Popular science physics and mathematics
quantum
2
1971
magazine
Academy
Sciences of the USSR
And
Academy of Pedagogical
Sciences of the USSR

In the room:

Paradox of S. I. Vavilov 1 V. A. Fabrikin
Two games with matches 4 I. M. Yael
Dialogue about temperature 11 M. Ya. Abzel

Math circle

Comparison of different averages of two positive numbers 20 Z. A. Skopets

Problem “Quenta”

Problems 24
Solutions to problems M21—M25; F29—F34 26
N. B. Vasiliev, I. Sh. SLOBODECHY

Applicant’s workshop

Two dozen progression problems 37 A. G. Mordkovich
Kinematika and communications 44 S. A. Bellev
Entrance exams in mathematics to the University of Other Nations 47 V. V. Ryzhkov

Reviews. Bibliography

Flight of Fire Arrows 52 Yu. M. Brook

Information

XII International Mathematical Olympiad for Schoolchildren 54 I. S. Petrakov
IV International Physics Olympiad for Schoolchildren 56 Yu. M. Bruk, N. A. Mini

Answers, directions, solutions 60

Crossword - 4th page of the cover

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PARADOX OF S. I. VAVILOV

In geometric optics, the concept of a beam of parallel light rays having a finite cross section is often and fruitfully used. Moreover, even in the theory of such a wave phenomenon as integration, in many cases it is permissible to use this concept.

In many cases, but not in all. In the book “Microstructure of Light” by the famous Soviet physicist S.I. Vavilov, a very instructive optical paradox in this sense is analyzed.

Let us briefly recall what the energy of the integration picture looks like. Integration, as we know, is the addition of oscillations. In our case, the directions of the electric and magnetic fields fluctuate in the waves. These directions at each point in space at each moment of time determine the energy of the electromagnetic field. An electromagnetic wave transfers energy, and the concept of energy flux density can be introduced. This is what we call quantities equal to the amount of field energy flowing per unit time through a unit surface.

Each wave is also characterized by a phase. If, when adding two light beams, the difference in their phases remains the same all the time, we say that we are dealing with correct beams.

In the integration picture, resulting from the addition of two correct beams, a redistribution of light energy occurs. In the light stripes the energy is greater than the sum of the energies of the folded beams; in dark stripes, on the contrary, it is less. The excess energy in the light stripes is just compensated by its lack in the dark ones. The total energy, distributed over the entire integration picture, is exactly equal to the sum of the energies of the two integrating beams.

figure 1 shows the dependence of the energy flux density in the integration picture on the displacement across the screen on which it is observed. The picture was obtained by adding two correct light beams of equal energy. Horizon-

Fig. 1.
1 Quantum No. 2

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Fig. 2.

The tall dotted line depicts the sum of the energy flux densities at the added points. The parts of the curve going above this straight line correspond to light interference fringes, and the parts of the curve lying under the dotted line correspond to dark ones. The total energy distributed in the interference cartene is represented by the plane under the curve. This area is exactly equal to the area under the dotted line. The requirements of the strict accountant of nature - the law of conservation of energy - are strictly fulfilled.

Let us now move on to the paradox of S.I. Vavilov. Imagine two completely identical correct sharply limited light points of width $a$, intersecting at a small angle $\alpha$. In the region $ABCD$ interference will occur (Fig. 2).

To observe the interference pattern, you can install a screen perpendicular to the drawing plane, passing through points $A$ and $C$. The interference pattern will consist of rectilinear alternating light and dark stripes filling the screen from point $A$ to point $C$ (Fig. 3).

The distribution of the energy flux density in the interference carten corresponds to the graph on figure 1. If the common beams have the same initial phases of light oscillations, then the difference in the phases of the light waves of the first and second beams at points lying on the straight line $BD$ will be equal to zero. It thus corresponds to the middle of the central light stripe. In the middle of the adjacent, dark stripe, the phase difference should be equal to $π$, in other words, the light vibrations in the common beams should be in antiphase. The phase difference $\Delta φ$ is equal to the path difference $\Delta L$ of both waves to a given location, divided by the wavelength and multiplied by $2π$:

$$\Delta φ = 2\pi \frac{\Delta L}{\lambda}.$$ (1)

An increase in the length of the beam traversed by the wave by $\lambda$ corresponds to a phase delay of $2π$. From formula (1) it follows that in the middle of the dark stripe closest to the center of the interference carten $\Delta L$ should be equal to $\frac{\lambda}{2}$.

Let’s calculate the path difference at point $A$. A parallel beam can be considered as a flow of plane waves perpendicular to the directions of light rays.

Let us draw through the point $D$ (Fig. 2) one of the wave surfaces (surfaces of equal phase) of the first beam. The points $D$ and $A'$ lie on this wave surface.

The paths taken by the bundles to the point $D$ are identical. In order to get to point $A$, the wave surface of the first beam passing through point $D$ must

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shift to the segment $A'A$, and the wave surface of the second beam, which previously also passed through the point $D$, must shift to the segment $DA$. As a result, a stroke difference will arise

$$\Delta L = DA - AA' = a \left( \frac{1}{\sin \alpha} - \frac{1}{\tg \alpha} \right) = \frac{2a \sin^2 \frac{\alpha}{2}}{\sin \alpha}.$$

(2)

It is clear that the same path difference, but with the opposite sign, will be at point $C$. Let’s now start decreasing the angle $\alpha$. If $\alpha$ is small enough, we can make $\Delta L$ equal to $\frac{\lambda}{4}$. Then all areas $AC$ will be filled with one bright interference fringe. Consequently, the energy will exceed the sum of the energies of two intersecting beams. There is no compensation due to the formation of dark stripes, since they are absent altogether!

It is also possible to obtain, so to speak, a negative result by forcing beams with an initial phase difference equal to $\pi$ to intersect. Then the region $AC$ will be filled with a dark interference fringe. In the first case, it is unclear where the additional energy comes from, in the second, it is unclear where the energy disappears.

Both cases clearly contradict the law of conservation of energy. Obviously, there is some defect in our reasoning that leads to the presence of one of the basic laws of nature. To understand what’s going on here, let’s write down formula (2) for the indicated case, when $\Delta L = \frac{\lambda}{4}$, and use the smallness of the angle $\alpha$ ($\sin \alpha \approx \alpha$ for small $\alpha$). We will show below that the angle $\alpha$ is indeed very small. Then we get

$$\frac{\lambda}{4} = \frac{2a \left( \frac{\alpha}{2} \right)^2}{\alpha} = \frac{a\alpha}{2},$$

(3)

or

$$\alpha = 0,5 \frac{\lambda}{a}.$$

(4)

Let’s take $a=1$ cm, $\lambda=5 \cdot 10^{-5}$ cm, then $\alpha=2,5 \times 10^{-5}$ radian. From (4) we see that problems with the law of conservation of energy arise when the angle between the beams is of the order of the relative wavelength to the cross section of the beam.

The solution to the paradox lies in the fact that at such small angles it is no longer possible to use the understandable ideal of a paradel beam of a finite cross-section.

Any attempt to implement such bundles will fail. By limiting the size of the beam, the phenomenon of diffraction necessarily leads to its transformation into a divergent beam. The beam divergence angle is determined precisely by formula (4). In this case, naturally, the angle at which the beams intersect can be determined only with an accuracy of the order of the divergence angle of the intersecting beams.

If diffraction had not yet been discovered, we, on the basis of the law of conservation of energy and formula (4), should not only agree on its existence, but also point out the basic pattern governing the magnitude of the diffraction angle. This is a good example of how the conservation law in physics can always serve as a reliable guiding star.

Taking light beams, we will never, of course, get a contradiction with the law of conservation of energy. In interference experiments of this type, it should always be reduced to the spatial redistribution of the energy flow.

There are, however, interference experiments where even more subtle energy paradoxes will arise, but that, as they say, is another story.

4

TWO GAMES WITH MATCHES

I. M. Yaglom

In this article we tell you about games “him”
and “Izyanshizians”. Both “nim” and “izianshiz” are Chinese folk games; “Izyanshiztsy” translated means “selecting stones” (of course, replacing stones with matches in games does not change anything in its essence). Subsequently, these games penetrated into Europe; At one time the game “nim” was popular in Western Europe.

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Here are the rules of this game.

Game “him”
There are three piles of matches on the table. Two tserayatsy take turns taking matches from these piles, and at one time the tserayatsy can take any number of matches from one (necessarily only from one!) pile. We will endure and quote whoever takes the last back.

