Basic Mathema,cs. Rende Steerenberg BE/OP. CERN Accelerator School Basic Accelerator Science & Technology at CERN 3 7 February 2014 Chavannes de Bogis

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1 Basic Mathema,cs Rende Steerenberg BE/OP CERN Accelerator School Basic Accelerator Science & Technolog at CERN 3 7 Februar 014 Chavannes de Bogis

2 Contents Vectors & Matrices Differen,al Equa,ons Some Units we use Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis

3 Contents Vectors & Matrices Differen,al Equa,ons Some Units we use Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 3

4 Scalars & Vectors Basics of Accelerator Science & Technolog at CERN Scalar, a single quan,t or value Vector, (origin,) length, direc,on Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 4

5 Coordinate sstems A vector has or more quan,,es associated with it (,) Cartesian coordinates r is the length of the vector r r + Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 5 θ (r, θ) Polar coordinates r θ gives the direc,on of the vector tan( θ ) & # θ arctan$! % "

6 Vector Cross Product a and b are two vectors in the in a plane separated b angle θ b θ a a b The cross product a b is defined b: Direcon: a b is perpendicular (normal) on the plane through a and b The length of a b is the surface of the parallelogram formed b a and b! a! b! a! b sin(θ) Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 6

7 Cross Product & Magne,c Field The Lorentz force is a pure magne,c field F e(! v! B) B field Direc,on of current Force The reason wh our par,cles move around our circular machines under the influence of the magne,c fields This aeernoon E.M. fields b Werner Herr Transverse Beam Dnamics b Bernhard Holzer Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 7

8 A Rota,ng Coordinate Sstem Horizontal Longitudinal Vertical θ It travels on the central orbit Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 8 s or z

9 Moving a Point To move from one point (A) to an other point (B) one needs control of both Length and Direcon. B ( new, new ) new new a c + + b d equa'ons needed!!! Rather clums! Is there a more efficient wa of doing this? Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 9 A (, )

10 Matrices & Vectors So, we have: new new a c Lets write this as one equa,on: Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis b d Rows A and B are Vectors or Matrices A and B have rows and 1 column M is a Matri and has rows and columns & $ % new new B MA #! " & $ % a c Columns b #&! $ d "% #! "

11 Matri Mul,plica,on This implies: new new a c + + b d This matri mul,plica,on results in: i k j l Equals This defines the rules for matri mulplicaon a c b e d g i ae + bg, j af + bh, k ce + dg, l cf + Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 11 new new dh f h a c b d Is this reall simpler?

12 Moving a point & Matrices Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis Lets appl what we just learned and move a point around:! " # $ % &! " # $ % & 1 1 M1! " # $ % &! " # $ % &! " # $ % & M M M! " # $ % &! " # $ % & M 3 M1 transforms 1 to M transforms to 3 This defines M3MM1

13 Matrices & Accelerators We use matrices to describe the various magne,c elements in our accelerator. The and co- ordinates are the posion and angle of each individual par,cle. If we know the posi,on and angle of an par,cle at one point, then to calculate its posi,on and angle at another point we mulpl all the matrices describing the magne,c elements between the two points to give a single matri Now we are able to calculate the final co- ordinates for an ini,al pair of par,cle co- ordinates, provided all the element matrices are known. See Bernhard Holzer s lectures Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 13

14 The Unit Matri There is a special matri that when mul,plied with an ini,al point will result in the same final point. The result is : new new X new X Y new Y 0 1 The Unit matri has no effect on and Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis

15 Going backwards What about going back from a final point to the corresponding inial point? new new A a c b d For the reverse we need another matri M M B such that B MM B The combina,on of M and M - 1 does have no effect 1 MM Unit Matri M - 1 is the inverse or reciprocal matri of M. Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 15 or B MA

16 Calcula,ng the Inverse Matri If we have a matri: Then the inverse matri is calculated b: M 1! M # a b " c d 1 (ad bc) d c b a The term (ad bc) is called the determinate, which is just a number (scalar). Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 16 " $ # $ & % % ' &

17 A Prac,cal Eample Changing the current in two sets of quadrupole magnets (F & D) changes the horizontal and ver,cal tunes (Q h & Q v ). This can be epressed b the following matri rela,onship: ΔQ ΔQ h v a c b ΔI d ΔI ΔI Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 17 F D M 1 or ΔQ ΔQ Change I F then I D and measure the changes in Q h and Q v Calculate the matri M Calculate the inverse matri M - 1 M ΔI Use now M - 1 to calculate the current changes (ΔI F and ΔI D ) needed for an required change in tune (ΔQ h and ΔQ v ).

