Astrophysical black holes

§ 1

Preliminary

Problems 4.5.1 – 4.5.26

4.5.1

Problem 4.5.1: Newtonian collapse time

Calculate in the frame of Newtonian mechanics the time of collapse of a uniformly distributed spherical mass with density $\rho_0$.

solution

Let there be $N$ identical pointlike masses $m$, distributed uniformly in a spherical volume of radius $R$ with density \[\rho _0 = Nm{3 \over {4\pi R_0^3 }}.\] We assume the system to be isolated, and consider the limit of ideal homogeneous system \[\lim\limits_{\substack{N \to \infty\\ m \to 0}} N m = const.\] Suppose in the same limit fluctuations are absent in the system. The mass contained within a ball $r_0 \le R_0 $ is \[M(r_0 ) = \int\limits_0^{r_0} 4\pi r^2 \rho _0 dr = \frac{4\pi }{3} r_0^3 \rho _0.\] Due to spherical symmetry the force acting on a particle of unit mass at distance $r_0$ from the center is determined only by the mass within the sphere: \[F(r_0 )= - \frac{GM(r_0 )}{r_0^2 }.\] This force is directed towards the center. Let us assume for simplicity that initial velocities of all particles are zero. The total energy in this case is equal to the potential energy and is negative. Let us use another simplifying assumption that different layers do not overlap during the collapse: a particle that is initially at $r_0$ will be attracted always by the same mass $M(r_{0})$ with force $ - GM(r_0 )/r^{2}(t)$ depending on time, so its equation of motion is \[\ddot r(t) =-\frac{GM(r_{0})}{r^2 (t)}.\] The first integral of this equation (energy conservation law) is \[\Big(\frac{dr}{dt}\Big)^2 = 2GM(r_0 ) \Big(\frac{1}{r}-\frac{1}{r_0}\Big).\] The solution in parametric form is \begin{align*} r(\tau) &= {{r_0 } \over 2}(1 + \cos \tau),\\ t(\tau ) &= \sqrt{\frac{3}{32\pi G\rho _0}}\;\cdot (\tau + \sin \tau ),\\ r(\tau = 0) = r_0 &\Rightarrow\; t(\tau = 0) = 0,\\ r(\tau = \pi) = 0 &\Rightarrow\; t(\tau = \pi ) =\sqrt {\frac{3\pi}{32G\rho _0}}\;\equiv t_{col}. \end{align*} Evolution takes place from radius $r_{0}$ at $t=0$ to zero radius (collapse) at $t = t_{col} $. Further dynamics in the frame of this primitive model will be periodic.

4.5.2

Problem 4.5.2: Enter white dwarfs: why not green giants?

Why are stars of a certain type called "white dwarfs"?

solution

White dwarfs are the product of evolution of stars of small masses. They belong to the class of so-called compact objects and are not stars in the strict sense, as there are no thermonuclear reactions going on anymore in their central parts. The masses of such objects are typically comparable with that of the Sun $(2\cdot10^{30} \mbox{kg})$; radii of the order of that of Earth (6400 km) and densities of the order of $10^6 \mbox{g/cm}^3$. The title "white dwarfs" is due to small sizes (on the scale of typical stellar dimensions) and white color of the first objects of this type to be discovered, which had high temperatures.

"I was visiting my friend and generous benefactor, Prof. Edward C. Pickering. With characteristic kindness, he had volunteered to have the spectra observed for all the stars—including comparison stars---which had been observed in the observations for stellar parallax which Hinks and I made at Cambridge, and I discussed. This piece of apparently routine work proved very fruitful---it led to the discovery that all the stars of very faint absolute magnitude were of spectral class M. In conversation on this subject (as I recall it), I asked Pickering about certain other faint stars, not on my list, mentioning in particular 40 Eridani B. Characteristically, he sent a note to the Observatory office and before long the answer came (I think from Mrs Fleming) that the spectrum of this star was A. I knew enough about it, even in these paleozoic days, to realize at once that there was an extreme inconsistency between what we would then have called "possible" values of the surface brightness and density. I must have shown that I was not only puzzled but crestfallen, at this exception to what looked like a very pretty rule of stellar characteristics; but Pickering smiled upon me, and said: "It is just these exceptions that lead to an advance in our knowledge", and so the white dwarfs entered the realm of study!"

