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бутузов

ф

математический

анализ

iii

скалярные

векторные

поля

часть

00:00:09

and an important object that is widely

00:00:12

used in mathematics and physics,

00:00:15

especially in theoretical physics, this

00:00:17

object is called the Laplace operator, the

00:00:21

Hamilton operator, or the

00:00:26

operator on the drone, it is denoted by a

00:00:29

triangle like this, 3rd down,

00:00:33

and by definition this is a vector

00:00:37

operator and dpd xd pdx is

00:00:43

partial derivative operator with respect to x plus g dpd y

00:00:50

plus to

00:00:53

dpdz Well, great as usual, this is a basis of

00:00:59

unit vectors directed

00:01:02

respectively along the x and y axes of the goat,

00:01:05

with the help of this operator the

00:01:08

operations of vector analysis are very conveniently written,

00:01:11

this is the operation of the

00:01:14

rotor divergence gradient, I will remind you of this we

00:01:17

wrote out the last time, let's say the gradient of the

00:01:20

scalar field y is just like

00:01:24

multiplication, but you can better look for the action of the

00:01:27

operator

00:01:29

on the function y,

00:01:33

so you need to judge under the sign of each of the

00:01:36

operators that are terms here to

00:01:39

write this function y this will be, by

00:01:43

definition, the gradient of the scalar field y divergence

00:01:47

vector field

00:01:49

well, if for a vector field we usually

00:01:57

denote the coordinates p q r

00:01:59

if we take in coordinates I won’t

00:02:02

do this now dppg x + d cube and up to

00:02:04

y + d rpd z then through the operator on the pla this is a

00:02:09

scalar

00:02:11

i by the operator on the oblo on the vector well

00:02:15

indeed, we know that the scalar product is

00:02:19

equal to the sum of the products of the

00:02:21

corresponding coordinates, but unable and

00:02:24

coordinates d pdx dpd y dpdz due to dpd

00:02:28

x will act on the n first

00:02:31

coordinate vector a plus

00:02:33

dpd y us and depot budget for r

00:02:37

and finally the operation rotor and this is the

00:02:44

vector product these two

00:02:48

vectors are

00:02:49

vectors on the bla, this is an unusual

00:02:53

operator vector on the vector, but in

00:02:56

coordinates, how they are written, you

00:02:57

know, I’m angry from and will continue to

00:02:59

consider several repeated

00:03:01

operations when the bla operator acts

00:03:05

on some function, either scalar

00:03:07

or vector.

00:03:08

then again it acts on the

00:03:10

result of the action

00:03:12

here 1, so to speak, the actions of the operator on

00:03:16

blan this or that scalar or vector

00:03:19

field in particular, we noted that the

00:03:25

divergence of the vector field gradient y

00:03:31

means the gradient of the scalar field st but the

00:03:34

gradient itself is a vector field,

00:03:37

which means it is the nabla operator in

00:03:41

squared and acting on the function, that

00:03:46

is,

00:03:48

[music]

00:03:49

this is the operator d2 pdx squared plus d2 where

00:03:56

y squared are the

00:03:57

operators of the second partial derivatives

00:04:00

plus d2 sub and z squared, this is the

00:04:04

operator that acts on the function and this

00:04:07

operator nabla squared, it is

00:04:09

denoted by the letter delta, as it were

00:04:12

flipped the symbol on the drone and the

00:04:16

delta operator is called the Laplace operator, here is the

00:04:20

delta operator, the operator of the question, well,

00:04:24

one thing I already talked about this last

00:04:26

time is one of the fundamental equations of

00:04:28

mathematical physics, you will

00:04:31

study this exactly in a year when he

00:04:32

takes a course in methods of mathematical physics,

00:04:35

this is the Laplace delta equation u is equal to

00:04:39

zero,

00:04:40

which means you need to find well.

