## Saturday, December 23, 2023

### Logic+ 02: UNIZOR.COM - Math+ & Problems - Logic

Notes to a video lecture on http://www.unizor.com

Logic 02

Problem A

The game starts with two sets of stones and two players.
At each turn a player can take any number of stones but only from one of the sets.
The winner is the player who takes the last stones.

The player who should start the game can yield this right of the first move to his opponent. After that to take some stones on each move is the rule.

What is the winning strategy for a person who starts the game?

Solution A

The strategy to win for the first player is on every move to equalize the number of stones in two sets.

If the initial number of stones is different in these two sets, the first move of the player who starts the game is to take a few stones from a bigger set to equalize the number of stones.

His opponent, always facing two equal numbers of stones in two sets, cannot take the last stone and inevitably will make the number of stones in two sets different.

If in the beginning the numbers of stones in these two sets are equal, the player who starts the game should yield the right of the first move to his opponent, achieving the same conditions as above.

Problem B

There is a set of six points, no three points of this set lie on a straight line.
For clarity we can assume that they lie on a circle, though it's not essential.

Consider all subsets of three points of this set of six (we will call these subsets triplets).
Let's call a triplet, that has one or more segments connecting its points, a good triplet.
If all three points of a good triplet are connected by three segments, we will call this triplet a triangle.

For example, on the picture below all triplets have at least one pair connected by a segment (triplet ABC has two segments AB and AC, triplet CDF has two segments CF and DF, triplet ABF has one segment AB etc.)

Let's try to connect some points with segments in such a way that in any triplet there would be a pair of points connected by a segment.
That is, let's try to make all triplets good.

Prove that, no matter how we try to accomplish this task of making all triplets good by connecting our six points, there would inevitably be a triplet with three segments connecting all its three points into a triangle (that is, there will be at least one triangle), like ΔACE or ΔCEF in the example above.

Solution B

First, let's prove the following auxiliary statement (lemma) using the following illustration.

Assume that all triplets are good (as assumed in the Problem B) and some point (for example, A on the picture above) has three segments joining at it and connecting it to three other points (like on the picture above, point A is connected to B, C and E).
Then there exist a triangle.

Proof:
Since a triplet of three connected points (BCE on the picture above) is, as all other triplets, assumed to be good, there exists an additional connecting segment between some of its points (between C and E on the picture above).
With three given segments going from a point with three connections (AB, AC and AE on the picture above) and one additional segment between some connected points (segment CE between C and E on the picture above), there is a triplet (in our example it is ACD) that forms a triangle.

Consequently, to prove that in the process of connecting pairs of points to make all triplets good there would eventually have to be formed a triangle, it's sufficient to prove that there would eventually have to be a point with three segments connecting it to other three points.

The plan is to
(1) count all the triplets that can be formed from the given six points and
(2) show that without connecting three points into a triangle or connecting some point to three others (which, as lemma above proved, leads to forming a triangle) it's impossible to make all counted in (1) triplets good.

The total number of triplets is, obviously, the number of combinations of our six points in groups of three, which is
C63 = 6!/(3!·3!) = 20

Now let's proceed to make all triplets good, but obeying the rule of having no more than two segments joining at any single point (since, as we have proved, when three segments join in one point, triangle is unavoidable).

We have to make 20 good triplets, but we will show that we cannot make more than 18 without breaking that rule, which would necessitate the existence of a point with three segments connecting it to three other points, which is sufficient condition for existence of a triangle.

Let's start the process of connecting pairs of points to make good triplets.

In the beginning there are no connecting segments, so we choose any two points (call them A and B) and connect them with a segment.
This is sufficient to make 4 new good triplets: ABC, ABD, ABE and ABF.

Now we have a choice of which points to connect next.
We can either
(a) choose a pair of new points to connect by a segment (for example, D and E) or
(b) choose one new (unconnected) point and one that already has a connecting segment (for example, B and C), or
(c) choose two points that already had connecting segments, which could happen on later steps.

In case (a) we have 4 more good triplets (DEA, DEB, DEC and DEF in our example).

In case (b) only 3 good triplets will be formed (BCD, BCE and BCF in our example).

In case (c) only 2 good triplets will be formed (if segments AB and CD exist, and we connect points B and C with a segment, new good triplets BCE and BCF are formed).

General consideration is that the more segments connect our six points - the more good triplets will be formed.

Let's start with the strategy that assures the maximum number of segments - the one when there are two segments joining at each point, so the rule of not having three segments at any point is preserved.

This can be accomplished if all points are connected into a closed chain:

Obviously, not all triplets are good in this case. For example, ACE or BDF.

Another chained connection
Here triplets ABD or CEF are not good.

Let's count good triplets using the first example of a chained connection above, starting from segment AB and moving clockwise (the order is unimportant, we can connect A-C-B-F-D-E-A, as exemplified on the second picture, and the result will be the same).
Segment AB, being the first, produced 4 good triplets, as was stated above.
Then four consecutive segments BC, CD, DE and EF formed 3 good triplets each.
Final segment FA added 2 more good triplets.
The total number of good triplets is
4+3+3+3+3+2=18
which is short of required 20, which means, to make all triplets good, we need more segments, which will result in some points to be connected by three segments to three other points and forming of a triangle will be guaranteed.

We can try to organize connections differently. Instead of putting all 6 points into a chain, we can do it with only 5 with points A-B-C-D-E, leaving point F outside.

Similar counting of good triplets will result in the following:
Segments AB, as before, created 4 good triplets.
Segments BC, CD, DE, as before, created 3 good triplets each.
Finally, segment EA created two new good triplets: EAC and EAF.
The total number of good triplets is:
4+3+3+3+2 = 15

The last case is to leave two points out of a closed chain as follows.

In this case the number of good triplets is:
from AB - 4,
from BC - 3,
from CD - 3,
from DA - 2 (DAE, DAF),
from EF - 4 (EFA, EFB, EFC, EFD)
The total number of good triplets is:
4+3+3+2+4 = 16
Still less than 20.

There are no logically different cases of connecting the points deserving a consideration, so we came to
CONCLUSION:
to have all 20 triplets as good, we need more that two segments joining at each point; it is impossible to avoid triangles.

## Saturday, December 9, 2023

### Arithmetic+ 02: UNIZOR.COM - Math+ & Problems - Arithmetic

Notes to a video lecture on http://www.unizor.com

Arithmetic 02

Problem

Prove that
SN = Σn∈[2,N]1/n
is not an integer number for any integer N.

Solution 1

Among fractional numbers in the sum SN there are those of type 1/(2k).
Let's choose among them the one with the maximum power k. This number is less than any other number of this type in a series and the closest to the end of a series among them. Let's call it the least binary fraction.

For example, in the sum
S10 = Σn∈[2,10]1/n
we choose 1/8=1/(2³).