Henry Norris Russell

Taken from here

4.5.3

Problem 4.5.3: Nature of WDs

What is the physical reason for the stopping of thermonuclear reactions in the stars of white dwarf type?

solution

Gravitational contraction in white dwarfs is opposed by the pressure of degenerate electron gas. This pressure does not depend on temperature, so for a white dwarf the loss of energy due to radiation does not lead to contraction of its core. Thus there is no production of thermal energy from the gravitational one. The temperature at the core of a degenerate white dwarf does not increase, and this leads to interruption of the chain of thermonuclear reactions.

4.5.4

Problem 4.5.4: Averages

Estimate the radius and mass of a white dwarf.

solution

The condition that gravitational pressure is compensated by the pressure of degenerate electron gas has the form \[P_G (R) + P_{e}(R) = 0.\] Let us consider first the gravitational part: \[P_G ( R) = \frac{F_G }{4\pi R^2 } =- \frac{dU_G / dR}{4\pi R^2} = - \frac{3GM^2 }{20\pi R^4 },\] where we have taken into account that gravitational energy of a spherically symmetric body is $U_G =-\frac{3}{5}\frac{GM^2 }{R}$. The energy of degenerate electron gas is known to be \begin{align} E& = \frac{3}{10}\left(\frac{6\pi^2}{g}\right)^{2/3}\frac{\hbar^2}{m}N\left(N \over V \right)^{2/3};\\ P &= - \frac{\partial E}{\partial V} = \frac{1}{5}\left(\frac{6\pi ^2}{g}\right)^{2/3}\frac{\hbar ^2}{m}\left(\frac{N}{V}\right)^{5/3}; \end{align} where $g=2$ is the degeneracy of the ground state with respect to spin's direction. From the condition of equilibrium then we obtain \[R_{eq} = \left(\frac{9\pi }{4}\right)^{2/3} \frac{\hbar ^2 }{G} \frac{N^{5/3}}{mM^2 }.\] As we are considering a model of white dwarf, we can assume that its mass is mainly due to nucleons (protons and neutrons). Their number is the same as that of electrons, so $M = N(m_p+m_n)\approx 2m_p$, therefore \[R_{WD} = \left( \frac{9\pi }4 \right)^{2/3} \frac{\hbar ^2 }{Gm_e} \frac 1 {m_p^{5/3} M^{1/3} }.\] Let us estimate a white dwarf's density. Evidently, by the order of magnitude \[\rho _{WD} = {{2m_p } \over {{4 \over 3}\pi \lambda _{k,e}^3 }} = \frac{3}{2}\frac{m_e^3 c^3 }{8\pi ^4 \hbar ^3 }m_p.\] On the other hand, \[\rho _{WD} = \frac{M}{\frac{4}{3\pi}R_{WD}^3}.\] Comparing the two expressions, we get \begin{align} M_{WD} &= {1 \over {\sqrt {2\pi } }}{9 \over 32}{{M_{Pl}^3 } \over {m_p^2 }} \approx 0.22M_ \odot;\\ R_{WD} & = \left({9 \over \sqrt {2\pi } }\right)^{1/3}\frac{M_{Pl}}{m_p }\lambda _{k,e} \approx 4.8 \cdot 10^4 \mbox{\it km}. \end{align}

4.5.5

Problem 4.5.5: Typical example

What is the average density of a white dwarf of one solar mass, luminosity one thousandth of solar luminosity and surface temperature twice that of the Sun?

solution

In accordance with the Stefan--Boltzmann law, the luminosity of a star with radius $R$ and temperature $T$ is \[L = 4\pi R^2 \sigma T^4 = L_ \odot \left( {{R \over {R_ \odot }}} \right)^2 \left( {{T \over {T_ \odot }}} \right)^4.\] Then the white dwarf's radius is \[ R = R_ \odot \left( {{L \over {L_ \odot }}} \right)^{1/2} \left( {{T \over {T_ \odot }}} \right)^{ - 2} = 10^{ - 3/2} 2^{ - 2} R_ \odot \approx 0.008R_ \odot \approx 5.6 \cdot 10^6\mathit{m},\] and its density \[\rho = \rho _ \odot \left( {{R \over {R_ \odot }}} \right)^{ - 3} \approx 1.4 \cdot 10^3 \times 0.008^{ - 3} \approx 2.7 \cdot 10^9 \mathit{kg/m^3}.\]

4.5.6

Problem 4.5.6: Explosion!

