Variational Principle in Physics: From Lagrange to Hamilton, Study notes of Physics

Lecture 5 on the variational principle in physics, covering lagrange examples, hamilton's variational principle, and the calculus of variations. The historical development of variational principles and their significance in minimizing the time integral of energy differences for physical systems.

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Pre 2010

Uploaded on 03/28/2010

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Lecture 5 Outline - Variational Principle
one final Lagrange example (Section 1.6)...
Hamilton’s Variational Principle (Section 2.1)
Calculus of Variations (Section 2.2)
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Lecture 5 Outline - Variational Principle

one final Lagrange example (Section 1.6)...

Hamilton’s Variational Principle (Section 2.1)

Calculus of Variations (Section 2.2)

Another Lagrange Example

Time-dependant rotating (straight) wire

space

Hamilton’s Variational Principle

derivable from scalar potentials (except constraints!)For monogenic systems in which all forces are

which may depend on

q i , dq

i

dt

, t

The motion of the system from time

t 1

to

t 2

is such that

I

t 2

t 1

L

dt

t 2

t 1 ( T − V )

(^) dt

has a stationary value for the actual path of motion:

δI

δ

t 2

t 1 L ( q 1

, q

2 ,... ,

dq

1

dt

dq

2

dt

,... , t

dt

Sometimes

I

is called the action (units of energy/time).

Variational principle’s appeared in different forms

Calculus of variations setup

Assume correct path

y ( x,

with set of paths

y ( x, α

y ( x,

αη

x

)

by defn

η ( x

1 ) =

η ( x

2 ) = 0

(nice functions

y ( x,

η ( x

)

)

J

α

) =

x 2

x 1

f (^) ( y ( x, α

dx dy

x, α

, x

dx

want stationary point when

dαdJ

α

since

α

is independent of

x

, differentiate under the sign.

dα dJ

x 2

x 1

dα df

dx

x 2

x 1

[

∂y ∂f

∂α∂y

∂ ∂f

˙y

˙y

∂α

∂x ∂f

∂α∂x

]

dx

x 2

x 1

[

∂y ∂f

∂α∂y

∂ ∂f

˙y

2 y

∂x∂α

]

dx

now second part we integrate by parts

Calculus of variations technique

integrating by parts:

vdu

uv

udv

with

u

∂α∂y

and

v

∂ ∂f

˙y

so

dxdv

d

dx

∂ ∂f (^) ˙y )

thus

dv

d

dx

∂ ∂f (^) ˙y )

dx

x 2

x 1

∂^ ∂f

˙y

2 y

∂x∂α

dx

[

∂ ∂f

˙y

∂α∂y

]

x 2

x 1 − ∫ x 2

x 1

d

dx

∂ ∂f

y

)

∂α ∂y

dx

but

∂α∂y

x

1 ) =

n

( x

1 ) =

∂α∂y

x

2 ) =

n

( x

2 ) = 0

dα dJ

x 2

x 1

[

∂y ∂f

d

dx

∂ ∂f

˙y )]

∂α ∂y

dx

thus at the stationary value of

α

dα dJ

α

x 2

x 1

[

∂y ∂f

d

dx

∂ ∂f

y

)]

∂α∂y

α

dx

but

∂α∂y

η ( x )

is an arbitrary function!

Simple c of v example

Shortest distance between two points (in a plane)