Hawking radiation: a weight-loss program for black holes – another black hole conundrum and a cute connection between black holes and quantum field theory

Coming on the heels of the apparent success of Wegovy and Ozempic for obese humans, a situation that I am still very skeptical about for its long-term effects, I thought I would write a post about how Hawking radiation causes black holes to lose mass. It is subtle and not something I have seen explained satisfactorily (to me at least). It also is enlightening in that it connects two disparate fields – black holes and quantum field theory.

The simplest explanation, also the one that avoids any details, is this. What else, except a decrease in mass, could happen when a black hole emits radiation? You see a black hole in empty space, while some “radiation”, i.e., photons, small sub-atomic particles and the like are emerging from its vicinity; what could you expect, except that the black hole’s mass is leaking away through this radiation?

Not as easy as that. If you say that this radiation is emerging from the quantum vacuum around the black hole, then one could just imagine that something unexpected is happening to the quantum field around the black hole – why should that have anything to do with the black hole itself?

Let’s analyze this situation, using two fundamental tools. First, we need some Quantum Mechanics.

Quantum Mechanics is one of the two things we need to understand Black Hole weight loss. Indeed, the phenomenon is quantum mechanical. All we need for now, is this : if you have a photon with frequency $\nu$ , it carries an energy equal to $h \nu$ , where $h$ is Planck’s constant. This formula was first postulated by Planck, in a slightly different context. It was then successfully used by Einstein to explain the photoelectric effect.

This photon would have a mass equivalent to this energy, of $\frac{h \nu}{c^2}$ , where $c$ is the speed of light.

Einstein was among the foremost contributors to foundational questions in quantum mechanics. He asked really perceptive questions that (though completely answered by quantum mechanics) are still perplexing. These were such serious issues, in his mind, that he ultimately gave up on quantum mechanics as being an acceptable complete theory of nature.

The second thing one needs to understand Black Hole weight loss, at least in this simple picture, is Newton’s theory of gravity. In particular, we need two items. First, the attractive force between two masses $m_1, m_2$ separated by a distance $r$ is $F = \frac{G m_1 m_2}{r^2}$ , where $G$ is Newton’s gravitational constant, roughly $10^{-10}$ in SI units. Next, let us define the energy of a system consisting of a stationary particle (of mass $m_2$ ) that is infinitely far away from $m_1$ as $0$ . Then the energy of the system, when these two particles are brought to a distance $r$ away, would become $- \frac{G m_1 m_2}{r}$ .

Suppose we have a photon, which has a frequency $\nu$ and hence an energy $h \nu$ (mass equivalent of $\frac{h \nu}{c^2}$ ). Suppose this photon were at a distance $R$ from the center of a heavy mass $M$ – the attractive force between the photon and this mass would be $G \frac{M h \nu /c^2}{R^2}$ . The total energy of the system would be $h \nu - G \frac{M h \nu/c^2}{R}$ . The energy is just the energy of the photon plus the gravitational potential energy of the mass-photon system.

Let’s say the photon is traveling away from the black hole. As it gets further away, its energy has to decrease. Think of a ball flying away from the earth’s surface – it slows down. A photon cannot slow down, however. If its energy reduces, its frequency has to, since its energy is proportional to its frequency. At most, its frequency could get down to $0$ .

So, if the photon were to ${\it just}$ escape from the black hole, we should have

$h \nu - G \frac{M h \nu/c^2}{R} = h \nu_{\infty} = 0$

This means that $R_c = \frac{G M}{c^2}$ – if the photon leaves from this distance from the mass $M$ , it would just about escape the baneful influence of the mass. If this mass $M$ were confined to this radius, even light would not escape from inside this radius. This is what a rather interesting genius (and gentleman scientist) named John Michell discovered in the 1760s. This is the classical (Newtonian) equivalent of what we would call the Schwarzschild radius. General relativity corrects this formula to $R_S = \frac{2 G M}{c^2}$ . But let’s stick with the Newtonian formula for now.

One of the things we learn from quantum mechanics is that the quantum vacuum can produce particles out of nothing – particle/anti-particle production occurs spontaneously. Such a pair of particles would, however, be short-lived. The higher the energy of the pair of particles produced, the shorter the life-time of the pair and they would recombine and return to the vacuum.

