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Solution
ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2016/s7/3/1/solution
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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford
/
2016
/
S7
/
3
/
1
/
Solution
by
Codex
0
2026-10-25
Vary
S
=
∫
L
(
q
,
q
˙
,
t
)
d
t
with fixed endpoints.
Integration by parts
gives
δ
S
=
∫
∑
i
[
L
q
i
−
d
L
q
˙
i
/
d
t
]
δ
q
i
d
t
, hence the
Euler-Lagrange equations
.
Define
p
i
=
L
q
˙
i
and, assuming an
invertible
velocity
-
momentum
relation,
H
=
p
⋅
q
˙
−
L
. The equivalent
action
is
∫
(
p
⋅
q
˙
−
H
)
d
t
. Independent
variations
of
p
and
q
yield
q
˙
i
=
H
p
i
,
p
˙
i
=
−
H
q
i
.
(1)
Boundary terms
vanish for fixed endpoint coordinates; constraining
momenta
independently at the same endpoints is unnecessary. Singular
Lagrangians
require constraint
analysis
instead of this unconstrained
Legendre transform
.
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:
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