957, 74.56<, 8794.28, 758.66<, 80., 774.966<, 8258.9, 79.962<, 8584.89, 809.54<, 8995.26, 826.54<, 825.89, 844.28<, 862.28, 86.296<< Only a mild variation for a very wide range of m. (caution: the version of Mathematica I have seems to have a strange memory effect that always gives the first entry wrong. I get the reasonable results by quitting the kernel once and re-run the command.) 2. W DM S-wave Once we have the yield, it is easy to convert it to the current energy density. r DM = m YH L s 0 = m Å HHmL shml s 0 yh L s 0 while r c = H 0 2 8 p G N = H 2 0 M 2 Pl, x f = yh L, and HHmL 2 = M 2 g p2 Pl * ÅÅÅÅÅÅ 0 m4, shml = g 2 p2 * Å 45 m, and hence W DM = m "################ g ######### * 45 ÅÅÅÅÅÅ ÅÅÅÅÅ ÅÅÅÅÅÅ ÅÅÅÅÅ Å p2 M 2 Pl 0 m4 x f s 0 = 0.252 ÅÅ ÅÅÅÅÅÅ g * ê2 M Pl H 0 2 s 0 In[50]:= NA ÅÅÅÅ i j ÅÅÅÅ k Out[50]= 0.25646 p 2 0 x f 2 p 2 g * m s 0 s 0 H 2 0 M 2 Pl y z { ê2 i 2 p 2 y j ÅÅ k 45 z { - E
HW5.nb We found x f º 2 for the choice of parameters. From HW #, we found s = s g + s n = H + ÅÅÅÅÅÅ 2 22 L s g = ÅÅÅÅÅÅ 4 22 2 2 p2 Å 45 T 0 = 2890 cm - i k j + 2 y z 2 ÅÅ 2 p2 22 { 45 hbarc- T 0 ê. 8hbarc Ø 0.97 0-4, T 0 Ø 2.725 * 8.67 0-5 < 2890.54 It is useful to re-express the Hubble constant in the GeV unit (sounds crazy): H 0 = h km sec - Mpc - h km sec- 0.97 GeV fm = ÅÅ ÅÅÅ = 2..00µ0 5 km sec - 0 6.086 0 6 m 0-42 GeV h -5 ÅÅÅÅÅ 0.97 0 ÅÅ ÅÅÅ 0 5 0 6.086 0 6 2. µ 0-42 Therefore W DM h 2 is given by In[77]:= 0.25646 x f s 0 hbarc ÅÅÅÅÅÅ g ê2 M Pl H 0 2 s 0 ê. Out[77]=.94925 8x f Ø 2, s 0 Ø 2890, hbarc Ø 0.97 0 -, M Pl Ø 2.44 0 8, H 0 Ø 2. 0-42, g Ø, s 0 Ø 0-0 < This is too big. To get the realistic value W DM h 2 = 0.2, we want s 0 =.6 µ 0-9 GeV -2. This is the range of cross sections of our interest. In[52]:=.9492509097788` ê 0.2 Out[52]= 6.248 P-wave This time, yh L is enhanced to 775. Therefore W DM h 2 is given by In[8]:= 0.25646 Out[8]= 65.684 y s 0 hbarc ÅÅÅÅÅÅ g ê2 M Pl H 0 2 s 0 ê. 8y Ø 775, s 0 Ø 2890, hbarc Ø 0.97 0 -, M Pl Ø 2.44 0 8, H 0 Ø 2. 0-42, g Ø, s 0 Ø 0-0 < This is too big. To get the realistic value W DM h 2 = 0.2, we want s 0 = 5.5 µ 0-8 GeV -2 would do much better. This is the range of cross sections of our interest.
HW5.nb 2 In[84]:= 65.68650262` ê 0.2 Out[84]= 547.45 In either case, the cross section is of the order of electroweak scale, s 0 ~ Å p a2 with m ~TeV. Optional m 2 To be followed.