## 04 Feb For supermassive black holes, this value is millions and billions of times less.

# For supermassive black holes, this value is millions and billions of times less.

An unexpected observation that significantly reduces the enumeration of options can be made in the following problem.

Problem 3. Mathematician C decided to test how smart his two colleagues A and B. He told them:- I thought of two different natural numbers, each of which is greater than 1, but less than 100. Now I will tell A their product, and B – their sum. Your task is to name the hidden numbers.

After that, With a whisper in his ear, A told the product, and B – the sum. The puzzled mathematicians started talking:A: – I cannot unequivocally determine what these numbers are.Q: – I might not have spoken. I knew that beforehand.A: – Then I know these numbers!Q: – Then I know too!

What are these numbers?

Note that the first replica of A serves only to build a dialogue, since in the informative sense it is overlapped by the first replica of B (by the way, in our problem, the analogous first replica of B also plays only an artistic role: B will not be able to determine the graph conceived by A, for any \ ( n \)). Therefore, the search for a solution begins with the analysis of the first comment of a mathematician who knows the sum.

First, note that this number should not be represented as the sum of two distinct primes. Otherwise, there would be a variant in which a mathematician who knows the work would easily determine the factors. On this basis, we can discard all even numbers. The reader may argue that the famous Goldbach conjecture that any even number greater than 4 is represented as a sum of two primes has not yet been proven. This is true. But it has been tested for all numbers that are many orders of magnitude greater than the sum of two-digit numbers. Therefore, we discard the even numbers at once. But many odd numbers can also be the sums of two primes. It is clear that these are exactly numbers, 2 larger odd primes. Therefore, we are only interested in odd numbers that are 2 more than composite ones. These are the numbers 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53 …

Yes, it turns out a bit too much! After all, on the basis of the condition, the search should be carried out until 197. But no! It turns out that we have already listed all possible amounts! Even the next odd number, 2 more than the composite number, cannot be the sum known to B. Indeed, if B knows the number 57, then he cannot rule out that A knows the number 212 = 53 · 4. But then A will be able to determine the hidden numbers. Indeed, in the alternative expansion of the number 212, three-digit factors arise. For the same reason, all numbers greater than 57 can be discarded.

So, based on the fact that each of the hidden numbers is less than 100, we were able to prove that their sum does not exceed 53! This is the promised unexpected conclusion.

Further steps, based on the analysis of the second remarks A and B, and leading to the only solution, the reader can do independently. We only note that in order to reduce the enumeration necessary in the problem, the representation of the possible numbers known to B as a sum of a prime and a power of two helps a lot. The fact is that the product of such numbers has a unique representation in the form of factors, the sum of which is odd.

An interesting modification of the argumentative essay introduction outline last problem was suggested by Max Alekseev. All conditions are the same, except for dialog A and B, which looks like this:

A: – I cannot unequivocally determine what these numbers are.Q: – I might not have spoken. I knew that beforehand.A: – And I knew in advance that you would know this in advance.Q: – But I don’t know what these numbers are.A: – But I know!

Such problems are characterized by recursion not only in the reasoning, but also in the condition. Let’s give an example of such a matryoshka problem.

Problem 4. The leader of the Math marathon came up with a problem. But, before placing it in the Marathon, he decided to test the problem and told it to his colleague.

– Former classmates Pyotr and Nikolai met at an event dedicated to the 40th anniversary of graduation from school, and got into a conversation.P: – Yes … life scattered us, for 40 years I have not seen you and have not heard anything about you. But before, they were not spilling water, they sat at the same desk. So how are you? Got a family?N: – But how! I have three handsome sons!P: – Well, you give! And how old are they?N: – I hope you still love puzzles? Then guess yourself. The sum of their ages is equal to the number of the apartment you lived in during your school years, and the product of the ages is …

… For the convenience of a colleague, the Marathon presenter wrote the required number on a piece of paper and continued:

– Peter took out a pen and plunged into calculations for a few minutes …P: – You know, these data are not enough for me.N: – Oh, yes! I forgot to add that I named the middle one after you.P: – Thank you! Now there is enough information.How old are the sons of Nikolai?

The colleague of the presenter plunged into calculations (more time-consuming than Peter from the problem). But his comment did not differ from Peter’s comment:“You know, these data are not enough for me,” he said to the Marathon host.- Oh yes! – the presenter realized the mistake. – Then let them call Peter not the middle, but the eldest son Nikolai.- Unfortunately, this will not save the task. But if Nikolai names his youngest son Peter, then the problem will have a unique solution!

What number did the host write?

And what about the graphs? For some reason, they are not frequent guests in such tasks. To the best of his ability, the author is trying to fill this gap. The results of his efforts can be found in the "Mathematical Marathon" competition, as well as in the article "Logic problems based on non-standard material".