Game “Izyanshiztsy”
There are two piles of matches on the table. Two tserayatsy take turns taking matches from these piles, and at one time the tserayatsy can take any number of matches from one pile, or an equal number (but any number!) of matches from both piles. We will endure and quote whoever takes the last back.

The task is to determine at what initial positions (that is, the number of matches in each of the piles) the beginner can win, and at which not, and indicate the method of correct (win-win) winning.

It is very convenient for practical players to play it or izyanshiztsy at the blackboard - without any matches, but with chalk and a tripod in hand. On the board you can draw, say, three rows of fields ending at the right edge of the board, and three checkers - one in each row of fields (Fig. 1). Each player on his turn destroys one of the checkers and “replays” it any number of fields to the right (here we are talking about playing with him); The winner is the one who makes the last move. You can play Izyanshizi in the same way - only now the game board contains two rows

Fig. 1.

Fig. 2.

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Unfortunately, here too there is no specific law for compiling this table.

Is it possible to write Tables I and II differently?

We were unable to understand how Tables I and II were compiled, formed in the first case by the crown, and in the second by two lines of numbers. Let us now think about the possible reasons for the failure that befell us.

We have already said that tables I and II are numerical tables; they are formed by a series of numbers. Naturally, we wrote down the numbers appearing in these tables in the usual way, using the generally accepted decimal number system; so, for example, the numbers 24 and 39 that make up the last column of Table II will read: “2 tens and 4 units” and “3 tens and 9 units.” However, the decimal number system itself, the origin of which is associated with the random circumstance that a person has 10 fingers on his hands, is in no way connected with the games we are considering; Therefore, it seems reductive to attempt to write our tables differently.

The further content of this article is related to the article “Number Systems” («Kvant» No. 6). We will need the following from this article.

There are many systems for writing positive numbers using a small number of symbols (digits). Positional number systems (more precisely, positional systems for writing numbers) are constructed as follows. A certain basis, or system of key numbers, is specified, which represents an infinitely possible sequence of integers:

$$1 = q_0 < q_1 < q_2 < q_3 \dots$$

After this, each number $N$ is represented as a sum of numbers of our basis, taken with certain integer coefficients:

$$N = a_m q_m + a_{m-1} q_{m-1} + \dots + a_2 q_2 + a_1 q_1 + a_0 q_0;$$

this expression is written like this:

$$N = a_m a_{m-1} \dots a_2 a_1 a_0,$$

where $a_m, a_{m-1}, \dots$ are the “digits” of the number $N$.

For example, if the base number system is:

$$1, 2, 3, 5, 7, 11, 13, 17, 19, 23, \dots$$

then the number 27 is written as follows:

$$27 = 1 \cdot 23 + 0 \cdot 19 + 0 \cdot 17 + 0 \cdot 13 + 0 \cdot 11 + 0 \cdot 7 + 0 \cdot 5 + 1 \cdot 3 + 0 \cdot 2 + 1 \cdot 1 = \text{«1000000101»}.$$

In this case, the $k$th digit $a_k$ in the notation of any number $N$ in the number system with the basis (*) will always be less than the relative $q_k + 1/q_k$.

If the basis of the number system has the form:

$$1 = q^0, q = q^1, q^2, q^3, q^4, \dots,$$

where $q$ is some integer, then the number system is called a “q-digit”, in which all numbers are represented by sums of the power of the number $q$ and all digits in the notation of the number $N$ are less than $q$ (less than 10 in the case of the generally accepted decimal number system).

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The simplest of all systems for writing whole numbers is the double number system, in which all numbers are represented by sums of powers of 2. If we use this number system, then our tables I and II will take a completely different form (numbers in the double number system will be written in red numbers):

Table Ia

| №№ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— |
| a | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 10 | 0 | 11 | 10 | 1 | 0 | 0 | 1 |
| b | 1 | 10 | 10 | 11 | 100 | 100 | 101 | 100 | 110 | 100 | 101 | 110 | 111 | 1000 | 1000 |
| c | 1 | 10 | 11 | 11 | 100 | 101 | 101 | 110 | 110 | 111 | 111 | 111 | 111 | 1000 | 1001 |

Table IIa

| №№ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :— | :— | :— | :— | :— | :— | :— | :— | :— |
| a | 1 | 11 | 100 | 110 | 1000 | 1001 | 1011 | 1011 |
| b | 10 | 101 | 111 | 1010 | 1101 | 1111 | 1111 | 1010 |

| №№ | 110 | 1000 | 10001 | 10011 | 10101 | 1101 | 1111 | 10101 | 1010 | 1010 | 1010 | 1010 | 1010 |
| :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— | :— |
| 1110 | 10000 | 10001 | 10011 | 10101 | 1101 | 1111 | 1011 | 1011 | 1011 | 1011 | 1011 | 1011 | 1011 |

**A careful study of Table Ia allows us to see in it a complete definition of the pattern: yes, of course, the disjunctive positions (a, b, c) in cerem “nim” are .me positions, meaning the sum of the digits as before the digit in the double notation of the numbers a, b, c cemena (that is, equal to 0 or 2)! However, table IIa turns out to be more insidious: it is no better than table II, and no definite pattern in the composition of losing pairs (a, b) can be discerned from it.

But why should the double number system be the key to finding a win-win system for playing Qianshi? True, we saw that this system seems to help in developing the theory of the game nim: but the Qianshizians are a completely different game, and the key that opens one door does not at all have to fit another. Is it possible to rewrite Table II in a different way: perhaps a different system of recording the image

Does this system of numbers shed more light on its structure?

A careful study of Table II allows us to notice that Fibonacci numbers often appear in it and see, for example, the 1st, 2nd, 5th and 13th columns of the table *). And this, in turn, may lead us to think about the advisability of trying to write Table II in the “Fibonacci number system”, in which all numbers are represented by sums of numbers in the Fibonacci sequence (we will write numbers in this number system in blue numbers). For example, the number 100 in Fibonacci

*) Fibonacci numbers are a sequence of numbers starting with 1 and 2 and then building according to the following law: which number of the Fibonacci sequence is equal to the sum of the two previous ones. Its first terms are: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, … A good brochure by N. N. Vorobyov “Fibonacci Numbers” (M., “Hayka”, 3rd ed., 1969) is dedicated to Fibonacci numbers.

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to the above conditions a) and 6)) in ceremonies, “it” is.me and only not the positions (a, b, c), for which the sums of all digits corresponding to one digit in the recording of numbers a, b, c in the double number system, for all digits of the cemen.

Theorem II. Procedural in ceremonies “izanyitsy” are.me and not positions (a, b), for which a, being written in the Fibonacci number system, ends with a whole number of zeros, and the number b is obtained from the number a and we add one more zero at the end.

Then ee. From Theorems I and II it follows that both games - “nim” and “izanyitsy” - are in a certain sense beneficial for the beginner: there are quite a few losers for the beginner by position (see Tables I and II) in them, that is, they are much less common than all other (winning for the beginner) positions.

The proofs of these theorems form the content of the exercises below. The answers to them are given on page 60 (however, try to find the solutions yourself). We advise you to first deal with exercises 1-4, since exercises 5-7 are much more difficult.

Exercises

1. Prove that if the position (a, b, c) in the game “him” is not a losing one, that is, does not satisfy the conditions specified in the formulation of Theorem I, then the player can make it a losing one with his own move.

2. Prove that if the position (a, b, c) in the game “nim” is not a losing one, that is, does not satisfy the conditions specified in the formulation of Theorem I, then the player can make it a losing one with a move.

3. Indicate the correct game in the game “nim” if the number of piles of matches is any predetermined number number n ≥ 2 (we leave the rest of the correct game the same). What is the evidence for a win-win game in the case of n = 2?

4. Describe the losing positions (a, b, c) in the game “nim”, writing the numbers a, b, c not in the double, but in the quaternary number system (that is, representing, say, the number a in the form a = aₙaₙ₋₁…a₂a₁a₀ = aₙ4ⁿ + aₙ₋₁4ⁿ⁻¹ + … + a₂4² + a₁4 + a₀; here 0 ≤ aᵢ ≤ 3 for all i = n, n − 1,…, 2, 1, 0 *))

5. Prove that for any positive integer p, there is one and only one losing position (a, b) in the game of odds (that is, a position that satisfies the conditions of Theorem II) such that b − a = p. Check this for the first 15 losing positions listed in Table II.

6. Prove that if position (b, a) in the game “Izyanytsy” does not satisfy the conditions of Theorem II, then the player with his own move can either immediately win or lead the position to a losing one.

7. Prove that if position (a, b) in the game “Izyanytsy” is losing, then after any move of the player it ceases to be so.

