Department

PhysicsCourse Code

PHYS 4150Professor

Sascha KempfStudy Guide

FinalThis

**preview**shows pages 1-2. to view the full**6 pages of the document.**PHYS4150 — P L A S M A P H Y S I C S

l e c t u r e 9 - m a g n e t i c m i r r o r m a c h i n e s ,a d i a b a t i c

m o m e n t s ,t r a p p i n g

Sascha Kempf∗

G135, University of Colorado, Boulder

Fall 2019

1a d i a b at i c i n va r i a n t s

The presence of adiabatic invariants is actually a common phenomenon, which has

been studied extensively in classical mechanics. Here we follow Landau & Lifschitz

and consider a one-dimensional ﬁnite motion, where

λ

is a parameter describing a

very slow change of the system. Here, slow means slow compared to the period

T

of

the cyclic motion, i.e.

T˙

λ≪λ

. Now, because

λ

is slowly changing, so is the energy

E

of the system, where

˙

E∼˙

λ

. This implies that the change of energy is a function

of λ, from what follows that there is a combination of Eand λ, a so-called adiabatic

invariant, which remains constant.

Now let

H(p,q;λ)

be the Hamiltonian of such a system, where again

λ

is the

parameter characterizing the slow change. Then,

dE

dt=∂H

∂t=∂H

∂λ

dλ

dt.

Now we average over one cycle

T

and assume that

˙

λ

does not change on this time

scale

dE

dt=dλ

dt

∂H

∂λ.

Now,

∂H

∂λ=1

T

T

Z

0

∂H

∂λdt,

∗sascha.kempf@colorado.edu

1

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L E C T U R E 9PHYS4150

and using that ˙q=∂H

∂pwe obtain

∂H

∂λ=1

TI∂H

∂λ∂H

∂p−1

dq.

By further noting that

T=

T

Z

0

dt=I∂H

∂p−1

dq

we get

∂H

∂λ=H∂H

∂λ∂H

∂p−1dq

H∂H

∂p−1dq

,

and thus

dE

dt=dλ

dtH∂H

∂λ∂H

∂p−1dq

H∂H

∂p−1dq

.

We have assumed that

λ

is constant along the integration path, which implies that

E=H(p,q;λ)is constant as well. Differentiating Hwith respect to λgives

0=∂H

∂λ+∂H

∂p

∂p

∂λ,

and thus

∂H

∂λ∂H

∂p−1

=−∂p

∂λ.

After substituting this expression into our expression for the change of the mean

energy we get

dE

dt=−dλ

dtH∂p

∂λdq

H∂p

∂Edq,

or

0=I∂p

∂E

dE

dt+∂p

∂λ

dλ

dtdq=d

dtIpdq.

This result implies that the adiabatic invariant

I=1

2πIpdq(1)

remains constant even when the parameter

λ

is changing slowly.

I

is actually the area

2

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