Kinematics of collision processes

1) Introduction – collision and decay processes

2) Rutherford scattering (Rutherford experiment from all sides).

3) Laws of energy and momentum conservation.

4) Laboratory and centre-of-mass frame.

5) Reaction energy, decay energy.

6) Collision diagram of momentum

7) Nonrelativistic, relativistic and ultrarelativistic approach.

8) Relativistic invariant kinematics variables.

9) Ultrarelativistic approach – rapidity

10) Transformation of kinematic quantities and cross sections from laboratory frame to centre-of-mass and vice versa

IntroductionStudy of collisions and decays of nuclei and elementary particles – main method of microscopic properties investigation.

Elastic scattering – intrinsic state of motion of participated particles is not changed during scattering particles are not excited or deexcited and their rest masses are not changed.

Inelastic scattering – intrinsic state of motion of particles changes (are excited), but particle transmutation is missing.

Deep inelastic scattering – very strong particle excitation happens big transformation of the kinetic energy to excitation one.

Nuclear reactions (reactions of elementary particles) – nuclear transmutation induced by external action. Change of structure of participated nuclei (particles) and also change of state of motion. Nuclear reactions are also scatterings. Nuclear reactions are possible to divide according to different criteria:

According to history ( fission nuclear reactions, fusion reactions, nuclear transfer reactions …)

According to collision participants (photonuclear reactions, heavy ion reactions, proton induced reactions, neutron production reactions …)

According to reaction energy (exothermic, endothermic reactions)

According to energy of impinging particles (low energy, high energy, relativistic collision, ultrarelativistic …)

Set of masses, energies and moments of objects participating in the reaction or decay is named as process kinematics . Not all kinematics quantities are independent. Relations are determined by conservation laws. Energy conservation law and momentum conservation law are the most important for kinematics.

Transformation between different coordinate systems and quantities, which are conserved during transformation (invariant variables) are important for kinematics quantities determination.

Nuclear decay (radioactivity) – spontaneous (not always – induced decay) nuclear transmutation connected with particle production.

Elementary particle decay - the same for elementary particles

Rutherford scattering

Target: thin foil from heavy nuclei (for example gold)

Beam: collimated low energy α particles with velocity v = v0 << c, after scattering v = vα << c

The interaction character and object structure are not introduced

tt0 vmvmvm

Momentum conservation law: .. (1.1)

and so: ….. (1.1a)

square: ….. (1.1b)

2

t

2

tt

t220 v

m

mvv

m

m2vv

Energy conservation law: (1.2a)2tt

220 vm

2

1vm

2

1vm

2

1 and so: .. (1.2b)

2t

t220 v

m

mvv

Using comparison of equations (1.1b) and (1.2b) we obtain: ………… (1.3) tt2

t vv2m

m1v

For scalar product of two vectors it holds: so that we obtain:cosbaba

tt

0 vm

mvv

If mt<<mα:

Left side of equation (1.3) is positive → from right side results, that target and α particle are moving to the original direction after scattering → only small deviation of α particle

If mt>>mα:

Left side of equation (1.3) is negative → large angle between α particle and reflected target nucleus results from right side → large scattering angle of α particle

Concrete example of scattering on gold atom:mα 3.7·103 MeV/c2 , me 0.51 MeV/c2 a mAu 1.8·105 MeV/c2

1) If mt =me , then mt/mα 1.4·10-4:

We obtain from equation (1.3): ve = vt = 2vαcos ≤ 2vα

We obtain from equation (1.2b): vα v0

Then for magnitude of momentum it holds: meve = m(me/m) ve ≤ m·1.4·10-4·2vα 2.8·10-4mv0

Maximal momentum transferred to the electron is ≤ 2.8·10-4 of original momentum and momentum of α particle decreases only for adequate (so negligible) part of momentum .

cosvv2vv2m

m1v tt

t2t

Reminder of equation (1.3)

2t

t220 v

m

mvv

Reminder of equation (1.2b):

Maximal angular deflection α of α particle arise, if whole change of electron and α momenta are to the vertical direction. Then (α 0):

α rad tan α = meve/mv0 ≤ 2.8·10-4 α ≤ 0.016o

2) If mt =mAu , then mAu/mα 49

We obtain from equation (1.3): vAu = vt = 2(mα/mt)vα cos 2(mαvα)/mt

We introduce this maximal possible velocity vt in (1.2b) and we obtain: vα v0

Then for momentum is valid: mAuvAu ≤ 2mvα 2mv0

Maximal momentum transferred on Au nucleus is double of original momentum and α particle can be backscattered with original magnitude of momentum (velocity).

