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University of the Witwatersrand
Advanced Techniques in Physics : 2002
Examination : June 2002


Instructions
This is the take - home question. It is to be handed out on Friday 8:30 21$^{th}$ June.
There will be 24 hours to answer question 6.
Time for Question 6 : One day (take home conditions) Total Marks (Q6) = 70

3.
This question requires some programming.
Please remember to hand in your diskette with your script.

White dwarfs are cold stars where the nucleo-synthesis process is completed. They consist of heavy stable nuclei and their electrons. In a simple model, they can be considered centrally symmetric and non-rotating (neglect magnetic fields). Their structure is then determined by the hydrostatic equilibrium between gravitational pressure seeking to compress the stellar material and Pauli pressure which resists this.

Depending on the mass of the star, the nucleo-synthesis process is assumed to have run to completion leaving the star dominantly composed of a single stable nucleus terminating a fusion cycle, such as $^{12}$C, or $^{56}$Fe. The electrons are modelled as a $T=0$ free Fermi gas, and they dominate the Pauli pressure term.

Under these conditions, we can find two coupled first order differential equations for the radial mass and density distribution of the white dwarf.

As in electrostatics, the gravitational force per unit volume at a given radial distance from the centre of the star is dependent on the amount of matter enclosed by a sphere of that same radius. Therefore

\begin{displaymath}
\frac{dP(r)}{dr} = \frac{dF(r)}{dV} = -\frac{Gm(r)}{r^2}\rho(r)
\end{displaymath} (1)

where
\begin{displaymath}
m(r)= 4\pi \int_0^r \rho(r')r'^2\, dr'
\end{displaymath} (2)

is the mass of the star to radius $r$. In differential form, this equation is
\begin{displaymath}
\frac{dm(r)}{dr}= 4\pi r^2 \rho(r).
\end{displaymath} (3)

This is our first differential equation mentioned above. By the chain rule, we would have our second differential equation,
$\displaystyle \frac{d\rho(r)}{dr}$ $\textstyle =$ $\displaystyle \left(\frac{dP(\rho)}{d\rho}\right)^{-1}
\cdot \frac{dP(r)}{dr}$ (4)
  $\textstyle =$ $\displaystyle - \left(\frac{dP(\rho)}{d\rho}\right)^{-1}
\cdot\frac{Gm(r)}{r^2}\rho(r)$  

if we could just find $P(\rho)$, the equation of state for the white dwarf.

The initial condition $\rho(r=0) = \rho_c$ will determine the final mass $M$ and radius $R$ of the star, $M=m(R)$, by integration of the two coupled first order differential equations for $m(r)$ and $\rho(r)$.

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Four pages - Page two

PHYS400
PHYSICS HONOURS
June 2001 Advanced Techniques in Physics




The equation of state, $P(\rho)$, is in fact the Pauli pressure term mentioned above. We begin with the energy density of the electrons in units where $\hbar = c = 1$
\begin{displaymath}
\frac{E}{V} = 2\int_0^{p_f} \frac{d^3p}{(2\pi)^3}(p^2 + m_e^2)^{1/2}
\end{displaymath} (5)

where $p_f$ is the Fermi momentum
\begin{displaymath}
p_f = (3\pi^2n)^{1/3}
\end{displaymath} (6)

and $n$ is the electron density
\begin{displaymath}
n = Y_e\frac{\rho(r)}{m_n}.
\end{displaymath} (7)

In the expression for the electron density, $Y_e$ is the number of electrons per nucleon and $m_n$ is the nucleon mass. Upon integration of equation 5, we have
\begin{displaymath}
\frac{E}{V} = n_0m_ex^3\epsilon(x)
\end{displaymath} (8)

where
\begin{displaymath}
\epsilon(x) = \frac{3}{8x^3}\left[x(1+2x^2)(1+x^2)^{1/2}-\log[x+(1+x^2)^{1/2}] \right]
\end{displaymath} (9)

and
\begin{displaymath}
x = \frac{p_f}{m_e} = \left(\frac{n}{n_0}\right)^{1/3} = \left(\frac{\rho}{\rho_0}\right)^{1/3}
\end{displaymath} (10)

with
\begin{displaymath}
n_0=\frac{m_e^3}{3\pi^2} = 5.89 \times 10^{29} \mbox{cm}^{-...
..._n n_0}{Y_e} = 9.79
\times 10^{5}Y^{-1} \mbox{gm\,cm}^{-3}.
\end{displaymath} (11)

Thus $x$ is the electron density in units of $n_0$, chosen so that $x=1$ corresponds to the Fermi momentum $p_f$ being equal to the electron mass $m_e$ (see equations 7 and 6).

