2 classical field theories

04/13/23 1

Classical FieldTheories

SOLO HERMELIN

Updated: 10.10.2013 16.11.2014 8.04.2015

http://www.solohermelin.com

http://www.solohermelin.com/

04/13/23 2

SOLO

Table of Content

Classical Field Theories

Generalized Coordinates

1. Newton’s Laws of Motion

Analytic Dynamics

Work and Energy

Basic Definitions

Constraints

The Stationary Value of a Function and of a Definite Integral

The Principle of Virtual Work

2. D’Alembert Principle

3. Hamilton’s Principle

4. Lagrange’s Equations of Motion

5. Hamilton’s Equations

Introduction to Lagrangian and Hamiltonian FormulationExtremal of the Functional .

2

1

,,t

t jj tdtqtqtLCI Second Method (Carathéodory)

Equivalent Integrals

Hamilton-Jacobi TheoryTheorem of Noether for Single Integral

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SOLO

Table of Content (continue – 1)


Fermat Principle in Optics

Four-Dimensional Formulation of the Theory of Relativity

Electromagnetic Field

Introduction to Lagrangian and Hamiltonian Formulation for Continuous Systems and Fields

Transition from a Discrete to Continuous Systems

Extremal of the Functional

2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

Hamiltonian Formalism

Theorem of Noether for Multiple IntegralsElasticity

Vibrating String

Vibrating Membrane

Vibrating Beam

Plate Theories

Structural Model of the Solid Body

The Inverse Square Law of Forces

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Table of Content (continue – 2)


Variational Principles of Hydrodynamics

Part I – Lagrange Interpretation

Part II: Euler’s Representation

Part III: Simpler Eulerian Variational Principle

Part IV: Problem of the Motion of Fluid Subjected to Gravity Forces and

Surface Tension Forces

Dynamics of Acoustic Field in Gases

References

Equation of Motion of a Variable Mass System – Lagrangian Approach

5

Analytic DynamicsSOLO

Newton’s Laws of Motion

“The Mathematical Principles of Natural Philosophy” 1687

First Law Every body continues in its state of rest or of uniform motion instraight line unless it is compelled to change that state by forcesimpressed upon it.

Second Law The rate of change of momentum is proportional to the forceimpressed and in the same direction as that force.

Third Law To every action there is always opposed an equal reaction.

constvF

0

vmtd

dp

td

dF

2112 FF

vmp

td

pdF

12F

1 2

21F

Return to Table of Content

6


Work and Energy

The work W of a force acting on a particle m that moves as a result of this alonga curve s from to is defined by:

F

1r

2r

2

1

2

1

12

r

r

r

r

rdrmdt

drdFW

r

1r

2r

rd

rdr

1

2

F

m

s

rd is the displacement on a real path.

rrmT

2

1

The kinetic energy T is defined as:

1212

2

1

2

1

2

12

TTrrdm

dtrrdt

dmrdrm

dt

dW

r

r

r

r

r

r

For a constant mass m

7


Work and Energy (continue)

When the force depends on the position alone, i.e. , and the quantityis a perfect differential

rFF

rdF

rdVrdrF

The force field is said to be conservative and the function is known as the Potential Energy. In this case:

rV

212112

2

1

2

1

VVrVrVrdVrdFWr

r

r

r

The work does not depend on the path from to . It is clear that in a conservativefield, the integral of over a closed path is zero.

12W 1r

2r

rdF

01221

21

1

2

2

1

VVVVrdFrdFrdF

path

r

r

path

r

rC

Using Stoke’s Theorem it means that SC

sdFrdF

0 FFrot

Therefore is the gradient of some scalar functionF

rdrVdVrdF

rVF

8


Work and Energy (continue)

and

rFdt

rdF

t

V

dt

dVtt

00limlim

But also for a constant mass system

rFrrmrrrr

mrrm

dt

d

dt

dT 22

1

Therefore for a constant mass in a conservative field

.0 constEnergyTotalVTVTdt

d


9


1.4 Basic Definitions

Given a system of N particles defined by their coordinates:

Nkktzjtyitxzyxrr kkkkkkkk ,,2,1,,

where are the unit vectors defining any Inertial Coordinate System kji

,,

The real displacement of the particle mk :

Nkktdzjtdyitdxrd kkkk ,,2,1

is the infinitesimal change in the coordinates along real path caused by all theforces acting on the particle mk .

The virtual displacements (Δxk , Δyk, Δzk, Δt) are infinitesimal changes in thecoordinates; they are not real changes because they are not caused by real forces.The virtual displacements define a virtual path that coincides with the real one atthe end points.

10


Basic Definitions (continue)

trk

1trk

2trk

krd

1

2F

km

),,,( dttdzzdyydxxPrdr kkkkkkkk

),,,( tzyxP kkk

),,,( ttzzyyxxP kkkkkk

),,,( tzzyyxxP kkkkkk

tvrd kk

i j

k

True (Dynamical or Newton) Path

Virtual Path


11


1.5 Constraints

If the N particles are free the system has n = 3 N degrees of freedom. Nkzyxr kkkk ,,2,1,,

The constraints on the system can be of the following types:

(1) Equality Constraints: The general form (the Pffafian form)

mldttradztradytradxtra lt

N

kk

lzkk

lykk

lxk ,,2,10,,,,

1

or

maaarankmldtarda lzk

lyk

lxk

lt

N

kk

lk

,,,,2,101

We can classify the constraints as follows:

(a) Time Dependency

(a1) Catastatic mla lt ,,2,10

(a2) Acatastatic mla lt ,,2,10

(1) Equality Constraints

(2) Inequality Constraints

12


Constraints (continue)

Equality Constraints: The general form (the Pffafian form) (continue)


lyk

lxk

lt

N

kk

lk

,,,,2,101

(b) Integrability

(b1) Holonomic if the Pffafian forms are integrable; i.e.:

mldtt

fzd

z

fyd

y

fxd

x

fdf

N

k

lk

k

lk

k

lk

k

ll ,,2,1

1

or mltzyxzyxf NNNl ,,2,10,,,,,,, 111

(b2) Non-holonomic if the Pffafian forms are not integrable

(b2.1) Scleronomic:

(b2.2) Rheonomic:

ml

l

t

f

,,2,1

0

or

mlzyxzyxf NNNl ,,2,10,,,,,, 111

ml

l

t

f

,,2,1

0

13



(2) Inequality Constraints:

(a) Stationary Boundaries (time independent):

(b) Non-stationary Boundaries (time dependent):

mlzyxzyxf NNNl ,,2,10,,,,,, 111

mltzyxzyxf NNNl ,,2,10,,,,,,, 111

14



Displacements Consistent with the Constraints:

The real displacement consistent with theGeneral Equality Constraints (Pffafian form) is:

The virtual displacement consistent with theGeneral Equality Constraints (Pffafian form) is:

dtkdzjdyidxrd kkkk ,

mldtardadtadzadyadxa lt

N

kk

lk

lt

N

kk

lzkk

lykk

lxk ,,2,10

11

tkzjyixr kkkk ,

mltaratazayaxa lt

N

kk

lk

lt

N

kk

lzkk

lykk

lxk ,,2,10

11

Dividing the Pffafian equation by dt and taking the limit, we obtain:

mlraaN

kk

lk

lt ,,2,1

1

Now replace in the virtual displacement equationlta

mltrraN

kkk

lk ,,2,10

1

Define the δ variation as:

td

dt

15



Displacements Consistent with the Constraints (continue):

Define the δ variation as:td

dt

trk

kr

km

),,,( dttdzzdyydxxPrdr kkkkkkkk

),,,( tzyxP kkk

),,,( ttzzyyxxP kkkkkk

),,,( tzzyyxxP kkkkkk

dtrrd kk

i j

k

True (Dynamical or Newton) Path

Virtual Path

kr

trr kk

Then: kkk rtd

dtrr

From the Figure we can see that δ variation corresponds to a virtual

displacement in which the time t is

held fixed and the coordinates varied

to the constraints imposed on the system.

mlraN

kk

lk ,,2,10

1

For the Holonomic Constraints: mltzyxzyxf NNNl ,,2,10,,,,,,, 111

mlrfN

kklk ,,2,10

1


16


1.6 Generalized Coordinates

The motion of a mechanical system of N particles is completely defined by n = 3N coordinates . Quite frequently we may find it more

advantageous to express the motion of the system in terms of a different set of coordinates, say . If we take in consideration the m constraints wecan reduce the coordinates to n = 3N-m generalized coordinates.

Nktztytx kkk ,,2,1,,

Tnqqqq ,,, 21

Nkktqzjtqyitqxtqqqrtqr kkknkk ,,2,1,,,,,,,, 21

Nkkdzjdyidxdtt

rdq

q

rrd kkk

kj

n

j j

kk ,,2,1

1

Nkt

rq

q

r

td

rdrv k

j

n

j j

kkkk ,,2,1

1

In the same way

Nkkzjyixtt

rq

q

rr kkk

kj

n

j j

kk ,,2,1

1

and

Nktt

rtq

q

rt

t

rq

q

rtrrr k

j

n

j j

kkj

n

j j

kkkk ,,2,1

11

17


Generalized Coordinates (continue)

Nkqq

rtqq

q

rr

n

jj

j

kjj

n

j j

kk ,,2,1

11

where tqqq jjj

The Generalized Equality Constraints in Generalized Coordinates will be:

mldtt

raadq

q

ra

dtt

raadq

q

radtarda

N

k

klk

lti

n

i i

kN

k

lk

N

k

N

k

klk

lti

n

i i

klk

lt

N

kk

lk

,,2,1011 1

1 111

If we define

N

k

N

k

klk

lt

lt

n

i i

klk

li t

raac

q

rac

1 11

&

we obtain mldtcdqc lti

n

i

li ,,2,10

1

and the virtual displacements compatible with the constraints are

mlqc i

n

i

li ,,2,10

1

18


Generalized Coordinates (continue)

The number of degrees of freedom of the system is n = 3N-m. However, when thesystem is nonholonomic, it is possible to solve the m constraint equations for thecorresponding coordinates so that we are forced to work with a number of coordinates exceeding the degrees of freedom of the system. This is permissibleprovided the surplus number of coordinates matches the number of constraintequations. Although in the case of a holonomic system it may be possible to solvefor the excess coordinates, thus eliminating them, this is not always necessary ordesirable. If surplus coordinates are used, the corresponding constraint equationsmust be retained.


19


1.7 The Stationary Value of a Function and of a Definite Integral

In problems of dynamics is often sufficient to find the stationary value of functionsinstead of the extremum (minimum or maximum).

Definition: A function is said to have a stationary value at a certain point if the rate of change inevery direction of the point is zero.

Examples:

(1) niu

fdu

u

fdfuuuf

i

n

ii

in ,,2,100,,,

121

By solving those n equations we obtain for which f is stationary

nuuu ,,, 21

20


The Stationary Value of a Function and of a Definite Integral(continue)

Examples (continue):

(2) nuuuf ,,, 21 with the constraints marankmldua lk

N

kk

lk

,,2,101

Lagrange’s multipliers solution gives:

01 1

i

n

i

m

l

lil

i

duau

fdf

By choosing the m Lagrange’s multipliers λl to annihilate the coefficients of them dependent differentials dui we have

equationsmn

mldua

niau

f

n

li

li

m

l

lil

i

,,2,10

,,2,10

1

1

21




(3) The functional

2

1

,,x

x

dxxd

xydxyxFI

We want to find such that I is stationary, when the end points and are given.

xy 1xy 2xy

xy

xxyxyxy

11, yx

22 , yx

x

y

The variation of is xy

021 xxxxyxyxyxy

and

2

1

2

1

,,,,x

x

x

x

dxxd

xd

xd

xydxxyxFdx

xd

xydxyxFI

2

0

2

2

02

10

dd

Idd

d

IdII

22




Continue: The functional

2

1

,,x

x

dxxd

xydxyxFI

The necessary condition for a stationary value is

00

12

0

2

1

2

1

xx

xdyd

Fdxx

xdyd

F

xd

d

y

F

dxxd

xd

xdyd

Fx

y

F

d

Id

x

x

nintegratio

partsby

x

x

Since this must be true for every continuous function η(x) we have

210 xxx

xd

yd

F

xd

d

y

F

Euler-Lagrange Differential Equation

By solving this differential equation, y(x),for which I is stationary is found.

JOSEPH-LOUISLAGRANGE1736-1813

LEONHARD EULER1707-1783

23

Leonhard Euler (1707-1783) generalized the brothers Bernoulli methods in“Methodus inveniendi lineas curvas maximi minimive proprietate gaudentes sive solutio problematis isoperimetrici latissimo sensu accepti” (“Method for finding plane curves that show some property of maxima and minima”) published in 1744. Euler solved the Geodesic Problem, i.e. the curves of minimum length constrained to lie on a given surface.

Joseph-Louis Lagrange (1736-1813) gave the first analytic methods of Calculus of Variations in "Essay on a new method of determining the maxima and minima of indefinite integral formulas" published in 1760. Euler-Lagrange Equation:

0,,,,

txtxtFdt

dtxtxtF

xx

SOLOCALCULUS OF VARIATIONS


24


1.8 The Principle of Virtual Work

This is a statement of the Static Equation of a mechanical system.

If the system of N particles is in dynamic equilibrium the resultant force on each

particle is zero; i.e.:0iR

01

N

iii rRW

Because of this, for a virtual displacement the Virtual Work of the system is

ir

If the system is subjected to the constraints:


lyk

lxk

lt

N

kk

lk

,,,,2,101

Then we denote the external forces on particle i by and the constraint’s forces

by . The resultant force on i is:

iF

iF '

0' iii FFR

25


The Principle of Virtual Work (continue)

We want to find the Virtual Work of the constraint forces.

There are two kind of constraints:

(1) The particle i is constrained to move on a definite surface. We assume that the

motion is without friction and therefore the constraint forces must be

normal to the surface. The virtual variation compatible with the constraint

must be on the surface, therefore .

iF

ir

0' ii rF

ir

iF '

(2) The particle i is acting on the particle j and the distance between them is l(t). .

iF '

i j

jF '

ir

jr

tl

26



tlrrrr jiji2

tllrrrr jiji

llrrrr jiji

ji

rr

jiji

jjiiji

rrrrrr

trrtrrrr

ji

0

0

ji FF ''

If we compute the virtual variation and differential and we multiply the secondequation by and add to the first we obtaint

Because is a real (not a generalized) force we can use Newton’s Third Law: i.e.:iF '

and the virtual work of the constraint forces of this system is:

0''''' rFrFrFrFW iijjii

We can generalized this by saying that:

0'1

N

iii rF

The work done by the constraint forces in virtual displacements compatible withthe constraints (without dissipation) is zero.

27



From equation we obtain:0' iii FFR

N

iii

N

iii

N

iii

N

iii rFrFrFrR

1

0

111

'0

or

01

N

iii rFW

The Principle of Virtual Work

The work done by the applied forces in infinitesimal virtual displacements compatible with the constraints (without dissipation) is zero

28



mjNimaaarank

mjra

jzi

jyi

jxi

N

kk

jk

,,2,1&,,1,,

,,2,101

We found that the General Equality Constraint the virtual displacement compatible with the constraint must be:

ir

Let adjoin the m constraint equations by the m Lagrange’s multipliers and add to thevirtual work equation:

j

01 11 11

N

ii

m

j

jiji

m

j

N

ii

jij

N

iii raFrarFW

There are 3N virtual displacements from which m are dependent of the constraint relations and 3N-m are independent. We will choose the m Lagrange’s multipliersto annihilate the coefficients of the m dependent variables:

j

iationsmNtindependenNmi

mtheofbecausemiaF

jm

j

jiji

var33,,1

,,2,10

1

29



From we obtain:0' iii FFR

m

j

jiji aF

1

'

where are chosen such that j mjforaFm

j

jiji ,,2,10

1

Since , we obtain:k

n

k k

ii q

q

rr

1

01 1 11

1 111 1

n

kk

m

j

N

i k

ijij

N

i k

ii

N

i

n

kk

k

im

j

jiji

N

ii

m

j

jiji

qq

ra

q

rF

qq

raFraFW

We define:

nkq

rFQ

N

i k

iik ,,2,1

1

nkcq

raQ

m

j

jkj

m

j

N

i k

ijijk ,,2,1'

11 1

nkq

rac

N

i k

iji

jk ,,2,1

1

Generalized Forces

Generalized Constraint Forces


30


2. D’Alembert Principle

Jean Le Rondd’ Alembert1717-1783

Newton’s Second Law for a particle of mass and a linear momentumvector can be written as

imiii vmp

D’Alembert Principle: 0'

iii pFF

D’Alembert Principle enables us to trait dynamical problems as if they were statical.Let extend the Principle of Virtual Work to dynamic systems:

0'1

N

iiiii rpFF

Assuming that the constraints are without friction the virtual work of the constraintforce is zero . Then we have

Generalized D’Alembert Principle: 01

N

iiii rpF

0'1

N

iii rF

The Generalized D’Alembert Principle The total Virtual Work performed by the effective forces through infinitesimal VirtualDisplacement, compatibile with the system constraints are zero.

0

ii pF

is the effective force.

“Traité de Dynamique”

1743


where and are the applied and constraint forces, respectively.iF

iF '

31


3. Hamilton’s Principle

William RowanHamilton1805-1865

Let write the D’Alembert Principle: in integral form01

N

iiii rpF

02

11

t

t

N

iiii dtrpF

But

N

iiii

N

iiii

N

iiii r

td

dvmrvm

td

drvm

111

Let find irtd

d

iiiii rtd

dttvrr

are the virtual displacements compatible with the

constraints mjraN

ii

ji ,,2,10

1

tri

ir

tvi

tPi tP i'

ttP i 'ir

Virtual Path True Path (P)Newtonian orDynamic Path

The ConstraintSpace at t

mjraN

ii

ji ,,10

1

0

0

0

21

21

21

tttt

trtr

trtr

ii

ii

1t

2t

32


Hamilton’s Principle (continue)Since

td

rdvv i

Pi i

ttd

dvr

td

dvt

td

dr

td

dv

ttd

d

rtd

d

td

rd

tdtd

rdrd

ttd

rrdvvv

iiiii

ii

iiiittPii i

11

'

tartd

dtatvr

td

dt

td

dvr

td

dv iiiiiiii

Therefore

ecommutativaretd

dand

rtd

dvv

td

dttavr

td

diiiiii

33


Hamilton’s Principle (continue)Now we can develop the expression:

tavmvvmrvmtd

dram

N

iiii

N

iiii

N

iiii

N

iiii

1111

But the Kinetic Energy T of the system is:

N

iiii vvmT

12

1

N

iiii vvmT

1

N

iiii

N

iiii

N

i

iii vFmavmvvmT111

Therefore

Trvmtd

d

tTTrvmtd

dram

N

iiii

N

iiii

N

iiii

1

11

34


Hamilton’s Principle (continue)From the integral form of D’Alembert Principle we have:

2

1

2

1

2

1

2

1

2

1

2

1

1110

11

1

0

t

t

N

iii

t

t

N

iii

N

i

t

tiii

t

t

N

iii

t

t

N

iiii

t

t

N

iiiii

dtrFTdtrFTrvm

dtrFTdtrvmtd

d

dtrFam

We obtained

02

1

2

11

dtWTdtrFTt

t

t

t

N

iii

Extended Hamilton’s Principle

35


Hamilton’s Principle (continue)If we develop and we can writetTTT tvrr iii

02

1

2

1111

dttvFTrFTdtrFTt

t

N

iii

N

iii

t

t

N

iii

and because

N

iii vFT

1

022

11

dttTrFTt

t

N

iii

The pair and is arbitrary but compatible with the constraints:ir

t

mjtara jt

N

ii

ji ,,2,10

1

36


Hamilton’s Principle (continue)For a Conservative System VF ii

VrVrFWN

iii

N

iii

11

We have 02

1

2

1

2

1

t

t

t

t

t

t

dtLdtVTdtWT

where WTVTL

NiVFVTLdtL ii

t

t

,,2,1;02

1

Hamilton’s Principle

forConservative Systems

Hamilton’s Principle for Conservative Systems: The actual path of a conservative system in the configuration space rendersthe value of the integral stationary with respect to all arbitraryvariations (compatible with the constraints) of the path between the twoinstants and provided that the path variations vanish at those two points.

2

1

t

t

dtLI

1t 2t


37


4. Lagrange’s Equations of Motion

Joseph LouisLagrange1736-1813

“Mecanique Analitique”

1788

The Extended Hamilton’s Principle states: 02

11

dtrFTt

t

N

iii

where are the virtual displacements compatible with theconstraints:

ir

mjqcqq

rara

n

kk

ki

n

kk

N

i k

iji

N

ii

ji ,,2,10

11 11

T the kinetic energy of the system is given by:

N

j

n

i

ji

i

jn

i

ji

i

jj

N

j

jjj tqqTt

rq

q

r

t

rq

q

rmrrmT

1 111

,,2

1

2

1

where is the vector of generalized coordinates. nqqqq ,,, 21

tqqTtt

Tq

q

Tq

q

TtqqTtqqTttqqqqTT

n

ii

ii

i

,,,,,,,,1

t

Tq

q

Tq

q

TT

n

ii

ii

i

1

n

ii

ii

i

n

iii

iii

i

qq

Tq

q

Ttqq

q

Ttqq

q

TtTTT

11

38


Lagrange’s Equations of Motion (continue)

n

ii

ii

i

n

iii

iii

i

qq

Tq

q

Ttqq

q

Ttqq

q

TtTTT

11

But because δ and are commutative and:td

d ii qdt

dq

n

ii

ii

i

qdt

d

q

Tq

q

TT

1

This is an expected result because the variation δ keeps the time t constant.

