Notation:

1. Do a rough sketch of the situation, using symbols you have chosen. It helps both you and the marker.
2. Derive relationships between physical quantities (equations). In other words, solve the problem.
3. Check units in the most important equations. This step is life-saving in case you made a mistake in step 1.
4. Substitute numerical values to evaluate the answers, remembering to first convert constants and inputs to a common system of units: SI (or cgs, if you want to impress astrophysicists.) Do not hurry to do this before step 4 - it's a common source of mistakes. Besides, it's always inelegant and sometimes inefficient, as the symbolic manipulation of expressions can lead to simplifications or variable cancellation before step 4.
5. Think if your results are plausible. You may know the plausible range of physical quantities you compute, or if you don't, maybe can do an order-of-magnitude estimate. Clearly, if (let's say) you get a temperature inside the star equal to a fraction of 1 K, or 1e+23 K, then something's wrong. Maybe a missing constant or wrong units. Go back to step 2 if needed.
6. This step is somewhat optional. If you have time, then try to find an alternative way of solving the problem, as it is a fantastic cross-check on your result. It's unlikely to obtain two very similar results, both wrong, using different methods. Alas, during the exam there may be too little time for this.
7. Write your solution legibly. As you write down equations, now and then interject a comment on what you do. (For instance: "I substitute eq. 1 into eq. 2, then find X".) Although we sometimes seem to have the ability to read your mind, we usually don't. We may guess wrongly and misunderstand your work, or worse -- start having doubts that this is your original work.

DAY 1

Blackboard notes in Lecture 1: Proof that E=mc²

0 = Δ(M x_M + m x_m) = M Δx_M + m Δx_m.

The coordinates of the box and the mass m shift during the experiment by: Δx_M = -E/(cM) Δt, and Δx_m = c Δt, accordingly. Therefore

0 = -(E/c) Δt + m c Δt

or simplifying:

E = m c².

I thought that you can deliver this derivation of the most famous science formula by Albert Einstein in under one minute. In the tutorial however I tried to explain everything in such a detail that it took at least 5 minutes.

There is an anecdote about how Einstein was passing his Habilitation exam (for the highest academic degree in Germany and other countries parts even these days):
- Dr. Einstein, could you please derive the formula E = m c²?
- I'm sorry I did not memorize the derivation, but I know exactly on which shelve I can find the book it's in. (Today, he might say "..but I can google it up for you" :-) Einstein always stressed that there is no need to memorize trivial facts, they can be found in books. American inventor Thomas Edison was of a different opinion.. but that's a different story.

Blackboard notes in Lecture 1: Classical mechanics as a limiting case of relativistic mechanics

1. Determination of distances and sizes of Moon and Sun by Aristarchus

SOLUTION

(1 - 2.6 R_M/R_E) = (R_S/R_E -1) d_M/d_S = (R_S/R_E -1) R_M/R_S
because d_M/d_S = R_M/R_S from equality of angular sizes of Moon and Sun during total solar eclipse (step 1).

Expanding the r.h.s., we have after a simplification
3.6 R_M/R_E = 1 + R_M/R_S = 1 + α = 1 +1/396 ≈ 1
or
R_M = R_E/3.6 = 1770 km (1/3.6 of Eratosthenes's R_E = 6366 km)

BTW, the approximation we made was valid, because we know from step 2 that deviation from 90 degrees of the angle between M & S during an exact half-Moon phase is very small, α ≪ 1. In fact α and its sine function are both approximately equal to 9 arcminutes (expressed in radians).

Next, knowing the angular size of the Moon (diameter 2φ = 0.5331° on average) it is straightforward to obtain the distance to the Moon
d_M = R_M/ tan φ ≈ R_M/ φ = 382600 km.

This is an important milestone, since we just need one more observation of angle α in order to get the all-important value of AU, the mean Sun-Earth distance. The small angle equals 9' (9 arcminutes), which after conversion to radians (you should know how to convert!) yields

AU = d_S = d_M / α = 382600 km / (9/60 π/180) = 149.2 mln km.

Not a bad precision, as it is 149.5 mln km in reality (in time-average sense).

2. Determination of the luminosity of the Sun

SOLUTION

In problem related to fluxes, we can draw an imaginary sphere (of radius r = 1 AU in this case) and compute its surface area. Flux times area = power emitted by a star in watts, a.k.a. bolometric luminosity.

L_sun = 4 π r² F = 3.86e26 W.

DAY 2

Blackboard note on day 2: Derivation of pressure formula

E² = (mc²)² + (pc)² where c = speed of light, p = momentum, E = total energy

This allows the following proof that pressure P is (2/3) of spatial energy density (energy of motion divided by the volume) of normal non-relativistic gas (atomic motion at v ≪ c), and (1/3) of spatial energy density of the ultra-relativistic particles (including as a limiting case photons). In the latter case the velocity v is either very, very, close to or exactly equal to c, when rest mass m = 0 (e.g., photons).

The proof goes like this but you don't need to memorize it, just understand well the result.

Take a volume of a cube of side length L, and place N particles bouncing around in the box with momentary velocity vectors (v_x,v_y,v_z). We'll estimate the push of the gas on one chosen wall of the cube. An individual particle bouncing around the cube keeps the same component of the velocity perp. to the wall, e.g., v_z, if our wall is in XY plane.

Frequency of bouncing off 1 particular wall is 1/(2L/v_z), corresponding to time period 2L/v_z, because the particle covers distance 2L between the bounces, in z-direction. The amount of momentum given to the wall in each bounce is 2 p_z (it would be 1 p_z if the particle got stuck in the wall, instead of elastically bouncing, or in case of photons being re-emitted in a mirror-like way).

Force of pressure is the rate of change of momentum of the wall, F = N * frequency * (2 p_z) = N/L < v_z p_z > where I have added the averaging signs to remind me that not all values of v_z and p_z are equal for N different particles. P = F / L² (pressure) P = n < v_z p_z >, where n = N/L³ = number density of gas, that is the # of particles per unit volume.

