\documentstyle[12pt,ay45defs]{article}
\def\figin{\medbreak \centerline{\small figure here}\medbreak}
\def\w{\omega}
\def\e{\epsilon}
\def\rad{\, {\rm rad}}
\def\beq{\begin{eqnarray*}}
\def\eeq{\end{eqnarray*}}
\def\radhr{\, {\rm rad/hour}}
\def\au{\, {\rm AU}}
\def\tot{{\rm Tot}}
\def\g{\rm g}
\def\cm{\rm cm}
\def\ergs{\rm ergs}
\def\const{{\rm const}}
\def\kms{\km/{\rm s}}
\def\cms{\cm/{\rm s}}
\def\rads{\rad/{\rm s}}
\def\lyr{\, {\rm lyr}}
\def\fe{\rm Fe}
\def\c{\rm C}
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\textheight=8.5in
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\begin{document}
{\it Astronomy 45 solutions to problem set 5}\hfill {\it Spring
1997}\newline
\hrule
\vskip12pt
\begin{center}
{\bf Astronomy 45.  Introduction to Astrophysics}
\end{center}
\medskip
\noindent{\bf \underline{Problem 1}}
\smallbreak
\noindent {\bf a)} For $n=0$ Lane-Emden equation takes the simple form
$$
\frac{1}{\xi^2} \frac{d}{d \xi} \left(\xi^2 \frac{d \phi}{d \xi}\right) = -1.
$$
After multiplying both sides by $\xi^2\,d \xi$, we can integrate it directly:
\beq
\int{d \left(\xi^2 \frac{d \phi}{d \xi}\right)} = -\int{\xi^2\,d \xi},\\
\xi^2 \frac{d \phi}{d \xi} = -\frac{\xi^3}{3} + c_1,
\eeq
where $c_1$ is a constant of integration.  

Now we have a first order differential equation.  To integrate it, we 
multiply both sides by $d \xi/\xi^2$.
\beq
\int{d \phi}= \int{\left(-\frac{\xi}{3} + \frac{c1}{\xi^2}\right) d \xi},\\
\phi = -\frac{\xi^2}{6} - \frac{c1}{\xi} + c_2,
\eeq
where $c_2$ is another constant of integration.

To determine the values for $c_1$ and $c_2$, we have to remember our boundary
conditions.  They are 
$$
\frac{d \phi}{d \xi}\ \ \ \ \ {\rm and}\ \ \ \ \ \phi(0) = 1
\Bigg\vert_{\xi=0} = 0.
$$
The first boundary condition is a bit tricky.  The derivative of $\phi$ with 
respect to $\xi$ is of course 
$$
\frac{d \phi}{d \xi} = -\frac{\xi}{3} + \frac{c_1}{\xi^2},
$$
this was in fact one of our intermediate steps.  Now if we set $\xi = 0$, the 
second term on the right hand side diverges.  The only way to keep it from 
exploding is by setting $c_1 = 0$.   Applying the second boundary condition
gives us $c_2 =1$.

Thus our final result is 
$$
\phi = -\frac{\xi^2}{6} + 1.
$$

As a check, we can compute $\xi_1$ and $\xi_1^2 \phi_1^{\prime}$:
$$
\xi_1 = \xi(\phi=0) = \sqrt{6} \simeq 2.449
$$
and
$$
\xi_1^2 \phi_1^{\prime} = \xi_1^2 \frac{d \phi}{d \xi}\Bigg\vert_{\xi=\xi_1} = 
-\frac{\xi_1^3}{3} = -2 \sqrt{6} \simeq 4.899.
$$
Both values are in agreement with the table on page 131.

