Transcript Snímek 1

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
Introduction
Study 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 …)
Nuclear decay (radioactivity) – spontaneous (not always – induced decay) nuclear transmutation
connected with particle production.
Elementary particle decay - the same for elementary particles
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.
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



Momentum conservation law: m v0  m v  m t v t .. (1.1)
and so:
square:


m 
v 0  v  t v t
m

m   m
v 02   v2  2 t v  v t   t

m
 m

Energy conservation law:
….. (1.1a)
2
 2 
 v t
 
….. (1.1b)
mt 2
2
2
1
1
1
2
2
2
v

v

vt
(1.2a)
and
so:
0

m  v 0  m  v  m t v t
m
2
2
2
 m
Using comparison of equations (1.1b) and (1.2b) we obtain: v 2t 1  t
 m
For scalar product of two vectors it holds:
.. (1.2b)

 
  2v  v t ………… (1.3)

  
a  b  a b cos  so that we obtain:
Reminder of equation (1.3)
If mt<<mα:
 m
v 2t 1  t
 m

 
 
  2v  v t  2  v  v t  cos

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
Reminder of equation (1.2b):
v 02  v2 
mt 2
vt
m
Then for magnitude of momentum it holds: meve = m(me/m) ve ≤ m·1.4·10-4·2vα  2.8·10-4mv0
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 .
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/mv0 ≤ 2.8·10-4  α ≤ 0.016o
Reminder of equation (1.3)
 m
v 2t 1  t
 m

 
 
  2v  v t  2  v  v t  cos

Reminder of equation (1.2b): v 02  v2 
2) If mt =mAu , then mAu/mα  49
mt 2
vt
m
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
2
mt 2
m t  2m α v α 
4m 2
2
2
2

  v2 
v

v

v

v

v  v2
because: 0

t

m
m  m t 
mt
Then for momentum is valid: mAuvAu ≤ 2mvα  2mv0
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 electrons
2)  scattered to large angles – scattering on massive nucleus
Attention remember!!: we assumed that objects are point like and we didn't involve force character.
Inclusion of force character – central repulsive electric field:
Thomson atomic model
Electrons
Thomson model – positive charged cloud with radius of atom RA :
Electric field intensity outside:
Electric field intensity inside:
Positive charged cloud
Rutherford atomic model
Electrons
Q
40 r 2
E(r  R A ) 
1
1
Qr
40 R 3A
The strongest field is on cloud surface and
2eQ
force acting on  particle (Q = 2e) is: FMAX  2eE(r  R A ) 
40 R 2A
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:
4eQ
p  FMAX t 
40 R A v 0
Maximal angle is:
Positive charged nucleus
E(r  R A ) 
tan   p /p  
4eQ
40 R A m v 02
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.
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 layer
3) 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):
2
N
 N 
    i   i2  N 2
i 1
 i1 
2

  N 
…..…. (1)
We deduce equation (1). Scattering takes place in space, but for simplicity we will show
problem using two dimensional case:
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:
N
N
i 1
i 1
  i  i  0
Multiple particle scattering
i  
the same type of scattering on each atomic layer:
i2   2
Then we can derive given relation (1):
2
N 1 N
N 1 N
N
 N 2
 N   N 2
  i     i  2  i j    i  2  i j   i2  N 2
i 1 ji 1
i 1 j i 1
i 1
 i1   i1
 i 1
Because it is valid for two inter-independent random quantities a and b with Gaussian distribution:
M
1 N
1 M
1 N
1 N M
abk  ab
a  b   ai  b j 
 ai  b j  N  M 
N i 1 M j1
N  M i 1 j1
k 1
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 ~100 o (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:
P    e   
2
We substitute:


P  90o  e 90
o
1o
2  e 8100  10 3500
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: Ur   4 r
0

Two charged particles with the charges Ze and Z‘e and the distance r  r
experience a Coulomb force giving rice to a potential energy :
Vr  
1
Ze
ZZe 2
40 r
1
Coulomb force is:
 
 
1) Conservative force – force is gradient of potential energy: F r   Vr 


2) Central force: Vr   V r   Vr 
Magnitude of Coulomb force is
ZZe 2
Fr  
and force acts in the direction of particle join.
40 r 2
1
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.
First we find relation between b and  :

Nucleus gives to the  particle impulse  Fdt  particle
momentum changes from original value p0 to final
value p:

 

p  p  p 0   Fdt
…………. (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 = mv0=mv
Geometry of Rutheford scattering.
We see from figure:
1
 
 
p  m v 0  sin    p  2m v 0  sin  
2
2
2
……….. (2)
Because impulse is in the same direction as the change of
momentum, it is valid:
…………… (3)
 F  cos   dt
Momenta in Rutheford scattering:


where  is running angle between F and p along particle trajectory.