Black holes are anomalies in the structure of space-time that contain a singularity (a point with infinite curvature). A black hole is characterized by the \ (M \) parameter, which shows how much space-time is curved around it. This parameter is conventionally called the "mass" of the black hole. This convention is due to the fact that in our usual "Newtonian" thinking, the curvature of space created by a black hole with the parameter \ (M \) and measured far enough from the singularity is similar to the curvature of space created by some object with mass \ (M \) … Further, we will say "mass" of a black hole, implying its dimensional parameter \ (M \), but it is important not to forget that this does not mean that the black hole really originated from matter with a total mass \ (M \).

Despite the fact that it is customary to speak of black holes as "massive" objects, this is not entirely correct. If a star with mass \ (M \) is compressed to a singularity, then a black hole with the parameter \ (M \) will indeed appear. But exactly the same black hole can arise without matter at all, for example, at the stage of the early expansion of the Universe. These two black holes with the same parameter, but different history of occurrence are absolutely indistinguishable from each other. This property of black holes is called the no hair theorem. It underlies the paradoxes of information loss in the Universe. But this is not our task.

There is a special spherical surface around the singularity – the event horizon. Its radius \ (r_g \) – aka the Schwarzschild radius (or gravitational radius) – depends only on the mass of the black hole. The curvature of spacetime by a black hole outside and inside the event horizon is fundamentally different.

Far above the event horizon, as already noted, space is curved as if instead of a black hole there was an ordinary massive object: if you measure the "acceleration" experienced by an object at rest in such a curved space, then we get the usual Newtonian \ (GM / R ^ 2 \), where \ (R \) is the distance from the singularity (it is assumed that \ (R \ gg r_g \)). As we approach the event horizon, we will feel that in order for the object to be at rest (hanging over the horizon), it will take more and more strong acceleration. And after crossing the event horizon, the trajectories of any particles – both massive and massless – begin and end strictly within the event horizon (this, in particular, means that even light cannot escape from under the horizon). This limitation, imposed by the very properties of space-time, is as strict and fundamental as the limitation on the speed of propagation of electromagnetic and gravitational waves.

Thus, a black hole is a surface that divides the universe into two asymmetric regions: what is inside the event horizon and what is outside it. And the size of this surface is determined by some parameter \ (M \), often conventionally called mass.

General theory of relativity (GTR) describes the interaction of curved spacetime with matter that has energy. In this theory, there is only one fundamental constant, \ (c \), which physically corresponds to the speed of propagation of electromagnetic waves and space disturbances (gravitational waves). The gravitational constant \ (G \), strictly speaking, is not part of general relativity, and comes into play only when we try to interpret the curvature of space as a force, and determine the so-called gravitational mass of objects.

## A task

Find the radius of the event horizon \ (r_g \) of a black hole with mass \ (M \) using dimensional analysis.

No signal from within the black hole’s event horizon can reach an observer outside. Therefore, it is logical to assume that the entropy ("disorder") of a black hole grows with an increase in the surface area of the event horizon: \ (S \ propto 4 \ pi r_g ^ 2 \). For any object with finite entropy and energy, you can determine the temperature (whether this corresponds to anything physically is another question). In black holes, this temperature is called the Hawking temperature and is denoted \ (T _ {\ rm H} \). Find its value for a black hole of mass \ (M \). Estimate Hawking temperatures for stellar mass black holes and supermassive black holes.

It is believed that due to the peculiarities of quantum fluctuations in the immediate vicinity of the event horizon (outside of it) in a strongly curved space-time, particle radiation (including electromagnetic radiation) can be generated from the outer surface of a black hole, which corresponds to radiation of an absolutely black body with a temperature \ (T _ {\ rm H} \). This radiation, as you might guess, is called Hawking radiation. Estimate the luminosity (in ergs per second) of a black hole due to Hawking radiation for stellar mass black holes and supermassive black holes.

Hawking radiation gradually "takes" the energy of the black hole, reducing its entropy, and, consequently, the area. This process is called "evaporation" of the black hole. How long does it take for a black hole of mass \ (M \) to evaporate? Can absolutely stable black holes exist in our Universe, the mass of which does not change with time (for example, the Universe can exist indefinitely and does not expand)? What mass do these black holes have?

## Hint 1

To estimate the Hawking temperature, we can assume that the energy of the black hole is \ (E = Mc ^ 2 \).

## Hint 2

What global process in a static Universe can “feed” black holes, constantly compensating for their evaporation?

## Decision

To display the radius of the event horizon, we have only two dimensional quantities: the parameter \ (M \) and the speed of light \ (c \). If we want to interpret the parameter \ (M \) as the mass from classical Newtonian physics, then instead of the parameter \ (M \) itself, we need to take the combination \ (GM \), where \ (G \) is the gravitational constant. So, we are looking for such degrees \ (\ alpha \) and \ (\ beta \) that the equality

\ [[\ textrm {length}] = [GM] ^ \ alpha [c] ^ \ beta, \]

where square brackets denote that the dimension of the quantity inside them is taken. It is easy to see that the only option is a combination:

\ [r_g \ sim \ frac {GM} {c ^ 2}. \]

To get an exact equality, there must be a factor of 2 in front of the fraction on the right side, which cannot be obtained by dimensional analysis. In what follows, we will omit all dimensionless coefficients.