Maximal angular deflection α of α particle will be up to 180o.

Full agreement with Rutherford experiment and atomic model:

1) weakly scattered - scattering on electrons2) scattered to large angles – scattering on massive nucleus

Attention remember!!: we assumed that objects are point like and we didn't involve force character.

Reminder of equation (1.3)

2t

t220 v

m

mvv

Reminder of equation (1.2b):

cosvv2vv2m

m1v tt

t2t

222

2

t

ααt22t

t220 vv

m

4mv

m

v2m

m

mvv

m

mvv

t

because:

Inclusion of force character – central repulsive electric field:

20

A r

Q

4

1)RE(r

Thomson model – positive charged cloud with radius of atom RA :

Electric field intensity outside:

Electric field intensity inside:3A0

A R

Qr

4

1)RE(r

2A0

AMAX R4

Q2)R2eE(rF

e

The strongest field is on cloud surface and force acting on particle (Q = 2e) is:

This force decreases quickly with distance and it acts along trajectory L 2RA t = L/ v0 2RA/ v0 . Resulting change of particle

momentum = given transversal impulse:

0A0MAX vR4

eQ4tFp

Maximal angle is: 20A0 vmR4

eQ4/pptan

Substituting RA 10-10m, v0 107 m/s, Q = 79e (Thomson model):rad tan 2.7·10-4 → 0.015o only very small angles.

Estimation for Rutherford model:

Substituting RA = RJ 10-14m (only quantitative estimation):tan 2.7 → 70o also very large scattering angles.

Thomson atomic model

Electrons

Positive charged cloud

Electrons

Positive charged nucleus

Rutherford atomic model

Possibility of achievement of large deflections by multiple scattering

Foil at experiment has 104 atomic layers. Let assume:

1) Thomson model (scattering on electrons or on positive charged cloud)2) One scattering on every atomic layer3) Mean value of one deflection magnitude 0.01o. Either on electron or on positive charged nucleus

Mean value of whole magnitude of deflection after N scatterings is (deflections are to all directions, therefore we must use squares):

N2N

1i

2

2N

1ii

2 N

i

…..…. (1)

We deduce equation (1). Scattering takes place in space, but for simplicity we will show problem using two dimensional case:

Multiple particle scattering

Deflections i are distributed both in positive and negative

directions statistically around Gaussian normal distribution for studied case. So that mean value of particle deflection from original direction is equal zero:

0N

1ii

N

1ii

the same type of scattering on each atomic layer: i 22 i

2N

1i

2i

1

1 11

2N

1i

1N

1i

N

1ijji

2

2N

1ii N22

N

i

N

ijji

N

iii

Then we can derive given relation (1):

ababMN

1ba

MN

1b

M

1a

N

1ba

MN

1kk

M

1jj

N

1ii

M

1jj

N

1ii

Because it is valid for two inter-independent random quantities a and b with Gaussian distribution:

And already showed relation is valid: N

We substitute N by mentioned 104 and mean value of one deflection = 0.01o. Mean value of deflection magnitude after multiple scattering in Geiger and Marsden experiment is around 1o. This value is near to the real measured experimental value.

Certain very small ratio of particles was deflected more then 90o during experiment (one particle from every 8000 particles). We determine probability P(), that deflection larger then originates from multiple scattering.

If all deflections will be in the same direction and will have mean value, final angle will be ~100o (we accent assumption each scattering has deflection value equal to the mean value). Probability of this is P = (1/2)N =(1/2)10000 = 10-3010. Proper calculation will give:

2 eP We substitute: 350081002

190 1090 eePooo

Clear contradiction with experiment – Thomson model must be rejected

Derivation of Rutherford equation for scattering:

Assumptions:1) particle and atomic nucleus are point like masses and charges.2) Particle and nucleus experience only electric repulsion force – dynamics is included.3) Nucleus is very massive comparable to the particle and it is not moving.

Acting force: Charged particle with the charge Ze produces a Coulomb potential: r

Ze

4

1rU

0

Two charged particles with the charges Ze and Z‘e and the distance rr

experience a Coulomb force giving rice to a potential energy : r

eZZ

4

1rV

2

0

Coulomb force is:

1) Conservative force – force is gradient of potential energy: rVrF

2) Central force: rVrVrV

Magnitude of Coulomb force is and force acts in the direction of particle join. 2

2

0 r

eZZ

4

1rF

Electrostatic force is thus proportional to 1/r2 trajectory of particle is a hyperbola with nucleus in its external focus.