We substitute (2) and (3) to (1):
 
2m v 0  sin     F  cos  dt
2 0
 
2m v 0  sin   
2
We change integration variable from t to :
………….....................……(4)
 1 2   
-1 2

  
F  cos  

dt
d
d
…. (5)
where ddt is angular velocity of  particle motion around nucleus. Electrostatic action of nucleus

on particle is in direction of the join vector  r  F  0  force momentum do not act 
angular momentum is not changing (its original value is mv0b) and it is connected with angular
velocity = d/dt  mr2 = const = mr2 (d/dt) = mv0b
2
then:
dt
r

d v 0 b
 1 2   
 
2m v b  sin   
Fr 2 cos   d

 2  1 2   
2
0
we substitute dt/d at (5):
We substitute electrostatic force F (Z=2):
We obtain:
 1 2   
2
2Ze 2
Fr cos   d 

40
1 2   
 1 2   
because it is valid:
1 2
 1 2   
1 2
2Ze 2
F
4 0 r 2
1
Ze 2
  d 
cos



  d sin   
cos

 1 2   
 1 2   

................................… (6)
0
 
cos 
2
  
 
 2 sin     2 cos 
2 2
2
2
   Ze
 
cos 
We substitute to the relation (6): 2m v b  sin   
 2  0
2
2
Scattering angle  is connected with
   20 m v0 b 40 E KIN
cotg   

b
collision parameter b by relation:
Ze 2
Ze 2
2
2
0
The smaller impact parameter b the larger scattering angle .
… (7)
Energy and momentum conservation law
Just these conservation laws are very important. They determine relations between kinematic
quantities. It is valid for isolated system:
nf
nf
E  E
Conservation law of whole energy:
k 1
 m c
ni
k
j
j1
k 1
 m c    E    m c    E 
nf
k 1
nf
2
0
k
k 1
nj
ni
KIN k
j1
j
j1
 E KIN
   m c
nj
k
j1
0
2
 E KIN
M f0 c 2  E fKIN  M i0 c 2  E iKIN
2
0
0
2
KIN j
Nonrelativistic approximation (m0c2 >> EKIN): EKIN = p2/(2m0)
M f0  M i0
M f0 c 2  M i0 c 2
Together it is valid for elastic scattering: E
f
KIN
E
ni
 p2 
 p2 

   


k 1  2m 0  k
j1  2m 0  j
nf
i
KIN
Ultrarelativistic approximation (m0c2 << EKIN): E ≈ EKIN ≈ pc
E E
f
i
E
f
KIN
E
i
KIN
nf
ni
k 1
j1
 p k c   p jc
nf
Conservation law of whole momentum:
ni


p

 k pj
k 1
j1
nf
ni
k 1
j1
 pk   p j

j
We obtain for elastic scattering:
Using momentum conservation law:
0  p1 sin   p2 sin 
and p1  p1 cos  p2 cos
We obtain using cosine theorem:
p22  p12  p12  2p1p1 cos 
Nonrelativistic approximation:
Using energy conservation law:
p12
p12
p22


2m1 2m1 2m 2
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:
 m 
 m 
m
p12 1  1   p12 1  1   2 1 p1p1 cos   0
m2
 m2 
 m2 
 m 
 m 
m
EKIN1 1  1   E KIN1 1  1   2 1 E KIN1EKIN1 cos   0
m2
 m2 
 m2 
Ultrarelativistic approximation:
Using energy conservation law:
p1  p1  p2  p2  p1  p1  2p1p1
2
We obtain using this relation and momentum conservation law:
2
2
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.
Remainder of centre-of-mass installation: we assumed two particles (with masses m1, m2 and
 
 
positions r1 , r2 ) interacting mutually only by central potential V r2  r1  :




rCM  rCM , r̂CM
r  r, r̂
r2  r2 , r̂2
r1  r1 , r̂1
coordinate
origin
Equations of motion can be written in form:



 

 


m1 r1  1V r2  r1 
m 2 r2   2 V r2  r1  …………(1)

where  has in spherical coordinates form ( rˆi , ˆi , ˆi are
appropriate unit vectors):

ˆi
 ˆi 

 i  r̂i


i = 1,2
ri ri i ri sin i i
Potential energy depends only on relative distance of particles.
Reminder of equations (1):
We define new coordinates:
  
r  r1  r2


 
m1r1  1V r2  r1 



m1 r1  m 2 r2
rCM 
m1  m 2


 
m 2r2   2 V r2  r1 
………………………. (2)