8. Indicate the correct game in the “Izyanytsy” ceremony if the number of piles is odd (the player with each of his moves has the right to take any number of matches from each one pile, or equally - and also any number! - from all piles).

Problem. Indicate the correct game in the “Izyanytsy” ceremony if the number of piles is four.

I would like to warn the reader that the author does not know the solution to the problem posed.

*) The line above the number indicates that this is a number written in digits aₙaₙ₋₁…a₂a₁a₀, and not a product of numbers.

DIALOGUE ABOUT TEMPERATURE

M. Ya. Azbel

A popular science article is always a kind of game between the Author and the Reader, in which the Author is believed at his words (hence the not so rarely realized possibility of desire), but he has no right to write incomprehensibly and cause boredom. Otherwise, the Author will lose: the article will not be read and, therefore, was written in vain.

A generally popular article looks like a long monologue. But the ancient Greeks already understood that scientific dialogue is much more interesting. Since this does not break the rules of the game, let’s follow their example. As a result, the Reader will have the opportunity, showing character, to first try to find answers to the questions posed, each of which is a physical task, and sometimes an entire problem.

So the curtain rises - we begin.

Author
Temperature, please, is the first understandable quantity that a person encounters in his life. A five-year-old child does not yet know how to count, but he already says: “I am healthy, I have 36.6,” without even understanding the meaning of the spoken words (for decimal fractions you need to double the size). This means that we are talking about one of the most important physical concepts. Throughout its history, humanity has been trying to understand things that amaze us even in childhood: space, time… For the more instilled the concept, the simpler the question is formulated, the deeper it is usually and the more difficult it is to answer. However, the more difficult it is to ask it - it seems so prominent. Thus, in one of the detective stories, detectives are knocked off their feet, looking for what is in front of their eyes.

Reader
God be with him, and with detectives, and with philosophical reasoning. Can’t you be more specific?

Author
Please. This is a question, the consistent clarification of which ultimately leads us to the basic concepts of the general theory of relativity.

All methods of measuring temperature are based on the expression of temperature, that is, on the fact that a hotter body is cooled and a colder body is heated. Isn’t it?

Reader
Certainly.

Author
Why is this happening? Why does heat always move from a more heated body to a less heated one?

Reader
Not always! A refrigerator, for example, cools even though there is no colder body near it. The heat here is transferred to a warmer body - the surrounding air and everything that is in the room.

Author
You were ahead of the course of my reasoning, making a very important point. Indeed, if necessary and expended work, you can heat a hot one with a cold body. No wonder the refrigerator turns on by itself. But in a system that does not interact with any other objects (physicists call such a system closed), the temperature always equalizes,

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the body is transferred from hot to cold. In all cases where this does not happen, you can find where the system is “open”. So, in the case of a refrigerator, its cooling occurs due to the energy received from a power plant, or, ultimately, from the cooling Sun.

Reader
Everything you say, in my opinion, is absolutely clear.

Author
I have already warned: the source of the deepest ones - each one is very much driven by the energy received from the power plant, or, ultimately, from the cooling Sun.

Reader
Everything you say, in my opinion, is absolutely very important - everyone is very much driven by the energy received from the power plant, or, ultimately, from the cooling Sun.

Usually, when the results of an experiment contradict a theory, the theory must be corrected. If the experiment contradicts the second law of thermodynamics, or there is an error in the experiment. And it always follows the so-called “second law of thermodynamics” (“the first law” is the law of conservation of energy), or the “law of increasing energy.”

Usually, when the results of an experiment contradict a theory, the theory must be corrected. And it always follows the so-called “second law of thermodynamics” (“the first law” is the law of conservation of energy), or the “law of increasing energy.”

The second law of thermodynamics states: “In a closed system of processes, temperatures are equalized.” In other words, heat transfer from a hot body to a cold one. The first is cooling, and the second is heating.

In order to respond to energy with a more heated body, work must be expended. This means that not only is the “perpetual motion machine of the first city”, which does work from “nothing”, impossible, but also the question of the first law of thermodynamics - the law of conservation of energy. It is also impossible and does not contradict this law of the “perpetual motion machine of the second city”, which we quickly formed by the second law of thermodynamics, did work, cooled the work, cooled the seas, oceans, materials. Their reserves of thermal energy are practically inexhaustible, but, alas, inaccessible. What own processes can be the second beginning!

So, in a closed system of processes, temperatures are equalized.

This process ends when the warm-eastern temperatures are the same. Then a state of equilibrium occurs, but not mechanical, but thermodynamic.

But this immediately leads to the conclusion about the “tellian death of the Universe”, which belongs to the German scientist Clausis. Sooner or later, all the stars will cool down, the same temperature will be established throughout the world, any processes will turn out to be impossible - the world will turn into a lifeless, empty place. All attempts to understand why this will not happen during the existence of the Universe were free until the advent of the general theory of relativity!

Reader
But according to the theory of relative temperature, does it not equalize?

Author
No, the answer is not so straightforward. But in order to understand how the general theory of relativity dealt with heat death, we need to talk a little about the basic concepts of this theory.

Reader
And this, you say now, is inaccessible to a schoolboy…

Author
On the contrary, I’m sure it’s accessible. I believe that the deeper the physical or mathematical idea, the easier it is to present it without using any special knowledge.

Reader
So please tell us about the general theory of relativity.

Author
This should be a topic for a separate conversation. Today we are talking about temperature, and it was important for me to show that by thinking about how a thermometer changes temperature, you can

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but come to Einstein’s theory; Show me how many problems and perspectives can be seen by reading school physics. As you can see, thinking about temperature turns out to be extremely fruitful. What is temperature?

Reader
From what you said, I understood that temperature is what equalizes when two bodies come into contact. At school we were told that temperature is a quantity proportional to the average kinetic energy of molecules:

$$\frac{mv^2}{2} = \frac{3}{2} kT,$$

where $k \approx 10^{-21} \text{ } \partial x/\partial p \text{ — постоянная Больцмана.}$

Temperature equalization occurs because during a collision, energy is transferred from particles with higher energy to particles with lower energy.

Author
What do you think is the potential energy of interaction between particles in a crystal at absolute zero temperature?

Reader
We offer this energy by turning the crystal into gas. This means that for each atom there is an energy equal to approximately $kT_k$, where $T_k$ is the boiling point.

Author
Absolutely correct assessment. So, in your opinion, a crystal located at absolute zero temperature can heat a gas having a temperature below the boiling point of the crystal?

Reader
Of course not. I did not mean total, but kinetic energy, which is higher the higher the temperature.

Author
Fine. Consider, for example, a piece of ice. Above it, of course, there is water vapor. Due to temperature equalization, ice and steam have the same temperatures at equilibrium. This means that the average velocities of molecules in vapor and ice crystal are the same.

Reader (puzzled).

No, this is something strange. After all, in melted ice the molecules are almost immobile, but they move around at enormous speeds. I remember well that when a liquid evaporates, it cools. There is even a way to obtain very low temperatures - evaporation of liquid gel. This is understandable. Only the fastest molecules can fly beyond the boundaries of molecular acceptance in a liquid. The remaining slow ones correspond to a lower temperature.

Author
But wouldn’t some of the energy be spent on pre-molecular absorption in the liquid? Will the molecules fly out of the liquid into vapor at the same speed with which they began their journey?

Reader
Honestly, it’s good to argue. I never thought that in a dispute the truth is actually born! I think I understood everything completely! Indeed, those molecules that were washed away quickly will evaporate, so first the liquid cools. Pre-acceptance, the molecules lose their speed, like a package flying towards the Moon. Therefore, gas also does not consist of record-breakingly fast molecules, and the average speeds of molecules in gas and liquid coincide. Thanks to evaporation, the overall temperature is established. If the escaping, formerly fast molecules are carried away, as during evaporation under pumping, a noticeable decrease in the temperature of the liquid can be achieved.

Author
What about the speeds of molecules in a crystal?

Reader
In a crystal, molecules move with the same average speeds as in a pair above it, but they do not perform a free translational motion, slightly distorted by rare molecules, as in a gas, but oscillate around standard places in the crystal lattice…

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Author
…which are called lattice nodes. Absolutely right.
But I will continue to pester you. Over time, you will learn to do this on your own. A graphomaniac color susceptible to any of his ideas - the scientist first begins in Japanese with possible refutations.
So, they said that temperature determines the average speed of molecules. This means that the average speeds of molecules in any body, regardless of their composition and state (water, iron, air), at the same temperature are the same. And at absolute zero the velocities are zero. But what then to do with the gel, which remains liquid even at absolute zero temperature?
Reader
But I read that absolute zero is unattainable. By the way, why?