Now **vectors** v and p are always parallel, so we can replace the dot product by the product of lengths of vectors, and write
< v p > = < v_x p_x + v_y p_y + v__z p_z > = < v_x p_x > + < v_y p_y > + < v_z p_z >.
But all three directions are equivalent, so < v_x p_x > = < v_y p_y > = < v_z p_z > = (1/3) < v p>.
Therefore, in physics we always have this expression for pressure of a gas of particles

P = (n/3) < v p >.

Here, to understand how much < v p > is, we finally will use our two separate cases:

1. Non-relativistic particles:
v ≪ c, E ~ mc² + p²/(2m) = mc² + (1/2)mv² = rest energy + translational motion energy (kinetic en.),
p = mv, and < v p > = < v mv > = m < v² > = twice the thermal energy of an average particle.
P = (n/3) < p v > = (2/3) (thermal energy / volume).

2. Ultra-relativistic particles, like a gas of photons:
v = c, while
E = p c is the translational motion energy, and p = E/c
< v p > = < c E/c > = E = thermal energy of a particle
P = (n/3) < v p > = (1/3) (thermal energy / volume).

This difference by a factor 2 in the two cases is going to be important. For now, the preceding explains why energy of photon gas per unit volume, equal to (aT⁴), is a factor of 3 different from the radiation pressure of photon gas, (aT⁴/3). Here "a" is a radiation constant. Photons have zero mass and are as relativistic as one can be. By the way, check that pressure and energy density have the same units!

Blackboard note on Day 2. Drunkard, diffusion and the stars

Solution:

All the symbols x_i or xi here are i-th vectors (not vector components), and * means the dot product of those vectors.

x = x1 + x2 + ... + xN
averge (over all possible orientations of vectors, which are random by assumption) of a sum is a sum of the averages. I'll denote averages as squre brackets
[x] = [x1] + [x2] + ... + [xN] = 0 + 0 + ... + 0 = 0 this just means that a clouds of points spreads out in time, always centered on zero (starting point). What's non-zero and actually grows with the length of walk is the always positive length of vector sum x. It's computed as square root of x*x. The mean squared value is:
[|x|^2] = [x*x]
let's write out x*x, using the above sum, term by term, as
x*x = x1*x1 + x2*x2 + ... + xN*xN + (these definitely don't vanish)
+x1*x2 + x1*x3 + ... + x1*xN +
+x2*x1 + x2*x3 + ... + x2*xN + .... (all these cross-terms vanish).

Why do dot products of different vectors vanish? Recall that the dot product is a product of lengths times the cosine of the angle between vectors. With large enough number of randomly oriented pairs of vectors, every cos(angle) will have a pair very closely representing one of the vectors having a switched direction. In the limit of infinite number of steps, cancellation of contributions will become perfect. (Directions of vectors do not even need to be uniformly distributed on the sphere, just bi-symmetrically!).

Since [x_i * x_i] = L² for all i, the sum averages over all possible directions of x_i's to
[x*x] = [x1*x1] + [x2*x2] + ... + [xN*xN] = L² + L² + ... = N L².

The square-root of this is called standard deviation or r.m.s-distance (remainder mean square distance). It has unit of length and value N^1/2 L. this quantity characterizes the half-width of the cloud of photons or drunkards, or diffusing molecules of perfume.

The significance the result is this: whereas a straight-line travel distance increases linearly with N, the expected distance (dispersion) of random walk grows much slower, ~N^1/2. For an average diffusing photon to reach distance τ L from the center, it needs to make τ² steps of random walk, which takes very much longer than to cover that same distance as the crow flies.

For instance, leaving a radius-100 obstacle course (obstacle spacing = 1) requires as many as N=100*100=10000 random-walk steps, on average. It isn't clear if a person under the influence can physically make that many steps..

Does our result depend on dimensionality of space? No. Check the derivation! How would the random walk on a plane (2-D walk) look vs. the 3-d walk. Would the r.m.s. distance be different after N steps? It would be more-dimensional but reaching to the same squared distance, on average. Is the textbook right when it discusses the influence of dimensionality on the average distance reached in a random walk after N steps? No. Was Einstein wrong? Probably not, someone misqueted something from his work, I think. [The book claims he proved a formula with a factor (1/3) that we do not find.]

Proof of the Virial Theorem for bodies in hydrostatic equilibium

It derives the relation between the volume-averaged pressure <P> and the gravitational energy of a star, E_grav = ∫[-G m(r)/r] dm.
<P> = (-1/3) E_grav/V
where V is the volume of the object. In other words, mean pressure is (-1/3) of the mean gravitational energy density of the object in hydrostatic equilibrium. The negative coefficient should not surprise us, gravitationally bound objects like stars have negative E_grav, so the pressure in them is positive.

Remember that we earlier proved that
<P> = (n/3) <p v> = (2/3) of mean thermal energy density of the star = (2/3) E_th/V.
even if we did not care to explicitly put averaging brackets around P. Eliminating the mean pressure from the two equations, we obtain

2 E_th = - E_grav

This proves one of the forms of the Virial theorem. The other two follow, if we invoke the total energy E = E_th + E_grav. Kinetic energy is the same as thermal energy, because we consider an object with no net momentum, so there is no ordered component of the kinetic energy. After simple algebra, we find:

E_th = - E
or
E = (1/2) E_grav   □

DAY 3

Exercise on day 2: When will we know if the sun suddenly stops to produce energy in nuclear reactions?

This was a quiz for the break between one lecture and tutorial. We did not discuss the solutions though.

SOLUTION
A previous hydrostatic balance of gravity and pressure forces will be perturbed in the energy-producing core. The inner core will cool rapidly (because its very small; it's hard to say exactly how quickly without more detailed considerations, but certainly much much quicker than the long diffusion time of photons through the whole star). Probably one day would be enough, rather than thousands of years necessary for random-walk diffusion (straight-line travel time at the speed of light is 2.33s, but random walk takes τ ~ 10¹¹ times longer, i.e. 7400 years).

A wave of decompression will move with the speed of sound toward the surface. If, as we shall show in the next tutorials, pressure and density are directly depending on each other like P ~ ρ^γ (gamma being the so-called adiabatic index, normally 5/3), then

v_s² = dP/dρ = γ P/ρ = γ kT/(μm_H)

This is (2/3)γ of the mean thermal energy per unit mass of gas. We could now just substitute the estimated virial temperature of 10 mln K.

