1549380323-Statistical Mechanics Theory and Molecular Simulation

20 Classical mechanics

+U(r 1 (q 1 ,...,q 3 N),...,rN(q 1 ,...,q 3 N)). (1.6.10)

Given the Hamiltonian (as a Legendre transform of the Lagrangian), one can obtain
the equations of motion for the system from the Hamiltonian using

q ̇α=

∂H

∂pα

, p ̇α=−

∂H

∂qα

. (1.6.11)

Eqns. (1.6.11) are known asHamilton’s equations of motion. Whereas the Euler–
Lagrange equations constitute a set of 3Nsecond-order differential equations, Hamil-
ton’s equations constitute an equivalent set of 6N first-order differential equations.
When subject to the same initial conditions, the Euler–Lagrange and Hamiltonian
equations of motion must yield the same trajectory.
Hamilton’s equations must be solved subject to a set of initial conditions on the
coordinates and momenta,{q 1 (0),...,q 3 N(0),p 1 (0),...,p 3 N(0)}. Eqns. (1.6.11) are com-
pletely equivalent to Newton’s second law of motion. In order to see this explicitly, let
us apply Hamilton’s equations to the simple Cartesian Hamiltonian of eqn. (1.6.3):

r ̇i=

∂H

∂pi

=

pi

mi

p ̇i=−

∂H

∂ri

=−

∂U

∂ri

=Fi(r). (1.6.12)

Taking the time derivative of both sides of the first equation and substituting the
result into the second yields

̈ri=

p ̇i

mi

p ̇i=mi ̈ri=Fi(r 1 ,...,rN), (1.6.13)

which shows that Hamilton’s equations reproduce Newton’s second law of motion. The
reader should check that the application of Hamilton’s equations to asimple harmonic
oscillator, for whichH=p^2 / 2 m+kx^2 /2, yields the equation of motionm ̈x+kx= 0.
Hamilton’s equations conserve the total Hamiltonian:

dH

dt

= 0. (1.6.14)

SinceHis the total energy, eqn. (1.6.14) is just the law of energy conservation. In
order to see thatHis conserved, we simply compute the time derivative dH/dtvia
the chain rule in generalized coordinates:

dH

dt

=

∑^3 N

α=1

[

∂H

∂qα

q ̇α+

∂H

∂pα

p ̇α

]