BiPolytrope with <math>n_c = 5</math> and <math>n_e=1</math>

Here we construct a bipolytrope in which the core has an <math>n_c=5</math> polytropic index and the envelope has an <math>n_c=1</math> polytropic index. This system is particularly interesting because the entire structure can be described by closed-form, analytic expressions. As far as I have been able to determine, this analytic structural model has not previously been published in a refereed, archival journal (author: Joel E. Tohline, March 30, 2013). In deriving the properties of this model, we will follow the general solution steps for constructing a bipolytrope that we have outlined elsewhere.

Steps 2 & 3

Based on the discussion presented elsewhere of the structure of an isolated <math>n=5</math> polytrope, the core of this bipolytrope will have the following properties:

<math> \theta(\xi) = \biggl[ 1 + \frac{1}{3}\xi^2 \biggr]^{-1/2} ~~~~\Rightarrow ~~~~ \theta_i = \biggl[ 1 + \frac{1}{3}\xi_i^2 \biggr]^{-1/2} ; </math>

<math> \frac{d\theta}{d\xi} = - \frac{\xi}{3}\biggl[ 1 + \frac{1}{3}\xi^2 \biggr]^{-3/2} ~~~~\Rightarrow ~~~~ \biggl(\frac{d\theta}{d\xi}\biggr)_i = - \frac{\xi_i}{3}\biggl[ 1 + \frac{1}{3}\xi_i^2 \biggr]^{-3/2} \, . </math>

The first zero of the function <math>\theta(\xi)</math> and, hence, the surface of the corresponding isolated <math>n=5</math> polytrope is located at <math>\xi_s = \infty</math>. Hence, the interface between the core and the envelope can be positioned anywhere within the range, <math>0 < \xi_i < \infty</math>.

Step 4: Throughout the core (<math>0 \le \xi \le \xi_i</math>)

Specify: <math>K_c</math> and <math>\rho_0 ~\Rightarrow</math>
<math>\rho</math>	<math>=</math>	<math>\rho_0 \theta^{n_c}</math>	<math>=</math>	<math>\rho_0 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2}</math>
<math>P</math>	<math>=</math>	<math>K_c \rho_0^{1+1/n_c} \theta^{n_c + 1}</math>	<math>=</math>	<math>K_c \rho_0^{6/5} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3}</math>
<math>r</math>	<math>=</math>	<math>\biggl[ \frac{(n_c + 1)K_c}{4\pi G} \biggr]^{1/2} \rho_0^{(1-n_c)/(2n_c)} \xi</math>	<math>=</math>	<math>\biggl[ \frac{K_c}{G\rho_0^{4/5}} \biggr]^{1/2} \biggl(\frac{3}{2\pi}\biggr) \xi</math>
<math>M_r</math>	<math>=</math>	<math>4\pi \biggl[ \frac{(n_c + 1)K_c}{4\pi G} \biggr]^{3/2} \rho_0^{(3-n_c)/(2n_c)} \biggl(-\xi^2 \frac{d\theta}{d\xi} \biggr)</math>	<math>=</math>	<math>\biggl[ \frac{K_c^3}{G^3 \rho_0^{2/5} } \biggr]^{1/2} \biggl( \frac{2\cdot 3}{\pi } \biggr)^{1/2} \biggl[ \xi^3 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \biggr]</math>

Step 5: Interface Conditions

			Setting <math>n_c=5</math>, <math>n_e=1</math>, and <math>\phi_i = 1 ~~~~\Rightarrow</math>
<math>\frac{\rho_e}{\rho_0}</math>	<math>=</math>	<math>\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{n_c}_i \phi_i^{-n_e}</math>	<math>=</math>	<math>\biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{5}_i </math>
<math>\biggl( \frac{K_e}{K_c} \biggr) </math>	<math>=</math>	<math>\rho_0^{1/n_c - 1/n_e}\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-(1+1/n_e)} \theta^{1 - n_c/n_e}_i</math>	<math>=</math>	<math>\rho_0^{-4/5}\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \theta^{-4}_i</math>
<math>\frac{\eta_i}{\xi_i}</math>	<math>=</math>	<math>\biggl[ \frac{n_c + 1}{n_e+1} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c}\biggr) \theta_i^{(n_c-1)/2} \phi_i^{(1-n_e)/2}</math>	<math>=</math>	<math>3^{1/2} \biggl( \frac{\mu_e}{\mu_c}\biggr) \theta_i^{2}</math>
<math>\biggl( \frac{d\phi}{d\eta} \biggr)_i</math>	<math>=</math>	<math>\biggl[ \frac{n_c + 1}{n_e + 1} \biggr]^{1/2} \theta_i^{- (n_c + 1)/2} \phi_i^{(n_e+1)/2} \biggl( \frac{d\theta}{d\xi} \biggr)_i</math>	<math>=</math>	<math>3^{1/2} \theta_i^{- 3} \biggl( \frac{d\theta}{d\xi} \biggr)_i</math>

Step 6: Envelope Solution

The most general solution to the <math>n=1</math> Lane-Emden equation is,

<math> \phi = A \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr] \, , </math>

where <math>A</math> and <math>B</math> are constants. The first derivative of this function is,

<math> \frac{d\phi}{d\eta} = \frac{A}{\eta^2} \biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr] \, . </math>

