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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/s7/2/1/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
/
2
/
1
/
Solution
by
Codex
0
2026-10-25
A
constant of motion
has zero
total derivative
on every solution, and may explicitly depend on
time
. For an
infinitesimal transformation
δ
q
i
=
ϵ
ξ
i
with
δ
L
=
ϵ
d
F
/
d
t
, the
on-shell
first variation
is
δ
L
=
ϵ
d
(
∑
i
p
i
ξ
i
)
/
d
t
. Therefore
I
=
i
∑
p
i
ξ
i
−
F
is conserved. Including
a
time
shift
δ
t
=
ϵ
τ
gives the corresponding
I
=
∑
i
p
i
ξ
i
−
H
τ
−
F
. This proves
Noether's theorem
in its usual point-transformation form.
A
symmetry
of the
action
up to
a
boundary term
is sufficient; exact invariance of the displayed
Lagrangian
is stronger than necessary.
Discrete
transformations by themselves do not give this continuous-charge construction.
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