I will give you a longer (-: way to get the value of v_s. The virial theorem says that the thermal energy of the whole star (3/2 kT M/μm_H) equals -1/2 or the gravitational binding energy of a star, let's say -αGM²/R, with α being some nondimensional number of order 1, so we can alternatively write the average soundspeed as
v_s² ≈ (1/3) γ α GM/R = (γ α/6) v_esc².
We used the fact that the escape speed from the surface of the star follows from: v_esc² =2 GM/R. The average soundspeed is somewhat smaller than the escape speed, which for the sun is about v_esc = 620 km/s. Suppose that γ=5/3 and α=1/3. Then v_s ≈ 0.3 v_esc ≈ 190 km/s. The time of travel of a decompression wave is just about R/v_s≈ 0.7e6 km/ 190 km/s ≈ 3700 s ≈ 1 hour, which is likely shorter than the cooling time of the inner core.

Observing big oscillation of the solar surface, we would notice the effects of the purely hypothetical sudden cessation of nuclear reactions much sooner (whithin a day?) than after the diffusion time of photons. That latter, incorrect, timescale can be found in some books as an answer to the problem's question.

DAY 4 (5 Oct)

From the Lecture: Simplified solution of ODEs

Example1 : Vertical fall of an object (with no air friction) is follwing two simultaneous ODES: (1) dv/dt = -g = const. (Galileo's law of gravity near earths surface), and (2) dz/dt = v (kinematical relationship). A sketch will show that v varies from v(0)=0 at time t=0, to some final negative speed v(T), at time t=T. The distance decreases from z=H to zero after time T. What is the approximate solution of those equations? It can be found approximation the derivatives slopes of assumed linear graphs, e.g.,
dv/dt = Δv/Δt ≈ [v(T) - v(0)]/(T - 0) = v(T)/T < 0.
Which simplifies (1) to the approximate form v(T)/T = -g, or v(T) = -gT.

Similarly, in eq. (2) we replace the l.h.s. derivative by a ratio of differences, dz/dt ≈ (0-H)/(T-0) = -H/T, and replace variable v(t) on the r.h.s. by its estimate equal to (1/2)v(T), halfway between v(0) and v(T). We get -H/T ≈ -gT/2. From this we obtain the following scaling of height with the time of flight:
H = gT²/2.
In this example, we happened to obtain the exact relationship for freefall, coincided with the honest solution of the set of ODEs. That's because the set had very simple right-hand sidesi and v(t) was indeed changing linearly, as we assumed. But, in a more general case, the numerical coefficients we get will not be so accurate. Scaling laws such as H ~ T² will always be correct.

In stellar structure equations, we use the linear approximation to first order derivatives. We'd also replace ρ by its mean value M/volume or, if we just want to obtain correct proportionalities and not necessarily correct coefficients, by ρ ≈ M/R³, which has a ~4 times higher value than the mean density.

From the Lecture: Wien's displacement law

x = hν/kT = (hc/k)/(λ T)

That's the exponent x in the fraction that looks like 1/[e^x -1]. Some power of x is also multiplying this factor. When we try to find the peak of radiation spectrum, say, as a function of λ, we equate to zero the derivative of Plank function over λ and get a condition obeyed by x, which is not a nice equation (we can't solve it analytically). We need to solve it on the calculator or computer, by bisection or just by trial and error. We find *some* concrete value of x_peak at the peak of radiation, a nondim. number of order 1. The definition of x then results in Wien's law:

λ = (hc/kx_peak) T = (1K / T) 0.29 cm

For instance, T = 2.9K of the cosmic background radiation yields the peak of radiation at 1 mm wavelength. Indeed we call the cosmic background the cosmic *microwave* background radiation.

Another example. Plugging in the temperature the sun (i.e., effective surface temperature T = 5780K), we get the peak wavelength λ = (hc/kx) T = 0.5 μm, right in the middle of the visible radiation range. That's of course not a coincidence, as animal eyes have evolved to see the wavelengths near the peak of the solar spectrum.

Problem 1: Radiation transfer in atmosheres and in stars

| object    |      r     |   ρ_w            |   λ_free  |
________________________________________
| clouds   |  5 μm   |  0.26 g/m³ |      ??   |
|    fog     |  8 μm   | 0.06 g/m³  |      ??   |
________________________________________

Complete the table by computing the mean free path of photons in the medium. Compute the average mean free path of photons inside the star (sun).

SOLUTION
Radiation transport in a parallel-ray beam is described by dF = - F dτ = - F κ ρ_w dx.
The solution by separation of variables (all terms with F to the l.h.s), all containing x on the r.h.s.) is

F = F₀ exp[-∫₀^x κ ρ_w dx] =: F₀ exp[-τ] .

What is the opacity coefficient κ in case of water droplets? The ratio of total cross-sectional area to total mass of droplets in a given volume. But that ratio is the same for one or for many droplets, so let's calculate it for one droplet of radius r:
κ = π r² / [(4/3)π r³ ρ_H2O]
where ρ_H2O = 1000 kg/m³ is the density of liquid water.
κ = 3 / (4r ρ_H2O)
Units of kappa are ok, m²/kg. Values of mean free path λ_free = 1/(κ ρ_w) are about 35m in clouds, and about 120m in dense fog.

A typical cloud layer might be 350 m thick. The optical depth is simply τ(clouds) = 350/35 = 10. Direct sunlight passing through a cloud is suppressed by a factor exp(-10) ~ 4.5e-5 after passage through the cloud. Why is it NOT getting very dark every time the sun hides behind a cloud? Because small water droplets efficiently remove the sun's photons from the beam but do not absorb them strongly. They scatter sun rays sideways and after multiple scattering (random walk inside the cloud) most light rays are getting out of the cloud in all possible directions. The day may be darker but definitely not e¹⁰ times darker!