00:04:44

x y z belongs to some area of life,

00:04:50

but if some function satisfies the

00:04:54

Laplace equation in the area, then it is

00:04:57

called harmonic in this area,

00:04:59

this is also one of the important concepts in

00:05:03

mathematical physics, the harmonic

00:05:05

function, but this is the function that does the

00:05:08

Laplace equation, we will now

00:05:12

look at several examples

00:05:15

examples examples first example

00:05:22

let's

00:05:24

say a vector field well, let's say a as before and

00:05:29

denote let a vector field a at

00:05:31

the same time potential

00:05:34

and salty distant and potentially

00:05:38

salty and distant that it is

00:05:44

potential means that it can be

00:05:47

represented as a gradient there is no

00:05:51

scalar field right this the fact that it is

00:05:55

potential, the fact that it is virgin and

00:05:59

distant means that the divergence and that

00:06:08

is, the divergence of the gradient y is equal to zero,

00:06:16

and the divergence of the gradient is the

00:06:21

Laplace operator, in accordance with the

00:06:23

equality my that is written at the top,

00:06:25

that is, if our vector field a and

00:06:32

potential and salty distant then we

00:06:35

see that the potential scalar potential

00:06:40

of this field satisfies the

00:06:43

Laplace equation and therefore it is a

00:06:46

harmonic function, which means that it is both

00:06:50

potential

00:06:53

and salt on and a given field, the scalar

00:06:56

potential is always a harmonic function the

00:06:59

second example the second example the

00:07:02

second example will be a specific example of

00:07:04

this first example let's

00:07:07

Let's consider the

00:07:09

electric field of a point charge, but I

00:07:12

will remind the defense against m, the strength of this

00:07:17

electric field is equal to k e divided

00:07:21

by r cube and multiplied by the vector r, this is how

00:07:25

this vector field itself is expressed, and

00:07:29

last time we said that it

00:07:31

is potential and the potential of this

00:07:35

field is y equal to ka divided by

00:07:40

bears p is the distance from the origin of

00:07:45

coordinates where the charge sits to the point m

00:07:49

with coordinates x y z that is, it is the

00:07:52

square root of x square plus y square

00:07:56

plus z square well, and we noted with you

00:08:01

that this field is also

00:08:03

potential its scalar

00:08:05

potential and salts on and far

00:08:08

divergence ex

00:08:12

means the divergence e is equal to zero zhurn

00:08:18

exists when p is not equal to zero

00:08:21

approximately equal to zero, that is, at all

00:08:23

points except the origin of coordinates

00:08:25

and if so, then by virtue of the first example,

00:08:31

this scalar potential cayenne p

00:08:35

satisfies the equation Laplace means

00:08:38

delta y is equal to I, of course, which constant

00:08:42

factor can be taken out, this will be the

00:08:45

Laplace operator of the applied function days

00:08:48

price r is equal to zero when r is unequal,

00:08:53

note that we got this without any

00:08:56

calculations directly of this

00:08:59

value delta from one to r

00:09:01

based on the first example, but leave

00:09:04

a little the place is interesting and

00:09:06

how to directly calculate it, calculate

00:09:09

the delta of one by r what does this mean

00:09:12

exactly already calculate the second derivative

00:09:14

with respect to x of one by r the second derivative with respect to

00:09:17

y add the second derivative with respect to z and

00:09:20

really get 0 so this is

00:09:23

such a small task, a small

00:09:25

task for you to make sure in this, directly

00:09:28

calculating the result of the action of the

00:09:31

Laplace operator on the function unit on

00:09:33

r

00:09:34

third example 3 examples let the vector

00:09:39

field and let us here its

00:09:43

coordinates with the pqr coordinates be important this

00:09:47

function x y z let the vector bullet be

00:09:51

salty far and without vortex

00:09:56

is salt on and far and without

00:09:58

vortex with she and distant means

00:10:02

divergence is zero without vortex

00:10:05

means that the company is zero well, in

00:10:08

particular, if it is potential we

00:10:11

know the potential field is always without

00:10:13

vortex and the gradient rotor if

00:10:17

the potential field of the vector is equal to the gradient

00:10:20

aru targaryen then it is always equal to 0

00:10:23

but there is even a more general the case when

00:10:26

the policy is almond and without vortex, the fact

00:10:29

that it is salty is distant, which means that

00:10:33

the divergence a and here we

00:10:36

will write it in the coordinates dpp dx

00:10:39

+ d cube and up to y + d rpd z is equal to zero, then the fact

00:10:51

that it is without vortex

00:10:56

means that the rotor of this field a is equal to the

00:11:00

zero vector and we will also write this in

00:11:04

coordinates in coordinates this is replace once we

00:11:07

have already written this is written like this dpp

00:11:10

to y is equal to deque pdx 3 the equality

00:11:16

further n goes into r&d and y goes into

00:11:21

z cyclic permutation drp z is equal to

00:11:27

oh excuse me and goes into q in the

00:11:31

numerator dpd efficiency z equals dr falls

00:11:37

and finally one more equality is dr

00:11:41

under and x equals dp lucky and so our field is

00:11:51

vector and with coordinates pqr salts on and

00:11:55

far

00:11:56

means the first equality is satisfied and without

00:11:59

vortex three more are satisfied equality and

00:12:01

now from these equalities we will extract

00:12:05

some other equalities, namely,

00:12:08

we will differentiate the first equality with respect to x,

00:12:12

we will differentiate 1 is equal to the soup from

00:12:14

we get d2 ppd x square plus,

00:12:20

then we take the derivative with respect to x from the deck

00:12:23

pads y means plus d2 will be xd y +

00:12:32

the last term tr pd z we take the

00:12:35

derivative with respect to x,

00:12:36

which means it will be d2r under and xd z is equal to 0, so

00:12:44

when differentiating 1 knapsack with respect to x,

00:12:46

now the second equality is this, here I will

00:12:51

underline it with

00:12:52

one line; the equality underlined by one line

00:12:54

will differentiate with respect to y,

00:12:58

we get d 2 n under and y square is equal to d

00:13:06

2 cube falls in

00:13:08

the game since we

00:13:10

have differentiated this equality in the game, so I

00:13:12

also underlined it with one line

00:13:14

and finally we will need the

00:13:17

last equality, let me

00:13:19

underline it with two lines, we

00:13:22

will differentiate it by z, which means it will be

00:13:27

in the forest on the left d2r by dxd z is equal to the right d2

00:13:35

ppd z in the city let's underline with two

00:13:40

lines so it will get and by

00:13:42

differentiation with respect to z

00:13:44

from that the equality pas cher with feet

00:13:47

outlines and now look d 2 cube and d

00:13:51

x on y this is 2y square means here in the

00:13:55

first equality we can replace

00:13:57

d2 falls behind and track by d2 ppd y

00:14:02

square and by virtue of the last equality we

00:14:06

can replace this is your rpd xd z with d2

00:14:09

hit z square and what then

00:14:12

happens

00:14:13

when we get d2 ppd x square plus

00:14:21

d2 ppd y square plus d2 p2p

00:14:30

go z square is equal to zero but after all on the left

00:14:36

is nothing more than the Laplace operator

00:14:39

acting on the function n,

00:14:41

which means the same thing can be written

00:14:45

delta p is equal to zero, similarly

00:14:51

it is proven that it is desirable that you

00:14:53

do similar calculations that

00:14:56

delta p is also equal to zero and delta p

00:15:00

is equal to zero

00:15:04

what is the conclusion from here the conclusion is if

00:15:07

the vector field and the salts are on and distant and without

00:15:11

vortex and then the coordinates p q r of

00:15:15

the scalar potential harmonic

00:15:17

functions you can smoke in this case,

00:15:21

well, what case at the very beginning we

00:15:23

said polis legendary without vortex

00:15:25

in this case the

00:15:26

coordinates of the vector and harmonic

00:15:31

functions and finally one last

00:15:37

example

00:15:38

it It’s interesting because we’re now

00:15:41

going to bridge the gap to another mathematical

00:15:44

course that you’re taking in the tfkp type in the

00:15:47

theory of functions of a complex variable,

00:15:49

consider a flat field, consider a

00:15:52

flat field means a vector and this is a

00:15:58

vector with coordinates p from x y z

00:16:02

answer there is no dependence on x and y 0 from

00:16:09

of this and there is no component due to the vector

00:16:12

lies in the x y plane

00:16:14

of this and there is no component and its

00:16:16

coordinate depends only on x and y and

00:16:19

again let this be a field, let this field be a salty long-range

00:16:24

one

00:16:25

and without a vortex and let this be a field

00:16:29

of a flax solenoid and without a vortex

00:16:36

well the fact that it is salty distant means

00:16:39

the divergence a is equal to zero and from here we

00:16:44

get dpp dx

00:16:48

+ d puppey d y

00:16:51

well and r is equal to zero means drp z is simply 0

00:16:55

equals 0 equals zero and therefore deku pada y is

00:17:03

equal to minus dpp dx

00:17:08

we emphasize this equality

00:17:11

now we will use the fact that our field

00:17:15

by condition without a vortex means the rotor a is

00:17:20

equal to the zero vector,

00:17:23

which follows from here those

00:17:27

three equalities that are now

00:17:29

written there are written at the top, but two of them

00:17:31

turn into simply 0 equals zero

00:17:34

since we have r equals zero

00:17:37

guarding does not depend on z,

00:17:40

well, let's take a look orally look at

00:17:46

this equality will remain to the navel dx

00:17:48

work ppd y to the navel dzt then 0 because

00:17:52

kuni depends on z or rpd y is 0

00:17:55

because r is equal to zero so this equality is

00:17:58

obviously fulfilled this also

00:17:59

holds means from the fact that the company is

00:18:03

equal to zero zero vector, more precisely,

00:18:06

we get that deco pdx I

00:18:10

will write in this order

00:18:11

equal to dpp to y, we will also emphasize this

00:18:18

equality

00:18:20

and so the underlined equality let's

00:18:24

write from are Riemann's porridge conditions are

00:18:30

Riemann's porridge conditions

00:18:36

for functions of a complex variable

00:18:43

underlined equality are

00:18:45

Riemann porridge conditions for functions of a

00:18:48

complex variable variable here is

00:18:50

f from z well z is a complex variable but

00:18:55

somewhere here on the side you can write

00:18:57

z is x plus

00:18:59

and y and imaginary unit what is it you

00:19:04

know and the facet is equal to q from x and y + and p from

00:19:12

x yrek well maybe you will have

00:19:15

other notations there, not q and p,

00:19:17

but any function, complex variables

00:19:20

can be represented in this form, the

00:19:22

real part is some function f of

00:19:25

x and y + and the imaginary part in this case 5

00:19:29

x y

00:19:30

you, under the Riemannian conditions, they consist

00:19:34

of two equalities, the derivative of coupe

00:19:37

x is equal to the derivative of g with respect to y, this is an

00:19:41

equality, and the derivative of coupe y is equal to

00:19:43

minus the derivative of ppx,

00:19:45

this is this equality, and if they are

00:19:48

satisfied, then the function f of z, this

00:19:51

function is an analytical function, an

00:19:56

analytical function, well, we

00:20:07

looked at several examples of related

00:20:09

with the Laplace equation and in very detail,

00:20:13

like the Laplace equation, you will, as

00:20:14

I have already said many times about this,

00:20:17

study in the fifth semester, that is, in the fall

00:20:20

of the third year,

00:20:25

well, we completed their paragraph, which

00:20:29

we called it, it was paragraph 4,

00:20:31

which was called the Hamiltonian operator,

00:20:33

and we move on to last fifth paragraph

00:20:36

in this chapter paragraph 5 operations of

00:20:42

vector analysis,

00:20:44

well, I mean operations gradient

00:20:46

divergence rotor operation of vector

00:20:49

analysis in curvilinear

00:20:52

orthogonal coordinates in

00:20:56

curvilinear

00:20:57

orthogonal coordinates operation of

00:21:02

vector analysis in curvilinear

00:21:05

orthogonal coordinates,

00:21:17

well, you and I when in the department gradient

00:21:21

then divergence rotor we introduced them in

00:21:24

Cartesian rectangular coordinates,

00:21:26

but they were later shown that in fact the

00:21:31

result does not depend on the choice of coordinate system,

00:21:34

that the gradient of the scalar

00:21:39

divergence field, the rotor of the vector field does not

00:21:41

depend only on the field itself, no matter in

00:21:44

which coordinate system we consider them, and

00:21:46

we even obtained some invariant

00:21:48

definitions of these operations, but in

00:21:52

mathematics, especially in physics, it is very often

00:21:55

convenient to use not

00:21:58

Cartesian rectangular coordinates at all, but

00:22:01

curvilinear coordinates, but in particular,

00:22:03

we are familiar with cylindrical

00:22:06

spherical coordinates, and then

00:22:09

formulas are required for the

00:22:12

rotor divergence gradients, in these

00:22:14

curvilinear coordinates, what is it?