We found that , therefore

n

ii

i

jj q

q

rr

1

2

1

2

1

2

1

2

111 11 11

t

t

n

iii

t

t

n

ii

N

j i

jj

t

t

N

j

n

ii

i

jj

t

t

N

jjj dtqQdtq

q

rFdtq

q

rFdtrF

where ForcesdGeneralizeniq

rFQ

N

j i

jji ,,2,1

1

Now

2

1

2

1

2

1

2

1

110

.int

11

0

t

t

n

iiii

ii

i

n

i

t

tii

partsby

t

t

n

iiii

ii

i

t

t

N

jjj

dtqQqq

Tq

q

T

td

dq

q

T

dtqQqq

Tq

td

d

q

TdtrFT

39



02

11

t

t

i

n

ii

ii

dtqQq

T

q

T

td

d

where the virtual displacements must be consistent with the constraints . Let adjoin the previous equations by the constraints multipliedby the Lagrange’s multipliers

iqmkqc

n

ii

ki ,,2,10

1

mkk ,,2,1

01 11 1

n

ii

m

k

kik

m

k

n

ii

kik qcqc

to obtain

02

11 1

t

t

i

n

i

m

k

kiki

ii

dtqcQq

T

q

T

td

d

While the virtual displacements are still not independent, we can chose the Lagrangian’s multipliers so as to render the bracketed coefficients of equal to zero. The remaining being independent can be chosenarbitrarily, which leads to the conclusion that the coefficients ofare zero. It follows

iq

iq

mkk ,,2,1

nmiqi ,,2,1 miqi ,,2,1

nicQq

T

q

T

dt

d m

k

kiki

ii

,,2,11

40



nicQq

T

q

T

dt

d m

k

kiki

ii

,,2,11

We have here n equations with n+m unknowns . To find all theunknowns we must add the m equations defined by the constraints, to obtain

mn tqtq ,,,,, 11

nicQq

T

q

T

dt

d m

k

kiki

ii

,,2,11

Lagrange’s Equations:

mkcqc kt

n

ii

ki ,,2,10

1

Let define

Generalized Constraint Forces: nicQm

k

kiki ,,2,1'

1

41


Lagrange’s Equations of Motion (continue) If the system is acted upon by some forces which are derivable from a potential function and some forces which are not, we can write: nn qqqVrrrV ,,,,,, 2121

n

jF

njjj FVF

n

iii

n

ii

N

j i

jnj

i

jj

N

j

n

ii

i

jnjj

N

jjj qQq

q

rF

q

rVq

q

rFVrF

11 11 11

But where

N

j i

jj

i q

rV

q

V

1

k

z

Vj

y

Vi

x

VV

jjjj

Therefore:

niQq

V

q

rF

q

rVQ in

i

N

j i

jnj

N

j i

jji ,,2,1

11

Generalized External Forces:

Generalized External Nonconservative Forces:

niq

rFQ

N

j i

jnjin ,,2,1

1

42



The Lagrange’s Equations nicQq

T

q

T

dt

d m

k

kiki

ii

,,2,11

Define: qVtqqTtqqL

,,,,

Because we assume that , we have i

q

qV

i

0

Lagrange’s Equations: nicQq

L

q

L

dt

d m

k

kikin

ii

,,2,11

mkcqc kt

n

ii

ki ,,2,10

1

We proved

NiVFVTLdtL ii

t

t

,,2,1;02

1

Hamilton’s Principle for Conservative Systems

Lagrange’s Equations for a Conservative System without Constraints:

0,,,,2,10

k

iiiii

cVTLVFniq

L

q

L

dt

d

If they are no constraints, from the Lagrange’s Equations, or from Euler-Lagrange Equation for a stationary solution of , we obtain:

2

1

t

t

dtLI



The Lagrangian gives unique Euler-Lagrange Equations, but the inverse is not true, there more than one Lagrangian that gives the same Euler-Lagrange Equations.

Example 1:

td

tqqdtqqLtqqL

,,,,:,,1

1

0

1

0

1

0

1

0

1

0

1

0

1

0

,,,,,,,,:,,

00

1

t

t

t

t

t

t

t

t

t

t

t

t

t

t

tdtqqLqq

qq

tdtqqLtqqtdtqqLtdtqqL

Example 2:

This is not the most general case

22

21

22

21221211 2

1:,,&:,, qqqqtqqLqqqqtqqL

0

0

112

1

2

1

221

1

1

1

qqq

L

q

L

dt

d

qqq

L

q

L

dt

d

0

0

222

2

2

2

111

2

1

2

qqq

L

q

L

dt

d

qqq

L

q

L

dt

d


44


5. Hamilton’s Equations

The Lagrange’s Equations nicQq

T

q

T

dt

d m

k

kiki

ii

,,2,11

can be rewritten as:

nicQq

T

tq

Tq

qq

Tq

qq

T

q

T

dt

d m

k

kiki

i

n

i ij

jij

jii

,,2,111

222

therefore consist of a set of n simultaneous second-order differential equations.

They must be solved tacking in consideration the m constraint equations.

mkcqc kt

n

ii

ki ,,2,10

1

A procedure for the replacement of the n second-order partial-differential equations by2n first-order ordinary-differential equations consists of formulating the problem in terms of 2n Hamilton’s Equations.

We define first: General Momentum: niq

Tp

ii ,,2,1

We want to find the transformation from the set of variables to the set by the Legendre’s Dual Transformation.

tqq ,,

tpq ,,

45


Hamilton’s Equations (continue)

Legendre’s Dual Transformation.

Adrien-MarieLegendre1752-1833

Let consider a function of n variables , m variables and time t.ix iy

tyyxxF mn ,,,,,, 11 and introduce a new set of variables defined by the transformation:iu

nix

Fu

ii ,,2,1

We can see that:

mmnn

m

nnn

n

n dy

dy

dy

yx

F

yx

F

yx

F

yx

F

dx

dx

dx

x

F

xx

F

xx

F

x

F

du

du

du

2

1

2

1

2

1

2

11

2

2

1

2

2

1

2

1

2

21

2

2

1

We want to replace the variables by the new variables .We can see that the new n variables are independent if the Hessian Matrix is nonsingular.

nidxi ,,2,1 ni

njji xx

F,,1

,,1

2

46



Legendre’s Dual Transformation (continue-1)

Let define a new function G of the variables , and t.iu iy

tyyuuGFxuG mn

n

iii ,,,,,, 11

1

Then:

dtt

Fdy

y

Fdx

x

Fudux

dtt

Fdy

y

Fdx

x

FdxuduxdG

m

jj

j

n

ii

iiii

n

i

m

jj

ji

i

n

iiiii

11

0

1 11

But because: tyyuuGG mn ,,,,,, 11

dtt

Gdy

y

Gdu

u

GdG

n

i

m

jj

ji

i

1 1

Because all the variations are independent we have:

t

F

t

Gmj

y

F

y

Gni

u

Gx

jjii

;,,1;,,1

47



Legendre’s Dual Transformation (continue-2)

Now we can define the Dual Legendre’s Transformation from

tyyxxF mn ,,,,,, 11 FxutyyuuGn

iiimn

111 ,,,,,, to

by using

nix

Fu

ii ,,2,1

niu

Gx

ii ,,2,1

End of Legendre’s Dual Transformation

48



Following the same pattern to find the transformation from the set of variables to the set , we introduce the Hamiltonian: tqq ,, tpq ,,

tqqTqpHn

iii ,,

1

whereni

q

Tp

ii ,,2,1

Then tpqHH ,,

dtt

Hdp

p

Hdq

q

Hdt

t

Tdq

q

Tdpq

dtt

Tdq

q

Tqd

q

TqdpdpqdH

n

ii

ii

i

n

ii

iii

n

ii

ii

iiiii

11

1

and

01

dtt

T

t

Hdpq

p

Hdq

q

T

q

Hn

iii

ii

ii

49



If the Hessian Matrix is nonsingular, all the are independent, but not the that must be consistent with the constraints:

ni

njji qq

T,,1

,,1

2

nidpi ,,2,1 nidqi ,,2,1

mjdtcdqc jt

n

ii

ji ,,2,10

1

Let adjoin the previous equations by the constraint equations multiplied by the m

Lagrange’s multipliers :j'0'''

11 11 1

m

j

jij

n

ii

m

j

jij

m

j

jt

n

ii

jij dtcdqcdtcdqc

We have

0''11 1

dtct

T

t

Hdpq

p

Hdqc

q

T

q

H m

j

jtj

n

iii

ii

m

j

jij

ii

50


Hamilton’s Equations (continue)By proper choosing the m Lagrange’s multipliers λ’j ,the remainder differentials and dt are independent and therefore we have:ii dpdq ,

ni

ct

H

t

T

cq

H

q

T

p

Hq

m

j

jtj

m

j

jij

ii

ii

,,2,1

'

'

1

1

Legendre’s Dual Transformation

By differentiating the General Momentum Equation and using Lagrange’s Equations we obtain:

m

j

jijji

i

m

j

jiji

iii cQ

q

HcQ

q

T

q

T

dt

dp

11

'''''

ni

cQq

Hp

p

Hq

m

j

jiji

ii

ii

,,2,1

1

mkcqc kt

n

ii

ki ,,2,10

1

Extended Hamilton’s Equations

Constrained Differential Equations

51



For Holonomic Constraints (constraints of the form )we can (theoretically) reduce the number of generalized coordinates to n-m and wecan assume that and n represents the number of degrees of freedom of the system(this reduction is not possible for Nonholonomic Constraints). Then:

mjtqqf nj .,10,,,1

0 jt

ji cc

ni

Qq

Hp

p

Hq

ii

i

ii

,,2,1

Extended Hamilton’s Equations for

Holonomic Constraints

niq

VQ

ii ,,2,1

qVtqqTtqqL

,,,,

niq

L

q

Tp

iii ,,2,1

Extended Hamilton’s Equations for


and a

Conservative System

Conservative

System

52



Define:

Hamiltonian for

Conservative Systems qVtqqTtqqLqptqqH

n

iii

,,,,,,1

Hamilton’s Canonical Equations for

Conservative Systems

with


ni

q

Hp

p

Hq

ii

ii

,,2,1

We have:


04/13/23 53

SOLO Classical Field Theories

Introduction to Lagrangian and Hamiltonian Formulation

Second Method (Carathéodory)

Constantin Carathéodory (1873-1950)

Carathéodory developed another approach to the Euler-Lagrange Equations

Covariant

k

l

l

jj

k

jj

k q

q

q

qqtL

q

qqtLp

,,,,:Define Canonical Momentum

We assume that is a one to one correspondence between the n components of pj and the n components of and from the definition of pk (by the Inverse Function Theorem) we have

jq

kkjj pqtx ,,

From we see that the one to one correspondence is possible only if

td

tq

qqtLqd

qq

qqtLqd

qq

qqtLpd

k

jjj

jk

jjj

jk

jj

k

,,,,,,

:222

0

,,det

2

jk

jj

qq

qqtL

Extremal of the Functional . 2

1

,,t

t jj tdtqtqtLCI

04/13/23 54




k

l

l

jj

k

jj

k q

q

q

qqtL

q

qqtLp

,,,,: Canonical Momentum

kjj ptq ,

Define kj

jikk

kk ptpptqtLpqtH ,,,,:,, Hamiltonian

Let compute

jq

q

Lp

j

j

k

kj

k

kj

qqp

pp

q

q

L

p

Hkk

kk

j

q

Lp

jjjjjj q

L

qp

qq

L

q

L

q

Hjj

t

L

tp

t

q

q

L

t

L

t

Hkk

kk

q

q

Lp

j

j

j

j


1

,,t

t jj tdtqtqtLCI


http://www.google.com/url?sa=i&source=images&cd=&cad=rja&docid=ycYx8RDDkTV4wM&tbnid=ZOpiSGx07xuEzM:&ved=0CAgQjRwwAA&url=http://www.thefamouspeople.com/profiles/sir-william-rowan-hamilton-552.php&ei=NGhrUtL9Csfk4QSXoYDACw&psig=AFQjCNHaqy50fW5WhL-5-j7VSNMvYnxITQ&ust=1382857140275958

04/13/23 55




k

jj

k q

qqtLp

,,

: Canonical Momentum

kj

jikk

kk ptpptqtLpxtH ,,,,:,, Hamiltonian

j

j

qp

H

jj q

L

q

H

t

L

t

H

Let compute j

j

jjj q

H

td

pd

q

L

q

L

td

dLE


1

,,t

t jj tdtqtqtLCI


04/13/23 56




Given a Scalar S = S (t,qk) є C2. Along a curve C: qj = qj (t) we can form the Total Derivative:

jj

qq

S

t

S

td

Sd

With the aid of this Scalar ,S, we may construct an alternative Lagrangian by writing

td

tqtSdtqtqtLcq

q

S

t

StqtqtLctqtqtL

jjjk

kjjjj ,

,,,,:,,*

We obtain a new Integral

12** 2

1

2

1

,,,, SSCIctdtd

SdtqtqtLctdtqtqtLCI

t

t

jjt

t

jj

21* SSCIcCI

where c > 0 is a constant.


1

,,t

t jj tdtqtqtLCI

04/13/23 57




Alternative Lagrangian

td

tqtSdtqtqtLctqtqtL

jjjjj ,

,,:,,*

12** 2

1

,, SSCIctdtqtqtLCIt

t

jj

We obtainThat is independent of the choice of the curve C joining the points P1 and P2.It follows that C will be an extremal of the integral I* , if and only if, it is an extremal of the integral I. Accordingly I and I* are called Equivalent Integrals.

21* SSCIcCI

Therefore the Lagrangian that gives the Equations of Motions is not necessarilyL = T – V but it can be any function L* = c (T – V) – dS/dt.


1

,,t

t jj tdtqtqtLCI

04/13/23 58




Alternative Lagrangian that gives the same equation of Motion doesn’t have necessarily to differ by a total time derivative. For example


1

,,t

t jj tdtqtqtLCI

22

21

22

21

2121

2

1qqqqL

qqqqL

b

a

0&0

0&0

2222

1111

1122

2211

qqq

L

q

L

td

dqq

q

L

q

L

td

d

qqq

L

q

L

td

dqq

q

L

q

L

td

d

bbbb

aaaa

221

2212

1qqqqLL ab


A Geodesic Field is defined as a Field of Contravariant Vectors , given at each point of a finite Region G in the n+1 Space (t, q1, q2,…,qn) by a set of n class C2 functions ψj (t,qk) that are such that, for a suitably chosen function S = S (t,qk), the following conditions are satisfied

while

One other condition (Jacobi) that the curves that define a Geodesic Field,is that they don’t intersect for t1 < t < t2.

kjj qtq ,

kjjkkkk qtqwhenevertd

SdqqtLqqtL ,0,,:,,*

otherwiseqqtL kk 0,,*

kjj qtq ,

04/13/23 59



Hamilton-Jacobi Theory

Geodesic Field

See “Calculus of Variation” for a detailed exposition.

Note: Since we see that it is equivalent to “Action”. t

t

kk tdqqtLtS1

,,


1

,,t

t jj tdtqtqtLCI

04/13/23 60




Geodesic Field (continue – 1)

For any other curve K in G that joins the points P1 and P2 we have

FieldGeodesictqqtdqqtL jjt

t

kk :0,,2

1

0

*

FieldGeodesictqqKtdqqtL jjt

t

kk

:0,,2

1

0

*

Therefore the Integral I* (Γ) is a local minimum, and since this is an Equivalent Integral to I (Γ), this is also a local minimum in G.

kjj qtq , is a System of n First-Order Differential Equations, that may be integrated to cover the region G. Let Γ: qj = qj (t) be a member of this family andLet P1 and P2 the points on Γ, corresponding to t1 and t2, It follows that


1

,,t

t jj tdtqtqtLCI

04/13/23 61




Basic Properties of Geodesic Field

Since FieldGeodesictqqqq

S

t

SqqtLqqtL jjl

lkkkk

:0,,:,,*

kjjjj

pq

L

jj

kk

j

kk

qtqwheneverq

Sp

qd

S

q

qqtL

q

qqtLjj

,,,,,

0*

Therefore FieldGeodesictqqq

Sp jj

jj

:

kjk

kjj qtpqtq ,,,

This defines the Covariant Vector Field pj = pj (t, qk) as a function of position in G. Assuming the one-to-one correspondence between pj and (j,k=1,2,…,n), we have

kq

We get ll

q

Sp

q

ll

kk pt

Sq

q

S

t

SqqtL

ll

ll

,,

k

kpqtq

q

Sp

ll

kkkk

q

SqtHpqqtLqqtH

kkkk

kk

,,,,,,,,

0,,

kk

q

SqtH

t

Sand


1

,,t

t jj tdtqtqtLCI

04/13/23 62





Carl Gustav Jacob Jacobi

(1804-1851)

0,,

kk

q

SqtH

t

S Hamilton-Jacobi Equation

Hamilton-Jacobi is a First-Order Nonlinear Partial Differential Equation for the Scalar Function S = S (t,qk), that defines the Geodesic Field through

from which a Unique Vector Field

is obtained

jj q

Sp

kjk

kjj qtpqtq ,,,


1

,,t

t jj tdtqtqtLCI

04/13/23 63






(1804-1851)

0,,

kk

q

SqtH

t

S

Start withjj q

Sp

lljjj

kj q

qq

S

xt

S

q

qtS

td

d

td

pd

22,

Partially Differentiate the Hamilton Jacobi Equation

with respect to qj

0

,,,, 222

lljjj

q

Sp

xp

Hjl

l

kk

jk

k

jq

qq

S

q

H

tq

S

q

p

p

pqtH

q

pqtH

tq

Sll

l

l

we obtainj

j

q

H

td

pd


1

,,t

t jj tdtqtqtLCI

04/13/23 64






(1804-1851)

we obtain

njq

H

td

pd

njp

H

td

qd

j

j

j

j

,,2,1

,,2,1

0

j

jtd

pd

q

L

q

H

q

Ljjj q

H

td

pd

q

L

q

L

td

dLE

jj

jj

A System of 2n First-Order Ordinary Differential Equations which the curve Γ must satisfy.This is equivalent to Euler-Lagrange n Second-Order Partial Differential Equations, since

k

jj

k q

qqtLp

,,

: Canonical Momentum

kkj

jiikk

kk pqtppqtqtLpqtH ,,,,,,:,, Hamiltonian

Hamilton’sEquations


1

,,t

t jj tdtqtqtLCI

04/13/23 65






(1804-1851)

Let return to the condition

0

,,,,,,*

k

k

jjjjjj q

q

qtS

t

qtSqqtLqqtL

where the equality holds for kjk

kjj qtpqtq ,,,

We want to eliminate S from this inequality.For this use llk

klljj

pq

S

jj

j

pqtppqtqtLq

SqtH

t

qtSjj

,,,,,,:,,,

0

,,,,,,,,,

k

k

j

llk

klljjjj q

q

qtSpqtppqtqtLqqtL

jjjj qqtE ,,, : Weierstrass Excess Function

Weierstrass Sufficiency

Condition fora Local

Minimum

0,,,,:,,,

kkk

jjjjjjjj q

q

LqtLqqtLqqtE

and

k

j

k

jj

k q

qtS

q

qqtLp

,,,:

Karl Theodor Wilhelm Weierstrass (1815-1897)


1

,,t

t jj tdtqtqtLCI


04/13/23 66





1

,,t

t

jj tdtxtxtLCI

Example: Recovering Newton Equation

mVF

r

r

VF - Force on Mass m, due to External Potential V

rrmT 2

1 - Kinetic Energy of Mass m

rV

- Potential Energy of the External Force

rmr

Lp

- Canonical Momentum

- HamiltonianEVTVrrmrrmVrrmxpLH :2

1

2

1:

VrrmVTL 2

1: - Lagrangian

04/13/23 67





1

,,t

t

jj tdtxtxtLCI

Example: Recovering Newton Equation (continue – 1)

Sr

Srm

r

Lp


- HamiltonianVTVrrmH 2

1

Hamilton-Jacobi Equation

where S is given by

mVF

r

r

VSSm

VrrmH 2

1

2

1

02

1

VSSmt


0,,

SrtHt

S

04/13/23 68





1

,,t

t

jj tdtxtxtLCI

Example: Recovering Newton Equation (continue – 2)

mVF

r

r

02

1

VSSmt


Let take the Gradient of this equation

Vrrr

mp

tVSS

mt

S rmpS

0

22

10

FVpt

ptd

d

Newton Equation

pt

rmr

rpt

ptd

d

0

04/13/23 69





1

,,t

t

jj tdtxtxtLCI

Example: Classical Harmonic Oscillator In Equilibrium

Displaced from Equilibrium

xmxkF - Force of Spring on Mass m

2

2

1xmT - Kinetic Energy of Mass m

2

2

1xkV - Potential Energy of the Spring

xmx

Lp


- HamiltonianEVTxkxmxxmxkxmxpLH :2

1

2

1

2

1

2

1: 2222

22

2

1

2

1: xkxmVTL - Lagrangian

02

2

mx

L

04/13/23 70





1

,,t

t

jj tdtxtxtLCI

Example: Classical Harmonic Oscillator (continue – 1)

In Equilibrium

Displaced from Equilibrium- Hamiltonian 2

2

2

1

2

1:, xk

m

ppxH

xm

k

td

pd

mtd

xd

m

p

p

H

td

xd

xkx

H

td

pd

12

2

Initial Conditions:

00

0

ttd

xd

Atx

m

k

tAmtxmtp

tAtx

:sin

cos

: Solution

On the minimizing curve Γ

.2

1cos

2

1sin

1

2

1

2

1

2

1:, 2222222

2

constAktAktm

kAm

mxk

m

ppxH

Hamilton’s Equations:

04/13/23 71





1

,,t

t

jj tdtxtxtLCI

Example: Classical Harmonic Oscillator (continue – 2)

In Equilibrium


td

SdtAktAktA

m

km

mxk

m

pVTpxL

2cos

2

1cos

2

1sin

1

2

1

2

1

2

1, 22

2

22

t

ttA

kttA

ktA

kS

cos

sincos

2

1cossin

2

12sin

4

1 222

t

tx

kxtS

cos

sin

2

1, 2

HAk

txk

t

xtS tAx

2

cos

22

2

1

cos

1

2

1,

ptA

k

t

tx

k

x

xtS tAmp

mk

tAx

sincos

sincos

sin,

mm

kmkm

mk

kk

/


04/13/23 72




1

,,t

t jj tdtqtqtLCI

Invariance Properties of the Fundamental Integral

Theorem of Noether for Single Integral

Amalie Emmy Noether

(1882 –1935)

Consider the Functional 2

1

,,t

t jj tdtqtqtLCI

Symmetry of Lagrangian is a Geometric Property in which quantities remain unchanged under coordinate transformation. Invariance is an Algebraic or Analytic Property in which “Integrals of Motion” are constant along System Trajectories..

