QMGreensite_merged

1.1. BASEBALL,F=MA,ANDTHEPRINCIPLEOFLEASTACTION 9

andintegralsbysums
∫t 0 +∆t

t 0

dtf(t)→

N∑− 1

n=0

!f(tn) (1.5)

wheref(t)isanyfunctionoftime. Asweknowfromelementarycalculus,theright
handsideof(1.4)and(1.5)equalsthelefthandsideinthelimitthat!→0,which
isalsoknownasthecontinuumlimit.
Wecannowapproximatethelawsofmotion,byreplacingtime-derivativesin(1.2)
bythecorrespondingfinitedifferences,andfind

xn+1 = xn+

(p

n

m

)

!

pn+1 = pn−

(

dV(xn)

dxn

)

! (1.6)

Theseareiterative equations. Givenpositionxandmomentumpattimet= tn,
wecan use(1.6)tofindthepositionandmomentumattimet=tn+1. Thefinite
differenceapproximationofcourseintroducesaslighterror;xn+1andpn+1,computed
fromxnandpnby(1.6)willdifferfromtheirexactvaluesbyanerrorof order!^2.
Thiserrorcanbemadenegligiblebytaking!sufficientlysmall.
Itisthenpossibleusethecomputertofindanapproximationtothetrajectoryin
oneoftwoways: (i)the”hit-or-miss”method;and(ii)themethodofleastaction.

• TheHit-or-MissMethod

Theequationsofmotion(1.2)require asinputbothaninitialposition,inthis
casex 0 =Xin,andaninitialmomentump 0 whichissofarunspecified.Themethod
istomakeaguessfortheinitialmomentump 0 =P 0 ,andthenuse(1.2)tosolvefor
x 1 ,p 1 , x 2 ,p 2 ,andsoon,untilxN,pN. IfxN ≈Xf, thenstop;the set{xn}isthe
(approximate)trajectory. Ifnot,makeadifferentguessp 0 =P 0 ′,andsolveagainfor
{xn,pn}. Bytrialanderror,onecaneventuallyconvergeonaninitialchoiceforp 0
suchthatxN≈Xf. Forthatchoiceofinitialmomentum,thecorrespondingsetof
points{xn},connectedbystraight-linesegments,givestheapproximatetrajectoryof
thebaseball. ThisprocessisillustratedinFig. [1.2].

• TheMethodofLeastAction

Letsreturntothe2nd-orderformofNewton’sLaws,writtenineq. (1.1). Again
using(1.4)toreplacederivativesbyfinitedifferences,theequationF=maateach
timetnbecomes
m
!

{x

n+1−xn

!

−

xn−xn− 1

!

}

=−

dV(xn)

dxn

(1.7)

Theequationshavetobesolvedforn= 1 , 2 ,...,N−1,withx 0 =XinandxN=Xf
keptfixed. Nownoticethateq. (1.7)canbewrittenasatotalderivative

d

dxn

{

1

2

m

(xn+1−xn)^2

!

+

1

2

m

(xn−xn− 1 )^2

!

−!V(xn)

}

= 0 (1.8)