Question

3. Tremaine et al. (2002) have shown that supermassive black holes in the nuclei of galaxies follow a relation between black hole mass, MB, and the velocity dispersion of the host galaxy bulge component, σ, such that log,o MBH8.0+40logo 200km/ The gravitational radius, Ra. (in a sense the "sphere of influence") of a black hole can be defined by the distance from the center at which the orbital speed of a star due to the black hole alone (i.e. neglecting the mass distribution of the host galaxy), vorb R), matches the velocity dispersion of the bulge, σ(R). a) Calculate Rc for an elliptical galaxy with an isothermal density profile and central velocity dispersion ơc-300 km/s. What is the mass of the black hole? Is the black hole likely to dominate the motions of stars in the bulk of the galaxy? (Recall that in the case of an isothermal gas 흘kT = my? = const., where u is the r.m.s. velocity of a typical particle in the gas. Also, G, evaluated in more convenient units, is given above.) b) What is the largest redshift for which Rc, in this particular case, could be resolved using STIS on HST (assume a resolution of ~0'1).

Answer 1

a) We have the velocity dispersion,
a300km.s1
. So, the black hole mass is given by
300 logioMBH 8.0+4.0lo8.7 200
. So, the mass of the BH is
8.704 10

Mo
.Now for $R_{G}$ we have,
GMBH muz GMBHm 2,2 RG
where $v = \sigma$.

putting the values we get .

RG = 7.5 × 101, m

b) If we assume a flat universe then we have the angular diameter distance,
1 + 2
where r is the comoving distance. Now, the angle subtented by the virial radius $R_G$ is given by, $\theta = \frac{R_G}{d_A}$ . Now,
0.1', = 2.78 × 10-"degree = 4.85 × 10-rad θ
.

So, $d_A=\frac{R_G}{\theta}=1.55\times10^{24} m = 50 Mpc$ .

Now, the comoving distance is given by $r = \frac{c}{H_0}\int_{0}^{z}\frac{dz'}{E(z')}$ where $E(z')=\sqrt{\Omega_{M}(1+z)^{3}+\Omega_r(1+z)^4+\Omega_\Lambda}$ . So, we need to solve $\frac{r}{1+z}=50Mpc$ . Solving this numerically we get z=0.118