Now let's bring all the numbers in a series to the least common denominator (LCD).
For an example of S10 the LCD=2520 and values of our fractions with this LCD are
1/2 = 1260/2520
1/3 = 840/2520
1/4 = 630/2520
1/5 = 504/2520
1/6 = 420/2520
1/7 = 360/2520
1/8 = 315/2520
1/9 = 280/2520
1/10 = 252/2520

Notice that the numerator of 1/8=315/2520 (a fraction of a type 1/(2k) with maximum power k) equals 315, which is odd, while all other numerators are even.
That means, the sum of numerators with this common denominator is odd, while the least common denominator 2520 is even, which means that the sum S10 cannot be an integer number.

Actually, this situation will be similar with any number of members in a series, that is for any N.
Here is why (and this is the required proof).

Let's analyze the process of determining the least common denominator (LCD).

Any number in the denominators of our series can be uniquely represented as a product of prime numbers.
Then LCD is formed from all the prime numbers in representation of all denominators with the quantity of each prime number being equal to the largest number of these primes among all denominators.

In our example of S10 we have the following representation of denominators:
2 = 2
3 = 3
4 = 2·2
5 = 5
6 = 2·3
7 = 7
8 = 2·2·2
9 = 3·3
10 = 2·5

As we see, the prime numbers we have to use are 2,3,5,7 and the number of these primes in the LCD should be:
prime 2 should be taken 3 times because of denominator 8.
prime 3 should be taken 2 times because of denominator 9.
prime 5 should be taken once.
prime 7 should be taken once.
Therefore, the LCD is
2³·3²·5·7=2520

In general case, in the representation of LCD the number of 2's is the largest in a member 1/(2k) with the largest k that we called above the least binary fraction.
Therefore, in the representation of the LCD as a product of prime numbers there are exactly k 2's and, obviously. other prime numbers, which are all odd, of course.

When we transform all the fractions in a series SN to common denominator, we should multiply numerators (which are all 1's) by all primes in the LCD that are not in a representation of its denominator:

In our example of S10
the numerator of a fraction 1/2 should be multiplied by 2³·3²·5·7/2 =
= 2²·3²·5·7 = 1260

the numerator of a fraction 1/3 should be multiplied by 2³·3²·5·7/3 =
= 2³·3·5·7 = 840

the numerator of a fraction 1/4 should be multiplied by 2³·3²·5·7/2² =
= 2·3²·5·7 = 630

the numerator of a fraction 1/5 should be multiplied by 2³·3²·5·7/5 =
= 2³·3²·7 = 504

the numerator of a fraction 1/6 should be multiplied by 2³·3²·5·7/(2·3) =
= 2²·3·5·7 = 420

the numerator of a fraction 1/7 should be multiplied by 2³·3²·5·7/7 =
= 2³·3²·5 = 360

the numerator of a fraction 1/8 should be multiplied by 2³·3²·5·7/2³ =
= 3²·5·7 = 315

the numerator of a fraction 1/9 should be multiplied by 2³·3²·5·7/3² =
= 2³·5·7 = 280

the numerator of a fraction 1/10 should be multiplied by 2³·3²·5·7/(2·5) =
= 2²·3²·7 = 252

As a result, the only numerator without any 2's in its representation as a product of primes will be the one in the one we called the least binary fraction.
That's why this numerator will be odd, while all others (with one or more 2's in their prime representation) will be even.
Therefore, the sum of all these numerators with the least common denominator will be odd and not divisible by the LCD, which is even.

Solution 2

Among the fractions of our series there are those with prime denominators, which we will call prime fractions.
Let's choose among all prime fractions the one with the largest prime denominator Pmax (hence, the smallest in value among all prime fractions) and call it the least prime fraction.
So, Pmax is the largest prime number not exceeding N.

At this point it's very important to state that all subsequent members of the series after 1/Pmax up to the last one 1/N have denominators not multiple of Pmax.
The reason why it is the case is based on so called Bertrand's postulate that states (and can be proven, but not at this moment) that for any integer M there exists a prime number p greater than M and less than 2·M−2.

If there is another fraction with denominator being a multiple of Pmax than its denominator is, obviously, not less than 2·Pmax and, according to the Bertrand's postulate, there must be another prime number between Pmax and 2·Pmax, so Pmax is not the largest prime number not exceeding N.

In our example with S10 Pmax=7 and the least prime fraction is 1/7.

Now let's bring all the numbers in a series to the least common denominator (LCD) in order to sum up all numerators.
Since this least prime fraction has the largest prime denominator, this prime denominator will be represented only once in the LCD.

In our example with S10 LCD is 2³·3²·5·7=2520 and, as we see, number 7 is presented only once.

The next step to calculate SN is to bring all fractions to the least common denominator by multiplying each numerator (all numerators are 1's) by all prime numbers in the LCD that are not in a prime representation of a corresponding denominator.

Because the largest prime denominator in a series occurs only once in the LCD and does not occur in prime representation of any denominator other than that of the least prime fraction, this largest prime denominator will be represented in all new numerators of all fractions with common denominator except in the numerator of the least prime fraction.

Consequently, all new numerators, except the one of the least prime fraction will be divisible by Pmax.

In our example with S10 and Pmax=7 the prime number 7 occurs exectly once in every new numerator except for 1/7.
Therefore, the sum of these new numerators will not be divisible by Pmax and the result of the summation cannot be an integer number.

## Wednesday, December 6, 2023

### Arithmetic+ 01: UNIZOR.COM - Math+ & Problems - Arithmetic

Notes to a video lecture on http://www.unizor.com

Arithmetic 01

Problem A

Given n pieces of rope.
Some of them are cut into n smaller pieces.
Then, again, some of the obtained pieces are cut into n smaller one.
This process is repeated a few times and, as a result, the final number of pieces is N.
Prove that N−n is divisible by n−1.

Solution A

Assume, out of initial n pieces of rope we have chosen k1 to cut each into n pieces.
Then, after the first cut we will have the total number of pieces equal to
N1 = n − k1 + n·k1 =
= n + (n−1)·k1

As we see, N1−n is divisible by n−1.

Assume, out of obtained after the first step N1 pieces of rope we have chosen k2 to cut each into n pieces.
Then, after the second cut we will have the total number of pieces equal to
N2 = N1 − k2 + n·k2 =
= N1 + (n−1)·k2 =
= n + (n−1)·k1 + (n−1)·k2 =
= n + (n−1)·(k1+k2)

As we see, N2−n is also divisible by n−1.

Obviously, it's easy to prove by induction that after sth step the number of obtained pieces of rope will be
Ns = n + (n−1)·(k1+k2+...+ks )
and Ns−n will be divisible by n−1, which is what's required to prove.

Problem B

One person thinks about some number from 0 to 1000.
What is the minimum number of questions needed to find this number, if the answers can be either YES or NO?

Solution B

Considering the answers are YES or NO, one of the ways is to find the binary expression for the number.
Since the number is less than 1024=210, it's sufficient to ask 10 questions about each binary digit.

So, the first question is: "Is the first digit in a binary representation of your number as 10 binary digits (including leading zeros) equal to 1? Based on the answer, we determine this first digit to be 1 or 0.
Then we repeat the same for the second binary digit etc.