We define:

Impact parameter b – minimal distance on which particle comes near to the nucleus in the case without force acting.

Scattering angle - angle between asymptotic directions of particle arrival and departure.

Geometry of Rutheford scattering.

Momenta in Rutheford scattering:

First we find relation between b and :

Nucleus gives to the particle impulse particle momentum changes from original value p0 to final value p:

dtF

dtFppp 0

…………. (1)

Using assumption about target fixation we obtain that kinetic energy and magnitude of particle momentum before, during and after scattering are the same:

p0 = p = mv0=mv

We see from figure:

2sinv2mp

2sinvm p

2

100

dtcosF

Because impulse is in the same direction as the change of momentum, it is valid:

where is running angle between and along particle trajectory. F

p

……….. (2)

…………… (3)

We substitute (2) and (3) to (1): dtcosF2

sinvm20

0

………….....................……(4)

We change integration variable from t to :

d

d

dtcosF

2sinvm2

21

21-

0

…. (5)

where ddt is angular velocity of particle motion around nucleus. Electrostatic action of nucleus on particle is in direction of the join vector force momentum do not act angular momentum is not changing (its original value is mv0b) and it is connected with angular velocity = d/dt mr2 = const = mr2 (d/dt) = mv0b

0Fr

then: bv

r

d

dt

0

2

we substitute dt/d at (5):

21

21

220 cosFr

2sinbvm2 d ................................… (6)

2

2

0 r

2Ze

4

1F

We substitute electrostatic force F (Z=2):

We obtain:

2cos

Zedcos

4

2dcosFr

0

221

210

221

21

2

Ze

because it is valid:

2

cos222

sin2sincos 2121

21

21

d

We substitute to the relation (6):

2cos

Ze

2sinbvm2

0

220

Scattering angle is connected with collision parameter b by relation:

bZe

E4

Ze

bvm2

2cotg

2KIN0

2

200

… (7)

The smaller impact parameter b the larger scattering angle .

Energy and momentum conservation law Just these conservation laws are very important. They determine relations between kinematic quantities. It is valid for isolated system:

Conservation law of whole energy:

Conservation law of whole momentum:

if n

1jj

n

1kk pp

if n

1jj

n

1kk EE

jf

n

1jjKIN

20

n

1kkKIN

20 EcmEcm

jif f

n

1jjKIN

n

1jj

20

n

1k

n

1kkKINk

20 EcmEcm i

KIN2i

0fKIN

2f0 EcMEcM

Nonrelativistic approximation (m0c2 >> EKIN): EKIN = p2/(2m0)

2i0

2f0 cMcM i

0f0 MM

Together it is valid for elastic scattering: iKIN

fKIN EE

if n

1j j0

2n

1k k0

2

2m

p

2m

p

Ultrarelativistic approximation (m0c2 << EKIN): E ≈ EKIN ≈ pc

if EE iKIN

fKIN EE

f in

1k

n

1jjk cpcp

f in

1k

n

1jjk pp

We obtain for elastic scattering:

Using momentum conservation law:

sinpsinp0 21 and cospcospp 211

We obtain using cosine theorem: cosp2pppp 1121

21

22

Nonrelativistic approximation:

Using energy conservation law:2

22

1

21

1

21

2m

p

2m

p

2m

p

We can eliminated two variables using these equations. The energy of reflected target particle E‘KIN

2 and reflection angle ψ are usually not measured. We obtain relation between remaining kinematic variables using given equations:

0cosppm

m2

m

m1p

m

m1p 11

2

1

2

121

2

121

0cosEE

m

m2

m

m1E

m

m1E 1 KIN1 KIN

2

1

2

11 KIN

2

11 KIN

Ultrarelativistic approximation:

112

12

12

2211 pp2pppppp Using energy conservation law:

We obtain using this relation and momentum conservation law: cos 1 and therefore: 0

Laboratory and centre-of-mass system

We are studying not only collisions of particle with fixed centre. Also the description of more complicated case can be simplified by separation of the centre-of-mass motion in the case of central potential. We solve problem using advantageous coordinate system.

