Homework 2 solutions

1. Solid or gas?

We want to determine if the Earth is solid. We start by writing the separation between atoms in the Earth as:

\[d \sim n^{-1/3} = \left ( \frac{\mu m_u}{\rho} \right )^{-1/3}\]

then the Coulomb energy is

\[E_c = \frac{Z^2 e^2}{d} = Z^2 e^2 \left ( \frac{\rho}{\mu m_u} \right )^{1/3}\]

Now the thermal energy is \(k T\), and we can take our estimate of \(T\) from the Virial theorem:

\[kT = \frac{1}{5} \mu m_u \frac{GM}{R}\]

the ratio is then

\[\Gamma = \frac{E_c}{kT} = 4 Z^2 e^2 \left ( \frac{\rho}{\mu m_u} \right )^{1/3} \frac{1}{\mu m_u} \frac{R}{GM}\]

and substituting in the average density, \(\rho \sim M / R^3\), we have

\[\Gamma \sim 5 Z^2 e^2 \frac{1}{(\mu m_u)^{4/3}} \frac{1}{GM^{2/3}}\]

putting in numbers, taking the core to be iron, \(Z = 26\), \(\mu = 56\), and the mass of the Earth as \(M_\oplus = 6\times 10^{27}~\mathrm{g}\), we have

\[\Gamma \sim 800\]

This is very large, so clearly the Earth is solid.

2. Dimensional analysis

We want to use the equations of stellar structure + Kramers’ opacity to predict main-sequence stellar properties.

We assume uniform composition and a fully-radiative star. Our equations of stellar structure are:

• radiation:

  \[L = -(4\pi r^2)^2 \frac{ac}{3\kappa} \frac{dT^4}{dM} \rightarrow L \sim \frac{T^4 R^4}{\kappa M}\]

• energy:

  \[\frac{dL}{dM} = \epsilon \rightarrow L \sim M \epsilon\]

• HSE:

  \[\frac{dP}{dM} = -\frac{GM}{4\pi r^4} \rightarrow P \sim \frac{GM^2}{R^4}\]

• continuity:

  \[\frac{dr}{dM} = \frac{1}{4\pi r^2 \rho} \rightarrow \rho \sim \frac{M}{R^3}\]

we also have our equation of state:

\[P = \frac{\rho k T}{\mu m_u} \sim \frac{\rho T}{\mu}\]

a.

We want to find the luminosity-mass relationship. We start by combining HSE + EOS:

\[\frac{\rho T}{\mu} \sim \frac{G M^2}{R^4} \rightarrow T \sim \frac{M\mu}{R}\]

and we use radiation, substituting this temperature expression:

\[L \sim \frac{T^4 R^4}{\kappa M} \sim \frac{R^4}{\kappa M} \left (\frac{M\mu}{R} \right )^4 \sim \frac{M^3\mu^4}{\kappa}\]

Now we assume an opacity of the form \(\kappa \sim Z (1 + X) \rho T^{-7/2}\), giving:

\[L \sim \frac{M^3 \mu^4}{Z (1+X)} \frac{1}{\rho} T^{7/2} \sim \frac{M^3 \mu^4}{Z(1+X)} \frac{R^3}{M} \left (\frac{M\mu}{R} \right )^{7/2} \sim \frac{M^{5.5} \mu^{7.5}}{Z (1+X)} R^{-0.5}\]

Now we start with the energy equation and substitute our form of nuclear energy generation, \(\epsilon \sim X^2 \rho T^4\):

\[L \sim M\epsilon \sim M X^2 \rho T^4 \sim M X^2 \left ( \frac{M}{R^3} \right ) \left (\frac{M\mu}{R} \right )^4 \sim \frac{M^6 \mu^4}{R^7} X^2\]

Equating these two expressions for \(L\), we find:

\[\frac{M^{5.5} \mu^{7.5}}{Z (1+X)} R^{-0.5} \sim \frac{M^6 \mu^4}{R^7} X^2\]

\[R \sim M^{1/13} \mu^{-7/13} \left [ X^2 (1 + X) Z \right ]^{2/13}\]

Now we can put this back into our expression for L:

\[L \sim \frac{M^{5.5} \mu^{7.5}}{Z (1+X)} M^{-1/26} \mu^{7/26} \left [X^2 (1 + X) Z \right ]^{-1/13} \sim M^{71/13} \mu^{101/13} \left [ Z (1 + X) \right ]^{-14/13} X^{-2/13}\]

From this point forward, we’ll ignore the \(\mu\) and \(X\) dependence. So we have:

\[L \sim M^{71/13} Z^{-14/13}\]

\[R \sim M^{1/13} Z^{2/13}\]

Note

This is a much steeper mass dependence for luminosity than we found with the constant electron scattering opacity.

b.

