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• Reading List
• Problem Set
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Week 9 - Stellar Black Holes

Problem Set

Problem 1 - A Simplified Accretion Disk

1. Let's divide the accretion disk into a series of rings each of mass dm. What is the total energy of a ring at a distance r from the central black hole of mass M?
2. Let's say that the ring shrinks by a distance dr. What is the change in the energy of the ring (dE/dr) ?
3. As the ring shrinks mass is moving toward the black hole. Divide both sides the answer to (2) by dt to get an equation for the energy loss rate per radial interval.
4. What is the energy loss rate per unit area?
5. Let's assume that this energy is radiated at the radius where it is liberated. Using the blackbody formula what is the temperature of the surface of the disk?
6. Let's assume that the disk extends from an outer radius r_A to an inner radius r₀. What is the total luminosity of the disk if the accretion rate is dm/dt? What and where is the peak temperature of the disk? What and where is the minimum temperature of the disk?
7. Sketch the spectrum from the accretion disk on a log-log plot. You can use temperature units for the energy axis (i.e. kT_max and kT_min). To do this you will have to think about the peak flux from a blackbody at a particular temperature and the size of the disk that radiates at T_max and T_min.
8. The accretion rate is determined by the evolution of the orbit of the black hole with its companion, so it doesn't know about the Eddington limit of the black hole. What do you suppose happens if the rate that matter falls onto the disk exceeds the Eddington limit?
9. What major bit of physics has been left out of this analysis?

Answer for Problem 1

1. The energy is given by

   dE =	G M dm 2 r

2. If you shift the ring inward by dr, the energy changes,

   dE dr

   = -

   G M dm 2 r2

3. Dividing by dt gives the power released as mass flows inward

   d2E dr dt

   =

   G M 2 r2

   dm dt

4. Dividing by the circumference of the rings gives

   d2E dA dt

   =

   G M 4 π r3

   dm dt

5. The flux emerges from both sides so F = (d² E/dA dt)/2

   σ T4 =

   G M 8 π r3

   dm dt

   Yielding,

   T =

   / | \

   G M 8 π σ r3

   dm dt

   \1/4 | /

6. The total energy released is

   dE dt

   = G M

   d m dt

   / | \

   1 r0

   -

   1 rA

   \ | /

   The peak temperature of the disk is at r₀ and the minimum temperature is at r_A.
7. This bit is a bit tricky. We just want a sketch of things so we will assume that the blackbody radiation comes out at a single energy kT to calculate the spectrum between kT_min and kT_max. Above kT_max we have the Wien tail and below kT_min we have the Rayleigh Jeans bit. We're ready to do the middle bit. I will drop all constants too to make it easier.

   F(E) =

   / rA | /r0

   δ (E - k T(r)) r-2 dr

   The delta function encapulates the assumption that the emission comes out at a single energy. To do this integral we have to change variables from r to T using r = T^-4/3 and dr = T^-7/3 dT

   F(E) =

   / kTmax | /kTmin

   δ (E - k T(r)) T1/3 dr

   = E1/3

Here is a sketch:

        |                           ***
        |            1/3         ***   *
        |           E         ***       *
        |                 ***           *
        |              ***               *                         
	|           ***                  *                        
	|        ***                     *                        
  log	|     ***                         *                       
  Flux	|    *                            *  Exponential                  
	|    *                            *    Cutoff                 
	| 2 *                             *                    
	|E  *                              *                   
	|  *                               *                   
	|  *                               *                   
	| *                                *                   
	| *                                *                   
	|*                                  *                  
	|*                                  *                  
	------------------------------------------------------
	       log kT              log kT      
	             min                 max   

8. Material gets blown away from the hole.
9. We lefts out viscosity and GR.

Problem 2 - Thinking about Instruments

1. The black hole in the center of our Galaxy has a mass of 10⁶ . Let us assume that it is a maximally rotating (a=M) Kerr black hole. How big is its horizon? How big is its ergosphere?
2. What angle does the horizon of the central black hole subtend in the sky?
3. I would like to build a telescope that can resolve the central black hole. What is the angular resolution of a telescope as a function of the wavelength of the light and the diameter of telescope. You can look up the formula, use dimensional analysis or the Heisenberg uncertainty principle.
4. What is the diameter of the telescope if you use 2 GHz radio waves?
5. What is the diameter of the telescope if you use 1 keV X-ray photons? Scaling from Chandra, what is the focal length of the telescope?

Answer for Problem 2

1. The horizon is M and the ergosphere is 2M which works out to 1.47 x 10⁶ km and 2.94 x 10⁶ km respectively.
2. The distance to the center of the Galaxy is 8.5 kpc = 8.5 (3.18 x 10²¹ cm) = 2.7 x 10²² cm. The angle the that black hole horizon subtends is 5.4 x 10^-12 radians or about 1 microarcsecond.
3. The angular resolution in radians is given by 1.22 λ/d where d is the diameter.
4. 2 GHz radio waves are about 15 cm, so we have d=(15cm)*1.22*(5.4 x 10^-12 which yields 3.3 x 10¹² cm or about 0.25 AU.
5. 1 keV photons have a wavelength of 12.4 Å giving 28000 cm or 280 m. Chandra has an f ratio of about 10/1.2 or 8.3, so the focal length is 280m*8.3 or 2.3 km.