Laboratory system – experiment is running in this system, all kinematic quantities are measured in this system. It is primary from the side of experiment. Target particle is mostly in the rest in this coordinate system (Experiments with colliding beams are exception).

Centre-of-mass system – centre-of-mass is in this system in the rest and hence total momentum of all particles is zero. Mostly we are interested in relative motion of particles and no motion of system as whole using of such coordinate system is very useful.

12 rrV

21 r,r

Remainder of centre-of-mass installation: we assumed two particles (with masses m1, m2 and

positions ) interacting mutually only by central potential :

111 r̂,rr

222 r̂,rr

r̂r,r

CMCMCM r̂,rr

Equations of motion can be written in form:

12111 rrVrm 12222 rrVrm

…………(1)

where has in spherical coordinates form ( are appropriate unit vectors):

iii ˆ,ˆ,ˆ r

iiiii

i

iii sinr

ˆ

r

ˆ

rr̂

i

i = 1,2

coordinateorigin

Potential energy depends only on relative distance of particles.

We define new coordinates:

21 rrr

21

2211CM mm

rmrmr

………………………. (2)

12111 rrVrm

12222 rrVrm

Reminder of equations (1):

Using relations (1) and (2) we obtain (it is valid ): rVrVrV 12

r̂

r

rVrV-rr

mm

mm

21

21

0rMr)m(m CMCM21 r̂constantrCM

where is the reduced and M the total masses of the system. In the case of central potential motion can be split by rewritten to relative distance and centre-of-mass coordinates:

CMCM rv

Centre-of-mass motion is uniform and straightforward, centre-of-mass moves in the laboratory with a constant velocity independently of specific form of the potential.

The dynamics is completely contained in the motion of a fictitious particle with the reduced mass and coordinate r. In the centre-of-mass system, the complete dynamics is described by the motion of single particle, with the mass , scattered by fixed central potential.

Kinetic energy splits into kinetic energy of centre-of-mass and into the part corresponding to relative particle motion (kinetic energy in centre-of-mass system).

Transformation relations between laboratory and centre-of-mass system for kinematic quantities:

We assume two particle scattering on fixed target (v2=p2=0):The centre-of-mass in the laboratory system moves in the direction of arrived particle motion with velocity:

21

11

21

2211CMCM mm

vm

mm

vmvmrv

Particles are moving against themselves in the centre-of-mass system with velocities:

21

12CM11 mm

vmvvv~

21

11CM22 mm

vmv-vv~

1111 vv~mp~ 1222 vv~mp~

and then:

(we see, that momenta have opposite directions and they have the same magnitude)

and 211 KIN

21

22 KIN1 KINKIN v

2

1E

mm

mE~

E~

E~

Laboratory coordinate system

Centre-of masscoordinate system

212 )~

cos21(

~cos

cos

We rewrite equation (3) to the form:

It is valid for elastic scattering:2

1

1

CM

m

m

v~v

1CMCM1

1

v~v~

cos

~sin

v~

cosv~

~sinv~

tan

..............…………………….. (3)

We divide these relations:

Derivation of relation between scattering angles in centre-of-mass and laboratory coordinate systems:

Relation between velocity components in direction of beam particle motion is:

~cosv~vcosv 1CM1 CM11 v

~cosv~cosv

Relation between velocity components perpendicular to the direction of beam particle motion:

~sinv~sinv 11

Laboratory coordinate system

Centre-of-masscoordinate system

Reaction energy, decay energy

Up to now we studied only elastic scattering. To extend our analysis on other reaction types (decays, nuclear reactions or particle creations), we introduce:

Reaction energy Q: is defined as difference of sums of rest particle energies before reaction and after reaction or as difference of sums kinetic energies after reaction and before it:

QEEcmcmQiffi n

1j

KINj

n

1k

KINk

f

n

1k

2k

i

n

1j

2j

Value of Q is independent on coordinate system. (Reminder: m indicates rest mass):

Exothermic reactions Q 0 energy is released (spontaneous decays of nuclei or particles, reactions are realized for any energy of arrived particle). We are talking about decay energy in such case.

Elastic scattering Q = 0

Endothermic reactions Q 0 energy must be delivered (reaction is not proceed spontaneously, it is necessary certain threshold energy of arrived particle to realize reaction).

