14 CHAPTER1. THECLASSICALSTATE
wherewehaveapplied(1.24).Next,differentiatingHwithrespecttox,
∂H
∂x
= p
∂x ̇(x,p)
∂x
−
∂L
∂x
−
∂L
∂x ̇
∂x ̇(p,x)
∂x
= −
∂L
∂x
(1.28)
UsingtheEuler-Lagrangeequation(1.21)(andthisiswheretheequationsofmotion
enter),wefind
∂H
∂x
= −
d
dt
∂L
∂x ̇
= −
dp
dt
(1.29)
Thus,withthehelpoftheHamiltonianfunction,wehaverewrittenthesingle2nd
orderEuler-Lagrangeequation(1.21)asapairof1storderdifferentialequations
dx
dt
=
∂H
∂p
dp
dt
= −
∂H
∂x
(1.30)
whichareknownasHamilton’sEquations.
Forabaseball,theLagrangianisgivenbyeq. (1.20),andthereforethemomentum
is
p=
∂L
∂x ̇
=mx ̇ (1.31)
Thisisinvertedtogive
x ̇=x ̇(p,x)=
p
m
(1.32)
andtheHamiltonianis
H = px ̇(x,p)−L[x,x ̇(x,p)]
= p
p
m
−
[
1
2
m(
p
m
)^2 −V(x)
]
=
p^2
2 m
+V(x) (1.33)
Note that the Hamiltonian forthe baseball issimply the kinetic energy plus the
potentialenergy; i.e. theHamiltonian isanexpressionforthe totalenergyofthe
baseball.SubstitutingHintoHamilton’sequations,onefinds
dx
dt
=
∂
∂p
[
p^2
2 m
+V(x)
]
=
p
m
dp
dt
= −
∂
∂x
[
p^2
2 m
+V(x)
]
=−
dV
dx
(1.34)
whichissimplythefirst-orderformofNewton’sLaw(1.2).