Lecture 2 Transverse Optics I Professor Emmanuel Tsesmelis Directorate Office, CERN Department of Physics, University of Oxford ACAS School for Accelerator Physics 2012 Australian Synchrotron, Melbourne November/December 2012

Contents – Lecture 2 Relativity, Energy & Units Accelerator Co-ordinates Magnets and their Configurations Hill’s Equation 2

CERN Accelerators The energies in the CERN accelerators range from 100 keV to 7 TeV. To do this increase the beam energy in a staged way using 5 different accelerators. 3

Relativity PSB CPS velocity energy c Newton: SPS / LHC Einstein: energy increases not velocity } 4

Energy & Momentum Einstein’s relativity formula: Rest mass Rest energy For a mass at rest this will be: as being the ratio between the total energy and the rest energy. Define: Then the mass of a moving particle is:, then can write: Define:, which is always true and gives: 5

Units: Energy & Momentum (1) Einstein’s relativity formula: Familiar with units Joules and Newton meter but for accelerators use eV…!? If we push a block over a distance of 1 meter with a force of 1 Newton, we use 1 Joule of energy. Thus : 1 Nm = 1 Joule The energy acquired by an electron in a potential of 1 Volt is defined as being 1 eV. 1 eV is 1 elementary charge ‘pushed’ by 1 Volt. Thus : 1 eV = 1.6 x Joules. The unit eV is too small to be used currently, we use: 1 keV=10 3 eV; 1 MeV=10 6 eV; 1 GeV=10 9 eV; 1 TeV=10 12 eV… 6

Units: Energy & Momentum (2) However: Therefore, the units for momentum are GeV/c…etc. Attention: when β=1 energy and momentum are equal, when β <1 the energy and momentum are not equal. Energy Momentum 7

Units: Example PS Injection Kinetic energy at injection E kinetic = 1.4 GeV. Proton rest energy E 0 = MeV. The total energy is then: E = E kinetic + E 0 = 2.338GeV. We know that,which gives γ = We can derive,which gives β = Using we get p = 2.14 GeV/c In this case: Energy ≠ Momentum 8

Accelerator Co-ordinates Refers to a: Rotating Cartesian Co-ordinate System Vertical Horizontal Longitudinal It travels on the central orbit 9

Beam Optics Physical fundamentals for beam steering & focusing. Fix particle trajectory and then repeatedly steer diverging particles back onto ideal trajectory. Performed by EM fields (E and B) satisfying Lorentz Force Lorentz force = centripetal force

Magnetic Rigidity The force evB on a charged particle moving with velocity v in a dipole field of strength B is equal to its mass multiplied by its acceleration towards the centre of its circular path. This is: Which can be written as: Momentum P=mv Bρ is called the magnetic rigidity, and in the correct units obtain: Bρ = ·p [KG·m] = ·p [T·m] (if p is in [GeV/c]) Like for a stone attached to a rotating rope e p e mv B  evBF   2 Radius of curvature 11

Multipoles in Beam Steering Magnetic field around beam is sum of multipoles.  Each has different effect on particle path. Linear beam optics refers to use of only dipoles and quadrupoles for beam steering.

Multipole Field Shapes

Dipole Magnets A ferromagnet consisting of two parallel Fe poles. Dipole strength is Limitation of useful field region due to edge field. Saturation effect:

Superconducting Magnets Disappearance of ohmic resistance in Hg at very low T. Dependence of superconductivity on the T and surrounding B-field.

Superconducting Magnets Construction of superconductor from niobium-titanium (NbTi)

Dipole Magnet A dipole with a uniform dipolar field deviates a particle by an angle θ. The deviation angle θ depends on the length L and the magnetic field B. The angle θ can be calculated: If  is small: So can write: 17

Two Particles in a Dipole Field Particle A Particle B Unfolding these circles…… What happens with two particles that travel in a dipole field with different initial angles, but with equal initial position and equal momentum ? Assume that Bρ is the same for both particles. 18

The Two Trajectories Unfolded Particle B oscillates around particle A. This type of oscillation forms the basis of all transverse motion in an accelerator. It is called ‘Betatron Oscillation’ The horizontal displacement of particle B with respect to particle A. Particle B Particle A 22 0 displacement 19