Author
In order to understand why, we need to talk about entropy. This will greatly increase us in the direction. And the fact that zero is unattainable does not exist in our case. In principle, we can get as close to it as we like. The only important thing for us is that at a temperature of several degrees Kelvin, the average speed of the atoms of the gel already weakly depends on temperature, and when the temperature changes from 1 to 10⁻⁶ degrees, it practically does not change, although, it would seem, it should decrease a thousand times.
Reader
Maybe this is due to the special properties of the gel, its superfluidity?

Author
There are some societies between these phenomena. However, remaining liquid at absolute zero is not only a property of a gel. As you know, the good electron conductivity of metals is due to free electrons. Their average kinetic energy practically does not depend on the temperature of the metal, and if we calculate it at the temperature using our formula $$\frac{mv^2}{2} = \frac{3}{2} kT$$, we will see that it corresponds to a temperature of 10⁵—10⁶ degrees, that is, the temperature inside the stars!
Reader
Maybe the temperatures of the metal and electrons do not match?
Author
No, in a state of equilibrium, temperatures, as we said, always level out. The point here is different. The definition of temperature that you shared is correct - but only in classical mechanics. Such a “simple”, well-known and fundamental quantity as temperature, strictly speaking, has a quantum nature! This is due to the superfluidity of the gel and the fact that it never replaces at atmospheric pressure - both of these properties are purely quantum. As you can see, in search of significant quantum effects it is not necessary to look for ways into the microworld - these effects are before your eyes.
Unfortunately, in the quantum case the meaning of temperature is not as clear as in the classical case. Let’s start at absolute zero temperature and consider a purely quantum effect - we determine the kinetic energy at absolute zero, that is, a quantity that in classical theory is exactly equal to zero. This allows us to find out whether all substances remain solid at zero temperature, whether an increase in pressure always promotes solidification, and what is the state of matter in the interior of the Earth and in the depths of stars.
All these conclusions can be derived from one very important quantum law—the uncertainty principle.
Reader
I know the principles of uncertainty as applied to the coordinate and momentum of a particle. Position and momentum cannot be measured accurately because they do not have a specific value at the same time. When measuring them

15

pulse particles $p$ and coordinates $x$ we find out the values ​​of these quantities with some errors $\Delta p$ and $\Delta x$.
The properties of these two quantities are that it is impossible to make the product of these errors (sometimes called “uncertainty”) less than $\hbar$ - Planck’s constant divided by 2π

$$\left( h = \frac{h}{2\pi} \approx 10^{-32} \frac{\partial \hbar}{\partial e_k} \right): \quad \Delta p \Delta x \geq \hbar.$$

Author
In an object consisting of identical atoms, all of them are also active, and therefore their position cannot be changed with an accuracy greater than the distance between them. If $\Delta x \sim a$, what is the smallest possible value of momentum?

Reader
Of course, $\frac{\hbar}{a}$! If I were to say that the momentum, for example, is less than 0.1 $\frac{\hbar}{a}$, then this would mean changing it with an accuracy of 0.1 $\frac{\hbar}{a}$, which is impossible!

Author
This means that the lowest possible kinetic energy…

Reader
…equal to $\frac{mv^2}{2} = \frac{(mv)^2}{2m} = \frac{p^2}{2m} = \frac{\hbar^2}{2ma^2}$, that is, the smaller the mass, the greater! I wonder what temperature this energy corresponds to in classical mechanics? The average distance between atoms and electrons in metals is $a \approx 10^{-10}$ m, the mass of an electron is $10^{-27}$ e, the mass of a helium atom is $10^{-23}$ e. From the formula

$$\frac{3}{2} kT = \frac{mv^2}{2} = \frac{\hbar^2}{2ma^2}$$

we find that for electrons the temperature is

$$T = \frac{\hbar^2}{3kma^2} = \frac{10^{-64}}{3 \cdot 1,38 \cdot 10^{-21} \cdot 10^{-30} \cdot 10^{-20}} \approx 2,5 \cdot 10^6 \text{°K}.$$

Big million degrees! For helium we would get temperatures $1^\circ \text{K}$ in the same way.

Author
Exactly the numbers that were discussed earlier. Now it is no longer difficult for you to find out what a substance will be like at zero temperature—solid or not. But first, think about why it is generally difficult for the parietal to harden at low temperatures.

Reader
Obviously, because at low temperatures the kinetic energy of the particle becomes small.

Author
What does small mean? Compared to what?

Reader
Apparently, in comparison with the potential energy of interaction between particles. When potential energy becomes significantly greater than kinetic energy, it forces atoms to occupy certain positions in the crystal lattice. Therefore, in order to say what a substance will be like at zero temperature, you need to know how the potential energy of interaction of particles depends on the distance between them.

Now it is clear to me that a liquid that does not freeze at absolute zero temperature should be sought among substances with light atoms and weakly interacting—further, increased kinetic energy of atoms and placed potential interaction energy $U$. Helium is the lightest inert gas, the atoms of which weakly interact with each other (as, indeed, with other atoms - this is their “inertness”). Obviously, it satisfies the general conditions.

In the case of an electron gas, the potential energy of interaction corresponds to Coulomb’s law: $U(a) = \frac{e^2}{4\pi\varepsilon_0}$, where $e$ is the electron charge, equal to approximately $1,6 \cdot 10^{-18} \text{K}$, and $\frac{1}{4\pi\varepsilon_0} \approx$

16

$$\approx 9 \cdot 10^9 \frac{m^2}{\kappa}.$$

At $a \approx 10^{-10} m$ the minimum kinetic energy of electrons

$$\frac{mv^2}{2} = \frac{\hbar^2}{2ma^2} \approx \frac{10^{-64}}{2 \cdot 9 \cdot 10^{-31} \cdot 10^{-20}} \approx 5 \cdot 10^{-15} \text{ дж}, a$$

$$U(a) \approx \frac{2,6 \cdot 10^{-36} \cdot 9 \cdot 10^9}{10^{-10}} \approx 2,3 \cdot 10^{-16} \text{ дж}.$$

$U(a)$ turns out to be approximately the same as the minimum kinetic energy of electrons! Therefore, electron gas cannot solidify.

Author

It is interesting that the uncertainty principle already implies the absence of electrons in the nucleus, the size of which $R$ is of the order of $10^{-13} \text{ cm}$. There it would have to have a speed of $v \geq \frac{\hbar}{mR} \sim 6 \times 10^{11} \frac{m}{\text{сек}}$, that is, 2000 times the speed of light!

Now you can trace what will happen to the substance at $T=0$ as the pressure increases, that is, as the mechanical distance decreases.

Reader

Well, yes, everything is determined by the correspondence between potential and kinetic energy: $U(a)$ and $\frac{\hbar^2}{2ma^2}$. If $2ma^2U(a) \gg \hbar^2$, you get a crystal; otherwise, you get a liquid or gas.

Author

If a substance is crushed enough to “crush” and lower atoms, then $U(a)$ is proportional to $\frac{1}{a}$.

Reader

Therefore, at $a \rightarrow 0$ it will become more and more gaseous! Pressure causes melting - very interesting!

Author

At very high pressures, when nuclei fall apart, the potential

$$E_2$$

$$E_1$$

$$E_0$$

Fig. 1.

cial energy increases rapidly with decreasing $a$, and the substance solidifies again. Then, in the region of ultra-high densities, corresponding directly to neutron stars, the energy of gravitational interaction becomes decisive, and new melting can occur.

There are stars that change their brightness and have a size of about 10 km and a mass comparable to the mass of the Sun. They have a solid bark (its density varies from $10^6$ to $5 \cdot 10^{13} \text{ e/cm}^3$) and a liquid core with a density of $5 \cdot 10^{14} \text{ e/cm}^3$.

Reader

All this is very interesting, but I still don’t understand what temperature is from the point of view of quantum mechanics.

Author

We have so far determined only the minimum energy of an atom or molecule. However, this does not yet determine the kinetic energy of the substance. Something to keep in mind. According to the laws of quantum mechanics, the energy of a particle cannot be any. It can only take on completely defined values, or, as we say, particles can be at certain energy levels. They are usually depicted in the form of horizontal lines located one above the other (Fig. 1). The energy level system is something like a bookcase with shelves. Books can be on shelves, but not between them.

Reader

Is the distribution of particles among the shelf levels purely random and disorderly?

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science, but also for technology. In famous quantum generators - lasers and masers - they specially create a negative temperature in exactly the sense that you were talking about, and then quickly “highlight” the accumulated energy.

Reader
Does temperature determine the distribution of particles by energy level only in quantum mechanics?