Consider also a continuous family of coordinate transformationqj (t) → qj (t,ε) parameterized by a single quantity ε.We assume that the Lagrangian is invariant to this parameter

which implies a Symmetry of the System. Thus

0,,0,,,,,, tqtqtLtqtqtL jjjj

njq

q

Lq

q

Lq

q

Lq

q

LtqtqtL

d

d j

j

j

j

n

j

j

j

j

jjj ,,100,,,,

1

http://www.google.com/url?sa=i&source=images&cd=&cad=rja&docid=Y1_Jw2lRr55tUM&tbnid=eVdccyVAxgfHYM:&ved=0CAgQjRwwAA&url=http://www.awm-math.org/noetherbrochure/AboutNoether.html&ei=01VmUqisJuXx4gTIjYDIAg&psig=AFQjCNHVX3VUH7MHrmNiLNyPFQy-pSbBBw&ust=1382524755684800

73




1

,,t

t jj tdtqtqtLCI


Theorem of Noether for Single Integral (continue – 1)

Amalie Emmy Noether

(1882 –1935)

njq

q

Lq

q

L j

j

j

j

,,10

Since qi(ε) is a solution of Euler-Lagrange’s Equation we have

njq

L

q

L

td

d

jj

,,10

Substitute this in previous equation we obtain

nj

tq

q

L

td

dtq

q

Ltq

q

L

td

d j

j

j

j

j

j

,,10,,,

The quantity is a Constant of Motion.

,, tq

ptq

q

L jj

j

j

Noether Theorem is not of great importance in Particle Dynamics.It becomes extremely important in Field Theory and in Quantum Mechanics.

njq

Lp

jj ,,1



74

SOLO Foundation of Geometrical Optics

Fermat’s Principle (1657) in Optics

1Q

1P

2P

2Q1Q

2Q

1S

SdSS 12

2PS

1PS

2'Q

rd

s

s

The Principle of Fermat (principle of the shortest optical path) asserts that the optical length

of an actual ray between any two points is shorter than the optical ray of any other curve that joints these two points and which is in a certain neighborhood of it. An other formulation of the Fermat’s Principle requires only Stationarity (instead of minimal length).

2

1

P

P

dsn

An other form of the Fermat’s Principle is:

Principle of Least Time The path following by a ray in going from one point in space to another is the path that makes the time of transit of the associated wave stationary (usually a minimum).

The idea that the light travels in the shortest path was first put forward by Hero of Alexandria in his work “Catoptrics”, cc 100B.C.-150 A.C. Hero showed by a geometrical method that the actual path taken by a ray of light reflected from plane mirror is shorter than any other reflected path that might be drawn between the source and point of observation.

75


Proof of Fermat’s Principle Using Calculus of Variations

We have:

constS constdSS

s

2

1

P

P

dsn

1P

2P

2

1

2

1

2

1

,,,,1

1,,1

,,1

0

22

00

P

P

P

P

P

P

xdzyzyxFc

xdxd

zd

xd

ydzyxn

cdszyxn

ctdJ

Let find the stationarity conditions of the Optical Path using the Calculus of Variations

xdxd

zd

xd

ydzdydxdds

22

222 1

Define:

xd

zdz

xd

ydy &:

22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF

76


Proof of Fermat’s Principle Using Calculus of Variations (continue – 1)

Necessary Conditions for Stationarity (Euler-Lagrange Equations)

22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF

0

y

F

y

F

dx

d

2/1221

,,

zy

yzyxn

y

F

y

zyxnzy

y

F

,,

1 2/122

011

,, 2/122

2/122

y

nzy

zy

yzyxn

xd

d

0

z

F

z

F

dx

d

011

2/1222/122

y

n

zy

yn

xdzy

d

77



Necessary Conditions for Stationarity (continue - 1)

We have

01

2/122

y

n

zy

yn

sd

d

y

n

sd

ydn

sd

d

In the same way

01

2/122

z

n

zy

zn

sd

d

z

n

sd

zdn

sd

d

78




Using xdxd

zd

xd

ydzdydxdds

22

222 1

we obtain 1222

sd

zd

sd

yd

sd

xd

Differentiate this equation with respect to s and multiply by n

sd

d

0

sd

zd

sd

dn

sd

zd

sd

yd

sd

dn

sd

yd

sd

xd

sd

dn

sd

xd

sd

nd

sd

zd

sd

nd

sd

yd

sd

nd

sd

xd

sd

nd

222

sd

nd

and

sd

nd

sd

zdn

sd

d

sd

zd

sd

ydn

sd

d

sd

yd

sd

xdn

sd

d

sd

xd

add those two equations

79




sd

nd

sd

zdn

sd

d

sd

zd

sd

ydn

sd

d

sd

yd

sd

xdn

sd

d

sd

xd

Multiply this by and use the fact that to obtainxd

sd

cd

ad

cd

bd

bd

ad

xd

nd

sd

zdn

sd

d

xd

zd

sd

ydn

sd

d

xd

yd

sd

xdn

sd

d

Substitute and in this equation to obtainy

n

sd

ydn

sd

d

z

n

sd

zdn

sd

d

xd

zd

z

n

xd

yd

y

n

xd

nd

sd

xdn

sd

d

Since n is a function of x, y, zx

n

xd

zd

z

n

xd

yd

y

n

xd

ndzd

z

nyd

y

nxd

x

nnd

and the previous equation becomes

x

n

sd

xdn

sd

d

80




We obtained the Euler-Lagrange Equations:

x

n

sd

xdn

sd

d

y

n

sd

ydn

sd

d

z

n

sd

zdn

sd

d

ksd

zdj

sd

ydi

sd

xd

sd

rd

kzjyixr

ˆˆˆ

ˆˆˆ

Define the unit vectors in the x, y, z directionskji ˆ,ˆ,ˆ

The Euler-Lagrange Equations can be written as:

nsd

rdn

sd

d

The equation is called Eikonal Equation Eikonal (from Greek έίκων = eikon → image) .

See “Geometrical Optics” Presentation

81


Proof of Fermat’s Principle Using Calculus of Variations (continue – 6)Hamilton’s Canonical Equations

Define

sd

zdzyxn

zy

zzyxn

z

Fp

sd

ydzyxn

zy

yzyxn

y

Fp

z

y

,,1

,,:

,,1

,,:

2/122

2/122

2222222 1 zynzypp zy Adding the square of twose two equations gives

2

222

2221

xd

sd

ppn

nzy

zy

from which

Substituting in 22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF

gives

222

2

,,,,zy

zy

ppn

nppzyxF

22

22

1,,1,,,,,, zyzyxnxd

zd

xd

ydzyxnzyzyxF

82



Hamilton’s Canonical Equations (continue – 1)

From

sd

zdzyxn

zy

zzyxn

z

Fp

sd

ydzyxn

zy

yzyxn

y

Fp

z

y

,,1

,,:

,,1

,,:

2/122

2/122

solve for

222

2

,,,,zy

zy

ppn

nppzyxF

and

222

222

zy

z

zy

y

ppn

pz

ppn

py

Define the Hamiltonian

sd

xdzyxnppzyxn

ppn

p

ppn

p

ppn

n

zpypppzyxFppzyxH

zy

zy

z

zy

y

zy

zyzyzy

,,,,

,,,,:,,,,

222

222

2

222

2

222

2

83




From

We obtain the Hamilton’s Canonical Equations

sd

xdzyxnppzyxnppzyxH zyzy ,,,,,,,, 222

222

222

zy

z

z

zy

y

y

ppn

p

p

H

xd

zdz

ppn

p

p

H

xd

ydy

222

222

zy

z

zy

y

ppn

z

nn

z

H

xd

pd

ppn

y

nn

y

H

xd

pd

84




From sd

xdzyxnppzyxnppzyxH zyzy ,,,,,,,, 222 222

zy ppn

n

sd

xd

By similarity with sd

ydzyxnpy ,,

define 222 ,,,,,,,,: zyzyx ppzyxnppzyxHsd

xdzyxnp

Let differentiate px with respect to x x

H

xd

Hd

ppn

xn

n

xd

pd

zy

x

222

Let compute

x

n

n

ppn

ppn

xn

n

sd

xd

xd

pd

sd

pd zy

zy

xx

222

222

85




and

x

n

n

ppn

ppn

xn

n

sd

xd

xd

pd

sd

pd zy

zy

xx

222

222

y

n

n

ppn

ppn

yn

n

sd

xd

xd

pd

sd

pd zy

zy

yy

222

222

z

n

n

ppn

ppn

zn

n

sd

xd

xd

pd

sd

pd zy

zy

zz

222

222

nsd

pd

xpnsd

xd 1

y

zy

zy

y pnn

ppn

ppn

p

sd

xd

xd

yd

sd

yd 1222

222

z

zy

zy

z pnn

ppn

ppn

p

sd

xd

xd

zd

sd

zd 1222

222

pnsd

rd ray

1

We recover the result from Geometrical Optics Return to Table of Content

04/13/23 86



The Two Body Central Force Problem

Assume a system of two mass points m1 and m2 located at , respectively, subject to an interaction potential V,where V is any function of the range vector between particles.

21 randr

Define:

212211

12

/

:

mmrmrmR

rrr

c

- range vector between particles

- system center of mass

21

12

21

21 ,

mm

rmRr

mm

rmRr cc

We obtain:

The kinetic energy of the system:

21

1

21

12

21

2

21

21222111 2

1

2

1

2

1

2

1

mm

rmR

mm

rmRm

mm

rmR

mm

rmRmrrmrrmT cccc

rrmm

mmRRmmT

M

cc

21

2121 2

1

2

1or

,,2

1

2

121 rrVrrMRRmmVTL cc

The Lagrangian of the system is

04/13/23 87



2121

222

/:2

1

2

1

mmmmM

rVrrMrVrrMVTL

When V is a function of r only like in classical gravitation and electromagnetic fields, we can write, in polar coordinates and ignoring the term describing the motion of center of mass

The Lagrange’s Equations are ,2,10

i

q

L

tq

L

td

d

ii

2. r - independent variable

r

VrM

r

LrM

r

L

2,

1. θ - independent variable

0,2

L

rML 02

rMtd

dLL

td

d

02

r

VrMrM

r

L

r

L

td

d

04/13/23 88



The Specific Angular Momentum and Angular Momentum are defined as

22 :&: rMhMlrh

02

rMtd

dLL

td

dThe Equation means that Angular Momentum is constant (Conservation of Angular Momentum )

This Equation can be rewritten as

drMdtl 2that implies

d

d

rM

l

d

d

rM

l

dt

d

d

d

rM

l

dt

d222

2

2

a

b

The Area swept out by one of the body when moving around the other (see Figure) is given by

drrAd2

1

It follows that

.22

consth

d

Ad

rM

l

td

Ad

This is Kepler Second Law.

04/13/23 89



02

r

VrMrM Return to the equation

2: rMl and

We have 022 2

2

2

2

3

2

rVrM

l

rd

drMrV

rM

l

rrM

r

V

rM

lrM

or

2

2

2

2

2:

2

r

lrVrV

rVrd

drV

rM

l

rd

drM

eff

eff

Multiplying the differential equation by we obtainr rV

td

drV

rd

d

td

rdrM

td

drrM effeff

2

2

1

or 02

1 2

energytotal

energypotential

eff

energykinetic

Etd

dVrM

td

d

The total Energy is conservedeffVrME 2

2

1:

04/13/23 90



The total Energy is conservedeffVrME 2

2

1:

Solving for we obtainr

2

2

2

22

rM

lVE

MVE

Mr eff

or

2

2

22

rMl

VEM

rdtd

Using we obtaindd

rM

l

dt

d2

2

222

122

1

r

rd

rlVM

lEM

d

Consider the case where V (r) = -k/r = - k u

ud

uul

kMl

EMd

222

22

1

Integrating this expression gives

12

11

sin28

22

sin

2

2

2

1

2

22

21

kMlE

rkMl

lkM

lEM

lkM

u

i

04/13/23 91



12

11

sin28

22

sin

2

2

2

1

2

22

21

kMlE

rkMl

lkM

lEM

lkM

u

i

Inverting this expression we obtain

ikM

lE

l

kM

rsin

211

12

2

2

This is usually written as

2

2

2

21:

sin11

kM

lEe

el

kM

r i

This is a equation of Conic Section:1.If e > 1 and E > 0, the trajectory is a Hyperbola.2. If e = 1 and E = 0, the trajectory is a Parabola.3. If e < 1 and E < 0, the trajectory is a Ellipse.4. If e = 0 and E = -mk2/(2l2), the trajectory is a Circle.


04/13/23 92

SOLO


Special Relativity Theory

We introduce a 4-dimensional space-time or four-vector x with components:

ctxxxxxxxx 003210 ,,,,:

The differential length element is defined as:

220232221202 : xdxdxdxdxdxds

or

dxdxgdxdxgds

summationsEimstein

convention

'3

0

3

0

2

The metric corresponding to this differential length is given by g:

gg

1000

0100

0010

0001

therefore:

0,1,1 33221100 ggggg

We can see that Igg

1000

0100

0010

0001

1000

0100

0010

0001

1000

0100

0010

0001

04/13/23 93

SOLO

Four-Dimensional Formulation of the Theory of Relativity (continue – 1)

Therefore since we have

gg

gg

If a 4-vector has the contravariant components A0,A1,A2,A3 we have

32103210 ,,,,,, AAAAwhereAAAAAAA

Using the g metric we get the same 4-vector described by the covariant components:

AA

A

A

A

A

A

A

A

A

AgA

,

1000

0100

0010

0001

0

3

2

1

0

3

2

1

0

The Scalar Product of two 4-vectors is:

ABgABgABABAABBABAABBAB

000000 ,,,,


04/13/23 94

SOLO


Therefore since we have

gg

gg

If a 4-vector has the contravariant components A0,A1,A2,A3 we have

32103210 ,,,,,, AAAAwhereAAAAAAA

Using the g metric we get the same 4-vector described by the covariant components:

AA

A

A

A

A

A

A

A

A

AgA

,

1000

0100

0010

0001

0

3

2

1

0

3

2

1

0

The Scalar Product of two 4-vectors is:

ABgABgABABAABBABAABBAB

000000 ,,,,


95

SOLOSpecial Relativity Theory


Let introduce the following definitions:

,,,,:

,,,,:

03210

03210

xxxxxx

xxxxxx

The 4-divergence of a 4-vector A is the invariant:

Ax

AAA

xAAA

xA

0

00

00

0,,,,

The four-dimensional Laplacian operator (d’Alembertian) is defined as:

220

2

0000,,,,:

xxxxx

04/13/23 96

SOLO


We have:

2

222222232221202 1

1dt

xd

cdtcxddtcdxdxdxdxds

We define

- the velocity vector of the particle in the inertial frame.td

xdu

21

1:&:

u

uu c

u

222222222

2222 /11 dcdtcdtc

c

udtcds uu

where

tdt

dtdu

u 21

is the differential of the proper time τ.


04/13/23 97

SOLO


4-Coordinates Vector

The 4-coordinates vector is:

xxxxxxxxxxxxxx ,,,,&,,,, 0321003210

23210000 22222

,, sdxdxdxdxdxdxdxdxdxdxdxdxdxd


04/13/23 98

SOLO


4-Velocity Vector

td

xduUUuc

d

dt

dt

xd

d

dt

dt

dx

d

xd

d

dxU uu

,,,,: 0

00

20

0

u1

1:,,,:

c

UUucd

xd

d

dxU uuu

The 4-Velocity vector is defined as:

22

22

2

2

222 1

1

1,, c

c

uc

cu

ucucucUU uuuuu

d

Ud

d

dUUUUU

d

Ud

d

dU

d

UdU

d

dUU

d

Ud

d

dUUUUU

d

Ud

d

dU

UUd

dUUUUUU

d

dUU

d

d

d

dc

,,2,,2

22,,,,

,,,,0

000

0

00

000

0

0000

2

0

UU

d

dU

d

dUU

d

dUUU

d

d


04/13/23 99

SOLO


4-Moment Vector

The 4-Momentum vector of the mass m is defined as: UUmUmp

,000

2

02

2

0000

0

1

1

:

cu

mmofEnergymcE

c

Emc

cu

cmcmUmp u

pumu

cu

mumUm u

2

000

1

Therefore: UUmUmpppUUmUmppp

,,&,, 000

0000

0

220

20

22

2

,, cmUUmpc

Ep

c

Ep

c

Epp

from which we get: 220

22222 cmcpcmE


100

SOLO


4-Force Vector

The 4-Force vector on the mass m is defined as:

d

pd

d

Ed

cp

c

E

d

dp

d

dF

,1

,:

From the relations:

uu

u d

dtdtdtd

21

c

uF

d

td

td

rd

c

F

d

Ed

cEdcmEdrdFdT u

12

0

Ftd

pd

d

td

d

pdu


2u

1

1:

c

u

Fc

uFF

c

uF

d

pdF

Fc

uFF

c

uF

d

pdF

uuu

uuu

,,:

,,:

101

SOLO

Four-Dimensional Formulation of the Theory of Relativity (continue –7)

4-Force Vector (continue – 1)

Assuming that the rest mass doesn’t change : 00 d

dm

d

dUmUm

d

dp

d

dF 00

Using the relation:

0

UU

d

dU

d

dUU

d

dUUU

d

d

we get:00

d

dUUmFU


102

SOLO



Assuming that the rest mass changes ,: 00 d

dm

FU

d

md

d

UdmUm

d

dp

d

d 000

Using the relation: 0

UU

d

dU

d

dUU

d

dUUU

d

d

where:

00

d

dUUmFU


U

d

md

d

UdmF 0

0 :,:

Fc

uF

d

pdF

Fc

uF

d

pdF

uu

constm

uu

constm

,:

,:

.

.

0

0

103

SOLO




dm


dt

xduUUuc

d

dt

dt

xd

d

dt

dt

dx

d

xd

d

dxU uu

,,,,: 0

00

20

0

/1/1:,,,: cuUUucd

xd

d

dxU uuu


Let define:

uuuu cc

,:,,:

02 :,,

uc

ucU uuuuu

FU

d

md

d

UdmU 0

0 02 0

0

0

0

UFUUU

d

md

d

UdUm

c

d

mdcU 020

The 4-Velocity Momentum vector that the particle loses per unit proper time τ.

104

SOLO




dm


D

R

Ud

mdF

d

Udm 0

0

D

d

Udm 0

U

cU

d

mdR

RFD

cd

md

2

00

020

:

:

Fμ – External ForceRμ – Reaction ForceDμ – Driving Force

0

0

00

2

0

20

RUFUDU

UUc

URU

c

F

cuF

cFp

c

E

d

dp

d

du ,

11, 0

uu

u d

dtdtdtd

21

Also

F

cuF

cp

c

E

td

dp

td

d,

11, 0

105

SOLO




dm


PcQdtcdtdt

cdQu

uu

dtd u

,/,/,/ 00

/

In special cases the rate of change of Πμ may be due to emission or absorption of Heat by the particle

where

tdP

tdQ

0

04/13/23 106

SOLO



Principle of Last Action for 4-Vector for a Free Particle (zero external forces)

We want to define the Action Integral for 4-Vectors, similar to Fermat Principle in Optics, to recover the 4-Momentum and 4-Force Vectors. Define

2

1

:t

t

b

atdLsdA α is a constant, to be defined

22 /1 cudtcsd Using we obtain 2

1

2

1

22 /1:t

t

t

ttdLtdcucA

Therefore the Lagrangian isc

uccucL

2/1

222

Since the constant α c in the Lagrangian doesn’t affect the equations of motion,we concentrate on the second term that must be equal to Energy. We choose α toobtain the non-relativistic kinetic energy m0u2/2 (m0 – rest mass), i.e.:

cm0222

0 /1 cucmL 2

1

2

10

2220 /1:

s

s

t

tsdcmtdcucmA

uzyxqzyxq ,,&,,

04/13/23 107

SOLO



Canonical Momentum

2220 /1 cucmL

umcu

umczyxcm

z

y

x

q

Lp

22

0222220

/1/1

/

/

/

0/1

/

/

/22222

0

czyxcm

z

y

x

q

L

Particle Equation of Motion

Qtd

pd

cu

um

td

d

q

L

q

L

td

d

22

0

/1

- are the External Forces acting on the particle (zero for free particle)Q

Principle of Last Action for 4-Vector for a Free Particle

22

0

/1:

cu

mm

- relativistic mass of Particle at velocity u

Free Particle Lagrangian

04/13/23 108

SOLO



2220 /1 cucmL

Principle of Last Action for 4-Vector for a Free Particle

The Free Particle Hamiltonian is defined as: LpH u:

umcu

umczyxcm

z

y

x

q

Lp

22

0222220

/1/1

/

/

/

where:

therefore:2

22

20222

022

0

/1/1u

/1u: cm

cu

cmcucm

cu

umLpH

We van see that the Hamiltonian H derived from the Lagrangian L is equal to the Total Energy E of the Free Particle

Ecmcu

cmLpH

2

22

20

/1u:

Free Particle Lagrangian

04/13/23 109

SOLO



UUcucmcucmL 22

0222

0 /1/1

Principle of Last Action for 4-Vector for a Free Particle (Covariant Treatment)

Let use: UαUα = c2 to write

2

1

2

1

2

100

2220 /1:

dUUcmdccmtdcucm

t

tA

We defined uu

u d

dtdtdtd

21 2

u1

1:

c

u

therefore the Action Integral is

To use this for a variational calculation we must add the constraint

2cUU 0

d

UdUor equivalently

This can be added using Lagrange’s multipliers technique, but here we use a different approach.

dcxdxdgxdxddd

xd

d

xddUU

xdgxd

04/13/23 110

SOLO



Principle of Last Action for 4-Vector for a Free Particle (Covariant Treatment)

2

1

2

1

2

100

2220 /1:

dUUcmdccmtdcucm

t

tA

Action Integral is

where s = s (τ) is any monotonically increasing function of τ.

dcsd

sd

xd

sd

xdgxdxdgdUU

2

10:

s

ssd

sd

xd

sd

xdgcm A

sd

xd

sd

xdgcmL

0:

gg

1000

0100

0010

0001

sd

xd

sdxd

sdxd

cm

sd

xd

sdxd

g

sd

xdg

sd

xdg

cm

sdxd

L

x

Ljj

2

2

2

,0 00

Euler-Lagrange Equation

sdd

c

sdxd

d

d

sd

dcm

sd

xd

sdxd

sdxd

sd

dcm

x

L

sdxd

L

sd

d

sd

d

000 0

2

2

0

d

xdm

4-Vector Free Particle Equation of Motion Return to Table of Content

04/13/23 111

SOLO

Classic Electrodynamics

Start with Microscopic Maxwell’s Equations in Gaussian Coordinates:

eE 4

Gauss’ Law (Electric)

jct

E

cB

41

Ampère’ Law (with Maxwell’s Extension)

01

t

B

cEFaraday’ Law of Induction

Gauss’ Law (Magnetic) 0 B

Lorentz Force Equation:

Electric Field Intensity [statV . cm-1]

E

Magnetic Induction [statV . sec . cm-2 = gauss]

B

ρe Charge density [statC . cm-3]

j

Current Density [statA . cm-3 ]

eF

Electromagnetic force [dynes]

u

Charge velocity [cm . sec]

1 V = 1/3x10-2 statV1 C = 3x109 statC1 A= 3x109 statA

B

c

uEqFe

Classical Electromagnetic Theory

q electric charge [statC]

04/13/23 112

SOLO

jct

E

cB

Etc e

41

41


Continuity Equation0

jt

e


04/13/23 113

SOLO

0111

t

A

cEA

tcE

t

B

cE

Therefore we can define a Potential φ, such that

We can define a vector such that0 B

0 AAB

A

01

t

A

cE

and φ are not uniquely defined sincewill give the same results for . Such a transformation that doesn’t change the results is called a “GaugeTransformation” on the Potentials.