Using relations (1) and (2) we obtain (it is valid  2 V r   1V r   V r  ):

V r 
  
m1m 2 
r    r  -V r   
r̂
m1  m 2
r


(m1  m2 )rCM  MrCM  0

rCM  constant  r̂
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:
Centre-of-mass motion is uniform and straightforward, centre-of-mass moves in the laboratory with


v CM  rCM
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:


m v m v
mv
v CM  rCM 
1 1
2
2
m1  m 2

1 1
m1  m 2
Particles are moving against themselves in the centre-of-mass system with velocities:

  
m 2 v1
~
v1  v1  v CM 
m1  m 2
and then:



~
p1  m1 ~
v1    v1


 
m1v1
~
v 2  v 2 - v CM  
m1  m 2



~
p2  m 2 ~
v 2     v1
(we see, that momenta have opposite directions and they have the same magnitude)
and
~
~
~
E KIN  E KIN 1  E KIN 2 
m2
1
E KIN 1    v12
m1  m 2
2
Laboratory
coordinate
system
Centre-of mass
coordinate
system
Laboratory
coordinate
system
Centre-of-mass
coordinate
system
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:
~
~
v1 cos  vCM  ~
v1 cos
v1 cos   ~
v1 cos   vCM
Relation between velocity components perpendicular to the direction of beam particle motion:
~
v sin   ~
v sin 
1
1
We divide these relations:
~
~v sin ~
sin 
1
tan   ~

~
~
v1 cos   v CM cos   v CM ~v1 
It is valid for elastic scattering:
v
..............…………………….. (3)
m
  ~CM  1
v1
m2
We rewrite equation (3) to the form:
~
cos   
cos  
~
(1  2 cos    2 )1 2
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:
 ni
  nf
 n f KIN   n i KIN 
2
2
Q    m jc    m k c     E k     E j   Q
 f  k 1
  j1
 j1
 i  k 1

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).
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:
 n f ~ 
 p k   0
 k 1  f
 n i ~ 
 p j   0
 j1  i
It is possible case, that all end particles have zero momentum and thus also their
individual kinetic energies are zero:  n ~ KIN 
 E k   0
 k 1
f
f
Thus threshold energy ETHR in ~
 n ~ KIN   n

 n
E

E

m

m
  j    k    j   Q  Q
THR
centre-of-mass coordinate system is:
j1
k 1
j1
i

f
i

i
f

i
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:
p12
~
E KIN 
2(m1  m 2 )
this energy is not usable for reaction. That means threshold energy must be: E THR
From definition ETHR is minimal EKIN 1:
Substituting p12 into previous equation:

p12
E THR 
2m1
 m 
E THR  Q 1  1 
 m2 

p12  2m1E THR

p12

Q
2m1  m 2 
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):
 m 
 m  2  m1m3E KIN1
Q  EKIN3 1  3   E KIN11  1  
m4
 m4 
 m4 
We often need relation:
EKIN 3 =f(EKIN 1,),
Solution is: E KIN3  r  r 2  s where r 
EKIN3 cos
we define x  √E‘KIN 3
m1m3E KIN1
cos
m3  m 4
and
s
m 4 Q  E KIN1(m 4  m1 )
m3  m 4
Inelastic scattering is always endothermic (where M0i M0f, EiKIN, EfKIN are total sums):
M0i  M0f  EiKIN  EfKIN  Q  0
Decay of particle at rest: Q = m0ic2 -M0fc2. Momenta of particles after two-particle decay have the
same magnitude but opposite direction. Isotropic distribution.
2m f01m f02Q
f
f
p

p

1
2
Momenta of products:
mf  mf
01
02
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:


 
 


m1 
m2 


p1  ~
p1 
p1
p2   ~
p1 
p1
v1  ~
v1  vCM
v2  ~
v2  vCM
m1  m 2
m1  m 2
  
(We obtain law of momentum conservation for studied case by sum of these equations: p1  p1  p2 )
Such relations are initial equations for construction of vector diagram of momenta:
1) Momentum p1 of impinging particle we represent by oriented abscissa AC .
2) We divide abscissa AC to two parts in the proportion
AO : OC  m1 : m 2
3) We describe circle around the point O passing through the point C. The circle radius is
~p  m 2 p . The circle
equal to magnitude of momenta p1 in the centre-of-mass system
1
1
m1  m 2
geometrical place of vertexes B of vector triangle of momenta ABC (represents law of
momentum conversation), which sides AB and BC represent possible momenta of particles
after collision in the laboratory system.
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
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  impinging particles are scattered to both hemispheres
In the laboratory system: m1 = m2  impinging particles are scattered to front hemisphere
m1 > m2  impinging particles are scattered to front hemisphere to cone
with top angle 2MAX (direction of impinging particles is axe of cone): sinMAX =m2/m1
Relation between scattering angle and reflection angle in the laboratory and the centre-of-mass
~
~
system (remainder of elastic scattering assumption):
sin

 
tg 
~

2
cos   m1 m 2 
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: E  p 2c 2  m02c 4
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):