Explain the mechanism of explosion of massive enough white dwarfs, with masses close to the Chandrasekhar limit.

solution

The pressure of degenerate electron gas does not depend on temperature. Therefore a temperature fluctuation leading to increase of thermonuclear reaction's rate will not be compensated by cooling due to expansion of matter. Unbounded rise of temperature leads to the white dwarf's explosion. For stars far away from the Chandrasekhar limit the cooling of hot gas, which produces work against the gravitational forces, ensures the damping of temperature fluctuations and therefore stable equilibrium.

4.5.7

Problem 4.5.7: No hydrogen in type I SNe.

Thermonuclear explosions of white dwarfs with masses close to the Chandrasekhar limit lead to the phenomenon of supernova explosions of type I. Those have lines of helium and other relatively heavy elements in the spectrum, but no hydrogen lines. Why is that?

solution

Whatever the history of the white dwarf (WD) reaching the Chandrasekhar limit, its composition is extremely poor in $H$, as a WD is formed when hydrogen and essentially most of $He$ is burnt into heavier elements within most of the star. If added due to accretion in any noticeable amount, $H$ and $He$ also burn into $C$, $O$ and heavier elements. Say, a shell of $H$ on the surface of a WD (say, accreted from a companion) would ignite when its mass reaches $\sim10^{-5}M_{\odot}$, which is the mechanism of a classical nova explosion. So, evolutionary, the composition of the SNIa progenitor is carbon, oxygen, and heavier elements. Lighter elements like $H$ and $He$ are not formed in any significant amounts during the explosion either -- nuclear reactions are dominated by the fastest alpha chain ($C+C\to Mg$ or $\to Ne+He$, $He$ is immediately consumed by fusion with $C$ and $O$, $\ldots$), mostly heavier elements up to ${}^{56}Ni$ are formed. These heavier elements are what is observed in the SNIa spectra.

Note on pre-explosion composition: If the WD was formed with large mass initially, before accretion added more mass, it has mostly $O+Ne+Mg$ composition -- these are the final products of nuclear reactions in the star of that mass. When such a WD reaches the Chandrasekhar mass and nuclear reactions ignite at its center, it collapses into a neutron star due to efficient neutronization and thus the results is not an SNIa. When the WD originally had somewhat lower mass, it is composed mostly of carbon and oxygen, and it is these white dwarfs that lead to normal SNIa.

4.5.8

Problem 4.5.8: SNe of type II

A supernova explosion of type II is related to the gravitational collapse of a neutron star. There are powerful hydrogen lines in their spectrum. Why?

solution

Supernovae of other than Ia types are the observed result of degenerate core collapse in the heavier stars (above the mass when a stable WD can be the end product of evolution). The lightest of these core-collapse supernovae progenitors have $O-Ne-Mg$ cores, the heavier ones have iron cores. The progenitors with masses $\geq 25 M_{\odot}$ have lost all of their $H$ envelope by this stage, so the explosion (due to $Fe$ core collapse) of such heavier progenitors leads to no $H$ in the spectra (supernovae of type Ib and Ic). Progenitors lighter than that have their $H$ envelope retained, or lost into massive wind right before the explosion (so a large mass of $H$ is still near the explosion site when the latter occurs) -- thus the $H$ lines are observed in the spectra of such supernovae -- these are classified as type II then.

4.5.9

Problem 4.5.9: Neutron star enters

Estimate the radius and mass of a neutron star.

solution

Using the results of problem Astrophysical_black_holes#BlackHole91 and assuming that all the mass of neutron star is due to neutrons $M_{NS}=Nm_n$, we get \[R_{NS} = \left(\frac 9 {8\sqrt {2\pi } }\right)^{1/3} \frac{M_{Pl}}{m_n }\lambda _{k,n} \approx 13.2\mathit{km}.\] Carrying out further calculations analogous to those of problem Astrophysical_black_holes#BlackHole91, we get \[M_{NS} = {9 \over 8}{1 \over {\sqrt {2\pi } }} {{M_{Pl}^3 } \over {m_n^2 }}\approx 0.7 M_ \odot.\] Note that the question discussed here is the equilibrium configuration of a neutron star. The theory of a neutron star's equilibrium was built by Landau, Oppenheimer and Volkov in 1938.

