Classical Mechanics Principles: D'Alembert, Hamilton, Lagrange Equations
Explore the fundamental principles of classical mechanics in Lecture 8, covering D'Alembert's principle, Hamilton's principle, and Lagrange equations in the presence of magnetic fields. Dive into the concepts of virtual work, generalized coordinates, and physical trajectories of generalized coordinates to minimize action. Connect Newton's laws with virtual work and delve into the details of these foundational principles.
Download Presentation
Please find below an Image/Link to download the presentation.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.
You are allowed to download the files provided on this website for personal or commercial use, subject to the condition that they are used lawfully. All files are the property of their respective owners.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author.
E N D
Presentation Transcript
PHY 711 Classical Mechanics and Mathematical Methods 10-10:50 AM MWF Olin 103 Plan for Lecture 8: Continue reading Chapter 3 & 6 1. D Alembert s principle 2. Hamilton s principle 3. Lagrange s equations in presence of magnetic fields 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 1
9/12/2014 PHY 711 Fall 2014 -- Lecture 8 2
Hamiltons principle: ( ) , q = Given the Lagrangian function: , , L L q t T U The physical trajectories of the generalized coordinates ( ) q t ( ) , q = Are those which minimize the action: , S L q t dt Euler-Lagrange equations: d dt L q L q d dt q L L q = = 0 for each : 0 q 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 3
DAlemberts principle -- notion of virtual work: Generalize coordinate d : s s d ( ) q ix Newton' a F laws s : ( ) = = F a s 0 0 -m -m d x = a s i m d m x q i q i x x d d = i i m x m x q i i dt q dt q i x x x dx x d = = Claim : and i i i i i q q dt q q dt q ( ) ( ) 2 2 m x m x d 1 1 = a s i i 2 2 m d q dt q q i 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 4
Some details -- x q ( ) ( ) t = a s where , m d mx q x q t i i i i x q x d dt d dt q = mx mx q i i i i i x q x q x dx dt x q d dt q = = Claim: and i i i i i q dx dt x q x q dq d x t x q x t = + = + Details: q i i i i i t x q = i i 2 2 x q x x x q d d + = = q i i i i ' q q q t t ' ' 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 5
DAlemberts principle -- notion of virtual work: Generalize coordinate d : s s d ( ) q ix Newton's laws: -m = F a ( ) = F a s 0 0 -m d ( ) ( ) x q 2 2 mx mx 1 2 1 2 d dt i i = a s m d q q q i ( ) T q q ( ) T q F d dt ( ) 2 = where q T mx 1 2 i i x q ) U x U q ) = = = F s d q q i i i i i i ( ( T U T U d dt ( ) = = F a s 0 -m d q q q 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 6
DAlemberts principle -- notion of virtual work: Generalize coordinate d : s s d ( ) q ix ( ) ( ) T U T U d dt ( ) = = F a s 0 -m d q q q U q = Provided that 0 Consistent with Hamilton's principle with = = ( ) , q , L T U L q t 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 7
Example simple harmonic oscillator 2 1 2 T mx = = 2 2 U m x 1 2 / ( ) = = = Assume 0 0 and 0 x( ) x( ) S T U dt 0 ( ( ) = = Trial functions ( ) sin 0 x t A t S 1 1 ) ) = A t ( = 2 2 ( ) 0.067 x t t S A m 2 2 = = 2 2 t ( ) sin 0.062 x t Ae t S A m 3 3 x1 x2 x3 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 8
Note: in proof of Hamiltons principle: d L L ( , q ) = = 0 for , L L q t T U dt q q It was d necessary U assume to that : contribute not does to the result. dt q How can we represent velocity - dependent forces? 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 9
