66 Chapter 3. Motion of a single plasma particle
3.3 Adiabatic invariant of a pendulum
Perfect symmetry is never attained in reality. This leads to the practical question of how
constants of the motion behave when space and/or time symmetries are ‘good’,but not
perfect. Does the utility of constants of the motion collapse abruptly when the slightest
non-symmetrical blemish rears its ugly head, does the utility decay gracefully, or does
something completely different happen? To answer these questions, we begin byconsid-
ering the problem of a small-amplitude pendulum having a time-dependent, butslowly
changingresonant frequencyω(t).Sinceω^2 =g/l,the time-dependence of the frequency
might result from either a slow change in the gravitational accelerationgor else from a
slow change in the pendulum lengthl.In both cases the pendulum equation of motion will
be
d^2 x
dt^2
+ω^2 (t)x= 0. (3.17)
This equation cannot be solved exactly for arbitraryω(t)but for if a modest restriction is put
onω(t)the equation can be solvedapproximatelyusing the WKB method (Wentzel 1926,
Kramers 1926, Brillouin 1926). This method is based on the hypothesis that the solution
for a time-dependent frequency is likely to be a generalization of the constant-frequency
solution
x= Re[Aexp(iωt)], (3.18)
where this generalization is guessed to be of the form
x(t) = Re
[
A(t)ei
∫t
ω(t′)dt′
]
. (3.19)
HereA(t)is an amplitude function determined as follows: calculate the first derivative of
Eq. (3.19),
dx
dt
= Re
[
iωAei
∫t
ω(t′)dt′+dA
dt
ei
∫t
ω(t′)dt′
]
, (3.20)
then the second derivative
d^2 x
dt^2
= Re
[(
i
dω
dt
A+ 2iω
dA
dt
−ω^2 A+
d^2 A
dt^2
)
ei
∫tω(t′)dt′]
, (3.21)
and insert this last result into Eq. (3.17) which reduces to
i
dω
dt
A+ 2iω
dA
dt
+
d^2 A
dt^2
= 0, (3.22)
since the terms involvingω^2 cancel exactly. To proceed further, we make an assumption
- the validity of which is to be checked later – that the time dependence ofdA/dtis
sufficiently slowto allow dropping the last term in Eq. (3.22) relative to the middle term.
The two terms that remain in Eq. (3.22) can then be re-arranged as
1
ω
dω
dt
=−
2
A
dA
dt
(3.23)
which has the exact solution
A(t)∼
1
√
ω(t)