7/25/2019 AD Homework 4 - Solution

http://slidepdf.com/reader/full/ad-homework-4-solution 1/2

Analytical Dynamics

Fall 2015 - Assignment 4 solutions

!"

We’re dealing with a central potential, so we can always choose a coordinate system

where the azimuthal equations are trivial. This reduces the Lagrangian to:

L = 12m ( !r 

2+ r 

2 !!2)+

k 

r e !r /a  

Energy and angular momentum are conserved:

d 

dt mr 

2 !! ! !l = 0 " E = 12m  !r 

2+

l 2

2mr 2#

k 

r e #r /a ,

So we obtain from the above:

2

m 

dr 

E  !(l 2/2mr 2)+(k /r )e !r /a 

= dt = (mr 2/l )d !

"  ! = !0+

l 

2m 

dr 

r 2 E  !

  l 2

2mr 2+

k 

r e 

!r /a #

$

%%%

&

'

(((

!1/2

) 

 

Now since: e !r /a = 1!r /a +O (r 2/a 2) ,

To first order in r /a, defining !! E "(k /a )  and taking !0= 0  we have:

! ! l 

2m 

dr 

r 2   "" l 

2

2mr 2+ k 

r #$%%%

&'(((

"1/2

)    "

Which is now reduced to the Kepler problem; the result can be expressed as:

r =a (1!e 

2)

1+e cos!, a "!

k 

2", e " 1+

2"l 2

mk 2

,

Where e < 1 for elliptic orbits; a circular orbit corresponds to:

e = 0!E =k 

a 

"

mk 2

2l 

2.

2.

From Homework 1, problem 3 we know that at the boundary:

sin!

sin"= 1+

V 0

E = n  

7/25/2019 AD Homework 4 - Solution

http://slidepdf.com/reader/full/ad-homework-4-solution 2/2

Physically, the situation is such as in the following diagram:

From which we immediately get:

! = 12!+ "" sin! = sin(1

2!+ ")= sin 1

2!cos"+ cos 1

2!sin",

  or: n #1= sin 1

2!cot"+ cos 1

2!

 

We also see that: s = a sin! . Substituting in the above and solving for s:

s =an sin 1

2!

1+n 2"2n cos 1

2!

,

From which we obtain:

s  ds 

d != 1

2

ds 2

d != 1

2(an )2

 d 

d !

sin2 12!

1+n 2"2n cos 1

2!

  =(an )2 sin 1

2!

2[1+n 2"2n cos 12!]2cos 1

2

! 1+n 2"2n cos 1

2

!

( )"n sin2 1

2

!

{ }  = sin 1

2!

(an )2(n cos 12!"1)(n "cos 1

2!)

2[1+n 2"2n cos 1

2!]2

# ! (!)=s 

sin!

ds 

d !=

(an )2(n cos 12!"1)(n " cos 1

2!)

4cos 12![1+n 

2"2n cos 1

2!]2

 

Extreme values correspond to ! = 0  and ! = 1

2 " , so that !

min= 0, n 

"1= cos 1

2!

max;

Defining y ! cos 1

2"# dy =$ 1

2sin 1

2"d " ,

We obtain for the total cross-section:

! T = 2" ! (!)sin!d !

0

!max

"  =#2"(an )2  (ny #1)(n #y )

(1+n 2 #2ny )2

dy 1

1/n 

"   

Or, with a final substitution x  ! 1+n 2"2ny # dx ="2ndy :

! T =   1

4 "a 

2 [(n 2 !1)2x !2 !1]dx (n !1)2

n 2!1

"  =! 14 "a 

2 (n 2 !1)2x !1+ x #

$%  &

'((n !1)2

n 2!1

= "a 2 .