Assignment-Stellar Polytropes (Solution)

Polytropic Models for Stars 1. Integrating the Lane­Emden Equation 1.1. Rearranging, we arrive at the desired coupled sys­ tem dφ = u (5) dξ du 2u (6) = −φn − ξ dξ Problem Integrate the Lane­Emden Equation � � 1 d 2 dφ ξ = −φn ξ 2 dξ dξ (1) Plotting the dimensionless temperature, φ(ξ) ver­ sus the dimensionless radius, ξ, for the n values of interest, we arrive at the graph below. Note that n increases from 1 to 3.5 as we move left to right. for polytropic indices of n = 1.0, 1.5, 2.0, 2.5, 3.0, and 3.5. Break up this second order differential equation into two first­order, coupled equations in dφ/dξ ≡ u and du/dξ. Use a 4th­order Runge­Kutta inte­ gration scheme or some other equivalent integra­ tion method to find φ(ξ). Recall the boundary conditions at the center: u(0) = 0 and φ(0) = 1. Use the analytic expression for φ(ξ) near the center: ξ2 φ(ξ) = 1 − (2) 6 to help start the integration. The surface is de­ fined by φ(ξ1 ) = 0. Now we similarly plot the dimensionless density, φ(ξ)n versus the normalized dimensionless radius, ξ. Note that n decreases from 3.5 to 1 as we move left to right. Plot the dimensionless temperature, φ(ξ), and the dimensionless density, φn (ξ), for all 6 values of n. It would be best to put all the temperature plots on one graph and all the density plots on another. 1.2. 1.2.1. Solution Coupled System of Differential Equations We take dφ/dξ ≡ u. Thus we can write 1 d � 2 � ξ u = −φn ξ 2 dξ By the product rule, � � 1 2 du 2ξu + ξ = −φn ξ2 dξ (3) More detailed versions of the plots are in the Ap­ pendix. (4) 1 2. Ω, the unitless potential (in −GM 2 /R), will then be given by Tabulating Some Physical Properties of Polytropes 2.1. Problem Ω= G 4πρ0 As the integrations in part 1 are underway, compute for each model the dimensionless poten­ tial energy, Ω (in units of −GM 2 /R), and the dimensionless moment of inertia, k (in units of M R2 ). Tabulate ξ1 , −(dφ/dξ)ξ1 , Ω, and k for each of the 6 polytropic models. 2.2. 2.2.1. Location of Stellar Surfaces 2.2.2. −U R �2 � ξ1 � )n ξ �2 dξ � φ(ξ 0 (10) These integrals do not appear to have a simple closed form. Thus, we set � �� � The location of the stellar surface, ξ1 , which is defined by φ(ξ1 ) = 0, can be numerically deter­ mined using the data from the RK4 integration in section 1. The values of −(dφ/dξ)ξ1 , which is just −u(ξ1 ), can then be found from examining the same RK4 data specifically at ξ1 . These data are tabulated below. ξ1 3.141 3.652 4.353 5.355 6.896 9.535 R3 ξ13 Plugging in U, and simplifying, we arrive at the expression �� � �ξ ξ ξ1 0 1φ(ξ)n ξ 0 φ(ξ � )n ξ �2 dξ � dξ Ω= (11) �2 �� ξ1 � )n ξ �2 dξ � φ(ξ 0 Solution n 1.0 1.5 2.0 2.5 3.0 3.5 � ξ1 U� = ξ φ(ξ � )n ξ �2 dξ � dξ φ(ξ)n ξ 0 and M� = −(dφ/dξ)ξ1 0.318 0.203 0.127 0.0763 0.0424 0.0208 (12) 0 � ξ1 φ(ξ � )n ξ �2 dξ � (13) 0 Numerically evaluating these integrals using a two­point Newton­Cotes method, and then plug­ ging the results into the expression for Ω, we find n 1.0 1.5 2.0 2.5 3.0 3.5 Dimensionless Potential Energy The gravitational potential energy of a sphere of radius r is given by the equation � r M (r� )ρ(r� )r�2 � U (r) = −4πG dr (7) r� 0 U� 2.355 1.727 1.335 1.070 0.885 0.749 M� 3.140 2.713 2.410 2.186 2.017 1.889 Ω 0.750 0.857 1.000 1.200 1.500 2.000 Ωf rac 3/4 6/7 1 6/5 3/2 2 From these results, we can see that Ω appears to obey the relation Thus, the first step in determining the potential is to find the mass as a function of ξ. To do so, we want to integrate the density using spherical shells. Let ρ0 be the central density of the object and let us take the radius of the object to be R. Then we can write � R3 ξ M (ξ) = 4πρ0 3 φ(ξ � )n ξ �2 dξ � (8) ξ1 0 Ω(n) = 3 5−n (14) The simplicity of this relation seems to imply that it is possible to find an elegant simplification for the integral formulation of Ω in equation 11. 2.2.3. This allows us to write the expression for potential as �� � � ξ −16π 2 Gρ20 R5 ξ1 n � n �2 � U= φ(ξ) ξ φ(ξ ) ξ dξ dξ ξ15 0 0 (9) An Analytical Expression for U This derivation is due to Chandrasekhar via Cox and Guili. 