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

One can derive a characteristic timescale for a process by dividing a given quantity by the time rate of change of that quantity. One important timescale for stars is the “dynamical” or “free-fall” timescale. Here our quantity of interest is the radius of a star. With that as a starting point,

a) Derive a simple relation for the dynamical timescale. Then, for parts b) and c) just below, order the objects from the shortest to the longest dynamical time.

b) Main Sequence Sun, Red Giant Sun, white dwarf Sun.

c) Main sequence stars: 0.2M\odot, 1M\odot, 5M\odot

d) Another timescale is the nuclear timescale, which is basically the main sequence lifetime. Where does the approximate main sequence lifetime relation, t_{MS} \propto M^{-2.5} , come from?

Solution(incomplete):

a) the most common way to derive the dynamical timescale is from the Kepler’s laws. But there is another shorter version based on the fact that the dynamical timescale is basically the ratio of a characteristic length and the characteristic speed: \tau_{dyn}=\frac{characteristic (R)}{characteristic (v)}=R/v_{esc} where the escape velocity is: v_{esc}=[2GM/R]^{1/2} so that \tau_{dyn}=[\frac{R^{3}}{2GM}]^{1/2}

b) We can assume that the Sun’s mass stays the same from the main sequence till the white dwarf phase(in reality it probably will eject significant amount of mass due to stellar winds during the red giant branch).  So the only varying parameter is the radius.

At the main sequence the radius is just 1R\odot . When the sun enters the red giant phase the sun will not be burning Hydrogen in the core anymore. The hydrostatic equilibrium will be lost so the core will contract heating up the shell outside it. And because of this shell burning the radius of the sun will swell up to around \sim 200R\odot in order to adjust to the increase in luminosity. When the sun is done with the giant phase most of the shell material will be blown into space leaving the hot Carbon-Helium core to cool down. The size of this core or the white dwarf Sun will be on the order of R_{whitedwarf Sun} \sim R_{earth} = 0.009R\odot . So the timescales are:

\tau_{dyn.WD} < \tau_{dyn.MS} < \tau_{dyn.RGB}

If you take the mass loss into account, the WD sun will retain roughly the half of the mass after RGB. But even with this mass, the result above will still hold.

c) There is an empirical relation ship between the radius and mass at the main sequence:

\frac{R}{R\odot}=(\frac{M}{M\odot})^{0.8}

So the radii for main sequence 0.2M\odot, 1M\odot, 5M\odot stars are 0.3R\odot,R\odot,3.6R\odot respectively, thus:

\tau_{dyn. 0.2M\odot} < \tau_{dyn. 1M\odot} < \tau_{dyn. 5M\odot}

d) The heat released by fusing a mass \Delta M of Hydrogen into Helium is approximately 0.007\Delta Mc^{2} . So for the Sun the time required to burn all its Hydrogen for the respective current solar luminosity is:

\tau_{nuc}=\frac{0.007M\odot c^{2}}{L\odot} \sim 10^{10} years

Another empirical relation between the luminosity and the mass at the main sequence is:

\frac{L}{L\odot}=(\frac{M}{M\odot})^{3.5}

Combining these 2 you get the main sequence time scale:

\tau_{MS} \sim (\frac{M}{M\odot})^{-2.5}

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1. 220A: Stellar Structure and Evolution – J. Fortney

In the diffusion approximation, the radiative temperature gradient is written:

\frac{dT}{dm}=\frac{-3}{64\pi^{2} ac}\frac{\kappa L_{r}}{r^{4}T^{3}}

a) In a post-main sequence Sun-like star, ascending the red giant branch, why does the helium core have a smaller temperature gradient than the hydrogen shell above it? Ignore any issue of energy loss due to neutrino emission.

b) What particular term is important for causing outer convective zones to grow in low-mass stars? What specifically is going on?

c) What particular term is important to causing inner convective zones to grow in high-mass stars? What specifically is going on?

Solution(incomplete):

a) another way to write equation above is:

F=\frac{\lambda c}{3}\nabla a T^{4}=\frac{ac}{3\rho\kappa}\nabla T^{4}

here you can see that in order to generate the same amount of Flux, for lower opacity \kappa you need lower temperature gradient. Since \kappa (He) < \kappa (H) the temperature gradient of the Helium core has lower temperature gradient corresponding to its opacity.

b) the growth of the outer convective zones of low-mass stars is due to the Hydrogen recombination. The photon sees a lot of ‘possibilities’ due to the availabilities of higher states, thus increasing the opacity. So the opacity is the important term in this situation.

c) For high-mass stars the inner convective zones grow in order to compensate the increasing luminosity. So the important term in this situation is the Luminosity.

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.

Solution (NOT AVAILABLE YET): 

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.

Solution (NOT AVAILABLE YET):

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,

Re=\frac{UL}{\nu}

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

Solution(incomplete):

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

\nabla\times{\bf u}=0

b)

c)

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.

g)

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.

Solution(incomplete):

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.

Solution:

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}