Lorentz forces: E r B r For particle of charge q electric an in field ( , magnetic and ) field ( , : ) t t ( q ) = + F E v B Lorentz force : q 1 c + ( ) ( ) = v B component : x F E 1 x x x c convenient is it case, In this = cartesian use to coordinate s ( m ) , , , , , , L L x y z + x y + z t T U ( ) = 2 2 2 T x y z 1 2 d L L = - component : 0 x dt x x U d U = + Apparently : F x x dt x q ( ) ( ) = r r A r Answer : , , U q t t c ( ) A r 1 , t ( ) ( ) ( ) ( ) t , = = E r r B r A r where , , , t t t t c 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 10
Lorentz forces, continued: ( ) ( ) = + v B component of Lorentz force : x F q E 1 x x x c q ( ) ( ) = r r A r Suppose : , , U q t t c U ) + d U = + Consider : F x x dt x ( x ) ( x ) ( ( x ) r , A t r r r , , , A t U t q A t y = + + x z q x y z U x x ) t c q ( , r = A x x c ( ) ( x ) ( y ) ( ) ( ) r r r r r , , , , , dA t A t A t A t A t d U q q = = + + + x x x x x x y z dt x c dt c z t 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 11
Lorentz forces, continued: ( x ( z ) ( x ( y ) ( ) ( x ( ) r , A t r r r , , , A t U t q A t y = + + + x z q x y z U A x x c ( x ) ) ) ) r r r r , , , , t A t A t A t d q = + + + x x x x x y z dt x c t U d U = + F x x dt , x ( x ) ( y ) ( y ) ( ) ( ) ( x ) r , A t r r r r r , , , , A t A t A t t q q A t q y = + + x x x z q y z x c c c t ( x ) ( ) ( y ) ( ) ( ) ( x ) r , A t r r r r r , , , , , A t A t A t t q q q A t y = + + x x x z q y z x c t c c z ( ) q q ( ) ( ) ( ) ( ) ( ( ) )x = + = + r r r r v B r , , , , , qE t y B t z B t qE t t x z y x c c 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 12
Lorentz forces, continued: Summary = of results (using cartesian coordinate s) ( ) , , , , , , L L x y z x y z t T U ( ) q ( ) ( ) = + + = 2 2 2 r r A r , , T m x y z U q t t 1 2 c ( ) A r 1 , t ( ) ( ) ( ) ( ) = = E r r B r A r where , , , , t t t t t c ( ) q ( ) ( ) t , = + + + 2 2 2 r r A r , L m x y z q t 1 2 c 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 13
Example Lorentz force L = ( ) q ( ) ( ) t , + + + 2 2 2 r r A r , m x y z q t 1 2 c B E B r B r z Suppose ( , ) y , 0 + x ( , ) t t 0 ( ) = A r x ) y ( , ) t 1 0 2 ( q ( ) = + + + + 2 2 2 L m x y z B y x x y 1 0 2 2 d c d L L q q = = 0 0 m x B y B y 0 0 2 2 q dt x x dt c c d L L d q = + + = 0 0 m y B x B x 0 0 2 2 dt y y dt c c d L L d = = 0 0 m z dt z z dt 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 14
Example Lorentz force -- continued ( ) q ( ) = + + + + 2 2 2 L m x y z B y x x y 1 0 2 2 c d q q q = = 0 0 m x B y B y m x B y 0 0 0 2 q 2 q dt c c c d q + + = + = 0 0 m y B x B x m y B x 0 0 0 2 2 dt c c c d = = 0 0 m z m z dt 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 15
Example Lorentz force -- continued ( 2 + = y x m L ) q ( ) + + + 2 2 2 z B y x x y 1 0 2 c q = + m x B y 0 c q = m y B x 0 c equations same that Note obtained are = 0 m z from direct applicatio q = n of Newton' laws s : r r 0z m B c 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 16
Example Lorentz force -- continued formulatio Consider ( ) r = A x n with different Gauge : B 0y ( ) q = + + 2 2 2 L m x y z B y x 1 0 2 c d q q = = 0 0 m x B y m x B y 0 0 dt c c d q q ( ) + = + = 0 0 m y B x m y B x 0 0 dt c c d = = 0 0 m z m z dt 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 17
Example Lorentz force -- continued Evaluation q x m of equations : = 0 B y ( ) 0 ( ) ( ) ( ) t ( ( c = + q sin x t V t 0 mc q ( ) + = 0 m y B x = + q cos y t V t 0 c 0 mc = z V = 0 m z 0 z ) ) ( ) ( ) ( ) t = + q cos x t x V t mc 0 0 q mc = + + q sin y t y V t mc 0 0 q mc = + z z V t 0 0 z 9/12/2014 PHY 711 Fall 2014 -- Lecture 8 18