4.5.10

Problem 4.5.10: Magnetic fields in NS

Why do neutron stars have to possess strong magnetic fields?

solution

Neutron stars are formed as the result of catastrophic contraction (collapse) of ordinary stars, which have exhausted their supply of thermonuclear energy. Stellar matter is hot plasma with high conductivity. The power lines of magnetic field in such plasma are attached to particles and move along with matter (this is called the freezing-in of the magnetic field). When a star collapses, the total number of power lines permeating the star is conserved. Therefore the number of power lines per unit area of the section increases, i.e. the field's intensity grows. Evidently, it grows inversely proportional to the star's radius squared. In this sense the magnetic field of the star increases. If we, however, measure the field's intensity at some fixed distance from the star, we will detect that it weakens instead. This can be understood if we recall that the field's intensity at a given distance from a system of currents is proportional to the system's magnetic dipole moment, which is in our case the product of magnetic flux permeating the star (which is conserved) and its radius.

4.5.11

Problem 4.5.11: Redshift

Find the maximum redshift of a spectral line emitted from the surface of a neutron star.

solution

$z=G\frac{M}{R{{c}^{2}}}\simeq 0.1$

4.5.12

Problem 4.5.12: Gravitational radius of the Universe

What is the gravitational radius of the Universe? Compare it with the size of the observable Universe.

solution

Consider the Schwarzschild radius $r_{g\odot}=3\;km$ and the physical radius $R_\odot =8\times 10^{5}\;km$ of the Sun. Their ratio is $(r_{g}/R)_{\odot}\approx 2.7\times 10^{5}$. One can readily check that for the Earth or for the Milky Way the analogous ratio is likewise much larger than $1$. Such objects are nowhere close to being black holes. Now consider the visible universe (VU), with mass $M_{VU}=10^{23}M_{\odot}$. It has $r_{gVU}=30\;Gly$ and $R_{VU}=48\;Gly$. Hence $(R/r_{g})_{UV}\approx 1.6$. The visible universe, within which we all live, is close to being a black hole. A formal computation, in which we do not especially care about strict definitions of $R_{VU}$ and $M_{VU}$, confirms this estimate (here $c=1$): \begin{align*} &{r_{gVU}} = 2G{M_{VU}},\qquad {R_{VU}} = {H^{ - 1}};\\ &{M_{VU}} = \frac{4\pi }{3}R_{VU}^3\rho ;\\ &{H^2} = \frac{8\pi G}{3}\rho \quad\Rightarrow\quad \rho = \frac{3{H^2}}{8\pi G};\\ &{M_{VU}} = \frac{4\pi }{3}R_{VU}^3\frac{3{H^2}}{8\pi G} = \frac{R_{VU}}{2G};\\ &{R_{VU}} = 2G{M_{VU}} = {r_{gVU}}. \end{align*}

4.5.13

Problem 4.5.13: Salpeter time

What is the time (the Salpeter time) needed for a black hole, radiating at its Eddington limit, to radiate away all of its mass?

solution

An important characteristic of accretion rate on a black hole of mass $M$ is the Eddington luminosity $L_{E}$. When luminosity reaches this value, radiation pressure on free electrons balances out their gravitational attraction to the center. As the radiation pressure depends on distance in the same way as gravitational force (in the nonrelativistic theory), $L_E$ does not depend on distance but is the function of a black hole's mass \[{L_E} = \frac{{4\pi GM{m_p}c}}{{{\sigma _T}}} \approx 1.3 \times {10^{38}} \frac{M}{M_{\odot}}\;erg/s. \] Here $\sigma$ is the Thompson cross section. If $L>L_{E}$, the source cannot sustain stable spherically symmetric accretion (though the picture may be quite different in the presence of the magnetic field). Then the Salpeter time is \[{t_S} = \frac{{{\sigma _T}c}}{{4\pi G{m_p}}} \approx 4.5 \times {10^7}yr.\]

4.5.14

Problem 4.5.14: AGN typical size

The time scales of radiation variability of active galactic nuclei (AGNs) are from several days to several years. Estimate the linear sizes of AGNs.

solution

As $L \le c{t_0}$, we have \[L \le 3 \times ({10^{15}} \div {10^{17}})\;cm \sim 1\;lyr.\] Note that typical dimensions of a galaxy is $\sim 10^{5}\;lyr$.

4.5.15

Problem 4.5.15: Rotation of SMBHs

What mechanisms can be responsible for the supermassive black hole (SMBH) in the center of a galaxy to acquire angular momentum?

solution

The SMBH at the center of a galaxy can acquire angular momentum due to continuous accretion or merger events.