00:22:16

curvilinear and,

00:22:18

moreover, orthogonal coordinates,

00:22:20

now we’re starting to introduce this concept, well,

00:22:23

just the concept of curvilinear coordinates, are

00:22:26

we already in the water? Was it in the second

00:22:28

semester in the chapter devoted to multiple

00:22:31

integrals? This was chapter 11,

00:22:33

so we’ll be on

00:22:36

this refer to the chapter and so let x y z be the

00:22:43

rectangular coordinates of the point m the

00:22:46

rectangular coordinates of the point m

00:22:51

and

00:22:55

we will denote the curvilinear coordinates

00:22:58

as q1 q2

00:23:02

q3 well, curvilinear coordinates are always

00:23:11

somehow related to rectangular coordinates of

00:23:13

some kind, the equality my in particular for

00:23:15

cylindrical and spherical coordinates

00:23:18

we will remember these today formulas and in the

00:23:21

general case we will write this means let the

00:23:24

curvilinear coordinates be related to the

00:23:26

rectangular ones the equality me

00:23:29

x is some kind of function it is convenient for him to

00:23:33

denote by the same letter x x from q1 q2 q3

00:23:41

y some function is convenient to denote by the

00:23:47

same letter y from q1 q2 q3

00:23:54

well and z is some function of z from q1 q2

00:24:02

q3

00:24:03

let's number these formulas of

00:24:07

formula-1 and so there are rectangular

00:24:13

coordinates x y z and some

00:24:14

curvilinear coordinates,

00:24:18

let's fix the values of coordinates q2 in

00:24:23

the morning and then give them some

00:24:25

numerical values

00:24:26

this is q2 and q3 we have fixed let's

00:24:34

then look at equation 1

00:24:36

if we have fixed given quite

00:24:38

definite numerical values u 2 sub 3

00:24:42

what do these equations turn into then

00:24:44

what do they describe then x and y and z are

00:24:48

functions of one variable q1 this is a

00:24:51

parametric equation of the curve hero the

00:24:54

parameter plays q1 and so then equation

00:24:56

1 from signs and describe a parametric

00:25:01

curve in brackets network level

00:25:05

y one parameter then equation 1

00:25:09

describes a parametric curve q1

00:25:13

parameter

00:25:17

which is called let’s even

00:25:21

draw it, so on this curve at

00:25:25

all points q2 and q3

00:25:27

have the same values change

00:25:29

only q-1 which is called

00:25:32

the coordinate q1 line q1 line and so

00:25:42

fixed q2 and q3 only q1 changes

00:25:45

and then these equations describe a

00:25:47

certain curve and a second line which is

00:25:49

called the coordinate q1 line well,

00:25:54

doublings are determined in the same way and q3 the line in the basement does not

