Mechanics Central Force, Lecture Notes - Physics, Study notes of Mechanics

Download Mechanics Central Force, Lecture Notes - Physics and more Study notes Mechanics in PDF only on Docsity!

Mechanics

Physics 151

Lecture 5

Central Force Problem

(Chapter 3)

What We Did Last Time „

Introduced Hamilton’s Principle

„

Action integral is stationary for the actual path

„

Derived Lagrange’s Equations

„

Used calculus of variation

„

Discussed conservation laws

„

Generalized (conjugate) momentum

„

Symmetry – Invariance – Momentum conservation

„

We are almost done with the basic concepts

„

One more thing to cover …

Energy Conservation „

Consider time derivative of Lagrangian

„

Using Lagrange’s equationone can derive

„

Conserved if Lagrangian does not depend explicitly on

t

j^

j

j^

j

j^

j

dq

dq

dL q q t

L

L

L

dt

q

dt

q

dt

t

∑

∑

 j^

j

L

d

L

q

dt

q

⎛

j

j^

j

d

L

L

q

L

dt

q

t

∑

^ Define this as energy function

h q q t

“Energy” Function? „

Does energy function represent the total energy?

„

Let’s try an easy example first

„

Single particle moving along

x

axis

„

How general is this?

j

j^

j L

h q q t

q

L

q ∂

∑

2

mx

L

V x

2 2

h

mx

L

mx

V x

T

V

Total energy

Energy Function „

Energy function equals to the total energy

T

V

if

„

st

condition is satisfied if transformation from

r

i^

to

qj

is

time-independent

„

nd

condition holds if the potential is velocity-independent „

No frictions

Æ

Friction would dissipate energy

„

Let’s look into the 1

st

condition

2

0

h q q t

L

L

L

T

V

2

T

L

0

V

L

and

Kinetic Energy

„

Using the chain rule

„

This wouldn’t work if

because

2

i

i

i

m

T

∑

 r

1 (

i^

i^

n

q

q

t

r

r

Time-independent

i^

i

j

j^

j

i

d

q

d

t

t

q

∑

r

r

r

2

,^

,

i^

i^

i^

i^

i^

i^

i

i^

j^

k^

j^

k

i^

i^

j k

j k

i

j^

k^

j^

k

m

m

m

q q

q q

q

q

q

q

⋅^

∑

∑

∑

∑

∑

r

r

r

r

 r

No

 q

nd

order homogeneous

i^

i

j

j^

j

d

q

dt

q ∂

∑

r

r

1 (

i^

i^

n

q

q

r

r

Central Force Problem „

Consider a particle under a central force

„

Force

F

parallel to

r

„

Assume

F

is conservative

„

V

is function of |

r

| if

F

is central

„

Such systems are quite common

„

Planet around the Sun

„

Satellite around the Earth

„

Electron around a nucleus

„

These examples assume the body at the center is heavyand does not move

O

m

F^ r

V r

F

Two-Body Problem „

Consider two particles without external force

„

r

1

and

r

2

relative to center of mass

„

Lagrangian is

1 m

2 m

2

2

2

1

2

1

i^ 2

i

i

m

m

m

L

V r

=

∑

r

 R

O

CoM

R

1 r

2 r

Motion of CoM

Motion of particles

around CoM

Potential is function of

| r|

r

2

r

1

Strong law of action and reaction

2

1

1

2

m

m

m

r

r

1

2

1

2

m

m

m

r

r

2

2

2

1

2

1

1

2

i^

i

i

m

m m m

m

=

∑

r

r

Hydrogen and Positronium „

Positronium is a bound state of a positronand an electron

„

Similar to hydrogen except

m

p

m

e

„

Potential

V

r

) is identical

„

Turn them into central force problem

„

Spectrum of positronium identical tohydrogen with

m

e

Æ

m

e

e

e

−

p

e

− 2

( )

q

V r

r

=

−

positronium

e^

e^

e

e^

e

m m

m

m

m

hydrogen

p

e

e

p

e

m m

m

m

m

Spherical Symmetry „

Central-force system is spherically symmetric

„

It can be rotated around any axis through the origin

„

Lagrangian

doesn’t depend on the

direction

„

Angular momentum is conserved

„

Direction of

L

is fixed

„

by definition

Æ

r

is always in a plane

„

Choose polar coordinates

„

Polar axis = direction of

L

const

×

L

r

p

r

L

O

r

L

,^

r

r

r

r

r

Azimuth

Zenith = 1/

π

2 (

L

T

V r

 r

Angular Momentum „

θ

is cyclic. Conjugate momentum

p

θ

conserves

„

Alternatively

„

Kepler’s 2

nd

law

„

True for any central force

2

2

2

m^2

L

T

V

r

r

V r

2

const

L

p

mr

l

θ

Magnitude of

angular momentum

2

const

dA

r

dt

d

r dA

Areal velocity

Radial Motion „

Lagrange’s equation for

r

Æ

„

Derivative of

V

is the force

„

Using the angular momentum

l

2

2

2

m^2

L

T

V

r

r

V r

2

d

V r

mr

mr

dt

r

V r

f

r

r

2

mr

mr

f

r

Central force

Centrifugal force

2

l

mr

2

3

l

mr

f

r

mr

We know how to integrate this.But we also know what we’llget by integrating this

Degrees of Freedom „

A particle has 3 degrees of freedom

„

Eqn of motion is 2

nd

order differential

Æ

6 constants

„

Each conservation law reduces one differentiation

„

By saying “time-derivative equals zero”

„

We used

L

and

E

Æ

4 conserved quantities

„

Left with 2 constants of integration =

r

0

and

0

„

We don’t have to use conservation laws

„

It’s just easier than solving all of Lagrange’s equations

Qualitative Behavior „

Integrating the radial motionisn’t always easy

„

More often impossible…

„

You can still tell general behavior by looking at

„

Energy

E

is conserved, and

E

V’

must be positive

„

Plot

V’

r

) and see how it intersects with

E

2

2

l

r

E

V r

m

mr

2

2

l

V

r

V r

mr

′^

Quasi potential including

the centrifugal force

E

V

r ′

2

mr

E

V

r ′

2

mr

E

V

r ′