la intro rced/plr1 least action, lagrangian & hamiltonian mechanics a very brief introduction to...

LA Intro RCED/PLR 1

Least Action, Lagrangian & Hamiltonian Mechanics

A very brief introduction to some very powerful ideas– ‘foolproof’ way to find the equations of motion for complicated

dynamical systems– equivalent to Newton’s equations– provides a framework for relating conservation laws to

symmetries– the ideas may be extended to most areas of fundamental

physics (special and general relativity, electromagnetism, quantum mechanics, quantum field theory…)

NOT FOR EXAMINATION

LA Intro RCED/PLR 2

Lagrange and Hamilton

• Joseph-Louis Lagrange (1736-1810) [Giuseppe Lodovico Lagrangia]

• Sir William Rowan Hamilton (1805-1865)

LA Intro RCED/PLR 3

Why?

• Newton’s laws are vector relations• For complicated situations maybe hard to identify all the

forces, especially if there are constraints

LA Intro RCED/PLR 4

Hints

• We have already exploited the energy conservation equation, especially for conservative forces

• Note that the energy equation relates scalar quantities

E

U(x)

xBxA

€

1

2m ˙ x 2 +U(x) = E

or

€

dx

dt=

2

m(E −U(x))

LA Intro RCED/PLR 5

More

PE

KE

t

PE

KE

t

falling – constant force

SHMFor another, more formal, approach see the appendix on D’Alembert & Hamilton

€

KE ≈ PEFor many simple systems

When averaged over a pathThis leads to the idea that

(the action)

evaluated along a pathmay take a minimum or stationary value

€

S = (T −V )dttA

tB

∫

LA Intro RCED/PLR 6

Least Action by brute force

x

y

O R

A

B

C

D

€

y =v0

u0R

⎛

⎝ ⎜

⎞

⎠ ⎟x(R − x)

Numerical integration…Fixed distance R and time tR

Horizontal speed u0=R/tR

Trajectory is of the form

Note

Vary v0 - calculate

For paths B, C, DA has v0 < 0 (!)C is path with S stationary

With

€

ymax =v0R

4u0

€

(T −V )dt∫

€

v0 =gR

2u0

LA Intro RCED/PLR 7

Path integrals

A

B

t

1 2 3 4 5 6 7 8

B

A t

€

x, ˙ x , t

L = T - Vis the Lagrangian, a functionof Minimise the action integral w.r.t. variations in the path?

Calculus of variations?- No, use Euler’s geometrical approach. Approximate S as sum

xi at start of each segment for segment

€

δ LdttA

tB

∫ ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟=δ(S) = 0

€

S = L(xi , ˙ x i , ti )i

∑ Δt

€

˙ x i ≡ x i = Δx /Δt

LA Intro RCED/PLR 8

Euler I

• Key observation – every section of action integral or sum must be stationary

• Consider two linearised sections of sum, with point at time t2 moved to N

M

O

N

1 2 3

x

t

€

SMNO(xN ) = L(xM , x M , t1)Δt + L(xN , x N , t2 )Δt

with x M =xN − xM

Δt, x N =

xO − xN

Δtand for simplicity, equal time steps

Δt = t2 − t1 = t3 − t2

LA Intro RCED/PLR 9

Euler II

€

Require SMNO to be stationary. It is a function of xN only, so

dSMNO

dxN

=∂L

∂x M

dx MdxN

+∂L

∂xN

+∂L

∂x N

dx NdxN

⎡

⎣ ⎢

⎤

⎦ ⎥Δt = 0

now dx MdxN

=1

Δt,

dx NdxN

= −1

Δt so the condition is

∂L

∂xN

−1

Δt

∂L

∂x N−

∂L

∂x M

⎛

⎝ ⎜

⎞

⎠ ⎟= 0.