4.5.16

Problem 4.5.16: The mass of Milky Way's SMBH

The Galactic Center is so "close" to us, that one can discern individual stars there and examine in detail their movement. Thus, observations carried out in 1992-2002 allowed one to reconstruct the orbit of motion of one of the stars (S2) around the hypothetical SMBH at the galactic center of the Milky Way. The parameters of the orbit are: period $15.2$ years, maximum distance from the black hole $120$ a.u., eccentricity $0.87$. Using this data, estimate the mass of the black hole.

solution

According to the third Kepler law, \[M = \frac{a^3}{T^2}\frac{4\pi^2}{G}.\] Here $a$ is the major semiaxis of the ellipse, related to the minimal distance to the black hole though \[a = \frac{1 + e}{1 - e^2}{r_{\min }}.\] Then \[{M_{SMBH}} \approx 3 \times {10^6}{M_ \odot }.\]

4.5.17

Problem 4.5.17: BHs are not necessarily "very dense"

Using the results of the previous problem, determine the density of the SMBH at the Galactic Center.

solution

As $r_{g}\sim M$, $\rho\sim M^{-2}$ and expressing $\rho$ through the density of a solar mass black hole, we get \[\rho_{BH} = \frac{M}{{\tfrac{4\pi}{3} r_g^3}} =\rho_{BH\odot} \left( {\frac{{{M_ \odot }}}{M}} \right)^2 \approx 2 \times {10^{16}} \left( {\frac{{{M_ \odot }}}{M}} \right)^2 \frac{g}{cm^3} \approx 2\;\frac{kg}{cm^3}.\]

4.5.18

Problem 4.5.18: BH in equilibrium with gas.

Show that for a black hole of mass $M$ the temperature of the surrounding hot gas in thermal equilibrium is proportional to \[T\sim {{M}^{-1/4}}.\]

solution

Assume the luminosity of the galactic nuclei (SMBH) is close to the Eddington limit. Then energy radiated from unit area per unit time is \[J = \frac{{{L_E}}}{{{S_{BH}}}} =\frac{{4\pi GM{m_p}c}}{{{\sigma _T}}} \frac{1}{4\pi r_g^2 } = \frac{1}{4}\frac{{{m_p}}}{M} \frac{{{c^5}}}{{G{\sigma _T}}}.\] On the other hand, the luminosity of a black body is \[J = \sigma {T^4},\] where $\sigma$ is the Stefan--Boltzmann constant, \[\sigma = \frac{{{\pi ^2}k_B^4}}{{60{\hbar ^3}{c^2}}} = 5.67 \times {10^{ - 5}}\frac{g}{s^3 {grad}^4}.\] Then \[T = M^{ - 1/4} \Big(\frac{m_p c^5}{4\sigma\sigma_{T}G}\Big)^{1/4}.\]

4.5.19

Problem 4.5.19: Why X-rays

Show that luminosity of a compact object (neutron star or black hole) of several solar masses is mostly realized in the X-rays.

solution

The accreting matter is accelerated to relativistic velocities, transforming the enormous gravitational energy in the vicinity of the compact object into the kinetic one. Assuming all of this energy eventually transforms into radiation, and using the result of the previous problem, one can calculate that for a compact object of several solar masses $T\sim 10^{7}\;K$. The radiation spectral density has the maximum at $\omega=\omega_{max}$ defined from $\hbar\omega_{max}/T \approx 2.822$. The corresponding frequency falls in the range of X-rays $10^{16}\div 10^{19}\;Hz$.

4.5.20

Problem 4.5.20: Masses of AGNs

In order to remain bound while subject to the rebound from gigantic radiative power, AGNs should have masses $M>{{10}^{6}}{{M}_{\odot }}$. Make estimates.

solution

nothing here yet

4.5.21

Problem 4.5.21: Another estimate

AGNs remain active for more than tens of millions of years. They must have tremendous masses to maintain the luminosity \[L\sim {{10}^{47}}\mathit{erg/sec}\] during such periods. Make estimates for the mass of an AGN.

solution

no hints here either

4.5.22

Problem 4.5.22: Energy release at BH merger.