18 Contents Vectors & Matrices Differenal Equaons Some Units we use Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 18

19 The Pendulum Lets use a pendulum as eample The length of the pendulum is L It has a mass m asached to it It moves back and forth under the influence of gravit Lets tr to find an equaon that describes the moon of the mass m makes. This will result in a Differenal Equaon Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 19 M L

20 Differen,al Equa,on L M The distance from the centre Lθ (since θ is small) θ Mg Restoring force due to gravit is - M g sinθ (force opposes mo,on) The velocit of mass M is: Mg sinθ d ( θ ) dt d( Lθ) dt a d Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 0 v M The acceleraon of mass M is: Newton: Force mass accelera,on d ( Lθ ) dt g + θ L 0 ( Lθ) dt θ is small L is constant

21 Solving a Differen,al Equa,on d ( θ ) dt d( θ ) dt g + θ L Find a solu,on Tr a good guess 0 Differen,ate our guess (twice) Differenal equaon describing the mo,on of a pendulum at small amplitudes. θ Acos( ωt) d ( θ ) Aω sin( ωt) and Aω cos( ωt) dt Put this and our guess back in the original Differen,al equa,on. g ω cos( ωt) + cos( ωt) L Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 1 0

22 Solving a Differen,al Equa,on Now we have to find the solu,on for the following equa,on: Solving this equa,on gives: Oscilla,on amplitude g ω cos( ωt) + cos( ωt) L ω The final solu,on of our differen,al equa,on, describing the mo,on of a pendulum is as we epected : θ Acos Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis g L g L t 0 Oscilla,on frequenc

23 Posi,on & Velocit The differen,al equa,on that describes the transverse mo,on of the par,cles as the move around our accelerator. 0 cos( ωt) d () dt + K ( ) 0 The solu,on of this second order describes oscillator moon For an sstem, where the restoring force is propor,onal to the displacement, the solu,on for the displacement will be of the form: 0 ω sin( ωt) Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 3 d dt

24 Phase Space Plot Plot the velocit as a func,on of displacement: d dt 0 cos( ωt) 0 ω sin( ωt) It is an ellipse. As ωt advances b π it repeats itself. This con,nues for (ω t + k π), with k0,±1, ±,..,..etc Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 4 d dt 0 ω 0

25 Oscilla,ons in Accelerators Under the influence of the magnec fields the parcle oscillate displacement angle d/ds Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 5 ds d s

26 Transverse Phase Space Plot Posi,on Angle This changes slightl the Phase Space plot ' d ds φ ωt is called the phase angle X- ais is the horizontal or ver,cal posi,on (or,me). Y- ais is the horizontal or ver,cal phase angle (or energ). Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 6 X φ

27 Contents Vectors & Matrices Differen,al Equa,ons Some Units we use Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 7

28 Rela,vit velocit c PSB Newton: 1 E mv CPS Einstein: energ increases not velocit SPS / LHC } E mc More about Rela,vit b Werner Herr energ Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 8

29 Units: Joules versus ev The unit most commonl used for Energ is Joules [J] In accelerator and par,cle phsics we talk about ev!? The energ acquired b an electron in a poten,al of 1 Volt is defined as being 1 ev 1 ev is 1 elementar charge pushed b 1 Volt. 1 ev Joules The unit ev is too small to be used currentl, we use: 1 kev 10 3 ev; 1 MeV 10 6 ev; 1 GeV10 9 ev; 1 TeV10 1 ev, Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 9

30 Energ versus Momentum E mc The ra,o between the total energ and the rest energ is γ E E 0 Then the mass of a moving par,cle is: We can write: Momentum is: β Einstein s formula:, which for a mass at rest is: mvc mc p mv m γm 0 E m pc β or p E c 0 0 The ra,o between the real velocit and the velocit of light is β Eβ c Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 30 v c

31 Energ versus Momentum Momentum p Eβ c Therefore the units for momentum are: MeV/c, GeV/c, etc. Energ are: MeV, GeV, etc. A[enon: when β1 energ and momentum are equal Energ when β<1 the energ and momentum are not equal Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 31

32 A Prac,cal Eample at PS Kine,c energ at injec,on E kine,c 1.4 GeV Proton rest energ E MeV The total energ is then: E E kine,c + E 0.34 GeV We know that γ E E, which gives γ We can derive β 1 1, which gives β γ Using p Eβ c we get p.14 GeV/c In this case: Energ Momentum Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 3

33 Pure mathematics is, in its wa, the poetr of logical ideas. Albert Einstein Rende Steerenberg (BE/OP) CAS - 3 Februar Chavannes de Bogis 33