00:25:57

change only q2 from q1 about the morning

00:25:59

constantly on q3 lines changes only q3

00:26:06

through each point in space

00:26:09

there are 3 coordinate lines,

00:26:12

well, somehow I conditionally pick up q1 line

00:26:18

q2 line y 3 line

00:26:24

here. m through each point m

00:26:29

space there are 3 coordinate

00:26:32

lines

00:26:35

curvilinear coordinates q1 q2 q3

00:26:38

are called orthogonal

00:26:44

curvilinear coordinates q1 q2 q3

00:26:47

are called orthogonal

00:26:52

if for each point

00:26:56

if for each point 3 coordinate

00:27:02

lines passing through this point 3

00:27:07

coordinate lines passing through the

00:27:09

point pairwise orthogonal pairwise

00:27:14

orthogonal

00:27:16

so curvilinear coordinates q1 q2

00:27:19

Potter are called orthogonal if

00:27:21

for any point for any point m

00:27:24

there are of course 3 coordinate lines

00:27:27

passing the point in pairs orthogonal the

00:27:29

singularity is pairwise orthogonal you need to

00:27:31

write that is, the tangents to these lines

00:27:35

at point m well, here I will conventionally depict

00:27:38

these tangents, that is, the tangents to

00:27:41

these the lines at point m are pairwise

00:27:45

perpendicular, what is the angle between

00:27:47

two intersecting curves, this is the angle in the

00:27:49

form of tangents at the point of intersection,

00:27:51

so if we bring the tangent of

00:27:53

each line to the point of intersection,

00:27:55

then they are pairwise perpendicular, that is,

00:27:58

they are located relative to each

00:27:59

other like a rectangular axis

00:28:02

rectangular coordinate system

00:28:04

Well, let's give examples of this and there will be

00:28:09

cylindrical and spherical coordinates that are closely familiar to you and me,

00:28:12

so examples

00:28:15

of the first example are

00:28:20

cylindrical coordinates cylindrical

00:28:24

coordinates

00:28:26

they are denoted by the usual ROC MP, let's remember

00:28:33

the equation of how Cartesian rectangular

00:28:35

coordinates x y z are expressed in terms of

00:28:38

cylindrical

00:28:40

x is equal to r cosine phi

00:28:47

r cosine phi y is equal to p

00:28:54

sine and nitrogen is the same as in

00:28:59

rectangular coordinates z is equal to z within

00:29:02

what limits do the cylindrical

00:29:05

coordinates change p is greater than or equal to zero fi

00:29:10

varies from 0 to 2n and sometimes

00:29:15

they write greater than or equal to 0 less than or

00:29:17

equal to 2 pet but it is enough to write less than

00:29:20

2 peak because prefer for Aries 2pi is

00:29:22

the same as prefer equal to zero and the

00:29:26

color changes from minus to plus

00:29:29

infinity,

00:29:38

which means these are cylindrical coordinates, well,

00:29:43

what about them are

00:29:45

orthogonal, this is the fastening of cards Natasha,

00:29:48

not orthogonal, we need to take any

00:29:50

point, we’ll do this a little later and

00:29:53

draw it draw the corresponding drawing

00:29:58

through this point p the Fellini line

00:30:02

y and z line what do you think what

00:30:06

is the p line

00:30:10

ray sirloin

00:30:15

for only the fairy a and since this

00:30:17

circle and the light isn’t the whole straight line

00:30:20

parallel to the z axis, well, we’ll

00:30:22

draw that again these are cylindrical

00:30:25

coordinates and the second example is spherical

00:30:30

coordinates p

00:30:33

you fe remember the formulas for the

00:30:38

connection between physical coordinates and

00:30:42

rectangular coordinates means x

00:30:45

equals x equals r sine theta cosine phi

00:30:55

y equals r sine theta sine x and z equals

00:31:06

who remembers r cosine theta from physics does not

00:31:11

depend on r cosine theta

00:31:13

within what limits the cylindrical

00:31:16

coordinates change p is again greater than or equal to 0

00:31:19

although this is not the same p we will also draw

00:31:22

a picture in due time

00:31:23

it means not the same not the same thing that in a

00:31:27

cylindrical coordinate system you

00:31:31

change from zero to peak and the

00:31:35

inequalities are not strict on both sides

00:31:38

nua fe as well as in cylindrical

00:31:41

coordinates greater than or equal to greater than or

00:31:43

equal to zero less than 2 pet, again in

00:31:47

order to make sure that these

00:31:48

curvilinear coordinates are orthogonal, you

00:31:51

need to draw a picture, draw, take an

00:31:56

arbitrary point, draw early newt

00:32:00

this line fi line and make sure that the

00:32:03

tangents to these lines at this point are

00:32:05

perpendicular in pairs, but now we

00:32:09

will return from this from these examples to the

00:32:12

general case, which means let the

00:32:18

curvilinear coordinates again be designated q1

00:32:23

q2 q3

00:32:31

consider the segment of the coordinate q1 and do not

00:32:36

consider the segment of the coordinate q1 line

00:32:40

what kind of cut is that it will become clear that

00:32:42

we have q1 line, which means this is q1

00:32:47

line, which means that only q1 q2

00:32:51

and q3 are fixed here, we will mark some

00:32:54

point m with curvilinear coordinates q1

00:32:59

q2 q3 and some point m 1 with

00:33:10

coordinates such that it is clear that q1

00:33:13

has changed since such 1 line we

00:33:16

will denote the first coordinate except q1

00:33:19

plus d q1

00:33:23

nuoku two 3 are the same as y at

00:33:26

point m because we have this q1 or not

00:33:29

and this one to the butterfly, that is, how far

00:33:33

we have moved the deck one will be considered

00:33:37

arbitrarily small and positive, that

00:33:41

is we have moved towards an increase in

00:33:43

q1, no matter how small in general, here are the words, no matter how

00:33:46

small, they are often used by physicists

00:33:49

and in many ways our further reasoning

00:33:52

will not be absolutely strictly

00:33:54

mathematical, and we don’t need this, but

00:33:56

as they say, this will be reasoning at the

00:33:59

physical level of rigor, let’s denote the

00:34:02

length of this segment q1 lines through dr1

00:34:09

through d or 1

00:34:15

nuuy points m m 1 there are also

00:34:19

rectangular coordinates

00:34:21

let's denote the rectangular coordinates of the point

00:34:24

m x y z

00:34:27

and the rectangular coordinates of the point m1

00:34:33

generally speaking when we look like a leaf there is

00:34:35

one there for two matins

00:34:38

change only q1 well x y z depends on

00:34:43

q1 This means they will also somehow change

00:34:45

coordinates. m1 we denote x + d x y +

00:34:51

d y z plus z what is

00:35:01

dx

00:35:04

this is how much it changed in its

00:35:06

coordinates when we moved from point to

00:35:09

point m 1

00:35:10

but we know we have equations 1

00:35:16

but they have already left there the power of

00:35:19

equation 1 dx equals x in point and at point

00:35:27

x1 plus d q1 q2 q3 that is, them at point m

00:35:34

1 minus x at point m that is, minus x from

00:35:40

q1 q2

00:35:42

q3 this is equal to well, look at the arguments q2

00:35:49

and q3 are the same and the first argument

00:35:53

has changed to deck 1 night we Let's apply the

00:35:57

Lagrangian formula of finite increments,

00:36:00

which means equal,

00:36:01

under this equal sign you can read it

00:36:03

in the Lagrangian form,

00:36:04

this will be the partial derivative dx under and

00:36:08

q1 multiplied by d q1 where the partial derivative is taken,

00:36:17

well, it is taken at some intermediate

00:36:20

point between the points m m 1

00:36:23

similarly similar to d y it will be d y

00:36:29

pdx 1 to y according to codec one excuse me on the

00:36:36

adding deck and even this will run under and

00:36:44

q1 on daewoo agent well here we will write

00:36:50

where the derivatives are taken at some

00:36:53

intermediate points between m&m one where the

00:36:56

derivatives are so the Lagrange formula

00:36:59

where the derivatives are taken at some

00:37:01

intermediate points between m m 1

00:37:08

but we believe that deku one is arbitrarily

00:37:11

small, so we will count and

00:37:15

here, since it’s not a certain rigor but a

00:37:19

physical level of rigor,

00:37:21

that means we will assume we will assume that the

00:37:24

derivatives are taken at point m we will

00:37:28

assume that they will be taken at point m

00:37:30

Thus, these equalities are true as they say.