This condition must be true for any point on the path, so taking Δt → 0

∂L

∂x−

d

dt

∂L

∂ ˙ x

⎛

⎝ ⎜

⎞

⎠ ⎟= 0; The Euler - Lagrange equation

LA Intro RCED/PLR 10

From action to Newton

€

An object of mass m moves in 1- D under the influence

of a conservative force with potential function V (x).

T =1

2m ˙ x 2 is the kinetic energy, so

L = T −V =1

2m ˙ x 2 −V (x) and

∂L

∂x= −

∂V

∂x,

∂L

∂ ˙ x = m ˙ x .

The E - L equation is

d

dt(m ˙ x )+

∂V

∂x= 0 or

m˙ ̇ x = −∂V

∂x which is NII, with the force F = −

∂V

∂x


Generalised coordinates

€

In general the Lagrangian function will be a function of more variables

(x, ˙ x , y, ˙ y or r, ˙ r ,θ , ˙ θ ,...), call these generalised coordinates, qi , ˙ q i{ },

So N × E - L equations d

dt

∂L

∂ ˙ q i

⎛

⎝ ⎜

⎞

⎠ ⎟−

∂L

∂qi

= 0 give the conditions for δ(S) = 0.

In many cases T =1

2m ˙ q i

2

i

N

∑ and V does not depend on ˙ q i .

Then ∂L

∂ ˙ q i= m ˙ q i ≡ pi is a momentum.

For any L, ∂L

∂ ˙ q i is defined to be the generalised momentum

corresponding to qi .


Example – Newtonian gravity

€

In polar coordinates T =1

2m ˙ r 2 +

1

2mr2 ˙ θ 2 and V = −

GMm

r∂T

∂r= mr ˙ θ 2 ,

∂V

∂r=

GMm

r2,

∂T

∂θ= 0,

∂V

∂θ= 0,

∂T

∂ ˙ r = m ˙ r ,

∂V

∂ ˙ r = 0,

∂T

∂ ˙ θ = mr2 ˙ θ ,

∂V

∂ ˙ θ = 0.

There are two E - L equations

r : m˙ ̇ r − mr ˙ θ 2 +GMm

r2= 0; and θ :

d

dtmr2 ˙ θ ( ) = 0.

The second equation gives the conservation of angular momentum

mr2 ˙ θ = mh. Use this in the r equation, giving

˙ ̇ r −h2

r 3+

GM

r2= 0; the radial equation of motion

NB. if L does not depend on qi , then ∂L

∂qi

= 0 and pi is a constant.


Sliding blocks

B

A

x

z

Mass A M, mass B mFind the initial acceleration of A

No friction

€

T =1

2M˙ z 2 +

1

2m( ˙ x 2 + ˙ z 2 + 2 ˙ z ̇ x cosα ), V = mgxsinα

E - L z : d

dt(M˙ z + m˙ z + m ˙ x cosα ) = 0 ⇐ cons. of horizontal momentum

x : d

dt[m( ˙ x + ˙ z cosα )]+ mgsinα = 0

from z : m˙ ̇ x = −(M + m)˙ ̇ z / cosα ,

hence ˙ ̇ z =mgsinα cosα

M + msin2 α


Pendulum with oscillating support

lθ

m

()cosxtAtω=

€

x(t) = Acosωt

€

θ

€

T =1

2m( ˙ x 2 + l2 ˙ θ 2 + 2 ˙ x l cosθ ˙ θ );

V = mgl(1− cosθ );

E - L equations

x : d

dt[m( ˙ x + l cosθ ˙ θ )] = 0; ⇐ horizontal momentum

θ : d

dt[m(l2 ˙ θ + l ˙ x cosθ )]+ ml ˙ x ˙ θ sinθ + mgl sinθ = 0;

The second equation gives

˙ ̇ θ +g

lsinθ =

Aω2

lcosωt cosθ; put ω0

2 = g / l & assume θ small

˙ ̇ θ +ω02θ =

Aω2

lcosωt; small angle approx. → driven oscillator.