\medbreak
\noindent {\bf b)} Let's start with the density as a function of radius.
By definition of $\phi$, $\rho(r) = \rho_c \phi^n = \rho_c$.  So the 
density is constant throughout the star and is equal to the average value
$$
\rho = \frac{M}{\frac{4}{3} \pi R^3}.
$$

There are two ways to compute $P(r)$.  We can use the polytropic equation of
state, even though it seems to diverge at a first glance,
$$
P(r) = K \rho(r)^{1+\frac{1}{n}} = K \rho_c^{1+\frac{1}{n}} \phi^{n+1} =
\left(K \rho_c^{1+\frac{1}{n}}\right) \phi.
$$
We know that $P$ must be finite at any point in the star, thus the quantity in 
parentheses is just a constant, which is equal to $P_c= P(0)$ 
(since $\phi(0) = 1$).  The expression for $P_c$ is derived in the notes (page
132):
$$
P_c = \frac{1}{4 \pi} \frac{1}{(\phi^{\prime}_1)^2} \frac{G M^2}{R^4} =
\frac{1}{4 \pi} \left(\frac{3}{\sqrt{6}}\right)^2 \frac{G M^2}{R^4} =
\frac{3}{8 \pi} \frac{G M^2}{R^4},
$$
where we substituted the numerical value for $(\phi^{\prime}_1)^2$ computed in
part (a).  So 
$$
P(r) = \frac{3}{8 \pi} \frac{G M^2}{R^4} \phi = \frac{3}{8 \pi} 
\frac{G M^2}{R^4} \left[1-\frac{1}{6} \left(\frac{r}{a}\right)^2\right].
$$
The expression for $a$ in terms of $K$ and $\rho_c$ (page 128) also
seems to diverge for $n=0$.  But we can easily eliminate this divergence:
$$
a^2 = \frac{K \rho_c^{\frac{1}{n}-1}}{4 \pi G} = 
\frac{K \rho_c^{\frac{1}{n}+1}}{4 \pi G \rho_c^2} =
\frac{3}{8 \pi} \frac{G M^2}{R^4} 
\frac{\left(\frac{4}{3} \pi R^3\right)^2}{4 \pi G M^2} = \frac{R^2}{6}.
$$
Finally, when we combine the results of the last two equations, we get
$$
P(r) = \frac{3}{8 \pi} \frac{G M^2}{R^4} (1-\frac{r^2}{R^2}) = 
\frac{3}{8 \pi} \frac{G M^2}{R^6} (R^2-r^2).
$$

Another and somewhat easier way to get the same result is by integrating the
equation for hydrostatic equilibrium.  This is easily done for $n=0$ model 
since the density is constant throughout the star.  Most of you used this 
method.   The main thing to remember here is that we need to use $M(r)$ and
{\em not} the total mass of the star:
$$
\frac{d P}{d r} = - \frac{G M(r) \rho(r)}{r^2} = -\frac{4 \pi r^3 G \rho_c^2}
{3 r^2} =   -\frac{3 G M^2}{4 \pi R^6} r.
$$
Now we multiply both sides by $d r$ and integrate (don't forget the 
integration constant which is equal to the central pressure):
$$
P = -\frac{3 G M^2}{8 \pi R^6} r^2 + P_c =\frac{3 G M^2}{8 \pi R^6} (R^2 -r^2).
$$.

Finally, the temperature can be determined from the perfect gas law equation
$$
T(r) = \frac{\mu m_p}{k} \frac{P(r)}{\rho(r)} = 
\frac{\mu m_p}{k} \frac{3 G M^2}{8 \pi R^6} \frac{\frac{4}{3} \pi R^3}{M}
(R^2 -r^2) = \frac{G M}{2 R^3} (R^2 -r^2),
$$

So $n=0$ polytrope model is just a uniform density sphere.

\bigskip
\noindent{\bf \underline{Problem 2}}
\smallbreak
\noindent {\bf a)} First of all, let's figure out what is the difference 
between $^{12}_6$C and $^{56}_{26}$Fe.  The only relevant quantity is the 
mean molecular weight per electron, $\mu_e$, which is exactly 2 for C and 
$56/26 \simeq 2.15$ for Fe, since iron has slightly more neutrons than protons.
 
