Solutions For No. 1 of Suggested Comprehension Exercises ('Construction of A Star', Part 2)

1) Using the appropriate stellar model parameter relations calculate the core values: T _c  ,  r _c ,  P_c  .


Soln. We have, for the three core parameters -

T  = T_c   q    ,    q  = (  r / r _c  )^1/n    And q =   (P/ P_c ) ^{(1+ 1/n)}

At the interface (core-radiative boundary)  one finds:  q  = 0. 7839768

Then, for the (dimensionless)  core temperature:   t _c     =  t /   q

And, since 0.6969 =    t _c   q   =     t _c    (0.7839768)  then:   t _c   =  0.8876

The dimensional form is:  T  =  T_c   q

Since we have, at r/R = 0.172, log T = 7.1643   then

antilog (7.1643)  =   T   =   1.46  x 10⁷  K

And:   T_c      =   T  /   q

For which we set up the ratio:

(0.6969)/  / (0.8876) =  (1.46  x 10⁷  )/   T_c 

Whence:  T_c     =   (1.46  x 10⁷  ) (0.8876)/  (0.6969)  = 1.86  x 10⁷  K

Take log T = log (1.86  x 10⁷  )  = 7.2695

Which is the core temperature log value we find at r /R  = 0.0, i.e., See Table in Part 1

For core density:

q  = (  r / r _c )^1/n   

Where:   The polytropic index n can be defined:   n = 1/ (g   - 1)

where g   is the ratio of specific heats. (g  =  C _p / C _v )

In a non-relativistic limit  one will have g = 5/3 and

n = 1 / (5/3 - 1)  =  1/ (2/3)  =  3/2

Then write the core density as:   r _c    =    r / q ^3/2

Now, take  q  =   0.7839  SO:   q ^3/2    =   0.6941

Since we have, at r/R = 0.172, log r = 1.4033   then

antilog (1.4033)  =   r   =   25.31 g/ cm³

Therefore:  r _c    =    r / q ^3/2  =   25.31 g/ cm³ / 0.6941

And:     r _c    =   36.46   g/ cm³ 

Now, take log  (r _c  ) =   1.5618

Which is the density log value we find at r /R  = 0.0, i.e. the core (See Table in Part 1)

For core pressure, we have:  q =   (P/ P_c ) ^{(1+ 1/n)}

Since:  q    =  (p/ p _c ) ^{(1+ 1/n)}     and n = 1.5 (assigned polytropic index - see top of Lane-Emden function table)

Then:   p _c   =     p / q ^5/2      and   q ^5/2 =  0.5441

And, since we have, from the model table,  at r/R = 0.172, log P = 16.6770   then

antilog (16.6770 )  =   P = 4.754  x 10¹⁶    dyne/ cm ²

Therefore:     P _c   =   P /   q ^5/2 =  (4.754  x 10¹⁶    dyne/ cm² )/  0.5441

P _c   =   8.737  x 10¹⁶    dyne/ cm²

and the log is: 16.9414 or the value shown at r/R = 0.0 (the core)