BE

,A

t

f

candfgradAA

1

11

t

A

cE

AB

1

We obtained


Classic ElectrodynamicsElectrodynamic Potentials and φA

04/13/23 114

SOLOClassical Electromagnetic Theory

jct

E

cB

41

From we obtain

t

A

cE

AB

1

jc

At

A

ctcA

tct

A

cA

41111 22

2

22

2

2

From we obtaineE 4

eAtc

41 2

Since and φ are not uniquely defined, let add the following relation:

A

01

tc

A

We obtain

Lorenz Condition

etc

jc

At

A

c

41

41

22

2

2

22

2

2

Waveform Equations for and φ.A

Ludwig Valentin Lorenz

1829-1891


04/13/23 115

SOLO

01

tc

A Lorenz Condition

etc

jc

At

A

c

41

41

22

2

2

22

2

2

Waveform Equations for and φ.A

If we define the new Potentials through the relations

t

f

candfAA

1

11

To satisfy the Lorenz Condition

0111

2

2

22

0

11

t

f

cf

tcA

tcA

01

2

2

22

t

f

cfTherefore f must satisfy



116

SOLO

Energy and Momentum

EJB

c

uEup e

Buu

Jue

ee

0

or

BEc

t

BBE

t

E

Et

EBEEB

c

Et

E

cB

cEJp

t

B

cE

BEBEEB

Jct

E

cB

e

e

44

1

4

1

4

1

4

1

41

0

4

41

1

BGM

EGE

Jct

E

cBA

t

B

cEF

e

e


BEcBBEE

tEJp e

4224

1

The power density of the Lorentz Force the charge ρe

and velocity isu

117

SOLO

Energy and Momentum (continue -1)

We identify the following quantities

EJ e

BBt

pBBw mm

8

1,

8

1

EEt

pEEw ee

8

1,

8

1

BEc

pR

4

- Magnetic energy and power densities, respectively (Energy transferred from the field to the particles)

- Electric energy and power densities, respectively- (Energy transferred from the field to the particles)

- Radiation power density (Power lost through the boundaries

EEtt

EE

2

1 BBtt

BB

2

1

-Power density of the current density eJ


BEcBBEE

tEJp e

4224

1

Note: The minus sign means transfer of power from (loss) the electromagnetic field .

BEc

S

4

: - Poynting power flux vector

118

SOLO

Energy and Momentum (continue – 2)

dve

E

B

eJv

,

V

FdF

Fd

Let integrate this equation over a constant volume V

VV td

d

t


BEcBBEE

tEJp e

488

PowerRadiationEM

V

PowerMagnetic

V

PowerEletrical

V

Powet

V

e vdc

BEcvd

BB

tvd

EE

tvdEJ

488

2

04/13/23 119

SOLOElectromagnetic Stress

Lorentz Force Density

0

4

41

1

BGM

EGE

Jct

E

cBA

t

B

cEF

e

e

Electric Field Intensity [statV . cm-1] E

Magnetic Induction [statV . sec . cm-2 = gauss] B

ρe Charge density [statC . cm-3]eJ

Current Density [statA . cm-3 ]

Maxwell Electromagnetic Stress

Start with Maxwell Equations in Gaussian Coordinates

uJ

BJc

Ef

ee

eee

1

ef

Electromagnetic force density [dynes . cm-3]

u

Charge velocity [cm . sec]

Bt

E

cBBBB

t

BE

cEEEE

Bt

E

cBBEEfe

1

4

1

4

11

4

1

1

4

1

4

1

4

1

0

0

t

BEB

t

E

cBBBBEEEEfe

1

4

1

4

1

4

1

0

1 V = 1/3x10-2 statV1 C = 3x109 statC1 A= 3x109 statA

uJ

BJc

Ef

ee

eee

1

04/13/23 120



t

BEB

t

E

cBBBBEEEEfe

1

4

1

4

1

4

1

Use to get BABABA B

EEEEEE E

EEIEEEEEEEEEEEE EE

�

and

EEIEEEEEEEEEEEE

�

2

1

2

1

therefore

where the unit dyadic.

100

010

001

I�

BBIBBBBBBBBBBBB�

2

1

2

1

On the same

04/13/23 121



c

BE

tBBIBBEEIEEfe 42

1

4

1

2

1

4

1

��

Define

EEIEETe

��

2

1

4

1:

BBIBBTm

��

2

1

4

1:

BBEEIBBEETTT me

��

2

1

4

1:

2cmdynesTm

�. - Magnetic Stress Tensor

2cmdynesTe

�. - Electric Stress Tensor

2 cmdynesTTT me

��. - Electromagnetic Stress Tensor

c

BEG

4:

213 secsec cmergcmdynesG

- Momentum Density Vector

t

STfe

�

Return to Relativistic EM Stress

04/13/23 122



Assuming a charge density ρ in a closed volume V.The Electromagnetic Force on the volume V is

VA

ThGauss

VVV ee VdGt

sdTVdGt

VdTVdfF�� .

BBEEIBBEETTT me

��

2

1

4

1:

c

BEG

4:

GcBE

cS

2

4:

Electromagnetic Stress Tensor

Momentum Density Vector -Poynting Power Flux Vector

04/13/23 123

SOLO

Classical Equations of Motion of Charge e and Mass m0 in an Electromagnetic Field

Start with Lorentz Force Equation on a charge e :

B

cEeFe

u

1


Equations of Motion Ac

ee

t

A

c

eB

cEe

dm

uu

1

td

u0

t

A

cE

AB

1and

AAAAAA

uuuuuUsing

trAtd

dtrAtutrA

t,,,

and

AuAuc

eeAuAu

t

A

c

edm

A

td

u0we obtain

or

Au

c

ee

td

Ad

c

e

td

dm

u0

Rearrange

Au

c

eeA

c

em

td

d u0

Define Ac

emp

u: 0

Au

c

ee

td

pd

04/13/23 124

SOLO



Equations of Motion

B

cEe

td

dm

u

1u0

Let derive the equation for the energy by multiplying (inner product) by u

uuuu1

uuu

00

t

A

c

eeB

cEe

td

dm

Using

uu

2

1u

u00

mtd

d

td

dm

trtd

dtrtutr

t,,,

and

we obtain

td

d

t

tA

c

eA

tc

e

te

td

de

t

A

c

e

te

td

dem

td

d

u

0u

0

uuuuu

2

1

ortd

dA

c

eA

c

ee

tt

A

c

e

teem

td

d uuuuu

2

10

04/13/23 125

SOLO



Summarize

B

cEe

td

dm

u

1u0

or

Ac

emp

u: 0

A

c

ep

mtd

rd

0

1u1

Au

c

ee

td

pd

2

Return to Hamiltonian

td

dA

c

eA

c

ee

tt

A

c

e

teem

td

d uuuuu

2

10

3

04/13/23 126

SOLO

Classical Lagrangian Equations of Motion of Charge e and Mass m0 in an Electromagnetic Field


B

cEeFe

u

1

t

A

cE

AB

1and

td

Ad

cA

ceA

ctd

Ad

ce

AAct

A

ce

Act

A

ceFe

1u

1u

11

uu11

u11

therefore

since At

A

td

Ad

u

x

V

x

V

td

dV

V

td

dFe

u

we have

Define Ac

eVA

ceV

uu

1: Electromagnetic Potential


04/13/23 127

SOLO

The Lagrangian in an externalElectromagnetic Field:

The Euler Lagrange Equations are: Qq

L

tdqd

L

td

d

Classical Lagrangian Equations of Motion of Charge e and Mass m0 in an Electromagnetic Field (continue – 1)


Let add the constraint by introducing the Lagrange multiplier utd

rd p

energypotentialneticelectromag

energykineticenergy

potentialenergykinetic c

emVTL

Au

1uu

2

1: 0

constraintMultiplier

sLagrange


energykinetic

td

rdp

cemL

0

'

0 uAu1

uu2

1:

Let now interpret the variables as independent variablesurp

,,

u,0

td

rd

p

L

tdpd

L

1. - independent variablep

0u

td

rdConstraint

04/13/23 128

SOLO


The Euler Lagrange Equations are:Q

q

L

tdqd

L

td

d



Aue

r

Lp

tdrd

L

c

e-,

2. - independent variabler

Aue

td

pd

c

e-

3. - independent variableu

Aumpu

L

tdud

L

c

e,0 0

Aump

c

e0

constraintMultiplier

sLagrange


energykinetic

td

rdp

cemL

0

'

0 uAu1

uu2

1:

04/13/23 129

SOLO

The Lagrangian in an externalElectromagnetic Field (without the constraint):




energykinetic

cemL

Au

1uu

2

1: 0

Canonical Momentum pAuu 0

c

em

L

Hamiltonian:

Au1

uu2

1-uAuL-u

u: 00

cem

c

em

LH

A-p

1u

0

c

e

m

ec

ep

mem

LH

2

00 A

2

1uu

2

1L-u

u:

Ac

ep

mp

H

td

rd

0

11

Au

c

ee

r

H

td

pd

2

We have: eA

c

epA

c

ep

memH

00 2

1uu

2

1

04/13/23

SOLOElectromagnetic Field

We want to obtain the EM Field Equations:

01

tc

A Classical Lorenz Condition

etc

jc

At

A

c

41

41

22

2

2

22

2

2

Classical Waveform Equations for and φ.A

Development of the Electromagnetic Field Equations from a Least Action Integral

Use as Action Integral of the form dzdydxVdtdVdBEt V

22

4

1

EMA

This Action Integral will give the Homogeneous Waveform Equations, but Lorenz Condition must be forced. To obtain the Lorentz Condition and the Nonhomogeneous Waveform Equation, additional terms must be added

dzdydxVdtVddAuctc

ABEt V

11

8

12

22 EMA

or dzdydxVdtVddAuctc

AAt

A

ct V

111

8

12

22

EMA

Classic Least Action Integral

04/13/23


Development of Equation of the Electromagnetic Field Equations from a Classic Least Action Integral

The Lagrangian Density for the field will be

c

uA

c

uA

c

uA

tcz

A

y

A

x

A

y

A

x

A

x

A

z

A

z

A

y

A

t

A

czt

A

cyt

A

cx

Auctc

AAt

A

c

zz

yy

xxe

zyx

xyzxyz

zyx

e

2

222

222

22

2

1

111

8

1

111

8

1

L

Euler-Lagrange Equations for φ equation

01

4

1

1

4

11

4

111

4

1

2

2

2

2

2

22

2

2

2

2

ez

yxzyx

zt

A

cz

yt

A

cyxt

A

cxtczt

A

yt

A

xt

A

c

z

z

y

y

x

x

t

t

LLLLL

etc

41 2

2

2

2

We obtain

04/13/23


Development of Equation of the Electromagnetic Field Equations from a Classic Least Action Integral


c

uA

c

uA

c

uA

tcz

A

y

A

x

A

y

A

x

A

x

A

z

A

z

A

y

A

t

A

czt

A

cyt

A

cx

Auctc

AAt

A

c

zz

yy

xxe

zyx

xyzxyz

zyx

e

2

222

222

22

2

1

111

8

1

111

8

1

L

Euler-Lagrange Equations for Ax computation

04

1

4

11

4

111

4

1

c

u

x

A

z

A

z

y

A

x

A

ytcz

A

y

A

x

A

xt

A

cxtc

A

z

Az

y

Ay

x

Ax

t

At

xe

zx

xyzyxx

xxxxx

LLLLL

ujjc

At

A

c e

41 2

2

2

2

In the same way we arrive to Ay and Az equations to obtain

xexxxx ujj

cA

t

A

c

41 2

2

2

2

04/13/23 133

SOLO

041

41

jt

jct

E

cB

Etc

e

e

Continuity Equation

jcJjcJ ee

,:&,:

Define the 4 Vectors Jα , Jα

We have

0,,

0,,

0

0

0

0

jt

jcx

J

jt

jcx

J

etcx

e

etcx

e

and the 4 Vectors ∂α , ∂α

,:&,:00 xx

Relativistic Electrodynamics


04/13/23 134

SOLO



Let define the 4 Vector Potential AAAA

,:&,:

01

,,&01

,,00

00

tcA

xA

tcA

xA

tcxtcx

0

AA Lorenz Condition

Jc

jc

Ax

AA

Jc

jc

Ax

AA

44,4,

44,4,

2

0

2

2

0

2

2

2

04/13/23 135

SOLO



2

1

1

2

1

3

3

1

3

2

2

3

0

0

0

x

A

x

A

x

A

x

A

x

A

x

A

y

A

x

A

x

A

z

A

z

A

y

A

A

A

A

xy

xz

yz

A

B

B

B

B

xy

zx

yz

z

y

x

z

y

x

with

t

A

cE

AB

1 3210 ,,,,,,,: AAAAAAAAA zyx

3

0

0

3

2

0

0

2

1

0

0

1

1

1

1

1

x

A

x

A

x

A

x

A

x

A

x

A

zt

A

c

yt

A

c

xt

A

c

t

A

cE

E

E

E

z

y

x

z

y

x

define

F

BE

E

BBE

BBE

BBE

EEE

AAF

T

xyz

xzy

yzx

zyx

0

0

0

0

0

:Field StrengthTensor

04/13/23 136

SOLO



FAA

BE

E

BBE

BBE

BBE

EEE

BEF

T

xyz

xzy

yzx

zyx

0

0

0

0

0

,

Let find

BE

E

BBE

BBE

BBE

EEE

BBE

BBE

BBE

EEE

FgF

T

xyz

xzy

yzx

zyx

xyz

xzy

yzx

zyx

0

0

0

0

0

0

0

0

0

1000

0100

0010

0001

BE

E

BBE

BBE

BBE

EEE

BBE

BBE

BBE

EEE

gFF

T

xyz

xzy

yzx

zyx

xyz

xzy

yzx

zyx

0

0

0

0

0

1000

0100

0010

0001

0

0

0

0

FAAAAg

BE

E

BE

E

IFgF

TT

00

0

01or

FAAgAA

BE

E

IBE

E

gFF

TT

0

0

010

137

SOLO



3,2,1,0,,0 FFF

0 AAAAAAFFF

3,2,1,0,,0 FFF

0 AAAAAAFFF

Properties of Fαβ and Fαβ

138

SOLO



3,2,1,0,,0 FFF

Properties of Fαβ and Fαβ

Let find the meaning of this result

01

,1,2,0

zxyz E

xyz

E

x

E

y

B

yx

y

E

x

E

t

B

ctc

A

xyytc

A

xx

A

y

A

tc

0,3,2,1

Bx

A

z

A

yy

A

x

A

zz

A

y

A

x

yzx B

zx

B

xy

B

yz

01

,2,3,0

xyzx E

yzx

E

y

E

z

B

zy

z

E

y

E

t

B

ctc

A

yzztc

A

yy

A

z

A

tc

01

,3,1,0

yzxy E

zxy

E

z

E

x

B

xz

x

E

z

E

t

B

ctc

A

zxxtc

A

zz

A

x

A

tc

3210 ,,,,,,,: AAAAAAAAA zyx

Use

t

B

cE

1

The equation on the top is equivalent to the two Maxwell Equations t

B

cEB

1&0

04/13/23 139

SOLO



Define the Totally Anti-Symmetric Fourth Rank Tensor

equalareindexestwoanyif

npermutatiooddanyfor

npermutatioevenanyandfor

e

0

1

,3,2,1,01

Define the Dual Field Strength Tensor

F

EB

B

EEB

EEB

EEB

BBB

Fe

T

xyz

xzy

yzx

zyx

0

0

0

0

0

2

1F

Examples

x

F

BFFeFeFe

32231

230132

1

32010101

32

2

1

2

1

F

y

F

BFFeFeFe

31131

130231

1

31020202

31

2

1

2

1

F

0

0

0

0

xyz

xzy

yzx

zyx

BBE

BBE

BBE

EEE

F

04/13/23 140

SOLO



Maxwell Equations in Covariant Form

Microscopic Maxwell’s Equations in Gaussian Coordinates:

4 E

Gauss’ Law (Electric)

jct

E

cB

41

Ampère’ Law (with Maxwell’s Extension)

Let compute

jc

cj

cB

t

E

cEB

x

EE

BE

E

xF

T

T

,44

,41

,,

0

,00

We obtain

J

cjc

cF

4,

4

jcJ

,: where

In the same way

Jc

jcc

F4

,4

jcJ

,: where

04/13/23 141

SOLO




Similarly the Microscopic Maxwell’s Equations in Gaussian Coordinates:

Let compute

Et

B

cBE

x

BB

EB

B

x

T

T

1,,

0

,00

F

We obtain 0 F

01

t

B

cEFaraday’ Law of Induction

Gauss’ Law (Magnetic) 0 B

3,2,1,0,,0 FFF

3,2,1,0,,0 FFF

We also have shownthat we obtain the twoMaxwell Equations from

0,0

04/13/23 142

x

y

z

x¶

y¶

z¶

SOLO

Relativistic ElectrodynamicsSpecial Relativity Theory

Transformation of Electromagnetic Fields

Assume a system of coordinates (O’x’y’z’) moving with velocity (without rotation – a Boost) relative to the system (Oxyz).We want to find the relation between .

u

BEandBE

,','

x

xF

x

xF

''

'

Let use the second order tensor Fαβ to write

T

TT

T

T

Tboostboost

IBE

E

I

LFLx

xF

x

xF

�

�

22

1

0

1

'''

2

22 1

1

1,

c

uwhere

04/13/23 143

T

T

zyzxzz

zyyxyy

zxyxxx

zyx

I

22

222

2

2

22

22

2

2

1

11

11

111

1

1111

222

22

2222

22

11

1

11

TTT

TTTT

T

T

T

T

II

I

II

��

�

�

�

2

22 1

1

1 2

2

I�

0

01

11

2

22222

T

01

111 2

232

32

22

TTTTTTTT I

II

III

T

TTTTTTT

�

�

��

0

222

1

2

422

222

1

1

1

2

11

2

112

2

SOLO


04/13/23 144

TBoost

T

T

zyzxzz

zyyxyy

zxyxxx

zyx

Boost L

I

L

22

222

2

2

22

22

2

2

1

11

11

111

1

1111

TT

TTT

T

T

T

TT

T

T

IBEBE

IEE

IIBE

E

I

�

�

�

�

�

2

2

2221

1

11

0

1

TTTTTTTT

TTTTTTTTT

IBIEIIEBEIE

IBEIEBEE

��

��

222222

22

2222

11111

11

SOLO



04/13/23 145

00

22 TT BB

�

TT

TBB

TT

EBEEBE

EBEBEIEE

11

11'

21

22

22

2

22

TT

E

TTTT

TTTTTTTT

TTTTTT

BBBEEB

BBBEEEE

IBIEIIE

��

111

1111

1111

0

2

2

2222

2222

�

BBI T

2

1

TTTTTTTT

TTTTTTTTT

IBIEIIEBEIE

IBEIEBEE

F

��

��

222222

22

2222

11111

11

'

022EE

TTTT

EE

SOLO



04/13/23 146

TTTTTTTT

TTTTTTTTT

IBIEIIEBEIE

IBEIEBEE

F

��

��

222222

22

2222

11111

11

'

��

BBEBBBEE

IBIEIIE

TT

E

TT

TTTTTT

11

11112222

BBBBBB

BBBBBB

TTTBBT

TTT

TTTTTTT

T

T

11

Since F’αβ is anti symmetric, i.e. F’αβ = - (F’αβ)T, we have

TT

E

TT

TTTTTT

BBBEE

IBIEIIE

��

11

11112222

SOLO




04/13/23 147

BEBEBE

EBE

BE

E

F

TT

T

T

T

11

10

''