What maximum energy can be released at the merger of two black holes with masses ${{M}_{1}}={{M}_{2}}=\frac{M}{2}$?

solution

When two black holes merge, the area of the resulting black hole has to be greater than the sum of the areas of the two colliding black holes. This enables us to estimate the released energy from below: \[4\pi {\left( {2{M_f}} \right)^2} \geq 4\pi {\left[ {2\left( {M/2} \right)} \right]^2} \times 2 \quad\Rightarrow\quad M_{f} \geq M/\sqrt 2.\]

4.5.23

Problem 4.5.23: BH do not multiply by fission

Show that it is impossible to divide a black hole into two black holes.

solution

todo

4.5.24

Problem 4.5.24: Entropy paradox

J. Wheeler noticed that in the frame of classical theory of gravity the existence of black holes itself contradicts the law of entropy's increase. Why is that?

solution

Imagine a black hole consuming a body which contains some amount of entropy. Evidently, the observer watching this process will detect a decrease of total entropy in the region available for his observation.

4.5.25

Problem 4.5.25: Entropy of the interior?

What is the reason we cannot attribute the observed entropy's decrease (see the previous problem) to the interior of the black hole?

solution

The attempt to contribute the entropy to the black hole's interior is unsatisfactory due to the fact that, according to the "no-hair theorem(s)", there are no experiments that would help an external observer determine the value of entropy that has been consumed by the black hole.

4.5.26

Problem 4.5.26: Smarr formula

Find the surface area of a stationary black hole as a function of its parameters: mass, angular momentum and charge.

solution

Using the explicit form of the Kerr-Newman metric, in the same way as was computed for the Kerr case (problem Kerr_black_hole#BlackHole61), one eventually obtains \[A = 4\pi \left( {2{M^2} - {Q^2} + 2M\sqrt {{M^2} - {Q^2} - {J^2}/{M^2}} } \right).\]

§ 2

Quantum effects

Problems 4.5.27 – 4.5.30

4.5.27

Problem 4.5.27: Densities of realistic BHs.

Estimate the maximum density of an astrophysical black hole, taking into account that black holes with masses $M<{{10}^{15}}g$ would not have lived to our time due to the quantum mechanism of evaporation.

solution

Maximum density corresponds to the minimum possible mass: \[\rho _{\max } = \frac{M}{\tfrac{4}{3}\pi r_g^3} \approx 2 \times {10}^{16} \left(\frac{M_\odot}{M_{\min}}\right)^2 \frac{g}{{cm}^3} \sim {10^{52}}\frac{\mathit{g}}{\mathit{cm^3}}.\]

4.5.28

Problem 4.5.28: BH lifetime

Determine the lifetime of a black hole with respect to thermal radiation.

solution

Hawking has shown that a black hole radiates as a black body with surface area $A$ and temperature $T_H$ given by \[{T_H} = \frac{{\hbar {c^3}}}{{8\pi GM{k_B}}}, \quad A = 4\pi r_g^2.\] Consequently \[\frac{dM}{dt} = \frac{\sigma T_H^4 \times 4\pi r_g^2}{c^2} = \frac{1}{512}\frac{\hbar c^4}{G^2} \frac{1}{M^2}.\] For simplicity we have omitted the factor depending on the number of states of particles that are radiated. Then for the lifetime of black hole we obtain \[{T_H} \approx 512\pi \frac{G^2}{\hbar c^4}{M^3}.\]

4.5.29

Problem 4.5.29: Typical temperatures

Determine the temperature of a black hole (Hawking temperature) of one solar mass, and the temperature of the supermassive black hole at the center of our Galaxy.

solution

Using the general formula from the previous problem, \[{T_{ \odot H}} \approx 6 \times {10^{ - 7}}K,\quad {T_{HG}} \approx 2 \times {10^{ - 13}}K.\]

4.5.30

Problem 4.5.30: Particle production limit

Particles and antiparticles of given mass $m$ (neutrinos, electrons and so on) can be emitted only if the mass $M$ of the black hole is less than some critical mass ${M}_{cr}$. Estimate the critical mass of a black hole $M_{cr}(m)$.

solution

A particle can be emitted only if its reduced Compton wavelength is greater than the Schwarzschild radius of the black hole. Hence \[\frac{\hbar }{mc} > \frac{2MG}{c^2} \quad\Rightarrow\quad M_{cr} = \frac{M_{Pl}^2}{2m}.\] For instance, thermal emission of electrons and positrons is possible only when $M < {10^{16}}\;kg$ (which is $16$ orders of magnitude smaller than the mass of the Sun).