Now we want to find effective temperature. We start with the blackbody relation:

\[L \sim R^2 T_\mathrm{eff}^4\]

so

\[T_\mathrm{eff} \sim \frac{L^{1/4}}{R^{1/2}} \sim \frac{M^{71/52} Z^{-14/52}}{M^{1/26}Z^{2/26}} \sim M^{69/52} Z^{-18/52}\]

or replacing \(M\) with \(L\), we have:

\[T_\mathrm{eff} \sim L^{69/284} Z^{-78/923}\]

and finally

\[L \sim T_\mathrm{eff}^{284/69} \left ( Z^{78/923} \right )^{284/69} \sim T_\mathrm{eff}^{284/69} Z^{8/23}\]

c.

Our \(Z\) dependence shows that for a smaller Z (e.g. pop II stars), a star is less luminous for the same effective temperature. This means that the low metalicity main sequence lies beneath the high metalicity main sequence.

3. Schönberg-Chandrasekhar limit

We want to compute the maximum mass of an isothermal core.

a.

Let’s rederive the Virial theorem without going all the way to the surface. Starting with HSE:

\[dP = - \frac{GM}{4\pi r^4} dM\]

we multiply by volume and integrate, giving:

\[\int_0^{P(r)} V dP = \frac{\Omega}{3}\]

We’ll integrate to the edge of the core, which we’ll denote as \(R_c\) and the pressure there as \(P_c = P(R_c)\) (this is not central pressure). Then, integrating by parts, our integral is:

\[\int_0^{P_c} V dP = \int_0^{P_c} d(VP) - \int_0^{V_c} P dV = V_c P_c - \int_0^{M_c} \frac{P}{\rho} dM\]

where \(M_c\) is the mass of the core. This is basically what we did originally when we derived the Virial theorem in class, except now we have the surface term, \(V_c P_c\).

b.

Now we take the core to be isothermal and of uniform composition and well-described by an ideal gas:

\[V_c P_c - \int_0^{M_c} \frac{k T_c}{\mu_c m_u} dM = \frac{\Omega_c}{3}\]

or

\[\frac{kT_c}{\mu_c m_u} M_c = V_c P_c + \frac{q}{3} \frac{G M_c^2}{R_c}\]

where \(q\) is the \(\mathcal{O}(1)\) constant in the gravitational potential energy.

Replacing the volume of the core, we have:

\[P_c = \frac{3}{4\pi} \frac{k T_c}{\mu_c m_u} \frac{M_c}{R_c}^3 - \frac{q}{4\pi} \frac{GM_c^2}{R_c^4}\]

c.

To find the radius where the pressure is maximum, we differentiate with respect to \(R_c\) and set it to zero:

\[\frac{dP_c}{dR_c} = -\frac{9}{4\pi} \frac{k T_c}{\mu_c m_u} \frac{M_c}{R_c^4} + \frac{q}{\pi} \frac{G M_c^2}{R_c^5} = 0\]

this gives

\[R_{c,\mathrm{max}} = \frac{4q}{9} \frac{\mu_c m_u}{k T_c} G M_c\]

and the corresponding pressure is

\[P_{c,\mathrm{max}} = P_c(R_{c,\mathrm{max}}) = \frac{3}{16\pi} \left (\frac{9}{4} \right )^3 \frac{1}{q^3} \left ( \frac{k T_c}{\mu_c m_u} \right )^4 \frac{1}{G^3 M_c^2}\]

d.

Requiring that

\[P_c > \frac{3}{8\pi} \frac{GM_\star^2}{R_\star^4}\]

and dropping constants, we have:

\[P_{c, \mathrm{max}}(M_c) \propto \left (\frac{kT_c}{\mu_c m_u} \right )^4 \frac{1}{G^3 M_c^2} \ge \frac{3}{8\pi} \frac{GM_\star^2}{R_\star^4}\]

and from the Virial theorem, we have

\[T_c = \frac{1}{2} \frac{GM_\star}{R_\star} \frac{\mu_\mathrm{env} m_u}{k}\]

giving:

\[\frac{M_c}{M_\star} \le \left (\frac{\mu_\mathrm{env}}{\mu_c} \right )^2 \cdot \mathrm{constant}\]