A fog layer may be somewhat thinner geometrically, perhaps 240m, and also thinner optically, τ = 240m/120m = 2. Exponential function exp(-2) = 0.135 ≪ 1. *Direct* light rays are suppressed but not very strongly, one order of magnitude in brightness/flux. That's why we can often still see the silhouette of the sun through the fog (or thin clouds of various kinds for that matter).

Inside the sun it's much foggier/cloudier. Minimum opacity is from Thomson scattering off free electrons in ionized gas, κ_Th = 0.4 cm²/g. It's not easy to decide what to take for the density of gas, the central density or the mean density perhaps. They differ by two orders of magnitude.
One usually quotes the mean density 1.4 g/cm³, then λ_free = 1cm/(0.4 * 1.4) ~ 2 cm. But in reality λ_free changer a lot from place to place, from much smaller than 2cm in the core of a star to much larger, near the photosphere (eventually photons are freed to travel in space, the mean free path is becoming infinite beyond the photosphere radius; the optical thicknass τ counted from photosphere to infinty is about 1, or if you want to be very precise and define photosphere as 50% chance of getting out, then τ obeys e^-τ = 1/2, i.e. τ = ln 2 = 0.693).

Problem 2: Using classical physics (electrostatics and thermodynamics) show that the sun cannot have thermonuclear fusion reactions, the stars don't shine, and the sun doesn't warm the Earth, therefore we cannot exist

The virial theorem says that the gas temperature inside the sun must be of order T ~ 10 million kelvin. We compared two energies of a proton as it enounters another proton in such plasma; electrostatic potential energy & the mean thermal energy.

To approach and touch another proton, at which point we might expect that the short-ranged strong forces take over and lead to a fusion of two protons, we need to supply energy E_pot = +A e² / (2 r_p), where e = +1.6e-19 C is the electric charge of the proton, and A is a constant: A = 1/(4πε) = 9e9 J m/C². Radius of a proton is r_p = 0.83 fm. In fact, four protons need to meet, to somehow convert to one alpha particle = nucleus of helium harboring two protons and two neutrons. In the reaction chain called pp reaction chain, 2 antielectrons called positrons, plus 2 electron neutrinos, are released. But if we show that 2 protons p⁺ cannot touch each other then there is no need to follow the more complicated scenario of the multi-step synthesis of helium. See this clickable image

We have thus shown that electrostatic energy barrier exceeds the sum of the thermal energies of protons several hundred times. A typical proton stops and turns back to depart from p⁺ without coming closer than r = several hundered r_p. The probability of having energy above E is proportional to exp(-E/kT), where the value of the exponent runs in the hundreds. It isn't enough that in a Maxwell velocity distribution we can occasionally find protons with high kinetic energy. Exceeding the electrostatic energy barrier happens with a vanishingly small probability. Even if all protons approach other protons simulatneously, there won't even be 1 themonuclear conversion of H nucleus into He nucleus. Therefore, we cannot exist □

DAY 5 (19 Oct)

And yet it shines!

Now prove that the quantum mechanics literally saves the day, allowing a p-p merger, based on the fact that particle in quantum mechanics are delocalized, i.e. fuzzy (described by a cloud of probability, high in the center but exponentially rapidly vanishing at large r).

Assume that the size of the cloud representing a proton has radius Δx and that the uncertainty of its momentum is of order of its linear momentum itself, call it Δp ~ p. Protons in the sun-like star are not ultrarelativistic so $p=mv$. (This can be checked separately, comparing kT with mc². Show that uncertainty relation which connects the position and momentum uncertainties
Δx   Δp ≥ h
h = 6.626e34 J s (Planck's constant). Show that $Delta;x is of order 1e3 times r_p, i.e. the overlap of quantum particles is not at all unlikely. You may assume that you know the temperature in sun's core from virial argument, or from detailed models, T = 15 mln K.

SOLUTION

Energy of an averge proton is mv²/2 = (3/2) kT, so

Δx ≥ h/√ (3 k T m)

Where m = mass of proton = 1.67e-27 kg. Units can be ckecke to agree, x is in meters. Δx numerically is much larger than 2rp = 1.7 fm, in fact some 400(?, I seem to rmember you telling me this number) times larger. This means that there is a fairly large overlap between the quntum-delocalized protons during their close encounter. In other words, internally they may be of approximate radius 0.84 fm, but the position of theor center can in actual experiment appear anywhere within a radius that is hundreds of time larger. This explains why the sun can shine. Quantum tunneling was first proposed as a solution to the hydrogen fusion paradox by George Gamov in Soviet U., and independently soon afterwards in America.

See this clickable images

   
Oh shoot! The pictures were a disaster, it's my slightly damaged iphone. Fell from car's rooftop where I left it ;-( onto the pavement during driving this summer, lost the protective plastic in front of the lens and right now decided to get foggy or unfocused. Fortunately, one problem was solved here explicity in this file, and the other (below) was part of set 2 of assignments, and as such is solved in appropriate file linked to our course page.

Polytropic gas law and polytropic index of normal and ultrarelativistic gas

Use the known connection between pressure P and the energy density U/V of gas to arrive at two different γ values in an adiabatic (dQ=0, no exchange of heat) gas change law, also sometimes called isentropic law, because dQ = 0 implies constant entropy. The experiment can be thought of as a gas cylinder with movable piston, filled with a const. mass of particles. Piston moves by dL, A=cross section area, therefore dV = A dL.

P ~ ρ^γ.

We know that P = q U/V from previous meetings. q = 2/3 for nonrelativistic gas, and 1/3 for ultrarelativistic. The difference is in the kinematics of particles, they hit the walls less often in the ultra-rel. case bacause they can't move faster than light.

dU/U + q dV/V = 0

Integration gives

ln UV^q = const.

Substituting P ~ U/V or U ~ PV, we get

P ~ V^-(1+q) ~ ρ^γ

where γ = 1 + q = 5/3 in normal gas, and 4/3 in ultra-relativistic gas. q.e.d.

DAY 6 - midterm

TUTORIAL 7 (2 Nov)

On the blackboard during Lecture: The nonrelativistic degenerate gas

The momentum space is spanned by three axes, p_x, p_y, and p_z. Uncertainties in these variables combine to Δp_x Δp_y Δp_z = (h/L)(h/L)(h/L) = h³/V. So says the quantum mechanics (to be consistent with Heisenberg's uncertainty relation, in each dimension Δx Δp ≥ h/2).

