Beam Optics
Manoel Couder
University of Notre Dame
Good aim is sometime important
Sometime it is critical!
Not Talking About Accelerators
Introduction to Accelerators: Evolution of Accelerators and Modern Day Applications
Sarah Cousineau, Jeff Holmes, Yan Zhang
USPAS January, 2011
The Plan: “Beamline”
• Type of ion optics equipment?
Don’t forget diagnostic!!
Outline
• Optics
• Charged particle optics
• Beam
– Collective description of particles
• Spectrometers/Separators
Straight Light Rays
• Z axis = Optics axis of a bundle of rays
𝑥 𝑧2 = 𝑥1 +
• Deviation of rays from bundle 𝑦(𝑧2) = 𝑦1 +
𝑧2 − 𝑧1 tan 𝛼1
𝑧2 − 𝑧1 tan 𝛽1
y
x
Will be using this a lot!!!
Optics of Charged Particles
By Hermann Wollnik
Straight Light Rays
• Z axis = Optics axis of a bundle of rays
𝑥 𝑧2 = 𝑥1 + 𝑧2 − 𝑧1 tan 𝛼1
• Deviation of rays from bundle 𝑦(𝑧2) = 𝑦1 + 𝑧2 − 𝑧1 tan 𝛽1
• Angular dependence
tan 𝛼(𝑧) = tan(𝛼 )
y
1
tan 𝛽(𝑧) = tan(𝛽1 )
Not really exiting …
x
Optics of Charged Particles
By Hermann Wollnik
Straight Light Rays
• Z axis = Optics axis of a bundle of rays
• Deviation of rays from bundle
• Angular dependence
𝑥 𝑧2 = 𝑥1 + 𝑧2 − 𝑧1 tan 𝛼1
𝑦(𝑧2 ) = 𝑦1 + 𝑧2 − 𝑧1 tan 𝛽1
tan 𝛼(𝑧) = tan(𝛼1 )
tan 𝛽(𝑧) = tan(𝛽1 )
𝑥2
1
=
tan(𝛼2 )
0
𝑙
1
𝑥1
tan(𝛼1 )
𝑦2
1
=
tan(𝛽2 )
0
𝑙
1
𝑦1
tan(𝛽1 )
Transfer matrix
from one profile
plane to another
In rotationally symmetric system
Both matrix are identical…
What is a Focal Point?
• https://phet.colorado.edu/en/simulation/lega
cy/geometric-optics
What is a Focal Point?
f
Rays that enter the system parallel to the optical axis are focused
such that they pass through the “rear focal” point.
Any ray that passes through it will emerge from the system
parallel to the optical axis.
Thin Lens
f
• transfer from 2 to 3
𝑥3 = 𝑥2
tan 𝛼3 = −𝑥2 𝑓
𝑥3
1
=
tan(𝛼3 )
−1 𝑓
0
1
𝑥2
tan(𝛼2 )
Profile plane
2 AND 3
𝑥3 = 𝑥2
tan 𝛼3 = tan(𝛼2 ) −𝑥2 𝑓
Optics of Charged Particles
By Hermann Wollnik
Transport through…
𝑥4
1
=
tan(𝛼4 )
0
𝑙2
1
1
−1 𝑓
𝑥4
1 − (𝑙2 𝑓)
=
tan(𝛼4 )
−1 𝑓
Optics of Charged Particles
By Hermann Wollnik
0
1
1
0
𝑙1
1
𝑥1
tan(𝛼1 )
𝑙1 + 𝑙2 − (𝑙1 𝑙2 𝑓)
1 − (𝑙1 𝑓)
𝑥1
tan(𝛼1 )
How do we get a proper picture?
𝑥4
𝑥1
1 − (𝑙2 𝑓) 𝑙1 + 𝑙2 − (𝑙1 𝑙2 𝑓)
=
tan(𝛼4 )
tan(𝛼1 )
−1 𝑓
1 − (𝑙1 𝑓)
Independence of final ray position on initial angle
𝑙1 + 𝑙2 − (𝑙1 𝑙2 𝑓) = 0
(1 𝑙1 ) + (1 𝑙2 ) = 1 𝑓
Under that condition there is a object-image relation
between profile planes 1 and 4
Additional Definitions
𝑥4
1 − (𝑙2 𝑓)
0
=
tan(𝛼4 )
−1 𝑓
1 − (𝑙1 𝑓)
𝑥4
𝑙2
= 𝑀 = 1 − = (− 𝑙2 𝑙1 )
𝑥1
𝑓
𝑥1
tan(𝛼1 )
M is called the magnification
Notation change to allow for generalization
𝑥(𝑧)
(𝑥2 |𝑥1 )
(𝑥2 |tan(𝛼1 ))
=
tan(𝛼(𝑧))
(tan(𝛼1 )|𝑥1 )) (tan(𝛼2 )|tan(𝛼1 ))
𝑥(𝑧)
(𝑥|𝑥)
=
𝑎(𝑧)
(𝑎|𝑥))
(𝑥|𝑎)
(a|𝑎)
𝑥1
𝑎1
Can be a simple drift or a very
complex system.