€

l


Symmetry and conserved quantities

• The Lagrangian approach provides a useful alternative to a direct formulation using Newton’s equations

• However it also provides the framework in which fundamental questions about the nature of forces and interactions can studied.

• In particular the very close relationship between a symmetry of the Lagrangian and a conserved quantity

• By symmetry we mean an operation – eg coordinate rotation – that leaves the Lagrangian unchanged

€

e.g. L =1

2m( ˙ x 2 + ˙ y 2 ) −

1

2k(x2 + y2 )

is unchanged by rotation about the Oz axis.


Noether’s theorem

€

Consider effect on L of an infinitesimal change in coordinates

qi → qi +εK i (q), where ε is small and K i is a function of qi{ }

L → L +εdL

dε where

dL

dε=

∂L

∂qi

∂qi

∂ε+

∂L

∂ ˙ q i

∂ ˙ q i∂ε

⎛

⎝ ⎜

⎞

⎠ ⎟

i

∑ .

Now ∂qi

∂ε= K i and

∂ ˙ q i∂ε

= ˙ K i; also E - L gives d

dt

∂L

∂ ˙ q i

⎛

⎝ ⎜

⎞

⎠ ⎟=

∂L

∂qi

, so

dL

dε=

d

dt

∂L

∂ ˙ q i

⎛

⎝ ⎜

⎞

⎠ ⎟K i +

∂L

∂ ˙ q i

d

dt(K i )

⎛

⎝ ⎜

⎞

⎠ ⎟=

d

dt

∂L

∂ ˙ q iK i

i

∑ ⎛

⎝ ⎜

⎞

⎠ ⎟

i

∑

hence if dL

dε= 0, then

∂L

∂ ˙ q iK i

i

∑ is conserved.


Examples

Free particle and translational invariance

€

L =1

2m( ˙ x 2 + ˙ y 2 + ˙ z 2 ) (since V = 0)

Consider a translation along one direction, e.g. x → x +ε , with

K x =1, K y = 0, K z = 0. L is clearly unchanged, so

∂L

∂ ˙ q iK i = m ˙ x is conserved, or px∑

Similarly, translations along Oy,Oz⇒ py, pz are conserved.

Thus

L invariant under a translation ⇔ conservation of momentum


Rotational Invariance

€

Consider 2 - D SHM L =1

2m( ˙ x 2 + ˙ y 2 ) −

1

2k(x2 + y2 )

L is unchanged by rotation about the Oz axis,

x → x +εy, y → y −εx, z → z [check this to first order in ε ]

K x = y, K y = −x, K z = 0

Now ∂L

∂ ˙ x = m ˙ x ,

∂L

∂˙ y = m ˙ y , so

∂L

∂ ˙ q iK i = (m ˙ x )y −(m ˙ y )x∑

Compare angular momentum : Jz = r × p( )z= (xpy − ypx )

so (up to a sign)

rotational invariance ⇔ conservation of angular momentum


Conservation of energy

€

Start from the total rate of change of L w.r.t. time

dL

dt=

∂L

∂qi

˙ q i +∂L

∂ ˙ q i˙ ̇ q i

⎧ ⎨ ⎩

⎫ ⎬ ⎭+

∂L

∂t, use E - L

d

dt

∂L

∂ ˙ q i

⎛

⎝ ⎜

⎞

⎠ ⎟=

∂L

∂qi

on 1st termi

∑

dL

dt=

d

dt

∂L

∂ ˙ q i

⎛

⎝ ⎜

⎞

⎠ ⎟˙ q i +

∂L

∂ ˙ q i

d

dt( ˙ q i )