Author
Of course not. This is also true in classical mechanics. For example, the energy distribution of molecules of a “classical” ideal gas is as shown in figure 4. It is determined by the formula
$$\frac{n}{n_0} = e^{-\frac{E}{kT}}.$$
Here $n_0$ is the number of particles in the state from which we start, there is no energy of particle $E$; $n$ is the number of particles with energy $E$; $e = 2,718$ is the base of natural logarithms, $k$ is Boltzmann’s constant, $T$ is the gas temperature.

Reader
If we transform this formula, continuing the rhymes of both sides of the equality, then it will take the form
$$\ln \frac{n}{n_0} = -\frac{E}{kI}.$$

Fig. 4.

Fig. 5.

From here
$$T = -\frac{E}{k(\ln n - \ln n_0)}.$$

If we supply energy to a gas, we will increase its temperature, transferring more and more of it from lower energy levels to upper ones. In doing so, we can ensure that $n$ becomes $n_0$ and $\ln n$ greater than $\ln n_0$. This will mean that the gas temperature is negative! It seems to me that we have reached the point of absurdity. Therefore, it makes no sense to talk about the negative temperature of an ideal gas.

Author
If we consider only a few energy levels, then we can make it so that the upper ones will have more gas molecules than the lower ones. However, it is impossible to ensure that the energy distribution of molecules is the same as shown in figure 5: the dependence of the number of a part on energy is exponential and there is more of a part at very high levels than at lower ones.

The whole point is that our conclusion about the possibility of negative temperatures is valid not for any system of particles, but only for such

19

which has a finite number of energy levels. After all, energy usually includes clinical energy as an integral part, and it can be much greater. In quantum language, this means that the number of energy levels is infinitely large. But in this case, it is impossible even to distribute a finite number of particles evenly over an infinitely large number of energy levels, let alone obtain a negative temperature. This means that our calculations are correct for systems whose energy is not related to clinical energy.

Reader
This is very difficult to understand. I have never heard of anything that someone held in their hands or saw a body with a negative temperature.

Author
This is not surprising. In a stable, equilibrium state, negative temperature is impossible. A state with a negative temperature is, as they say, a metastable state. Once it has arisen, it usually does not exist for long (although there are also very long-lasting metastable states, for example glass, which in the equilibrium state is an opaque crystal). For example, when a negative temperature is created in the working substance of a laser, it very quickly transforms into a state with a positive temperature: particles from high energy levels move, as if “falling” to lower levels, and the energy lost in this case is released in the form of a laser beam.

Reader
But I would like to return to the temperature. If I understand you correctly, temperature is a concept that has many different sides. On the one hand, temperature is a quantity that characterizes the state of the body, the replacement of the body, such that when a system transitions to equilibrium, it turns out to be the same in all parts of the system. On the other hand, temperature is related to the average clinical energy of the particles that make up the body. Finally, in quantum mechanics it acts as a quantity that determines the distribution of particles by energy state. It is easy to notice that in all three manifestations, temperature is most closely connected with another physical quantity - energy. But earlier you mentioned another quantity that was completely unfamiliar to me - entropy. Does it also have something to do with temperature?

Author
The most direct thing. And a story about temperature cannot be complete without talking about entropy. But entropy should always take up no less space than the one that is now time to finish.

20

COMPARISON
DIFFERENT MEDIUMS
TWO POSITIVE
NUMBERS

3. A. Skopets

It is well known that two positive numbers $a$ and $b$ or two segments of lengths $a$ and $b$ are associated with their arithmetic mean $\frac{a+b}{2}$ and geometric mean $\sqrt{ab}$, with $\sqrt{ab} \leq \frac{a+b}{2}$ (the equality holds only for $a=b$). The algebraic proof of this inequality is extremely simple:

$$(a-b)^2 \geq 0, (a+b)^2 \geq 4ab,$$
$$ab \leq \left(\frac{a+b}{2}\right)^2 \text{ и } \sqrt{ab} \leq \frac{a+b}{2}.$$

The geometric proof is based on one metric relationship between the elements of a right triangle: the height of a right triangle, lowered to the hypotenuse, is the geometric mean of the projections of the legs onto the hypotenuse. This height does not exceed the radius of the circle circumscribed about the triangle, while the arithmetic mean of the segments of the hypotenuse is exactly equal to the radius of this circle.

In addition to the indicated two interseno means, two more frequently occurring means are considered: the quadratic mean $\sqrt{\frac{a^2+b^2}{2}}$ and the harmonic mean $\frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}$.

Let us give an algebraic and then a geometric proof that

$$\frac{2ab}{a+b} \leq \sqrt{ab} \leq \frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}},$$

and in all cases the equal sign holds if and only if $a=b$. Let us denote the averages written out in (1) alternately by the letters $M, N, K, L$. We have already checked the inequality $N \leq K$. It remains to prove that 1) $K \leq L$; 2) $M \leq N$.

1. $K^2 - L^2 = \left(\frac{a+b}{2}\right)^2 - \frac{a^2+b^2}{2} = -\frac{1}{4} (a-b)^2 \leq 0,$

that is, $K \leq L$, and $K=L$ only for $a=b$;

2. $M^2 - N^2 = \frac{4a^2b^2}{(a+b)^2} - ab = -\frac{ab (a-b)^2}{(a+b)^2} \leq 0,$

that is, $M \leq N$, and $M=N$ at $a=b$.

The validity of inequalities (1) has been proven *).

We will now look at the geometric proof of these inequalities, which may turn out to be

*) These inequalities have already been discussed in the article by A. P. Savin “Maximum, minimum and the theorem on averages” («Kvant» No. 11, 1970).

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Useful in solving geometric problems.

Let the given two segments be $a$ and $b$. If $a=b$, then by checking we are convinced of the validity of inequalities (1). Assume $a>b$ and construct a trapezoid $ABCD$ with bases $AB$ and $CD$ equal to the correspondence of $a$ and $b$ (Fig. 1). Let us draw a secant line $l$ so that it is parallel to the bases of the trapezoid and cuts it into two equal trapezoids $T_1(ABL_2L_1)$ and $T_2(L_1L_2CD)$. Let us calculate the length of the segment $L_1L_2$, which we denote by $x$. From the equivalence of trapezoids $T_1$ and $T_2$ it follows that $(a+x)h_1=(x+b)h_2$, or

$$h_1:h_2=(x+b):(a+x),$$

(2) where $h_1$ is the height of trapezoid $T_1$, and $h_2$ is the height of trapezoid $T_2$. From the similarity of triangles $AL_1U$ and $DL_1V$ it follows that

$$h_1:h_2=(a-x):(x-b).$$

(3) Comparing (2) and (3), we obtain

$$\frac{x+b}{a+x}=\frac{a-x}{x-b},$$

where is that $x^2-b^2=a^2-x^2$ from, and finally

$$x=\sqrt{\frac{a^2+b^2}{2}}.$$

So, it has been proven that $L_1L_2=\sqrt{\frac{a^2+b^2}{2}}.$

Considering (2), we have $h_1<h_2$, so the segment $L_1L_2$ lies closer to the larger base $AB$ of the trapezoid than its midline $K_1K_2$, which is equal to $\frac{a+b}{2}$. The length of the segment $x$, parallel to the bases of the trapezoid, the ends of which lie on the lateral sides of the trapezoid, is enclosed between the lengths of its bases (this follows from the properties of similarity). So, at $a>b$ we have $\frac{a+b}{2}<\sqrt{\frac{a^2+b^2}{2}}$. This inequality remains in force for $a<b$ (it is enough to change the notations: $a$ to $b$ and $b$ to $a$). Therefore, for any $a$ and $b$ we have $\frac{a+b}{2}\leq\sqrt{\frac{a^2+b^2}{2}}.$

Let’s move on to the comparison of $N$ and $K$. To do this, we draw a secant line $n$ so that it cuts the given trapezoid into two similar (homothetic) trapezoids $T_1'(ABN_2N_1)$ and $T_2'(N_1N_2CD)$ (Fig. 2). Let us denote the length of the segment $N_1N_2$ by $y$. From the similarity of trapezoids $T_1'$ and $T_2'$ it follows that $a:y=y:b$, or

$$y=\sqrt{ab}.$$

Therefore, $N_1N_2=\sqrt{ab}$. From the similarity of trapezoids (taking into account (4)) it also follows that

$$AN_1:N_1D=a:y=\sqrt{\frac{a}{b}}>1.$$

This means that line $n$ is further from the base $AB$ of the trapezoid,

22

Than the middle line is $K_1K_2$, and therefore $N_1N_2 < K_1K_2$. So, for any $a$ and $b$ we have

$$\sqrt{ab} \leq \frac{a+b}{2}.$$

Finally, let us draw a line $m$ through the intersection point $O$ of the diagonals of this trapezoid, parallel to its bases and intersecting the lateral sides at points $M_1$ and $M_2$ (Fig. 3). Obviously $AM_1:M_1D = AO:OC = a:b$.