From Step 5, above, we know the value of the function, <math>\phi</math> and its first derivative at the interface; specifically,

<math> \phi_i = 1~~~~\mathrm{and} ~~~~\biggl( \frac{d\phi}{d\eta}\biggr)_i =3^{1/2} \theta_i^{- 3} \biggl( \frac{d\theta}{d\xi} \biggr)_i~~~~ \mathrm{at}~~~~\eta_i =3^{1/2} \xi_i \biggl( \frac{\mu_e}{\mu_c}\biggr) \theta_i^{2}</math>

From this information we can determine the constants <math>A</math> and <math>B</math>; specifically,

<math> \eta_i - B = \tan^{-1}(\Lambda_i^{-1}) = \frac{\pi}{2}- \tan^{-1}(\Lambda_i) \, , </math>

<math> A = \frac{\phi_i \eta_i}{\sin(\eta_i - B)} = \phi_i \eta_i (1 + \Lambda_i^2)^{1/2} \, , </math>

where,

<math> \Lambda_i = \frac{1}{\eta_i} + \frac{1}{\phi_i} \biggl(\frac{d\phi}{d\eta}\biggr)_i \, . </math>

Step 7

The surface will be defined by the location, <math>\eta_s</math>, at which the function <math>\phi(\eta)</math> first goes to zero, that is,

<math> \eta_s = \pi + B = \frac{\pi}{2} + \eta_i + \tan^{-1}(\Lambda_i) \, . </math>

Step 8: Throughout the envelope (<math>\eta_i \le \eta \le \eta_s</math>)

Knowing: <math>K_e</math> and <math>\rho_e ~\Rightarrow</math>
<math>\rho</math>	<math>=</math>	<math>\rho_e \phi^{n_e}</math>
<math>P</math>	<math>=</math>	<math>K_e \rho_e^{1+1/n_e} \phi^{n_e + 1}</math>
<math>r</math>	<math>=</math>	<math>\biggl[ \frac{(n_e + 1)K_e}{4\pi G} \biggr]^{1/2} \rho_e^{(1-n_e)/(2n_e)} \eta</math>
<math>M_r</math>	<math>=</math>	<math>4\pi \biggl[ \frac{(n_e + 1)K_e}{4\pi G} \biggr]^{3/2} \rho_e^{(3-n_e)/(2n_e)} \biggl(-\eta^2 \frac{d\phi}{d\eta} \biggr)</math>

Step 9

Step 1: Choose <math>n_c</math> and <math>n_e</math>.
Step 2: Adopt boundary conditions at the center of the core (<math>\theta = 1</math> and <math>d\theta/d\xi = 0</math> at <math>\xi=0</math>), then solve the Lane-Emden equation to obtain the solution, <math>\theta(\xi)</math>, and its first derivative, <math>d\theta/d\xi</math> throughout the core; the radial location, <math>\xi = \xi_s</math>, at which <math>\theta(\xi)</math> first goes to zero identifies the natural surface of an isolated polytrope that has a polytropic index <math>n_c</math>.
Step 3 Choose the desired location, <math>0 < \xi_i < \xi_s</math>, of the outer edge of the core.
Step 4: Specify <math>K_c</math> and <math>\rho_0</math>; the structural profile of, for example, <math>\rho(r)</math>, <math>P(r)</math>, and <math>M_r(r)</math> is then obtained throughout the core — over the radial range, <math>0 \le \xi \le \xi_i</math> and <math>0 \le r \le r_i</math> — via the relations shown in the <math>2^\mathrm{nd}</math> column of Table 1.
Step 5: Specify the ratio <math>\mu_e/\mu_c</math> and adopt the boundary condition, <math>\phi_i = 1</math>; then use the interface conditions as rearranged and presented in Table 3 to determine, respectively:
- The gas density at the base of the envelope, <math>\rho_e</math>;
- The polytropic constant of the envelope, <math>K_e</math>, relative to the polytropic constant of the core, <math>K_c</math>;
- The ratio of the two dimensionless radial parameters at the interface, <math>\eta_i/\xi_i</math>;
- The radial derivative of the envelope solution at the interface, <math>(d\phi/d\eta)_i</math>.
Step 6: The last sub-step of solution step 5 provides the boundary condition that is needed — in addition to our earlier specification that <math>\phi_i = 1</math> — to derive the desired particular solution, <math>\phi(\eta)</math>, of the Lane-Emden equation that is relevant throughout the envelope; knowing <math>\phi(\eta)</math> also provides the relevant structural first derivative, <math>d\phi/d\eta</math>, throughout the envelope.
Step 7: The surface of the bipolytrope will be located at the radial location, <math>\eta = \eta_s</math> and <math>r=R</math>, at which <math>\phi(\eta)</math> first drops to zero.
Step 8: The structural profile of, for example, <math>\rho(r)</math>, <math>P(r)</math>, and <math>M_r(r)</math> is then obtained throughout the envelope — over the radial range, <math>\eta_i \le \eta \le \eta_s</math> and <math>r_i \le r \le R</math> — via the relations provided in the <math>3^\mathrm{rd}</math> column of Table 1.

Example Solutions

Analytic solution for <math>n_c = 5, ~n_e = 1</math>.

Related Discussions

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