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**Molekuláris bionika és Infobionika Szakok tananyagának komplex fejlesztése konzorciumi keretben

***A projekt az Európai Unió támogatásával, az Európai Szociális Alap társfinanszírozásával valósul meg.

Development of Complex Curricula for Molecular Bionics and Infobionics Programs within a consortial* framework**

Consortium leader

PETER PAZMANY CATHOLIC UNIVERSITY

Consortium members

SEMMELWEIS UNIVERSITY, DIALOG CAMPUS PUBLISHER

The Project has been realised with the support of the European Union and has been co-financed by the European Social Fund ***

(2)

PHYSICS FOR NANOBIO-TECHNOLOGY

Principles of Physics for Bionic Engineering

Chapter 2. Classical mechanics

(A nanobio-technológia fizikai alapjai )

(Klasszikus mechanika)

Árpád I. CSURGAY,

Ádám FEKETE, Kristóf TAHY, Ildikó CSURGAYNÉ

(3)

Table of Contents

2. Classical Mechanics

1. Basic Concepts of Analytical Mechanics 2. Mechanics of Many Point-like Bodies 3. Particle in Electric and Magnetic fields

(4)

1. Basic Concepts of Analytical Mechanics

 Classical Physics is based on four fundamental notions: Space, Time, Body and Force.

 The “Theatre of Physics” is geometrical Space r and Time t.

 On the theater Bodies are moved by Forces.

 Bodies are composed of undividable bodies (a-tomos).

 Forces follow the vector superposition rules.

(5)

CLASSICAL BODIES

“COLLISION”

CLASSICAL FORCES

“INTERFERENCE”

Trajectory Velocity Momentum )

(t ) r

(t z

) (t x

z

m ) (t y

x y

t t t

d ) ( ) d

( r

v

) (t r

) ( )

(t m v t

p  

Dynamics of a Single Point-like Body

(6)

From given initial position and velocity the trajectory can be determined by solving an ordinary differential equation.

Two fundamental principles of Mechanics

Least action Analytical principle Calculus of Variations Differential (local)

equation of motion (Newton equations)

I. II. III.

If F = 0, then

momentum is constant

 

0

2 2

d ,

d 1 /

m m

t m v c

 

F v F12 F21

(7)

Newton’s nonrelativistic dynamics

In case of conservative force field (potential)

The force depends only on r. It does not depend on velocity.

Total energy (which is conserved)

d ( )

d

m m t

tv  

r F

).

pot(r V

)

pot(r F  V

) ( )

( pot r

r t V

m  

) ( )

2 ( 1

pot

2 r

r t V m

E   

Kinetic energy

Potential energy

d 0

d m

t

  

 

 

(8)

Newton equation is second order, thus two initial conditions are to be given: initial position and initial momentum.

Instead of initial position and momentum, let us fix the initial and final positions.

A question:

which will be the trajectory of the particle ?

) (t1 r

) (t2 r

(9)

Two new notions are introduced:

Lagrangian Action

 Action is the time-integral of the difference between kinetic and potential energies.

 Proposition: Along the real trajectory the Action will be extremum (minimum or maximum).

 

2 2

1 1

2

pot

( ) d 1 ( ) d .

2

t t

t t

S tL t   mV  t

 

 

r r r

2

pot

1 ( ),

L  2 mrV r

(10)

Proof of the Proposition

 Action is a functional (a function of a function). The extreme value of this functional can be calculated by calculus of

variations.

Consider a given path . We ask how the action changes when we change the path slightly

(we keep the end points of the path fixed).

) (t r

), ( )

( )

(t r t r t

r   r( )t1 r( )t2  0

  

2

( ) d ,

2

2 1

1

pot 2

 2 

t

t

t V

r m

S rr rrr rr

 

.

) ( )

( pot pot 2

pot rr V r Vr Or

V    

(11)

Which should hold for all changes that we make to the path.

The only way this can happen if the expression in is zero.

This means

We recognize this as Newton’s equations. Requiring that the action is extremized it is equivalent to requiring that the path obeys Newton’s equations. Q.e.d.