‘Stable’ or ‘Unstable’ Motion ? What can we say about the vertical motion in the same simplified accelerator ? Is it ‘stable’ or ‘unstable’ and why ? Since the horizontal trajectories close, the horizontal motion in simplified accelerator with only a horizontal dipole field is ‘stable’ What can we do to make this motion stable ? Need some element that ‘focuses’ the particles back to the reference trajectory. This extra focusing can be done using: Quadrupole magnets 20

Quadrupole Magnet A Quadrupole has 4 poles, 2 north and 2 south. There is no magnetic field along the central axis. They are symmetrically arranged around the centre of the magnet. Magnetic field Hyperbolic contour x · y = constant 21

Quadrupole Fields Magnetic field The ‘normalised gradient’, k is defined as: On the x-axis (horizontal) the field is vertical and given by: B y  x On the y-axis (vertical) the field is horizontal and given by: B x  y The field gradient, K is defined as: 22

Types of Quadrupoles It focuses the beam horizontally and defocuses the beam vertically. Force on particles This is a: Focusing Quadrupole (QF) Rotating this magnet by 90º will give a: Defocusing Quadrupole (QD) 23

Quadrupole Magnets Quadrupole magnet with hyperbolic pole surface. Field shape determined by Calculation of field gradient g of quadrupole.

Focusing and Stable Motion Using a combination of focusing (QF) and defocusing (QD) quadrupoles solves problem of ‘unstable’ vertical motion. It will keep the beams focused in both planes when the position in the accelerator, type and strength of the quadrupoles are well chosen. By now accelerator is composed of: Dipoles, constrain the beam to some closed path (orbit). Focusing and Defocusing Quadrupoles, provide horizontal and vertical focusing in order to constrain the beam in transverse directions. A combination of focusing and defocusing sections that is very often used is the so-called: FODO lattice. This is a configuration of magnets where focusing and defocusing magnets alternate and are separated by non-focusing drift spaces. 25

FODO Cell The ’FODO’ cell is defined as follows: ‘FODO’ cell QF QD QF or like this…… Centre of first QF Centre of second QF L1 L2 26

The Mechanical Equivalent The gutter below illustrates how the particles in our accelerator behave due to the quadrupole fields. Whenever a particle beam diverges too far away from the central orbit the quadrupoles focus them back towards the central orbit. How can we represent the focusing gradient of a quadrupole in this mechanical equivalent ? 27

The Particle Characterized A particle during its transverse motion in accelerator is characterized by: Position or displacement from the central orbit. Angle with respect to the central orbit. x = displacement x’ = angle = dx/ds This is a motion with a constant restoring force, like the pendulum ds x’ x dx x s 28

Hill’s Equation (1) The betatron oscillations exist in both horizontal and vertical planes. The number of betatron oscillations per turn is called the betatron tune and is defined as Qx and Qy. Hill’s equation describes this motion mathematically If the restoring force, K is constant in ‘s’ then this is just a S imple H armonic M otion. ‘s’ is the longitudinal displacement around the accelerator. 29

Hill’s Equation (2) In a real accelerator K varies strongly with ‘s’. Therefore, we need to solve Hill’s equation for K varying as a function of ‘s’ Remember what we concluded on the mechanical equivalent concerning the shape of the gutter…… The phase advance and the amplitude modulation of the oscillation are determined by the shape of the gutter. The overall oscillation amplitude will depend on the initial conditions, i.e. how the motion of the ball started. 30

Solution of Hill’s Equation (1) ε and  0 are constants, which depend on the initial conditions.  (s) = the amplitude modulation due to the changing focusing strength.  (s) = the phase advance, which also depends on focusing strength. Remember, this is a 2 nd order differential equation. In order to solve it try to guess a solution: 31

Solution of Hill’s Equation (2) Define Twiss parameters: In order to solve Hill’s equation differentiate guess, which results in: ……and differentiating a second time gives: …and let Remember  is still a function of s Now need to substitute these results in the original equation. 32

Solution of Hill’s Equation (3) So need to substitute and its second derivative into initial differential equation This gives: Sine and Cosine are orthogonal and will never be 0 at the same time 33

Using the ‘Sine’ terms, which after differentiating gives Defined Combining and gives: since Which is the case as: So guess seems to be correct! Solution of Hill’s Equation (4) 34

For x’ have: Thus, the expression for x’ finally becomes: Solution of Hill’s Equation (5) Since solution was correct, have the following for x:      2 ' ds d 35