0p~

i

n

1jj

i

Threshold energy in centre-of-mass coordinate system:

Using definition of centre-of-mass system we obtain for beginning state:

We obtain from momentum conservation law: 0p~

f

n

1kk

f

It is possible case, that all end particles have zero momentum and thus also theirindividual kinetic energies are zero:

0E~

f

n

1k

KINk

f

Thus threshold energy ETHR in centre-of-mass coordinate system is: QQmmE

~E~

i

n

1jj

f

n

1kk

i

n

1j

KINjTHR

ifi

Threshold energy in laboratory system:

Usually we need to know reaction threshold in laboratory system. We assume nonrelativistic reaction of two particles with rest masses m1 a m2. The target particle is in the rest in the laboratory system. Centre-of-mass is moving in the laboratory system, it has momentum p1 and equivalent kinetic energy:

)m2(m

pE~

21

21

KIN

Qmm2

pE

21

21

THR

this energy is not usable for reaction. That means threshold energy must be:

From definition ETHR is minimal EKIN 1: 1

21

THR 2m

pE

THR121 E2mp

Substituting p12 into previous equation:

2

1THR m

m1QE Case m1 << m2 leads to ETHR = |Q|

Relation between reaction energy and kinematic variables of arrived and scattered particle can be written (we use the same procedure as for similar relation for elastic scattering):

cosEm

Emm2

m

m1E

m

m1EQ KIN3

4

KIN131

4

1KIN1

4

3KIN3

We often need relation: EKIN 3 =f(EKIN 1,), we define x √E‘KIN 3

Solution is: srrE 2KIN3 where cos

mm

Emmr

43

KIN131

43

14KIN14

mm

)m(mEQms

Inelastic scattering is always endothermic (where M0i M0

f, EiKIN, Ef

KIN are total sums):

M0i M0

f EiKIN Ef

KIN Q 0

and

Decay of particle at rest: Q = m0ic2 -M0

fc2. Momenta of particles after two-particle decay have the

same magnitude but opposite direction. Isotropic distribution. Momenta of products: f

02f01

f02

f01f

2f1 mm

Qm2mpp

Collision momentum diagram

We assumed again, that target nucleus is in the rest and no relativistic approximation. We write relations between momenta of particles before and after collision:

CM11 vv~v

121

111 p

mm

mp~p

CM22 vv~v

121

212 p

mm

mp~p

(We obtain law of momentum conservation for studied case by sum of these equations: ) 211 ppp

Such relations are initial equations for construction of vector diagram of momenta:

1p

AC1) Momentum of impinging particle we represent by oriented abscissa .

AC 21 m:mOC:AO 2) We divide abscissa to two parts in the proportion

121

21 p

mm

mp~

3) We describe circle around the point O passing through the point C. The circle radius is equal to magnitude of momenta p1 in the centre-of-mass system . The circle geometrical place of vertexes B of vector triangle of momenta ABC (represents law of momentum conversation), which sides and represent possible momenta of particles after collision in the laboratory system.

AB BC

m1 < m2 :

m1 = m2 : m1 > m2 :

The point A can be inside given circle, on it or outside dependent on ratio of particle masses. A scattering angle in centre-of-mass system can take all possible values from 0 do . Allowed values of scattering angle in the laboratory system and reflection angle in the laboratory system are in the table:

~

m1 m2 m1 = m2 m1 m2

v1 > vCM v1 = vCM v1 < vCM

+ /2 + = /2 + /2

= <0,> = <0,/2> = <0,MAX >

= <0,/2> = <0,/2> = <0,/2>

m1 < m2 : m1 = m2 : m1 > m2 :

In the laboratory system:

m1 < m2 impinging particles are scattered to both hemispheres m1 = m2 impinging particles are scattered to front hemisphere m1 > m2 impinging particles are scattered to front hemisphere to cone with top angle 2MAX (direction of impinging particles is axe of cone): sinMAX =m2/m1

Relation between scattering angle and reflection angle in the laboratory and the centre-of-mass system (remainder of elastic scattering assumption):

2

~ 21 mm

~cos

~sin

tg

Vector momentum diagram provides full information given by conservation laws of energy and momentum. It shows possible variants of particle fly away but it has no information about probabilities of realisation of particular possible variants.

