16—Calculus of Variations 398This produces an expression for∆S
∆S=L
(
(tb,y(tb),y ̇(tb)
)
∆tb−L
(
(ta,y(ta),y ̇(ta)
)
∆ta
+
∂L
∂y ̇
(tb)δy(tb)−
∂L
∂y ̇
(ta)δy(ta) +
∫tbtadt
[
∂L
∂y
−
d
dt
(
∂L
∂y ̇
)]
δy
(16.41)
Up to this point the manipulation was straight-forward, though you had to keep all the algebra
in good order. Now there are some rearrangements needed that are not all that easy to anticipate,
adding and subtracting some terms.
Start by looking at the two terms involving∆tbandδy(tb)— the first and third terms. The
change in the position of this endpoint is not simplyδy(tb). Even ifδyis identically zero the endpoint
will change in both thet-direction and in they-direction because of the slope of the curve (y ̇) and the
change in the value oftat the endpoint (∆tb).
The total movement of the endpoint attbis horizontal by the amount∆tb, and it is vertical by
the amount
(
δy+y ̇∆tb
)
. To incorporate this, add and subtract this second term,y ̇∆t, in order to
produce this combination as a coefficient.
y
t
a b
y(t)
y+δy δy
y ̇∆tb
∆tb
L
(
(tb,y(tb),y ̇(tb)
)
∆tb+
∂L
∂y ̇
(tb)δy(tb)
=
[
L(tb)∆tb−
∂L
∂y ̇
(tb)y ̇(tb)∆tb
]
+
[
∂L
∂y ̇
(tb)y ̇(tb)∆tb+
∂L
∂y ̇
(tb)δy(tb)
]
=
[
L−
∂L
∂y ̇
y ̇
]
∆tb+
∂L
∂y ̇
[
y ̇∆tb+δy
]
(16.42)
Do the same thing atta, keeping the appropriate signs. Then denote
p=
∂L
∂y ̇
, H=py ̇−L, ∆y=δy+y ̇∆t
His the Hamiltonian and Eq. (16.41) becomes Noether’s theorem.
∆S=
[
p∆y−H∆t
]tbta+
∫tbtadt
[
∂L
∂y
−
d
dt
(
∂L
∂y ̇
)]
δy (16.43)
If the equations of motion are satisfied, the argument of the last integral is zero. The change inSthen
comes only from the translation of the endpoint in either the time or space direction. If∆tis zero, and
∆yis the same at the two ends, you are translating the curve in space — vertically in the graph. Then
∆S=p∆y
∣∣
∣∣
tbta