⎧ ⎨ ⎩

⎫ ⎬ ⎭i

∑ +∂L

∂t

=d

dt

∂L

∂ ˙ q i˙ q i

i

∑ ⎧ ⎨ ⎩

⎫ ⎬ ⎭+

∂L

∂r, rearrange to give

∂L

∂t=

d

dtL −

∂L

∂ ˙ q i˙ q i

i

∑ ⎧ ⎨ ⎩

⎫ ⎬ ⎭

. Now ∂L

∂ ˙ q i= m ˙ q i , so

∂L

∂ ˙ q i˙ q i =

i

∑ m ˙ q i2

i

∑ = 2T

so { } → T −V − 2T = −(T +V ), hence

∂L

∂t= −

d

dtT +V{ } =

dE

dt, so E = constant if

∂L

∂t= 0

L invariant under displacement in time ⇔ conservation of energy


The Hamiltonian & Hamilton’s Eqs of Motion

€

The quantity E is also identified with a special function - the Hamiltonian

which has a special significance both for Classical Mechanics, and as a link

to the formulation of quantum mechanics. H is thus defined as

H =∂L

∂ ˙ q i˙ q i − L = pi ˙ q i − L = T +V

i

∑i

∑

where pi is a generalised momentum and defined by pi =∂L

∂ ˙ q i.

The Hamiltonian can be used in an alternative formulation of the equations

of motion, though these are less commonly used in Classical Mechanics

other than in formal treatments. First form the implicit differential of H

dH = pid ˙ q i + ˙ q idpi −∂L

∂qi

dqi −∂L

∂ ˙ q id ˙ q i −

∂L

∂tdt

i

∑


€

But from the definition of pi above and the E - L equations, this simplifies to

dH = ( ˙ q idpi − ˙ p idqi

i

∑ ) −∂L

∂tdt

From this, we can obtain 2N 1st order differential equations of motion

˙ q i =∂H

∂pi

; ˙ p i = −∂H

∂qi

⇔ Hamilton' s Canonical Equations of Motion

(cf the N 2nd order differential equations of motion from the Euler- Lagrange

approach). Note that the second set of Hamilton's equations

˙ p i = −∂H

∂qi

≡ −∂V

∂xi

is equivalent to NII....

Finally, note that a function analogous to the Hamiltonian forms the core

of the Schrödinger Equation in quantum mechanics....


Summary

• Lagrangian L=T-V plus Euler-Lagrange equations give a convenient way of generating equations of motion for complicated dynamical problems

• Noether’s theorem provides a mechanism for finding conserved quantities that follow from symmetry of the Lagrangian

• The Hamiltonian is an alternative formulation, useful in formal treatments, and with an analogue in Schrödinger’s equation of quantum mechanics.

• The method of ‘least action’ can be extended to many other areas of physics

to learn more… try

• Classical Mechanics Short Option – S7 – next year


Sources and further reading

• Any advanced text on classical mechanics– Kibble & Berkshire 5th Ed, IC Press 2005– Goldstein, Poole & Safko 3rd Ed, AW 2002– Fowles & Cassidy, Harcourt Brace 1993– Cowan, RKP 1984

• Feynman Lectures on Physics, Vol II 19-1– Feynman, Leighton & Sands, AW 1964

• For many interesting, almost evangelical, articles on the principles of least action

http://www.eftaylor.com/leastaction.html (Prof Edwin Taylor (MIT) and collaborators)

http://www.eftaylor.com/leastaction.html


Appendix: D’Alembert, Hamilton & Least ActionB

A

tBtA

variation of pathat fixed t

Hamilton considers variationin paths between fixed points at A and B such that variedcoordinates are evaluated atthe same time, so δt = 0.Note that all paths must have the same start and end pointsin space and time.

€

To find the integral to be varied, start from D'Alembert's principle of

'virtual work' - the inertial forces of a system are in equilibrium with the

forces applied to the system - an inertial force is defined by F* = −˙ p (NII)

so (Fi + Fi* ) ⋅δsi = 0, where

i

∑ δs i are virtual displacements compatible with

any inertial constraints of the system.