𝑥1
tan(𝛼1 )
𝑎 = 𝑣𝑥 𝑐 ≃ tan(𝛼)
𝑏 = 𝑣𝑦 𝑐 ≃ tan(𝛽)
Examples
𝑥𝑎 =0
𝑥(𝑧)
(𝑥|𝑥)
=
𝑎(𝑧)
(𝑎|𝑥))
(𝑥|𝑎)
(a|𝑎)
𝑥1
𝑎1
𝑥𝑥 =0
𝑎𝑥 =0
𝑎𝑎 =0
What constraints on which matrix
element needs to explain the plots?
Optics of Charged Particles
By Hermann Wollnik
Charged Particle Optics
• Same idea
• 𝐹 = 𝑞(𝐸 + 𝑣 × 𝐵)
• In a uniform magnetic field charged particles
have a circular motion
– 𝑚𝑣 2 𝑅 = 𝑞𝑣𝐵
– 𝐵𝜌 = 𝑝 𝑞
Magnetic Rigidity
• Equivalent for Electric field
– 𝐸𝜌 = 2𝐾 𝑞
Electric Rigidity
Approximation
Quadrupole Lenses
Focus only in one direction
Electric
Magnetic
The motion of charged
particles in any of the
element presented here
is known.
The transformation matrix
is known as well.
You can solve the equation
of motion to find them.
Combination of
Quadrupole Lenses
Point to parallel
Point to point
With appropriate diagnostic
tune by slowly changing
one field and correcting
with the other
Optics of Charged Particles
By Hermann Wollnik
Symmetric Focusing
A Single tuning parameter.
Maximize transmission to
appropriate diagnostic.
Works at relatively low energies ~100 keV Max
WHY?
Bending Charged Particles
+V
-V
B⊗
The Color Twist
d could represent any variable
They could be more
Actually, most transport code
Use at least 6 dimensions.
Most of the matrix element
are zero
𝑥(𝑧)
(𝑥|𝑥)
=
𝑎(𝑧)
(𝑎|𝑥))
(𝑥|𝑎)
(a|𝑎)
𝑥1
𝑎1
𝑥(𝑧)
(𝑥|𝑥) (𝑥|𝑎)
𝑎(𝑧) = (𝑎|𝑥) (𝑎|𝑎)
𝑑(𝑧)
(𝑑|𝑥) (𝑑|𝑎)
(𝑥|𝑑)
(𝑎|𝑑)
(𝑑|𝑑)
𝑥1
𝑎1
𝑑1
Dispersion
𝐵𝜌 = 𝑝 𝑞
p+dp
.
Dipole field B
perpendicular
to paper plane
Object (size x0)
x
Radius
r
p
Dispersion dx/dp
used in magnetic analysis
Beam?
Pedro Castro
Introduction to Particle Accelerators
DESY, July 2010
Beam?
Same for the vertical plane
Pedro Castro
Introduction to Particle Accelerators
DESY, July 2010
Concept of emittance
Same for the vertical plane
The emittance is conserved as long as the forces to which the
beam is subjected to are conservative.
Pedro Castro
Introduction to Particle Accelerators
DESY, July 2010
Concept of emittance
Concept of emittance
𝜖𝑁 = 𝜖. 𝛽. 𝛾 =constant
The normalized emittance is conserved during acceleration
Pedro Castro
Introduction to Particle Accelerators
DESY, July 2010
Emittance and
Radioactive Beam Production Method
• Depending on the production method
• Emittance of RI Beam can be much worse than
the one of stable beam
– Equipment have to be designed to accommodate
• Large energy and angular acceptance
• Larger bore
• ReA3(,6,12), Caribu
29
Equivalence of Transport of
ONE Ray Ellipse
Defining the s Matrix representing a Beam
The 2-dimensional case ( x, Q )
s= s11 s21
s21 s22
Real, pos. definite
symmetric s
𝜖=
𝜎11 𝜎22 −
2
𝜎12
= det 𝜎
1/2
Matrix
s-1 =1/e2 s22 -s21
-s21 s11
Equivalence of Transport of
ONE Ray Ellipse
Defining the s Matrix representing a Beam
The 2-dimensional case ( x, Q )
2-dim. Coord.vectors
(point in phase space)
X=
x 
Q
X T = (x Q)
Ellipse in Matrix notation:
X T s-1 X = 1
Exercise: Show that Matrix notation
is equivalent to known Ellipse equation:
s22 x2 - 2s21x Q+s11Q2 = e2
Beam Sigma Matrix and Transfer Matrix
Ray X0 from location 0 is transported by a 6 x 6 Matrix R to location 1 by:
Note: R maybe a matrix representing a complex system is :
from which we derive
X0 T s0-1 X0 = 1
(2)
X0 T (RTRT-1) s0-1(R-1 R) X0 = 1
(RX0)T (Rs0 RT)-1 (RX0) = 1
The equation of the new ellipsoid after transformation becomes
where
(1)
R = Rn Rn-1 … R0
Ellipsoid in Matrix notation, generalized to e.g. 6-dim. using s Matrix:
Inserting Unity Matrix I = RR-1 in equ. (2) it follows
X1 = RX0
X1 T s1-1 X1 = 1
s1 = Rs0 RT
(3)
Conclusion: Knowing the TRANSPORT matrix R that transports one ray
through an ion-optical system using (1) we can now also transport
the phase space ellipse describing the initial beam using (3)
Beam/Ion Transport Calculation
• Now we have a “framework” to calculate either
individual particle trajectories or a full beam…
– How do we use it? Do we have to calculate individual
transfer matrix?
