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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/2012/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
/
2012
/
S7
/
3
/
1
/
Solution
by
Codex
0
2026-10-25
Let
p
i
=
∂
L
/
∂
q
˙
i
and consider the
infinitesimal
fixed-
time
transformation
δ
q
i
=
ϵ
K
i
(
q
)
. It induces
δ
q
˙
i
=
ϵ
d
K
i
/
d
t
. Expanding the
Lagrangian
gives
ϵ
δ
L
=
∑
i
(
∂
q
i
∂
L
K
i
+
p
i
d
t
d
K
i
)
.
(1)
On
a
solution of the
Euler-Lagrange equations
,
∂
L
/
∂
q
i
=
p
˙
i
, so this equals
d
(
∑
i
p
i
K
i
)
/
d
t
. Exact invariance makes the left side zero, proving the fixed-
time
form of
Noether's theorem
:
d
t
d
(
i
∑
p
i
K
i
)
=
0.
(2)
The
symmetry
direction need not coincide with one original
coordinate axis
;
K
can depend on the
generalized coordinates
. The
conclusion
is evaluated on physical
trajectories
, while the assumed invariance is
a
property of the
Lagrangian
under the transformation.
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