Astroparticle Physics

17.9 Chapter 9 365which is the Schwarzschild radius. The result of this classical treatment of the problem(even though classical physics does not apply in this situation) accidentially agrees withthe outcome of the correct derivation based on general relativity.By definition the Schwarzschild radius is R S = 2GM and the calculation givesR S ={ 8.9mmforEarth2.95 km for the Sun.c 26.7.GM 2R > 3 2 kT M µM> 3 kT2 µG R = 3 kT2 µG, µ – mass of a hydrogen atom ,( ) 3M 1/3;4πϱ( 32since M = 4 3 πϱR3 : M 3 >ϱ> 34πkTµG) 33M4πϱ ,( )1 3 kT 3M 2 ≈ 3.9 × 10 −10 kg/m 32 µG(k = 1.38 × 10 −23 J/K, M = 2 × 10 30 kg, µ = 1.67 × 10 −27 kg).starγ v(t+ d 1✻c) = dDd 1d 2td✻❅❅❅❅❅❅❄D − d 1❄❄observer attime t 0❄observer attime t 0 + tThe first observation is made at time t 0 . The star is at a distance D from the observer.It moves at an angle γ to the line of sight. During t it has moved d 1 = d cos γ closerto the observer. The first light beam has further to travel. The light was emitted t + d 1cearlier compared to the second measurement. The apparent velocity is( )t + d 1csin γv ∗ = d 2t = d sin γ v=( t= v 1 + d )2ct cot γ sin γ = vt(= v 1 + d )1sin γct(1 + v∗c cot γ )sin γ = v sin γ + v c v∗ cos γ.