Fundamentals of Plasma Physics

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)

. (3.24)