00:37:33

accuracy up to infinitesimals of the second

00:37:36

order, but if we expand there according to the

00:37:40

Taylor formula, then the center of embedding is at point m,

00:37:43

these derivatives will not do this, then the

00:37:46

main term

00:37:47

is the value of the derivative at point m plus an

00:37:52

addition of

00:37:53

order deco 1 and this is also multiplied by

00:37:57

dy q1 so that these formulas are correct up

00:38:01

to infinitesimals of the second

00:38:03

order

00:38:05

so we have calculated dxd y de cette and then

00:38:10

we can find del 101 this is clear the

00:38:14

root of b x squared plus y

00:38:19

squared plus d z squared this is clear

00:38:30

why it is clear it means this is actually the

00:38:33

Pythagorean theorem

00:38:34

three-dimensional Pythagorean theorem if you, if

00:38:38

we draw the x y z axes there, respectively,

00:38:42

or more precisely, not the axes themselves, but the straight lines

00:38:45

parallel to the axes, then this

00:38:49

segment of the coordinate lines will be

00:38:53

the diagonal of a rectangular parallelepiped

00:38:55

with sides up to exo y and z, we will equally

00:38:59

substitute these expressions for dxd y and

00:39:03

z in each term we will have a

00:39:07

multiplier for q 1 squared, we will take it out of

00:39:10

brackets and extract the root, taking into account that before

00:39:13

q1 we took positive, which means the root

00:39:15

will be plus

00:39:16

this will be the square root of d xp deco

00:39:21

1 squared plus d and flu deco 1

00:39:31

squared plus d z feat u1 squared

00:39:39

well, we took out deco 1 as the sign of the root on

00:39:43

deco one, this square root we will

00:39:49

denote by the letter h the big first

00:39:54

since the derivatives are taken at point m

00:39:58

with curvilinear coordinates q1 q2 q3

00:40:00

then this h1 depends of course on q1 q2 q3

00:40:06

but in order not to clutter up the record, we

00:40:09

will often not write arguments this way and that way,

00:40:12

as if an element of length up to or one

00:40:18

when we say such words element of

00:40:20

length when we consider that this is this

00:40:23

piece of a coordinate line, no matter how

00:40:25

small, we have only one arbitrarily small

00:40:28

and then we say that this element is a

00:40:30

long element of length on the line fu1

00:40:33

expresses with the following formula this and went to the

00:40:37

deck alone similarly similarly to the

00:40:45

q2 line on the q2 line the element of length I will

00:40:54

designate it as dale 2 will be equal to as much as with

00:40:58

index 2

00:40:59

on deck 2 Now we’ll write down what h2 is,

00:41:02

although it’s clear that if h1 has

00:41:05

the sum of squares at the root of the argument q1, then h2

00:41:10

will have everything the same, only the derivative

00:41:12

will be dad q2,

00:41:13

and on y, there are three lines at morning

00:41:26

d3, this will be h3

00:41:30

on deadpool 3, well, I’ll write immediately the general

00:41:36

expression for h

00:41:37

we have three quantities h1 h2 h3 as much as

00:41:41

this is me this is the square root of dx by deco and

00:41:46

t squared plus d y by 9 squared

00:41:55

plus d z by deco and t squared and

00:42:04

takes the value 1 2 3 these

00:42:10

quantities are night a special name for the

00:42:13

quantity as much as I am called parameters of the

00:42:16

lama

00:42:23

parameters of the

00:42:24

lama by the name of one of the previously

00:42:30

living mathematicians and physicists parameter

00:42:33

lamin what they actually show is

00:42:36

if we again look at this segment

00:42:40

of 1 line, then the parameter shows you

00:42:45

when we moved from the point m sq . m 1

00:42:48

how much the length has changed,

00:42:51

well, here we are going from some point, we have reached

00:42:54

point m, this is some length of ours, have

00:42:59

n’t we moved to point m 1 how much has the

00:43:02

length changed if the parameter q1

00:43:04

has changed to the deck one here is

00:43:06

proportional to the deck a dance

00:43:08

coefficient respectively h1 and on the

00:43:11

lines q2 and q3 with coefficients h2 h3, well,

00:43:16

sometimes they are called scale

00:43:17

factors night lama parameters or

00:43:20

scale factors

00:43:24

let's see what these

00:43:28

scale factors are equal to in cylindrical and

00:43:31

spherical coordinates

00:43:33

at night cylindrical coordinates here

00:43:43

we have them, let's raise them so that they are

00:43:49

before our eyes formulas connecting

00:43:59

here are formulas connecting x y z with

00:44:05

cylindrical coordinates our father

00:44:08

means h1 h1 and the

00:44:13

coordinate reverently t.r. father means

00:44:17

h1 this will be the square root of d

00:44:22

expo d p squared plus d and flu d p

00:44:27

squared plus d z pd squared

00:44:41

well, counting muse on

00:44:43

and take a break dx pd what is the

00:44:49

cosine equal and look at the formula to y pd

00:44:52

sine x even pd zero setter does not depend

00:44:55

on the root cosine x plus sine squared

00:44:58

this unit

00:44:59

means age1 is equal to one

00:45:02

who calculated h2

00:45:08

h2 this refers to fair

00:45:12

this is r of course, etc. I won’t calculate it according to

00:45:16

the formula, we will now, and finally,

00:45:18

h3

00:45:19

is this one, these are the same fortune tellers,

00:45:25

these same exact values can be

00:45:28

obtained in pure geometric terms, what we

00:45:31

will do with you, let’s draw such

00:45:36

a picture, which means this is a rectangular

00:45:40

coordinate system x y z

00:45:43

and we took some point now I

00:45:50

’ll mark it a little later and through this point we’ll draw

00:45:54

three coordinate lines

00:45:56

p the fillet line and the z line, well, to

00:46:01

do this I’ll first build a

00:46:05

circle in a plane perpendicular to

00:46:08

tim 10 and z

00:46:09

is x and this is y let’s draw a

00:46:14

circle in a plane perpendicular

00:46:17

axis z and on this circle I will mark a

00:46:21

point m

00:46:24

a little second here, well, here

00:46:31

through this point let's draw yes

00:46:35

at this coordinate point and r z

00:46:41

cylindrical where here and where here

00:46:46

fig where here z let's denote the origin of the

00:46:50

Cartesian system with the letter o

00:46:53

center circle is designated by the letter c

00:46:56

through point m we will draw a straight

00:46:59

line parallel to the z axis, here it is somewhere here that

00:47:03

will intersect at a point as much as the plane y x y but

00:47:11

we will draw this r line

00:47:15

r line here c

00:47:21

m.a. the ray,

00:47:25

well, what can I say, is not a straight line, but the ray cm is

00:47:28

early not an action, then here and or changes,

00:47:30

here we have cm equals r on this line p

00:47:37

changes, if we understand we are moving, the color does not

00:47:40

change because this line is

00:47:41

parallel to the plane x and y and fi also does not

00:47:44

change angle fi

00:47:46

it is best visible here, here it is from the

00:47:52

x axis,

00:47:53

this angle fi is plotted and the

00:47:59

circle is a Fellini circle and the

00:48:03

taffy line on this line r does not change

00:48:07

and the native one

00:48:08

is also a circle of radius r z also does not

00:48:10

change and the z-line is what a straight line this is a

00:48:14

straight line mash here z changes air does not

00:48:18

change efi does not change means this is the

00:48:20

z-line and now let us go a little from point

00:48:27

m on each coordinate line

00:48:31

and take another

00:48:34

point well on the r line we will take Odessa I will

00:48:37

not designate m1 on Fellini and so

00:48:41

fi increases in this direction, so I’ll

00:48:45

highlight it so it’s fatter, this will be it.

00:48:47

m2 and back lines from point m we will move to

00:48:52

point m 3

00:48:56

then

00:48:59

d or 101 well this is m m 1 the

00:49:09

length of this segment is r line

00:49:12

and if we are at point m 1 here our first

00:49:20

coordinate is equal to r and here we count r

00:49:23

+ dr

00:49:25

means this is simply equal to the increment of the

00:49:28

coordinate p this is equal to dr well, let’s emphasize the

00:49:33

left right part

00:49:36

here is the coefficient that stands in front of dr this is

00:49:39

h1

00:49:40

and the coefficient is one, it’s clear

00:49:43

from here we see it follows that age1

00:49:47

is equal to one

00:49:49

now ds-2 segment on Fellini,

00:49:55

we moved from point m to point m 2,

00:49:58

which means this is the length of the arc, the length of the arc is mm2,

00:50:03

well, I’ll just write mm2, although here we need to

00:50:06

have the length of the

00:50:07

segment, mm2, the length of the arc, what is

00:50:10

this arc length, if we consider that at

00:50:14

point m 2

00:50:15

phi has changed on the def this is this corner

00:50:19

delphi

00:50:21

circle of radius r

00:50:23

central angle defy and what is the arc the

00:50:27

length of the others erna dayhi equals pir on

00:50:32

defy

00:50:33

again we look at the left right participation

00:50:36

here is the coefficient for dpr this is h2o

00:50:41

h2 equal to p

00:50:43

well and quite simply on the z-line

00:50:47

we moved from point m to a pile of m3, we consider

00:50:51

that there was for this and the coordinate is

00:50:53

equal to z and at point m 3 changed for

00:50:56

dessert, which means dm3 dm3

00:51:03

is m m 3

00:51:06

this is equal to z knives and before to z

00:51:09

one

00:51:10

means h3 is equal to one so pure

00:51:15

geometrically, without any formulas, you can

00:51:19

determine the parameters of the lama for

00:51:21

cylindrical coordinates;

00:51:24

the same

00:51:26

for spherical coordinates means

00:51:29

now let’s consider the spherical coordinates

00:51:35

p you taffy,

00:51:39

well, I’ll only depict the lines themselves, but

00:51:43

then you’ll do it in detail yourself,

00:51:46

due to our rectangular

00:51:48

coordinate system and

00:51:50

x y z well . and

00:52:01

we’ll give the origin of coordinates here, so I’ll continue down and z and

00:52:07

draw an arc of a circle of radius r,

00:52:16

but we believe that if this point

00:52:20

is projected onto the

00:52:23

x y plane, then this circle

00:52:27

lies here, well, let’s

00:52:30

assume that this projection is

00:52:33

our

00:52:34

corner and We count hey from the x axis

00:52:38

and the angle theta is the angle we count hey from the

00:52:44

z axis nuage.

00:52:47

m here we have .

00:52:51

m with coordinates rtt fi means p is the

00:52:56

distance a.m.

00:52:57

you, this is this angle here, I’ll mark it with

00:53:00

two arcs,

00:53:01

and fi, this is the one that is depicted here

00:53:05

for both spherical and also for

00:53:10

cylindrical coordinates. It’s easy to look

00:53:13

at the picture to make sure that these are

00:53:14

orthogonal coordinates,

00:53:16

let’s look at the cylindrical one, for example.

