Physical Chemistry , 1st ed.

Since we are confining motion to one dimension, only one momentum needs

to be defined. Using the Lagrangian defined above, we can substitute into the

one-dimensional Hamiltonian

Hpx ̇L

(See equation 9.11.) Substituting for pand L:

Hmx ̇x ̇^12 mx ̇^2 ^12 kx^2 

^12 mx ̇^2 ^12 kx^2

where in the last equation we have combined the first two terms. Because we

will need to solve the differential equations given by equations 9.14 and 9.15,

it will be easier for the first derivative if we rewrite the Hamiltonian as

H

2

1

m

p^2 

1

2

kx^2

Applying equation 9.14 to this expression, we get







H

p

 (^2) 
2

1

m

p

m

1

mx ̇x ̇

which is what this derivative should be. We have not gotten anything new out

of this expression. However, upon evaluating the derivative in 9.15 using the

rewritten form of the Hamiltonian, we find:







H

x

kx

which, by equation 9.15, must equal p ̇:

kxp ̇

or

kx

d

d

t

p

kx

d

d

t

mx ̇

kxmd

dt

(^2) x
 2
This can be rewritten as

d
dt
(^2) x
 2 m
k
x 0
which is the same differential equation that we found upon applying both
Newton’s and Lagrange’s equations of motion.
Example 9.1 illustrates that the three different equations of motion produce
the same description for the motion of a system, only by different routes. Why
present three different ways of doing the same thing? Because all three ways are
not equally easy to apply to all situations! Newton’s laws are most popular for
straight-line motion. However, for other systems (like systems involving revo-
lution about a center) or when knowledge of the total energy of a system is
9.2 Laws of Motion 247