7.3 Solutions
329
Using (2), (3) and (4) in (1)
dL −
n
r =1
( ˙
p r dq r + p r d ˙
q r ) +
∂ L
∂t
dt
(5)
Equation (5) can be rearranged in the form
d
n
r =1
p r ˙
q r − L
=
n
r =1
( ˙
q r d p r − ˙
p r dq r ) −
∂ L
∂t
dt
(6)
The Hamiltonian function H is defined by
H =
n
r =1
p r ˙
q r − L(q 1 , . . . , q n , ˙
q 1 , . . . , ˙
q n , t)
(7)
Equation (6) therefore my be written as
dH =
n
r =1
( ˙
q r dp r − ˙
p r dq r ) −
∂ L
∂t
dt
(8)
While the Lagrangian function L is an explicit function of q 1 , . . . , q n ,
˙
q 1 , . . . , ˙
q n and t, it is usually possible to express H as an explicit function only
of q 1 , . . . , q n , p 1 , . . . , p n , t, that is, to eliminate the n generalized velocities
from (7). The n equation of type (3) are employed for this purpose. Each
provides one of the p’s in terms of the ˙
q s. Assuming that the elimination of
the generalized velocities is possible, we may write
H = H (q 1 , . . . , q n , p 1 , . . . , p n , t)
(9)
H now depends explicitly on the generalized coordinates and generalized
momenta together with the time. Therefore, taking the differential dH , we
obtain
dH =
n
r =1
∂ H
∂q r
dq r +
∂ H
∂ p r
d p r
+
∂ H
∂t
dt
(10)
Comparing (8) and (10), we have the relations
∂ H
∂ p r
= ˙
q r ,
∂ H
∂q r
= − ˙
p r
(11)
∂ H
∂t
= −
∂ L
∂t
(12)