10 2017apho理论第二题答案解析

1、Evolution of Supermassive Black Holes BinarySolutionA. Dynamical Frictionpy =, assuming that 1. One can ndA1. The de ection angle is de ned from: tan Rpxpy=Fy dt, and according to Newtons gravitylawGMmb2Fy=cos3 bdv cos2 Changing the variable dt =we haveZ/2GMbv2GMmbvpy =cos d =./2Here we assume that

2、the body moves along the stright line, due to 1, see Fig 1. Sopy2GM2b1 . =pbv2bA2. During the transit of a massive body, stars energy remaconstant: p2 + p2 = const.xyHence222(p p ) + p =y p .xWe know that py p, so the momentum change along the x-axis:p222G2M 2mpx = y = p = .2p2b2v3A3. To calculate n

3、et force we might integrate over stars with di erent impact parameters. The number of stars transits during the time t equals N = 2bvndb t, so force, decelerating the object along the x-axis,Z 1 t2 nm Zbmax db 222FDF(1)px dN = 4G M= 4G Mlog =v2bv2bminThe above formulas are true only for b b1, so the

4、 lower integration limit is bmin = b1, and the upper limit is determined by the galaxy size bmax = R. So we have2 2(2)FDF = 4G Mlog v2where = R/b1.GMv2A4. We calculate:b1 = 11 pc, log = 7.6.B. Gravitational slingshotB1. Fromthesecond Newtons lawmv2GMm ,=4a2212q GM . The system energy isand we have f

5、or the orbital velocity v bin =4aMv2GM 2GM 2 .E = Ekin + U = 2 = 22a4aB2. From angular momentum conservation lawb = rmv0,express v0. Write down the energy conservation law2v2GM2 0=rm22q 2GM2 .and derive b = rm1 +2rmB3. To estimate the time between collisions let us use an analogy with the gas. As kn

6、own from the molecular kinetic theory, given that molecules have radii r, thermal velocities v, and the molecular concentration is n, time t between collisions of one molecule with the others can bebmax1estimated from the relation r2vtn = 1. In our problem radius, therefore for estimation it can be

7、written t =stands in place of the molecule.2bnmaxbmax,Estimate the maximal impact parametercorresponding to the star collision with thebinary system. The star should reach the distance of a to the binary system to collide. Whenthe star is at large distances from the SBH binary, it interacts with it

8、as with a poqint object 4GM .of mass M2 = 2M . From the results of B.2, assuming rm = a, we obtainbmax= a1 +2aTaking into account that , simplify: bmax =GMa, so we haveGMa2mt =.GMaB4. During the one act of gravitational slingshot, star energy increases at average by Estar =mv2m.2 bin 22So after the

9、one collision the energy of the binary system decreases by the same magnitude.m 2vbin, we derive E = v.Taking into account that 2bin2 2dE EG M 4dEGM 2 ,= . In respect to=Average binary system energy loss rate equals=dttdt4a2orbit radius variation rate can be estimated asGa2 .dadt= (3)B5. Obtained eq

10、uation can be easilyintegratedda a2G= dt.(4) Ga1To reduce the radius twice it takes time TSS = 0.0048 Gy.3C. Emission of gravitational waves?1. Usingpreviousresults(B) one can obtain:GM 2 da4a2 dt ;dGM 2dE(5)=dtdt4aThis leads to the di erential equation:GM 2 da1024 G6M 2a4(6)= ,4a2 dt5c5r GMwhere an

11、gular velocity is known from part B: =The desirable result is:4a3 .G3M 3da dt(7)= c5a3?2. Solving the equation one can obtain:G3M 3a4G3M 3da2565 2 a3 = (8)The nal result for time: TGW ;dt4c5c5a4c552.TGW =(9)?3. Solvingtheprevious equation:1024G3M 3s 1024 tHG3M 34(10) 0.098 pc.aH =5c5D. Full evolutio

12、nD1. The gRalaxy is spherically symmetric, so mass enclosed within a sphere of radius r equalsm(r) =r 4x2(x) dx = 2r .Thus the free fall acceleration of the body equals in the0GGm(r)2 .gravitational eld of stars is g(r) =Therefore the body velocity is determinedr2rby relation= g = 2 , which means v

13、= =const.v2rrD2. The energy of SBH in this gravitational eld isM2E =+ U2=and= g(r)M = M2 . SodUdEdU dtdU da da dt M2 da .=daadta dtUsing the result of A2 we havedEdtGM 2 log ,2 (r)2= Fv = 4G Mlog = Dfva2and the answer is(11)da dtGM log = a4D3. To estimate one can assume that SBHs form a binary when

14、the mass of stars sphere of radius a equals to M :2aide them(a) = M,GGM2so a1 = 11 pc.D4. Integrating the equation (11) we have22a aGM log 01=T12and using that a1 a0 we havea2 02GM ln T1 = 0.12 Gy.D5. Total energy losses are caused by gravitational slingshot and gravitational waves emission, sodEG2M

15、 264G4M 5(12)=dt4G2M 2145c5a564G4M 55c5a5Energy losses due to GW dominates wheni.e. a a where2256 G2M 3 .5a2 =5c51Numerical answers are 1 = 6 103MS/pc3 and a2 = 0.026pc.D6. For rough approximation it can be considered that at the slingshot stage losses are due to slingshot only, so T2 is calculated analogiously to B5:T2 = 0.27 GyGa2And at the GW emission stage losses ar