But $\frac{a}{b} > \sqrt{\frac{a}{b}}$, because $a > b$. Therefore, taking into account (5), we conclude that the segment $M_1M_2$ is close to the smaller base of the trapezoid than the segment $N_1N_2$, and therefore $M_1M_2 < N_1N_2$. On the other side,

$$M_1O:a = DO:DB = CO:CA = OM_2:a.$$

This means $M_1O = OM_2$. Next,

$$\frac{M_1O}{a} + \frac{OM_2}{b} = \frac{DO}{DB} + \frac{BO}{DB} = 1.$$

From this it follows that $M_1M_2 = 2M_1O = 2\frac{1}{a} + \frac{1}{b}$, and therefore $M_1M_2 = \frac{2ab}{a+b}$. So, $M < N$ if $a > b$. Similarly, it is proved that $M < N$ if $a < b$. Therefore, for any $a$ and $b$ we have

$$\frac{2ab}{a+b} \leq \sqrt{ab}.$$

This correctly proves inequality (1).

So all four are average

$$M_1M_2 = M = \frac{2ab}{a+b}; \quad N_1N_2 = N = \sqrt{ab};$$

$$K_1K_2 = K = \frac{a+b}{2}; \quad L_1L_2 = L = \sqrt{\frac{a^2+b^2}{2}}.$$

below the two segments $a$ and $b$ received a geometric origin: if these two segments serve as the bases of a trapezoid, then close to the larger base of the trapezoid is the quadratic mean, followed by the arthmetic mean, then comes the geometric mean and, finally, the harmonic mean (Fig. 4). Obviously, the lengths of these segments do not depend on the shape of the trapezoid - for all trapezoids with given lengths $a$ and $b$ bases, the indicated averages of the corresponding ones are equal. When $a = b$, the trapezoid turns into a parallelogram and all four middle ones coincide and become equal to $a$.

It remains to construct these four averages over the given segments $a$ and $b$. We give such constructions so that one can be convinced of the truth of inequalities (1) in another way.

Let us plot on the straight line on one side of the point $O$ the segments $OA = a$, $OB = b$, where $a < b$. Let $S$ be the midpoint of the segment $AB$ (Fig. 5). Then

$$OS = a + \frac{b-a}{2} \text{ или } OS = \frac{a+b}{2} = K.$$

23

Fig. 5.

$$OC = M = \frac{2ab}{a+b};$$
$$OP = N = \sqrt{ab};$$
$$OS = K = \frac{a+b}{2};$$
$$OE = L = \sqrt{\frac{a^2+b^2}{2}};$$
$$M < N < K < L.$$

Let us construct a circle on the segment $AB$, as on a diameter, and draw tangents $OP$ and $OQ$ ($P, Q$ are points of tangency) to it from point $O$. Let us use the relationship between the secant and tangent segments drawn from the point $O$ to the circle:
$$OP^2 = ab.$$ Therefore, $OP = \sqrt{ab} = N$, and $N < K$. Next, let $PQ$ intersect $OS$ at $C$. Then $OP^2 = OS \cdot OC$ and
$$OC = ab: \frac{a+b}{2} = \frac{2ab}{a+b} = M,$$
and $M < N$.

Finally, we construct the radius $SE$ perpendicular to $AB$. We have
$$OE^2 = \left(\frac{a+b}{2}\right)^2 + \left(\frac{a-b}{2}\right)^2,$$
$$OE = \sqrt{\frac{a^2+b^2}{2}} = L,$$
and $K < L$. So, if $a \neq b$, then $M < N < K < L$.

Exercises

1. On a straight line on one side of the point $O$, the segments $OA = a, OB = b, a < b,$ are laid out and a circle $k$ of diameter $AB$ with center $S$ is constructed. Next, $k_1$ with a center $O$ of radius $OP$ and $k_2$ with a diameter $OS$ are constructed.

Prove that if the line passing through $O$ intersects part of the lines $g, k_1$ and $k_2$ located inside $k$ at points $C_1, P_1, S_1$, and the circle $k$ at points $A_1$ and $B_1$, then
$$OC_1 = \frac{2a_1b_1}{a_1+b_1},$$
$$OP_1 = \sqrt{a_1b_1},$$
$$OS_1 = \frac{a_1+b_1}{2} \quad (a_1 = OA_1, b_1 = OB_1).$$

2. Find even ratios connecting each three averages of two numbers $a$ and $b$:
   $$M = \frac{2ab}{a+b},$$
   $$N = \sqrt{ab},$$
   $$K = \frac{a+b}{2},$$
   $$L = \sqrt{\frac{a^2+b^2}{2}}.$$

3. Prove the truth of the following inequalities:
   a) $N - M \leqslant K - N,$
   b) $K - M \leqslant L - N,$
   b) $L - K \leqslant K - N.$

4. A circle with center $S$ on the x-axis of the right angular coordinate system intersects the axis at points $A(a, 0)$ and $B(b, 0), a > 0, b > 0, a < b$. A secant line passing through the origin $O$ intersects the circle at points $A_1$ and $B_1$. Write an equation of the lines to which those points $C_1(x, y)$ on the secants belong, for which
   $$(OC_1)^2 = \frac{1}{2} \left[ (OA_1)^2 + (OB_1)^2 \right].$$

5. On the continuation of side $AB$ of triangle $ABS$ beyond point $A$ there is a bottom point $O$, through which a line is drawn intersecting $AS$ and $BS$ corresponding at points $B_1$ and $A_1$. Through the intersection point $P$ of lines $AA_1$ and $BB_1$, a line $SP$ is drawn, intersecting $AB$ at point $C$. Prove that $c = \frac{2ab}{a+b}$, where $OA = a, OB = b, OC = c$.

6. Given a parallelogram $OABC$ and a midline $MN$ of a triangle $ABC$ parallel to $AC$. A straight line is drawn through vertex $O$; intersecting the lines $BC, CA, AB, MN$ of the corresponding one at points $A_0, B_0, C_0, D_0$. Prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{3}{d}$, where $OA_0 = a, OB_0 = b, OC_0 = c, OD_0 = d$; point $A_0, B_0, C_0, D_0$ is located on one side of $O$ ($d$ is the harmonic mean of the segment $a, b, c, d$).

How does the stated equality change if the points $A_0, B_0, C_0, D_0$ do not lie on the same side of $O$?

26

SOLUTIONS

In this section of the magazines we continue to publish solutions to problems from the puzzle «Kvant». Some of these problems are difficult even for mathematicians and physicists. Many of them give rise to a conversation about a whole class of problems, about methods of approaching certain phenomena. Each solution is a short separate article, and we recommend that you get acquainted with it even for those who have not tried to solve problems before.

In most cases, we indicate the names of the readers who sent us the most successful solutions, but, of course, we will not be able to answer everyone who sent us letters.

M21
Inside the square, on side 1, there are several circles, the sum of their lengths is 10. It is impossible to prove that there is a straight line. Then each point on side AB is covered by at most three projections of circles (Fig. 1), otherwise a line passing through the point and perpendicular to AB would intersect more than three circles. Consequently, the sum of all projections, that is, the sum of the lengths of the diameters of the circles, does not exceed 3; therefore, the sum of the lengths of the circles does not exceed 3π, while according to the condition it is equal to 10>3π. The resulting contradiction proves the statement of the problem.

This decision was added by D. Grigoriev from Leningrad and N. Rasshin from Moscow.

The following more general statement can be proven in the same way. Let us call the diameter of a convex figure the width of the narrowest strip in which this figure can be placed. (It can be shown that this definite is valid for any bounded figure.) Then, if inside a figure with diameter h there are n figures whose sum of diameters is greater than kh, then there is a straight line that intersects at least k + 1 of these figures.

