Lagrangian Mechanics and Hamilton's Principle in Classical Mechanics

PHY 711 Classical Mechanics and Mathematical Methods 10-10:50 AM MWF in Olin 103 Notes on Lecture 5 Chap. 3&6(33) in F&W Lagrangian mechanics 1. Lagrange s equations in the presence of velocity dependent potentials such as obtained when a charged particle moves in a magnetic field. 2. Effects of constraints 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 1

4 PM Olin 101 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 2

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Comment on single and multiple coordinates -- Hamilton's principle for optimization for a single trajectory ( ): q t t f = = ( , , ) L q q t dt where ( , , ) Kinetic energy-Potential energy S L q q t t i Hamilton's principle for optimization for a multiple trajector ies { ( )}: q t t f = = ({ },{ }, ) where ({ },{ }, ) Kinetic energy-Potential energy S L q q t dt L q q t t i This "works" provided that the variation of each trajectory ( ) can be analyzed. q t Note that the trajectory components can be independent (as in the case of cartesian coordinates and/or multiple particles or can be dependent in which case we can use the trick of Lagrange multipliers. 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 5

Previously derived form for the Lagrangian -- Generalize coordinate d : s s d ( ) q ix U q d dt q T T q ( ) = = F a s 0 -m d q q ( ) ( ) T U T U d dt = = 0 q q q d dt q ) = L L q = = 0 q Note q only true is This : if ( , q ; L q t T U U = 0 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 6

Generalize coordinate d : s s d ( ) q ix Lagrangian - - Define : L , T U ( dt ) = , L L q q t d L L ( ) = = F a s 0 -m d q q q t f ( q , ) i t = Minimizati integral on : , S L q t dt Hamilton s principle from the backwards application of the Euler-Lagrange equations to Define -- Lagrangian: = L T U ( ) , q , L L q t 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 7

Summary Hamilton s 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 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 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? Why do we need velocity dependent forces? a. Friction is sometimes represented as a velocity dependent force. (difficult to treat with Lagrangian mechanics.) b. Lorentz force on a moving charged particle in the presence of a magnetic field. 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 9

Some details -- d dt q L L q ( ) , q = = 0 for , L L q t T U It was necessary to assume that: d U dt q does not contribute to the result. This comes from D'Alembert's analysis which gave us: U q d dt q T T q ( ) = = F a s 0 -m d q q ( ) d dt q T T U ( ) = = F a s 0 -m d q q ( ) ( ) d dt T U T U = while we want to use: 0 q q q 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 10

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 In this case, it is convenient to use cartesian coordinates , , , , , , L L x y z x y z t T U = ( ) Note: Here we are using cartesian coordinates for convenience. ( ) = + + 2 2 2 T m x y z 1 2 d dt x L L x d dt = -component: 0 x U x U x = + Apparently: F x q c ( ) r A( ) = r r Answer: , , U q t t ( ) A r , t 1 c ( ) ( ) ( ) ( ) = = E r r B r A r where , , , , t t t t t 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 11

More details -- ( ) ( ) d dt T U T U = Consider: 0 q q q 1 2 = + + 2 2 2 Suppose ( ) T m x y z ( ) ( ) d dt T U T U d dt d dt U x U x = + 0= mx x x d dt U x U x = = mx F x 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 12

Units for electromagnetic fields and forces cgs Gaussian units -- (as used your textbook) E B and fields as related to vector and scalar potentials: , 1 , , t t c t ( ) A r t ( ) ( ) ( ) ( ) = = E r r B r A r , , t t Corresponding Lagrangian potential: q U q t c SI units -- ( ) ( ) = r r A r , , t E B and fields as related to vector and scalar potentials: , , , t t t ( ) A r t ( ) ( ) ( ) ( ) = = E r r B r A r , , t t Corresponding Lagrangian potential: , U q t q = r r A r ( ) ( ) , t 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 13

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 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 14

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 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 15

Some details on last step: U x d dt r U x = + F x ( ) ( ) ( ) ( ) ( ) ( ) r r r r r , A t , , , , , t A t A t A t A t q c q c q c y x z x x = + + q y z x x y x y t ( ) ( ) ( ) ( ) ( ) ( ) r r r r r r , A t , , , , , t A t A t A t A t q c q c q c y x x z x = + + q y z x t x y x z ( ) A r , t 1 c E r ( ) ( ) ( ) ( ) = = r B r A r Note that: , , , , t t t t t So that: q c q c ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = + = + r r r r r v B r , , , , , , F t qE t yB t zB t qE t t x x z y x x ( ) ( ) r r It follows that similar analyses can be applied to , and , . t F t F y z 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 16

Lorentz forces, continued: Summary of results (using cartesian coordinates) , , , , , , L L x y z x y z t T = ( ) U q c ( ) ( ) ( ) = + + = 2 2 2 r r A r , , T m x y z U q t t 1 2 ( ) A r , t 1 c ( ) ( ) ( ) ( ) = = E r r B r A r where , , , , t t t t t q c ( ) ( ) ( ) = + + + 2 2 2 r r A r , , L m x y z q t t 1 2 = + Note that, more generally, U U U mechanical EM q c ( ) ( ) = r r A r , , U q t t EM 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 17

