276
An Introduction to Beam Physics
where ˆ
M is a 4 × 4 real matrix. The eigenvalue of ˆ
M can be obtained through
solving
det
ˆ
M − λ ˆ
I
= 0.
(11.5)
Since ˆ
M is real,
ˆ
MM v = λλ v =⇒ ˆ
M
† v
† = λ
† v
† =⇒ ˆ
MM v
† = λ
† v
† .
So λ
† is also an eigenvalue of ˆ
M and v
† is the eigenvector. Keep in mind that
det( ˆ
A ˆ
B) = det( ˆ
A) det( ˆ
B), det( ˆ
A
T ) = det( ˆ
A).
With these identities, we can transform eq. (11.5).
det
ˆ
M − λ ˆ
I
= 0 =⇒ det ˆ
J · det( ˆ
J ˆ
M − λ ˆ
I) = 0 =⇒ det( ˆ
J ˆ
M − λ ˆ
J ) = 0
=⇒ det ˆ
M
T
· det( ˆ
J ˆ
M − λ ˆ
J) = 0 =⇒ det( ˆ
J − λ ˆ
M
T ˆ
J) = 0
=⇒ det( ˆ
I − λ ˆ
M
T ) = 0 =⇒ λ
2n det(λ
−1 ˆ
I − ˆ
M
T ) = 0
=⇒ det( ˆ
M
T
− λ
−1 ˆ
I) = 0.
Therefore λ
−1 is also an eigenvalue. Together with λ
† , we reach the conclusion that if ˆ
M has an eigenvalue λ then λ
† , λ
−1 , λ
†−1 are also eigenvalues.
We know that for ˆ
M to be stable, |λ| has to be smaller or equal to 1 for
all eigenvalues of ˆ
M . As a result, all eigenvalues of ˆ
M lie on the unit circle.
Apparently, we have, for this case, λ = λ
†−1 and λ
−1 = λ
† .
The next question we can ask is that supposing ˆ
M is stable and every
eigenvalue is reasonably far away from each other, what happens when ˆ
M
is perturbed? In terms of betatron motion, it means that μ x and μ y are
reasonably far away.
To frame it in a more mathematical way, we say that there is a neighborhood
around each eigenvalue so that there is only one eigenvalue in it. When ˆ
M
is perturbed its eigenvalues will move. They may all stay on the unit circle,
or some of them may move away from it. Let us say one of them, λ, moves
away from the unit circle, then λ
†−1 will also move away from it and be in
the same neighborhood that λ is in, making the total number of eigenvalues
greater than 4, which is impossible. So the conclusion is that every one of
them will stay on the unit circle (see Fig. 11.1). This is consistent with our
experience, which is that when we change quadrupole strength, μ x and μ y
change, but the motion is stable.
The question that relates to linear coupling is what happens if two or more
of the eigenvalues get close to each other? The answer is more complicated.
It has been proven that when two colliding eigenvalues have the same sign of
phase, motion remains stable after collision. Otherwise, instability may occur.
In other words, difference resonance μ x − μ y does not lead to instability, sum
resonance μ x + μ y does.
An Introduction to Beam Physics
where ˆ
M is a 4 × 4 real matrix. The eigenvalue of ˆ
M can be obtained through
solving
det
ˆ
M − λ ˆ
I
= 0.
(11.5)
Since ˆ
M is real,
ˆ
MM v = λλ v =⇒ ˆ
M
† v
† = λ
† v
† =⇒ ˆ
MM v
† = λ
† v
† .
So λ
† is also an eigenvalue of ˆ
M and v
† is the eigenvector. Keep in mind that
det( ˆ
A ˆ
B) = det( ˆ
A) det( ˆ
B), det( ˆ
A
T ) = det( ˆ
A).
With these identities, we can transform eq. (11.5).
det
ˆ
M − λ ˆ
I
= 0 =⇒ det ˆ
J · det( ˆ
J ˆ
M − λ ˆ
I) = 0 =⇒ det( ˆ
J ˆ
M − λ ˆ
J ) = 0
=⇒ det ˆ
M
T
· det( ˆ
J ˆ
M − λ ˆ
J) = 0 =⇒ det( ˆ
J − λ ˆ
M
T ˆ
J) = 0
=⇒ det( ˆ
I − λ ˆ
M
T ) = 0 =⇒ λ
2n det(λ
−1 ˆ
I − ˆ
M
T ) = 0
=⇒ det( ˆ
M
T
− λ
−1 ˆ
I) = 0.
Therefore λ
−1 is also an eigenvalue. Together with λ
† , we reach the conclusion that if ˆ
M has an eigenvalue λ then λ
† , λ
−1 , λ
†−1 are also eigenvalues.
We know that for ˆ
M to be stable, |λ| has to be smaller or equal to 1 for
all eigenvalues of ˆ
M . As a result, all eigenvalues of ˆ
M lie on the unit circle.
Apparently, we have, for this case, λ = λ
†−1 and λ
−1 = λ
† .
The next question we can ask is that supposing ˆ
M is stable and every
eigenvalue is reasonably far away from each other, what happens when ˆ
M
is perturbed? In terms of betatron motion, it means that μ x and μ y are
reasonably far away.
To frame it in a more mathematical way, we say that there is a neighborhood
around each eigenvalue so that there is only one eigenvalue in it. When ˆ
M
is perturbed its eigenvalues will move. They may all stay on the unit circle,
or some of them may move away from it. Let us say one of them, λ, moves
away from the unit circle, then λ
†−1 will also move away from it and be in
the same neighborhood that λ is in, making the total number of eigenvalues
greater than 4, which is impossible. So the conclusion is that every one of
them will stay on the unit circle (see Fig. 11.1). This is consistent with our
experience, which is that when we change quadrupole strength, μ x and μ y
change, but the motion is stable.
The question that relates to linear coupling is what happens if two or more
of the eigenvalues get close to each other? The answer is more complicated.
It has been proven that when two colliding eigenvalues have the same sign of
phase, motion remains stable after collision. Otherwise, instability may occur.
In other words, difference resonance μ x − μ y does not lead to instability, sum
resonance μ x + μ y does.