Noting that $n=N/V$, we see that

\begin{displaymath}
\frac{\partial x}{\partial V} = - \frac{x}{3V}
\end{displaymath} (12)

According to thermodynamics, the pressure is the change in energy with volume, so that
$\displaystyle P$ $\textstyle =$ $\displaystyle -\frac{\partial E}{\partial V}$ (13)
  $\textstyle =$ $\displaystyle -\frac{\partial E}{\partial x} \frac{\partial x}{\partial V}$  
  $\textstyle =$ $\displaystyle \mbox{$\frac13$}n_0m_ex^4\frac{d\epsilon(x)}{dx}$  

Because the electron density scales with the density, $x \equiv x(\rho)$, equation 13 is indeed the equation of state. We can now evaluate
$\displaystyle \frac{dP(\rho)}{d\rho}$ $\textstyle =$ $\displaystyle \frac{dP(x)}{dx}\frac{dx(\rho)}{d\rho}$ (14)
  $\textstyle =$ $\displaystyle Y_e \cdot \frac{m_e}{m_n} \cdot \frac{x^2}{3(1+x^2)^{1/2}}$  

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PHYS400
PHYSICS HONOURS
June 2002 Advanced Techniques in Physics




The two coupled first order differential equations then become :
$\displaystyle \ \qquad \qquad \qquad
\frac{dm(r)}{dr}$ $\textstyle =$ $\displaystyle 4\pi r^2 \rho(r)$ (15)
$\displaystyle \mbox{and} \qquad \qquad \qquad$      
$\displaystyle \ \qquad \qquad \qquad
\frac{d\rho(r)}{dr}$ $\textstyle =$ $\displaystyle - \frac{G}{Y_e(m_e/m_n)}
\cdot\frac{m(r)}{r^2}\frac{\rho(r)}{\gamma(x(\rho))}$  

where
\begin{displaymath}
\gamma(x) = \frac {x^2} {3(1+x^2)^{1/2} }.
\end{displaymath} (16)

Converting to dimensionless variables $\bar r$, $\bar\rho$ and $\bar m$ for the stellar radius, density and mass,

$\displaystyle \ \qquad \qquad \qquad
\frac {d\bar m}{d\bar r}$ $\textstyle =$ $\displaystyle \bar r^2 \bar \rho$ (17)
$\displaystyle \mbox{and} \qquad \qquad \qquad$      
$\displaystyle \ \qquad \qquad \qquad
\frac {d\bar\rho}{d\bar r}$ $\textstyle =$ $\displaystyle - \frac {\bar m \bar \rho}{\gamma \bar r^2}$  

where
\begin{displaymath}
r = R_0 \bar r, \ \ \ \rho = \rho_0 \bar\rho, \ \ \ m = M_0\bar m,
\ \ \ x=\bar\rho^{1/3}
\end{displaymath} (18)

and the scale factors have been chosen to achieve unit coefficients in equation 17,
$\displaystyle R_0$ $\textstyle =$ $\displaystyle \left [ \frac{Y_e(m_e/m_n)}{4\pi G\rho_0} \right] ^{1/2}
= 7.72 \times 10^8 Y_e \mbox{cm},$ (19)
$\displaystyle M_0$ $\textstyle =$ $\displaystyle 4\pi R^3_0\rho_0 = 5.67 \times 10^{33} Y_e^2 \mbox{gm}$  
$\displaystyle \rho_0$ $\textstyle =$ $\displaystyle \frac {M_p n_0}{Y_e}
= 9.79 \times 10^5 Y_e^{-1} \mbox{gm\,cm}^{-3}.$  

Note that solutions for different values of $Y_e$ may all be scaled from $Y_e$ = 1.

a)
As a partial approximate analytic solution, find the leading terms in the $\epsilon(x)$ and $\gamma(x)$ expressions for the non-relativistic $(x\ll 1)$ and the relativistic $(x \gg 1)$ limits and discuss the plausibility of the results. Using the result for small $\bar\rho$ with $\bar m$ and $\bar r$ finite, show the white dwarf has a well defined surface and find the functional dependence of $\bar\rho$ near the surface. (10 marks)
b)
Solve numerically for $\bar\rho (\bar r)$ and $\bar m (\bar r)$. Mention any steps you could take, of a numerical nature, that would generate confidence in the numerical result. Calculate the stellar total mass and radius for centre densities $\bar\rho_c = \bar\rho(0)$ in the range $10^{-1}$ to $10^6$. Does the behaviour of the solution correspond to your approximate analytical solution ? (explain).
(35 marks)
c)
Deduce the composition of Sirius B, $R=0.0074(6) R_0$, $M = 1.05(3) M_0$. (5 marks)

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PHYS400
PHYSICS HONOURS
June 2002 Advanced Techniques in Physics




d)
Investigate the behaviour of the mass and radius of the white dwarf for large central densities. You will find the mass approaches a limit and the radius collapses. Deduce the Chandrasekhar limit for the mass and central density of the white dwarf star? The destiny of such a large star would be a neutron star. (10 marks)
e)
Explain the previous result using a simple model where the density profile is constant $d\bar\rho/d\bar r = k$. Calculate the total energy $U+W$ of the star where
\begin{displaymath}
U = \int_0^R\left( \frac{E}{V}\right)4\pi r^2 dr
\end{displaymath} (20)

arises from the repulsive Pauli pressure and
\begin{displaymath}
W = -\int_0^R\frac{Gm(r)}{r^2}\rho(r)4\pi r^2 dr
\end{displaymath} (21)

arises from the attractive gravitational potential. Show that for a given total mass $M$ and at large densities (extreme relativistic limit) then both terms have a $1/R$ dependence and predict the Chandrasekhar collapse for large enough $M$. (10 marks)

(Your script will take the form of a report that addresses the questions. It will be accompanied by a diskette containing a well documented tested source code file and a graphic file for display of the theory curves.)

Total Question 6 : [70 marks]





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Connell 2004-04-15