'0

'

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Finally we obtain

Therefore

BEBB

EBEE

1'

1'

x

y

z

x¶

y¶

z¶

For we obtain 0,0,/ cuT

yyy

zyy

yx

yzz

zyy

xx

EBB

EBB

BB

BEE

BEE

EE

'

'

'

'

'

'

y

z

x¶

y¶

z¶

x

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Let write the Lorentz Force Equation in Covariant Form:

Lorentz Force Equation:

B

cE

td

pdf ee

u

1

F

c

uF

d

pd

d

Ed

cp

c

E

d

d

d

pdF u

,,1

,:

Force 4 Vector was defined as:

Therefore the 4 Vector Lorentz Force Equation is:

BUUEUEc

Bc

EEcd

pdf e

uuuee

0,u

1,u

1

dt

xduUUuc

d

dt

dt

xd

d

dt

dt

dx

d

xd

d

dxU uu

,,,,: 0

00

20

0

/1/1:,,,: cuUUucd

xd

d

dxU uuu


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Therefore the 4 Vector Lorentz Force Equation is:

BUUEUEcd

pdf e

e

0,

One other way to obtain the 4 Vector Lorentz Force Equation is:

e

Te

T

eee fBUUEUEc

U

U

BE

E

cd

xdAA

cUF

c

0

0

,

0

The covariant form is

eTe

T

eee fBUUEUEc

U

U

BE

E

cd

xdAA

cUF

c

0

0

,

0

BUUEUEcd

pdf e

e

0,

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dt

xduUUuc

d

dt

dt

xd

d

dt

dt

dx

d

xd

d

dxU uu

,,,,: 0

00

20

0

/1/1:,,,: cuUUucd

xd

d

dxU uuu


The Lorentz 4 Vector Force Equation is:

JFc

UFc

f

JFc

UFc

f

uee

uee

The Electromagnetic Energy-Momentum Tensor

UUUucJ

UUUucJ

u

e

u

eee

u

e

u

eee

,,

,,

0

0

The 4 Vectors Jα , Jα for a charge e and velocity are:u

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FFFF

FFFFFFJFc

f

uuFF

uuu

Fc

J

ue

44

444

4

FFFFFJFc

uu

2

1

4

FFFFFFFFJFc

uuu

2

1

42

1

4

The Electromagnetic Energy-Momentum Tensor (continue – 1)

FFFFFFFF

develop

FFFFFFFF

0use

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FFFF Since FαβFαβ = Fαβ Fαβ is a scalar we have

FFFFFFFFFFFFFFFF

FFFFFFFFFFFF

2

1

2

1

FFFFFFFFJFf uuue 4

1

42

1

44

TFFFFf ue

4

1

4

FFFFT u

4

1

4:

where

Electromagnetic Stress Tensor

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222

222

222

222

00

yxzzyzyzxzxxyyx

zyzyzxyyxyxzxxz

zxzxyxyxzyxyzzy

yxxyxzzxzyyzzyx

T

TTTT

BBEBBEEBBEEEBEB

BBEEBBEBBEEEBEB

BBEEBBEEBBEEBEB

EBEBEBEBEBEBEEE

BBEEEB

BEEE

BE

E

BE

E

FF

222

222

222

222

yxzzyzyzxzxz

zyzyzxyyxyxy

zxzxyxyxzyxx

zyxzyx

BBEBBEEBBEEBE

BBEEBBEBBEEBE

BBEEBBEEBBEBE

BEBEBEEEE

IBBEEIBBBIEEEI

BBEBBEEBBEEEBEB

BBEEBBEBBEEEBEB

BBEEBBEEBBEEBEB

EBEBEBEBEBEBEEE

trFF zyxzyx

yxzzyzyzxzxxyyx

zyzyzxyyxyxzxxz

zxzxyxyxzyxyzzy

yxxyxzzxzyyzzyx

��

222 222222

222

222

222

222

The Electromagnetic Energy-Momentum Tensor(continue – 3)

xyyx

zxxz

yzzy

z

y

x

xy

xz

yz

BEBE

BEBE

BEBE

B

B

B

EE

EE

EE

BE

0

0

0

0

0

0

0

,

0

0

0

0

xyz

xzy

yzx

zyx

xyz

xzy

yzx

zyx

BBE

BBE

BBE

EEE

F

BBE

BBE

BBE

EEE

F using and

04/13/23 154

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22

222

222

222

222

1000

0100

0010

0001

2

1

4

1BE

BBEBBEEBBEEBE

BBEEBBEBBEEBE

BBEEBBEEBBEBE

BEBEBEEEE

FFFF

yxzzyzyzxzxz

zyzyzxyyxyxy

zxzxyxyxzyxx

zyxzyx

2/

2/

2/

2/

44

1

4:

2222

2222

2222

22

BEBEBBEEBBEEBE

BBEEBEBEBBEEBE

BBEEBBEEBEBEBE

BEBEBEBE

FFFFT

zzzyzyzxzxz

zyzyyyyxyxy

zxzxyxyxxxx

zyx

uu

The Electromagnetic Energy-Momentum Tensor(continue – 4)

BBEEIBBEEc

BE

tf

BEcBE

tcf

ue

ue

�

2

1

4

1

4

48

22

0

T

tcTfff eee

,1

,0

The Lorentz 4 Vector Force Equation :

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t

GTS

c

BE

tcfff ueee

�

,1

8,

22

0

t

GTf

SBE

tcf

ue

ue

�

8

22

0

We can see that the 4 vector density force fe α satisfies the rules of relation between relativistic 4 vector to non relativistic counterpart.

Fc

uF

d

pdF u

constm

,:

.0

4-Force Vector

BBEEIBBEET

GcBEc

SBEc

G

��

2

1

4

1:

4:,

4

1: 2

We get the result of Classical EM Stress

BBEEIBBEEc

BE

tf

BEcBE

tcf

ue

ue

�

2

1

4

1

4

48

22

0

04/13/23 156

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Relativistic Equations of Motion of Charge e and Rest Mass m0 in an Electromagnetic Field


B

cEeFe

u

1

t

A

cE

AB

1and

td

Ad

cA

ceA

ctd

Ad

ce

AAct

A

ce

Act

A

ceFe

1u

1u

11

uu11

u11

therefore

since At

A

td

Ad

u


x

V

x

V

td

dV

V

td

dFe

u

we have

Define Ac

eVA

ceV

uu

1: Electromagnetic Potential

04/13/23 157

SOLO


potentialneticelectromag

kineticc

ecucmL

Au

1/1: 222

0


The Euler Lagrange Equations are:td

qdQ

q

LL

td

d

u,u

A/1

1/1

/

/

/

u 22

2022222

0

c

e

cu

cmAzAyAx

ceczyxcm

z

y

xL

zyx

uAeuA

/

/

/

/

/

/

,Au1

,/1

/

/

/

q222

0

c

e

z

y

x

c

e

z

y

x

eqtc

qtecucm

z

y

xL

td

Ad

c

e

cu

cm

td

dL

td

d

22

20

/1u

Qtd

Ad

cA

ce

cu

cm

td

dA

ctd

Ad

ce

cu

cm

td

dLL

td

d

ee FF

1

u1

/1u

11

/1qu 22

20

22

20

– External Forces (not electromagnetic) acting on the particleQ

eFQtd

pd

cu

cm

td

d

22

20

/1

Relativistic Equations of Motion of Charge e and Rest Mass m0 in an Electromagnetic Field (continue – 1)

04/13/23 158

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The Lagrangian in an external Electromagnetic Field:

Au

1/1: 222

0

c

ecucmL


The Canonical Momentum is: ApA/1u 22

20

c

e

c

e

cu

cmLP

The Hamiltonian is defined as: LPH u:

Let eliminate from the Hamiltonian such that it will a function only of Canonical Momentum and position . Start from

u

P

q

22

0

/u-1

up&A-Pp

c

m

c

e

2

22

220

/u-1

u

A

c

eP

c

m 2

22

022

22022

022

22022

0

2

/u-1/u-1/u-1

u

c

cm

c

cmcm

c

mcmA

c

eP

A

/u-1p

/u-1u

0

22

0

22 c

e-P

m

c

m

c

220

2

Au

cmAce

P

-ePc


04/13/23 159

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Au

1/1: 222

0

c

ecucmL


22

0

/u-1

up&A-Pp

c

m

c

e

ec

cmecucm

c

m

cecucmA

c

epLPH

22

20222

022

0

2220

/u-1/1u

/u-1

u

Au1

/1uu:

2

22

022

22022

022

22022

0

2

/u-1/u-1/u-1

u

c

cm

c

cmcm

c

mcmA

c

eP

The Hamiltonian is defined as:

Therefore the Hamiltonian is :

ecmAc

ePcH

22

0

2

Relativistic Equations of Motion of Charge e and Rest Mass m0 in an Electromagnetic Field (continue 3)

04/13/23 160


The Hamiltonian is : ecmAc

ePcH

22

0

2


We can write the Hamilton-Jacobi Equation in S (t,x,y,z) by defining :

PS

Ht

S

:

:

01 22

0

2

2

2

cme

t

S

cA

c

eS

Hamilton-Jacobi Equation

We solve the Hamilton-Jacobi Partial Differential Equation for S (t,x,y,z), and then we compute H and . P

04/13/23 161

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The Lagrangian is:


Relativistic Equations of Motion of Charge e and Rest Mass m0 in an Electromagnetic Field (Covariant Treatment)

td

xduUUuc

d

dt

dt

xd

d

dt

dt

dx

d

xd

d

dxU uu

,,,,: 0

00

2

20

0

/1/1:,,,:

cUUUU

cuUUucd

xd

d

dxU uuu


The 4 Vector Potential is defined as: AAAA

,:&,:

22

20

222220

/1

Au/1Au

1/1:

cu

c

c

ecmcu

cecucmL

or:

2

10

s

ssd

sd

xdA

c

e

sd

xd

sd

xdgcm A


The Action Integral is:

2

1

22

2

10

/122

22

20 /1

/1

Au

dUA

c

eUUcmtdcu

cu

c

c

ecm

tdcudt

t

A

04/13/23 162


Relativistic Equations of Motion of Charge e and Rest Mass m0 in an Electromagnetic Field (Covariant Treatment) (continue – 1)


2

10

s

ssd

sd

xdA

c

e

sd

xd

sd

xdgcm A


The Lagrangian corresponding to the Action Integral is:

sd

xdA

c

e

sd

xd

sd

xdgcmL 0:

Ac

e

sd

xd

sdxd

sdxd

cmAc

e

sd

xd

sdxd

g

sd

xdg

sd

xdg

cm

sd

xd

L

sd

xd

x

A

c

e

x

Ljj

2

2

2

, 00


00 00

x

A

d

xd

sd

d

c

e

d

Ad

sd

d

c

e

sdd

c

sdxd

d

d

sd

dcm

x

A

sd

xd

c

e

sd

Ad

c

e

sd

xd

sdxd

sdxd

sd

dcm

x

L

sdxd

L

sd

d

sd

d

UF

c

e

d

xd

x

A

x

A

c

e

x

A

d

xd

c

e

d

Ad

c

e

d

xdm

2

2

0

4-Vector Particle in External EM Field Equation of Motion

(Lorentz Force)

04/13/23



The Lagrangian corresponding to the Action Integral is:

AUc

ecmA

sd

xd

c

e

sd

xd

sd

xdgcmL 2

00:

Ac

eUmA

c

e

d

xd

dxd

dxd

cmAc

e

d

xd

d

xdg

d

xdg

dxd

gcm

dxd

LP

c

U

jj

000

2

2

2

:

Conjugate Momentum 4 Vector

20

20:ˆ cmA

c

ePUAU

c

ecmUPLUPH

Hamiltonian corresponding to the Action Integral

Define the Hamiltonian

Using Conjugate Momentum 4 Vector we get

Ac

eP

mUA

c

eP

mU

00

1&

1

200

20

0

1:ˆ cmUUmcmA

c

ePA

c

eP

mLUPH

Note: we can see that since Uαβ Uαβ = c2 we have Ĥ = 0.

04/13/23



Hamiltonian corresponding to the Action Integral

The Hamiltonian

200:ˆ cmUUmLUPH

Ac

eP

mUA

c

eP

mU

00

1&

1

Ac

eUmPA

c

eUmP 00 &Calculate

d

Pd

x

H

ˆ

x

AU

c

eA

c

eP

mxUm

x

UUm

0

00

1

x

A

d

xd

c

e

d

xd

d

dm

x

AU

c

e

UU

0

x

A

x

AU

c

e

d

xdm

2

2

0


(Lorentz Force)

Axd

xd

c

e

d

Udm

d

Pd

x

UUm

x

H

d

d

00

ˆ

04/13/23




2

10

s

ssdA

sd

xd

c

e

sd

xd

sd

xdgcm A


Let rewrite the Action Integral as:

b

axdA

c

exdxdcm

0A

Development of Equation of Motion from The Principle of Least Action

The Principle of Least Action states

000

b

a

b

axdA

c

exdA

c

exdxdcmxdA

c

exdxdcm

A

xdsd

xd

xdxd

xdxd

xdxd

xdxdxdxdxdxd

2

1

b

axdx

x

A

c

exdA

c

e

sd

xdcm 00

A

x

x

AA

04/13/23



Development of Equation of Motion from The Principle of Least Action (continue -1)

According to the Principle of Least Action

b

axdx

x

A

c

exdA

c

e

sd

xdcm 00

A

Integration by parts of the first part

b

a

b

a

xAc

e

sd

xdcmxdx

x

A

c

exdA

c

e

sd

xddcm 0

0

00

A

Since the boundaries a, b are fixed δ(a) = δ(b) = 0, the second term is zero.

b

asd

sd

xdx

x

A

c

esdx

sd

dA

c

e

sd

xd

sd

dcm 00

A

sd

xd

x

A

sd

Ad

sdsd

xdx

x

A

c

esd

sd

xdx

x

A

c

e

Write and

b

asdx

sd

xd

x

A

x

A

c

e

sd

xd

sd

dcm 00

A

04/13/23



Development of Equation of Motion from The Principle of Least Action (continue – 2)

According to the Principle of Least Action

b

asdx

sd

xd

x

A

x

A

c

e

sd

xd

sd

dcm 00

A

2

222

22

2

222

2

22 1

c

sdxdxddtc

ctd

xd

td

xdc

c

dtuc

c

dtd

Let change the integration variable ds to dτ, where

b

adcx

dc

xd

x

A

x

A

c

e

dc

xd

dc

dcm 00

A

Since the variation δxα is arbitrary the integral is zero only if

00

d

xd

x

A

x

A

c

e

d

xd

d

dm

xdxdtdcxdxdxdxdxdxdxdxdxdsd

220002 2

,,

dcsd

UF

c

e

d

xd

x

A

x

A

c

e

d

xdm

2

2

0

We recovered

d

xdU

x

A

x

AF

,:


(Lorentz Force)

04/13/23


We want to obtain the EM Field Equations:

01

tc

A Classical Lorenz Condition

etc

jc

At

A

c

41

41

22

2

2

22

2

2

Classical Waveform Equations for and φ.A

J

cjc

cF

4,

4

jcJ

,: where

Jc

jcc

F4

,4

jcJ

,: where

Relativistic Waveform Equations:

0

AA Relativistic Lorenz Condition

AAF

AAF

Development of the Electromagnetic Field Equations from a Least Action Integral

04/13/23


Development of Equation of the Electromagnetic Field Equations from a Relativistic Least Action Integral

Max Born in 1909 and Hermann Weyl in 1918 proposed the use as Action Integral of the form

dzdydxdtcddFFdBE

8

1

16

1 22EMA

This Action Integral will give the Homogeneous Waveform Equations, but Lorenz Condition must be forced. To obtain the Lorentz Condition and the Nonhomogeneous Waveform Equation, additional terms must be added

dzdydxdtcddJAc

AFF

1

8

1 2

1EMA

It appears in the definition of the Electromagnetic Energy-Momentum Tensor Tαλ


FFFFT u

4

1

4:

JAc

AAAAA

FF

1

8

1

8

1:

2 1EML

The expression FαβFαβ is an Invariant equal to 2 (E2-B2).

04/13/23



The Lagrangian Density for the field is

01

4

1

J

cF

A

L

A

L

J

cF

4

JAc

AAAAA

FF

1

8

1

8

1:

2 1EML

0. constA

08

1

A

AA

LL

Euler-Lagrange Equations

FFFA

FF

4

1

8

1

8

1

1EML

JcA

1

L

1

AA

8

1L0

A

L2

04/13/23



In the same way to obtain the conjugate equations we use

dzdydxdtcddJAc

AFF

1

8

1

8

1 2

2EMA

The Lagrangian Density for the field is

JAc

AFF1

8

1

8

1 2

2EML

J

cF

4

01

4

1

J

cF

AA22 EMEM LL

0. constA 08

1

A

AA22 EMEM LL

AA

8

12EML0

A2EML

2

FFFA

FFEM

4

1

8

1

8

12

L

JcA

1

2EML

1


04/13/23


Particles in an Electromagnetic Field Equations from a Relativistic Least Action Integral

It was shown that when a Charged Particle in a Electromagnetic Field the Action Integral must consist of three partsAmech – depends on properties of the particle (mass, velocity)Ainter – depends on the interaction between particle and the fieldAEM – depends on the properties of the EM field in the absence of the charge

2

10

s

ssd

sd

xd

sd

xdgcm

mechA

2

1

s

ssd

sd

xdA

c

e interA

dJAc

AFF

sdsd

xdA

c

esd

sd

xd

sd

xdgcm

s

s

s

s

1

8

1

8

1 2

0

2

1

2

1EMmfm AAAA

therefore

Defines the motion of the Free Particle

Defines the motion of the Particle due toLorentz Force

Defines the Electromagnetic Field

dJAc

AFF

1

8

1

8

1 2

1EMA


04/13/23 173

SOLOClassical Field Theories



Consider a Discrete Infinite One-DimensionalSystem composed of equal masses m, at a distance a, at equilibrium, from each other and connected by massless spring with spring constant k. Define by ψi the position of the i mass.

The Kinetic Energy of the System is

i

i

td

dmT

2

2

1

The Potential Energy of the System is i

iikV 212

1

The Lagrangian for the System is

iii

i ktd

dmVTL 2

1

2

2

1

or

ii

i

iii aLaa

aktd

d

a

mL

2

1

2

2

1

The Equation of Motion

022

1

aak

aak

a

mLL

td

d iiiiii

ii

In Equilibrium


04/13/23 174

In Equilibrium




Transition from a Discrete to Continuous Systems (continue -1)

To go from Discrete to Continuous we use:

xxi

iii xdxdx

Yt

Laa

aktd

d

a

mL L

222

1

2

2

1

2

1

0lim02

2

2

2

02

2

221

2

2

xY

txd

xxY

taak

aak

ta

m xdxx

xd

iiiiii

xda

txt

Yak

am

xda

i

i ,

/

The Lagrangian for the Continuous System is

where

The Equation of Motion

22

2

1

x

Yt

:L Lagrangian Density

04/13/23 175

In Equilibrium




Transition from a Discrete to Continuous Systems (continue – 2)

The Canonical Momentum is:

i

iii

i

ii a

aak

td

d

a

m

td

dpLH

2

1

2

2

1

td

dm

tdd

Lp i

ii

/

The Hamiltonian is

and

For Continuous System

t

xt

ta

pi

a

,

/:lim

0

L

xdxdt

xdx

Yt

Hi

HL

22

2

1

t

/

:

L Canonical Momentum Density

i t

L:H Hamiltonian Density

04/13/23 176


Introduction to Lagrangian and Hamiltonian Formulation for Continuous Field Systems


For Continuous Field Systems, the General Form

t x

n

x n

mm

nmn

t

tdxdxdttxxxx

xxttdLIn1

111

1

1

1

111 ,,,,,,,,,,,,,,

L

Define txxxxx ni :,,,: 010

njxdxdxdVdtdddx

,,xIVd

n

tdR j

kkj ,,1,010

L

Use the following shorthand notations

L - Lagrangian Density

1

11

1

1

1

111 ,,,,,,,,,,,,,,,:

x

n

x

m

n

mm

nmn xdxd

ttxxxxxxtL

n

L

L - Lagrangian

njmkx

mkxxx

j

k

nkk

,,1,0,,,2,1:

,,2,1,,,,: 10

or

nitdVdtx

,,xtIt V

k

i

kki ,,1,,

L


04/13/23 177



The Integral is a Function of the Trajectory C between the Initial (P1) and Final (P2) points. We want o find an Extreme Value (Extremal) of I (C).

nkmjtx

xtxt j

k

jkjk ,,2,1,,,2,1,,,,,

Assume that we found such a trajectory defined by

A small variation to this trajectory is given bywhere ε is a small parameter and η (t) are class C1 functions for t1 ≤ t ≤ t2, and suchthat η (t1)= η (t2)=0 .

kjkjkjkjk xtxttxtxtxt ,,,,,,,

2

1

,,,,,,t

tV

jj

k

j

k

jkjkjk tdVd

ttxxxtxtxtI

L

IIId

Id

d

IdII 2

0

2

22

0

02

10

where0

:

d

IdI - First Variation

0

2

222

2

1:

d

IdI - Second Variation

Extremal of the Functional .

2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

178

SOLO Classical Field TheoriesIntroduction to Lagrangian and Hamiltonian Formulation for

Continuous Systems and Fields

0//

02

11 1

0

t

tV

m

j

j

j

n

kk

j

kjj

j

tdVdttxxd

IdI

LLL

IIId

Id

d

IdII 2

0

2

22

0

02

10

Now suppose that an extreme value (extremal) of I (C) exists for ε = 0.This implies that δ I =0 is a necessary condition.