Relativistic description – nonrelativistic and ultrarelativistic approximations

Total energy is connected with momentum by relativistic relation: 420

22 cmcpE

We label rest mass m m0. Rest masses and rest energies are invariant under Lorentz transformation (they are the same in all inertial coordinate systems) and then invariant is also quantity (optimal coordinate system can be chose for its calculation):

42222 cmcpE

It is valid not only for single particle but also for particle system in the given time:

2n

1i

2i

2

2n

1ii

2n

1ii cmcpE

222

22 cm

c

Ep

We express kinetic energy and momentum:2

KIN mcEE

... (1)

2

f

n

1j

2j

221

2221

f

cmcp)cm(E

Threshold energy in centre-of-mass system leads to zero sum of kinetic energies of system in ending state. We express invariant (1) for beginning state of system in laboratory and for ending state in centre-of-mass systems:

We substitute p2: 2

f

n

1j

2j

421

21

2221

f

cmcmEcmE

2

f

n

1j

2j

422

421

221

f

cmcmcmcmE2

and EKIN 1:2

f

n

1j

2j

422

421

421

221 KIN

f

cmcmcmcmm2cmE2

We obtain:2

2

n

1j

22

21

2j

n

1j

22

21

2j

KIN1THR c2m

cmcmcmcmcmcm

EE

ff

Because:

fn

1j

22

21

2j cmcmcmQ

we obtain

222

12

2

22

21

THR cm

Q

m

m1Q

c2m

Qc2mc2mQE

2

1THR m

m1QE

In no relativistic approximation (Q<<m2c2) we obtain known relation.

In ultrarelativistic approximation (Q>>m1c2 a Q>>m2c2): ETHR = Q2

22

422

421

421

2

f

n

1j

2j

1 KIN c2m

cmcmcmm2cm

E

f

We express EKIN1:

Lorentz transformation of momenta and energy from centre-of-mass system to laboratory system is (centre-of-mass moves to the direction of axis y):

2

CM

2CM

x

x

c

v1

c

E~

vp~

p

yy p~p 2

CM

xCM

c

v1

p~vE~

E

We use polar coordinate system: pcospx psinpy ~cosp~p~x ~sinp~p~y and

We derived relation for angle :CM

2

CM

2CM

2

CM

2CM

x

2

CMy

x

y

v~

cosv~c

v1

~sinv~

cE~

v~cosp~

cv

1~

sinp~

cE~

vp~

cv

1p~

p

ptan

In nonrelativistic approximation, where vCM << c we obtain known relation, which we already derived.

Relativistic relation between scattering angle in centre-of-mass and laboratory system

In common practice, kinetic energy of impinging particle is used instead centre-of-mass velocity:

Centre-of-mass velocity in the laboratory system is given by ratio between total momentum and total energy of the system in the laboratory system:

22

21KIN1

1CMCM cmcmE

cp

c

v

We use relation between kinetic energy and momentum:

21

221

421KIN1 cmcpcmE 2

1KIN12KIN11 cm2EEcp

We obtain:2

22

1KIN1

21KIN1

2KIN1

CM cmcmE

cm2EE

This relation can be substitute to the relation for scattering angle. We will show special case, when scattering angle in the centre-of-mass system is π/2:

2

1KIN12KIN1

22KIN1

4221

CM

2CM

cm2EE

cm2Ecmm

c

v~

v

1v~tan

In ultrarelativistic approximation (EKIN 1 >> m1c2 and EKIN 1 >> m2c2) we obtain:

0E

c2m

c

v~tan

KIN1

22

In the laboratory system, particles are produced to the very small angle.

Relativistic invariant variablesWe can obtain velocity of centre-of-mass during scattering of two particles with the rest masses m1 and m2 by total relativistic momentum and total relativistic energy:

21

21CM

CM

EE

c)pp(

c

v

(1.a)

m1 refers to the projectile mass and m2 to target mass. We use laboratory kinematic variables and we obtain:

22

421

221

12

21

1

cmcmcp

cp

cmE

cp

CM …………… (1.b)

Nonrelativistic approximation (m1c2 p1c):)cm(m

vm

cmcm

cvm

21

112

22

1

11

CM …………….. (1.c)

Ultrarelativistic approximation (m1c2 p1c a m2c2 p1c):

