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ASTR 498Problem Set 2Due Thursday, February 281. Low-energy cosmic rays. The radius of curvature R (in centimeters) of a particle of energy E(in ergs) and nuclear charge Ze (where e = 4.8 × 10−10esu) in a magnetic field of strength B (inGauss) is R = E/(ZeB). Consider a supernova that occurs a distance d = 1000 pc from us, andassume that the average strength of the interstellar magnetic field is B = 3 × 10−6Gauss.(a) 2 points What is the energy in GeV needed for an iron nucleus so that R = d? As this energyand above, the nucleus will basically come directly at us.(b) 2 points If R ¿ d, the tangled nature of the interstellar magnetic field means that the nucleuswill basically undergo a random walk. Note, though, that the nucleus will not lose significantenergy from the magnetic deflections. In this random walk, the nucleus will effectively take (d/R)2steps of length R each to go a net distance d. Use this, plus special relativistic principles, tocompute the energy Emin(in GeV) below which most10Be nuclei (with rest-frame half-life of3.9 × 106yr) will decay in transit from the distance d = 1000 pc.2. 4 points Dr. I. M. N. Sane has resolved the origin of cosmic rays with energies above the“ankle” at 1019eV. Like many fundamental advances in human thought, the basic idea is simple:ultra-high-energy cosmic rays (UHECRs) are neutrons. Dr. Sane argues that these are producedin active galactic nuclei (AGN); the acceleration to such high energies actually happens to protons,but these convert to neutrons (which have a rest-frame half-life to decay of about 1000 seconds)before moving on. Dr. Sane points out that neutrons aren’t deflected, and that this is consistentwith directions of UHECRs being close to many known AGN, the closest of which is about 15 Mpcaway. Your favorite candidate for President wants to discuss this discovery at a fundraiser as anexample of American ingenuity, but is consulting you first to determine your thoughts. What isyour evaluation of the idea?3. Gravitational redshift.(a) 2 points Use a thought experiment to derive the gravitational redshift of photons one wouldexpect in weak (i.e., Newtonian) gravity. To start, suppose you are at a radius R from the center ofa spherically symmetric mass M, where GM/(Rc2) ¿ 1 for weak gravity. You take two photons ofidentical energy Einitand let them go to some radius r > R. At this radius r the photon energiesare now each Ered= mec2, where meis the mass of an electron. You then cause the photons toproduce an electron and positron with zero kinetic energy. The electron and positron are thendropped back to radius R, where they annihilate to form two photons, and each of the two hasenergy Efinal. Use the energy conservation condition Efinal= Einitto derive the gravitationalredshift factor Ered/Einitas a function of M, R, and r. Use Newtonian potential energies.(b) 2 points Suppose that you can measure the frequency of laser light to one part in 1015. Ifyou shine light straight upwards from the surface of the earth (mass M = 6 × 1027g, radiusR = 6400 km), how high does the light have to get so that you can measure the change infrequency?4. 4 points Properties of the event horizon.In the notes we showed that a particle dropped from rest at infinity will approach the eventhorizon at a speed that approaches the speed of light as seen by a local static observer very closeto but outside the horizon. But what if the particle is dropped from a closer radius?Demonstrate that regardless of the initial radius r > 2M, at the horizon the particle wouldappear to go at the speed of light. Do this in the following way. Consider a particle of nonzerorest mass in a Schwarzschild spacetime. Release it from rest at some radius r > 2M. Then, provethat for any 0 < ² ¿ 1 there is a static observer at some radius R = 2M(1 + δ) with 0 < δ ¿ 1who measures the radial speed of the particle to be c(1 − ²). Hint: remember that the specificenergy −utis conserved for the particle; that’s −utin the global frame, of course, not the locallymeasured specific energy


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