Calculating Accretion Rate and Luminosity of Black Holes

[bowlbase]

Homework Statement

This is the original problem for part 1
A)
Suppose that a black hole of mass M accretes mass at a rate $\dot{M}$. Further suppose that accretion of mass Δm leads to the radiation of energy $\Delta E= \eta Δmc^2$, for some effeciency of energy conversion, $\eta$. What is the luminosity of emitted radiation in terms of $\dot{M}$ and $\eta$?

B)For what accretion rate $\dot{M}$ does this luminosity equal the Eddington luminosity
for the black hole? Leave your answer as an expression, without plugging in
numerical values.

C) In terms of $\eta$ what is the shortest amount of time that a black hole could increase
its mass by a factor of e≈2.71828, assuming that all of its mass growth occurs
through accretion?

D) Supermassive black holes of mass M≈10⁹M_sun have been detected as ultra luminous quasars in the very early universe, roughly 780 million years after the
Big Bang. Assuming that these black holes have been growing at the maximal
rate possible that you computed in part (c), since the beginning of the universe,
what is the smallest initial mass they have begun with to reach 10⁹M_sun after 780 million years? Assume that $\eta$ =0.1.

Homework Equations

L=η$\dot M$c^2 For A

The Attempt at a Solution

For B I set L=L_e. Where L_e= 10³¹ watts $\frac{M}{M_{sun}}$

\eta \dot M c^2=10^{31} \frac {M}{M_{sun}}
\dot M=\frac{10^{31} \frac {M}{M_{sun}}}{\eta c^2}

Now C) I took $\dot M$ = $\frac{dm}{dt}$ and moved that dt to the right hand side and integrated from M→2.71828M and 0→t respectively.

This leaves an M on both sides of the equation that cancels and only the $\eta$ variable remains (M_sun is known of course). The units work out to seconds as I assumed and I'm left with:
t=1.03x10¹⁶$\eta$

Part D) For this I just took the 2.71828M and divided by the time and got 2.64x10^-16.

So, the initial mass plus the mass rate times initial mass times 780 million years equals 10⁹M_sun.

M+2.64(10^{-16})M (7.8(10^8))=10^9(M_{sun})

Pulling out M and solving I get what amounts to M=10⁹M_sun because I'm basically dividing by 1. This does not make sense. SO for 780 million years the mass didnt change. I must be doing something wrong.

Anyone have a good idea?? My rate must be wrong but I don't know where my mistake is. Thanks in advance.

 

[fzero]

I think that for part C, the net mass increase of the black hole should be the mass accreted minus the mass equivalent of the energy radiated away. I have not checked that this leads to something reasonable for part D, but it should be easy for you to do so.

 

[bowlbase]

I've been staring at what you wrote for a while now but I'm not following what you mean. At least not how I can set up the equation to reflect what you're saying.

 

[Dick]

I've been staring at what you wrote for a while now but I'm not following what you mean. At least not how I can set up the equation to reflect what you're saying.

It's what I was missing last night when I made my simplistic answer. I knew I was missing something. If a mass Δm falls into the black hole and radiates away ηΔmc^2 in energy then the black hole doesn't gain mass Δm. It gains Δm-(the mass equivalent of ηΔmc^2)=Δm-(ηΔmc^2)/c^2. So $\dot M=\dot m(1-\eta)$ And luminosity is $L=\dot E=η {\dot m}c^2$. You'll have to eliminate $\dot m$ from the last two equations to get a correct answer to last nights question. Sorry.

 

[bowlbase]

Okay, I did the problem using these equations. We were certainly right to be suspicious of our answer last night. This makes a lot more sense to me now.

Our new luminosity would then be $L=\eta \frac{\dot M c^2}{1-\eta}$. Setting this equal to the L_E: $\frac{\eta}{1-\eta} \dot M c^2= 10^{31} \frac{M}{M_{sun}}$ so that $\dot M= 1.1(10^{14})\frac{1-\eta}{\eta}\frac{M}{M_{sun}}$.

Doing the integral as before I get pretty much the same answer but I realize I made a mistake with the time before. $t=3.12(10^{16})\frac{\eta}{1-\eta}$

Now, as before I take the rate: $\frac{2.71828M}{3.12(10^{16})\frac{\eta}{1-\eta}}=6.95(10^{-17})M\frac{1-\eta}{\eta}$

Beginning mass plus rate times beginning mass and time(*converting time to seconds*):
$M+6.95(10^{-17})M\frac{1-\eta}{\eta}(2.46(10^{16}))=2(10^{39})$

$\eta$=.1

$M=\frac{2(10^{39})}{1+6.95(10^{-17})\frac{1-.1}{.1}(2.46(10^{16})}$

M=1.2205(10³⁸)

So, if I did everything correct, I hope, then it only grew in one order of magnitude in that 780 million years. I guess that may be correct...

 

[fzero]

It's what I was missing last night when I made my simplistic answer. I knew I was missing something. If a mass Δm falls into the black hole and radiates away ηΔmc^2 in energy then the black hole doesn't gain mass Δm. It gains Δm-(the mass equivalent of ηΔmc^2)=Δm-(ηΔmc^2)/c^2. So $\dot M=\dot m(1-\eta)$ And luminosity is $L=\dot E=η {\dot m}c^2$. You'll have to eliminate $\dot m$ from the last two equations to get a correct answer to last nights question. Sorry.

I didn't give an equation because I didn't want to give too much of the answer and because I hate the notation. We have $\dot{M}$ the accretion rate, $\Delta m$ the amount of mass accreted in a time $\Delta t$ and then we should really use a different symbol for the net mass. So call the net mass $\mu$. Then in a time $dt$, the amount of matter that accreted is $\dot{M} dt$, while the mass-equivalent energy radiated away is $\eta \dot{M} dt$. So the net mass gain is given by

$\dot{\mu} = (1 - \eta ) \dot{M}.$

Since this is the net mass gain, it is physically important that the coefficient here $1-\eta < 1$ appears in the right way.

 

[fzero]

Our new luminosity would then be $L=\eta \frac{\dot M c^2}{1-\eta}$. Setting this equal to the L_E: $\frac{\eta}{1-\eta} \dot M c^2= 10^{31} \frac{M}{M_{sun}}$ so that $\dot M= 1.1(10^{14})\frac{1-\eta}{\eta}\frac{M}{M_{sun}}$.

The luminosity is always $L=\eta \dot M c^2$. It is defined in terms of the total matter being accreted, not the net mass gain of the black hole.

 

[Dick]

I didn't give an equation because I didn't want to give too much of the answer and because I hate the notation. We have $\dot{M}$ the accretion rate, $\Delta m$ the amount of mass accreted in a time $\Delta t$ and then we should really use a different symbol for the net mass. So call the net mass $\mu$. Then in a time $dt$, the amount of matter that accreted is $\dot{M} dt$, while the mass-equivalent energy radiated away is $\eta \dot{M} dt$. So the net mass gain is given by

$\dot{\mu} = (1 - \eta ) \dot{M}.$

Since this is the net mass gain, it is physically important that the coefficient here $1-\eta < 1$ appears in the right way.

I gave a lot more of the answer than I usually would, because I felt like such an idiot for botching this question before. I think the books notation is calling the net mass M and sort of implying you should use $\dot m$ for the infall rate.

 

[bowlbase]

I appreciate the help and like to work through problems myself. For this I was just unable to make the connection to what you were saying to how I could implement it.