Let’s suppose we have such a pair of particles (mass $m$ ) produced near this classical black hole radius, where the mass $M$ is stuffed inside the radius $R_c = \frac{G M}{c^2}$ . The total energy of this system is the sum of two things, namely

the energy of the pair of particles, which is $2 mc^2$ and
the gravitational energy of the system, which would be

$- G \frac{M (2 m)}{R_c} = - G \frac{M (2 m)}{G M / c^2} = - 2 m c^2$

Surprise : the sum of these two is zero.

In this Newtonian picture, pair creation can happen without bound at this boundary – it is the ultimate free lunch. You can even create pairs of particles that are as massive as the black hole without fear of creating a problem.

However, if one of these particles gets pulled into the black hole and the other works its way out of the black hole, what an observer infinitely would see is a particle of mass $m$ coming out. If energy were conserved, we should expect that there is a negative mass $m$ that went into the black hole, so its mass would drop would drop by $m$ .

This argument is cute but full of holes. First, the Schwarzschild radius is the real black hole radius and it has a pesky factor of $2$ , i.e., $R_S = \frac{2 GM}{c^2}$ . That is the real radius of the black hole.

Next, where did the negative mass $m$ come from? There were only positive mass particles created in the pair.

Also, if we really could create pairs of particles as a free lunch, what’s to prevent us from creating particles that are more massive than the black hole? Could the black hole’s mass become negative? Huh?

There is a better argument. And it gives more useful results.

Let’s go back to the pair creation that produced two masses $m$ . Their mass energy is $2 m c^2$ . The gravitational potential energy of the system with the two particles created around the Schwarzschild radius $R_S$ is equal to

$- G \frac{M (2 m)}{R_S} = - G \frac{M (2 m)}{2 G M/c^2} = - m c^2$

This means the packet has the total energy $2 m c^2 - m c ^2 = m c^2$ .

Now, the trick.

Suppose, in addition, the black hole’s mass went from $M$ to $M - m$ .

Then, we’d have a free lunch : the black hole’s mass reduces by $m$ , an extra $m$ appears to be left over from the pair creation and the gravitational energy and we have recreated the situation from before with two positive mass particles. In addition, in most ways in which the pairs are created, the particles in the pair would go off in opposite directions. Only one of the two particles would emerge and be seen far away from the black hole. The other would be eaten by the black hole.

We can now answer the question of why the masses created in this pair production process cannot be bigger than the black hole. The energy of the biggest vacuum fluctuation in a quantum field can only be the Planck energy – this is usually a cutoff above which quantum field theory is not valid. The Planck energy is $E_{Pl}= \sqrt{\frac{h c^5}{G}}$ : the corresponding Planck ${\bf mass}$ is $M_{Pl} = \sqrt{\frac{h c}{G}}$ . Therefore,the pair of particle created with mass $2 m$ could at most be as heavy as the Planck mass $M_{Pl}$ , i.e., $m = \frac{M_{Pl}}{2}$ .

We can also re-write the Schwarzschild radius $R_S$ in terms of the Planck mass and a corresponding Planck length ( $L_{Pl} = \sqrt{\frac{h G}{c^3}}$ ) as

$R_S = 2 \frac{M}{M_{Pl}} L_{Pl}$

Were the fluctuation big enough to wipe out the black hole, i.e., $M = m= \frac{M_{Pl}}{2}$ , this would mean that $R_S = L_{Pl}$ and indeed the smallest possible black hole could only be this big. Such a black hole would have surface area ${\cal A} = 4 \pi R_S^2 = 4\pi L^2_{Pl}$ . And this would be the smallest possible black hole. Interestingly, the condition that determines its size and mass comes from an argument about the largest possible energy fluctuation in a quantum field. This is a neat connection between black holes and quantum physics.

Such extremal black holes are studied in physics, though this connection is one I haven’t seen mentioned before, undoubtedly it is well-known.

The featured image was taken by the Event Horizon telescope and is reproduced under their open access CC by 4.0 terms.

Hawking radiation: a weight-loss program for black holes – another black hole conundrum and a cute connection between black holes and quantum field theory

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