Alternatively, the number can be guessed by multiple steps of dividing the group of all possible numbers into two approximately equal parts.
Then the first question should be: "Is your number greater than 500?".
If the answer is YES, the next question is "Is it greater than 750?", but if the answer is NO, the next question is "Is it greater than 250?".
And so on.

Problem C

There is a set of 100 numbers, each from 1 to 9.
The sum of these numbers is 789.
Is it possible to choose 70 numbers out of this set with a sum less than 500?

Solution C

If we select 70 numbers with a total sum of less than 500, the remaining 30 numbers must have sum not less than 789−500=289.
But the maximum sum of 30 numbers, each of which is from 1 to 9, is 9·30=270, which is insufficient to cover 289 needed to complete the total sum to 789.

It is impossible to choose 70 numbers out of this set with a sum less than 500.

Problem D

A regular clock with large round face and 12 numbers for 12 hours is used for this problem.
A coin is placed onto each number of this clock - a total of 12 coins.
In one move two coins should be moved from their numbers into a neighboring ones, but one of these coins should move clockwise, while another - counterclockwise.

Is it possible to collect all coins on one number on the clock by performing th described moves?

Solution D

Assign a value to each position of coins as follows.
If there are ni coins on number i, the value of this coin position is
V = Σi∈[1,12] i·ni

Initially, when there is one coin per number on a clock, the total value is
V=1+2+3+4+5+6+
+7+8+9+10+11+12=78
.

In one step one coin that is on number n and steps clockwise will increase the value V of a position by 1 except when it moves from number n=12 to n=1, in which case the value of a position will decrease by 11.

In one step one coin that is on number n and steps counterclockwise will decrease the value V of a position by 1 except when it moves from number n=1 to n=12, in which case the value of a position will increase by 11.

The combination of these two steps, which constitutes a single move, will either retain the same value of a position (for example, 3→4, 10→9) or increase the value by 12 (for example, 3→4, 1→12) or decrease the value by 12 (for example, 4→3, 12→1).

In any case, no matter how many moves we make, the value of a position V is changing by a number that is a multiple of 12.

But, if we consider the value of a position when all 12 coins are on the same clock number n, the value of a position is V=12·n - a multiple of 12.

Starting from an initial position value of 78 (not a multiple of 12) and changing this value by a number that is a multiple of 12 on each move, we will never come to a position whose value is a multiple of 12.

This proves the impossibility to gather all coins on the same clock number.

## Tuesday, December 5, 2023

### Logic+ 01: UNIZOR.COM - Math+ & Problems

Notes to a video lecture on http://www.unizor.com

Logic 01

Problem A

Given a rectangular matrix with m rows and n columns.
It's filled with real numbers, and all m·n numbers are different.

First, we go through all rows and choose the smallest number in each row, getting m numbers.
Then we choose the largest among these chosen m numbers X and call it minimax.

Similarly, we go through all columns of this matrix and choose the largest number in each column, getting n numbers.
Then we choose the smallest among these chosen n numbers Y and call it maximin.

Which number is larger, maximin Y or minimax X?

Solution A

Assume. X is from row i and Y is from column j If X and Y happened to be from the same row i, X≤Y, because X is chosen as the smallest number in this row.

If X and Y happened to be from the same column j, X≤Y, because Y is chosen as the largest number in this column.

If X and Y do not share the same row nor the same column, consider an element Z from the same row iY as X and the same column j as Y.

X is smaller than Z, because they share a row i and X is the smallest in this row.
Y is larger than Z, because they share a column j and Y is the largest in this column.
Since X≤Z≤Y, X is smaller or equal to Y.

Under any circumstances X≤Y.

Examples A
A case when MaxiMin equals to MiniMax.
 Col1 Col2 MaxiMin Row1 1 2 1 Row2 3 4 3 MiniMax 3 4

A case when MaxiMin is larger than MiniMax.
 Col1 Col2 MaxiMin Row1 1 3 1 Row2 4 2 2 MiniMax 4 3

Problem B

There was a crime committed by someone and three suspects, X, Y and Z are interrogated. One of them was the person who committed this crime.

Here is the protocol of interrogations.
X stated:
(a) I did not do it and
(b) Y did not do it.
Y stated:
(a) X did not do it and
(b) Z did it.
Z stated:
(a) I did not do it and
(b) X did it.

Subsequent interrogations found that one of the suspects lied twice, another told truth twice and yet another one told one truth and one lie.

Who committed the crime?

Solution B

To solve this problem, we have to consider different cases.

Case 1: X said the truth twice.
It follows then that Z is the criminal. But that means that Y said the truth twice, which cannot be because the problem specifies that only one suspect said truth twice.
So, Case 1 is not what really took place.

Case 2: X lied twice.
This is impossible because it means that both X and Y committed this crime, but there should be only one criminal.

Therefore, what really took place was: X said one lie and one truth.
But which is which?

Assume, his (a) statement was the truth. Then his (b) statement is the lie and Y is a criminal.

Both statements of Y must be either truths or lies.
But both Y's statements being truths implies that Z committed a crime, not Y, which contradict our last assumption.
And both Y's statements being lies implies that X committed a crime, not Y, which also contradicts that same assumption.

The only remaining choice is that X's (a) statement was a lie and his (b) statement is the truth.
That implies that X committed a crime.
Now from this follows that both statements of Y are lies and both statements of Z are the truth.
It fully corresponds to conditions of the problem.

X has committed a crime.

Problem C

After work one person likes to have dinner in one of two of his favorite restaurants, let's call them A and B.
These restaurants are located on opposite ends of the city from this person's place of work.

His work ends at some random time between 5 and 6 o'clock. Then he goes down to a train station, where on the same platform two trains going in opposite directions stop.

When this person is at the platform, he takes the first train that comes and, subsequently has dinner in either restaurant A or B, depending on which train came first.

Trains in each direction come with periodicity of 5 minutes.

At the end of a year this person discovers that he attended restaurant A 4 times more than restaurant B.
What is an explanation for such a significant discrepancy?

Solution C

The timing between trains coming in the same direction is 5 minutes, but the timing between trains going in opposite direction is not specified.
Apparently, the timing between train going in the direction of restaurant A and that going towards B is a quarter of the time between a train towards B and towards A.

Assume the following train schedule.
Trains towards A come at 5:00, 5:05, 5:10, 5:15 etc.
Trains towards B come at 5:01, 5:06, 5:11, 5:16 etc.

If our person comes after 5:00 but before 5:01, he would take the train towards B, but after 5:01 but before 5:05 he would go to restaurant A.
Analogously, between 5:05 and 5:06 he would take the train towards B, but from 5:06 to 5:10 he would go to A.

As we see, he has an opportunity to go to B during 1 minute intervals, after which he would go to A during the next 4 minutes intervals.
That's the explanation of such a significant discrepancy in attendance of these two restaurants.

## Sunday, December 3, 2023

### Who Needs Problems? UNIZOR.COM - Math+ & Problems

Notes to a video lecture on http://www.unizor.com

Who Needs Problems?

Nobody.
Nobody needs problems.
People need solutions.

But how to derive with a solution for a problem that nobody before you had solved?

