Current Research in High-Energy AstrophysicsHigh-energy astrophysics involves phenomena in rather extreme physical situations.These include ultrastrong magnetic fields (up to 1015G), strongly curved spacetime (suchas near the horizon of a black hole), extremely dense matter (up to several times nucleardensity), and particles with energies a billion times what can be achieved in terrestriallaboratories. This means that high energy astrophysics can reveal not just interesting thingsabout astronomy, but can help us probe fundamental physics as well.In this class we will explore some of the phenomena that generate high energy photonsand particles, meaning relativistic particles and photons in the X-ray range and above.These include black holes, neutron stars, gamma-ray bursts, and the generation of ultra highenergy cosmic rays. To understand these we will first get a grounding in the relevant basicphysics, including the interactions and generation of high-energy photons and particles,their detection, and general relativity. We will then go over the sources themselves andsome of the phenomena observed from them. These include black holes, neutron stars, andclusters of galaxies. As we discuss each source, we will first go through the basics and thendiscuss current frontier areas of research. For example, active research is proceeding onhow to detect signatures of strong gravity from black hole sources; how is this done? Whatare people doing with respect to detecting evidence of ultrastrong magnetic fields? Whatcauses gamma-ray bursts?Incidentally, throughout this course we will follow the normal astronomical conventionand use cgs units instead of SI (note, though, that solar and planetary scientists often useSI). For many applications there is no real preference, but for electromagnetic calculationsthe use of SI requires the introduction of the constants ²0and µ0, and means that electricand magnetic fields are measured in different units. We will also, of course, use other unitsnot in either system, such as parsecs or eV. It’s just one of those things.Developing Astrophysical Reasoning SkillsAs discussed in detail in the “Hints about doing research in astrophysics” file on theclass web page, there’s quite a transition between classwork and research. In this course Iwill encourage development of research-oriented skills. One of these is the ability to size upa problem and determine how best to approach it, given the goal of the research and theneeded accuracy. Some things are best solved analytically and some with a computer; somerequire great accuracy and some are best done with order-of-magnitude estimates; and soon. In all cases, though, you’ve got to be able to sit back and ask yourself “Does this makesense?” so that a programming bug doesn’t convince you that energy isn’t conserved!One aspect of “does this make sense” is that you need to be able to look at a resultand determine if it satisfies several “common-sense” criteria, from simple to complex. Doesit have the right units? Is it correct in limits that I can check easily? Does it possessthe appropriate symmetries? Does it depend on what it should depend on, and no more?Ideally, you should do this before you embark on a calculation, and also afterwards, to checkyour result. You’d be surprised at how often you can catch errors this way or sharpen yourintuition. Here’s an example, due to Doug Hamilton:Units, Limits, and Common SenseYou launch a rocket straight up from the Earth’s North pole, and it rises up to amaximum height H, then falls back to Earth. The maximum height above the Earth isgiven by one of the expressions below. Here REis the Earth’s radius, X = v2RE/GME, Gis the gravitational constant, MEis the Earth’s mass and v is the launch velocity. Withoutsolving the problem, rule out as many of the incorrect equations as possible using simplephysical arguments.A) H = REX/(1 +√X)B) H = REX/(1 − X)C) H = REX/(2 − X)D) H = RE(1 − X)/(2 −X)E) H = vX2/(2 − X)F) H = REX/2G) H = REX2/(2 − X)H) H = REX|1 − X|/(2 − X)You can rule out all but one of these possible answers using relatively simple arguments.This means a huge savings in time, and if you get in the habit of thinking about answers inthis way your intuition will improve dramatically.Simplification of equationsA separate skill that is often useful is the ability to examine a problem and determinewhat complications can be dropped and yet retain the essence of the physics or at leastthe desired accuracy. Here’s one example. Suppose you have a photon with energy ¯hωscattering off an electron at rest. The total cross section, with x ≡ ¯hω/mec2, isσ =34σT(1 + xx3"2x(1 + x)1 + 2x− ln(1 + 2x)#+12xln(1 + 2x) −1 + 3x(1 + 2x)2). (1)Looking at such an equation, I don’t feel any surges of intuition! Moreover, in manycircumstances where this result might apply, there are other uncertainties in the problemthat make unnecessary accuracy superfluous. In such cases, you could approximate byassuming that for low-energy photons, x < 1, the cross section is the low-energy limit ofthis expression, which is just σ = σT= 6.65 × 10−25cm2, the Thomson cross section, andthat for x > 1 the cross section is the high-energy limit, σ ≈38σTx−1(ln 2x + 0.5) .Here’s another example, from the 2000 ASTR 688R midterm:Suppose that low-mass stars (M < M¯) have a uniform density equal to that of theSun (ρ=1.4 g cm−3) and a central temperatureTc= 2 × 107ÃMM¯!ÃRR¯!−1K . (2)Suppose also that the stars are pure hydrogen (X = 1), and that the main reaction burninghydrogen to helium has an energy generation rate of²eff≈2.4 × 104ρX2T2/39e−3.38/T1/39erg g−1s−1, (3)where T9= Tc/109K. Stars stay on the main sequence until they have exhausted most ofthe hydrogen in their cores, which we will assume have a uniform temperature equal toTc. To within a factor of 2, calculate the mass of the lowest-mass stars, which have mainsequence lifetimes of ∼ 1013yr.Answer: Now, this looks like a killer equation. If you tried to solve it exactly you’d needa computer. However, if you have the insight that the burning rate is extremely low andthat this implies that the exponential must be very small, you have a dramatic simplificationopen to you. In particular, to the required accuracy you can drop the power-law prefactor(!) and treat it as simply an exponential equation, which is trivial to solve. To get thenumerical answer you