So when we draw the distribution of N particles in 3-d momentum space, it consistes of a large number elementary unit cells of volume h³/V each. Every cell has zero, one or maximum two particles, let's say electrons. Two is a maximum, when we pack one spin-up and one spin-down particle into a momentum cell. What's a spin? It is an intrinsic angular momentum of a particle, electrin has a 1/2 spin which can either be "down" or "up" and up-down pair of electrons differs in that spin orientation quantum number, so two two particles are distingushable by spin direction and can co-exist in one momentum-space cell. OK?

Then as we add more and more particles to volume V, they occupy first the cells near p=0 center of the space, but then gradually more and more distant empty cells. For N ≫ 1 we get a ball of cells accupied each by 2 particles. (Why a ball and not e.g. a cube?)

Let's say that particle on the surface of the ball, those with maximum momentum value, have p = p_F, the so-called Fermi momentum. That's the radius of the ball in momentum space, in other words. The number of particles N equals the volume of the ball divided by the volume of an elementary cell (h³/V), times 2 for 1 up and 1 down spin particle in each momentum cell.

N = 2V h^-3 (4π/3)p_F³

n = N/V = (8π/3)h^-3 p_F³

p_F ~ n^1/3.

So far so good. If we increase the number density n by a factor of 8, Fermi momentum will double. Now let's see how that affects the pressure P.

In nonrelativistic gas, P is (2/3) of the energy density (which is total kinetic energy devided by V). And the total energy is just N times the mean value of p²/(2m), since that's how we can express kinetic energy mv²/2 of a particle of mass m. In a ball of radius p_F, the mean value of p² can be computed like so:

<p²> = [∫ (4π p²) p² dp] / [∫ (4πp²) dp]
The integrals are from 0 to p_F, and (4π p²) dp is the volume element in 3-d momentum space (surface of a thin shell of radius p, times its thickness dp).
The average <p²> = (3/5) p_F² (convince yourself by doing the simple calculation). Therefore the total kinetic energy of N particles is equal (3N/5) p_F²/(2m), and the degeneracy pressure, being 2/3 of energy density, is

P = (2/5) n p_F²/(2m)

Since the inverse of mass m is almost 2000 times larger for electrons than for protons, while their number densities n are comparable in stars, we can neglect the contribution to degeneracy pressure from protons, and only consider the degeneracy pressure of electron gas.

Earlier we established that p_F ~ n^1/3, therefore the electron degeneracy pressure scales with their number density like

P ~ n^1+2/3 ~ n^5/3.

On the blackboard during Lecture: The ultrarelativistic degenerate gas

E = p c

for every ultrarelativistic particle. Reminder: particles are ultrarelativistic their energy of motion much exceeds the rest energy mc². The classical mechanics formula E = p²/(2m) no longer applies.

The energy density of relativistic gas can be computed as N times c times the mean momentum <p>, which by integration similar to the one done above becomes <p> = [∫ (4π p²) p dp] / [∫ (4πp²) dp] = (3/4) p_F (convince yourself by doing that calculation).

This modifies the final result of the ultrarel. case to:

P ~ n^1+1/3 ~ n^4/3.

On the blackboard during Lecture: R vs M relationaship for quantum stars

(1/ρ) dP/dr = -GM(r)/r²

taking into account P~ρ^γ, gives

(M/R³)^γ-1 ~ M/R

M^γ-2 ~ R^3γ-4.

This is an important scaling, and it is the same for degenerate or non-degenerate adiabatic objects that also obey P~ρ^γ.

For instance, adiabatic objects such as super-jovian planets may have γ ≈ 2. Then R = const. follows. They are the largest globes of gas not undergoing thermonuclear reactions.
Adding more mass to such objects results in γ < 2 and approaching 5/3. Then, like for nonrelativistic White Dwarfs,
R ~ M^-1/3.
Radius decreases slowly with increasing mass in such objects. The density of object increases rapidly with M.
That's what Ralph Fowler and Subrahmanian Chandrasekhar found around 1930 by combining the astrophysics and quantum physics of dense gas. Chandrasekhar on his long journey to England by steam ship also discovered that as the gas becomes dominated by particles moving practically at the speed of light, radius R shrinks faster and faster with increase of M, and eventually, when γ becomes equal to 4/3, the mass cannot exceed M = const. (substitute γ=4/3 into the scaling relation to see this!). The so-called Chandrasekhar mass is the limiting mass white dwarfs can have. If you add more mass to a white dwarf, it loses stability and initially implodes rapidly, then bounces back to explode as a supernova type Ia.

Tutorial Problem 1. Gravitational redshift

1. Sun, M = 1 M_sun, R = 700000 km
2. White dwarf, M = 1 M_sun, R = 10000 km
3. Neutron star, M = 2 M_sun, R = 11 km

What is the gravitational (red)shift Δλ = λ - λ₀
(λ₀=656 nm is the wavelength of red hydrogen emission line (Hα), emitted at the surface. λ is the wavelength seen by observer at infinity. Also, what is the relative redshift Δλ /λ₀?

Judge the observabily of gravitational redshift, assuming that today the best spectrographs have resolving power δλ/λ₀ ~ 10^-6.

SOLUTION

λ = λ₀ / [1 - 2GM/(Rc²)]
This may be re-written as   λ = λ₀ / (1 - R_Sch/R), where R_Sch = 2GM/c² = 3 km (M/M_sun) [see next problem below]. Since λ₀ = 656 nm,
λ ={656.003, 656.197, 1443} nm for stars 1..3.
□ The 1st star, our sun, shows extremely small increase of wavelength by a factor 1.0000043 which, however, is detectable with the most modern spectroscopes with resolving power 10^-6.
□ The 2nd, a white dwarf star, has a shift of wavelength equal to 0.197 nm = 1.97 Å (relative to λ₀ = 656 nm, the observed λ is 1.0003 times longer. That's easily detectable today, and was detectable even 100+ years ago.
□ The 3rd, a neutron star (NS), has huge wavelength shift from 656 nm to 1483 nm, placing the light in the invisible infrared range. The wavelength is 2.2 times longer than the emitted one. NS is an extremely small and dense object, almost a black hole (BH). If it could be shrunk by a factor of 4 in size, it would in fact become one. Therefore, light loses most of its energy while climbing out of the deep gravitational potential well of NS and departing to infinity. Light emitted from the surface of a BH has according to the gravitational redshift formula an infinite wavelength. That's correct because then uses all its initial energy to climb out of the potential well. Zero energy corresponds to infinite wavelength (E = hν = hc/λ).

