Woosley: Astr220C – Advanced Stages of Stellar Evolution

If the neutron star formed in the core collapse of a massive star can be thought of as a sphere of constant density, and if the neutrino path length is, on average, 10cm, estimate:

a) the total energy in ergs that comes out in all forms of neutrinos (6 species) as the protoneutron star (1.4M\odot ) contracts to its final radius (10km)

b) the time scale of the neutrino burst

c) the average luminosity in each flavor (assuming each carries 1/6 of flux), and

d) assuming blackbody emission, the temperature of each neutrino species in either MeV or K

Don’t worry about the time dependence of the structure – treat the neutron star as a sphere of constant size and density. Of course, in reality it must contract, and its density is higher in the center.


a) binding energy

b) different t for each species

Astr233 Cosmology – J. Primack

The nuts and bolts of the Universe.

a) Show that if the curvature k=0 and the scale factor a grows as t^{2/3} , the apparent angular size of distant objects of the same linear size has a minimum at z=1.25.

b) Under the same assumptions, suppose that a galaxy is observed at z=1.25. For what fraction of the Hubble time has its light been traveling towards us?

11. Astr 233 Cosmology – J. Primack

A two-part problem dealing with the Big Picture.

a) Explain what the critical density \rho_{cr} is, and show using the Friedmann equation that \rho_{cr}=3H^{2}/8\pi G

b) Suppose (incorrectly) that the Hubble parameter H scales as temperature squared all the way back untill the temperature of the universe was the Planck energy 1.22\times 10^{19}GeV (i.e., suppose that the universe was dominated by radiation all the way back to the Planck time). Also suppose that the dark energy is in the form of cosmological constant \Lambda , such that \rho_{Lambda} today is equal to 0.7\rho_{cr} , and \rho_{Lambda} remains constant throughout the evolution of the universe. Calculate the value of \rho_{Lambda}/\rho_{cr} at the Planck time, and comment on why this is interesting.


3. Astr204 Physics of Astrophysics I – G.Laughlin

In this question, we develop a scaling law for incompressible turbulence.

In the absence of viscosity (or any other form of dissipation) Newton’s law for a fluid takes the form:

\frac{\partial {\bf u}}{\partial t} +{\bf u}\cdot \nabla {\bf u} = -\nabla \Phi - \frac{1}{\rho}\nabla P

a) If the fluid velocities are very small in comparison to the speed of sound (but nevertheless nonzero) what important approximation can we make to the continuity equation? Write down an expression for this simplified form.

b) With the simplification from part a) in hand, show that Newton’s law (the momentum equation shown above) takes the form:

\frac{\partial \omega}{\partial t}+\nabla \times (\omega \times {\bf u})=0

where \omega=\nabla\times{\bf u}

c) nviscid (or nearly inviscid) flows are generally turbulent. Imagine that the flow described by the equation in part b) is typified by a velocity scale, U, and then succumbs to instability at its largest length scale, L. To order-of-magnitude approximation, what is the rate,\epsilon , at which energy is fed into the largest body?

d) The so-called turbulent cascade occurs because turbulent energy cannot remain only at the largest scale L if steady-state is maintained. If energy is deposited into a smaller eddy scale, \lambda , what is the typical velocity of an eddy at this scale?

e) In the absence of viscosity, energy would be continually dumped into ever-smaller eddys. Any realistic fluid, however, will have a finite Reynold’s number,


In terms of physical quantities, what is \nu ?

f) If the turbulence is fully subsonic, what can we say about the smallest eddy size, \lambda_{0} , in comparison to the particle mean free path l ?

g) Show, drawing on arguments developed above, that \lambda_{0} \sim Re^{-3/4}L


a) When the fluid velocities are smaller than the speed of the sound in the medium, you can assume:

\nabla\times{\bf u}=0



d) v(L)^{2}\frac{v(L)}{L} \sim \mbox{the rate of Energy flow, thus:}

v(L)^{2}\frac{v(L)}{L} \sim \frac{v(\lambda)^{3}}{\lambda}

so v(\lambda)=[v(L)^{3}\frac{\lambda}{L}]^{1/3}

e)  \nu=\mbox{kinematic viscosity}=\mu/\rho [m^{2}/s] where \mu is the dynamical viscosity.

In terms of microphysical quantities, the kinematic viscosity is the thermal velocity times the mean free path: \nu=v_{thermal}l_{mfp}

f) The eddy size is independent from velocity.


