ASTR 680Problem Set 1Due Thursday, February 81. Cosmic Microwave BackgroundThe cosmic background radiation that we see comes from the era when the universe firstbecame transparent to light. At very early stages, a typical photon would scatter many times asit crossed the universe. However, as the universe became larger and cooler, photons could travelfarther. At some point, a typical photon could travel across the entire universe without scattering,and that’s the radiation we see.(a) To make a first simple guess as to when the universe became transparent, let’s do the following.The most important interaction of light with matter at that stage was Thomson scattering. The“cross section” for scattering (i.e., the effective area of an electron for scattering by a photon) wasthenσ = 6.65 × 10−25cm2. (1)The universe is mostly ionized at the present time (that is, most electrons are free rather thanbound in atoms). The number density of electrons at the current time is n0≈ 2 × 10−7cm−3.At a redshift z, the number density is n(z) = n0(1 + z)3. The age of the universe wast(z) = 1.4 × 1010yr(1 + z)−3/2(actually, it’s somewhat different than this, because of thedominance of dark energy in the last half of the age of the universe), hence the radius of a causallyconnected part of the universe wasR(z) = ct(z) ≈ 1.4 × 1028(1 + z)−3/2cm . (2)The mean free path of a photon (i.e., the typical distance one would travel) isL(z) = 1/[σn(z)] . (3)You should find that at large redshift, L(z) < R(z), so that a photon scatters before it crosses theuniverse, whereas at smaller redshift, L(z) > R(z), so that a typical photon crosses the universewithout scattering.Using these assumptions, calculate the redshift when L(z) = R(z).(b) Now let’s do things more carefully, distinguishing between cross section and opacity. Forsimplicity, we will pretend that the universe was pure hydrogen instead of about 25% helium bymass. The Saha equation then tells us that the fractional ionization y ≡ ne/n (i.e., the numberdensity of free electrons divided by the total number density) is given byy21 − y=4 × 10−9ρT3/2e−1.579×105/T, (4)where ρ is the total mass density (including protons) in g cm−3and T is the temperature in K:T = 2.725(1 + z).With these assumptions, compute the redshift at which L(z) = R(z). You should find a valuesomewhat larger than the current best estimate z = 1089 from the third year WMAP data. Thisisn’t part of your grade on this problem, but what would you speculate is the main reason for thediscrepancy?2. Neutrino background.At redshifts so high that the temperature of the universe is T = 2.7(1 + z) KÀ mec2, thenumber density of electrons, positrons, photons, neutrinos, and antineutrinos (electron flavor inboth cases) are about equal, and in thermodynamic equilibrium. Use this to compute, to withina factor of 2, the redshift at which neutrinos decouple from the electrons and positrons (using acriterion similar to that in problem 1a above). Note that everything is ionized, and that the totalnumber of electrons is dominated by electron-positron pairs, rather than by the small numberof electrons we have today. The total number density of blackbody photons at temperature Tis n ∼ 20T3cm3, where T is in Kelvin. When integrating over the blackbody spectrum theelectron-neutrino scattering cross section is roughly hσi ≈ 4 × 10−44(T/1010K)2cm2. In addition,since this was the radiation-dominated part of the evolution of the universe, the age at redshift zwas approximately t(z) ≈ 0.3(z/1010)−2s.3. High energy particles from neutron stars.Suppose you want to generate high-energy protons from potential drops along neutron starmagnetic fields. We assume that the star has a dipole field, so that the strength at radiusr is B(r) = B∗(R/r)3, where B∗is the surface field strength and R = 106cm is the stellarradius. We will say that neutron stars can have surface field strengths up to 1015G and angularfrequencies up to Ω = 4000 rad s−1. The electric field produced by a rotating magnetic field isE = (v/c)B = (Ωr/c)B, and to compute the potential drop we assume that this field persists for adistance ∼ r, although this is not precisely correct.(a) Given the above assumptions, what is the maximum Lorentz factor achievable by a protonaccelerated along a field line? Calculate this to within a factor of 2. Assume that the radius ofcurvature is ∼ r everywhere. You may consider the field to be dipolar from the stellar radius Rout to the “light cylinder” rL= c/Ω, which is where particles rotating with the field would moveat the speed of light.(b) Suppose instead that the proton is accelerated to a potential Er perpendicular to the fieldlines, in the magnetic equator (we assume an aligned rotator, so this is also the spin equator).To within a factor of 5, what is the maximum Lorentz factor that the proton can have when itreaches infinity? Note that you can change the surface magnetic field, rotation rate, and radius atwhich you start the proton to maximize the Lorentz factor. Hint: synchrotron losses are greatestclose to the star, but far away the potential drop is small, so you need to determine which wins.Again, you can start anywhere between the star and the light cylinder.4. Detection of gravitational waves.Dr. I. M. N. Sane, noted astrophysical gadfly, has had a startling revelation: laser interferometerscannot possibly detect gravitational waves! He has realized that a gravitational wave will stretchor shrink all “measuring sticks” in the same way. One consequence of this is that if an arm of agravitational wave detector is changed by some fractional amount ² ¿ 1, then the wavelength ofthe laser light in the cavity is also changed by precisely that same factor. Therefore, the length ofthe cavity as measured in laser light wavelengths is unchanged by the gravitational wave, leadingDr. Sane to conclude that the interference patterns will be unchanged. Dr. Sane is consideringthis in the LIGO-like limit that the frequency of the gravitational wave is much less than c/L,where L is the unstretched length of an arm.You have been approached by the National Science Foundation to evaluate this argument.The future of LIGO depends on your