  

2

1

d

pot ...

t

t

t V

m S

S

S rr r rrr

pot. V mr  

 

 

2

1

pot d

t

t

t V

mrr

 

 

d

12.

2

1

pot

t t t

t

m t

V

mr rrr

Vanishes because we

fixed the end points.

(12)

Hamilton‟s Principle (Principle of Least Action)

Hamilton’s Principle

„Least Action Principle”

Newton’s Equations of Motion

2 2

1 1

2

pot

( ) d 1 ( ) d 0,

2

t t

t t

S L t m V t

r  

 

 rr  

   

2 d extremum.

1

t

t

t L t

S r

(13)

The principle of least action requires a new attitude!

In Newton’s mechanics the particle is thinking in each moment:

where should I go?

He looks around, evaluates the local potential (the forces):

“Ah … this is the right direction”, and makes a small step into that direction, and evaluates the new potential step by step.

In case of the principle of least action he chooses all possible

paths one by one, and evaluates the actions, and chooses the path of least action and says

“YES, I WILL TAKE THIS PATH”.

(14)

The Advantages of the Analytical („Variational‟) Approach

1. It is independent of the coordinates we choose to work in. The idea of minimizing the action holds in any system of

coordinates you choose to work in. This can be very useful.

2. 2. It is easy to implement constraints in this set-up. This means that we can solve rather tricky problems, such as the double pendulum, or the strange motion of spinning tops, with ease.

3. Unification of Physics: All fundamental laws of physics can be expressed in terms of a least action principle (e.g. Mechanics and Electrodynamics can be unified).

(15)

2. Mechanics of Many Point-like Bodies

 Generalized coordinates

 Trajectory

are given as a function of time.

In case of n free particles  3n coordinates.

In most cases particles are not free (molecule, rigid body, pendulum, etc).

m2

r2

m1

r1

mi mn rn

x

y z

ri

. ,..., 2 , 1 ,

, , 2 3

1 q q i f

q

) ( ),...,

( ),

( 2

1 t q t q t

q f

(16)

If a mechanical system consists of n particles, and there are m kinematical conditions imposed, the configuration can be

characterized uniquely by coordinates:

f is called “the degree of freedom”.

“Configuration space” is an f dimensional abstract space.

Pictorial language: Motion (dynamics) of a complex mechanical system: a “POINT” moves in the configuration space.

, 3n m f   ,

,..., ,...,

, 2

1 q qi qf

q

(17)

The positions of the particles are

The potential energy of the mechanical system depends only on the position coordinates, thus

The time-derivatives of the generalized coordinates are called

“generalized velocities”

where

. , ...

, 2 , 1 ),

, ...

, ,

(q1 q2 qf i n

i

ir

r

1, 2, ,

.

p

pot E q q qf

E  

, ,

...

, , 2

1 q qf

q   , 1, 2,..., .

d

d i f

t

qiqi

(18)

The kinetic energy of the mechanical system depends on generalized velocities and on position coordinates

The Lagrangian = kinetic energy – potential energy Principle of least action (Hamilton’s principle)

The Euler–Lagrange equations for an n-particle system

are f second order ordinary differential equations.

1, 2, , , 1, 2, ,

.

k

kin E q q qf q q qf

E      

p.

k E

E L  

. 0 d 

L t

f q i

L q

L

t i i 0, 1, 2, , d

d 

  

 

(19)

Generalized momentum coordinates Hamiltonian

Hamilton equations of motion

are 2 f first order ordinary differential equations.

   

, , , ,

i i i i

i i i

i p

q p q H

q q p p H

 

 

 

i 1, 2,..., f

i .

i L q

p   

i, i

i, i .

H q p

q pL

(20)

Example 1: Equation of motion of the pendulum in a plane

Energies Lagrangian

Hamilton’s Principle

Lagrange equation

equation of motion

 

;

1 cos

;

2 1

pot 2

kin m E mg

E

  

1 cos

;

2

1 2

m mg L

  

1 cos

d 0;

2

2 1

1

2





t

t

t mg

m

S

; sin

mg L

;

d d d

d 2 2

t m m L

t

( )t gsin ( ) .t

(21)

Example 2: A pendulum: point-like body is hanged on a rubber string The length of the string in force-less state is

The string constant is k,

We are looking for the equations of motion.