Now consider a non-relativistic white dwarf star.  First thing we need to 
compare is $\rho(r)$ for two stars of {\em the same} $\rho_c$ (this point is 
very important, in fact most of you were confused exactly what quantity to hold
constant).  Since $\phi(\xi)$ is fixed for a given polytropic index and 
$\rho_c$ is constant, $\rho(\xi)$ will be the same for both Fe and C white 
dwarfs.  But not $\rho(r)$!  Because $r = \xi a$, where $a \propto \sqrt{K} 
\propto \mu_e^{-5/6}$ for fixed $\rho_c$ (see pages 128 and 141).  Thus, 
$a_{\fe} < a_{\c}$, and for a given $\xi$, $r_{\fe} < r_{\c}$.  Since density 
decreases monotonically with radius, this implies that $\rho_{\fe} (r) < 
\rho_{\c} (r)$ at a fixed $r> 0$.  Physically this just means that in the iron 
white dwarf density falls off faster with radius than in the carbon white 
dwarf.  An immediate consequence of this result is that Fe white dwarf will 
have a smaller radius.

To compare $M(r)$ in the two cases remember that $M(r) = 
\int_0^r{4 \pi r^2 \rho(r) d r}$.  Since $\rho_{\fe} (r) < \rho_{\c} (r)$
for all $r > 0$, this clearly implies that $M_{\fe} (r) < M_{\c} (r)$.

To compare pressure is a little bit more tricky.  In the center, since $\rho_c$
is the same in both cases, $P_{\fe}$ is clearly less than $P_{\c}$ (just 
because we have less electrons in the center of the iron star).  At $r>0$ we
have 
$$
P(r) \propto \left(\frac{\rho(r)}{\mu_e}\right)^{5/3},
$$
and since $\rho_{\fe} (r) < \rho_{\c} (r)$ and $\mu_e(\fe) > \mu_e(\c)$, we 
must have $P_{\fe}(r) < P_{\c}(r)$.  Alternatively, we can argue that since 
pressure 
has to support the weight of the outer layers of the star, and for a given $r$ 
there is less material on top in the iron white dwarf than in the carbon one,
pressure in the former is less.

\medbreak
\noindent {\bf b)} Clearly, since iron is able to provide less electron 
degeneracy pressure per unit mass, $M_{\rm Ch}$ must be smaller.  To compute 
the fractional difference remember that $M_{\rm Ch} \propto \mu_e^{-2}$.  Then
$$
\frac{M_{\rm Ch,C}}{M_{\rm Ch,Fe}} = \left(\frac{\mu_e (\fe)}{\mu_e (\c)} 
\right)^2 = \left(\frac{2.15}{2}\right)^2 \simeq 1.16.
$$
 
\bigbreak
\noindent{\bf \underline{Problem 3}}
\smallbreak
\noindent {\bf a)} Number of neutrinos created during the inverse 
$\beta$-decay is
$$
N_{\nu_e} = \frac{1.4 M_{\odot}}{m_p + m_e} \simeq \frac{1.4 M_{\odot}}{m_p}
= 1.8\times 10^{57}.
$$
Each one carries an energy of $\sim 1\,{\rm MeV} = 10^6\,{\rm eV} \times
1.6 \times 10^{-12}\,{\rm ergs/eV} = 1.6\times 10^{-6} \ergs$, so the
total energy loss is
$$
E_{\nu_e} = 1.8\times 10^{57} \times 1.6\times 10^{-6} \ergs = 2.9 \times 
10^{51} \ergs.
$$
\medbreak
\noindent {\bf b)} The initial radius of the white dwarf (assuming 
non-relativistic) equation of state can be easily computed using the equation 
on page 142:
$$
R_{\rm WD}  = 10^9\,\cm \left[\frac{0.7011 M_{\odot}}{1.4 M_{\odot}} 
\left(\frac{2}{\mu_e}\right)^5\right]^{1/3} = 7.9 \times 10^8\,\cm.
$$
Note that this is rather inexact, since we know that $1.4 M_{\odot}$ white 
dwarfs {\em are} relativistic.  However, as we will see later, the exact number
doesn't really matter here.

