Astroparticle Physics

17.6 Chapter 6 3555. Measured flux R = source flux R ∗ × efficiency ε × solid angle Ω;ε = 1 − e −µx = 1 − e −125×5.8×10−3 ≈ 51.6% , Ω = 104 cm 24π(55 kpc) 2 ≈ 2.76 × 10−44 ,R ∗ =RεΩ ≈ 1.95 × 1040 /s .6. In contrast to Compton scattering of photons on an electron target at rest, all threemomentaare different from zero in this case. For the four-momenta k i , k f (photon) andq i , q f (electron) one has k i − k f = q f − q i . Squaring this equation gives k i k f = q i q f − m 2 e .On the other hand, rewriting the four-momentum conservation as q f = q i + k i − k f andmultiplying with q i leads to q i q f − m 2 e = q i(k i − k f ), yieldingk i k f = q i (k i − k f ).Let ϑ be the angle between k i and k f , which givesω i ω f (1 − cos ϑ) = E i (ω i − ω f ) −|q i |(ω i cos ϕ i − ω f cos ϕ f ).Solving for ω f leads to1 − √ 1 − (m e /E i )ω f = ω 2 cos ϕ ii1 − √ .1 − (m e /E i ) 2 cos ϕ f + ω i (1 − cos ϑ)/E iThis expression is still exact. If the terms ω i /E i and m e /E i are neglected, the relationquoted in the problem is obtained.7. Starting from the derivative dP/dν,dP/dν ∼ 3ν2 (e hν/kT − 1) − ν 3 kT h ehν/kT(e hν/kT − 1) 2 = ν 2 3 (ex − 1) − x e x(e x − 1) 2 ,and the condition dP/dν = 0 one obtains the equatione −x = 1 − x 3 , x = hνkT .An approximated solution to this transcendental equation gives x ≈ 2.8, leading to afrequency, resp. energy of the maximum of the Planck distribution ofhν M ≈ 2.8 kT .This linear relation between the frequency in the maximum and the temperature is calledWien’s displacement law.Forhν M = 50 keV a temperature ofT ≈ 2 × 10 8 K .is obtained.