2 We begin with the differential of potential en­ ergy due to a spherical shell of mass Mr GMr dMr dU = − r From this we can the write � � GMr2 GMr2 dU = d − − dr 2r 2r2 2.2.4. Dimensionless Moment of Inertia The moment of inertia of a body is given by the formula � 2 I= ρ(r)r⊥ dV (24) (15) V Therefore, the moment of inertia will be propor­ tional to the following integral. (16) � ξ1� Applying hydrostatic equilibrium, 2π� π I∝ 0 2 dP/dr = −(GMr /r )ρ 0 φ(ξ)n ξ 4 sin3 (θ) dθ dφ dξ (25) 0 Simplifying and inserting the correct constants, we arrive at the expression we find � � GMr2 1 dP dU = d − + Mr ρ 2r 2 (17) 8πρ0 R5 3ξ15 I= From the relation � ξ1 φ(ξ)n ξ 4 dξ (26) 0 The unitless moment of inertia (in M R2 ) will thus be given by P ∝ ρ(n+1)/n we can show that dP = (n + 1)d ρ which allow us to write � � GMr2 dU = d − + 2r � P ρ � equation 17 as � � 1 P (n + 1)Mr d 2 ρ (18) k= 0 � 4πρ0 R5 ξ 0 ξ13 (19) I� = φ(ξ)n ξ 4 dξ φ(ξ � )n ξ �2 dξ � (27) � ξ1 φ(ξ)n ξ 4 dξ (28) 2I � 3M � ξ12 (29) 0 We can write k= We again use a two­point Newton­Cotes method to evaluate the integrals and plug in the results to find k. We apply the virial theorem, concluding that � � P 4 3 dMr − 3 d P πr3 + dU = 0 (21) ρ 3 n 1.0 1.5 2.0 2.5 3.0 3.5 Now we eliminate (P/ρ)dMr in equation 20. Solv­ ing for dU we find � � � � � 3 GMr2 Mr P dU = d − + (n + 1)d 5−n r ρ � �� 4 −(n + 1)d P πr3 (22) 3 3. If we integrate this from r = 0 to r = R, we can see that the last two terms vanish, giving us 3 GM 2 5−n R � ξ1 Putting M � as above and setting We can then write this as � � � � GMr2 1 Mr P dU =d − + (n + 1)d 2r 2 ρ 1 P − (n + 1) dMr (20) 2 ρ U =− 8πρ0 R5 3ξ15 3.1. I� 12.152 11.116 10.607 10.511 10.843 11.737 k 0.261 0.205 0.155 0.112 0.0754 0.0456 Model of the Sun Problem Use an n = 3 polytropic model to represent the internal structure of the Sun. The two parameters to fix are M = M� and R = R� . (23) This is what we found numerically. 3 (a) Plot the physical temperature (in K) vs. ra­ dial distance in units of r/R. Plot log T vs. r/R. Do the same for the physical density (g/cm3 ). Again, plot log ρ vs. r/R. Instead of using the values for the central density ρ0 , and central tem­ perature T0 , deduced for an n = 3 polytrope with M = M� and R = R� , take the known values of ρ0 = 158 g/cm3 and T0 = 15.7 × 106 K. (b) Compute the nuclear luminosity of the sun using the above temperature and density profiles. Take the nuclear energy generation rate to be 3.2.2. −2/3 (−33.81T6−1/3 ) �(ρ, T ) = 2.46 × 106 ρ2 X 2 T6 e We plot the density against normalized radius which is in ergs cm−3 sec−1 , where ρ is in g/cm3 , T6 is the temperature in units of 106 K, and X is the hydrogen mass fraction. (take X = 0.6). Re­ duce the problem to a dimensioned constant times an integral involving only functions of φ and ξ. (There will also appear a T0 inside the integral for which you can plug in the value of 15.7 × 106 ) Show the value of your constant and the form of the dimensionless integral. Evaluate the nuclear luminosity of the Sun in units of ergs sec−1 . 3.2. 3.2.1. Solar Density Plots Solutions Solar Temperature Plots Now we plot the logarithm of density We first plot temperature against normalized radius 3.2.3. Now we plot the logarithm of temperature Solar Nuclear Luminosity We have both density and temperature for an n = 3 polytrope as a function of ξ. Thus we can write, � ξ1 R3 ξ 2 �(ρ(ξ), T (ξ))dξ (30) L� = 4π 3� ξ1 0 4 Using the unitless, polytropic density and temper­ ature, we can write ξ1 � L� = Lc −1/3 ξ 2 φ(ξ)16/3 e−13.5φ(ξ) dξ (31) 0 where, −2/3 Lc = 2.46 × 106 × ρ20 × X 2 × T0 × 4π 3 R� (32) ξ13 Thus, Lc is our dimensioned constant, which must be in ergs/s, and we can evaluate it, giving Lc = 4.546 × 1040 (33) Numerically evaluating the integral using a two­ point Newton­Cotes method, we arrive at the re­ sult � ξ1 −1/3 ξ 2 φ(ξ)16/3 e−13.5φ(ξ) dξ = 3.26 × 10−7 0 (34) And thus, L� = 1.48 × 1034 ergs/s (35) This is within an order of magnitude of the ac­ tual solar luminosity, which seems reasonable for a simple polytropic model. 5 A. 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