1549380323-Statistical Mechanics Theory and Molecular Simulation

12 Classical mechanics

In order to verify thatEis a constant, we need only show that dE/dt= 0. Differen-
tiating eqn. (1.4.11) with respect to time yields

dE

dt

=

∑N

i=1

mir ̇i· ̈ri+

∑N

i=1

∂U

∂ri

·r ̇i

=

∑N

i=1

r ̇i·

[

mi ̈ri+

∂U

∂ri

]

=

∑N

i=1

r ̇i·[mi ̈ri−Fi]

= 0, (1.4.12)

where the last line follows from the fact thatFi=mi ̈ri.
The power of the Lagrangian formulation of classical mechanics lies inthe fact that
the equations of motion in an arbitrary coordinate system, which might not be easy to
write down directly from Newton’s second law, can be derived straightforwardly via the
Euler–Lagrange equation. Often, the standard Cartesian coordinates are not the most
suitable coordinate choice for a given problem. Suppose, for a givensystem, there exists
another set of 3Ncoordinates,{q 1 ,...,q 3 N}, that provides a more natural description
of the particle locations. These coordinates are known asgeneralized coordinates, and
they can be related to the original Cartesian coordinates,r 1 ,...,rN, via acoordinate
transformation

qα=fα(r 1 ,...,rN), α= 1,..., 3 N. (1.4.13)

Thus, each coordinateqα is generally a function of theN Cartesian coordinates,
r 1 ,...,rN. It is assumed that the coordinate transformation eqn. (1.4.13) has a unique
inverse

ri=gi(q 1 ,...,q 3 N), i= 1,...,N. (1.4.14)

In order to determine the Lagrangian in terms of generalized coordinates, eqn.
(1.4.14) is used to compute the velocities via the chain rule:

r ̇i=

∑^3 N

α=1

∂ri

∂qα

q ̇α, (1.4.15)

where∂ri/∂qα≡∂gi/∂qα. Substituting eqn. (1.4.15) into eqn. (1.4.4) gives the kinetic
energy in terms of the new velocities ̇q 1 ,...,q ̇ 3 N:

K ̃(q,q ̇) =^1

2

∑^3 N

α=1

∑^3 N

β=1

[N

∑

i=1

mi

∂ri

∂qα

·

∂ri

∂qβ

]

q ̇αq ̇β