00:53:20

picture for a cylindrical

00:53:22

coordinate system means at this point here we have an

00:53:26

r line, a Fellini circle and a

00:53:30

straight z-line parallel to the z axis, it is

00:53:33

absolutely clear that this is a z-line, it is perpendicular to the plane of the

00:53:36

circle

00:53:38

perpendicular to the

00:53:40

straight line, that is, a seni care line, if we

00:53:43

draw a tangent to the circle,

00:53:45

it is perpendicular as you know from the

00:53:48

school geometry course, the

00:53:49

point as is drawn to the radius, they locked it, not that

00:53:52

they are

00:53:55

orthogonal curvilinear coordinates,

00:53:57

obviously the same thing, look

00:53:59

carefully at the drawing, so to speak, and

00:54:01

see for yourself that the spherical

00:54:04

coordinates are also curvilinear

00:54:06

orthogonal coordinates,

00:54:08

but the parameters of the lama h1 I give the answer and you

00:54:16

will draw the conclusion of this form yourself

00:54:19

corresponding to the p coordinate, this unit

00:54:22

h2 is the lama parameter for the theta coordinate,

00:54:26

this is r and h3 for the fi coordinate,

00:54:35

what is r equal to sine theta, r sine theta,

00:54:40

which means 2 assignments for you, here you see

00:54:46

these forms, firstly, according to the formulas

00:54:48

here we have there there was a written formula,

00:54:51

let's say in the first place what

00:54:53

h2 is, this is the square root of the sum of the

00:54:59

squares of the derivatives dx pd theta

00:55:02

squared plus d y according to the accident, this will happen

00:55:05

in the square, you will see that it will be

00:55:07

similar for the other two,

00:55:09

and in addition, draw the

00:55:14

actual p line here You can already see

00:55:16

in this picture this is the

00:55:18

AM beam. this line is also visible this is a

00:55:22

semicircle

00:55:25

offe line where offe line is we must

00:55:34

draw a plane through point m

00:55:36

perpendicular to the z axis and in this plane

00:55:40

and in this plane draw a circle of this

00:55:44

radius

00:55:47

this radius will be r multiplied by

00:55:51

sine theta from here there and it will appear here is this

00:55:54

p for sine theta1 in a word, this is

00:55:56

a task for you that you should be able to derive and

00:55:58

geometrically, since we just did this

00:56:01

for cylindrical coordinates

00:56:03

and according to the formulas that we received,

00:56:05

now let’s return again to these general

00:56:07

formulas, which means that from the formulas we obtained it

00:56:09

follows that d or one for house-2

00:56:16

model 3 this is h1 h2 h3

00:56:23

deco one deco 2d q3

00:56:30

well since before or one it was already 11

00:56:33

per deco 1 and so on multiplied

00:56:36

but now let's look at this

00:56:39

equality from this point of view

00:56:42

12 13

00:56:44

these are elements or small cups,

00:56:50

small segments on lines on q1 q2 q3

00:56:56

aline, these are orthogonal, thus in fact

00:57:01

one d2 and d3 are the long sides of a

00:57:06

rectangular parallelepiped since the

00:57:08

coordinate lines are orthogonal

00:57:10

and that means what is on the left is a

00:57:15

volume element in Cartesian

00:57:17

rectangular coordinates, I’ll

00:57:19

write it like this ICE index x y z is the volume of a

00:57:25

small rectangular

00:57:28

parallelepiped with sides equal to d1

00:57:32

d2 d3 and dk1 of type 2d q3 is the

00:57:37

volume element dv in curvilinear coordinates,

00:57:41

so here we put

00:57:43

q2 q3 as an index,

00:57:45

which means we can

00:57:47

rewrite this feeling as the volume element in

00:57:53

rectangular coordinates

00:57:54

d with index x y z is equal to the product of

00:57:59

all three lam parameters by the

00:58:05

volume element in curvilinear orthogonal

00:58:08

coordinates, and on the other hand, if you

00:58:17

remember the chapter on multiple integrals,

00:58:21

then when we studied the change of

00:58:26

variables in the triple integral

00:58:29

and passed to some actually

00:58:33

curvilinear coordinates, we had

00:58:36

such a formula there volume element dv in

00:58:43

rectangular coordinates as expressed

00:58:47

through the volume element in curvilinear

00:58:50

coordinates what appeared there

00:58:53

as a multiplier as a waiter

00:58:55

stretching no and bean Jacobi module it is

00:58:59

by the way in my my books there is no additional

00:59:02

typo typo no typo there you

00:59:04

need to put the module this is the Jacobi module

00:59:08

it d x y z pads q1 q2 q3 on deck one

00:59:22

deck 2d q3

00:59:26

well, it would be better if I should write d.v.

00:59:30

y 1 to 2 q3 this is what you have here, the first

00:59:37

line is the partial derivatives dx pdk

00:59:40

1dx in spirit 2 dx pd q3 the

00:59:42

second line is the partial derivatives with respect to the

00:59:45

game from x to q2 from x y z and the third,

00:59:50

similarly, compare the two formulas, we

00:59:53

get that the wi-fi module the biana of the transition

00:59:57

from curvilinear orthogonal coordinates

01:00:08

is equal to the product of your parameters

01:00:11

h1 h2 h3

01:00:16

well, so far we have had a

01:00:19

preliminary conversation about what

01:00:21

curvilinear orthogonal coordinates are and

01:00:24

our goal is the following: we want to

01:00:28

derive a formula for the

01:00:29

divergence gradient of the router in the Kremlin

01:00:32

orthogonal coordinates we are this Let’s do it

01:00:34

only for the gradient, but the conclusion is not complicated,

01:00:38

it’s there, well, it can be deduced in different ways

01:00:43

if you open the textbook Alina prazdnaya,

01:00:46

in my opinion, it’s very mathematical,

01:00:48

more rigorous, but very mathematical and the

01:00:53

physical meaning of everything

01:00:54

that is being done is not very visible, unfortunately, but there is also

01:00:58

I’ll introduce you textbook booths fomin they

01:01:01

wrote a wonderful textbook it is very

01:01:03

written specifically for physicists so if

01:01:08

you take this textbook then there is such a

01:01:12

good conclusion for the operations gradient

01:01:16

divergence rotor in curvilinear

01:01:18

orthogonal coordinates the same conclusion is

01:01:21

in this one of my lectures on mathematical analysis but

01:01:24

for the gradient we are now

01:01:26

Let's derive this expression and so we can

01:01:29

write the gradient operation in

01:01:32

rectilinear orthogonal coordinates; the

01:01:35

gradient operation in curvilinear

01:01:37

orthogonal coordinates;