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 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 18

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 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 19

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 = 0 m z equations same that Note obtained are from direct applicatio q = n of Newton' laws s : r r 0z m B c 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 20

Example Lorentz force -- continued Evaluation q x m of equations : ( ) ( ) ( ) ( ) z t = 0 qB mc B y = + sin x t V t 0 0 c 0 ( ) q qB mc = + cos y t V t + = 0 m y B x 0 0 0 c = = V 0 m z 0 z ( ( ) ) ( ) ( ) ( ) z t qB mc = + Note: 3 second order differential equations need 6 constants for specific solution. cos mc qB x t x V t 0 0 0 0 qB mc = + + sin mc qB y t y V t 0 0 0 0 = + z V t 0 0 z 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 21

Some details Evaluation q x m of equations : qB mc qB mc qB mc qB mc d dt d dt dz dt = 0 B y = = 0 0 0 x y x y K 0 1 c q + = + = 0 0 0 y x y x K + = 0 m y B x 2 0 c = = 0 z K = 0 m z 3 How can you solve coupled differential equations? How can you determine the constants K1, K2, K3? 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 22

Example Lorentz force -- continued Consider formulation with different Gauge: ( ) = A r x 0 B y ( ) 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 Does it surprise you that the same equations of motion are obtained with a different Gauge? 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 23

How do these two different forms of A correspond to the same B? ( ) ( ) ( ) ( ) Note that , t A r = B r A r , , t t ( ) A r ( ) + A r A r r Consider ' , = , , t t f t ( ) = ' , t 1 ( ) + A r x y In our case, ( , )=2 '( , )= What is ( , )? f r t B y x 0 A r x t B y 0 t 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 24

Now consider formulation of motion with constraints -- Comments on generalized coordinates: ( , ) q ) = ( ( , ) t L L q t q t d L L = 0 dt q Here we have assumed that the generalized coordinates q are independent. Now consider the possibility that the coordinates are related through constraint equations of the form: ( ( ) 0 , ) ( : s Constraint = = j j t t q f f ) Lagrange multipliers = Lagrangian : ( , ) t ( , ) t L L q q t f d L L j j + = Modified Euler - Lagrange equations : 0 j dt q q q 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 25

Some details -- ( f , ) ( t 0 ) = Lagrangian : ( , ) t ( , ) = L L q q t t ( ) = Constraint : s f q t j j f d L L j j + = Modified Euler - Lagrange equations : 0 j dt q q q T hi S s a = moun 0 and fo ts t modifying our optimization problem - i f = + o - r e a h : = c 0 i ( ) 0, introducing the ne w const ant s . W S f i i i i 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 26

Simple example: = + 2 ( ( ), ( )) sin L u t u t m u mgu 1 2 u y ( ) = x + 2 2 ( , , , ) L x y x y m x y mgy 1 x 2 = + = sin ( , ) u sin = cos s 0 f x y x y y Note that: co 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 27

Case 1: ( ( ), ( )) L u t u t d L dt u Case 2: = + 2 sin 1 2 mu mgu L u = = = = 0 sin 0 sin mu mg u g ( ) = Which method would you use to solve the problem? Case 1 Case 2 = + 2 2 ( , , , ) L x y x y m x y mgy 1 2 = + + ( , ) f x y d dt x d dt y sin L x L y cos 0 x y L f x f y = + = = + 0 sin mx L + = = + + 0 cos my mg sin cos mg x 0 x = y Force of constraint; normal to incline cos ( ) = cos sin sin y g 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 28

Rational for Lagrange multipliers Recall Hamilton's principle: t f ( ) = ( ) , ( ) , S L q t q t t dt t i t ) t ( d dt q f L L q q t f = = ( ) , 0 S q dt t i = = With constraints: 0 f j j Variations are no longer independent. q f q j = = 0 at each f q t j Add 0 to Euler-Lagrange equations in the form: f q q j j j 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 29

Euler-Lagrange equations with constraints: ( f , ) ( t 0 ) = Lagrangian : ( , ) t ( , ) = L L q q t t ( ) = Constraint : s f q t j j f d L L j j + = Modified Euler - Lagrange equations : 0 j dt q q q Example: ( ) = + + 2 2 2 Lagrangian : cos L m r r mgr 1 2 r r = = Constraint : s 0 f mg 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 30

Example continued: ( ) = + + 2 2 2 Lagrangian : cos L m r r mgr 1 2 r = = Constraint : s 0 f dmr dt dmr dt r = + = 2 cos 0 mr mg + = 2 sin 0 mgr = = 0 r r g = sin = + 2 cos m mg 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 31

Another example: = + + + 2 1 2 2 Lagrangian : L m m m g m g 1 1 1 2 1 1 2 2 2 2 = + = Constraint : s 0 f d 1 2 + = 0 m m g 1 1 1 dt d + = 0 m m g 2 2 2 dt + = = + 0 1 2 1 2 2 m m = 1 2 g + m m 1 2 m m = = 1 2 g 1 2 + m m 1 2 9/4/2024 PHY 711 Fall 2024 -- Lecture 5 32