Astroparticle Physics

374 17 SolutionsThe axion mass does not enter the calculation, i.e., the escape velocity does not dependon the mass.A relativistic treatment, (γ − 1)m 0 c 2 = Gγ m 0 M/R, leads to a similar result:√ √2GMv = βc = 1 − GMR 2Rc 2 .Since this expression is not based on general relativity, it is of limited use.5. Spherically symmetric mass distribution.∫ ra) Mass inside a sphere: M(r) = ϱ(r ′ ) dV ′ = 4πpotential energy of mass m: E (m)pot (r) = Gm ∫ rverification: − ∂E(m) pot (r)= F =−G mM(r)∂rr 2 .0∞∫ r0M(r ′ )r ′2 dr ′ ,ϱ(r ′ )r ′2 dr ′ ;b) Potential energy of mass shell dM = M ′ (r) dr:dE pot = GM ′ (r) drtotal potential energy: E pot = G∫ ∞0∫ rM ′ (r) dr∫ rM(r ′ ∣) ∣∣∣integration by parts: E pot = GM(r)∞ r ′2 dr ′ ∞∞M(r ′ )r ′2 dr ′ ;r=0− G∫ ∞∫ ∞M 2 (r)margin term vanishes for finite masses: E pot =−G0 r 2 dr.c) Mass M on a shell, i.e., M(r < R) = 0andM(r > R) = M:E (shell)pot=−GM 2 ∫ ∞R1dr = GM2r2 r∣∞r=R=−G M2R .0∫ r∞M 2 (r)r 2M(r ′ )r ′2 dr ′ ;d) Homogeneous mass distribution: M(r < R) = Mr 3 /R 3 and M(r > R) = M:∫ E pot(hom) =−G M2 RR 6 r 4 dr + E pot(shell) =−G M2 r 5RR 6 − G M25 ∣ R=−6 5 GM2 R .06. n max m ν = m4 ν v3 f3π 2 ¯h 3 . (*)If n max m ν is assumed to be on the order of a typical dark-matter density (Ω = 0.3) onecan solve (*) for m ν ,r=0dr;m ν =( ) 1/43π 2 ¯h 3 n max m ν.v 3 f