2

1

,,,,,,t

tV

jj

k

j

k

jkjkjk tdVd

ttxxxtxtxtI

L

Using the Divergence Theorem we can transform the Volume Integral to the BoundarySurface Integral


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

0//

0

11

Sj

S

n

k

kk

kjj

V

n

kkj

jk

Sdnx

Vdxx

LL

nk are the Direction Cosines of the outdrawn normal to the Boundary Surface S.

Integrate by parts

V

n

kkjk

j

V

n

kkj

jk

V

n

k jkkj

Vdxx

Vdxx

Vdxx

11

1

//

/

LL

L

04/13/23 179



0//

02

11 1

0

t

tV

m

j

j

j

n

kk

j

kjj

j

tdVdttxxd

IdI

LLL


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

In the same way

V

n

kkjk

j

V

n

k jkkj

Vdxx

Vdxx 11 //

LL

2

1

1

2

2

1

2

1

2

1 ////

0

0

t

tj

j

t

t

t

tj

j

t

tjj

t

t jj

tdtt

tdttt

tdtt

j

j

LLLL

0//

2

11 1

t

tV

m

jj

n

kkjkj

j tdVdttxx

I

LLL

If δ I = 0 for arbitrary δψj then

0//1

ttxx j

n

kkjkj

LLLEuler-Lagrange Equation

180



V

m

j

j

jj

n

kkjkj

j

k

jkjk Vd

ttxxtxxtxtL

1 1 //,,,,,

LLL


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

Using 2

1

t

ttdLCI

V

j

k

jkjk Vd

txxtxtL

,,,,,: L

Computing the Functional Derivative of L with respect to ψj, ∂ψj/∂t we obtain

n

kkjkjj xx

L1 /

LL

n

kkjkjj xtxtt

L1

0

2 ///

LL

0/2

kj xtL

since does no depend on

txxtt j

k

jkj

,,,,L kj xt /2

Therefore

V

m

j

j

jj

j

j

k

jkjk Vd

tt

LL

txxtxtL

1 /,,,,,

Functional Derivative Definition:

n

kkjkjj xx1 /

:

181




2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

the condition δ L = 0 becomes

V

m

j

j

jj

j

j

k

jkjk Vd

tt

LL

txxtxtL

1 /,,,,,

2

1

1

2

2

1

2

1

2

1 ////

0

0

t

tj

j

t

t

t

tj

j

t

tjj

t

t jj

tdt

L

ttd

t

L

tt

Ltd

tt

L j

j

Since

0/

t

L

t

L

jj


j = 1,2,…,m

Leonhard Euler

(1707-1783)

Joseph-Louis Lagrange

(1736-1813)

Functional Derivative Definition:

n

kkjkjj xx1 /

:


182




2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

Hamiltonian Formalism

In this case the free variables are t and xk (k=1,2,..n), we have

tx

xtxtt

j

k

jkjk

i

,,,,,

/

L:i

m

iij

k

jkjk ttx

xtxt1

,,,,,

iL:H

m

iV

i

i

L

V

j

k

jkjk

V

Vdtt

Vdtx

xtxtVdH1 /

,,,,,

LLH

V

m

ii

ii

i

V

m

j

j

jj

jV

m

ii

ii

i Vdtt

Vdtt

LLVd

ttLH

111 /

tt

L

t

L i

ii

/ iii tt

L

//

Lwe obtained and

therefore

V

m

i ii

ii Vd

ttH

1

183




2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

Hamiltonian Formalism (continue – 1)

We have

tx

xtxtt

j

k

jkjk

i

,,,,,

/

L:i

Assume that we have a one-to-one correspondence between πi (i=1,..,n) and linear combinations of ∂ψj/∂t (j=1,…,n) , or since

m

j

m

i

j

iji

m

i

n

kk

i

iki

n

k kiki t

dttx

dtx

xdtx

tdtt

d1 11 11 //////

LLLL:i

The one-to-one correspondence between πi and linear combinations of ∂ψj/∂t is possible only if

nkVxandtttfortttt k

njniji

njniiji

,,1,,0//

det//

det 21

,,1,,1

2

,,1,,1

LL

then

k

jkjkj

k

jkjk

j

k

jkjk x

xtxtx

xtxttx

xtxt

,,,,,,,,,,,,,, LL

We want to derive another interpretation of δH

184




2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L


We want to derive another interpretation of δ H

since

k

jkjkj

k

jkjk

j

k

jkjk x

xtxtx

xtxttx

xtxt

,,,,,,,,,,,,,, LL

jkjk

m

i

jkjki

k

jkjkj

k

jkjk xtxt

t

xtxt

xxtxt

xxtxt

,,,,

,,,,,,,,,,,,,

1HL i

V

n

i

n

kk

i

ikii

ii

ijkjk Vd

xtxxtxtH

1 1 //,,,,

HHHwe have

V

iikikV

iikikV k

i

iki

Vdtxx

Vdtxx

Vdxtx

//////

0

HHH

0////

0

S Son

iiki

TheoremDivergence

V

iikik

Sdtx

Vdtxx

HH

V

n

i ii

i

n

kikiki

jkjk Vdtxx

xtxtH1 1 //

,,,,

HHH

185




2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L


The previous interpretations for δ H was

V

n

i ii

i

n

kikiki

jkjk Vdtxx

xtxtH1 1 //

,,,,

HHH

V

m

i ii

ii Vd

ttH

1

Using Functional Derivative Definition:

n

kkjkjj xx1 /

:

n

kikikii txx1 //

HHH

i

n

kikikii txx

HHHH1

0

//

V

n

i ii

ii

jkjk VdxtxtH1

,,,,

HH

i

i

i

i

t

t

H

H

186




2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L


The condition for an extremum are

We used Functional Derivative Definition:

n

kkjkjj xx1 /

:

i

i

i

i

t

t

H

H

The result Closely Resemble Hamiltonian’s Equations in Classical Dynamics

We have

VV

n

ii

i

i

i

Vdt

Vdttttd

Hd HHHH1

0

If H is not a Function of Time and 0

t

H

.0 constHtd

Hd


04/13/23 187




Theorem of Noether for Multiple IntegralsAmalie Emmy

Noether (1882 –1935)

Consider the Functional

mjnknid

x

xxxI

R i

ijiji ,,1;,,1;,,1,0,,

L

Assume that we are given some variation of coordinates that changes also the domain of integration from R to R’, with boundaries S and S’, respectively.

nixxx iii ,...,1,0'

R i

ijiji

R i

ijiji d

x

xxxd

x

xxx

,,'

'

'','','

'

LL

We are looking for variations such that the Integral remains unchanged, i.e.


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

ni xxtxx ,,,: 10


04/13/23 188




Theorem of Noether for Multiple Integrals (continue -1)

In the First Integral x’i, represent dummy variables, therefore we can replace them by xi like in the Second Integral, but we still have two different regions of Integration R and R’ (see Figure). Let compute the Variation

RR i

ijiji

R i

ijiji

R i

ijiji d

x

xxxd

x

xxxd

x

xxx

'

,,,,,,

LLL


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

R i

ijiji

R i

ijiji d

x

xxxd

x

xxx

,,'

'

'','','

'

LL

Amalie Emmy Noether

(1882 –1935)

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&docid=kmTYCKBGsORbzM&tbnid=FAydf5YlVd9LiM:&ved=0CAUQjRw&url=http://www.findingpatterns.info/scrapbook/emmy-noether-einsteins-appreciation.html&ei=5AttUviRJcua0QWn_oDQBg&bvm=bv.55123115,d.d2k&psig=AFQjCNH3ZC6IkI_0qAAkuFBapdjP9TII3Q&ust=1382964496683652

189





2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

RR i

ijiji

R i

ijiji

R i

ijiji d

x

xxxd

x

xxxd

x

xxx

'

,,,,,,

LLL

Let find the meaning of δψi since we have a change in coordinates from t, xi to t’,x’i

i = 0,1,…,n

n

r rr

ijijijijijij x

x

xtxtxtxtxtxt

0

,,,,'','','

iii xxx '

ijijij

ijijij

xtxtxt

xtxtxt

,,,'

,,',''

n

r rr

kjijij x

x

xtxtxt

0

,,,

Amalie Emmy Noether

(1882 –1935)


190



Invariance Properties of the Fundamental IntegralExtremal of the Functional .

2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

RR i

ijiji

R i

ijiji

R i

ijiji d

x

xxxd

x

xxxd

x

xxx

'

,,,,,,

LLL

0

/,,

1 0

V

m

j

n

ii

j

ijj

jR i

ijiji d

xxd

x

xxx

LL

L


Integrate by parts

R

n

iiji

j

R

n

iij

ji

R

n

i jiij

dxx

dxx

dxx

00

0

//

/

LL

L

S

n

i

ii

ijj

R

n

iij

ji

Sdnx

dxx 00 //

LL

ni are the Direction Cosines of the outdrawn normal to the Boundary Surface Si.

Amalie Emmy Noether

(1882 –1935)


191




Amalie Emmy Noether

(1882 –1935)


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

RR i

ijiji

R i

ijiji

R i

ijiji d

x

xxxd

x

xxxd

x

xxx

'

,,,,,,

LLL

S

m

j

n

i

ii

ijj

S

m

j

n

i

ii

ijj

R

m

j

LagrangeEuler

n

iijij

j

R i

ijiji

Sdnx

Sdnx

dxx

dx

xxx

1 01 0

1

0

0

//

/,,

LL

LLL

S

n

i

iii

RR i

ijiji Sdnxd

x

xxx

0'

,,

LL

Therefore

S

n

i

ii

m

jij

ji

R i

ijiji Sdn

xxd

x

xxx 0

/,,

0 1

L

LL


192





2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

R i

ijiji

R i

ijiji

R i

ijiji d

x

xxxd

x

xxxd

x

xxx

,,'

'

',',',,

'

LLL

0// 0 10 1

R

n

i

m

jij

jii

TheoremDivergence

S

n

i

ii

m

jij

ji xx

xSdn

xx

L

LL

L

This is true if 0

/0 1

n

i

m

jij

jii x

xx

LL

n

r rr

kjkjkj x

x

xtxtxt

0

,,,

Substitute

0//0 11

n

i

m

jij

ji

n

rr

j

kji xx

xxx

LLL

Amalie Emmy Noether

(1882 –1935)


04/13/23 193





The Noether Theorem allows to construct quantities, which are constant along any extremal, that is, a Trajectory which satisfies theEuler-Lagrange Equations Ei (L) = 0. In the case of physical applicationsone obtain Conservation Law relations:-Conservation of Energy-Conservation of Linear Momentum-Conservation of Angular Momentum


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L


04/13/23 194



Let choose

2

1

2

1

2

1

2

1

:

00

t

t jjj

t

t

jj

t

t

jt

t jj

tdx

L

x

L

tdtdtdx

L

x

L

d

Id

0:2

1

2

112

t

t jjj

t

t j tdcttctd

0: 211

tttdct jjt

t jjj

Therefore 02

10

t

t jjj tdcd

Id

Add to this relation , to obtain , or 02

1

t

t jjj tdcc jj

t

t jj ctdc 01

2

Differentiate this equation, with respect to time:

0:

jjj x

L

x

L

td

dLE


In order that the curve C is an extreme value of Integral I (C), it is necessary that the functions xj(t) that define C are such that the Euler-Lagrange Equations Ej (L) = 0 are satisfied along C.

j

t

t jjj ctdx

L

x

L

2

1

: Du Bois Reymond Relation

Paul David Gustav Du Bois-Reymond

(1831-1889)

Leonhard Euler

(1707-1783)


(1736-1813)


1

,,,,t

tV

kjkj tdVdxtxttCI L

04/13/23 195



nkjxxxxxx kjjj ,,2,1,,, 21

nkjxxtxxx

xx

td

xdx kkjk

k

kjjj ,,2,1,,,

nlkjxxx

x

x

x

nkjx

x

x

x

llk

j

k

j

k

j

k

j

,,2,1,,

,,2,1,

2

We assume that L is invariant to a change of coordinatesa Scalar Transformation.

kkjkjjj xxxxxtLtxtxtL ,,,,,

We have

k

j

jk

j

jk

kkjkj

x

x

x

L

x

x

x

L

x

xxxxxtL

,,,

Let compute:

llk

j

jk

j

jk

j

jk

j

jkx

xx

x

x

L

x

x

x

L

td

d

x

x

td

d

x

L

x

x

x

L

td

d

x

L

td

d

2

llk

j

jk

j

jk

j

jk

j

jk

kkjkj

xxx

x

x

L

x

x

x

L

x

x

x

L

x

x

x

L

x

xxxxxtL

2,,,

k

j

jjkkj x

x

x

L

x

L

td

d

x

L

x

L

td

dLE

Ej(L) constitute the Components of a Covariant Vector (Euler-Lagrange Vector of L)


1

,,,,t

tV

kjkj tdVdxtxttCI L


196

A Tension Force T [N] is applied at theend points of a String, of density ρ [kg/m].Also applied is a normal load f (t,x) Δs [N],where Δs [m] is the arc length. If u (t,x) defines the string position

xx

ux

x

us

x

u

212

2

111

2

Vibrating String

The sum of forces in the u direction is12 sinsin TTsf

x

xu

xux

xu

sd

xd

x

txu

sd

txud

21

1

1,,sin

xxxx

x

u

x

xu

xu

xx

u

x

xu

sd

xd

x

xxu

sd

xxud

2

2

22

2

2

1

1sin

xx

uTxfTTsf

2

2

12 sinsin

SOLO Elasticity

197

The Inertial Force exerted by the same element of length Δs is given by

Vibrating String

2

2

2

2

t

ux

t

us

Equate the External Forces to the Inertial Force exerted on the element of length Δs we obtain

xx

uTxf

t

ux

2

2

2

2

Tacking Δ x → 0 and accordingly ξ → x we obtain

fx

uT

t

u

2

2

2

2

Vibrating String Equation

SOLO Elasticity

198

Vibrating String

xufxx

uT

xufxx

uTuxfxsTV

2

2

2

1

11

Let compute the Potential Energy of the String Element Δs

The Kinetic Energy of the String Element Δs2

2

1

t

uxK

The Extremal Function of a String of length L is

2

1

2

1

2

1

2

1 00

22

2

1

2

1t

t

Lt

t

L

LoadsExternaltodueEnergy

ForcesInternaltodueEnergyPotentialEnergyKinetic

t

t

t

t

tdxdtdxdufx

uT

t

utdWVKtdLCI L

The Euler-Lagrange Equation is

0

u

xu

tutd

d LLL

fx

uT

t

u

2

2

2

2

Vibrating String Equation

SOLO Elasticity


199

A Membrane is an Elastic Skin (h << L) which does not resist bending (zero shear)but does resist stretching. We assume that such a Membrane is stretched over a certain simple connected planar region R (x-y plane) bounded by a rectifiable curve C. We assume a constant tension τ on the boundary curve, normal to C in the Membrane plane.

The Potential Energy of the Membrane is

RR

ydxdSdV

Since

2222

2

111

y

u

x

u

y

u

x

u

ydxd

S

Letbe the parametric representation of γ, where s stands for the arc length on γ.

,,,: suusyysxx

2/1222 zdydxdsd

Let consider an arbitrary Membrane Surface Element Δ S, encompassed by a closed curve γ, and its projection on x-y plane is the Surface Element Δ A.

τ - the external tension on the Membrane Boundaryf – force per unit surface normal to Membrane

Vibrating Membrane

SOLO

The Energy due to xternal Loads on the Membrane is R

ydxdufW

Elasticity

200

We obtain, by neglecting all terms higher than second order,the Membrane Potential Energy

R

ydxdy

u

x

uV

22

The Total Kinetic Energy of the Membrane is

R

ydxdt

uT

2

2

RR

ydxdydxdufy

u

x

u

t

uWVTL L2

2

1222

The Lagrangian is

ufy

u

x

u

t

u2

2

1222

:L


The Kinetic Energy of the Surface ΔS is22

22

t

uydxd

t

uST

ρ [kg/m2] is the constant density of the Membrane

Vibrating Membrane

0

u

yu

xu

tutd

d LLLLf

y

u

x

u

t

u

2

2

2

2

2

2

Vibrating

MembraneEquation

SOLO

R

ydxdufW 2The Membrane Energy due to External Loads is

Elasticity


SOLO

(1) Longitudinal Tension/Contraction

x

uE

A

PE

x

u

A

Pxxxxxxxx

,,,

xA

zdydA :

The sum of forces in the x direction, on Beam Element Δx is

xx

uAEx

x

PPx

x

PP

2

2

The Inertial Force exerted by the same element of length Δx is given by

2

2

t

uxA

Equate the External Forces to the Inertial Force exerted on the element of length Δx we obtain

xx

uAE

t

uxA

2

2

2

2

Therefore

2

2

2

2

x

uE

t

u

Longitudinal Beam Vibration

Vibrating BeamElasticity

SOLO

(1) Longitudinal Tension/Contraction (continue – 1)

The Virtual Work of forces in the x direction, on Beam Element Δx is

2

2

1

t

uxAK

xx

uExExW xxxxxx

2

2

2

1

2

1

2

1

The Kinetic Energy of the Beam Element Δx is

The Total Energy of a Beam of length L is

LL

xdxdAx

uE

t

uWKE

00

22

2

1

2

1L


0

xu

tutd

d LL

Therefore

2

2

2

2

x

uE

t

u

Longitudinal Beam Vibration

ElasticityVibrating Beam

SOLO

Energy Equation for Longitudinal Tension/Contraction Beam

x

uE

A

PE

x

u

A

Pxxxxxxxx

,,,

AE

LPld

x

uAEldAd

x

uEldAdEldAdV

LL

A

L

A

xx

L

A

xxxx

EnergyPotential

2

0

2

0

2

0

2

0 2

1

2

1

2

1

2

1

2

1

xA

zdydA :

(1) Longitudinal Tension/Contraction (continue – 2)


04/13/23 204

Rayleigh Bending Beam Model

Shear Bending Beam Model

Euler-Bernoulli Bending Beam Model

Timoshenko Bending Beam Model

Rotating Timoshenko Beam

The assumptions made by all models are as follows.

1. One dimension (the axial direction) is considerably larger than the other two.2. The material is linear elastic (Hooke’s Law).3. The Poisson effect is neglected.4. The cross-sectional area is symmetric so that the neutral and centroidal axes coincide.5. The angle of rotation is small so that the small angle assumption can be used.

Bending Beam Models

SOLO ElasticityVibrating Beam

SOLO

Energy Equations for Pure Bending Beam

(2) Pure Bending

ldd

Q

M

M� P�

P

N

N�

Q�

A B

SR

dx

x

y

z

S� R�

A�B�

bM M

y

ρ

y

d

ddyxx

y

EE xxxx zxz IE

AdyE

AdyM

22

2

ld

dIE

ld

dIEM zzz

ld

d v

yI

M

z

zxx

L L

zz

L

z

L

z

L

z

zL

A z

zL

A

xxL

A

xxxx

EnergyPotential

ldld

dMld

ld

dMld

ld

dIEld

ld

dIE

ldIE

MldAdy

IE

MldAd

EldAdV

0 02

2

0

2

2

2

0

2

0

2

0

22

2

0

2

0

v

2

1

2

1v

2

1

2

1

2

1

2

1

2

1

2

1

xA

z zdydyI 2:

Euler-Bernoulli Bending Beam Model


04/13/23 206

Dynamic Lateral Beam Equation

Finite element method model of a vibration of a wide-flange beam (I-beam).

The dynamic lateral beam equation is the Euler-Lagrange equation for the following action

2

1

2

1 0

2

2

22

2

2

0

,vx

v

2

1v

2

1

x

v,

vv,,,

t

t

L

xqLoadsExternaltodueEnergy

ForcesInternaltodueEnergyPotential

z

EnergyKinetic

t

t

L

tdxdtxxqIEt

tdxdt

xt

L

Euler-Lagrange

0x

v

x

v

vvv

2

2

2

2

2

2

2

22

2

xqIEt

x

x

t

td

d

z

LLL

xqt

IE z

2

2

2

2

2

2 v

x

v

x Dynamic Beam Equation


04/13/23 207

Boundary v’’’ v’’ v’ v

Clamp Δv’=0 Δv=0

Simple support Δv’’=0 Δv’=0 Δv=0

Point force Δv’’’=λ Δv’’=0 Δv’=0 Δv=0

Point torque Δv’’’=0 Δv’’=0 Δv’=0 Δv=0

Free end v’’’=0 v’’=0

Clamp at end Δv’ fixed Δv fixed

Simply supported end

v’’=0 Δv fixed

Point force at end v’’’=±λ v’’=0

Point torque at end v’’’=0 v’’=±τ

Assuming that the product EI is a constant, and defining where F is the magnitude of a point force, and where M is the magnitude of a point torque, the boundary conditions appropriate for some common cases is given in the table below.