2. A220A: Stellar Structure and Evolution – J.Fortney

Let’s think about three stars: 10M\odot,1M\odot,0.1M\odot . Please make one plot for each: For each star, plot X, the hydrogen fraction vs. mass shell. First do this at the zero-age main sequence, and then for each of our three stars, overplot how X vs. mass shell evolves with time until the center becomes pure helium. Describe how differences in the hydrogen fusion pathway and the interior temperature structure contribute to differences between the three stars.


I have run models of 10M\odot,1M\odot,0.1M\odot stars with the solar composition (Y=0.2485 & Z=0.0122) till the hydrogen at the center gets below 1%(end of main sequence) using the MESA stellar evolutionary code:

Hydrogen fractions on the mass coordinates for the 10(top),1(middle) and 0.1(bottom) times the solar Mass models during the main sequence evolution. Computed by the MESA Stellar Evolutionary Code.

Since I assumed the solar composition, the starting Hydrogen fractions are around ~0.7 for all 3 models and the dashed lines represent the evolution of hydrogen fractions till the end of the main sequence.

Massive stars around like our 10M\odot get their energy through the CNO cycle, which has a very steep temperature dependense (\epsilon\propto T^{17} ), while 1M\odot and 0.1M\odot stars get their energy through the ppI chain with a lot ‘milder’ temperature dependency.

Note on 0.1M\odot model: stellar objects at this low mass are fully convective.  Therefore when the hydrogen is burned into helium in the core, it gets mixed with the rest of the stellar material through convection, so the hydrogen fractions should be roughly flat all the time. But on the plot above generated form the stellar evolutionary model MESA you see the hydrogen goes to zero at the center and 10% near the surface. This is an unrealistic ‘consequence’ of the model due to the diffusive mixing. So just be aware that’s just the defect of the unrealistic mixing model in the stellar codes.

1. A220A: Stellar Structure and Evolution – J.Fortney

Given an equilibrium configuration (spherical) of mass M and radius R , assume that the density distribution is given by \rho(r)=\rho_{c}(1-r^{2}/R^{2}) , where \rho_{c} is the central density. Compute the central pressure and temperature, in terms of M and R , but not in terms of \rho_{c} . Assume an ideal gas and a surface pressure of zero.


Mass continuity: dm=4\pi r^{2}\rho(r)dr

m(r)=4\pi\int r^{2}\rho(r)dr=4\pi\rho_{c}\int(r^{2}-r^{4}/R^{2})dr=4\pi\rho_{c}(\frac{r^{3}}{3}-\frac{r^{5}}{5R^{2}} )=\pi\rho_{c} r^{3}(4/3-\frac{r^{2}}{5R^{2}})

(the integration constant is ignored)

Then the total mas of the star would be: M(R) =\pi\rho_{c} R^{3}(4/3-1/5)=17\pi\rho_{c} R^{3}/15 and using this you can express the central density in terms of Mass and Radius:

\rho_{c}=\frac{15M}{17\pi R^{3}} (*)

Hydrostatic equilibrium: \frac{\partial P}{\partial r}=-g\rho => dP=-G\frac{m(r)}{r^{2}}\rho(r)dr=-G\pi\rho_{c}^{2}r(4/3-\frac{r^{2}}{5R^{2}})(\frac{R^{2}-r^{2}}{R^{2}})

P(r)=\int dP=-\frac{\pi G \rho_{c}^{2}}{15r^{4}}[20R^{4}r^{2}/2-23R^{4}r^{3}/4+3r^{6}/6]+P_{0}

Here you use the boundary condition for the pressure (at the surface P(r=R)=0 ):

P_{0}=\frac{\pi G\rho_{c}^{2}}{15R^{4}}[10R^{6}-\frac{23R^{6}}{4}+\frac{R^6}{2}]=\frac{19\pi GR^{2}\rho_{c}^{2}}{60}

Substituting the central density by (*):

P_{0}=\frac{19\pi GR^{2}}{60}\rho_{c}^{2}=\frac{19\times 15GM^{2}}{4\times 17^{2}\pi R^{4}}

To find the temperature you use the ideal gas approximation (you basically ignore the photon pressure):

P V = n k T  <=>   T=\frac{PV}{nk}=\frac{P m_{H}\mu}{\rho(r) k}

where m_{H} is the mass of the Hydrogen and \mu is the mean molecular weight.

So the central Temperature would be:

T_{c}=\frac{P_{c}}{\rho_{c}}\frac{m_{H}\mu}{k}=\frac{19m_{H}\mu GM}{68kR}