PROBLEM SET #9 SOLUTIONS AST142 1. Quasar luminosity

PROBLEM SET #9 SOLUTIONS
AST142
1. Quasar luminosity
Suppose that a quasar is apparently as bright as a solar-type star, but that the quasar
is a factor of a million further away than the star.
(a) What is the quasar’s luminosity?
(b) Assuming an efficiency of 0.1 what is the accretion rate in solar masses per year?
———————–
Solution
The ratio of their luminosities is L/L = (d/d0 )2 . If the quasar is a million times
further away then it has a luminosity of 1012 L . That’s quite a large luminosity – our
Galaxy only puts out a total of about 2 × 1010 L .
M˙
L
1012 × 3.9 × 1033 erg s−1
=
c2 × 0.1
(3 × 1010 cm/s)2 × 0.1
3 × 107 s
M
= 4.3 × 1025 g s−1 ×
×
yr
2 × 1033 g
= 0.6M /yr
=
2. Eddington Luminosity
A quasar has an average luminosity of 3 × 1013 L , and an X-ray brightness that can
vary substantially in as little as three hours. Assume that the quasar’s black-hole engine
is accreting at the Eddington rate, and show that these two findings are consistent with
one another.
————–
Solution
The light travel distance in 3 hours is
3 × 60 × 60s × 3 × 1010 cm/s = 3 × 1014 cm
1
2
PROBLEM SET #9 SOLUTIONS
AST142
This is only 22 AU. The X-ray variability suggests the accretion region is less than this
size .
If the central black hole is accreting and producing luminosity at the Eddington rate,
its mass is
2e4 L
= 2 × 1042 g = 109 M
M=
3Gmp m2e c5
The Schwarzschild radius for a billion solar mass black hole is
2GM
RSch =
= 3 × 1014 cm = 2.8 light hours
c2
about the same as the quoted ”variability size” from the light travel time of the X-rayemitting region. The X-rays would arise in the hottest part of the accretion disk, and
that could be closest to the event horizon.
Figure 1. Angle and lengths for a relativistically moving gas blob. The
blob moves at speed v with respect to the galaxy and at an angle θ with
respect to the line of sight to a distant observer that is distance d from the
galaxy.
3. Apparent Velocity for a Relativistic Object (Superluminal Motion)
Consider a blob of gas ejected from the center of an active galaxy, moving at speed
v with respect to the galaxy’s nucleus, at angle θ with respect to the line of sight of an
observer a distance d away. Working the galaxy’s reference frame, and assuming that
the galaxy is at rest with respect to the observer, at t = 0 the blob emits a pulse that is
detected at the observer location at time t0 = d/c. At time t = T later, the blob emits
another pulse that is detected at the observer location at time t1 . See Figure 1.
Show that
t1 = T +
d vT cos θ
−
c
c
PROBLEM SET #9 SOLUTIONS
AST142
3
Show that the apparent motion of the blob (as seen from the observer) is
vapparent =
v sin θ
1 − β cos θ
————————————Solution
After time T the blob has moved a distance of vT . The blob emits at time T in the
galaxy frame, and it is emitted a distance d − vT cos θ nearer the observer. So it is
received at time
t1 = T + (d − vT cos θ)/c
The second pulse is received later than the first pulse by this
∆t = t1 − t0 = T + (d − vT cos θ)/c − d/c = T (1 − β cos θ)
The apparent distance on the sky as seen by the observer is
∆R = vT sin θ
The apparent velocity is the
vapparent =
vT sin θ
v sin θ
∆R
=
=
∆t
T (1 − β cos θ)
1 − β cos θ
4. Superluminal motion
The largest apparent superluminal motions seen so far in quasar jets are about 20c.
What would this imply for the ejection speed v (in units of c, to three significant
figures) and jet angle with the line of sight θ? (in degrees), if the jet is oriented for
maximum apparent superluminal motion?
—————————
Solution
Using the relation for maximum apparent motion
(v⊥,apparent )max = vγ = 20c
β(1 − β 2 )−1/2 = 20
β 2 (1 − β 2 )−1 = 400
401β 2 = 400
p
β = 400/401 = 0.9987
v = 0.9987c
4
PROBLEM SET #9 SOLUTIONS
AST142
θ = arccos β = 2.86◦
5. Sphere of Influence
M87 has a massive black hole of mass M = 3.5 ± 0.8 × 109 M . The black hole lies
inside a galaxy. Suppose the stars in the center of the galaxy (in the bulge) have velocity
dispersion 300 km/s.
(a) Within what radius around the black hole does a circular orbit have to be in order
to have a circular velocity that is larger than the random motions of the bulge
stars? This radius is sometimes called the sphere of influence. Inside this radius
motions are dominated by the gravity of the black hole.
(b) M87 has a radial velocity of 1307 km/s. Using a Hubble constant of 70 (km/s)
Mpc−1 how far away is M87?
(c) At this distance how larger is 1 arc second?
(d) What distance in arc seconds is the sphere of influence radius?
———————Solution
r=
GM
6.67 × 10−8 × 3.5 × 109 × 2 × 1033
=
= 5 × 1020 cm = 180pc
v2
(3 × 107 )2
The distance is 1307/70 = 18.7 Mpc.
1” gives 18.7Mpc × 5 × 10−6 = 93 pc.
The sphere of influence is about 1.5”