To answer this really difficult and complicated question is the purpose of Math+ & Problems part of this Web site.

First of all, let's clarify what we mean saying "Math+".
We all know what Math is - a set of subjects like Algebra, Geometry, Combinatorics etc.
Math+ is, basically, all of these and some less typical, unusual, tricky, unexpected aspects of all these subjects, usually presented in a form of problems that require less typical approach to solving them.

The purpose of presenting these less typical and often more difficult problems is to introduce students to thinking outside the box, when it comes to solving problems. This quality will be extremely helpful in practically all aspects of real life.

Stepping outside the usual approaches to solving practical problems is how real progress is achieved.
Math+ & Problems is a training ground for students' minds to develop the ability to invent, find, connect different pieces of information into a solution of important problem.

The problems presented in Math+ & Problems part of this Web site are, mostly, taken from books mentioned in the References page of this site. Major sources are books by Shklyarsky, Yaglom and Chentsov and also from a book by Modenov.

## Tuesday, November 28, 2023

### Newton+Maxwell=mc2:UNIZOR.COM - Relativity 4 All - Conservation

Notes to a video lecture on UNIZOR.COM

Newton + Maxwell = m·c²

In this lecture we will use the laws of classic Newtonian mechanics and the properties of light we discussed in the chapter Waves - Electromagnetic Field Waves of this course to derive the famous Einstein's equation E=m·c².

In particular, we will use the results presented in the lecture Momentum of Light of the above mentioned chapter of this course, which we recommend to review prior to studying materials presented in this lecture.

In the Momentum of Light lecture we have derived the relationship between the amount of work W performed by light moving an object with absorbing surface and the momentum p lost by this light during this process of light absorption:
p = W/c (c is the speed of light)
which, basically, means that the energy E contained in certain amount of light and this light's momentum are related as p=E/c.

Now consider the following thought experiment.
Initially, two identical objects, α and β, of mass M each are at rest and positioned on a distance 2L from each other along X-axis with coordinates −L for α and L for β.
We assume that no external forces are acting on these objects.

Since no external forces act on these objects and they are at rest, the total momentum of motion is zero and the center of mass is at X-coordinate x=0.

At some moment an object α sends a short light signal towards object β.
As we know, this light carries some energy E and momentum p=E/c.

For this totally closed isolated system the total momentum must be preserved.
In particular, it means that the center of mass should stay at the original location x=0.

It also necessitates that immediately after issuing a light signal object α must acquire a momentum −p to neutralize the momentum p of the light signal, that is, object α recoils after issuing this light signal.

This momentum of object α means that it has acquired some speed −V (the value V is presumed positive) and moves towards negative direction of the X-axis, while a light signal with speed c moves towards object β in the positive direction of the X-axis.

If light, together with its energy, does not carry some portion of object α's mass, the center of mass would shift with object α's movement, which should not be the case.
So, light signal carries not only some energy E and momentum p=E/c, but also some part of object α's mass m, leaving object α with mass M−m.

But, according to classic Newtonian Mechanics, if the light signal carries mass m moving with speed c, its momentum must be p=m·c (mass times speed of light).
Since, as we stated above, p=E/c, we have a relationship between light signal's mass and energy E it carries:
p=m·c = E/c
from which follows
E = m·c²

The last equation means that energy E lost by an object α by issuing a light signal equals to its lost mass m times a square of the speed of light.

We came to the same Einstein's relationship between mass and energy using a combination of the Maxwell theory of electromagnetic field, which gave us a relationship between light energy E and its momentum p as p=E/c, and Newtonian Mechanics with its definition of a momentum p as mass m times speed that in the case of light equals to c.

Let's continue our analysis of this experiment and find the behavior of all components involved in it.

Obviously, the light signal moves with speed c from α towards β.
It's initial coordinate is x=−L, so its movement can be described as
xsig(t) = −L + c·t
for all t from 0 to T=2L/c - the moment light hits object β.
Then it's absorbed by object β and disappears as an independent entity (see below the analysis of movement of β object.

Object α is losing mass m and moves to the negative direction of X-axis with speed Vα.
To conserve the momentum, this speed must satisfy the equation
(M−m)·Vα = m·c
Therefore,
Vα = m·c/(M−m)
and the equation of motion for α object is
xα(t) = −L − m·c·t/(M−m)

Object β stays in place at coordinate x=L up until it gets hit by a light signal at time T=2L/c.
Then β gets hit by a light signal and absorbs it energy, momentum and mass.
From the Law of Conservation of Momentum follows that the momentum of an object that absorbed the light signal, increasing its mass by m, should be equal to a momentum of this signal:
(M+m)·Vβ = m·c
Therefore,
Vβ = m·c/(M+m)
and the equation of motion for β object, considering it starts by time T later than α, is
xβ(t) = L + m·c·(t−T)/(M+m)

Let's check if the center of mass is not changing the position.
There are two stages of this experiment: before the light signal hit β object and after.

1. For time t less than T we have three objects - α, light signal and β.
Their combined "mass times X-coordinate" is
(M−m)·xα(t)+m·xsig(t)+M·L =
= −L·(M−m) − m·c·t −
− L·m + m·c·t + M·L = 0

as it should.

2. For time t equal or greater than T we have only two objects - α and β.
Their combined "mass times X-coordinate" is
(M−m)·xα(t) + (M+m)·xβ(t) =
= −L·(M−m) − m·c·t +
+ (M+m)·L+m·c·t−m·c·2L/c =0

as it should.

As we see, the center of mass is always at x=0, which is expected from a closed system.

## Friday, November 17, 2023

### Momentum of Light: UNIZOR.COM - Physics 4 All - Waves - Electromagnetic ...

Notes to a video lecture on http://www.unizor.com

Momentum of Light

In this lecture we will deal with electromagnetic field in vacuum with a flat wave front consisting of sinusoidal monochromatic (same frequency) synchronous (same phase) oscillations, which we will simply refer to as light.

We know that light carries energy.
More precisely, as we have established earlier (lectures Electric Field Energy and Magnetic Field Energy of topic Energy of Waves of the current part Waves of the course), the total electric + magnetic energy density of electromagnetic field (that is, an amount of energy per unit of volume) is
PE+M = ½·[ε·E²+(1/μ)·B²]
where E(t,x,y,z) is an intensity of electric component and B(t,x,y,z) is a magnetic component's intensity of the field.

In our case of monochromatic sinusoidal oscillations in vacuum this energy density expression is simplified to (see lecture Electric Flux Density, where we used B=E/c relation)
PE+M = ε0·E²

When the light hits an absorbing surface of some object, this energy is transferred into this object causing moving its electrons, heating and other manifestations, even mechanical movement of an object.
Let's explain the nature of this transfer of energy.

On the picture above the light propagates along X-axis, electric component E of the electromagnetic field oscillates along Y-axis and magnetic component B oscillates along Z-axis.

As vector E oscillates up and down, negatively charged electrons on an object's surface move down and up.
The Lorentz force on these electrons caused by their movement in the magnetic field B pushes them forward along X-axis.
When the direction of vector E changes to opposite, the direction of electrons' movement and vector B also change to opposite. As a result, the Lorentz force will still push the electrons forward along X-axis in the same direction.