E 2  p 2c 2  m 2c 4
2
2
 n
  n   2  n
2
E
It is valid not only for single particle but also for particle   i     pi  c    mi c 
 i 1   i 1 
 i 1

system in the given time:
We express kinetic energy and momentum: E KIN  E  mc
2
2
... (1)
 2 E2
p  2  m 2c2
c
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:
2
n
 f


(E1  m 2 c 2 ) 2  p12 c 2    m jc 2 
 j1
f
We substitute
p2 :
E
1
 m 2c

2 2
2
2
 nf

2
2 4
 E1  m1 c    mjc 2 
 j1
f
 nf

2
2 4
2 4
2E1m 2 c  m1 c  m 2 c    m jc 2 
 j1
f
2
and EKIN 1:
 nf

2
4
2 4
2 4
2E KIN 1m 2c  2m1m 2c  m1 c  m 2 c    mjc 2 
 j1
f
2
 nf

  mjc 2   2m1m 2c 4  m12c 4  m 22c 4


j1

f
E KIN 1 
2m 2c 2
We express EKIN1:
We obtain:
E THR
 nf
 n f

2
2
2
  mjc  m1c  m 2c   mjc 2  m1c 2  m 2c 2 



j1
j1




 E KIN1 
2
2m 2c
nf
Because:
Q   mjc 2  m1c 2  m 2 c 2
j1
we obtain
E THR
Q  2m c


 m1
 2m 2c 2 Q
Q 


Q
1


 m m c 2 
2m 2c 2
2
2


1
2
In no relativistic approximation (Q<<m2c2) we obtain known relation.
 m 
E THR  Q1  1 
 m2 
In ultrarelativistic approximation (Q>>m1c2 a Q>>m2c2): ETHR = Q2
Relativistic relation between scattering angle in centre-of-mass and laboratory system
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):
~
v CM E
~
px  2
c
px 
2
 v CM 
1 

 c 
We use polar coordinate system:
px  pcos
p y  psin 
2
We derived relation for angle :
E
py  ~
py
and
~
E  v CM ~
px
v 
1   CM 
 c 
~
~
px  ~
p cos
2
2
~
~
py  ~
psin 
~
~
v 
v 
v 
~
~
~
p y 1   CM 
p sin  1   CM 
v sin  1   CM 
py
 c 
 c 
 c 
tan  

~ 
~ 
~
~
px
~ v E
v E
v cos   v CM
~
~
px  CM2
p cos   CM2
c
c
2
In nonrelativistic approximation, where vCM << c we obtain known relation, which we already
derived.
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:
 CM 
v CM
p1c

c
E KIN1  m1c 2  m 2 c 2
We use relation between kinetic energy and momentum:
E KIN1  m12 c 4  p12 c 2  m1c 2
We obtain:
 CM
p1c  E 2KIN1  2E KIN1m1c2
E 2KIN1  2E KIN1m1c 2

E KIN1  m1c 2  m 2c 2
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
~
~
v 1   CM
v
tan  

v CM
c
m1  m 2 2 c 4  2E KIN1m 2c 2
E 2KIN1  2E KIN1m1c 2
In ultrarelativistic approximation (EKIN 1 >> m1c2 and EKIN 1 >> m2c2) we obtain:
~
v
  tan  
c
2m 2 c 2
0
E KIN1
In the laboratory system, particles are produced to the very small angle.
Relativistic invariant variables
We 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: v CM    (p1  p 2 )c
(1.a)
CM
E1  E 2
c
m1 refers to the projectile mass and m2 to target mass. We use laboratory kinematic variables and


we obtain:

p1c
p1c
…………… (1.b)
 CM 

2
2 2
2 4
2
E1  m 2 c
Nonrelativistic approximation (m1
c2
p1 c  m1 c  m 2 c

p1c):  CM


m1v1c
m1v1


m1c 2  m 2 c 2 (m1  m 2 )c
…………….. (1.c)
Ultrarelativistic approximation (m1c2  p1c a m2c2  p1c):

 CM   CM
1  ((m1c 2 ) (p1c)) 2  (m 2 c 2 ) (p1c)



2
2
2
2
2
2
2
1

(
(
m
c
)
(
p
c))