01:01:48

well, as before, we denote

01:01:51

curvilinear orthogonal coordinates

01:01:53

q1 q2 q3; we consider the scalar field y; the

01:02:09

scalar field y; well, it depends on x y z

01:02:13

in curvilinear coordinates from q1 q2 q3

01:02:15

we are considering a scalar field and our

01:02:21

task is to obtain an expression for the gradient

01:02:24

y

01:02:26

here in curvilinear orthogonal

01:02:28

coordinates q1 q2 q3

01:02:32

but what does the gradient y mean in curvilinear

01:02:35

orthogonal coordinates and the

01:02:38

vector means when we introduced it in

01:02:41

Cartesian rectangular coordinates,

01:02:44

we introduced it as if laying out the

01:02:49

hedgehogs to the basis associated with

01:02:52

the rectangle, we have the coordinates on which

01:02:55

basis we will be decomposing here,

01:02:57

so first we need to enter the basis on

01:03:00

which we will be decomposing, so we will

01:03:03

consider the gradient y, well, at some

01:03:06

point m at this point, as we noted at

01:03:13

this point,

01:03:14

3 coordinate lines pass through this point

01:03:17

in pairs orthogonal,

01:03:19

here we introduce a basis associated with point m,

01:03:24

which means we take as a basis

01:03:28

associated with point m, we take as a

01:03:31

basis associated with point m the unit

01:03:37

vectors e1, e2, e3

01:03:48

directed along the tangent coordinate

01:03:52

lines at point m, which means we introduce the base

01:03:57

associated

01:03:59

with point m here in as a basis, we take

01:04:03

the vector and 1 and 2 or directed along the

01:04:05

tangent coordinate lines at point

01:04:07

m and while the pressure can be directed in one direction

01:04:12

and the other in the direction of increasing the

01:04:14

corresponding coordinates in the direction

01:04:16

of increasing the corresponding coordinates

01:04:24

what is the peculiarity of such a basis the

01:04:26

peculiarity is that if we take

01:04:30

another point m, then we draw a

01:04:36

basis in this way at this point, then it will be somehow

01:04:39

rotated in relation to this,

01:04:41

generally speaking, because the tangent is

01:04:44

there at different points in different

01:04:46

directions, so this distinguishes the

01:04:50

basis along which we will now lay out

01:04:52

the gradient at point m from the basis of

01:04:54

the language at whatever point we have the vectors and burned

01:04:58

them, they remain, so to speak, well,

01:05:02

of course, unit vectors just like

01:05:04

these we consider unit vectors,

01:05:06

but they are also directed equally at all points,

01:05:08

and these at different points

01:05:11

we will write the gradient decomposition differently y at

01:05:15

point m according to the base one d2 d3

01:05:19

associated with point m means the gradient at

01:05:25

point m and that is, some coefficient

01:05:28

one on e1 plus some coefficient

01:05:32

c2 leg 2 + some coefficient c3 on

01:05:37

e3, this is our task to calculate these

01:05:41

coefficients are easy to do,

01:05:44

let's scalarly multiply this equality by the vector

01:05:49

e 1, multiply the scalar traverses by

01:05:53

vectors 1, we get the scalar

01:05:58

product of the gradient y at point m by e1

01:06:07

is equal to,

01:06:08

well, let's immediately write with us, because the

01:06:12

use of orthogonal coordinates

01:06:15

means I money 2 and 3 are unit

01:06:18

vectors pairwise orthogonal so

01:06:22

when we multiply c 1 d 1 scalarly by e1

01:06:26

what happens modulo 1 squared is

01:06:30

the unit of

01:06:31

the nation 10 is equal to c1 and I am alone on e2 as long as it is

01:06:38

not orthogonal this is 0103 this is also 0

01:06:42

so we found c1 but on the other side on the

01:06:47

other side well, similarly, although

01:06:50

similarly, we write a little later on the other

01:06:53

side, this is the scalar product of

01:06:57

the vector gradient by the unit vector, this is

01:07:01

what the derivative with respect to the direction is,

01:07:05

coordinate lines in the direction of Gaia 1

01:07:07

means that with from here we get that

01:07:11

c1 and that is,

01:07:15

dpdz one at point m similarly

01:07:21

similar to c2 this will be Daewoo

01:07:26

victory 2 at point m and c3 this will be before the

01:07:32

fall of e3 at point m but that’s not all,

01:07:40

transfer here we will draw a segment, well,

01:07:42

for simplicity I will make it rectilinear

01:07:44

even though it is generally speaking a curved

01:07:46

segment of the coordinate line and 1 means here

01:07:52

we have point m here is some kind of .

01:07:57

m1 due to m this is a point with coordinates q1

01:08:02

q2 q3

01:08:05

and let’s even say there is more than one vector

01:08:11

shown here m1 we took a segment of

01:08:14

the coordinate line q 10 m 1 will have

01:08:18

coordinates q1 plus d q1 q2 q3 the same well

01:08:23

and let’s denote by d or one the

01:08:30

length of this piece of this segment of the curve

01:08:33

we know that d or one is already one

01:08:37

parameter of the llama

01:08:39

on gpu 1 therefore

01:08:44

before e1 falls at point m and that is, the limit,

01:08:53

remember the definition of the derivative with respect to

01:08:55

direction, this is the limit of the

01:08:58

fraction in the numerator we take y at point m

01:09:02

one minus y at point m is the increment of

01:09:06

the function when we moved from point m to

01:09:08

point m 1 in the denominator we divide by d1 and the

01:09:13

limit one tends to zero

01:09:17

but we can write it like this it is

01:09:23

equal to the limit dr1

01:09:28

this is h1 on deck 1 means that in fact one

01:09:31

tends to zero it is necessary so that deku one

01:09:33

tends to zero, this will be the limit with

01:09:36

dp1 tending to zero, we

01:09:39

take h1 at point m, let's take here

01:09:43

duel one, we replace deco 1

01:09:49

unit with h1, let's take our one out of the

01:09:52

limit sign and under the limit sign there will be y at

01:09:58

point 1 + d q1 q2 q3

01:10:06

minus y at point q1 q2 putri

01:10:12

and this is divided by deck 1, well, let's look at

01:10:19

this fraction, this is the increment of the function y with respect to the

01:10:23

argument q1 divided by this very

01:10:28

increment of the argument, it is clear that

01:10:30

this limit is an action that is nothing more than the

01:10:34

derivative of y with respect to spirit one at point m

01:10:40

so we get that

01:10:45

c1 which was equal to the water or one at

01:10:52

point m give e1 excuse me at point m and that

01:10:56

is, units on h1 d

01:11:01

where q1 at point m so we calculated the kai in the

01:11:06

center 1 here at the very top of we

01:11:09

have written this decomposition of the gradient y according to the

01:11:12

basis associated with point m, we found c1

01:11:15

quite clearly that with absolutely the same

01:11:17

formulas we

01:11:18

similarly obtain that c2 is one on

01:11:22

h2 before the fall of q2 at point m and c3

01:11:31

is one on h3 two below and q3

01:11:39

at point m

01:11:45

Well, all that remains is to substitute these values c

01:11:49

1123 into the equality for the gradient and we

01:11:52

finally get an expression for the

01:11:55

gradient in curvilinear orthogonal

01:11:57

coordinates, we get that the gradient y at

01:12:04

point m is equal to one on h1b fall fu1

01:12:12

at point m plus one on h2 dewoo

01:12:19

type 2 at point m plus one our 3d upa

01:12:29

deco 3 at point p here is the expression for which

01:12:38

I didn’t write the most important thing and there are

01:12:40

even more people there one forgive me

01:12:42

for God’s sake this is multiplied by e1 this is

01:12:46

multiplied by e2 I only put the

01:12:51

coefficients themselves

01:12:52

this is multiplied by e3 and So in the

01:12:58

decomposition of the gradient,

01:13:00

that is, the expression of the gradient in

01:13:02

orthogonal coordinates, and we have

01:13:04

decomposed the vector gradient y at point m according to the

01:13:08

basis of the connected muses. m if we go to

01:13:11

another point there will be its own basis there the

01:13:13

vector convectors will somehow turn

01:13:15

and

01:13:16

well, so to speak, here are the derivatives if we

01:13:22

take another point what will change and

01:13:24

laugh vectors d1 d2 d3 will change

01:13:28

of course these derivatives will remain but

01:13:30

they will be taken in this new points and

01:13:33

h1 h2 h3

01:13:35

means somehow note to yourself

01:13:37

that the value h1 h2 h3

01:13:40

they also depend on the point are taken at

01:13:43

point m h1 h2 h3 are taken at point m well,

01:13:47

as an example, let's write out the

01:13:51

expression for the gradient y

01:13:54

in cylindrical coordinates

01:13:57

example example in cylindrical

01:14:03

coordinates

01:14:04

bredent q is equal to h1, we were equal to

01:14:09

one in cylindrical coordinates and

01:14:12

1 by 1 it was p, so it will be d y

01:14:18

cdr I will denote here not one not two e3

01:14:23

but like this e p is the unit vector in

01:14:28

coordinate early not

01:14:30

plus one on h2 these are units on r dow

01:14:35

according to delphi

01:14:37

and the unit vector according to Fellini will be denoted by

01:14:42

diffe

01:14:43

and plus the hole codes

01:14:47

z on

01:14:50

not with the sign z let's draw

01:14:54

these coordinate vectors associated with

01:14:57

point m, which means this is almost the same

01:15:02

drawing that we drew, but it will be

01:15:04

less loaded this means these are rectangular

01:15:07

coordinates and again, let's draw

01:15:12

this very circle here, let's mark here

01:15:17

point m

01:15:20

point m here we have r line here we have this point at

01:15:25

the beginning c

01:15:26

this is our upward axis z this x this y is the beginning of the

01:15:32

carnot.