Tutorial Problem 2. Gravitational radius, a.k.a. Schwarzschild radius

SOLUTION

Let the test particle have a mass m. If released from distance R_Sch, its total energy is zero, because the kinetic energy ½ mv_esc² exactly matches the absolute value of gravitational binding energy -GMm/R_Sch. Substituting c for v_esc we have
R_Sch = 2GM/c² = 3 km (M/M_sun).
The radius is a black hole horizon radius of a non-rotating black hole, or the Schwarzschild radius. It is named after Jewish-German scientist Karl Schwarzschild [pron.: shwarts-shild, literal English transl.: Blackshield], who found an exact solution of the fresh-from-the-press Einstein's equations and thus solving the structure of a non-rotating BH, in the trenches of the Russian front of World War I, then died of autoimmune desease he developed there.

The CMBR problem

Consider CMBR as a gas of photons having Planck spectrum (today with temperature T = 2.9 K, but much higher T in the early universe). As the universe expands, the spacing between the photons increases and the spatial number density n decreases. Show that n ~ R^-3. Knowing that the mean energy of a photon is equal hν = kT, where ν is photon's frequency, and that energy density of photon gas (i.e. energy per unit volume) is equal aT ⁴ (a is Stefan's constant), demonstrate that:
(i) n ~ T³
(ii) T ~ 1/R
(iii) mean wavelength λ = c/ν scales like λ ~ R.
Thus the wavelength of CMBR photons grows in step with the scaling factor R, as if the photon's wave was somehow pinned at the ends to the expanding space. It isn't, but you will prove that λ grows in step with the vacuum of the universe.

NO SOLUTION was provided in T6. You'll provide it as part of Problem Set #3!

TUTORIAL 8 (16 Nov)

Relationships between basic quantities in a galaxy

The snapshots (left-center-right part) shown below illustrate relationships between the above variables. Formulae for spherically symmetric galaxies are useful beyond the exact domain of application, since even somewhat flattened objects will not deviate completely from those formulae.

clickable images:

Invoking Newtons Theorem 2 and proving Newton's Theorem 1, we have assembled on the blackboard the general formula for spherically symmetric potential at radius x

Φ(x) = -GM(x)/r + ∫₀^x r ρ(r) dr
[where M(r) is total mass inside radius x], or

Φ(x) = (-4πG/x) ∫₀^x r² ρ(r) dr +(-4πG) ∫₀^x r ρ(r) dr.

BTW, this is one place in our course where we should clearly distinguish r from x (the same meaning, radius) since x in this tutorial means radius of the point of observation, while r is a dummy integration variable running from 0 to x, or x to infinity.

On the rightmost blackboard, there is a tiny confusion of signs. Can you spot it? It does not affect the result much (two terms of f cancel, one has a wrong sign, the final -GM(x)/x^2 has the correct sign.)

How to compute the velocity curve of a disk galaxy

f(x) = - dΦ(x)/dx = -GM(x)/r².

We have used this equation in stellar structure equations. Indeed one difference of what we do now and what we did earlier in stellar stucture work is that we pay more attention to explicit Φ(r) calculation, task also known as potential-density pair calculation.

The second difference is that we previously de-emphasized the idea of circular speed around a star. One gets the circular rotation speed v_c easily, by assuming that gravitation in direction of the equatorial plane of the galaxy is balanced by the centrifugal acceleration at distance R from the axis of rotation, which in circular motion equals v_c²/r (you can derive that, but you won't be the first. Huyghens did it even before Newton).

v_c²(R) = GM(R)/R.

M(R) is total mass inside radius R. Remember that we deal with spherically symmetric density distribution, so why do not use variable r? The formula is written with R, the radial coordinate in cylindrical system (R,φ,z), not with r (the 3-D radial coordinate in spherical coordinate system) to emphasize that v_c is only defined in z=0 plane; it also does not depend on azimuthal angle φ by assumption of galaxy being axisymmetric). But you will also often see the speed written as v_c(r), which is not incorrect! When z=0, then r=R.

The conclusion is that radial velocity curve can be obtained directly from &rho(r); by doing one radial integral, to find M(r). Use that direct relationship when asked about v_c.

Poisson equation

∇²Φ = 4πGρ

We will use it later, just a note on the basics of nabla squared, also known as div grad, also known as Laplace operator.
∇² = div grad = ∇·∇, where the central dot has a special meaning and belongs to the left-hand nabla operator. ∇² takes a scalar field such a Φ and performs two spatial differentiations, first a gradient ∇Φ then a divergence on that gradient, so the result again is a scalar field.
In Cartesian coordinates, ∇Φ(x,y,z) = (∂Φ/∂x, ∂Φ/∂y, ∂Φ/∂z), a vector field. Divergence ∇·A is defined for vector field A as ∇·A(x,y,z) = ∂A/∂x + ∂A/∂y + ∂A/∂z. Therefore in Cartesian coordinates,
∇²Φ = ∇·∇Φ = ∂²Φ/∂x² + ∂²Φ/∂y² + ∂²Φ/∂z².
In curvelinear coordinate systems (cylindrical, spherical) the expression looks a little more complicated than just the sum of all second derivatives along the n dimensions. (See the link below, last four pages). The above equation is useful, it makes computation of unknown ρ from known Φ rather automatic (we can always do derivatives analytically, there's an algorithm for it; integrals - only when we are lucky).
Lecture notes (listed on course web page) gives much more detail. Look there! Notice that the designations of x and r for radial distance are swapped there in the formula for Φ on the last page. Indeed, it is more common to name observation point r, and integration variable something else. Anybody can use the symbols they like, the best is when they are clearly defined and suggestive of being a distance.

