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Lets suppose we have a BH with mass $M$ and surface area $A$. From the Schwarzschild raidus we can say that the Area of the BH is proportional to its mass $$A=4\pi r^2=\frac {16 \pi G^2M^2} {c^4}$$ or simply let us say

$$A(M)=\alpha M^2$$

Now let us suppose this black hole gains mass $m$ per second. We can easily see that the gained mass will be proportional to the surface Area of the Black hole. Hence we can write

$$m\propto A$$ or $$m(A)=\beta A$$ where $\beta$ is constant.

Now I want to find that the gained mass by the black hole per second ($M(t)=?$).

Lets think like this at $T$ The surface area is $$A=\alpha M^2$$; $$T:A=\alpha M^2$$ then it gaines mass by amount of $dm$ and it increases the area by $dA$.

So at $$T+dt:A+dA=\alpha (M+dm)^2$$

And the gained mass at this $dt$ time will be equal to $dm=dA\beta$ or can we say $dm=A\beta$

Edit: Lets thing mass $m$ around the surface of the black hole. The mass has a density $p_m$. By using this we can write such equation. $dm=Ap_mdr$ or $m=\int Ap_mdr=\int 4\pi r^2dr$ for $r>R$ where $R$ is the radius of the black hole. I was thinking that the $dM/dt=m$ but I thought later that units dont match. But I cannot stop thinking that the logical answer should be like this since the "change in the black hole mass" should be equal to the gained mass right ?

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    $\begingroup$ First of all, the volume of a black hole is not a well-defined quantity in general. What you're considering is the volume that the event horizon of a black hole appears to enclose from the perspective of a faraway, stationary observer. Second, $dM/dt$ depends on both $dV/dt$ and either $M$ or $V$ itself. In order to get a nontrivial answer to this question, you're going to have to constrain your problem more, by choosing some form for either $dM/dt$ or $dV/dt$ (for example, you could require that $dM/dt$ is constant, or linearly increasing, or some other functional form). $\endgroup$ – probably_someone Nov 17 '18 at 11:59
  • $\begingroup$ Then is it a good approximation to use the volume equation? Let's say $dM/dt$ increases linearly. I was thinking that "we can figure out $dM/dt$ by using these equations. $\endgroup$ – Reign Nov 17 '18 at 13:05
  • $\begingroup$ It's not an issue of a good or bad approximation, it's just that the thing you called $V$ here is not necessarily the volume of a black hole, because the concept of "the volume of a black hole" itself isn't well defined. You can choose to define it that way, sure, but my point was to let you know that you are choosing to define $V$ as the volume of the black hole, and it may have a different expression under other definitions. In particular, it may not behave like the quantity we usually call volume. $\endgroup$ – probably_someone Nov 17 '18 at 14:59
  • $\begingroup$ In any case, if you want to "figure out" $dM/dt$, and you want anything other than a trivial answer, then you need to tell us more information. After all, not every black hole of the same mass and volume is going to acquire mass at the same rate; some may be surrounded by more dust and gas than others, for example. In order to figure out how much mass the black hole is gaining, you have to know the rate at which the "volume" of the black hole is expanding. So what is that rate? $\endgroup$ – probably_someone Nov 17 '18 at 15:03
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    $\begingroup$ @ArthurMorgan And what we've been telling you is that you haven't provided enough information/assumptions to do that. $\endgroup$ – probably_someone Nov 18 '18 at 15:53
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The quantity $A$ is actually the surface area of the event horizon of the black hole, as seen from a faraway, stationary observer; as such, it may not behave like the quantity that we traditionally call surface area.

If we differentiate

$$A=\frac{16\pi G^2 M^2}{c^4}$$

with respect to time, we get

$$\frac{dA}{dt}=\frac{32\pi G^2 M}{c^4} \frac{dM}{dt}$$

which tells us that the rate at which the surface area of the event horizon grows, $\frac{dA}{dt}$, is dependent on both the current mass $M$ of the black hole and the rate at which the black hole accretes mass, $\frac{dM}{dt}$. If you know the rate at which the surface area of the event horizon grows, and you know the mass of the black hole, then you can determine how much mass the black hole is accreting. If you do not know the rate at which the surface area of the event horizon is growing, then you also do not know the rate at which the mass is growing.

As to your claim that the rate of mass gain is proportional to the surface area of the event horizon, that is incorrect in several ways. To see why, let's rearrange the first equation:

$$M=\frac{c^2}{4G}\sqrt{\frac{A}{\pi}}$$

and differentiate with respect to time:

$$\frac{dM}{dt}=\frac{c^2}{8G}\sqrt{\frac{1}{A\pi}}\frac{dA}{dt}$$

We can see, firstly, that your assertion is incomplete; the rate of mass gain is dependent on both the area of the event horizon and the rate at which that area is growing, neither of which are necessarily constant. Secondly, your assertion that the rate of mass gain is proportional to the area of the event horizon is incorrect. Rather, it's proportional to $\frac{1}{\sqrt{A}}$. If we say that the rate of mass gain is $m$ and the rate that the area is growing is $a$, then we can fix your statement in the following way:

$$m(A,a)=\beta\frac{a}{\sqrt{A}}$$

If you want to know $M(t)$, though, all of this isn't really helping you. All you need to find $M(t)$ is $A(t)$. Plug $A(t)$ into the first equation and you have $M(t)$.

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  • $\begingroup$ I find also another thing. I would like to share. Should I edit my post for that ? $\endgroup$ – Reign Nov 21 '18 at 18:18
  • $\begingroup$ Edited my question. Could you look at it ? $\endgroup$ – Reign Nov 22 '18 at 15:05
  • $\begingroup$ Nothing to say about it ..? $\endgroup$ – Reign Nov 26 '18 at 11:11
  • $\begingroup$ @ArthurMorgan I have no idea what you're trying to ask. Maybe asking another, separate question would be clearer. $\endgroup$ – probably_someone Nov 26 '18 at 12:32
  • $\begingroup$ I shared it you guys can check $\endgroup$ – Reign Nov 26 '18 at 15:37

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