2

12

22

12

1

12

22

12

1

12

22

12

1

CMCM )c)p()cm((c))p()cm((1

c)(p)c(mc))(p)c(m(1

)cp()c(m))c(p)cm((1

1

2

1

1

1

21

22

21

21 p

cm

2

1

p

cm1)cp()cm()c)p()cm((1

)pcm1( 12CMFor m1 m2: and c2mppcm21)1()1( 212/1

122/1

CMCM2/12

CMCM

We obtain general relativistic relation for CM : using equation (1.b): 2221

2212

cmE

cpCM

2221

221

422

421

2221

221

422

221

212

CM )cm(E

cm2Ecmcm

)cm(E

cpcmcm2EE1

So that (m1

2c4 = E12-p1

2c2):

and we obtain 1/22

2142

242

1

2212/12

CMCM )cm2Ecmc(m

cmE)1(

… (2)

Equation is reduced for limits E1 p1c m1c2 and p1c m2c2 to formerly given ultrarelativistic limit:

Quantity in the divisor (2) is invariant scalar. We prove this using the square of following four-vector in the laboratory frame (p2 = 0):

221

2221

2221

221 cp)cm(Ec)pp()E(Es

221

422

421

221

221

422

21 cm2Ecmcmcpcm2EcmE

This scalar has the same value in arbitrary reference frame. It has simple interpretation in the centre-of-mass reference frame (total momentum in this reference frame is zero):

2221

221

221

422

421 c)p~p~()E

~E~

(cm2Ecmcm

s 2TOT

221 E

~)E

~E~

(

and s is square of total energy accessible in centre-of-mass system. Then TOT

TOT

TOT

221

CME~E

E~

cmE

Invariant variable s is often used for description of high-energy collisions. The quantity s is very useful in the case of colliders.

Invariant variable t is also often used – square of the four-momentum transfer in a collision (square of the difference in the energy-momentum four-vectors of the projectile before and after scattering):

22i1

f1

2i1

f1 c)pp()E(Et

……………………. (3a)

c2mp 21CM

Energy and momentum conservation laws are valid and we can express t also in target variables:

22i2

f2

2i2

f2 c)pp()E(Et

……………….. (3b)

variable t is invariant and it can be calculated in arbitrary coordinate system.

We add yet variable u:22i

1f2

2i1

f2 c)pp()E(Eu

22i

2f1

2i2

f1 c)pp()E(Eu

or

Variables t, u and s are named as Lorentz invariant Mandelstam variables, which sum generally satisfy equation: f42

242

1i42

242

1 cmcmcmcmuts

p~p~p~p~ fi

fi E~

E~

In the case of elastic scattering in the centre-of-mass system (for both particles and ) )

~cos1(cp~2cp~p~2p~p~t 222i

1f1

2i1

2f1

Such diagrams were pioneered by R. Feynman in the calculation of scattering amplitudes in QED and they are referred to as Feynman graphs. Let us define a variable q2 ( q2c2 = -t ), which is equal to square of momentum transferred to target nucleus q2 (m2v2)2 in no relativistic approximation.

Feynman graph:

Because –1 ≤ cos ≤ 1 it is valid t 0. Using (3a,b) we can look on variable t as on mass-square of exchanged particle (with energy and momentum ). Imaginary mass virtual particle.

i2

f2 EE i

2f2 pp

Ultrarelativistic approximation -rapidity

In high-energy physics (ultrarelativistic collisions velocity of beam particles v c) new kinematic variable – rapidity – is useful to introduce (usually we have c=1, m is total mass):We choose beam direction as axe z, thus we can write total energy and momentum of particle as:

E = mTc2cosh y, px, py a pz = mTc sinh y

2

eecosh(y)

yy

1e

1e

ee

eetanh(y)

2y

2y

yy

yy

Reminder:2

eesinh(y)

yy

2y

2x

2222T ppcmcm We introduced transversal mass mT :

and rapidity y:

z

z

pcE

pcE

ln2

1y and thus:

cos1

cos1ln

2

1

mvcosmc

mvcosmcln

2

1y

For nonrelativistic limit (β → 0): y = β

For ultrarelativistic limit (β → 1): y → ∞

Rapidity using leads to very simple transformation from one coordinate system to another: 2112 yyy where y21 is rapidity of the coordinate system 2 in the system 1. Thus we write for transformation from the laboratory to the centre-of-mass systems :

CMyyy~ Examples: GSI Darmstadt ( ELAB = 1GeV/A y=0.458 β=0.875 ) SPS CERN ( ELAB = 200GeV/A y=6.0 β=1.000 ) LHC CERN ( ELAB=3500+3500GeV/A y=17.8 β=1.000 )

Relation between transversal component of velocity and

rapidity

cos21ln2

1

cos1

cos2

cos1

cos1ln

2

1)0y(( )