The "temperature" of a black hole can be estimated as follows: \ (T _ {\ rm H} \ sim E / S \), where \ (S \) is its entropy, and \ (E \) is the total energy equal to \ (M c ^ 2 \). As it was said in the condition, \ (S \ propto r_g ^ 2 \), therefore, it is necessary to find such a constant \ (\ eta \) so that the value \ (k _ {\ rm B} M c ^ 2 / \ eta r_g ^ 2 \) had the dimension of energy (since \ (k _ {\ rm B} T _ {\ rm H} \) has the dimension of energy). Boltzmann’s constant \ (k _ {\ rm B} \), in fact, determines the temperature on the Kelvin scale, so we do not need it. You can simply find a certain constant \ (\ tilde {\ eta} \) for which \ (M c ^ 2 / \ tilde {\ eta} r_g ^ 2 \) has the dimension of energy (then the "temperature" of the black hole will not be expressed in Kelvin, and, conventionally, in ergs).

It is impossible to compose such a combination of numbers from the constants we have (\ (G \) and \ (c \)). For this we need one more constant – Planck’s constant, \ (\ hbar \). In fact, this simple fact means that general relativity alone is not enough to describe the thermodynamic properties of black holes – in addition, some other theory is needed that sets another constant and describes the interaction of radiation with the black hole horizon. In this case, it is quantum mechanics (or quantum field theory). How exactly this interaction occurs, and how exactly quantum mechanics and general relativity are combined in a single formalism – we do not yet know (not only within the framework of this problem, but in general). However, we come to a simple conclusion about the need to connect them (at least in the context of black holes) even by such simple "dimensional" reasoning.

It is not hard to see that \ (\ tilde {\ eta} \ sim c ^ 3 / G \ hbar \) gives us the correct dimension. It is more convenient to rewrite it as follows:

\ [k _ {\ rm B} T _ {\ rm H} \ sim M c ^ 2 \ frac {\ hbar G / c ^ 3} {r_g ^ 2} = M c ^ 2 \ left (\ frac {l_P} { r_g} \ right) ^ {2}, \]

where the value of the dimension of length \ (\ sqrt {\ hbar G / c ^ 3} \) we denoted as \ (l_P \). This parameter is also called the Planck length; its value is numerically \ (l_P \ sim 1 {,} 6 \ cdot 10 ^ {- 33} \) see.

The same formula can be rewritten in a more convenient form for calculations:

\ [k _ {\ rm B} T _ {\ rm H} \ sim \ frac {c ^ 3 \ hbar} {GM}. \]

Note that the more massive the black hole, the lower its temperature. This can be understood intuitively by remembering that "acceleration" in the classical sense near the event horizon also falls with mass \ (GM / r_g ^ 2 = c ^ 4 / GM \).

A black hole of 10 solar masses has a temperature of approximately \ (1 {,} 3 \ cdot 10 ^ {- 11} \) eV, or \ (1 {,} 5 \ cdot 10 ^ {- 7} \) K. For supermassive black holes, this value is millions and billions of times less. This, of course, completely kills all hopes of experimentally measuring this temperature for any of the black holes we know.

From the work of Hawking in the 70s (for example, S. W. Hawking, 1975. Particle creation by black holes), we know that this temperature actually corresponds to some effective "radiation" of a black hole, the energy (and mass) of which, respectively , will be carried away to her. To find the luminosity of Hawking radiation from black holes, it is enough to use the Stefan – Boltzmann law:

\ [L _ {\ rm H} \ sim \ sigma T _ {\ rm H} ^ 4 r_g ^ 2 = \ frac {c ^ 6 \ hbar} {G ^ 2 M ^ 2}. \]

Thus, as strange as it sounds, the less massive the black hole, the greater its total luminosity. For a typical black hole of stellar masses (10 solar masses), the luminosity is \ (4 \ cdot 10 ^ {- 19} \) erg / s. Such luminosity will be difficult to measure even if you fly extremely close to the horizon.

The radiation energy cannot come from nowhere, and in fact, due to Hawking radiation, the black hole will gradually lose energy, and, consequently, mass. It is easy to estimate the characteristic evaporation time of a black hole:

\ [t _ {\ rm H} \ sim \ frac {M c ^ 2} {L _ {\ rm H}} = \ frac {G ^ 2 M ^ 3} {c ^ 4 \ hbar}. \]

As you might guess, for standard black holes this time is \ (10 ^ {66} \) years or more, which is many orders of magnitude longer than the lifetime of the Universe. In other words, for ordinary black holes (and even for fantastic "dwarf" black holes with planetary mass), Hawking radiation cannot be measured technically and its effect on the black hole itself is negligible.

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