01:15:33

it was a p line, well, the tangent of the line

01:15:38

coincides with this line itself, but we

01:15:41

imposed one more condition: we take the

01:15:44

corresponding coordinates in the direction of increasing,

01:15:46

which means from point m we need to direct the

01:15:48

circle inward or outward outward, of course, this

01:15:52

means this is the vector we will have a

01:15:56

unit vector r e

01:16:00

m tangent to the circle the unit

01:16:06

vector z and again in the direction of

01:16:09

increasing angle fi and parallel to the z axis

01:16:14

and codirectional with the z axis vector

01:16:17

z here is the basis, notice how

01:16:23

this basis will change if we move from

01:16:25

point to point

01:16:26

only one vector will not change

01:16:27

which and z is clear, it will remain equal at

01:16:32

all points they are equal a and e r e figs

01:16:35

generally speaking they change the task for you is

01:16:38

small write a decomposition write an

01:16:41

expression for gorin then y in spherical

01:16:43

coordinates write an expression for the

01:16:45

gradient y in spherical coordinates

01:16:47

and draw this basis e e t tf

01:16:51

and

01:16:52

for spherical coordinates for an

01:16:53

arbitrary point m, well now about

01:17:03

divergence and rotor

01:17:05

for arbitrary orthogonal

01:17:07

curvilinear coordinates, the idea of the derivation itself is

01:17:11

very simple, we already drew through the

01:17:15

point m there are 3 coordinate lines, you

01:17:17

need to take pairs of orthogonal ones,

01:17:21

you need to take pieces of these lines as

01:17:23

small as you like complete or construct, as it

01:17:27

were, a rectangular parallelepiped on these three segments

01:17:30

and use the

01:17:32

invariant definition of divergence

01:17:35

divergence according to the

01:17:36

invariant definition that we

01:17:38

derived in previous lectures this is the limit of the

01:17:41

ratio of the flow through the surface through the

01:17:46

outer side of the surface why to the

01:17:49

volume of this area there we will have a

01:17:51

small arbitrarily small with

01:17:54

arbitrarily small sides almost a

01:17:56

rectangular parallelepiped, well, at

01:17:58

least tangents to its

01:17:59

curvilinear sides

01:18:02

perpendicular in pairs perpendicular

01:18:04

but to calculate it is very easy to do in

01:18:06

one word, I’ll write it in the formula itself, I’ll

01:18:10

write it now, of course you need to know it at

01:18:13

least in the exam for a four

01:18:17

and for you need to be able to get a five and the conclusion is

01:18:20

very simple, once you read it

01:18:22

you will understand the idea and then you don’t even need to

01:18:25

memorize all these if you suddenly

01:18:27

get such a ticket and so now

01:18:30

divergence means the divergence of a

01:18:35

vector field and at point m it is expressed by the following

01:18:42

formula: one on h1 h2 h3

01:18:54

further in brackets,

01:18:55

of course, there will be three terms as well

01:18:59

as in rectangular coordinates,

01:19:02

Cartesian coordinates,

01:19:03

since we have three independent

01:19:06

variables, these are derivatives dpd fu1

01:19:12

from yes, we need to write the decomposition

01:19:17

of the vector, and so let the decomposition of the

01:19:20

vector from m along the basis be related muses.

01:19:24

m here, according to the very base court 1 g 2 and 3

01:19:27

that we wrote when we derived

01:19:29

the gradient, let’s briefly denote them one on e1

01:19:34

plus a-two on e 2 plus 3

01:19:41

on e3 so 1 a 23

01:19:46

are the coefficients of the decomposition of the vector a

01:19:48

tempo to the basis associated with the point m there and

01:19:53

the gradient was told and here here is the

01:19:55

vector itself a team

01:19:56

is dpd q1 from the product and one h2 h3

01:20:03

plus dpd q2 from the product and two

01:20:10

h3 h1

01:20:13

plus dpd q3 from a3

01:20:20

page 1 of 2

01:20:24

well, and I’ll write all this on the right side in the

01:20:27

same way as in the left one is taken at point m and

01:20:30

these values h1 h2 h3 are taken at point m

01:20:34

and of course these coordinates a1 a2 a3

01:20:37

are the coordinates of the vector a from m in the basis

01:20:41

associated with the point, so where do you think

01:20:44

this factor

01:20:46

denominator h1 h2 comes from? h3

01:20:49

of course it comes from that very same

01:20:52

formula for volume,

01:20:53

well, not from this one, but where we

01:20:57

wrote df x y z above, well, in a

01:21:02

word, it comes from the volume,

01:21:03

see for yourself this conclusion,

01:21:05

let's write out the divergence from m in

01:21:10

cylindrical coordinates

01:21:12

cylindrical coordinates divergence

01:21:15

a from m well vector a tempo in cylindrical

01:21:20

coordinates just like we wrote here here

01:21:21

we will denote its base with e r g

01:21:26

visit means yes and the remainder must be written

01:21:33

means a m because we don’t need a basis but

01:21:37

we will expand the coordinates according to the basis

01:21:42

verdi fio z and the coefficient r is filled in but than

01:21:46

a.r.

01:21:47

to the vector e r plus offe

01:21:52

to the vector f and and plus z to the vectors z that

01:22:01

is, by a r a fio z we denote

01:22:03

the coordinates of the vector a in the basis associated

01:22:07

with the cylindrical coordinates plus z on

01:22:11

e z and so the divergence from m is equal to what is

01:22:19

equal to h1 h2 h3 products in

01:22:21

cylindrical coordinates h 1 unit

01:22:24

h2l as many as three units are

01:22:26

not the price r and all this is one per r as in the

01:22:30

common factor further dpd ra1 is a.r.

01:22:36

h2 h3 is h2, etc. as many as three units, the

01:22:40

derivative of power is taken from r a r + q 2

01:22:48

this is fi and two is a with the index fe

01:22:54

this is this very coefficient h3 h1

01:22:58

this is units,

01:23:00

which means it will be d a fee

01:23:04

by defy plus the last term here I have

01:23:12

it here ter plus dpd q3 was here a3 h1

01:23:20

h2 means 3 this is our z means plus

01:23:27

dpdz

01:23:30

from a3 this is a with the index z h1 h2 and three

01:23:37

times then well, we can simplify this a little

01:23:42

because when we take the

01:23:44

derivative of this last

01:23:46

term according to z p you can take it out

01:23:50

and shorten it with someone, but these are already

01:23:51

details and so the divergence in

01:23:55

cylindrical coordinates leave

01:23:58

a little space and you yourself write

01:24:00

the divergence in spherical coordinates,

01:24:03

and the last

01:24:07

rotor means for the rotor for the

01:24:17

rotor of the vector field and at point m

01:24:22

it turns out like this the expression,

01:24:24

again, is the factor one by

01:24:27

h1 h2 h3 multiplied by the

01:24:33

third-order determinant, we know how it

01:24:36

looked in Cartesian coordinates,

01:24:40

rectangular coordinates, here it’s not

01:24:43

just the basis vectors d1 d2 d3, but

01:24:47

with the multipliers the parameters are cute h11 h2

01:24:55

e2 h3

01:25:00

e3 the second operator line dpd q1 dpd

01:25:12

q2 dpd q3

01:25:17

well, below, again, not just

01:25:22

vector coordinates, but well, we believe that they are the same

01:25:27

as we wrote before a1 a2 a3

01:25:30

and there are already 11 two of two h3o trips, here is the

01:25:42

expression for the rotor and in the given

01:25:45

orthogonal coordinates, again you

01:25:47

will have this formula using the

01:25:49

invariant definition of the rotor and

01:25:51

the conclusion is simple, look it up in

01:25:53

the textbook yourself,

01:25:54

well, with this we have completed the chapter of this one,

01:25:59

write it out for cylindrical and spherical

01:26:00

coordinates, so to speak, you need to be able

01:26:03

to do this, we completed the chapter on scalar

01:26:08

vector fields

Description:

0:00:20 1. Оператор Гамильтона 0:20:34 2. Операции векторного анализа в криволинейных ортогональных координатах 0:32:09 3. Параметры Ламе

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