€

It is more convenient to write D'Alembert's equation using cartesian coordinates

Xi − mi ˙ ̇ x i( )δxi + Yi − mi ˙ ̇ y i( )δyi + Zi − mi˙ ̇ z i( )δzi{ }i

∑ = 0 (1)

Consider just one term in the sum

˙ ̇ x δx =d

dt( ˙ x δx) − ˙ x

d

dt(δx) =

d

dt( ˙ x δx) − ˙ x δ ˙ x =

d

dt( ˙ x δx) −

1

2δ( ˙ x 2 )

Note that d

dt(δx) =δ ˙ x because δt = 0

Clearly similar relations will hold for ˙ ̇ y δy, ˙ ̇ z δz, so eqn(1) becomes

d

dtm ˙ x δx + ˙ y δy + ˙ z δz( ){ } =

1

2δ m ˙ x 2 + ˙ y 2 + ˙ z 2( ){ }+(Xδx +Yδy + Zδz)

On the RHS, 1st is variation in KE (δT ), 2nd is virtual work (δW ). Integrate w.r.t. time

m ˙ x δx + ˙ y δy + ˙ z δz( )[ ]tA

tB = δT +δW( )dt (2)tA

tB

∫

The LHS is zero because δx etc. are zero at the end points; on the RHS

δWdt = − δVdt = −δ Vdt, provided a potential function exists,∫∫∫

so overall (2) gives δ (T −V )dt = 0tA

tB

∫

This identifies L = T −V as the function whose path is to be minimised.


Problems

rθ

€

1. Action integral (p.6). Parameterise the parabola using x = u0t, y = v0t −(v0u0 / R)t 2, and show that

the action integral S = (T −V )dt =1

2mRu0 +

mv02

6u0

−mgv0R2

6u02

.0

tR

∫

Identify the KE and PE terms and discuss how they contribute to S for different values of v0 .

Find the value of v0 that maximises S.

€

2. An object of mass m slides without friction on the surface

of a hemisphere of radius R. Construct the Lagrangian for

motion in plane polar coordinates, as shown, with gravity

acting downwards. Include the reactive force

between hemisphere and object as a potential V (r). Using

the Euler - Lagrange equations, derive the equations of

motion and add the constraint that r = R fixed. Show that if

the object starts from rest at the highest point it will leave

the surface when cosθ =2

3.


x

z

y

r

θ

€

3. An object of mass m can move without friction

on the inner surface of an inverted cone with

half - angle α as shown. In cylindrical polar

coordinates the cone has equation z = r cotα .

Show that the Lagrangian may be written in terms

of the variable θ and the coordinate η = r /sinα -

the distance of the object from the apex of the cone.

L =1

2m ˙ η 2 +( ˙ θ η sinα )2

( ) − mgη cosα .

€

Find the equations of motion in terms of η and θ . Describe qualitatively

possible motions. Show that circular motion is possible with η (or r) fixed

and find the angular frequency ω0. The object is moving in such an orbit

with η =η 0 when it is perturbed slightly so that η =η 0 +ξ . Using the

appropriate original equation of motion, show that the new orbit is SHM

about the radius r0 with angular frequency Ω. Show that

Ω2 =ω02 sin2 α , independent of r0 .


ax

y θ

b

xB

xA

mA

mB

O

€

4. Double pendulum

Masses mA, mB are suspended from O on light

inextensible rods of lengths a, b as shown. Find

the equations of motion for oscillations under

gravity in terms of the angles θ and φ. For

small oscillations - both θ and φ are small,

cos(φ −θ ) =1, sin(φ −θ ) = 0 - it is more

convenient to work in terms of xA and xB . Show

that the equations reduce to

€

mA ˙ ̇ x A + mB ˙ ̇ x B +(mA + mB )xA

g

a= 0, and ˙ ̇ x B + xB

g

b− xA

g

b= 0.

For the case when a = b = l, find the angular frequencies of oscillation

when (a) mB >> mA; (b) mA >> mB . For (b) show that

ω2 =g

l1± mB / mA[ ] (to first order small). If the mass at A is disturbed

from its equilibrium position, describe the subsequent motion.

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