So, electrons inside the surface layer of an object are always pushed in the same direction with pulsating force, exerting a radiation pressure on the entire object.
Let's analyze the quantitative characteristic of this pressure.

Assume the flat surface of some object is perpendicular to the direction of light propagation and its area is A.
Let the density of electric charge on this surface be σ, so the total charge on this surface is q=σ·A.

The sinusoidal electric component E(t) acts on this charge q, causing electrons to move up and down with velocity v(t).
The Lorentz force on this charge consists of electric component moving electrons along Y-axis and magnetic component pressing electrons perpendicularly to their velocity vector along X-axis.
The total force on the electrons in vector form is, therefore,
F = q·E + q·(vB)

Since only B component pushes electrons in the direction of light propagation and, as we mentioned above, B = E/c, the force that pushes object along the X-axis is
Fx(t) = q·v(t)·E(t)/c
Notice that v(t) is the speed of electrons along the Y-axis, the same axis the electric component of the field E(t) acts.

Let's deviate for a moment from the electromagnetic field and consider classical Newtonian Mechanics.
Recall the Newton's Second Law of Mechanics connecting the force F, mass m and acceleration a:
F(t) = m·a(t)

Since acceleration a(t) is a derivative of speed v(t) by time, the above can be transformed into a relation between an impulse of force and an increment of an object's momentum
F(t) = m·[dv(t)/dt] =
=
d
[m·v(t)]/dt
from which follows
F(t)·dt = d[m·v(t)] = dp(t)
where p(t) is a momentum of an object.

Using the above relation between an increment of impulse F(t)·dt and an increment of momentum dp(t), we can state that during an infinitesimal time from t to t+dt the force Fx(t) gives an object a push forward in the X-direction quantified as impulse Fx(t)·dt which is converted into an increase of momentum of an object px(t):
Fx(t)·dt = dpx(t)

Using the formula for Fx(t) above, the expression for an increment of a momentum of an object is
dpx(t) = (1/c)·q·v(t)·E(t)·dt

Let's analyze and interpret the right side of the equation above.

First of all, a product of a charge q and the intensity of an electric component E(t) is the electric force the field exerts upon a charge:
Fe(t) = q·E(t)

Secondly, expressing the speed v(t) as a derivative of the distance of electrons' movements along Y-axis, that is v(t)=dy(t)/dt, we can write Fe(t)·v(t) = Fe(t)·dy(t)/dt =
=
dW(t)/dt

where W(t) is work performed by the field to move electrons along Y-axis.

Substituting all the obtained equalities into the formula for an increment in momentum of an object above, we obtain
dpx(t) = (1/c)·[dW(t)/dt]·dt =
= (1/c)·
dW(t)

The equation above relates increment of momentum of an object in the direction of the light propagation (X-axis) and amount of work done by this light.

The Law of Conservation of Energy then dictates that the light by its electric intensity component E(t) has performed this work by transferring its own energy to move the electrons on an object's surface along Y-axis.
That, in turn, caused these electrons to push along X-axis because of action of light's magnetic intensity component B(t).
Finally, this caused a movement of an object along X-axis and increased its momentum in the direction of propagation of light.
The amount of this lost energy of light, divided by the speed of light, equals to an increment of a momentum of a movement of an object along X-axis.

But there is also the Law of Conservation of a Momentum. Therefore, that increment of a momentum of an object must be equal to a decrement in light's momentum.
That means:
(a) light carries an energy and a momentum
(b) when light is completely absorbed by a surface of some object, its energy and momentum are transferred to this object and the decrement of light's momentum equals to the decrement of light's energy divided by the speed of light.
dpx(t) = (1/c)·dW(t)
(c) an object absorbing light absorbs its energy and momentum in the same proportion

Incidentally, since
Fx(t)·dt = dpx(t)
we can express the force exerted by light on an object fully absorbing this light as
Fx(t) = dpx(t)/dt =
= (1/c)·
dW(t)/dt

The expression dW(t)/dt represents amount of energy carried by light in a unit of time.

Dividing both sides by the area A on which light falls, we will get a radiation pressure P(t)=Fx(t)/A on the left side and energy flux density divided by the speed of light on the right side:
Fx(t)/A = (1/c)·[dW(t)/dt]/A

As presented in the previous lecture Electromagnetic Energy Flux Density, the energy flux density can be expressed as Poynting vector S(t)=(1/μ)·E(t)B(t).
Therefore, in terms of Poynting vector, a radiation pressure on an object fully absorbing the light is
Pabs(t) = S(t)/c

Everything above was about the case of full absorption of the light by an object (the object fully absorbing the light is called blackbody).

Let's assume now, we have a fully reflective object.
That means that, if an incident light had momentum p, the reflected light will have momentum −p.

From the Law of Conservation of Momentum follows that, if the momentum of the light was p and became −p, the increment of the momentum of an object should be 2p.

That means, the force of radiation pressure in case of fully reflecting object should be twice as strong as in case of fully absorbing object, that is
Pref(t) = 2·S(t)/c
So, a light reflective object feels twice as strong radiation pressure than a light absorbing object.

The following experiment confirms this.

Four small squares with one side black and another being a reflective mirror are arranged like a propeller that can freely rotate on a needle inside a sphere with vacuum inside.

As soon as light falls on this propeller, it starts rotating because the mirror side has more radiation pressure than the black one.

## Sunday, November 5, 2023

### E-M Energy Flux Density: UNIZOR.COM - Physics 4 Teens - Waves - Electrom...

Notes to a video lecture on http://www.unizor.com

Electromagnetic
Energy Flux Density

As we have established earlier (lectures Electric Field Energy and Magnetic Field Energy of topic Energy of Waves of the current part Waves of the course), the total electric + magnetic energy density of electromagnetic energy (that is, an amount of energy per unit of volume) is
PE+M = ½·[ε·E²+(1/μ)·B²]
where E(t,x,y,z) is an intensity of electric component and B(t,x,y,z) is a magnetic component's intensity of the field.

In the previous lecture Energy Continuity of the current topic Electromagnetic Field Waves we discussed a concept of electromagnetic field energy flux density vector, which, based on a concept of energy continuity, we have identified as Poynting vector
S = (1/μ)·(EB)

Using the Poynting vector, we have represented the rate of change of electromagnetic energy flux as
PE+M /t = ·S
In this form this expression consttutes the continuity equation of the flux density rate of change of electro-magnetic energy.

Let's examine the correspondents of both expressions, PE+M and S, in a simple case of flat sinusoidal monochromatic electromagnetic waves.

In the lecture E-M Waves Amplitude of the current topic we came up with a simple relationship between electric and magnetic components of the plane electromagnetic waves in vacuum.