28

Let’s rewrite the square of this expression as follows:

$$P_n^2 = \frac{1^2}{2^2} \cdot \frac{3^2}{4^2} \cdot \frac{5^2}{6^2} \cdot \dots \cdot \frac{(2n-1)^2}{(2n)^2} = \frac{1}{2} \cdot \frac{3^2}{2 \cdot 4} \cdot \frac{5^2}{4 \cdot 6} \cdot \dots \cdot \frac{(2n-1)^2}{(2n-2) \cdot 2n} \cdot \frac{1}{2n}.$$

Since for any integer $k > 1$

$$\frac{(2k-1)^2}{(2k-2) \cdot 2k} = \frac{(2k-1)^2}{(2k-1)^2 - 1} > 1,$$

That

$$P_n^2 > \frac{1}{2} \cdot \frac{1}{2n} \text{ и } P_n > \frac{\sqrt{2}}{2} \cdot \frac{1}{\sqrt{2n}}.$$

On the other hand, since always

$$\frac{(2k-1)(2k+1)}{(2k)^2} = \frac{(2k)^2 - 1}{(2k)^2} < 1,$$

That

$$P_n^2 = \frac{3}{2^2} \cdot \frac{3.5}{4^2} \cdot \frac{5.7}{6^2} \cdot \dots \cdot \frac{(2n-1)(2n-3)}{(2n-2)^2} \cdot \frac{2n-1}{(2n)^2} < \frac{3}{4} \cdot \frac{1}{2n} \text{ и}$$

$$P_n < \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2n}}.$$

A slightly different solution was proposed by D. Gritsoriev (this solution was also leaked by other readers of the magazine). Let us consider together with this expression

$$P_n = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \dots \cdot \frac{2n-3}{2n-2} \cdot \frac{2n-1}{2n}$$

the following:

$$Q_n = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \dots \cdot \frac{2n-4}{2n-3} \cdot \frac{2n-2}{2n-1}.$$

Since each of the factors of the product $Q_n$ is not greater than the corresponding factor $P_n$, then $P_n > Q_n$ and, consequently, $P_n^2 > P_nQ_n = \frac{1}{4n}$, that is, $P_n > \frac{\sqrt{2}}{2} \cdot \frac{1}{\sqrt{2n}}.$

On the other side,

$$R_n = \frac{1}{2} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \dots \cdot \frac{2n-2}{2n-1} \cdot 1 > P_n,$$

poseable $P_n^2 < P_nR_n = \frac{3}{8n}$ and $P_n < \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2n}}.$

It would seem more natural to find $R_n$ not with 1/2, but with 2/3. But then for $P_n$ the result will be a rougher estimate than required in the conditions of the problem. On the contrary, if we leave in $R_n$ not one, but more factors equal to the first factors $P_n$, then for $P_n$ we get a more accurate estimate of the form $P_n < \frac{c}{\sqrt{n}}$, where $c$ is some number less than $\sqrt{\frac{3}{8}}$. The same remark applies to evaluation from the other side.

There is also a solution based on the method of mathematical induction. It was on the example of this problem that E. G. Nikolaev’s note “The Case with the Method of Mathematical Induction” (see «Kvant» No. 7, 1970, p. 37 and «Kvant» No. 12, 1970, p. 58) said that sometimes a stronger inequality is easier to prove by induction than a weaker one.

29

From the proven inequalities it follows that the value $P_n$ tends to zero as $n$ increases, and the value $P_n \sqrt{n}$ lies between $\frac{1}{2}$ and $\frac{1}{2} \sqrt{\frac{3}{2}}$. It can be proven (see, for example, problems 145, 160, 161 in the book by A. M. Yaglom and I. M. Yaglom “Non-elementary problems in an elementary presentation”, GITTL, M., 1954) that as $n \to \infty$ $P_n$ tends to a limit equal to $\sqrt{\frac{1}{\pi}}$.

M24

Prove that any fraction $\frac{m}{n}$, where $0 < \frac{m}{n} < 1$, can be represented as

$$\frac{m}{n} = \frac{1}{q_1} + \frac{1}{q_2} + \frac{1}{q_3} + \dots + \frac{1}{q_r},$$

where $0 < q_1 < q_2 < \dots < q_r$ are integers and each $q_k$ ($k=2, 3, \dots, r$) is divisible by $q_{k-1}$.

Which fraction $\frac{m}{n}$ can be divided and the denominator by their most common divisor, replaced by an irreducible fraction equal to it. For example, $\frac{288}{504} = \frac{4.72}{7.72} = \frac{4}{7}$. In what follows we will consider only such irreducible fractions.

Let us prove the statement of the problem by induction on $m$. For $m=1$ it is obvious: the fraction $\frac{m}{n}$ itself already has the desired form. Now we will prove that if the statement of the problem is true for all fractions with numerators less than $m$, then it is also true for fractions with a numerator equal to $m$. Let $\frac{m}{n}$ be such a fraction ($1 < m < n$). Divide $n$ by $m$ with remainder; you get the quotient ($d_0-1$) and the remainder ($m-k$), that is

$$n = m (d_0-1) + (m-k) = md_0-k,$$

where $d_0 > 1$ and $0 < k < m$. Let’s rewrite (1) like this: $md_0=n+k$, or

$$\frac{m}{n} = \frac{1}{d_0} \left( 1 + \frac{k}{n} \right).$$

Since $0 < k < m$, the fraction $\frac{k}{n}$ can be represented in the desired form:

$$\frac{k}{n} = \frac{1}{d_1} + \frac{1}{d_1 d_2} + \dots + \frac{1}{d_1 d_2 \dots d_r},$$

where $d_1, d_2, \dots, d_r$ are some natural numbers greater than 1. From (2) and (3) we get

$$\frac{m}{n} = \frac{1}{d_0} + \frac{1}{d_0 d_1} + \frac{1}{d_0 d_1 d_2} + \dots + \frac{1}{d_0 d_1 d_2 \dots d_r}.$$

The fraction $\frac{m}{n}$ is presented in the required form.

Note that from our solution to the problem it is not difficult to extract a simple algorithm - the rule of how to represent any given fraction as a sum (3). Let’s demonstrate it with one example. Let us be given the fraction $\frac{5}{7}$:

$$7 = 2.5 - 3; \quad \frac{5}{7} = \frac{1}{2} \left( 1 + \frac{3}{7} \right);$$

$$7 = 3.3 - 2; \quad \frac{3}{7} = \frac{1}{3} \left( 1 + \frac{2}{7} \right);$$

$$7 = 4.2 - 1; \quad \frac{2}{7} = \frac{1}{4} \left( 1 + \frac{1}{7} \right).$$

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So,

$$\frac{5}{7} = \frac{1}{2} + \frac{1}{2.3} + \frac{1}{2.3.4} + \frac{1}{2.3.4.7} = \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{168}.$$

Of course, they can find several representations in the form (3), for example:

$$\frac{3}{8} = \frac{1}{4} + \frac{1}{8} = \frac{1}{3} + \frac{1}{24}.$$

The solution to this problem was sent by D. Gritsorye.

M25

In a set consisting of $n$ elements, $2^{n-1}$ subsets are selected, each three of which have a common element. Prove that all these subsets have a common element.

In total, in the set $A$, consisting of $n$ elements, there are $2^n$ different subsets, including the set and the set itself $A$ (see the article by N. Ya. Vilenkin “Combinatorics”, «Kvant» No. 1, 1971). We will denote different subsets $A$ by the letters $B, C, \dots$, the intersection of the set $B$ and $C$, that is, a set composed of common elements $B$ and $C$, is usually denoted as follows: $B \cap C$; the additional subset $B$ to the entire set $A$, that is, the set composed of all those elements $A$ that are not included in $B$, is denoted as follows: $\bar{B}$.

By condition, $2^{n-1}$ subsets were selected, that is, half of all possible ones, and each of the selected subsets, the intersection of any two (and even three) of them is not empty. Therefore, from each pair of subset $(B, \bar{B})$, complementary to each other up to $A$, exactly one is chosen. Further, if $B$ and $C$ are chosen, then $D = B \cap C$ is also chosen, since $\bar{D}$ cannot be chosen (all three subsets $\bar{D}, B$ and $C$ cannot have common elements, since the common elements $B$ and $C$ are contained in $D$). Using this very strong indication, we can prove that if $B_1, B_2, \dots, B_k$ are chosen, then the intersection of all these subsets $B_1 \cap B_2 \cap \dots \cap B_k$ is also chosen. Therefore, the intersection of all $2^{n-1}$ chosen subsets also belongs to the chosen subset, which means it is not empty.

Thus, we have proven that only such a subset system satisfies the conditions of the problem: some element $a$ of the set $A$ is fixed and all subsets containing this element are selected.

N. B. Vasiliev

F29

Find the rate of evaporation per unit surface area of ​​water into vacuum at temperature $20^\circ\text{C}$. (The saturated vapor pressure at this temperature is 17.5 mm Hg.) How long will it take for the water in the room to evaporate if filled to the top in an ordinary tea saucer? The evaporation of a small amount of water practically does not change the air humidity in the room, which is equal to 70%.

