Homework 7 solutions

1. Electron heat capacity

We want to find the specific heat of degenerate non-relativistic electrons and compare to that of the ions in a white dwarf to see if we were justified in neglecting their contribution when we did white dwarf cooling.

import numpy as np

We’ll use unyt to keep track of units for us. This uses SymPy underneat to symbolically track the units we attach to quantities.

import unyt as u

In a previous homework, we found:

(8)\[\begin{equation} P_e = \frac{h^2}{20m_e} \left ( \frac{3}{\pi} \right )^{2/3} n_e^{5/3} \left [ 1 + \frac{40\pi^2}{(3/\pi)^{4/3}} \frac{m_e^2}{h^4} \frac{(kT)^2}{n_e^{4/3}} \right ] \end{equation}\]

We’ll write this as:

\[P_e = \alpha n_e^{5/3} \left [ 1 + \beta \frac{T^2}{n_e^{4/3}} \right ]\]

with

\[\alpha = \frac{h^2}{20m_e} \left ( \frac{3}{\pi} \right )^{2/3}\]

and

\[\beta = \frac{40\pi^2}{(3/\pi)^{4/3}} \frac{m_e^2}{h^4} k^2\]

Let’s compute these

h = 6.63e-27 * u.erg * u.s
m_u = 1.67e-24 * u.g
m_e = 9.11e-28 * u.g
k = 1.38e-16 * u.erg / u.K

alpha = h**2 / (20 * m_e) * (3 / np.pi)**(2./3.)
alpha

unyt_quantity(2.33951736e-27, 'erg**2*s**2/g')

beta = 40 * np.pi**2 / (3 / np.pi)**(4./3.) * m_e**2 / h**4 * k**2
beta

unyt_quantity(3.43404019e+21, 'g**2/(K**2*erg**2*s**4)')

Now, we know that for non-relativistic electrons,

\[\rho e_e = \frac{3}{2} P_e\]

and

\[n_e = \frac{\rho}{\mu_e m_u}\]

so

\[e_e = \frac{3}{2} \alpha \frac{n_e^{5/3}}{\rho} \left [ 1 + \beta \frac{T^2}{n_e^{4/3}} \right ]\]

and

\[c_v = \left . \frac{\partial e_e}{\partial T} \right |_\rho = 3 \alpha \beta \frac{T}{\rho^{2/3}(\mu_e m_u)^{4/3}}\]

If we take typical conditions in a white dwarf, we can evaluate this

rho = 1.e7 * u.g / u.cm**3
T = 1.e7 * u.K
mu_e = 0.5

cv_e = 3 * alpha * beta * T / rho**(2./3.) / (mu_e * m_u)**(1./3.)
cv_e

unyt_quantity(551430.44751741, 'cm**2/(K*s**2)')

cv_e.to("erg/g/K")

unyt_quantity(551430.44751741, 'erg/(K*g)')

We see that for the electrons,

\[c_{v,e} = 5.5\times 10^5~\mathrm{erg / g / K}\]

For the ions, in class we found that

\[c_v = \frac{3}{2} \frac{k}{\mu_I m_u}\]

we can evaluate this for a carbon white dwarf (\(\mu_I = 12\)):

mu_I = 12
cv_ions = 1.5 * k / (mu_I * m_u)
cv_ions

unyt_quantity(10329341.31736527, 'erg/(K*g)')

We see that we have

\[c_{v,ions} = 10^7~\mathrm{erg/g/K}\]

so \(c_{v,ions} \gg c_{v,e}\), justifying our assumption in class.

2. WD cooling

Another assumption we made when we did white dwarf cooling was to assume that the radiative layer was thin. Here we’ll check that.

We showed that the radiative layer behaves like a polytrope with an equation of state:

\[P = K^\prime T^{1+n_\mathrm{eff}}\]

with \(n_\mathrm{eff} = 13/4\).

Let’s start with hydrostatic equilibrium:

\[\frac{dP}{dr} = -\rho \frac{G M_\star}{r^2}\]

where \(M_\star\) is assumed to be the constant mass of the star (the layer does not have significant masss).

Using the chain rule, we can write:

\[\frac{dP}{dr} = \frac{dP}{dT} \frac{dT}{dr} = \frac{1}{\nabla} \frac{P}{T} \frac{dT}{dr}\]

and from our EOS, we see \(\nabla^{-1} = 1 + n_\mathrm{eff} = 17/4\).