If a sinusoidal in time (t) electromagnetic field propagates along Z-axis with speed c according to these equations
E(t,z) = E0·sin(Ï‰·(t−z/c))
B(t,z) = B0·sin(Ï‰·(t−z/c))
where electrical component E oscillates along X-axis and magnetic component B oscillates along Y-axis, then
E0 /c = B0
and, therefore,
E(t,z)/c = B(t,z)

It's quite appropriate now to check if our formula for Poynting vector as an energy flux density checks in this simple case.

Replacing with E²/c² in the expression for PE+M and taking into consideration that the speed of light c in terms of electrical permittivity ε and magnetic permeability μ is expressed as c=√1/(ε·μ) (see lecture Speed of Light of the current topic), the equation for total energy density can be simplified as
PE+M = ½·[ε·E²+(1/μ)·E²/c²] =
= ½·
[ε·E²+(1/μ)·E²·ε·μ] =
= ε·E²

Consider a unit area of 1m² and light flowing perpendicularly through it with speed c.
During the unit time interval of 1s the light going through this area will fill the volume
1(m²)·c(m/s)·1(s) = c(m³)

So, the amount of electromagnetic energy flowing through a unit area during a unit of time (that is, energy flux density j) is this volume times the calculated above density of the energy ε·E²
j = c·ε·E²
The above is the magnitude of an electromagnetic energy flux density vector directed along Z-axis.

In our special case vectors E (oscillating along X-axis) and B (oscillating along Y-axis) are perpendicular to each other.
Therefore, the magnitude of their vector product equals to a product of their magnitudes and the magnitude of Poynting vector is
|S| = (1/μ)·|E|·|B| =
= (1/μ)·E·B = (1/μ)·E·E/c =
= (1/μ)·E²/c

Since c²=1/(ε·μ) and 1/μ=c²·ε,
|S| = c·ε·E²
which is exactly the same as j above calculated based on energy density.

The direction of Poynting vector is perpendicular to both electrical and magnetic components and, therefore, is along Z-axis.

As we see, Poynting vector in this special case fully corresponds in magnitude and direction to the energy flux density obtained by direct calculation based on energy density and speed of light.

This confirms (at least, in this simple case) that Poynting vector correctly represents the electromagnetic energy flux density.

## Saturday, October 21, 2023

### Electromagnetic Energy Continuity: UNIZOR.COM - Physics4Teens - Waves - ...

Notes to a video lecture on http://www.unizor.com

Electromagnetic Energy Continuity

Let us recall the definition of the divergence of a time-dependent vector field V(x,y,z,t) in three-dimensional Cartesian space (X,Y,Z), with three components Vx(x,y,z,t), Vy(x,y,z,t) and Vz(x,y,z,t) along each space coordinate presented in the lecture Divergence of the topic Electromagnetic Field Waves of the Waves part of this course:

divV(x,y,z,t) = · V(x,y,z,t) =
=
Vx(x,y,z,t)/x +
+
Vy(x,y,z,t)/y +
+
Vz(x,y,z,t)/z

In particular, in that lecture we used an example of magnitude and direction of air blown by the wind in a unit of time, a time-dependent vector field V(x,y,z,t), and its relationship with air density, a time-dependent scalar field ρ(x,y,z,t).

The final conclusion we came up with was an equation that connects the divergence of the above vector field at any point and time with the rate of change of the air density there as follows
· V(x,y,z,t) = −ρ(x,y,z,t)/t

Constant Flux Density

Consider a trivial example.
If oil of density 800 kg/m³ uniformly moves along a pipeline with speed 6 m/s and the pipe cross section is 0.2 m², the rate of oil flow is
q = 800 kg/m³ · 6 m/s · 0.2 m²

In this case the flux density j, by definition, is the rate of oil flow per unit of time per unit of area of a pipe cross section, which equals to
j = 800 kg/m³ · 6 m/s =
= 4,800 (kg/s)/m²

For an area A = 0.2 m² of a pipe cross section the total rate of flow of oil through a pipe in a unit of time is q = j·A:
q = 4,800 (kg/s)/m² · 0.2 m² =
= 960 kg/s

Flux Density

For any such "flowing" substance at any point inside this substance we define flux density vector field (usually denoted j) as an amount of this substance "flowing" through a unit of area perpendicular to its direction during a unit of time or, in other words, the rate of change of the density of this substance at a given point.

Then, given some surface and knowing the value of flux density at each point of this surface, we can evaluate the quantity of a substance "flowing" through this surface per unit of time.

In the oil flow of the example above the flux density vector at any point inside a pipe at any moment in time is a vector directed along a pipe of a magnitude j=4,800(kg/s)/m² independent of space position inside a pipe and time.
Assuming the pipe is stretched along X-axis, this flux density vector is
j(x,y,z,t) = {j,0,0}
and its divergence is
·j(x,y,z,t) = 0
because all components of the flux density vector are constants.
Therefore, according to the formula about relationship between divergence and density
the derivative of oil density by time is zero, which means that the density of oil is constant.

The above examples of an air flow because of a wind or oil flow in a pipe can be generalized to other measurable substances that can "flow" - number of molecules, mass, electric charge, momentum, energy, etc.

Assume, ρ(x,y,z,t) represents a time-dependent density in space of some substance.
Further assume that a time-dependent vector field j(x,y,z,t) represent the flux density of this substance, that is an amount of substance going through a unit of area perpendicular to the flux vector in a unit of time.

Then an equation similar to the above air flow equation
· j(x,y,z,t) = −ρ(x,y,z,t)/t
constitutes the continuity equation for this substance.

Electric Charge Flux Density

Let's examine the flow of electricity in terms of the flux density, which somewhat similar to an air wind.

Assume, we have certain time-dependent continuous distribution of electric charge in space with charge density ρ(x,y,z,t).

The amount of electricity going through some area in a unit of time is an electric current
I = dq/dt
Therefore, the electric current density (electric current per unit of area) is
J = dI/dA
where A is a perpendicular to electric current unit area, through which the current I is flowing.

The electric current density is an example of a flux density applied to electricity. Therefore, according to the same divergence theorem, the following continuity equation holds:
· J(x,y,z,t) = −ρ(x,y,z,t)/t

The equation above represents a local law of conservation of electric charge.
The charge does not appear from nowhere, neither it disappear without a trace, but gradually increases or decreases with electric current density delivering additional charge into an area or taking charge from an area.

This law is stronger than a statement that the total amount of charge in a closed space remains constant because the latter does not exclude instantaneous transfer of charge from one point to another within a closed space without any trace in between.

As such, the equation above represents a continuity of electric charge.

Electromagnetic
Field Flux Density

The next step after analyzing an electric current flux density and its relation to electric charge continuity and local law of conservation is to discuss electromagnetic field energy flux density and corresponding continuity and local conservation law.

The first difficulty is to realize that electromagnetic field is not similar to air or electric charge - those are real material substances, while electromagnetic field is not "material" in the same sense as air molecules or electrons.

Nevertheless, electromagnetic field carries energy (see lectures Electric Field Energy and Magnetic Field Energy in the topic Energy of Waves of the Waves part of this course), electromagnetic field energy is transferred in space with some speed (with speed of light in vacuum), so it must have a flux.

