1549380323-Statistical Mechanics Theory and Molecular Simulation

(jair2018) #1
The action integral 31

satisfy
δQ(t 1 ) =δQ(t 2 ) = 0, δQ ̇(t 1 ) =δQ ̇(t 2 ) = 0. (1.8.3)


The variation in the action is defined to be the difference


δA=

∫t 2

t 1

L(Q(t) +δQ(t),Q ̇(t) +δQ ̇(t)) dt−

∫t 2

t 1

L(Q(t),Q ̇(t)) dt. (1.8.4)

This variation must vanish to first order in the path differenceδQ(t). Expanding to
first order, we find:


δA=

∫t 2

t 1

L(Q(t),Q ̇(t)) dt+

∫t 2

t 1

∑^3 N


α=1

[


∂L


∂qα
δqα(t) +

∂L


∂q ̇α
δq ̇α(t)

]


dt


∫t 2

t 1

L(Q(t),Q ̇(t)) dt

=


∫t 2

t 1

∑^3 N


α=1

[


∂L


∂qα

δqα(t) +

∂L


∂q ̇α

δq ̇α(t)

]


dt. (1.8.5)

We would like the term in brackets to involve onlyδqα(t) rather than bothδqα(t) and
δq ̇α(t) as it currently does. We thus integrate the second term in brackets by parts to
yield


δA=

∑^3 N


α=1

∂L


∂q ̇α

δqα(t)



∣∣


t 2

t 1

+


∫t 2

t 1

∑^3 N


α=1

[


∂L


∂qα


d
dt

(


∂L


∂q ̇α

)]


δqα(t) dt. (1.8.6)

The boundary term vanishes by virtue of eqn. (1.8.3). Then, sinceδA= 0 to first
order inδqα(t) at a stationary point, and each of the generalized coordinatesqαand
their variationsδqαare independent, the term in brackets must vanish independently
for eachα. This leads to the condition


d
dt

(


∂L


∂q ̇α

)



∂L


∂qα

= 0, (1.8.7)


which is just the Euler–Lagrange equation. The implication is that thepath for which
the action is stationary is that which satisfies the Euler–Lagrange equation. Since the
latter specifies the classical motion, the path is a classical path.
There is a subtle difference, however, between classical paths that satisfy the end-
point conditions specified in the formulation of the action and those generated from
a set of initial conditions as discussed in Sec. 1.2. In particular, if an initial-value
problem has a solution, then it is unique, assuming smooth, well-behaved forces. By
contrast, if a solution exists to the endpoint problem, it is not guaranteed to be a
unique solution. However, it is trivial to see that if a trajectory withinitial conditions
Q(t 1 ) andQ ̇(t 1 ) passes through the pointQ 2 att=t 2 , then it must also be a solution
of the endpoint problem. Fortunately, in statistical mechanics, this distinction is not
very important, as we are never interested in the unique trajectory arising from one

Free download pdf