04/13/23 208

1st lateral bending1st vertical bending

2nd lateral bending2nd vertical bending

http://en.wikipedia.org/wiki/Bending



04/13/23 209

Rayleigh Beam Model

Shear Beam Model

Euler-Bernoulli Beam


04/13/23 210

Timoshenko Beam Model

Rotating Timoshenko Beam


SOLO

Energy Equations for a Pure Shearing Beam

(3) Pure Shearing

GE

EG

12

12Ak

V

ldAdldAdUd 2

1

2

1

ld

d

kAG

V

G

LL

LL

A

L

A

L

A

ldld

d

k

Vld

ld

dAGk

ldAGk

VldAd

AGk

VldAd

GldAdU

00

2

0

2

02

2

0

2

0

2

1

2

1

2

1

2

1

2

1

2

1


SOLO

Energy Equations for a Beam(4) Pure Torsion

ldd ld

dGG

xLdx

ld

d

ld

dJGAd

ld

dGAd

ld

dGAdFdTx

22

AdJ 2:

JTx

L

x

L

Lx

L

A

xL

A

L

A

ldld

dTld

ld

dJG

ldJG

TldAd

JG

TldAd

GldAdU

00

2

0

2

02

22

0

2

0

2

1

2

1

2

1

2

1

2

1

2

1


SOLO

(4) Pure Torsion

xLdx

The Kinetic Torsional Energy of the Beam of Length L

L

p ldt

JT0

2

2

1

The Total Energy of a Beam of length L is

L

p ldld

dJ

ld

dJGUTE

0

22

2

1

L r

p ldAdrJ0 0

2

1st torsional

2nd torsional


0

xt

td

d

LL

AdJ 2:

Therefore

2

2

2

2

xJ

JG

t p

Torsional Beam Vibration

Elasticity


Vibrating Beam

04/13/23 214

SOLO

DeformedMidsurface

OriginalMidsurface

ydxdyd

y

wxd

x

wwd xy

xy

x

w

y

wyx

,

DeformedMidsurface

OriginalMidsurface

DeformedMidsurface

OriginalMidsurface0

0

:,22

0

:,

:,

22

2

2

2

2

2

2

2

2

2

2

y

w

y

w

y

u

z

u

x

w

x

w

x

u

z

u

yx

wkkz

yx

wz

x

u

y

u

z

wz

z

u

y

wkkz

y

wz

y

u

x

wkkz

x

wz

x

u

zyyz

zxxz

xyxyyx

xy

zzz

yyyyy

yy

xxxxx

xx

wuzy

wzuz

x

wzu zxyyx

,, Small Displacements

Strain

Plate Theories

Kirchhoff Plate Theory (Classical Plate Theory)Elasticity

04/13/23 215

SOLO

xyxy

yyxxzzyyxxyyyy

yyxxzzyyxxxxxx

E

EEE

EEEzz

zz

12

11

11

0

0

yx

wzE

yx

wzGG

E

y

w

x

wzEE

y

w

x

wzEE

xyxyxy

yyxxyy

yyxxxx

22

2

2

2

2

22

2

2

2

2

22

12

12

11

11

0

:,22

:,

:,

22

2

2

2

2

2

2

2

2

zzyzxz

xyxyyx

xy

yyyyy

yy

xxxxx

xx

yx

wkkz

yx

wz

x

u

y

u

y

wkkz

y

wz

y

u

x

wkkz

x

wz

x

u

Strain

Stress-Strain

Plate Theories

Kirchhoff Plate Theory (Classical Plate Theory)

Elasticity

04/13/23 216

SOLO

Deformation Energy

The Virtual Work due to External Loads q [N/m2] and Discrete Forces Fi [N] is

Plate Theories


ydxdyx

w

y

w

x

w

y

w

y

w

x

w

x

whE

ydxdzdzyx

w

y

w

x

w

y

w

y

w

x

w

x

wE

ydxdzdy

wz

y

w

x

wz

yx

wz

x

wz

y

w

x

wz

E

zdydxdzdydxdU

S

h

hS

S

h

h

V

yyyyxyxyxxxx

V

T

22

2

2

2

2

2

2

2

2

2

2

2

2

2

3

2/

2/

2

22

2

2

2

2

2

2

2

2

2

2

2

2

2

2/

2/2

2

2

2

2

2222

2

2

2

2

2

2

2

121122

1

1212

1

1212

1

22

1~~2

1

ydxdyyxxtyxwFydxdtyxwqWi S

iii

S

,,,,

Kinetic Energy

S

ydxdt

tyxwhT

2,,

2

Total Energy

i S

iii

S

S

D

S

ydxdyyxxtyxwFydxdtyxwq

ydxdyx

w

y

w

x

w

y

w

y

w

x

w

x

whEydxd

t

tyxwhWUTL

,,,,

121122

1,,

2

22

2

2

2

2

2

2

2

2

2

2

2

2

2

32

Elasticity

SOLOPlate Theories


2

1

2

1

2

1

2

1

2

1

2

1

,,,,,,

122

1,,

2

22

2

2

2

2

2

2

2

2

2

2

2

22

t

t Si

t

t S

iii

t

t S

t

t S

t

t S

t

t

tdydxdtdydxdyyxxtyxwFtdydxdtyxwtyxq

tdydxdyx

w

y

w

x

w

y

w

y

w

x

w

x

wDtdydxd

t

tyxwhtdL

L

Euler-Lagrange:

02

2

2

22

2

2

22

2

w

yx

wyx

y

wy

x

wx

t

wtd

d LLLLL

iiii yyxxFq

yx

w

yxx

w

y

w

y

w

x

w

yy

w

x

w

y

w

x

w

xD

t

wh

22

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

142

1

02

2

2

2

2

2

2

2

2

2

iiii yyxxFq

y

w

x

w

yxD

t

wh Plate Vibration Equation

024

4

22

4

4

4

2

2

i

iii yyxxFqy

w

yx

w

x

wD

t

wh

iiii yyxxtyxwFtyxwtyxq

yx

w

y

w

x

w

y

w

y

w

x

w

x

wD

t

tyxwh

,,,,,,

122

1,,

2:

22

2

2

2

2

2

2

2

2

2

2

2

22

L

Elasticity


218

SOLO


Assume that the elastic deformations are small, and can berepresented in terms the normal un-damped modes of vibration.

1i

ii tRe

- are mode shape functions that depend on the position of the mass element of the system. Ri

- are generalized coordinates giving the magnitude of the modal displacements and are functions of time.

ti

Structural Dynamic Analysis (e.g. final element method) provides the mode shapefunctions component of each element of the system, as well as the vacuo modalfrequencies ( ) , for a selected number of modes.

Ri

i

iii

td

d 2

2

2

The mode shape functions are orthogonal.

ji

jiMMmd i

iji

m

ji 0

SECOND ELASTIC MODE

FIRST ELASTIC MODE

Elasticity

219

SOLO

Structural Model of the Solid BodySECOND ELASTIC MODE

FIRST ELASTIC MODE

1

2

1 1

1 1

2

1

2

1

2

1

2

1

i

ii

i j

jiiji

tm

j

i j

iji

tm

td

dM

td

d

td

dM

mdtd

d

td

dmd

td

ed

td

edT

Kinetic Energy

Elastic Deformation Potential eU

1i

iii tle

iii

td

d 2

2

2

i

m

ii Mmd

0ji

m

ji md

m iiii

jjj

m B

e mdmdtd

edeU

1

2

12

2

2

1

2

1

1

22

1

22

1 1

2

2

1

2

1

2

1

iiii

iii

m

iij i

jii

m

ij Mmdmd

From

We obtain

Elasticity

220

SOLO


SECOND ELASTIC MODE

FIRST ELASTIC MODE

1

22

2

1

22

1

2

11 2

1

2

1

2

1:,,,,,

iii

ii

iiii

i

iie

ii td

dMM

td

dMUT

td

d

td

dL

Lagrangian

Euler-Lagrange Equations:

,2,12

2

2

i

tM

L

t

L

td

dii

ii

ii

Vibration Mode Equation,2,102

2

2

it ii

i

We recover:

Elasticity


SOLO



1736-1813 Leonhard Euler

1707-1783 FIXED IN SPACE

(CONSTANT VOLUME)

EULER

LAGRANGE

MOVING WITH THE FLUID(CONSTANT MASS)

1e

3e

2e

u

The phenomena considered in Hydrodynamics are macroscopic and the atomic or molecular nature of the fluid is neglected. The fluid is regarded as a continuous medium. Any small volume element is always supposed to be so large that it still contains a large number of molecules.

There are two representations normally employed in the study of Hydrodynamics:

- Euler representation: The fluid passes through a Constant Volume Fixed in Space

- Lagrange representation: The fluid Mass is kept constant during its motion in Space.

Hydrodynamic Field

SOLO


Material Derivatives (M.D.)

Vector Notation Cartesian Tensor Notation

1e

2e

3e

r

u

b

rd Frddtt

FtrFd

,

d

dtF r t

F

t

dr

d tF

,

d

dtF r t

F

tb F

b

,

rdanyfor d F r t

F

tdt d r

F

xi ki

ki

k

,

d

d tF r t

F

t

d r

d t

F

xi ki k i

k

,

d

d tF r t

F

tb

F

xb

i ki

ki

k

,

vectoranybbtdrd

Fut

FF

tD

DtrF

td

d

u

,

k

ik

iki

u x

Fu

t

FF

tD

DtrF

td

d

,velocityfluiduu

td

rdIf

uuu

t

u

uut

uu

tD

D

2

2

k

ik

i

jj

ji

i

k

ik

ii

x

uu

x

uu

uxt

u

x

uu

t

uu

tD

D

2

2

1

Acceleration Of The Fluid

1e

2e

3e

r

u duu

dr

Material Derivatives = = Derivative Along A Fluid Path (Streamline) tD

D

Hydrodynamic Field

SOLO


Hydrodynamic Field

In Eulerian representation the fluid is defined by 5 quantities:• 3 components of velocity vector•2 thermodynamic quantities, most commonly used are - Pressure [N/m2] - Density [kg/m3]

txu ,

txp ,

tx,

All fluids must satisfy: vF (t)

m

SF (t)

O

x

y

z

r u,OConservation of Mass (C.M.)

Control Volume attached to the fluid (containing a constant mass m) bounded bythe Control Surface SF (t).

tvF

trVtru OfluidO ,, ,,

Flow Velocity relative to a predefined Coordinate System O (Inertial or

Not-Inertial) [m/sec]

Because vF(t) is attached to the fluid and there are no sources or sinks in this volume,the Conservation of Mass requires that: d m t

d t

( )0

)(

,,

1

)(

,

)( )(

)(0

tv

OO

GAUSS

tS

O

tv tv

REYNOLDS

FFF F

vdut

sduvdt

dvtd

d

td

tmd

Since this relation holds for any Control Surface SF (t) attached to the fluid, we must have

0,, OO ut

Conservation of Mass Equation (C.M.)

224

CONSERVATION OF ENERGY (C.E.) – THE FIRST LAW OF THERMODYNAMICS (DIFFERENTIAL FORM)

- Fluid mean velocity [m/sec] u r t,

- Body Forces Acceleration (gravitation, electromagnetic,..) [m/sec2]

G

- Surface Stress [N/m2]T

nnpnT ˆ~ˆˆ~

mV(t)

G

q

T n ~

d E

d t

Q

t

uu

d s n ds

- Internal Energy of Fluid molecules (vibration, rotation, translation) per mass [J/kg]

e

- Rate of Heat transferred to the Control Volume (chemical, external sources of heat) ) [W/m3]

Q

t

- Rate of Work change done on fluid by the surrounding (rotating shaft, others) (positive for a compressor, negative for a turbine) [W]td

Ed

SOLO

Consider a volume vF(t) attached to the fluid, bounded by the closed surface SF(t).

Hydrodynamic Field

- Rate of Conduction and Radiation of Heat from the Control Surface (per unit surface) ) [W/m3]

q

p - Pressure [N/m2]

- Stress Tensor [N/m2]~

225

(2.3) CONSERVATION OF ENERGY (C.E.) – THE FIRST LAW OF THERMODYNAMICS (CONTINUE)

mv(t)

Q

t

uq

u

S(t)

td

Wd

dsnsd ˆ

nT ˆ~

dm

G

- The Internal Energy of the molecules of the fluid plus the Kinetic Energy of the mass moving relative to an Inertial System (I)

The FIRST LAW OF THERMODYNAMICS

CHANGE OF INTERNAL ENERGY + KINETIC ENERGY =CHANGE DUE TO HEAT + WORK + ENERGY SUPPLIED BY SUROUNDING

SOLO

The energy of the constant mass m in the volume vF(t) attached to the fluid, bounded by the closed surface SF(t) is

This energy will change due to

- The Work done by the surrounding

- Absorption of Heat

- Other forms of energy supplied to the mass (electromagnetic, chemical,…)

Hydrodynamic Field

systementeringChangeHeat

tSv

systemontnmenenvirobydonetd

Wd

shaft

tSvv

REYNOLDS

KineticInternal

tv FFFFFF

sdqvdt

Q

td

Wd

ForcesSurface

sdTu

ForcesBody

vdGuvduetD

Dvdue

td

d

)()(

2)3(

)(

2

2

1

2

1

226

(2.3) CONSERVATION OF ENERGY (C.E.) – THE FIRST LAW OF THERMODYNAMICS (CONTINUE)

SOLO

THERMODYNAMIC PROCESSES

1. ADIABATIC PROCESSES

2. REVERSIBLE PROCESSES

3. ISENTROPIC PROCESSES

No Heat is added or taken away from the System

No dissipative phenomena (viscosity, thermal, conductivity, mass diffusion, friction, etc)

Both adiabatic and reversible

2nd LAW OF THERMODYNAMICS

Using GAUSS’ THEOREM

0)()(

tStv FF

AdT

qvds

td

d

00)(

)1(

)()(

tv

GAUSS

tStv FFF

vdT

q

tD

sDAd

T

qvd

tD

sD

- Change in Entropy per unit volumed s

- Local TemperatureT K- Rate of Conduction and Radiation of Heat from the System per unit surface [W/m2]q

For a Reversible Process

dp

dsTdvpdsTwqedv

2

/1

Hydrodynamic Field Return to Table of Content

SOLO


Part I - Lagrange’s Representation: The fluid Mass is kept constant during its motion in Space.

Hydrodynamic Field

Consider a small fluid mass dm that started at (x10, x20, x30) at t0 and reaching (x1, x2, x3) at t. We have

3020100302010321321 ,,,,,, xdxdxdtxxxxdxdxdtxxxdm

or

20

3

20

3

10

3

20

2

20

2

10

2

20

1

20

1

10

1

302010

321 det:

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

xdxdxd

xdxdxdJ

where is the Jacobian

So the Conservation of Mass in the Lagrange’s Representation is

00302010321 :,,,,,, txxxJtxxx

0302010302010

321321 ,,,,,, txxx

xdxdxd

xdxdxdtxxx

SOLO


Hydrodynamic Field

Since the fluid is ideal the Entropy s (x1, x2, x3,t) is also conserved, i.e., for the small fluid mass dm we have

00302010321 :,,,,,, stxxxstxxxs

Since the fluid is ideal (a reversible process) the relation between Entropy s and Internal Fluid Change in Energy e is given by Gibbs relation

dp

sdTsed2

,

where T – Temperature [°K] p – Pressure [N/m2]

Therefore

s

eT

ep

s

,2

Part I - Lagrange’s Representation

SOLO


Hydrodynamic Field

Consider an Ideal (Isentropic = Adiabatic and Reversible) Compressible Fluid

– Fluid Density [kg/m3] – Fluid Velocity [m/sec] tru ,

tr ,

– Internal Fluid Energy (per unit mass) [J/kg] – Potential Energy (per unit mass) due to external forces acting on the Fluid

trU

trse

,

,,,

Kinetic Energy K (to distinguish from Temperature T):

V V i

i

i

i xdxdxdt

xxdxdxd

t

xK 302010

3

1

2

0

0321

3

1

2

2

1

2

1

Potential Energy VP:

VV

P xdxdxdUexdxdxdUeV 3020100321

Define a Variation Integral that takes into consideration the Conservation of Mass and the Conservation of Entropy constraints by using Lagrange Multipliers trtr ,,,

1

0

1

0

1

0

1

0

32103210

t

t V

t

t V

t

t

P

t

t

tdxdxdxdsstdxdxdxdJtdVKtdL


SOLO


Hydrodynamic Field

We restrict ourselves to variations that vanish at the boundary of the region of integration (t =t0, t=t1, S the surface enclosing the volume V), the variations with respect to γ, ρ and .u

0,,0,,0

0,,0,,0 1

0

1

0

1

0

SrSrSri

t

t

t

t

t

ti

trstrx

trstrx

0

1

0

2

1

0

1

0

321

3

1

/

302010

3

1

3

10

0

0

t

t V ii

i

Tp

t

t V i ii

i

ii

t

t

tdxdxdxdsxx

JJs

s

ee

tdxdxdxdxx

U

t

x

t

xtdL

V

t

tii

t

t V

ii

t

t V ii

ii

i

t

t

xdxdxdxt

xtdxdxdxdx

x

JsT

tdxdxdxdxx

J

x

U

t

xJ

ptdL

321

00

321

302010

3

10

0

2

2

0

1

0

1

0

1

0

1

0

Rearranging

Using Hamilton’s Principle , we have01

0

t

t

tdL


SOLO


Hydrodynamic Field

0

1

0

1

0

1

0

321302010

3

10

0

2

2

0

t

t V

ii

t

t V ii

i

i

t

t

tdxdxdxdxx

JsTtdxdxdxdx

x

U

t

xJ

ptdL

Since the Variations δρ, δxi, δs are independent this relation is satisfied if

0 Jp

0 T

1

0 0

1

0

1

0

302010/

321321

t

t V

ii

t

t V

ii

t

t V

ii

tdxdxdxdJxx

ptdxdxdxdx

x

ptdxdxdxdx

x

J

011

0

00

1

0

1

0

302010

3

100

0

0

:

321302010

3

1 0

2

2

t

t V ii

ii

it

xu

t

t V

i

x

p

i

t

t V ii

i

i

tdxdxdxdxx

p

x

U

t

u

tdxdxdxdxx

Jtdxdxdxdx

x

U

t

x

ii

i

Therefore

pJp

J

p

&

0

T

This relation is satisfied if 3,2,11

0

ix

U

x

p

t

u

ii

i


SOLO


Hydrodynamic Field

We found 3,2,11

0

ix

U

x

p

t

u

ii

i

In Vector Notation FUpuut

u

1

Euler’s Equation


Using uut

u

t

u

x

uu

t

u

t

u Vector

Notationk k

ik

ii

0

3

10

3,2,113

1

ix

U

x

p

x

uu

t

u

iik k

ik

i

we obtain

We can see that the introduction of Entropy Conservation constraint does not influence the results.


SOLO


Consider an Ideal (Isentropic = Adiabatic and Reversible) Compressible Fluid

Kinetic Energy K (to distinguish from Temperature T): V

zdydxduuK

2

1

Potential Energy : V

zdydxdUe

Conservation of Fluid Mass constraint 0

ut

Rudolf Friedrich Alfred Clebsch

( 1833 – 1872)

0

u

ttD

D

tddzdydxut

ut

UeuutdLt

t V

t

t

1

0

1

02

1

– Fluid Density [kg/m3] – Fluid Velocity [m/sec] tru ,

tr ,

– Internal Fluid Energy (per unit mass) [J/kg] – Potential Energy (per unit mass) due to external forces acting on the Fluid

trU

tre

,

,,

To enable non-zero vorticity Clebsch (1859) introduced Lagrange Coordinate Equation constraint

0

u

tr ,

To include the constraints we introduce Lagrange Multipliers trtr ,,,

Hydrodynamic Field


SOLO


We restrict ourselves to variations that vanish at the boundary of the region of integration (t =t0, t=t1, S the surface enclosing the volume V), the variations with respect to γ, ρ and .u

tddzdydxut

uut

UeuutdLt

t V

t

t

1

0

1

02

1

0,,0,,0,

0,,0,,0, 1

0

1

0

1

0

SrSrSr

t

t

t

t

t

t

trutrtr

trutrtr


0

t

t

tdL

0

2

1

1

0

1

0

0

tddzdydxuut

ut

tddzdydxuuuut

uue

Ueuu

t

t V

t

t V

Hydrodynamic Field


SOLO


1

0

1

0

1

00

t

t VV

t

t

t

t V

tdzdydxdt

zdydxdt

tdzdydxdt

VS

SVVV

zdydxduSduzdydxduzdydxduzdydxdu

0

02

11

0

tdzdydxdut

uuuut

Ue

euut

t V

Let perform integration by part of some of the expressions

VS

SVVV


0

VS

SVVV


0

1

0

1

0

1

0 0

t

t VV

t

t

t

t V

tdzdydxdt

zdydxdt

tdzdydxdt

0

2

1

1

0

1

0

0

tddzdydxuut

ut

tddzdydxuuuut

uue

Ueuu

t

t V

t

t V

Therefore we obtain

Hydrodynamic Field


SOLO


We restrict ourselves to variations that vanish at the boundary of the region of integration, the variations with respect to γ, ρ and .u

0

2

1

ut

Ue

uu

0 u

0

u

t

0

ut

0

ut

02

11

0

tdzdydxdut

uuut

Ue

euut

t V

Since by introduction the Lagrange Multipliers the variations are independentto have we need

,, u

01

0

t

t

tdL

We have two additional constraints:

Conservation of Fluid Mass

Lagrange Coordinate Equation

Hydrodynamic Field


SOLO


We obtained: u

0

ut

Therefore :

This is the Clebsch representation (1859) of the velocity field. It allows for fluids with non-zero vorticity Rudolf Friedrich Alfred

Clebsch ( 1833 – 1872)

00

: u

or u

:

We can see that using Clebisch representations the Vortex Lines are the intersections of the surfaces β = constant and γ = constant. Those surfaces are defined by the equations

0

0

u

tu

tu

t

0

ut

u

0tr

rd

0t

11 cr

22 cr

Streamline

consttzyx

consttzyx

,,,

,,,

Hydrodynamic Field


SOLO


0

2

1

ut

Ue

uu

0 uLet start with

0

tttt

u

t

Using: sdTdp

ed 2

ppp

ep

ee

ee

sdfor

00

2 FluidIsentropicsd 0

02

1

uuUe

uuttt

u

0

2

1

uuU

puu

ttt

u

uu

uuuuuuu

uuuuuuuuuuu

FUp

uuuut

u

2

1

02

1

ut

Ue

uu

Hydrodynamic Field


SOLO


FUp

uuuut

u

2

1

We obtain:

Using the Vector Identity: we obtain uuuuuu

2

1

FUp

uut

u

Euler’s Equation (1755) for Compressible, Isentropic, Rotational Fluid

Leonhard Euler (1707-1783)