So, our task is to express the electromagnetic field energy flux density vector in terms of characteristics that define this field - time-dependent vector fields E(x,y,z,t) of electric intensity and B(x,y,z,t) of magnetic intensity.

In the above mentioned lectures we have derived two formulas for electric PE and magnetic PB energy densities of electromagnetic field with electric intensity E and magnetic intensity B:
PE = ½ε·E²
PM = ½(1/μ)·B²
where ε is electric permittivity and μ is magnetic permeability of the medium where electromagnetic field exists (correspondingly, ε0 and μ0 for vacuum).

The total energy density of electromagnetic field is, therefore,
PE+M = ½·[ε·E²+(1/μ)·B²]
Strictly speaking, we have to use vectors E and B instead of scalars E and B. Then the energy density would be
PE+M =
= ½·
[ε·(E·E)+(1/μ)·(B·B)]

Here are the steps that we will follow to accomplish our task of expressing the energy flux density vector field S(x,y,z,t) in terms of vector fields E(x,y,z,t) of electric intensity and B(x,y,z,t) of magnetic intensity.
.

The continuity equation relates the flux density with the rate of change of the density of whatever substance we talk about.
In our case we are analyzing the electromagnetic energy as such a substance and know how its density expressed in terms of intensities of electromagnetic field E and B.

Using the Maxwell equations, we can express the rate of E and B change (that is, their time derivative) in terms of divergence of B and E (see the Faraday Law as the equation #3 and Amper-Maxwell Law as the equation #4).

Finally, comparing the expression for the rate of change of energy density in term of divergence of E and B with an expression in terms of divergence of energy flux density vector, we will express the energy flux density vector in terms of vectors E and B.

To determine the rate of electromagnetic field energy density change, we have to differentiate the above expression for the energy density in terms of E and B by time:
PE+M /t =
= ε·(E·
E/t)+(1/μ)·(B·B/t)

Let's use the Faraday's Law (Maxwell equation #3 in this course) to express B/t in terms of E
E = −B/t

Let's use the Amper-Maxwell Law (Maxwell equation #4 in this course) to express E/t in terms of B, assuming a simple case of the vacuum (no electric current within a field)
B = μ·ε·E/t

Now the rate of change of the energy density looks like
PE+M /t =
= ε·(E·(
B))/(μ·ε) −
− (1/μ)·(B·(
E)) =
= (1/μ)·
[E·(B)−B·(E)]

At this point we can use a vector identity
Q·(∇⨯P)−P·(∇⨯Q) = ∇·(PQ)
(see the proof of this identity in Problem 3 of Vector Field Identities notes from topic Electromagnetic Field Waves of the Waves part of this course)

Using this identity for
E=Q and
B=P,
we obtain an expression for the rate of change of the energy density in terms of a divergence of vector BE:
PE+M /t = (1/μ)··(BE)

Standard continuity equation has negative rate of change equated to divergence of the flux density.
Therefore, we can rewrite the above equation in a standard form, changing the sign on the left and order of vectors on the right, getting

PE+M /t = (1/μ)··(EB)

The equation above implies that vector
S = (1/μ)·(EB)
represent the electromagnetic field energy density rate of change.
This vector S is called Poynting vector.

Poynting vector represents the direction and magnitude of the electromagnetic field energy flow.
It's perpendicular to both electric and magnetic intensity vectors.

The equation above represents the continuity of electromagnetic field energy and local Energy Conservation Law.
Not only the total amount of energy in a closed system remains constant, but also the energy movement is continuous, it gradually flowing from one location to its immediate neighborhood.

## Friday, October 13, 2023

### Amplitude of Electromagnetic Waves: UNIZOR.COM - Physics 4 Teens - Waves...

Notes to a video lecture on http://www.unizor.com

Electromagnetic Waves Amplitude

In the previous lecture Speed of Light of the current topic Electromagnetic Field Waves we discussed the relationship between electric permittivity of a medium ε (ε0 for vacuum), its magnetic permeability μ (μ0 for vacuum) and the speed of light in this medium (in vacuum speed of light is c and 1/c² equals to ε0·μ0).

In the same lecture we derived an expression for electric component of the simplest electromagnetic field oscillations in vacuum - monochromatic plane waves - as a function of time t and distance z from the source of oscillations along the Z-axis (direction of propagation of waves)
E(t,z) = E0·sin(Ï‰·(t−z/c))
where we assumed that the electric component E of an electromagnetic field regularly oscillates in the direction of X-axis, magnetic component B regularly oscillates in the Y-direction, electromagnetic waves propagate along Z-axis with speed c with angular frequency ω.

The expression for magnetic component B(t,z) of an electromagnetic field oscillations is similar and its derivation follows along exactly the same steps as the one above leading to the same formula:
B(t,z) = B0·sin(Ï‰·(t−z/c))

In this lecture we will derive a simple relationship between amplitudes of electric (E0) and magnetic (B0) amplitudes of the simplest electromagnetic field oscillations - monochromatic plane waves.

The third Maxwell equation that relates induced electric field to a changing magnetic field - the Faraday's Law - is

E = −B/t

Recall the definition of a vector (cross) product of a pseudo-vector
∇={∂/∂x,∂/∂y,∂/∂z}
by any vector
V(x,y,z) = {Vx,Vy,Vz}
using unit vectors i, j and k along the coordinate axes:
V =
= (
Vz/y − Vy/z)·i +
+ (
Vx/z − Vz/x)·j +
+ (
Vy/x − Vx/y)·k

(you can refresh this in the lecture "Curl in 3D" of this chapter of a course on UNIZOR.COM)

Considering a special characteristics of vector E during the simplest oscillations of electromagnetic field under consideration with only Ex(t,z)≠0,
E =
= (
Ez/y − Ey/z)·i +
+ (
Ex/z − Ez/x)·j +
+ (
Ey/x − Ex/y)·k =
= (
Ex/z)·j

Similarly, considering a special characteristics of vector B with only By(t,z)≠0,
B/t = −(By/t)·j

Therefore, the third Maxwell equation in this case of a simple electromagnetic field takes the form
(Ex/z)·j = −(By/t)·j

Hence,
Ex/z = −By/t

Let's use the expressions for E(t,z) and B(t,z) as sinusoidal functions above (we dropped subscripts for brevity, as other components of the field electric and magnetic intensities are zero).
E(t,z) = E0·sin(Ï‰·(t−z/c))
B(t,z) = B0·sin(Ï‰·(t−z/c))

These functions should satisfy the above differential equation for the Faraday's Law.
E/z = −B/t

Therefore,
E/z=E0·cos(Ï‰·(t−z/c))·(−ω/c)
B/t = −B0·cos(Ï‰·(t−z/c))·ω
from which follows the relationship between amplitudes of electric and magnetic components of these simple electromagnetic field oscillations

E0 /c = B0

## Wednesday, October 11, 2023

### Doppler Effect for Sound: UNIZOR.COM - Physics 4 Teens - Waves - Waves i...

Notes to a video lecture on http://www.unizor.com

Doppler Effect for Sound Waves

In this lecture we will consider longitudinal sound waves in the air and dependence of their frequency at the source of sound with a perceived frequency by an observer moving relatively to this source with a speed less than the speed of sound waves propagation.

