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• Problem Set
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Week 11 - The Gamma-Ray Burst Controversy

Problem Set

Problem 1 - Isotropy

1. Suppose that GRBs lie in a spherical distribution of radius R, and we lie a distance R₀ away from the center. Furthermore suppose that we can see every GRB in the volume and we have seen a total of N GRBs. How many GRBs will we have seen in the hemisphere of sky toward the center of the distribution versus the opposite hemisphere?
2. Using the results from number (1), what is the difference in the number of GRBs in each hemisphere to lowest order in R₀/R?
3. Let's compare this result with observations. The one-sigma error in the difference between number of objects in the hemispheres is given by approximately by N^1/2. If you have observed N objects without detecting an difference between the two hemispheres at the two-sigma level, what is the minimum acceptable value of R/R₀?
4. At the time of the Paczynski-Lamb debate BATSE had observed about 600 bursts and R₀ was assumed to be 10kpc, what was the minimal acceptable value of R at the two-sigma level? Does this distance sound familiar?

Answer for Problem 1

1. Let x=R₀/R. Toward the center of the sphere we would see N (0.25 x³-0.75 x + 0.5). In the opposite direction we would see N (-0.25 x³+0.75 x + 0.5).
2. The difference is N (1.5 x - 0.5 x³).
3. The signal-to-noise ratio is 1.5 x N^1/2, so for N objects R/R₀ > 0.75 N^1/2
4. 183.6 kpc c.f. the Galactic corona extends to 100 kpc.

Problem 2 - Boosting

1. Use the Minkowski metric to figure this out. I measure a photon to have an energy E. What is the four-momentum of the photon?
2. My pal is travelling toward me in the opposite direction of the photon at a velocity β c. What is his four-velocity? Use the definition γ = (1-β²)^-1/2 to simplify the expression. What energy would he measure for the photon? What does the expression look like as γ gets much larger than one?
3. If my pal observes the photon to have an energy of 100 MeV while I say its energy is less than 500 keV, what is the minimal value of γ for my pal (take β=1 to make life easier)?
4. My pal is still coming toward me at a velocity β c. When he is a distance r away from me (at a time t₀) he emits a photon toward me. How long does it take this photon to reach me?
5. From his point of view a short time Δt later he emits another photon toward me. How long is Δt in my frame and when do I receive the second photon? What is the difference in time between when I receive the first and second photons? What does the expression look like as γ gets much larger than one? Compare it with you answer to (2).
6. Explain two ways that relativistic motion (big values of γ) can relieve the compactness problem.

Answer for Problem 2

1. p^μ = ( E, E, 0, 0)
2. u^μ = ( γ, -γβ, 0, 0)
   E_pal = (γ + β &gamma) E ≈ 2 γ E
3. γ > 100
4. t₁ = r/c + t₀
5. Δt_me = γ Δ t_pal
   t₂ = t₀ + γ Δ t_pal+ (r-c β γ Δt_pal)/c=t₁ + &Delta t;_pal γ (1-β)
   t₂-t₁ = Δt_pal γ (1-β) = Δt_pal / ( γ (1+β) ) ≈ Δt_pal / (2 γ )
   It is the reciprocal of the energy ratio. This is not too surprising because the energy of a photon is inversely proportional to the period of the wave.
6. First, the relativistic effect reduces the energy of the photons in the frame where they are emitted so fewer photons lie above the pair-production threshold. Second, the time compression means that from in the frame of emission one millisecond lasts 2 γ milliseconds so the emission region can be much larger reducing the total optical depth.

Problem 3 - Fermi Process

1. He is still toward me at a high value of γ. He serves a photon with energy E in his frame toward me. What is its energy when I receive it (take the high γ limit)?
2. I reflect the photon back toward him with a mirror. What is the energy of the photon in my pal's frame when he receives it again? What is its energy after another complete volley? After n complete volleys?
3. Let's say that we're not so good and that the probability that we successfully reflect the photon is p where p is much smaller than unity. What is the probability of completing n complete volleys?
4. Take the logarithm of both sides of the energy of the photon after n complete volleys and solve for n.
5. Take the logarithm of both sides of the probability of completing n complete volleys and substitute the value of n from the previous part. Rearrange to obtain an expression for the probability as a function of the photon energy -- it should be a power-law.
6. Take p to be 10^-5 (we're not very good) and γ = 100. What is the exponent of the power-law?

Answer for Problem 3

1. E₁ = 2 γ E₀
2. E₂ = (2 γ)² E₀
   E₄ = (2 γ)⁴ E₀
   E_2n = (2 γ)²ⁿ E₀
   
3. p_2n = (p)^2n-1
4. Rearranging yields

   2n =	ln E2n E0 ln (2γ)

5. Here it is,

   f2n =

   1 p

   / | \

   E2n E0

   \[ln p/ln (2γ)] | /

6. The exponent is -2.17.