Let the generalized coordinates be and

The Lagrangian of this system is

The equations of motion are Equilibrium

0

 

.

Force k 0 )

(t

2 2 2

2 0

2 cos 1

2

1 m  mg k k

L

2 0)

(

cos



m g k

2 gsin

mg/ k

0

0 equ

equ

,

)

(t

 

;

2

1 2 2 2

kin m 

E pot

0

2

 

0 2

2 1 2

cos 1

k k

mg

E

m

0 φ

(22)

Example 3: a double pendulum in a plane.

Consider a double bob pendulum with masses attached by rigid mass- less wires of lengths

Further, let the angles the two wires make with the vertical be denoted by The positions of the bobs are given by

The potential energy of the system is

The kinetic energy of the system is

x y

2 1, m m

1, 2 . l l

. , 2

1

1 1sin 1, x l

1 1cos 1,

y  l

2 1sin 1 2 sin 2 , x l l

2 1cos 1 2 cos 2.

y  l l

2.

2 1

1gy m gy

m

V

l1

m1

1

l2

m2

2

   

.

2 1 2

1 2

1 2

1 2

2 2

2 2 2

1 2

1 1 2

2 2 2

1

1v m v m x y m x y

m

T

(23)

The Lagrangian is

The Euler-Lagrange equations of motion (two second order differential equations)

From the Lagrangian the generalized momentums the Hamiltonian

The Hamiltonian equations of motion (four first order differential equations) .

V T L

, d 0

d

1 1

L L

t 0.

d d

2 2

L L t

,

1

1

p L ,

2

2

L p

2 .

2 1

1p p L

L p

q

H i

i

i

,

1

1 p

H

,

2

2 p

H

,

1

1

H

p .

2

2

H

p

(24)

The Euler-Lagrange equations of motion

   

 

 

2 2 2 2

1 1 1 2 2 2 1 1 2 2

2 2 2 2 2 2

1 1 1 2 2 2 2 1 1 1 2 1 2 1 2

1 2 1 1 2 2 2

1 1

2 2

1 1 1

2 cos

2 2 2

cos cos .

L T V m x y m x y m gy m gy

m l m l m l l l

m m gl m gl

   

  

, d 0

d

1 1

L L

t 0.

d d

2 2

L L t

(25)

1 2

sin

1 2

 

1 2

1sin 1.

2 1 2 1 2 1

L m l l m m gl

     

 

sin .

sin d cos

d

1 2

1

2 1

2 2 2 2 2

1 2

2 2 1

2 1 2 1

1 1

g m m

l m l

m l

m L m

L t

 

,

cos 1 2

2 2 1 2 1

2 1 2 1

2 1 1 1

L ml m l m ll

 

cos

   

sin

 

,

d d

2 1

2 1

2 2 1 2 2

1 2

2 1 2 1

2 1 2 1

1

L m m l m l l m l l

t

, d 0

d

1 1

L L t

(26)

m1 m2

l121 m2l22 cos

1 2

m2l2 22 sin

1 2

 

m1 m2

gsin1 0.

 

sin

 

sin 0.

cos 1 2 2 1 12 1 2 2 2

1 1 2 2

2

2l m l m l m g

m

), ( ),

( 2

1 t y t

y

 

cos

sin

1 2

sin 1 0 2

2 2

1 2

2 2

1 Ay By y y By y y C y

y

 

sin

 

sin 0.

cos 1 2 22 1 2 2

1

2 Dy y y Dy y y E y

y

(27)

3. Particle in Electric and Magnetic fields The Force acting on a particle (Lorentz force)

F in newton [N], E in [V/m],

q in [As], v in [m/s], B in [Vs/m2].

, B v

E

Fqq

) (t

r d ( )

( ) ( ) ,

d t m t m t

  rt

p v

 

d d

d d ,

m

t t

pr

F d

 

d .

m q q

tv   

E v B

) (t r ) (t z

) (t x

z

q m,

) (t y

x y

,

(28)

The law of motion is valid in case of large velocities as well, (relativistic case)

If m

is approximately the rest mass non-relativistic case

In general In static case

 

.