To compute the radius of the neutron star we need to go back to the equation 
of state on page 141.  It is derived for electron degeneracy pressure, but the
result for neutron degeneracy pressure must be practically the same, except
we have to substitute $\mu_n \rightarrow \mu_e$ and $m_n \rightarrow m_e$.  
Then
$$
P_n = \left[\frac{1}{20} \left(\frac{3}{\pi}\right)^{2/3} \frac{h^2}
{m_n m_p^{5/3} \mu_n^{5/3}}\right] \rho^{5/3} \simeq
\left[\frac{1}{20} \left(\frac{3}{\pi}\right)^{2/3} \frac{h^2}
{m_p^{8/3}}\right] \rho^{5/3},
$$
where we set $m_n \simeq m_p$, and of course mean molecular weight per neutron,
$\mu_n$, is unity.  

Now we follow the same sequence of steps as we did for a non-relativistic 
white dwarf (page 142) with $n= 3/2$. Since $P_n = K_n \rho^{1+\frac{1}{n}}$,
$$
K_n = \left[\frac{1}{20} \left(\frac{3}{\pi}\right)^{2/3} \frac{h^2} 
{m_p^{8/3}}\right].
$$
And using the expressions for $R$ and $M$ derived from polytrope theory,
$$
R_{\rm NS}  = \xi_1 \left[\frac{5 K_n}{8 \pi G}\right]^{1/2} \rho_c^{-1/6} =
1.45 \times 10^6\,\cm \left(\frac{10^{15}\,\g/\cm^3}{\rho_c}\right)^{1/6}
$$
and
$$
M_{\rm NS} = -\xi_1^2 \phi_1^{\prime}\, 4 \pi\left[\frac{5 K_n}{8 \pi G}
\right]^{3/2} \rho^{1/2} =
1.11 M_{\odot} \left[\frac{\rho_c}{10^{15}\,\g/\cm^3}\right]^{1/2}.
$$
After eliminating $\rho_c$, we have an expression for $R$ in terms of $M$
$$
R_{\rm NS} = 1.5 \times 10^6 \left(\frac{M_{\odot}}{M_{\rm NS}}\right)^{1/3}.
$$
For $M_{\rm NS} = 1.4 M_{\odot}$, $R_{\rm NS} = 13.4\,\km$.

The gravitational binding energy for a star of mass $M$ and radius $R$ is
roughly $\sim G M^2/R$, so energy released in the collapse is 
$$
E_g \sim -\frac{G M_{\rm WD}^2}{R_{\rm WD}} + 
\frac{G M_{\rm NS}^2}{R_{\rm NS}} = G (1.4 M_{\odot})^2  
\left(\frac{1}{R_{\rm NS}}- \frac{1}{R_{\rm WD}}\right) \simeq
\frac{G (1.4 M_{\odot})^2}{R_{\rm NS}},
$$
since $R_{\rm NS} \ll R_{\rm WD}$.  So
$$
E_g \sim 3.9 \times 10^{53}\,\ergs.
$$

Thus, the energy lost through neutrinos is on the order of $1\%$ of the total 
energy released in the collapse.

\medbreak
\noindent{\bf c)} The mass of the envelope is $M_{env} = 8.6 M_{\odot}$.
Since $E_{\nu_e} \ll E_g$, 
$$
\frac{M_{env} v^2}{2} = E_g
$$
and the velocity of the  ejected envelope is 
$$
v = \sqrt{\frac{2 E_g}{M_{env}}} = 6.7 \times 10^9\,\cms = 0.22 c,
$$
fairly relativistic.

\medbreak
\noindent {\bf d)} Both protons and electrons have the same velocity because
the plasma must be neutral.  Since protons have larger mass, they have larger
kinetic energy and correspondingly larger temperature.
$$
T_p \simeq \frac{m_p v^2}{2 k} = 2.7 \times 10^{11}\,\K,
$$
and 
$$
T_e \simeq \frac{m_e v^2}{2 k} = 1.5 \times 10^8\,\K.
$$

Radiation is emitted by the electrons, so $T_e$ is the relevant temperature.
Thermal radiation is always emitted at a characteristic blackbody frequency
$$
\nu_{max} \sim 5.9 \times 10^{10} T_e\,{\rm Hz}=8.8 \times 10^{18} \,{\rm Hz}.
$$
Thus, observations in X-rays are most appropriate for studying these shock 
fronts.
\end{document}