Obviously, the analysis is applicable to other types of waves and media where these waves propagate.

We will consider the situation when the source of sound is at some fixed position and an observer is moving to or from this source with certain speed v.

The case when observer is at rest and the source of sound moves to or from an observer is no different than the above.

The case when both the source of sound and an observer are moving is just a combination of the above, as only the relative speed of an observer relative to a source of sound is really important.

For simplicity, we will consider only a one-dimensional case when both a source of a sound and an observer are always on the X-axis of some Cartesian coordinate system.
More complicated case of three-dimensional movement involves more calculations but is not important from the conceptual viewpoint.

Let's assume that a source of sound is at position x=0 and the speed of sound waves propagation in the air is u.

Let's assume further that the frequency of oscillations produced by a source of sound is f0 and waves propagate along X-axis in both direction with the same speed u mentioned above.

Then the standard relationships among parameters of oscillation are:
period T0 = 1/f0
wavelength λ0 = u·T0 = u/f0
frequency f0 = u/λ0

Case 1 - Observer moves towards a source of sound with speed v.

Let's take some relatively long time interval t substantially greater than a period of oscillations T0.
If an observer stands still (v=0) during this time interval t, considering the period of oscillations is T0, the number of waves passing the observer would be N0=t/T0.

If during time t there are N0 waves passing an observer, the perceived frequency of sound would be
N0 /t = (t/T0)/t = 1/T0 = f0
which is the original frequency of sound produced by a source.

So, a standing still observer perceives the same sound frequency as was emitted by a source of sound.

Assume now that an observer at an initial distance a (greater than zero) is moving with speed v (less than the speed of sound u) towards the source of sound.

Again, let's choose some time interval t, not very small (substantially larger than period of oscillations T0) but not too long, so an observer will not reach the source of sound (that is, t should be less that a/v).

If an observer is moving towards a source of sound with speed v during time t, he will cover the distance l=v·t and will come to a point at distance a−l from the source of sound.

Let's calculate how many waves of sound will pass an observer during this time interval.
In other words, how many waves will fall into a zone of perception of an observer.

The first wave to cross will be where an observer started, that is at distance a from the source.
Obviously, all waves that are already in the interval from distance a−l to distance a will be crossed.

In addition, all those sound waves, moving with speed u and positioned at the beginning of time closer to a source of sound at a distance from a−l−u·t to a−l by the time t will come to a distance of a−l or larger from the origin of sound, thereby crossing an interval where an observer can hear them.

Therefore, all waves that at time t=0 are at a distance from a source of sound from a−l−u·t to a will cross ways with an observer during time t.
The number of these waves is
N = (l+u·t)/λ0

If an observer hears N sound waves during time t, the perceived frequency of sound is
f = N/t = (u·t+l)/(λ0·t) =
= (u+v)/λ0 = (u/λ0)·(1+v/u) =
= f0·(1+v/u)

As we see, the perceived frequency f by an observer moving towards a fixed in the medium source of sound is greater than the frequency emitted by this source f0 by a factor 1+v/u, where u is the speed of sound emitted by a source and v is the speed of an observer.

Case 2 - Observer moves away from a source of sound with speed v.

Now the zone of an observer's perception is from a distance a from a source of sound to distance a+l, where l=v·t.

Initially, in this zone of perception there were l/λ0 waves of sound. But, as the time goes and an observer gets closer to the end of his zone of perception at distance a+l, all these waves will escape this zone because they move quicker than an observer.

Instead, some waves that are behind an observer (closer to a source of sound) will be able to be in the zone of perception and be heard by an observer because they will overcome him by moving faster.

The first wave an observer perceives is the one at his initial position at distance a from a source of sound.

The last wave he perceives is the one that by time t is at the end of his perception zone at distance a+l from the source, that is that at time t=0 was at distance a+l−u·t from a source.

Therefore, all waves that at time t=0 are at the distance from a+l−u·t to a will be perceived by an observer moving from distance a to distance a+l from a source.

The interval from a+l−u·t to a is u·t−l long.
It contains all the waves heard by an observer while he moves from distance a to a+l.
The number of these waves is
N = (u·t−l)/λ0
The perceived frequency of sound is
f = N/t = (u·t−l)/(λ0·t) =
= (u−v)/λ0 = (u/λ0)·(1−v/u) =
= f0·(1−v/u)

As we see, the perceived frequency f by an observer moving away from a fixed in the medium source of sound is smaller than the frequency emitted by this source f0 by a factor 1−v/u, where u is the speed of sound emitted by a source and v is the speed of an observer.

Incidentally, this formula shows why we consider only a case of a speed of an observer to be smaller than a speed of sound. If that condition is not met, we would have a negative frequency, so formula would not be correct.

## Monday, October 9, 2023

### Energy-Momentum Ratio: UNIZOR.COM - Relativity 4 All - Conservation - Pr...

Notes to a video lecture on UNIZOR.COM

Problem on Relativistic Momentum and Energy

Recall an effect of photoelectricity discussed in Waves - Photoelectricity chapter of UNIZOR.COM course Physics 4 Teens - a ray of light, directed towards a metal plate, kicks electrons from a metal surface.

Electrons, kicked out of metal by a ray of light, have energy and momentum.
Therefore, according to the laws of conservation of energy and momentum, light must carry energy and momentum.

In the previous two lectures of this chapter of the course we have derived formulas for relativistic momentum p and relativistic kinetic energy K of an object. These formulas depend on the object's rest mass m0 and its speed u:
p =
 m0·u √1−u²/c²
= γ·m0·u
K =
 m0·c² √1−u²/c²
− m0·c² =
= m0·c²(γ−1)
where c is the speed of light in vacuum and γ is Lorentz factor
γ =
 1 √1−u²/c²

The problem to apply these formulas to light is that, on one hand, light has zero rest mass m0=0 and, on the other hand, Lorentz factor for light (u=c) is not defined since it turns into division by zero.

In spite of all these difficulties, here is a problem.

Problem

What is the ratio of relativistic kinetic energy of light to its momentum K/p?

Solution

Obviously, we cannot substitute light characteristics m0=0 and u=c directly into formulas for energy and momentum because of undefined Lorentz factor.
Instead, let's consider an object of some rest mass m0 and very high speed u.
Then for this object
K/p = (c²/u)·
 γ−1 γ
or
K/p = (c²/u)·(1−1/γ)

Now we can substitute parameters for light:
u = c
1/γ = √1−c²/c² = 0

The result of this substitution is
K/p = c

For light in vacuum the ratio of relativistic kinetic energy to its momentum is
K/p = c
Consequently, the momentum of light can be expressed in terms of its kinetic energy
p = K/c

Historical Note

The formula for ratio of energy of a photon to its momentum was derived from Maxwell equations long before the Theory of Relativity was invented by Einstein.
Compton effect and Poynting Theorem were the bases for this derivation.
We have derived this ratio from relativistic standpoint without directly resorting to Maxwell equations and properties of electromagnetic waves.