1

/ 2

0 v c

m

m   /

c v 

d .

d v E v B

q q m t

 

r,t , B B

 

r,t .

E

E  

 

,

 

.

 

E E r B B r

(29)

Example 4: Determine the trajectory!

A charged particle ( m, q ) from a given initial position [ r (t0) ], with initial velocity [ v (t0) ], moves in a static, homogeneous electric field E0.

2 2 0

d d

d d ,

m q q m q

t t

v r

E v B E

0 0

2

0 0 0 0

2

d d

d d ( ) ,

d d

t t

t t

m t q t m q t t

t t

r

E r v E

 

,

2 ) ) (

( d

) (

d d

d 0 2

0 0

0 0

0 0

0

0 0

t q t

t t m m

t t

t q

t t m

t

t t

t

r v E r r v E

. )

2 ( ) ) (

( 0 0 0

2 0

0 v r

r E

t t t t

m t q

(30)

0, 0, 0

E v r

r

E0

 

2

0

0

1

2 q t t

m

E

0

0 t t v

q 0

Em a

r0

t0

t v0

t t .

) 2 (

) ) (

( 0 0 0

2 0

0 v r

r E

t t t t

m t q

(31)

Example 5: Determine the trajectory!

A charged particle ( m, q ) from a given initial position, with given initial

velocity moves in a static, homogeneous magnetic field B0. Let us choose the z coordinate as then, the initial conditions are

The equations of motion is

0 ,

0 k

B

B B

constant.

, 0 ,

0 0 0 0 0

0 r v v xiv zk

t

d . d d

d

B0

v v B

v

v E q m t

q t q

m

, 0

d 0 d d

d

0 0

B v v

v q q

v v v m t

m t x y z

z y

x i j k

B

v v

,

0 d

d d

d d

d

0 0

B v

B v q t v

m t v m

t v m

x y

z y x

(32)

0 0 0

d d

0 ( ) ,

d z z z d z z

v v v z v z t v t

t    t

d , d d

d d d

2 0 2

0 2

2

0 0

y y

y

x y y

x

v m v

qB t

v m v

qB m v qB

t v

t v





2

2

0 1 0 2 0

2

d sin cos ,

d

y

y y

v v v C t C t

t  

( 0) 2 0,

v ty C

x

y v

t v d 0

d vx  C1cos0t v tx( 0)   C1 v0x vx v0x cos0t.

(33)

The trajectory in the x-y plane is a circle with radius

in the z direction linear constant velocity.













0 0

0 0

0 0

0 0

0 0

0 0

cos sin sin

cos d

d d d

y v t

x v t

y x t

v

t v

t y t x

x x

x x

0 0 0

0

0 0

0

0 0

0 0 )

0 (

) 0 (

x x

y v x v y

x t

y t x













.

2

0 0 2

0 2 0









x x

v y v

x

,

0 0

qB m r v x

z

x

y

v0

B

(34)

If The trajectory is a circle.

Period in time and frequency:

Angular velocity:

0 B,

v ,

2

B r q

mv v .

qB r mv

2 , 2

qB m v

T r

1 .

f T 2 .

m qB T

v0

B

B v q r m 0

++

(35)

Application: Mass spectroscopy

After ionization, acceleration, and velocity homogenization the particles, the ions move into a mass spectrometer region where the radius of the path and thus the position on the detector is a function of the mass.

+ +

+ + + +

+ +

+ +

+

+ - + + + +

- - - -

Ionization Velocity

selector Acceleration

voltage applied

magnetic field region +

Bs

v Es

B out toward viewer

2r detector

B B

E q m B

v q r m

s

s

(36)

Radius of path produced by magnetic field

If the velocity v is produced by an accelerating voltage V:

Substitution gives:

2

mv mv. r qvB qB

1 2 2

2 .

mv qV v qV

m

1 2 mV .

r B q +

magnetic field region

B out toward viewer

+ v

r

m 2

q r

v

F v B

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