PHYS 705: Classical Mechanics - George Mason Universitycomplex.gmu.edu/ Central Force I.pdf · PHYS 705: Classical Mechanics Central Force Problems I 1. ... First Step: Central force

PHYS 705: Classical MechanicsCentral Force Problems I

1

Two-Body Central Force Problem

- Based his 3 laws on observational data from Tycho Brahe

- Formulate his famous 3 laws:

- Orbit of each planet is an ellipse with sun at one of its foci

- Equal areas swept out in equal time by an orbit

- The ratio is the same for all planets, where is the period and R

is the semi-major axis

-All these results were obtained through amazing sheer mathematical efforts.

Historical Background: Kepler’s Laws on celestial bodies (~1605)

2 3R

Before we get back to this important celestial application, we will address

the general problem…

2

Kepler’s Notes

3


Set up of the general problem: 2 masses interact via forces directed along

the line that connects them (central force): strong form of 3rd law

First Step: Central force problems can be reduced to an effective 1-body problem:

1r2r

r

2 1 r r r' '2 1 r r r

R

'1r

'2r

CM

'1 1

'2 2

r R r

r R r

Instead of using , use (CM & relative position)1 2&r r &R r

Change to generalized coordinates: CM position R and relative position r,

4


From def. of CM:

1 1 2 2

' '1 1 2 2

i ii

M m m m

m m

M

R r r r

R r R r

R 1 2m m R ' '1 1 2 2m m r r 1r

2rr

R

'1r

'2r

CM' '

1 1 2 2 0m m r r

Combining it with , we have' '2 1 r r r

' ' '2 21 2 1

1 1

'2 21

1 1

' 21

1 2

1

m mm m

m mm m

mm m

r r r r

r r

r r ' ' 12 1

1 2

mm m

r r r rand

5


Now, form the Lagrangian:

(express in terms of only)

2 22 ' '1 21 22 2 2

CM aboutCM

T T T

m mM

R r r

We have a central force so that U(r) depends on the relative distance r only.

&R r

6

2 22 21 2 2 1

2 21 2 1 2

2 21 2

1 2

12 2

12 2

m m m mMm m m m

m mMm m

R r

R r

' 21

1 2

mm m

r r

' 12

1 2

mm m

r r


So,2 21 2

1 2

1 ( )2 2

m mML T U U rm m

R r

Note that R does not appear in L (cyclic) so that EL equation for R will only give:

M constR

CM is either stationary or moving uniformly.

Pick an inertial ref. frame (CM frame) in which CM is not

moving and we can ignore the 1st term in L.

The result is then:

2 1 2

1 2

1 ( ), where 2

m mL U rm m

r Same as a single particle

with mass moving in U(r).

0R

7

Motion of and in CM frame

Reference Frames for Central Force Problems

Motion in the relative coordinate frame with (3 dofs) :

2m

(fixed),R r

Effective one body problem with reduced mass and relative position (dist from to ).2m1m

r

1m

r

' 12

1 2

mm m

r r

' 21

1 2

mm m

r r

1m

2m

2 'r

CM1 'r

1m

8

Reference Frames for Central Force Problems

Motion in original space with 6 dofs (not all independent):

R

r

1m

2m

CM

2r

1r

1 2,r r

and circle each

other in space as their

CM move with constant

velocity along a fixed

direction.

2m1m

9


The reduced problem is a much simpler problem with 3 dofs instead of 6.

Also, we often have (e.g., sun-earth)1 2m m

the CM is so close to and

Then,1 2 1 2

21 2 1

m m m m mm m m

1m '2r r

10


Further reduction of the two-body central force problem:

So, we have,

0 0danddt

LN N

Since U(r) is central, i.e., F always directs toward CM along r,

constLor

11

L will points in a constant direction fixed by initial condition.

Then, =const, and r & p must be to L always, i.e., motion

has to be planar.

If (trivial case), then the trajectory is a 1D motion through

the origin.

L r p

0L

/ /

(Alternatively, as we will see later, is cyclic !)


With all these considerations, it is natural to use spherical coordinates

and to orient the polar axis with L.

Then, we can analyze the problem entirely in a 2D polar plane .

(Recall, in polar coord.) ˆˆr r v r θ

2 2 2 ( )2mL r r U r (we will simply call m= from now on)

12

Lastly, using the fact that U(r)= U(r) depends only on the magnitude of the

distance, we can show that the Lagrangian is effectively one-dimensional.

To see that, start with L in :

L

,r


Since U(r) is central, is cyclic (does not appear) in L.