Now we use the ideal gas law to replace \(P/T = \rho k /(\mu m_u)\), giving

\[\frac{dT}{dr} = -\frac{\mu m_u}{k} \nabla \frac{GM_\star}{r^2}\]

We can integrate this from the transition radius (where \(T = T_c\)) to the surface of the star

\[\int_{T_c}^{T_\star} dT = - \frac{\mu m_u}{k} \nabla G M_\star \int_{r_\mathrm{tr}}^{R_\star} \frac{1}{{r^\prime}^2} dr^\prime\]

which gives

\[T_\star - T_c = \nabla \frac{\mu m_u}{k} G M_\star \left ( \frac{1}{r_\mathrm{tr}} - \frac{1}{R_\star} \right )\]

we can take the surface temperature to be 0, giving

\[\frac{k T_c}{\mu m_u} = \nabla \frac{GM_\star}{R_\star} \frac{R_\star - r_\mathrm{tr}}{r_\mathrm{tr}} \sim \nabla g l\]

with \(l = R_\star - r_\mathrm{tr}\) as the layer thickness.

This gives

\[l \sim \frac{17}{4} \frac{k T_c}{\mu m_u g}\]

Let’s put in numbers

M = 0.6 * u.solar_mass_cgs
R = 0.012 * 7.e10 * u.cm
G = 6.67e-8 * u.dyne * u.cm**2 / u.g**2

g = G * M / R**2
g.to("cm/s**2")

unyt_quantity(1.12778349e+08, 'cm/s**2')

Tc = 1.e8 * u.K
mu = 12

l = 17./4. * k * Tc / (mu * m_u * g)
(l/R).to("dimensionless")                      

unyt_quantity(0.03089337, 'dimensionless')

We see that for \(T_c = 10^7~\mathrm{K}\) the layer is indeed thin.

We also see that if we increase \(T_c\) by a factor of 10, then \(l \sim R_\star\), so our approximation is no longer as good.

3. Hot-CNO

We want to estimate the energy release from hot-CNO burning. As discussed in class, the rate becomes limited by the decays of \({}^{14}\mathrm{O}\) and \({}^{15}\mathrm{O}\). These have half-lives of

\[\tau_{1/2}({}^{14}\mathrm{O}) = 71~s\]

\[\tau_{1/2}({}^{15}\mathrm{O}) = 122~s\]

we’ll denote these as \(\tau_{14}\) and \(\tau_{15}\) here.

a.

In equilibrium, we have:

\[\frac{dY({}^{14}\mathrm{O})}{dt} = \frac{dY({}^{15}\mathrm{O})}{dt}\]

and a \(\beta\)-decay reaction has the form:

\[\frac{dY}{dt} = -\lambda Y = - \frac{\log(2)}{\tau_{1/2}} Y\]

so equilibrium means that

\[\lambda_{14} Y_{14} = \lambda_{15} Y_{15}\]

and since \(Y = X/A\), we have

\[X_{14} = X_{15} \frac{A_{14}}{A_{15}} \frac{\tau_{14}}{\tau_{15}}\]

We define \(Z_\mathrm{CNO} = X_{14} + X_{15}\), so \(X_{15} = Z_\mathrm{CNO} = X_{14}\), giving:

\[X_{14} = (Z_\mathrm{CNO} - X_{14} ) \frac{14}{15} \frac{\tau_{14}}{\tau_{15}}\]

solving for \(X_{14}\), we have

\[X_{14} = Z_\mathrm{CNO} \frac{ \frac{14}{15} \frac{\tau_{14}}{\tau_{15}} } {1 + \frac{14}{15} \frac{\tau_{14}}{\tau_{15}} } \approx \frac{Z_\mathrm{CNO}}{2.84}\]

b.

Again using equilibrium, we have

\[\frac{dY({}^{1}\mathrm{H})}{dt} = +4 \frac{dY({}^{14}\mathrm{O})}{dt} = - 4 \frac{\log(2)}{\tau_{14}} \frac{X_{14}}{14}\]

where the “+4” comes from the fact that 4 H are consumed each time a single \({}^{14}\mathrm{O}\) is consumed.

Using the result above, we see

\[\frac{dY({}^{1}\mathrm{H})}{dt} = -4 \frac{\log(2)}{\tau_{14}} \frac{Z_\mathrm{CNO}}{2.84 \cdot 14}\]

This reaction also produces a single \({}^{4}\mathrm{He}\), so

\[\frac{dY( {}^{4}\mathrm{He} )}{dt} = -\frac{1}{4}\frac{dY( {}^{1}\mathrm{H} )}{dt} = \frac{\log(2)}{\tau_{14}} \frac{Z_\mathrm{CNO}}{2.84 \cdot 14} \approx 2.46\times 10^{-4} Z_\mathrm{CNO} / \mathrm{s}\]

The energy release per reaction is \((28.3 - 2.03)~\mathrm{MeV}\) or \(4.2\times 10^{-5}~\mathrm{erg}\)

Therefore the total energy rate is

\[\epsilon = N_A \frac{dY( {}^{4}\mathrm{He} )}{dt} Q = N_A Q \cdot 2.6\times 10^{-4} Z_\mathrm{CNO} / \mathrm{s} = 6.2\times 10^{15} Z_\mathrm{CNO} \mathrm{erg/g/s}\]