TUTORIAL 9 (23 Nov)

Relaxation in Pleiades

The formula for t_cross has correct units. The value for R = 17.5/3.26 pc, is
t_cross ≈ (0.206e6*17.5/3.26)^1.5/800.^0.5 ≈ 41 Myr.
It's so long, because there are relatively few stars, and the cluster size is rather large, hence the mean velocity is low.

This formula with N = 3000 gives relaxation time
t_rlx ≈ 70 t_cross ≈ 2.8 Gyr.

That's much longer than the current age of the cluster, ~100 Myr. This result seems to support a long future life for M45.

Correction: I have actually erred in giving you the diameter of the cluster, instead of core *radius*, which is R = 8 kpc. Then we'd get a bit more vigorous motions, shorter crossing time equal 12.6 Myr etc. Relaxation would still take 70 crossing times, but would evaluate to a smaller time
t_rlx ≈ 0.9 Gyr.
Same conclusion, though. Relaxation has not happened in Pleiades yet.

On the other hand, something else could work to dissolve this open cluster. It would be premature to conclude that Pleiades will keep together for another billion years or so, since we haven't yet considered the influence of galactic force field (tidal interaction) on the cluster. In fact, astronomrs thing the cluster will be dessolved in 250 Myr.

Problem 2. Which power-law of density vs. radius produces a flat rotation curve?

SOLUTION

First, find mass inside radius r:

M(r) = 4π ∫ r² ρ(r) dr

M(r) ~ [1/(α+3)] r^3+α,
unless α = 3 (special value of integral, which is not power-law, it's a logarithm if α = 3. Let's skip that case.).

Force law is F(r) ~ GM(r)/r² ~ r^1+α,
and the velocity curve satisfies
v_c² ~ F(r) r ~ r^2+α.

Evidently, a flat rotation curve requires α=-2, an unrealistic object of infinite extent and mass, known as singular isothermal sphere. But over a finite range of radii it is a decent approximation to real objects, such as dark haloes.

Problem 3. Dark halo of NGC 6503

Observations of NGC 6503 show velocities v_c = {120, 115} km/s at radii r = {4, 20} kpc, correspondingly. The visible components of the galaxy: stars and gas, contribute the following amounts to the rotation curve (values it would have if the medium was the only one present, all in km/s): {88,40} and {15,25} km/s.

Clickable image:

A. Derive the contribution to velocity curve at those two radii from Dark Matter (DM).

SOLUTION

If nothing else but one type of medium causes gravity (forces of gravity can be arithmetically added/superposed so they can be considered separately), then
v_c,i² = G M_i(R)
where index i assumes the value {*, gas, DM, total (all densities superposed)}. Summing the masses we see that
v_c,tot² = v_c,*² + v_c,gas² + v_c,DM².
i.e. the circular velocities sum up quadratically. Therefore,
v_c,DM² = v_c,tot² - (v_c,*² + v_c,gas²).

Numerically, dark matter contributes {80,100} km/s at 4 and 20 kpc, correspondingly.

Now we can model the dark halo, and from the two data points derive the two free parameters of the model:
ρ(r) = ρ / (1 + r²/r₀²)
It's left as an exercise for you...

Comment 1: this type of functional aproximation is providing the halo with constant density inner region (core). At large radii, the density law becomes the same as in the singular isothermal sphere, thus assuring an asymptotically constant rotation curve.
Comment 2: When you know core size and density, you can find the mean speed of the particles making it, and the density at any point in space out to infinity. The speed is particularly important since non-relativistic speed case is the successful CDM model (Cold Darm Matter), while ultra-relativistic particles would make up a Hot Darm Matter object, but do not fiollow from calclulations. Can you compute the typial speed? If yes, you'll know quite a bit about the otherwise unknown particles making up the Dark Matter Halos (mot only the spatial distribution but kinematics as well).

TUTORIAL 10 (30 Nov)

Problem 10.1: Apparent superluminal motions in AGN jets

SOLUTION

From the drawing (see lecture 23) of light emitted straight at us and after time Δt by a blob of jet material moving at physical speed V < c,

observed speed = observed transverse distance/observed time difference

Transverse distance is equal V Δt sin θ. Because material moved in time Δt by   Δx = V Δt cos θ in our direction, the time between the detection of the two pictured photons is equal
(c Δt - Δx)/ c = Δt [1 - (V/c) cos θ].

V_obs = V sin θ / (1 - V cos θ)

Clickable image:

You can prove the facts stated on the margin of the picture, in partcular that superluminal speeds are observed only when V > √2 c. To do this, you first need to treat the jet speed V as constant, and derive the maximum observed speed at any angle (by calculus). Then you just need to show that V_obs > c if and only if V > c / √2. Have fun.

Pictures of the blackboard with your work. Thanks Ming!   Pic1 , Pic2 .

Problem 10.2: Review the solutions to set #3 of assignments

Pictures of the blackboard:   Pic3 , Pic4 , Pic5 .

Notice that I only sketched the rosettes in the constant speed region. Near the center of a galaxy, where solid body rotation of stars takes place (Ω = const.), we have κ = 2 Ω. This means there are two full radial oscillations in one azimuthal period. The orbit closes on itself and is an ellipse, with galaxy center in the symmetry center (not the focal point) of the ellipse. In other words, there is a big precession amounting to the maximum radius point travelling backward exactly 180 degrees in one full radial period.

Problem 10.3: SMBH in a faraway galaxy

1. Given the observed speeds, how many Schwarzschild radii R_s away from the black hole is the radius R?

2. What is the mass M inside the radius R?

3. Could this be a cluster of sun-like stars and not a SMBH?

4. Is the observation described above plausible, given that the resolution of the largest interferometers spanning the whole Earth is given by the diffraction limit θ ≈ λ/D (D = diameter of antenna or mirror, λ ~ 0.6 m is the wavelength of EM radiation used)? In other words, can we resolve a region of half-width 0.06 pc, at the distance of 6 Mpc from us?