Replacing with the constant in the Lagrangian, we then have an effectively

1D system in r only:

2Lp mr l const

2l mr

13

22

2 ( )2 2m lr U r

mr

And, the generalized momentum with respect to is conserved,

l

2 2 2 ( )2 2m mL r r U r

Note: is ignorable but it is NOT a constant. r and change while l is fixed

in time. Motion is still in 2D in the CM frame*.


[One can recognize this as the constant ang momentum: .]

Now, we calculate the EOM for both r and :

This gives:

0d L Ldt

(L is cyclic in )

2 2 2 ( )2mL r r U r

0L

2L mr

2mr l const

2( )l I mr

14

(This is the equivalent statement about the conservation of L as before.)


Note that area swept out by r with an

infinitesimal rotation d is

Connecting back to Kepler’s 2nd law…

212

dA r d

Thus, Eq (*) establishes Kepler’s 2nd Law!

r drd

And, 2 21 12 2

dA dr rdt dt

From EL equation, we have 2 0d mrdt

21 0 (*)2

d rdt

15

Combining with the equation , we have,


Now, we are back to the EOM for r:

r

gives0d L Ldt r r

2L dUmrr dr

L mr

r

2l mr

2 0dUmr mrdr

2

2

2

3

0l dUmr mrmr dr

l dUmrmr dr

2 2 2 ( )2mL r r U r

16

U doesn’t depend on and does not depend on t explicitly

so that h (Jacobi Integral/energy function) is conserved.


We have seen that the equation give one constant of motion l.

0Lt

jq

2 2 2 2

22 2 2 2

2

12

1 1 12 2 2 2

jj j

Lh q Lq

r mr mr m r r U

lmr mr U mr U Emr

We can get another constant of motion E since:

( )jqr

so that h = E total energy is conserved.

check:

17


Here is another way to argue that E is conserved. Let start with the r EOM,

2 2

3 22dU l d lmr Udr mr dr mr

Multiplying on both sides,r2

222 2

d m d lmrr r drUdt dr mr dt

So,2

22 ( ) 0

2 2d m lr U rdt mr

E

E is conserved!

(reverse chain rule)

18

22

22 2dd m lr U

t dtd mr


Summary: We get 2 EOMs and 2 integrals of motion (l, E) for this problem.

2

3

dU lmrdr mr

2

22 ( )

2 2m lE r U r

mr

Note: The E equation effectively is the 1st integral of the equation. We can

rewrite it explicitly as,

2mr l

r

2

2

2 ( )2

lr E U rm mr

2 2 2 ( )2mL r r U r

19


Integrating once more time gives:

0

2

2

'

2 ( ')2 '

r

r

drtlE U r

m mr

With initial condition

r0 at t=0

Similar, we can integrate the equation and get,

02

0

( )( )

t

ldttmr t

With initial condition

0 at t=0

So, the problem is completely solved with t(r) and (t). The problem has been

reduced to a quadrature (doing the integrals).

20

Graphical Analysis of Central Force Problem

Using the concept of an effective potential, one can get an useful

qualitative understanding of the problem without actually integrating!

Let consider the r equation:

The last two terms combined can be

considered as an effective force

This looks like a 1D problem: a single particle moving in 1 dimension under the

influence of an effective force,

2

3

dU lmrdr mr

2

3'( ) dU lf rdr mr

'( )f r

21

There is now a “effective” extra force term that is not directly associated with

the two objects:


No angular momentum, then, it is really a 1D problem. (uninteresting case)

Case 1 :

2

3

dU lmrdr mr

ˆ ˆˆwhere is the component of velocity v v r r θ r θ

0l

Case 2 :0l

2 2 22 2 2 22

3 3 3

mr m r r r vl m mmr mr mr r r

So, we recognize this as the centrifugal acc. term in a co-rotating ref. frame.

22

We will consider three examples:


For the case, we can combine the RHS into an effective potential term:

3

2

2

2 '( )2

dU l d dmrdr mr

l U rUdr dmr r

where

0l

2

3 4

2

.

. 3

. / 2

a U k r f k rb U a r f a rc U kr f kr

Note: zero point for U(r) is different in these three cases.