– Code with various approach and level of details
•
•
•
•
•
TRANSPORT
COSY INFINITY
GIOS
GICOSY
…
– SIMION, OPERA, …
Resolving Power
𝑥(𝑧)
(𝑥|𝑥) (𝑥|𝑎)
𝑎(𝑧) = (𝑎|𝑥) (𝑎|𝑎)
𝑑(𝑧)
(𝑑|𝑥) (𝑑|𝑎)
𝑥1
𝑎1
𝑑1
(𝑥|𝑑)
(𝑎|𝑑)
(𝑑|𝑑)
Your favorite spectrometer/separator
𝑥 𝑧 = 𝑥 𝑥 𝑥1 + 𝑥 𝑎 𝑎1 + 𝑥 𝑑 𝑑1
Half size of your target
2 𝑥 𝑥 𝑥1
d represent a relative
difference.
For example: 𝑑𝑝/𝑝0 ,
𝑑𝑚/𝑚0 , 𝑑𝐸/𝐸0 or 𝑑𝑄/𝑄0
2 𝑥 𝑥 𝑥1
𝑥 𝑑 𝑑1
x
0
Resolving Power
𝑥 𝑧 = 𝑥 𝑥 𝑥1 + 𝑥 𝑎 𝑎1 + 𝑥 𝑑 𝑑1
Half size of your target
2 𝑥 𝑥 𝑥1
2 𝑥 𝑥 𝑥1
𝑥 𝑑 𝑑1
x
0
The resolving power is the inverse value of d1 that still provide a separation
2 𝑥 𝑥 𝑥1 2 𝑥 𝑥 𝑥1
𝑥 𝑑 𝑑1 = 2 𝑥 𝑥 𝑥1
𝑥 𝑑 𝑑1
x
1 𝑑1 =
(𝑥|𝑑)
2 𝑥 𝑥 𝑥1
Resolving Power
The resolving power is the inverse value of d1 that still provide a separation
2 𝑥 𝑥 𝑥1 2 𝑥 𝑥 𝑥1
𝑥 𝑑 𝑑1 = 2 𝑥 𝑥 𝑥1
𝑥 𝑑 𝑑1
x
1 𝑑1 =
18O(a,g)22Ne
dM
M
(𝑥|𝑑)
2 𝑥 𝑥 𝑥1
@ 2. MeV
(𝑥| 𝑑𝑚 𝑚) = 0.52
( x | m)
𝑀= 𝑥𝑥 =1
0.52
𝑅 = 1 𝑑1 =
= 173
2 × 1 × 0.0015
Recoil
Beam
MDM @ Texas A&M
What Did we Ignore
Truncation of equation of motion
to allow for linear combination
and usage of matrix formalism.
Truncation is similar to selection of
order of a Taylor expansion.
Missing parts are called “Higher
Order”
Element “limitation”
Beam Space Charge Effects
1 nA
Space charge in the accelerating lens
can cause broadening of the beam,
which will affect all of the ions in the
beam, independent of mass.
100 nA
300 nA
39
Higher Order
Designed to have a momentum resolving
power of ~1000. But the measurement
at the focal plane only achieves ~65.
Higher order 𝑥 𝑎𝛿𝑝 is responsible.
Fortunately 𝑥 𝑎𝛿𝑝 is easy to correct
because it only causes a tilt of the focal
plane.
If you take a particle leaving from 𝑥1 = 0
𝑥𝑓𝑝 = 𝑥, 𝛿 𝛿 + 𝑥 𝑎𝛿 𝑎0 𝛿
𝑎𝑓𝑝 = 𝑎 𝑎 𝑎0 + 𝑎 𝛿 𝛿
If another particle leave from the same
place with 𝑎 = 𝑎0 + Δ𝑎0
Δxfp = x a𝛿 Δa0 𝛿
Δ𝑎𝑓𝑝 = 𝑎 𝑎 Δa0
Δ𝑥𝑓𝑝 /Δ𝑎𝑓𝑝 does not depend on Δ𝑎0