SOLUTION OUTLINE

1. Think about the maximum speed a particle (star) can have at distance R from SMBH: it's the escape speed, equal to (2GM/R)^1/2. Objects with higher speeds would have left the surroundings of a SMBH already. Making that equal to v_obs = c/30, we get
R = 2GM/(c²/900) = 900 R_s.

2. From (2GM/R)^1/2 = c/30, it follows that M = c²R/(1800 G). Another, simpler way to evaluate M is by recalling that the sun's R_s is 3 km and, in general, R_s = 3km (M/M_sun). Then we have from pt. 1:
R = 2700 km (M/M_sun).
M = (R/2700km) M_sun = (0.06 pc/2700 km) M_sun = (3.09e13 km * 0.06/2700 km) M_sun ~ 6.6e8 M_sun . The suspected SMBH is almost a billion solar masses.

3. This is an interesting question that requires some work. First, one could check that, yes, there is a space within a ball of radius 0.06pc, for a billion volumes of the sun. The real question is, however, how long could such a cluster survive? To compute the nean time between physical collisions of the stars, one could derive an optical density of the cluster: the sum of stellar cross sections divided by the cros secitional area of the cluster. That is a measure of the probabilty of a given star striking some other star as it moves through the cluster once. It turns out that this calculation shows a rather short collisional destruction time, much shorter than the likely old age of the active galactic nucleus of the faraway galaxy. Compute the optical thickness and the crossing time, and do the calculation of collisional time scale! You should see why the object can be a stellar cluster for a very short period of time.

4. This is a typical question in observational astronomy. θ ≈ λ/D > (0.6m/6000km) rad = 10^-7 radians, because the largest D on Earth is of order 6000km. At a distance of 6 Mpc, this minimum angle spans a transverse distance of 0.6 pc. Ans: No, a diameter of 2*0.06 pc = 0.12 pc cannot be resolved with the largest interferometers on Earth at 6 Mpc distance from us. It's about 5 times too small.

Problem 10.4: Tidal radius in a flat rotation curve.

SOLUTION

First, let us establish the function describing the galactic forces and centrifugal force. We'll be talking about accelerations corresponding the these forces (they differ from forces by a constant factor m, which is not worth repeating.)

In a given rotation curve v_c(r), gravity matches the required centrifugal force at every radius r . For instance, in a flat rotation curve where v_c is constant, acceleration of galactic gravity is equal
f_g(r) = -v_c²/r ~ -1/r.

The crucial idea in this problem is to choose a frame of reference corotating with the mass m. That is, we are going into a noninertial frame that has a rigid body rotation with linear speed v_c at radius R and angular speed everywhere equal Ω = v_c/R. Then depending on location r in the rotating frame there is a ficticious centrifugal force (acceleration): f_cen = +Ω² r = +(v_c/R)² r.

The sum of the two accelerations equals

f = -v_c²/r + (v_c/R)² r = (v_c²/R) [r/R - R/r].

We can easily check by substituting r=R that at the orbit of mass m the sum f(R) = 0, allowing mass m to be stationary in the chosen reference frame.

Let's now expand f(r) up to linear order around r=R, by writing
r = R + x,     |x| ≪ R.

f ≈ (v_c²/R) [ 1 + x/R -1 +x/R]
f ≈ +2(v_c/R)² x.
This differential acceleration is called tidal acceleration. At tidal radius (distance x = r_t away from mass m), it is exactly cancelled by the attraction of mass m, -Gm/x²:

+2(v_c/R)² r_t - Gm/r_t² = 0.

Solving for tidal radius, we obtain two alternative expressions, one directly following from the above and one substituting v_c² = GM(R)/R,

r_t = R (Gm/2Rv_c²)^1/3 = R [(1/2) m/M(R)]^1/3.

Tidal radius of M15

SOLUTION

In our Galaxy at about 10 kpc we can assume v_c = 220 km/s. At current distance, tidal radius of M15 is

r_t = R (Gm/2 R v_c²)^1/3 = 135 pc = 5 r_M15.

Currently the tidal radius is 5 times larger than the mean radius of the cluster, so the cluster's stars are safely attached to it. (The actual rule of thumb is that a star is on a stable orbit and won't escape if it is within 1/4 of the tidal radius, the stability is precarious or non-existent between 1/4 and 1 tidal radius).

But even a moderate decrease of distance R from 10 kpc to 5 kpc from the center (which is quite possible: M15 travels on an elongated orbit, it's in the halo) could make the physical radius about 1/2 of the tidal radius. That would likely result in a dissolution of M15 in the Galaxy, since the loss of stars would tend to decrease m, shrink r_t (cf. the formula), and further accelerate the escape of stars. A vicious circle!

Problem 10.5. Hot or cold dark matter?

Consider a halo of dark matter inferred from a rotation curve of a typical disk galaxy, for instance NGC 7331. It has a large core radius, say, R_c ~ 10 kpc. The total mass within that core is M_c ~ 10¹¹ M_sun or slightly more. Is the DM in such galaxies a CDM or HDM? Use the virial relation between mass, radius and speed. Obtain an order-of-magnitude estimate of velocity dispersion σ in the core of the halo of DM particles.

SOLUTION

The DM in galaxies is CDM. We can prove it by setting up virial relationship for the velocity dispersion σ in a system of DM particlessubject to their own gravitation. We will skip factors of order unity, and provide an order-of-magnitude estimate.
σ² ~ G M_c /R_c
This gives σ ~ 250 km/s, which is very non-relativistic (σ ~ c/1000). BTW, the same order of magnitude as V_c, the circular speed in disk galaxies. This is not a coincidence. DM halos are more massive than luminous parts of galaxies and so dominate the rotation curve in the outer parts of galaxies (DM halo radii are larger than luminous matter part). There, DM, barions and fermions all move at similar speeds, even if not necessarily on orbits of the same shape. In the inner parts (a few kpc), however, normal matter may dominate the density and rotation, although there are some examples of galaxies whose rotation everywhere is governed by DM. But as long there are no collisions or any interactions between the 2 types of particles and they contribute similarly to total rotation curve (and mass density), one and the same quantity determines how fast both are moving around the galaxy: the depth of its combined gravitational potential well. The speeds should therefore be similar.