2

2'( ) ( )2

lU r U rmr

(gravitational, EM)

(Hooke’s law: HMO)

23

Central Force Problem: Inverse Square Force

Example a: inverse-square force 2

2'( )2

k lU rr mr

For a fixed l, we can plot:

2

22lmr

kr

'( )U r

r

, 'U U

2

2

1. , ' 02. , ' 0

3. 1 dominates 1

4. 1 dominates 1 0

U U as rU U as r

r r as r

r r as r

Note the asymptotic:

2 2

'' 2a well in the m ddle

0i

U U k r as rU l mr as r

This gives,

24


Now, for a fixed value of total E > 0 & consider what kind of orbit is possible?

22 2

2

1 1'( )2 2 2

k lE mr U r mrr mr

Recall

Notice that

21 '( ) 02

mr E U r since the radial KE can’t be negative.

min mins.t. '( ), 0r E U r r

r can’t go beyond rmin, the turning point [where ]min'( )E U r

The orbit is to the line from the origin to the orbit:r

v

0 herer

2since 0r and r

25


'( )U r

r

E

min ( )r turning pt

The system becomes less

influence by U(r) and more

like free particles.

(far away)

' 0' ( )

as rUE U E const

r const

rE > 0 (unbounded orbit)

(close by)At turning pt, there is no

instantaneous radial motion.

All motion is angular.

21 '( )2

mr E U r

26


'U

r

E

(close by)

gets big gets big

All motion is angular here

U

minr r

2r

(far away)

and and

r

20r 0

All motion is radial here

2

2 2

' 2

' 2

E U mr

U U mr

(radial KE)

(angular KE)

E

'UU

2r

2 2r

27

r const


In the relative position space r, the orbit will look like:

0 and r const (far away)

0r

here is the point of closest approach (turning point)

origin

minr

is large

as if it were a free particle

28


- As we have seen, if E > 0, for r large, the particles act like free particles

with T = E.

- What happens to the orbit when E < 0? orbits will be bounded

- Now, if E = 0, for r large, the particles will be stopped with T = E = 0.

'U

rE

1r 2r

turning pts again 0r -Orbit is bounded between r1 & r2

-These distances are called

“apsides” or “apsidal distances”

-Similar to previous turning

points, orbits are tangent to

circles: r = r1 & r = r2

- is largest at inner radius (U’-U

is largest at r1)

U

29

r

E

0r


So, for bounded orbits (E < 0), the orbits in r space look like:

1r 2r

(for the inverse-square force)

We will talk more about bounded

orbits later.

In particular, cond for closed orbits(fig. 3.7 is wrong, curvature must be concave for

attractive forces)

If E is at the minimum (r= r0) , the bounded orbit will

have a constant radius r0 and it will be a circular orbit.

30


'U

r

turning pts (apsides) 0r

U

2 0 ( )E E unbounded

origin

minr

minr2E

1E

1r 2r

1 0 ( )E E bounded

1r 2r

0E

0 ( )E E circular

31

Central Force Problem: Stronger Attractive U

Example b: Stronger Attractive Potential2

3 2'( )2

a lU rr mr

3 43U a r f a r

This gives,

'U

r

1E

1r

U

2

22lmr

2E

3E

2r

3

1 dominatesr

2

1 dominatesr

1. E=E1: orbit is unbounded

2. E=E2: orbit is bounded away

from r2 or bounded within r1

particles can collide

3. E=E3: orbit is bounded

within r3 and it can go thru

origin particles can collide

3r

32


What is the condition for the particle to go thru the origin?

22

2

1 '( ) ( )2 2

lmr E U r E U rmr

From the total energy equation, we have,

Let, ( ) nU rr

For the origin to be accessible, near the origin. Thus, we must have,

2

2 02n

lEr mr

0r 2

22 2n

lErr m or

33


As the orbit gets closer to the origin , LHS 0

Then, this is true if2 2

0, . .,2 2l li em m

For this inequality to be satisfied in the limit , RHS must be negative for

r small.

0r

22

2 2n

lErr m

0r

This can certainly be true for , attractive force

stronger than inverse-square type.

2n

If n = 2, then must hold at the limit 2

2

2lErm

0r

34

'U

r

EU

2

22lmr

Central Force Problem: Hooke’s Law

Example c: Hooke’s Law/SHO

2 2

2'( )2 2

kr lU rmr

2 2U kr f kr

U’ is the green line

Motion is 1D on a line and

is simple harmonic

0 : 'l U U

0 :l Motion is on a 2D plane but

orbit will be bounded and

typically elliptic

(Homework problem)

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PHYS 705: Classical Mechanics - George Mason Universitycomplex.gmu.edu/ Central Force I.pdf · PHYS 705: Classical Mechanics Central Force Problems I 1. ... First Step: Central force