In order to solve the problem, you need to find the number of liquid molecules leaving 1 cm$^2$ of its surface in 1 second. By multiplying it by the mass of water molecules, we can find out the evaporation rate, that is, the mass of water evaporating per unit surface area of ​​​​water in 1 second.

Bydem reason like this. If there were saturated steam above the liquid, then the number of liquid molecules leaving it in 1 second would be the same as in the absence of steam. At the same time, one day, exactly the same number of molecules entered the liquid as came out of it. This allows us to calculate

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how many molecules fly out of water in 1 second, since finding the number of molecules entering the liquid is quite simple.

Let’s use the following model of an ideal gas: all gas molecules have the same speed $v$, and each of the molecules can move only in one of three mutually perpendicular directions along the coordinate axes (Fig. 4). Moreover, the number of molecules moving in each of these three directions is the same. If one of the coordinate axes is perpendicular to the liquid, then in time $\tau$ a mass of vapor enters the liquid, which is located at a distance $l = v \cdot \tau$ from it. Let there be $n$ vapor molecules in a unit volume, then $N = \frac{1}{6} nv\tau S$ vapor molecules fall on a surface area with an area $S$: along the coordinate axis perpendicular to the surface of the liquid, $\frac{1}{3}$ part of the vapor molecules located in the volume $v\tau S$ moves, and the speed of half of them is directed away from the liquid. If the mass of vapor molecules is $m$, then in time $\tau$ a mass of vapor $M = \frac{1}{6} nv\tau Sm$ enters the liquid. $n \cdot m$ is the vapor density $\rho$. That’s why

$$M = \frac{1}{6} v\tau S\rho.$$

The speed of steam molecules can be expressed in terms of its pressure and density. Consider a cubic vessel with edge $l$ and faces perpendicular to the coordinate axes (Fig. 4). During the elastic collision of vapor molecules with a wall, its momentum changes by $2mv$ - before the collision the momentum of the molecule is $mv$, and after the collision it is $mv$: the molecule moves away from the wall. Since time $t = \frac{2l}{v}$ passes between two successive collisions of a molecule with the same wall, then, in accordance with Newton’s second law, we can assume that an average force $f = \frac{2mv}{2l} = \frac{mv^2}{l}$ acts on the molecule from the side of the wall. According to Newton’s third law, a force of the same magnitude acts on the wall. Since along each of the axes of motion there are $N = \frac{1}{3} nl^3$ molecules, each of which contributes to the pressure on the wall, the total force acting on the wall is equal to

$$F = \frac{mv^2}{l} \cdot \frac{1}{3} nl^3 = \frac{1}{3} nml^2v^2,$$

and the vapor pressure on the wall is equal to

$$P = \frac{F}{l^2} = \frac{1}{3} nmv^2 = \frac{1}{3} \rho v^2.$$

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Therefore
$$v = \sqrt{\frac{3P}{\rho}}.$$

Substituting this expression for $v$ into formula (*), we get
$$M = \frac{1}{6} \tau S \rho \sqrt{\frac{3P}{\rho}}.$$

Let us also express the density of the vapor in terms of its pressure. They are related by the Clapeyron equation:
$$P = \frac{\rho}{\mu} RT,$$

where $\mu$ is the mass of one gram-molecule of vapor, $T$ is its temperature, and $R = 8,3 \frac{\partial \kappa}{\partial \rho} - \text{газовая постоянная}.$

From this formula we find that $\rho = \frac{P \mu}{RT}$. Therefore $M = \frac{1}{6} \tau S \frac{P \mu}{RT} \sqrt{\frac{3RT}{\mu}}.$

Thus, if there is steam above a liquid, then per unit of its surface in 1 second a mass of steam equal to
$$\frac{M}{\tau S} = \frac{1}{6} P \sqrt{\frac{3\mu}{RT}}.$$

This means that the rate of evaporation of liquid into vacuum is equal to
$$\frac{1}{6} P \sqrt{\frac{3\mu}{RT}}.$$

Substituting here the numerical values ​​$T = 293^\circ \text{K}$, $\mu = 18 \frac{\kappa_2}{\kappa_0 \text{м}^2}$ and $P = 2,3 \cdot 10^3 \frac{\mu}{m^2}$ ($P$ is the saturated vapor pressure at $T = 293^\circ \text{K}$), we obtain that the evaporation rate is $14,8 \cdot 10^{-8} \frac{\kappa_2}{m^2 \cdot \text{сек}}.$

We could solve the problem more accurately, taking into account that molecules move at different speeds and in all possible directions *). This, however, would not change the qualitative result that we obtained. Only the numerical factor in the last formula would change.

Now it is not difficult to find the time during which the water in the room will evaporate after being poured into a tea saucer. In this case, with uniform surfaces of the liquid, molecules with a total mass equal to $\frac{1}{6} P_0 \sqrt{\frac{3\mu}{RT}} \left( P_0 = 10^{-5} \frac{\mu}{m^2} \right)$ fly out in 1 sec, and a vapor mass equal to $\frac{1}{6} P \sqrt{\frac{3\mu}{RT}}$, where $P = \eta \cdot P_0 = 0,7P_0$, enters the liquid. Thus, the rate of liquid evaporation is
$$\frac{1}{6} \left( P_0 - P \right) \sqrt{\frac{3\mu}{RT}}.$$

If the surface area of ​​water is $S$ and its mass is $M$, then all the water will evaporate in time
$$t = \frac{M}{\frac{1}{6} S \left( P_0 - P \right) \sqrt{\frac{3\mu}{RT}}}.$$

*) Such a calculation is given in the article by Ya. A. Smorodinsky “Ideal Gas” («Kvant» No. 10, 1970).

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Assuming that the saucer contains 100 $e$ of water, its diameter is 10 $cm$, the air temperature in the room is $17^\circ C=290^\circ K$ (at this temperature $P_0 = 2,3 \cdot 10^3 \frac{H}{M^2}$ and $P = 1,6 \cdot 10^3 \frac{H}{M^2}$), we find $t=1$ sec.

The result was paradoxical. Where did we go wrong? The magnitude of the vapor pressure near the surface of the water. Here the steam pressure is significantly greater than in the room. In a thin layer near the surface, the vapor is almost saturated. Only thanks to this does the water evaporate slowly enough.

The example we looked at shows how important it is when solving a physical problem to understand the “physics” of the phenomenon, that is, to understand what can and cannot be neglected.

F30

The film camera and film projector produce 8 frames per second. On the movement screen there is an automatic vehicle with wheels, the actual diameter of which is 1 m. The images of the wheels make 2 revolutions per second. What is the speed of the automatic?

Since the images of the wheels are rotated by 2 turns during the time during which 8 frames pass through the projector, then on each frame the wheel must be rotated by $\frac{2}{8} = \frac{1}{4}$ turns compared to the previous one. Moreover, the images of the automatic vehicle wheels on the screen can rotate both “forward” and “backward,” although the automatic vehicle itself moves in both the first and the second, serving in the same direction (Fig. 5). The wheels on the screen rotate “forward” if the speed of movement of the automatic vehicle is such that during the time between frames $\tau = \frac{1}{8} c$ the wheels of the automatic vehicle make $n$ full and another $\frac{1}{4}$ revolutions around the axis. If in $t$ the wheels make $n$ full and $\frac{3}{4}$ revolutions around the axis, then the images of the wheels on the screen will rotate “backwards”. Thus, the angular speed of the automaton is either $\omega_{\text{вп}} = \frac{\left(n + \frac{1}{4}\right) 2\pi}{\frac{1}{8}} \frac{pa\partial}{c} = 16 \left(n + \frac{1}{4}\right) \pi \frac{pa\partial}{c}$ or $\omega_{\text{н}} = 16 \left(n + \frac{3}{4}\right) \pi \frac{pa\partial}{c}$. This means that the wheel axles, and with them the automatic vehicle, moves at a speed of $v = 16 \left(n + \frac{1}{4}\right) \pi R \frac{m}{c}$ (in which case the wheel images rotate “forward”) or at a speed of $u = 16 \left(n + \frac{3}{4}\right) \pi R \frac{m}{c}$ (in this case the wheel images rotate “backward”). Substituting $n = 1, 2, 3, \dots$ into these formulas, we will get the answers

$$v_1 \approx 12,6 \frac{m}{c} = 45 \frac{km}{час}, \quad v_2 \approx 223 \frac{km}{час}, \quad u_1 \approx 136 \frac{km}{час}, \quad u_2 \approx 316 \frac{km}{час}, \dots$$