(
(
m
c
)
(
p
c)
)
1  ((m1c ) (p1c))  (m 2 c ) (p1c)
1
1
2
1
1
m c 1m c
 1  ((m1c ) (p1c))  (m 2 c ) (p1c)  1  2   1 
p1
2  p1 
2
2
2
2
1/ 2
1/ 2
2
For m1  m2:  CM  (1  m 2c p1 ) and  CM  (1  CM
) 1/ 2  (1  CM )1  CM   2m2c p1   p1 2m2c
We obtain general relativistic relation for CM :
So that (m12c4 = E12-p12c2):
1 
2
CM
using equation (1.b): 
2
CM

p12c 2
E  m c 
2 2
1
2
E12  2E1m 2 c 2  m 22 c 4  p12 c 2 m12 c 4  m 22 c 4  2E1m 2 c 2


(E1  m 2 c 2 ) 2
(E1  m 2 c 2 ) 2
and we obtain
 CM  (1  
2
1 / 2
CM
)
E1  m 2 c 2

(m12 c 4  m 22 c 4  2E1m 2 c 2 )1/2
… (2)
Equation is reduced for limits E1  p1c  m1c2 and p1c  m2c2 to formerly given ultrarelativistic
limit:
 CM  p1 2m2c
Quantity in the divisor (2) is invariant scalar. We prove this using the square of following four-vector
in the laboratory frame (p2 = 0):
 
s  (E1  E 2 ) 2  (p1  p 2 ) 2 c 2  (E1  m 2c 2 ) 2  p12c 2 
 E12  m 22c 4  2E1m2c 2  p12c 2  m12c 4  m22c 4  2E1m 2c 2
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):
~ ~ 2 ~ ~ 2 2 ~ ~ 2 ~ 2
2 4
2 4
2
s  m c  m c  2E m c  (E  E )  (p  p ) c  (E  E )  E
1
2
1
2
1
2
1
2
1
2
TOT
and s is square of total energy accessible in centre-of-mass system. Then  CM 
E1  m 2c 2 E TOT
~
~
E TOT
E TOT
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):
 
t  (E1f  E1i ) 2  (p1f  p1i ) 2 c 2
……………………. (3a)
Energy and momentum conservation laws are valid and we can express t also in target variables:
 
t  (E f2  Ei2 ) 2  (pf2  pi2 ) 2 c 2
……………….. (3b)
variable t is invariant and it can be calculated in arbitrary coordinate system.
 
 
u  (E f2  E1i ) 2  (pf2  p1i ) 2 c 2 or
u  (E1f  E i2 ) 2  (p1f  pi2 ) 2 c 2
We add yet variable u:
Variables t, u and s are named as Lorentz invariant Mandelstam variables, which sum
generally satisfy equation:
s  t  u  m 2 c 4  m 2 c 4   m 2 c 4  m 2 c 4 
1
2
1
i
2
f
In the case of elastic scattering in the centre-of-mass system (for both particles
~
~
Ei  Ef )
f i 2
~
f 2
i 2
~
~
~
t   p1   p1   2p1 ~
p1  c  2~
p 2c2 (1  cos )




~
pi  ~
pf  ~
p ~
p

Because –1 ≤ cos ≤ 1 it is valid t  0. Using (3a,b) we can look on variable t as on mass-square
f i
of exchanged particle (with energy
E f2  E i2 and momentum p 2  p 2 ). Imaginary mass 
virtual particle.
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:
and
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
Reminder:
e y  e y
sinh(y) 
2
ey  ey
cosh(y) 
2
e y  e  y e 2y  1
tanh(y)  y

e  e  y e 2y  1
2 2
2 2
2
2
We introduced transversal mass mT : m T c  m c  p x  p y
E

 pz 
1 c

and rapidity y: y  ln  E
2  p 

z 
c

1
mc  mvcos   1  1  cos 

  ln 
2  mc  mvcos   2  1  cos 

and thus: y  ln 
For nonrelativistic limit (β → 0): y = β
(
y(   0) 
1  1  cos 2cos  1
  ln 1  2cos   
ln 

2  1  cos 1  cos  2
)
For ultrarelativistic limit (β → 1): y → ∞
Rapidity using leads to very simple transformation from one
coordinate system to another:
y2  y1  y21
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 :
y  y  yCM
Examples: GSI Darmstadt ( ELAB = 1GeV/A
y=0.458 β=0.875 )
SPS CERN
( ELAB = 200GeV/A
y=6.0
β=1.000 ) Relation between transversal
LHC CERN ( ELAB=3500+3500GeV/A y=17.8 β=1.000 ) component of velocity and
rapidity