Hydrodynamic Field



SOLO


We used the Action Integral:

tddzdydxut

uut

UeuutdLt

t V

t

t

1

0

1

02

1

That has 4 unknowns that define the fluid : , and 3 more unknown potentials ϕ, β, γ.Also the Conservation of Mass was introduced as a constraint and not derived from a Variation Principle. Let find a a Simpler Eulerian Variational Principle

u

,

Rearrange the terms in the Action Integral to separate the velocity terms.:u

1

0

1

0

1

02

1t

t V

t

t V

t

t

tddzdydxuuuutddzdydxtt

UetdL

Using: we have uuu

1

0

1

0

1

0

1

0

22

22

2

1

t

t S

t

t V

t

t V

t

t V

tdSdutddzdydxu

tddzdydxutddzdydxuuu

1

0

1

0

1

0

t

t VV

t

t

nintegratio

partsby

t

t V

tddzdydxt

dzdydxtddzdydxt

We have

Asher Yahalom, “A Simpler Eulerian Variational Principle for Barotropic Fluids”, February 2, 2008,http://arxiv.org/pdf/physics/9906050.pdf


Hydrodynamic Field

SOLO


1

0

1

0

1

0

1

0

1

0

2

2

2

2

1

t

t S

t

t V

V

t

t

t

t V

t

t

tdSdutddzdydxu

dzdydxtddzdydxUett

tdL

Therefore after rearranging to separate terms we obtain:u

Define:

V

t

t

t

t V

t

t

r dzdydxtddzdydxUett

tdL 1

0

1

0

1

0

2

2

1:

1

0

1

0

1

0

2

2:

t

t S

t

t V

t

t

v tdSdutddzdydxutdL

We restrict ourselves to variations that vanish at the boundary of the region of integration, the variations with respect to ϕ, γ, ρ and .u


0

1

0

t

t

v

t

t

r tdLtdL

Hydrodynamic Field


SOLO


V

t

t

t

t V

t

t V

t

t

r

dzdydxtddzdydxttt

tddzdydxUe

ett

tdL

0

2

1

0

1

0

1

0

1

02

1:

1

0

1

0

1

0 0

t

t VV

t

t

t

t V

tdzdydxdt

zdydxdt

tdzdydxdt

Let perform integration by part of some of the expressions

1

0

1

0

1

0 0

t

t VV

t

t

t

t V

tdzdydxdt

zdydxdt

tdzdydxdt

VVV

zdydxdzdydxdzdydxd

VS

SzdydxdSd

0

VVV

zdydxdzdydxdzdydxd

VS

SzdydxdSd

0

tddzdydxttt

t

t V

1

0

tddzdydxttt

t

t V

1

0

Hydrodynamic Field


SOLO


tddzdydxttt

tddzdydxUe

ett

tdL

t

t V

t

t V

t

t

r

1

0

1

0

1

0

2

2

1:

We obtained:

0

0

0

02

1 2

t

t

t

Ue

ett

0

0

tt

0

0

0

02

1 2

t

t

t

Uhtt

enthalpyhp

ee

e

Since the variations δρ, δϕ, δγ are independent if 01

0

t

t

r tdL

Hydrodynamic Field


SOLO


0

0

0

02

1 2

t

t

t

Uhtt

We obtained:

Define: :v

v:v

ttD

D

We obtain:

0v

0v

0vvvv

0vv2

1

v

v

v

tD

D

t

tD

D

t

tD

D

tt

Uhtt

Hydrodynamic Field


SOLO


In the same way:

1

0

1

0

1

0

1

0

1

0

22 v22

:t

t SS

t

t V

t

t SS

t

t V

t

t

v tdSdutddzdydxutdSdutddzdydxutdL

1

0

1

0

2v2

t

t V

t

t

v tddzdydxutdL

v0v2

1

0

2

utddzdydxu

t

t V

If the volume V extends to infinity we have ϕ|S=0 and the second Integral vanishes. Under those conditions the Variation of is given by

1

0

t

t

v tdL

We have

We can see that if we define Clebsch Representation, we don’t need the

:u

1

0

t

t

v tdL

Hydrodynamic Field


SOLO


Summary:

tddzdydxUett

tdLt

t V

t

t

1

0

1

0

2

2

1:

This Action Integral has only four unknowns: Density ρ and Potentials ϕ, β and γ, defined by the four differential equations

EquationsBernoulliUhuutt

tD

Du

t

tD

Du

t

EquationContinuitytD

Duu

tu

t

u

u

'02

1

0

0

0vv

u

:

Define

and :u

Clebisch Representation

Hydrodynamic Field


SOLO Hydrodynamic Field


Summary:

u

02

1

Uhuutt

u

Using the Vector Identity: we obtain uuuuuu

2

1

tttt

u

u

t

0

2

1

Uhuuttt

u

02

1

Uhuuuut

u

uuu

ppp

ep

ee

ee

h

sdfor

00

2

enthalpyhp

ee

e

FUp

uut

u

Euler’s Equation (1755)

We see that using this Action Integral with 4 unknowns and Clebsch Representation we recover Continuity Equation, Bernoulli’s Equation and Euler’s Equation.



248

SOLO

Part IV: Problem of the Motion of Fluid Subjected to Gravity Forces and Surface Tension ForcesLet develop the boundary conditions at the flow surface using the Hamilton’s Principle.

From the molecular theory we now that the action of particles on one other is determined by the pressure and by the binding forces acting between the particles. In interior of the fluid the binding forces cancel one another as a result of their average uniform distribution. On the surface, however, they don’t cancel. In most of the cases when the liquid surface is sufficient large the surface tension forces are negligible in comparison with body forces. There are cases (in capillaries or for weightless conditions) when the surface forces can not be neglected.

n nS

1

2

Container

Fluid

SS Sn

Define:

Σ1 - part of container internal surface, that is not in contact with the fluid

Σ2 - part of container internal surface, that is in contact with the fluidS - free fluid surface

The tension forces per unit area between the fluid and the container internal surfaces are:

α1 - between fluid and the container internal surface, that is not in the contact with the fluid (Σ1)

α2 - between fluid and the container internal surface, that is not in the contact with the fluid (Σ2)

Α - between fluid and the air in the container (S)

Hydrodynamic FieldVariational Principles of Hydrodynamics

249

Problem of the Motion of Fluid Subjected to Gravity Forces and Surface Tension Forces

Assume a motionless container.

To find the equations of motion of the liquid let use the Hamilton’s Principle, for virtual displacements:

n nS

1

2

Container

Fluid

SS Sn

t

dtTL0

Action Integral

V

II

dVtd

Rd

td

RdT

2

1Kinetic Energy of the Fluid

The variation of T due to virtual displacement isR

t

V I

t

V IV

t

I

const

t

V I

t

VII

t

V II

t

dVdtRtd

RddVdtR

td

RddVR

td

Rd

dVdtRtd

RddtdVR

td

Rd

td

ddtdV

td

Rd

td

RddtT

02

2

02

2

0

0

02

2

000

SOLO

Part IV: Problem of the Motion of Fluid Subjected to Gravity Forces and Surface Tension Forces


250

n nS

1

2

Container

Fluid

SS Sn

Π = ΠB+ΠS - the work done by the forces acting on the fluid

ΠB - the work done by the body forces (gravitation,

electromagnetic,…) acting on the fluid

ΠS - the work done by the surface tension forces acting on the fluid

The variation of ΠB is V

BB dVRf

- body force per unit mass (for gravitation ) Bf

gf B

The variation of ΠS is SS 2211

The Action Integral L is defined for a set of motions that transfer the system from one fixed state to another during time t. This motion must be constrained by the conservation of mass condition:

0

000

RconstR

Rtd

d

td

Rd

td

Rd

tII

const

I

To account for this constraint we will multiply this equation by the Lagrange multiplier

integrate over the volume V and time t and add to the variation of the Action Integral to obtain:

000

t

V

t

dtdVRpdtTL

tRp ,

SOLO



251

Using the facts that and , let develop02

R

nRS

VS

R

nR

VS

Gauss

VVV

dVRpdSnp

dVRpSdRpdVRpdVRpdVRp

S

02

2

n

1ld

'1ld'2ld

2ldSd

'Sd

cotn

2

21 Container

Surface

Element ofFluid

Surface

Variation ofElement of

FluidSurface

1d2d

2R1R

'2ld

Snn

sin

n

We obtain

0000

t

V

t

S

t

dVRpdSnpdtTL

Let find the variations in the surfaces δΣ1, δΣ2, δS

Assume a surface element dS=dl1.dl2 (see Figure) that is

virtually displaced by δn and becomes dS’=dl1’’.dl2 ‘ .

Assume that the surface element dS=dl1.dl2 has the

Radiuses of Curvature R1 and R2 corresponding to l1 and l2, respectively. They are regarded positive when the center of curvature lies on the same side as the fluid and negative otherwise. We can write

thereforedRldanddRld ,222111

222111 '' dnRldanddnRld

SdnRR

ddRRR

n

R

nldldldldSdSd

K

Sd

111121

212121

11111'''

SOLO



252

n nS

1

2

ContainerSurface

Fluid

SS Sn

n

1ld

'1ld'2ld

2ldSd

'Sd

cotn

2

21 Container

Surface

Element ofFluid

Surface


FluidSurface

1d2d

2R1R

'2ld

Snn

sin

n

21

11

RRK Mean Curvature of the Fluid Surface

To this we must add the fluid surface variation due to container wall given by 'cot 2dln

where θ is the angle between (normal to Σ ) and (normal to S ), at the contact line Г, between them.

nSn

From this we have

dlnSdnKdlnSdSdSSS

cotcot'

From the Figure bellow we can see that 221 sindldlwheredl

n

SOLO



253

n nS

1

2

ContainerSurface

Fluid

SS Sn

n

1ld

'1ld'2ld

2ldSd

'Sd

cotn

2

21 Container

Surface

Element ofFluid

Surface


FluidSurface

1d2d

2R1R

'2ld

Snn

sin

n

Therefore

dlnSdnKdln

SS

S

cotsin212211

Let put together the results obtained

0000

t

V

t

S

t

dVRpdSnpdtTL

0

sincos

00

0

12

002

2

t

V

t

S

tt

V

B

t

V I

tddVRptdSdnKp

tdldn

tddVRfdVdtRtd

RdL

SOLO



254

0sin

cos00

12

02

2

t

S

tt

V

B

I

tdSdnKptdldn

dVdtRp

ftd

RdL

p

ftd

RdB

I

2

2

12cos

Euler’s Equation (Ideal Liquid)

The Contact Angle θ of the Liquid Surface Contour Г

α1 – Tension forces per unit area between fluid and the container internal surface, that is not in the contact with the fluid (Σ1)

α2 – Tension forces per unit area between fluid and the container internal surface, that is not in the contact with the fluid (Σ2)

n nS

1

2

Container

Fluid

SS Sn

α - Tension force per unit area between fluid and the air

21

11

RRK Mean Curvature of the Fluid Surface

Since the variations are arbitrary and independent from each other, if and only if

landnR ,

011

21

RRpKp Laplace’s Formula

The physical interpretation of the Lagrange’s multiplier must be defined. tRp ,

SOLO




SOLO


Use the Field approach to describe the state of the Gas in Motion. The relevant Fields are:1.Density Field2.Velocity Field3.Pressure Field

trp

tr

tr

,

,v

,

The equations defining the gas that we are using are:•Conservation of Mass (C.M.)•Conservation of Linear Momentum (C.L.M.)•Constitutive Relations

Conservation of Mass (C.M.)

)(

1

)()( )(

vv)(

0tV

GAUSS

tStV tV

REYNOLDS

FFF F

Vdt

SdVdt

dVtd

d

td

tmd

Consider any Volume VF (t) attached , and moving with the fluid, enclosing a constant mass m, bounded by a Surface SF (t), then:

Since this is true for any Volume VF attached to the fluid, we must have

0v

t(C.M.)

Acoustic Field

SOLO


)(

v

tVF

dVtd

d

Consider any Volume VF (t) attached , and moving with the fluid, enclosing a constant mass m, bounded by a Surface SF (t), then the Linear Momentum of this mass is:

Assuming that the fluid is viscous free, and only forces acting on the volume VF (t) are do to external pressure p acing normal to the surface SF (t), the total force acting on the fluid is

0v

ptd

d

Conservation of Linear Momentum (C.L.M.)

)()(

ˆtV

Gauss

tS FF

VdpSdnp

Therefore

)( )(

v

tV tVF F

VdpdVtd

d

)(

0v

tVF

dVptd

d

Since this is true for any Volume VF attached to the fluid, we must have

Euler’s Equation(1755)

(C.L.M.) Leonhard Euler (1707-1783)

Acoustic Field

SOLO


fp

Constitutive Relations (C.R.)

To define the problem completely, we must relate the Pressure p to the Density ρ, through a Equation of State

Because the range of variation of p in a Sound Wave is very small, the exact expression of f is not omportant.

Let look first at a Static, in Equilibrium atmosphere, for which the Density is ρ0 and the Pressure p0. The Density variation from Equilibrium will be:

110 ss

s – called the “condensation” is the fractional variation from Equilibrium of Density.

000

12

0

''

2

2

0

'

0 '2

1

20

0

00

0

00

fspd

fd

d

fdfp

s

s

f

s

f

p

Taylor

Acoustic Field

SOLO


0,,v

trptd

trd(C.L.M.)

0

vvvv2

1vv

000 ' fspp

Irotational 0v

vv2

1

t

tr

t

tr

,vvv

2

1,v

s 10 0',v

1 00

1

0

sft

trs

0v

t

(C.M.) 0v11

1

0

s

t

s

0v

t

s

0'v

0

sft

0v

2

2

tt

s

t

sft

s 202

2

'

0'v 0

sft

(C.R.)

trtrt

tr

td

trd,v,v

,v,v

v

ϕ – Velocity Potential of Irotational Flow

Acoustic Field

SOLO


sft

s 202

2

'

Acoustic Wave Propagation Equation

soundofspeedaad

pdf 2

0 :'0

where

also from

0'v

2

0

sft

a

0v

t

s

t

vvv 22

vvv

0v

22

2 2

aa

t

v

2

2

22 1

ta

tas

2

1

by choosingϕusing

02

sa

t

2

2

2

1

tat

s

t

Acoustic Field

SOLO


sat

s 222

2

Acoustic Wave Propagation Equations

soundofspeedad

pdfa

0

02 ':

vvv 22

vvv

0v

22

2 2

aa

t

2

2

22 1

ta

The Following Equations are equivalent:

Acoustic Field

SOLO


Energy

Potential Energy VConsider a given mass element of the fluid ΔmF , that occupies a volume ΔVF , and has a density ρ

sΔVs

ΔV

s

ΔmΔmV F

sF

ΔmΔVFF

F

FF

111 0

10

/

0

00

The Work done to compress the volume from ΔVF0 to ΔVF is

0

20

2

0

002

0 2

102

0

00

F

s

F

sapp

ΔVsdΔVd

V

ΔV

F ΔVsadssΔVaΔVdppFF

F

F

Using we obtain, the Potential Energy in a given fluid

Volume VF0 is

tas

2

1

00

0

2

20

02

02

2

1

2

1

FF V

F

V

F dVta

dVsaV

Acoustic Field

SOLO


Energy

Kinetic Energy TConsider a given mass element of the fluid ΔmF, that occupies a volume ΔVF, and has a density ρ

00

00 2

1vv

2

1

FF V

F

V

F dVdVT

Lagrangian L

0

0

2

20 1

2:

FV

FdVta

VTL


01

2

2

20

ta

t

t

LL

2

20 1

2:,

tat

LLagrangian Density L

We recovered the Acoustic Wave Propagation Equation

Acoustic Field


04/13/23 263

Equation of Motion of a Variable Mass System – Lagrangian Approach

SOLO Variable Mass System


The subject is developed in a separate presentation:

v(t)

I

R

CR

dm

C tS

2openS

1openS

g

OR

O

Or

OCr ,

Bx

Bz

shaftr

rotorr

By

Ix

Iy

Iz At a given time t the system has

v (t) – system volume.

m (t) – system mass.

S (t) – system boundary surface.

The mass enters and leaves the

System through the openings Sopeni (i=1,2,..) in the system boundary surface.

We want to derive the Equation of Motion of the system using the Lagrangian approach.

“Equation of Motion of a Variable Mass System 3”

The System contains moving parts

solid (rotors, pistons,..) and elastic.

04/13/23 264

SOLO

References


Sin-itiro Tomonaga, “The Story of Spin”, University of Chicago Press, 1997

A.O. Barut, “Electrodynamics and Classical Theory of Particles”, Dover, 1964

A. Beiser, “Perspective of Modern Physics”, McGraw-Hill, International Student Edition, 1969

B.L. Moiseiwitsch, “Variational Principles”, Dover, 1966, 2004

D.E. Soper, “Classical Field Theory”, Dover, 1976, 2008

L.D. Landau, E.M. Lifshitz, “The Classical Theory of Fields”, 4th Revised English Edition, Pergamon Press, 1975

H. Goldstein, “Classical Mechanics”, 2nd Ed., Addison Wesley, 1980

M. Gourdin, “Lagrangian Formalism and Symmetry Laws”, Gordon and Breach, 1969

J.V. José, E. J. Saletan, “Classical Dynamics – A Contemporary Approach”, Cambridge University, 1998

04/13/23 265

SOLO

References


D. Lovelock, H. Rund, “Tensors, Differential Forms, and Variational Principles”, Dover, 1975, 1989

I.M. Gelfand, S.V.Fomin, “Calculus of Variations”, Prentice-Hall, 1963


266

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

04/13/23 267




Theorem of Noether for Single IntegralsAmalie Emmy

Noether (1882 –1935)

Consider the Functional

mjnknitdVd

t

xt

x

xtxtxtd

x

xxxI

t V

kj

k

kjkjk

R i

ijiji ,,1;,,1;,,1,0

,,

,,,,,,,

LL

Assume that we are given some variation of coordinates that changes also the domain of integration from R to R’, with boundaries S and S’, respectively.

kkk xxx

ttt

'

'

R i

ijiji

R i

ijiji d

x

xxxd

x

xxx

,,'

'

'','','

'

LL

We are looking for variations such that the Integral remains unchanged, i.e.


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

04/13/23 268




Theorem of Noether for Single Integrals (continue -1)Amalie Emmy

Noether (1882 –1935)

In the First Integral x’i, represent dummy variables, therefore we can replace them by xi like in the Second Integral, but we still have two different regions of Integration R and R’ (see Figure). Let compute the Variation

RR i

ijiji

R i

ijiji

R i

ijiji d

x

xxxd

x

xxxd

x

xxx

'

,,,,,,

LLL


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

R i

ijiji

R i

ijiji d

x

xxxd

x

xxx

,,'

'

'','','

'

LL

269



Invariance Properties of the Fundamental Integral (continue -2)Amalie Emmy

Noether (1882 –1935)


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

RR i

ijiji

R i

ijiji

R i

ijiji d

x

xxxd

x

xxxd

x

xxx

'

,,,,,,

LLL

Let find the meaning of δψi since we have a change in coordinates from t, xk to t’,x’k

k = 1,…,n

n

k kk

kjkjkjkjkjkj x

x

xtxtxtxtxtxt

1

,,,,'','','

kkk xxx '

kjkjkj

kjkjkj

xtxtxt

xtxtxt

,,,'

,,',''

n

k kk

kjkjkj x

x

xtxtxt

1

,,,

270




Noether (1882 –1935)


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

RR i

ijiji

R i

ijiji

R i

ijiji d

x

xxxd

x

xxxd

x

xxx

'

,,,,,,

LLL

0

//,,

2

11 1

t

tV

m

j

j

j

n

kk

j

kjj

jR i

ijiji tdVd

ttxxd

x

xxx

LLLL


Integrate by parts

V

n

kkjk

j

V

n

kkj

jk

V

n

k jkkj

Vdxx

Vdxx

Vdxx

11

1

//

/

LL

L

S

n

k

kk

kjj

V

n

kkj

jk

Sdnx

Vdxx 11 //

LL

nk are the Direction Cosines of the outdrawn normal to the Boundary Surface Sk.

2

1

2

1

2

1 ///

t

tj

j

t

tjj

t

t jj

tdttt

tdtt

LLL

271




Noether (1882 –1935)


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

RR i

ijiji

R i

ijiji

R i

ijiji d

x

xxxd

x

xxxd

x

xxx

'

,,,,,,

LLL

2

1

2

1

2

1

2

1

2

1

////

//,,

1 11 1

1

0

1

t

tjj

t

tS

m

j

n

k

kk

kjj

t

tjj

t

tS

m

j

n

k

kk

kjj

t

tV

m

j

LagrangeEuler

j

n

kkjkj

j

R i

ijiji

ttdSdn

xttdSdn

x

tdVdttxx

dx

xxx

LLLL

LLLL

2

11

'

,,t

tS

n

k

kkk

RR i

ijiji tdSdnxd

x

xxx

LL

Therefore

2

1

2

1

0//

,,1 1

t

tS

n

k

kk

m

j

t

tjkjjk

R i

ijiji tdSdn

txxd

x

xxx

LL

LL

272




Noether (1882 –1935)


2

1

,,,,,t

tV

j

k

jkjk tdVd

txxtxtCI

L

R i

ijiji

R i

ijiji

R i

ijiji d

x

xxxd

x

xxxd

x

xxx

,,'

'

',',',,

'

LLL

2

1

2

1

2

1

0//// 1 11 1

t

tS

n

k

m

j

t

tjkjjk

lt V

TheoremDivergence

n

k

kk

m

j

t

tjkjjk tdVd

txx

xtdSdn

txx

LL

LLL

L

This is true if

0/1 1 1

n

k

m

j

n

kkj

jkk x

xx

LL

n

r rr

kjkjkj x

x

xtxtxt

1

,,,

Substitute

0////1 11

2

1

2

1

n

k

m

j

t

tjkjjk

n

r

t

t

j

jr

j

kjk txx

ttxxx

LLLLL

04/13/23 273

http://en.wikipedia.org/wiki/Momentum

A Newton's cradle demonstrates conservation of momentum

Two-dimensional elastic collision. There is no motion perpendicular to the image, so only two components are needed to represent the velocities and momenta. The two blue vectors represent velocities after the collision and add vectorially to get the initial (red) velocity.

Science

2 classical field theories