Fundamentals of Plasma Physics

3.2 Hamilton-Lagrange formalism v. Lorentz equation 63

1. containsallinformation about the particle dynamics for a given situation,


2. depends only on generalized coordinatesQi(t),Q ̇i(t)appropriate to the problem,


3. possibly has an explicit dependence on timet.
   If such a functionL(Qi(t),Q ̇i(t),t)exists, then information on particle dynamics is
   retrieved by manipulation of theaction integral

S=

∫t 2

t 1

L(Qi(t),Q ̇i(t),t)dt. (3.2)

This manipulation is based on d’Alembert’s principle of least action. According to this
principle, one considers the infinity of possible trajectories a particlecouldfollow to get
from its initial positionQi(t 1 )to its final positionQi(t 2 ), and postulates that the trajectory
actuallyfollowed is the one which results in theleastvalue ofS. Thus, the value ofS
must be minimized (note thatShere is action, and not entropy as in the previous chapter).
MinimizingSdoes not give the actual trajectory directly, but rather gives equations of
motion which can be solved to give the actual trajectory.
MinimizingSis accomplished by considering an arbitrary nearbyalternativetrajectory
Qi(t) + δQi(t)having thesamebeginning and end points as the actual trajectory, i.e.,
δQi(t 1 ) =δQi(t 2 ) = 0.In order to make the variational argument more precise,δQiis
expressed as
δQi(t) =ǫηi(t) (3.3)
whereǫis an arbitrarily adjustable scalar assumed to be small so thatǫ^2 <ǫandηi(t)is a
function oftwhich vanishes whent=t 1 ort=t 2 but is otherwise arbitrary. Calculating
δSto second order inǫgives

δS =

∫t 2

t 1

L(Qi+δQi,Q ̇i+δQ ̇i,t)dt−

∫t 2

t 1

L(Qi,Q ̇i,t)dt

=

∫t 2

t 1

L(Qi+ǫηi,Q ̇i+ǫη ̇i,t)dt−

∫t 2

t 1

L(Qi,Q ̇i,t)dt

=

∫t 2

t 1

(

ǫηi

∂L

∂Qi

+

(ǫηi)^2

2

∂^2 L

∂Q^2 i

+ǫη ̇i

∂L

∂Q ̇i

+

(ǫη ̇i)^2

2

∂^2 L

∂Q ̇^2 i

)

dt. (3.4)

Suppose that the trajectoryQi(t)is the one that minimizesS.Any other trajectory must
lead to a higher value ofSand soδSmust be positive for any finite value of ǫ. Ifǫis
chosen to be sufficiently small, then the absolute values of the terms oforderǫ^2 in Eq.(3.4)
will be smaller than the absolute values of the terms linear inǫ.The sign ofǫcould then
be chosen to makeδSnegative, but this would violate the requirement thatδSmust be
positive. The only way out of this dilemma is to insist that the sum of the terms linear inǫ
in Eq.(3.4) vanishes so thatδS∼ǫ^2 and is therefore always positive as required. Insisting
that the sum of terms linear inǫvanishes implies

0 =

∫t 2

t 1

(

ηi

∂L

∂Qi

+ ̇ηi

∂L

∂Q ̇i

)

dt. (3.5)

Usingη ̇i= dηi/dtthe above expression may be integrated by parts to obtain

0 =

∫t 2

t 1

(

ηi

∂L

∂Qi

+

dηi

dt

∂L

∂Q ̇i

)

dt

=

[

ηi

∂L

∂Q ̇i

]t 2

t 1

+

∫t 2

t 1

{

ηi

∂L

∂Qi

−ηi

d

dt

(

∂L

∂Q ̇i

)}

dt. (3.6)