3
$\begingroup$

We predict that black holes will lose mass over time due to Hawking radiation. I want to find an equation for the mass over time $M(t)$ of a black hole with initial mass $M_0$ in an otherwise empty universe, meaning no matter or energy enters the black hole and the Hawking radiation is the only factor affecting its mass. For further simplification, let us assume the black hole is a non-rotating, non-charged Schwarzschild black hole that acts as a perfect black body radiator, that all energy emission is in the form of photons, and that the event horizon is the radiating surface.

With these simplifications/assumptions, Wikipedia gives the total time for evaporation in seconds as $t_{ev}(M) = \frac{5120 π G² M³} {ħ c⁴}$, where $ħ = 1.054571817 × 10⁻³⁴ J \, s$ is the reduced Planck constant, $c = 299792458 \frac{m} {s}$ is the speed of light in a vacuum, and $G = 6.674 × 10⁻¹¹ \frac{m³} {kg \, s²}$ is the gravitational constant. Thus the equation for mass as a function of time, $M(t)$, should equal $M_0$ at time $t = 0$ and should equal zero at time $t = t_{ev}(M_0)$.

Wikipedia also states that the power $P$ in Watts (Joules per second) emitted by a black hole of mass $M$ in kilograms is given by the equation $P(M) = \frac{ħ c⁶} {15360 π G² M²}$. Note that the mass term is in the denominator, so the power emission (and thus the rate of evaporation) increases as the mass decreases, approaching infinity as the mass approaches zero. Thus the derivative of $M(t)$ should be some negative value at time $t = 0$ and should approach negative infinity as $t$ approaches $t_{ev}(M_0)$.

However, I don't know how to go from the power output as a function of mass function $P(M)$ to a function for mass as a function of time $M(t)$. Because this is math dealing with a rate of change (which is itself changing) I assume it requires calculus, but I don't know enough of calculus to even know how to search for a method to find the answer I seek.

$\endgroup$

1 Answer 1

4
$\begingroup$

The power emitted translates into mass loss, because power is the (negative) time derivative of energy and because $E = M c^2$.

So, we can write the evolution equation as $$ \frac{{\rm d} M}{{\rm d} t} = - \frac{P(M)}{c^2} = - \frac{k}{M^2} $$ for a constant $k$ which encapsulates all the ones you have written, plus the extra $c^2$: $$ k = \frac{\hbar c^4}{15360 \pi G^2} $$

Now, this is what's called a separable ordinary differential equation; it can be solved analytically to yield $M(t)$.

The typical way to do so involves writing it as $$ M^2 {\rm d} M = -k {\rm d} t $$ and integrating from now ($t=t_0$ and $M = M_0$) to when the BH evaporates ($t = t_1$ and $M=0$), to get $$ \frac{M_0^3}{3} = k (t_1 - t_0) $$ which can be rearranged into $$ t_1 - t_0 = t_{\text{evap}} = \frac{M_0^3}{3 k} $$ which is the result you mentioned.

If, on the other hand, we want to compute $M(t)$ for some arbitrary time $t$, we can apply the same procedure, only now we integrate up to $M_1$ and $t_1$: $$ \frac{M_0^3}{3} - \frac{M_1^3}{3} = k (t_1 - t_0) $$ or $$ M_1(t_1) = \left( M_0^3 - 3k (t_1 - t_0) \right)^{1/3} $$ where $M_0$ and $t_0$ are our initial conditions, while $t_1$ is the time at which we want to compute the mass.

Substituting the value of $k$ and simplifying produces the full equation for mass as a function of time:

$$M(t) = \sqrt[3]{M_0^3 - \frac{ħ \; c⁴ \; t} {5120 π \; G²}}$$

$\endgroup$
7
  • $\begingroup$ What do you mean by "a constant $k$ which encapsulates all the ones you have written"? $\endgroup$
    – Lawton
    Apr 24, 2022 at 22:19
  • $\begingroup$ I've looked into separable ordinary differential equations, and it seems like I would need to have a function in terms of $t$ in order for this to count. Is that correct? Both equations I have are in terms of $M$, which doesn't seem to fit the criteria. $\endgroup$
    – Lawton
    Apr 24, 2022 at 22:42
  • $\begingroup$ @Lawton as for the second point, yes you do need a function of $t$, but a constant function is indeed a (very simple) function of $t$ or of any other variable you like! I'll include some more details. $\endgroup$ Apr 25, 2022 at 9:51
  • 1
    $\begingroup$ Right! I hadn't properly parsed that part of the question. Addressed it now. $\endgroup$ Apr 25, 2022 at 13:44
  • 1
    $\begingroup$ That's it! I made a small edit to the answer to include the complete equation. Once the edit is approved I'll accept this as the correct answer. Thank you for your help! $\endgroup$
    – Lawton
    Apr 25, 2022 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.